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122 files changed, 59335 insertions, 9 deletions
diff --git a/Chemical_Engineering_Thermodynamics/screenshots/T-x-y_diagram.png b/Chemical_Engineering_Thermodynamics/screenshots/T-x-y_diagram.png Binary files differnew file mode 100755 index 00000000..f231eb76 --- /dev/null +++ b/Chemical_Engineering_Thermodynamics/screenshots/T-x-y_diagram.png diff --git a/Elementary_Fluid_Mechanics/README.txt b/Elementary_Fluid_Mechanics/README.txt new file mode 100755 index 00000000..d40ffd56 --- /dev/null +++ b/Elementary_Fluid_Mechanics/README.txt @@ -0,0 +1,10 @@ +Contributed By: jaykishan mehta +Course: btech +College/Institute/Organization: CSPIT +Department/Designation: CS +Book Title: Elementary Fluid Mechanics +Author: J. K. Vennard +Publisher: John Wiley & Sons, USA +Year of publication: 1961 +Isbn: 9780471905851 +Edition: 4
\ No newline at end of file diff --git a/Elementary_Fluid_Mechanics/ch1.ipynb b/Elementary_Fluid_Mechanics/ch1.ipynb new file mode 100755 index 00000000..d7bfb12e --- /dev/null +++ b/Elementary_Fluid_Mechanics/ch1.ipynb @@ -0,0 +1,253 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:93240c9461961fb318dd0bfb1fd759f5d83f741dfd8a4a76ea2b51b81a776bec" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 1 : Fundamentals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.1 Page No : 8" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# variables\n", + "T = 80.;\t\t#temperature of chlorine gas in degree F\n", + "p = 100.;\t\t#pressure in psia\n", + "W = 2*35.45;\t\t#molecular weight of chlorine \n", + "\n", + "# calculations \n", + "R = 1545/W;\t\t#specific gas constant in ft-lb/lb-degreeR\n", + "gam = p*(144/R)*(1/(460+T));\t\t#specific weight of chlorine in lb/cuft\n", + "Spec_vol = 1/gam;\t\t#specific volume in cuft/lb\n", + "rho = gam/32.2;\t\t#density of chlorine in slug/cuft\n", + "\n", + "# results \n", + "print 'Specific weight = %.3f lb/cuft \\nSpecific volume = %.3f cuft/lb \\ndensity = %.4f slug/cuft'%(gam,Spec_vol,rho);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Specific weight = 1.224 lb/cuft \n", + "Specific volume = 0.817 cuft/lb \n", + "density = 0.0380 slug/cuft\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.2 Page No : 12" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# variables\n", + "gamma = 1.4;\n", + "T1 = 60.;\t\t#temperature of air in degree F\n", + "p1 = 14.7;\t\t#pressure in psia\n", + "k = 0.5;\t\t#(final volume/initial volume) = k\n", + "R = 53.3;\t\t#Engineering gas constant\n", + "\n", + "# calculations \n", + "gam1 = p1*(144/R)*(1/(460+T1));\t\t#lb/cuft\n", + "gam2 = gam1/k;\t\t#lb/cuft\n", + "p2 = (p1/(gam1**(gamma)))*(gam2**(gamma));\t\t# in psia\n", + "T2 = p2*(144/R)*(1/gam2);\t\t#in degree F\n", + "a1 = math.sqrt(gamma*32.2*R*(460+T1));\t\t# in fps\n", + "a2 = math.sqrt(gamma*32.2*R*(T2));\t\t# in fps\n", + "\n", + "# results \n", + "print 'Final pressure = %.1f psia \\\n", + "\\nFinal temperature = %d degreeR \\\n", + "\\nSonic velocity before compression = %d fps \\\n", + "\\nSonic velocity after compression = %.f fps'%(p2,T2,a1,a2);\n", + "\n", + "#the answers differ due to rounding-off errors" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "0.152749314475\n", + "Final pressure = 38.8 psia \n", + "Final temperature = 686 degreeR \n", + "Sonic velocity before compression = 1117 fps \n", + "Sonic velocity after compression = 1284 fps\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.3 Page No : 17" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "from scipy.integrate import quad \n", + "\n", + "# variables\n", + "r1 = 0.25;\t\t# radius of cylinder in feet\n", + "l = 2.;\t \t#length of cylnider in feet\n", + "r2 = 0.30;\t\t# radius of co-axial cylinder in feet\n", + "mu = 0.018;\t\t#lb-sec/ft**2\n", + "torque = 0.25;\t\t# in ft-lb\n", + "dv_dy1 = torque/(4*math.pi*mu*r1**2);\t\t#velocity gradient at radius = 0.25 in fps/ft\n", + "dv_dy2 = torque/(4*math.pi*mu*r2**2);\t\t#velocity gradient at radius = 0.30 in fps/ft\n", + "\n", + "# calculations \n", + "def f4(r): \n", + "\t return -torque/(4*math.pi*mu*r**2)\n", + "\n", + "V1 = quad(f4,r2,r1)[0]\n", + "\n", + "rpm1 = V1*60/(2*math.pi*r1);\n", + "V2 = torque*(r2-r1)/(4*math.pi*mu*r1**2);\t\t#in fps\n", + "rpm2 = V2*60/(2*math.pi*r1);\n", + "hp = 2*math.pi*r1*(rpm1/(550*60));\n", + "\n", + "# results \n", + "print 'Velocity gradient at the inner cylinder wall is %.1f fps/ft and at the outer cylinder wall is %.1f fps/ft'%(dv_dy1,dv_dy2);\n", + "print 'rpm = %.1f and approximate rpm = %.1f, hp = %.5f '%(rpm1,rpm2,hp);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Velocity gradient at the inner cylinder wall is 17.7 fps/ft and at the outer cylinder wall is 12.3 fps/ft\n", + "rpm = 28.1 and approximate rpm = 33.8, hp = 0.00134 \n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.4 Page No : 20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables\n", + "T = 70. \t\t#degreeF\n", + "del_p = 0.1;\t\t# in psi\n", + "sigma = 0.00498;\t\t# lb/ft\n", + "\n", + "# calculations \n", + "R = (sigma*2)/(del_p*144);\t\t#in ft\n", + "d = 12*2*R;\t\t# in inches\n", + "\n", + "# results \n", + "print 'Diameter of the droplet of water = %.4f in'%(d);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diameter of the droplet of water = 0.0166 in\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.5 Page No : 20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# variables\n", + "l = 12.;\t\t# length of the cylinder\n", + "T = 150.;\t\t#temperature of water in degreeF\n", + "p1 = 14.52;\t\t#atmospheric pressure in psia\n", + "p2 = 3.72;\t\t#the pressure on the inside of the piston in psia\n", + "\n", + "# calculations \n", + "F = 0.25*(p1-p2)*math.pi*l**2;\t\t#Force on the piston in lb\n", + "\n", + "# results \n", + "print 'Minimum force on the piston to be applied is, F = %d lb'%(F);\n", + "\n", + "#incorrect answer given in textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum force on the piston to be applied is, F = 1221 lb\n" + ] + } + ], + "prompt_number": 12 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_Fluid_Mechanics/ch10.ipynb b/Elementary_Fluid_Mechanics/ch10.ipynb new file mode 100755 index 00000000..1f2ac4f3 --- /dev/null +++ b/Elementary_Fluid_Mechanics/ch10.ipynb @@ -0,0 +1,291 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 10 : Liquid Flow in Open Channels" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.1 Pageno : 353" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import *\n", + "\n", + "# variables\n", + "yo = Symbol(\"yo\")\n", + "A = 20*yo\n", + "P = 20 + 2*yo\n", + "Rh= 20*yo/(20+2*yo)\n", + "n = 0.017 # ft\n", + "\n", + "# calculations\n", + "ans = solve((1.49/n)*A*Rh**(2./3)*(0.0001)**(1./2) - 400) #*yo*Rh**(2./3)\n", + "\n", + "# result\n", + "print \"Yo = %.2f ft\"%ans[0]\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Yo = 8.32 ft\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.2 pageno : 356" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# calculations\n", + "Rh = 20*8.34/(20+2*8.34)\n", + "\n", + "# result\n", + "print \"Rh = %.2f ft\"%Rh" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rh = 4.55 ft\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.3 pageno : 362" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import *\n", + "\n", + "# variables\n", + "yo = Symbol(\"yo\")\n", + "n = 0.017 #ft 1/6\n", + "A = 2*yo**2\n", + "Rh = yo/2\n", + "\n", + "# Calculations\n", + "ans = solve((1.49/n)*A*Rh**(2./3) * (0.0001)**(1./2) - 400,yo)\n", + "Yo = ans[0]\n", + "B = 2*Yo\n", + "\n", + "#result\n", + "print \"Yo**(8/3) = %.f \"%(Yo**(8./3))\n", + "print \"Yo = %.2f ft \"%(Yo)\n", + "print \"B = %.2f ft \"%(B)\n", + "\n", + "# note : answer are slightly differenet because of solve method of sympy" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Yo**(8/3) = 362 \n", + "Yo = 9.11 ft \n", + "B = 18.22 ft \n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.4 Page no : 367" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from sympy import Symbol, solve\n", + "\n", + "# variables\n", + "q = 500./40 # cfs/ft\n", + "ye = ((12.5)**2 / 32.2)**(1./3)\n", + "n = 0.017\n", + "So = Symbol(\"So\")\n", + "\n", + "# calculations\n", + "Sc = 32.2 * n**2 / (2.2 * ye**(1./3))\n", + "ans = solve((1.49/n)*160*(3.33)**(2./3)*So**(1./2) - 500)\n", + "So = ans[0]\n", + "\n", + "# result\n", + "print \"Sc = %.4f\"%Sc\n", + "print \"So = %f\"%So\n", + "print \"So is a mild slope since it is less than Se\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Sc = 0.0035\n", + "So = 0.000256\n", + "So is a mild slope since it is less than Se\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.5 page no : 369" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import solve,Symbol\n", + "import math\n", + "\n", + "# variables\n", + "y = Symbol(\"y\")\n", + "A = 10*y + 2*y**2\n", + "n = 0.017\n", + "b = 10 + 4*y\n", + "P = 10 + 2*math.sqrt(5)*y\n", + "\n", + "# calculations\n", + "ans = solve(A**3/b *32.2/1000**2 - 1)\n", + "yo = ans[1]\n", + "Ae = 97.3 # sqft\n", + "be = 29.65 # ft\n", + "Rhe = Ae/32 # ft\n", + "Sc = 32.2*n**2/2.2 * (Ae/(be*Rhe**(4./3)))\n", + "\n", + "# result\n", + "print \"Yo = %.2f ft\"%yo\n", + "print \"Sc = %.5f\"%Sc" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Yo = 4.91 ft\n", + "Sc = 0.00315\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.6 page no : 373" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import Symbol,solve\n", + "\n", + "# variables\n", + "Q = round((1.49/0.015)*50*(50./20)**(2./3)*(0.001)**(1./2)) # cfs\n", + "E = 5.52 # ft\n", + "\n", + "# calculations and results\n", + "Ye = ((Q/10)**2/32.2)**(1./3)\n", + "Emin = 3 * Ye/2\n", + "\n", + "print \"Ye = %.2f ft\"%Ye\n", + "print \"Minimum height of hump : Emin = %.2f ft\"%Emin\n", + "\n", + "Ye = 2 * 5.52/3\n", + "b = Symbol(\"b\")\n", + "ans = solve( ( (289/b)**2 / 32.2 )**(1./3) - Ye)\n", + "b = ans[1]\n", + "\n", + "print \"Ye = %.2f ft\"%Ye\n", + "print \"Maximum width of contraction : b = %.1f ft\"%b" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ye = 2.96 ft\n", + "Minimum height of hump : Emin = 4.44 ft\n", + "Ye = 3.68 ft" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "Maximum width of contraction : b = 7.2 ft\n" + ] + } + ], + "prompt_number": 37 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_Fluid_Mechanics/ch11.ipynb b/Elementary_Fluid_Mechanics/ch11.ipynb new file mode 100755 index 00000000..8703141c --- /dev/null +++ b/Elementary_Fluid_Mechanics/ch11.ipynb @@ -0,0 +1,247 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:0e31426ac89cda24a76ca914e9a6ea49437d729b6061fc8d885e61953cdd31e2" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 11 : Fluid Measurements" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.1 Page No : 409" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "C_I = 0.98;\t\t#coefficient of pitot tube\n", + "d = 3.;\t\t#in\n", + "\n", + "# calculations \n", + "del_p = (d/12)*(13.55-0.88)/0.88;\n", + "v_c = C_I*math.sqrt(2*32.2*del_p);\n", + "\n", + "# results \n", + "print 'The velocity at the centerline of the pipe = %.1f fps'%(v_c);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The velocity at the centerline of the pipe = 14.9 fps\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.2 Page No : 411" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "P = 25.;\t\t#in. of mercury\n", + "p = 18.;\t\t#in. of mercury\n", + "T = 150.;\t\t#degreeF\n", + "\n", + "# calculations \n", + "k = P/p;\n", + "if k < (1.893) :\n", + " V = math.sqrt(2*32.2*186.5*(T+460)*(1-(1/k)**0.286));\n", + "\n", + "print 'The local velocity just upstream from the pitot static tube = %d fps'%(V);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The local velocity just upstream from the pitot static tube = 810 fps\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.3 Page No : 411" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "P = 20.;\t\t#in. of mercury\n", + "p = 5.;\t\t#in. of mercury\n", + "T = 150.;\t\t#degreeF\n", + "\n", + "# calculations \n", + "k = P/p;\n", + "\n", + "if k >1.893:\n", + " M_0 = 1.645;\n", + "\n", + "V_0 = math.sqrt(2*32.2*186.5*(T+460)/(1+ (2*186.5)/(53.3*1.4*M_0**2)));\n", + "\n", + "print 'The speed of this airplane = %d fps'%(round(V_0,-1));\n", + "\n", + "\t\t" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The speed of this airplane = 1600 fps\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.4 Page No : 420" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "b = 6.;\t\t#in\n", + "d = 3.;\t\t#in\n", + "p = 20.;\t\t#psi\n", + "del_p = 6.;\t\t#in. of mercury\n", + "p_bar = 14.70;\t\t#psia\n", + "T = 60.;\t\t#degreeF\n", + "\n", + "# calculations \n", + "k = (p + p_bar - b*(p_bar/29.92))/(p+p_bar);\n", + "gam1 = (p+p_bar)*144/53.3 /(T+460);\n", + "A2 = 0.0491;\t\t#sqft\n", + "Y = 0.949;\n", + "Cv = 0.98;\n", + "G = Y*Cv*A2*gam1*math.sqrt(2*32.2*b*10*A2*144/gam1) /(math.sqrt(1-0.25**2));\n", + "Cv_true = 0.981;\n", + "G_true = G*Cv_true/Cv;\n", + "\n", + "# results \n", + "print 'G = %.2f lb/sec'%(G);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "G = 3.31 lb/sec\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.5 Page No : 422" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "d = 3.;\t\t#in\n", + "l = 6.;\t\t#in\n", + "h = 6.;\t\t#in\n", + "T = 60.;\t\t#degreeF\n", + "\n", + "# calculations \n", + "Cv= 0.99;\n", + "A1 = 0.25*math.pi*(d/12)**2;\n", + "Q = Cv*A1*math.sqrt(2*32.2*(h/12)*(13.55-1)) /(math.sqrt(1-0.25**2));\n", + "Cv_true = 0.988;\n", + "Q_true = Q*Cv_true/Cv;\n", + "h_L = 3.8;\n", + "\n", + "# results \n", + "print 'Q = %.3f cfs'%(Q);\n", + "print 'True Q = %.3f cfs'%(Q_true);\n", + "print 'Total head loss is about %.1f ft of water'%(h_L);\n", + "\n", + "#there are small errors in the answer given in textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Q = 1.009 cfs\n", + "True Q = 1.007 cfs\n", + "Total head loss is about 3.8 ft of water\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_Fluid_Mechanics/ch12.ipynb b/Elementary_Fluid_Mechanics/ch12.ipynb new file mode 100755 index 00000000..7e205823 --- /dev/null +++ b/Elementary_Fluid_Mechanics/ch12.ipynb @@ -0,0 +1,65 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:3092cd2ae5a90e09126e9d647e026620efd1b57b1d7964dfe20aa5cd324764f3" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 12 : Elementary Hydrodynamics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.1 Page No : 490" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "Q = 0.00010;\t\t#cfs\n", + "t = 0.1;\t\t#ft\n", + "h = 3.;\t\t#ft\n", + "d = 6.;\t\t#in\n", + "\n", + "# calculations \n", + "K = Q*h/(t*0.25*math.pi*(d/12)**2);\n", + "\n", + "# results \n", + "print 'K = %.4f fps'%(K);\n", + "\n", + "\t\t" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "K = 0.0153 fps\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_Fluid_Mechanics/ch13.ipynb b/Elementary_Fluid_Mechanics/ch13.ipynb new file mode 100755 index 00000000..99c09079 --- /dev/null +++ b/Elementary_Fluid_Mechanics/ch13.ipynb @@ -0,0 +1,192 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:bd33818d3448353cd987bb363aa4b45c6fc5ac2971d0999934915bf83d11a9cd" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 13 : Fluid Flow about Immersed Objects" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.1 Page No : 502" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "b = 50.;\t\t#ft\n", + "c = 7.;\t\t #ft\n", + "CL = 0.6;\t\t#lift coefficient\n", + "CD = 0.05;\t\t#drag coefficient\n", + "alpha = 7.;\t\t#degrees\n", + "V = round(150/0.681818);\t\t#coverting mph to fps\n", + "H = 10000.;\t\t#ft\n", + "rho = 0.001756;\t\t#slug/cuft\n", + "\n", + "# calculations \n", + "D = CD*b*c*rho*0.5*V**2;\n", + "hp = D*V/550;\n", + "L = CL*b*c*rho*0.5*V**2;\n", + "mu = 3.534*10**-7;\t\t#lb-sec/sqft\n", + "R = V*c*rho/mu;\n", + "a = math.sqrt(1.4*32.2*53.3*(23.4+459.6));\n", + "M = V/a;\n", + "\n", + "# results \n", + "print 'hp = %.f hp, L = %.2f lb, R = %d, M = %.3f'%(hp,L,round(R,-4),M);\n", + "\n", + "# note : answer is different because of rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "hp = 297 hp, L = 8923.99 lb, R = 7650000, M = 0.204\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.2 Page No : 511" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "l = 5.;\t\t#ft\n", + "d = 0.5;\t\t#ft\n", + "v = 1.;\t\t#fps\n", + "T = 60.;\t\t#degreeF\n", + "D = 0.04;\t\t#lb\n", + "k = 1./64;\t\t#model scale\n", + "\n", + "# calculations \n", + "nu = 0.00001217;\n", + "R = round(v*l/nu,-4);\n", + "Cf1 = 0.0020;\n", + "Cf2 = 0.0052;\n", + "Dx1 = round(2*Cf1*l*d*1.938*0.5*v**2,4);\n", + "Dx2 = round(2*Cf2*l*d*1.938*0.5*v**2,3) \n", + "delta1 = round(l*5.20/math.sqrt(R),2);\n", + "delta2 = l*0.38/(R**0.2);\n", + "V_0 = math.sqrt((v**2 /l)*(l*(1/k)));\n", + "R_p = V_0*l*(1/k)/nu;\n", + "Cf = 0.00185;\n", + "Dx = 2*Cf*l*d*(1/k)**2 *1.938*0.5*V_0**2;\n", + "Dw = D-Dx2;\n", + "Dw_p = (1/k)**2 *d*l*V_0**2 *Dw/(l*d);\n", + "D = round(Dw_p + Dx,-1);\n", + "\n", + "# results \n", + "print 'Total drag of the prototype = %d lb'%(D);\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Total drag of the prototype = 6280 lb\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.3 Page No : 524" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# variables\n", + "c = 6.;\t\t#ft\n", + "b = 36.;\t\t#ft\n", + "AR1 = 6.;\t\t#aspect ratio\n", + "Cd = 0.0543;\t\t#drag coefficient\n", + "Cl = 0.960;\t\t#lift coefficient\n", + "alpha1 = 7.2;\t\t#degrees\n", + "AR2 = 8.;\n", + "\n", + "# calculations \n", + "#for aspect ratio = 8\n", + "CL = 0.960;\t\t#negligible change of lift coefficient\n", + "#for aspect ratio = 6\n", + "C_Di = Cl**2 /(math.pi*AR1);\n", + "#for aspect ratio = infinity\n", + "C_D0 = Cd - C_Di;\n", + "#for AR = 8\n", + "C_D = C_D0 + Cl**2 /(math.pi*AR2);\n", + "#for AR = 6\n", + "alpha_i = (180/math.pi)*Cl/(math.pi*AR1);\n", + "#for AR = infinty\n", + "alpha_0 = alpha1 - alpha_i;\n", + "#for AR = 8\n", + "alpha = alpha_0 + Cl/(AR2*math.pi) *(360/(2*math.pi));\n", + "\n", + "# results \n", + "print 'Lift coefficient = %.3f (negligible change of lift coefficient)'%(CL); #incorrect answer in the textbook\n", + "print 'Drag coefficient = %.4f'%(C_D);\n", + "print 'Angle of attack = %.1f degress'%(alpha);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Lift coefficient = 0.960 (negligible change of lift coefficient)\n", + "Drag coefficient = 0.0421\n", + "Angle of attack = 6.5 degress\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_Fluid_Mechanics/ch2.ipynb b/Elementary_Fluid_Mechanics/ch2.ipynb new file mode 100755 index 00000000..5fbe2709 --- /dev/null +++ b/Elementary_Fluid_Mechanics/ch2.ipynb @@ -0,0 +1,505 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:fae9d4c3d571f5f27d1259174500abbced27ed669ce1d759f1518557eb996672" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 2 : Fluid Statics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.1 Page No : 32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "# variables\t\t\n", + "T = 68;\t\t#degreeF\n", + "p = 10;\t\t# psi\n", + "d = 15;\t\t# feet\n", + "rho = 1.59;\t\t#specific gravity\n", + "\n", + "# calculations \n", + "gam = rho*62.4;\t\t#lb/cuft\n", + "p1 = gam*d + p*144;\t\t#psf\n", + "\n", + "# results \n", + "print 'p1 = %d psf = %.1f psi '%(p1,p1*0.00694);\n", + "\n", + "#incorrect answer given in the textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "p1 = 2928 psf = 20.3 psi \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.2 Page No : 32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables\t\t\n", + "h = 35000;\t\t# feet\n", + "p1 = 14.7;\t\t# psia\n", + "T1 = 519;\t\t# degreeR\n", + "gam1 = 0.0765;\t\t# lb/cuft\n", + "p2 = 504;\t\t# psfa\n", + "\n", + "# calculations \n", + "T2 = T1 - h*0.00356;\t\t# degreeR\n", + "gam2 = p2/(53.3*T2);\t\t# lb/cuft\n", + "\n", + "# results \n", + "print 'p2 = %d psfa = %.2f psia \\nspecific weight = %.3f lb/cuft'%(p2,p2*0.00695,gam2);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "p2 = 504 psfa = 3.50 psia \n", + "specific weight = 0.024 lb/cuft\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.3 Page No : 35" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\t\t\n", + "# variables\n", + "h1 = 12.5;\t\t# inches\n", + "p1 = 14.50;\t\t# psia\n", + "\n", + "# calculations \n", + "p = p1 - h1*(14.70/29.92);\t\t#absolute pressure in psia\n", + "\n", + "# results \n", + "print 'Absolute pressure = %.2f psia'%(p);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Absolute pressure = 8.36 psia\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.4 Page No : 37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables\t\t\n", + "gam1 = 0.9*62.4;\n", + "gam2 = 13.55*62.4;\n", + "l1 = 10;\t\t# feet\n", + "l2 = 15./12;\t\t# feet\n", + "\n", + "# calculations \n", + "p_x = gam2*l2 - gam1*l1;\t\t# psf\n", + "\n", + "# results \n", + "print 'The gauge reading = %d psf = %.2f psi'%(p_x,0.00694*p_x);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The gauge reading = 495 psf = 3.44 psi\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.5 Page No : 42" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "from scipy.integrate import quad \n", + "\t\t\n", + "# variables \n", + "l1 = 4.;\t\t# feet\n", + "b1 = 6.;\t\t# feet\n", + "b2 = 6.;\t\t# feet\n", + "l2 = 2.55;\t\t# feet\n", + "t = 1.; \t\t# feet\n", + "\n", + "# calculations \n", + "F1 = 0.5*l1*b1*62.4*(0.5*l1 + t) ;\t\t# lb\n", + "F2 = 0.25*math.pi*b2**2 *62.4*(l2 + t);\t\t# lb\n", + "a1 = l1*b2**3 /(36*0.5*b2*0.5*l1*b1);\t\t# feet\n", + "a2 = 70/((0.5*l2 + t)*28.3);\t\t# feet\n", + "l_p = (F1*(0.5*l1 + a1)+F2*(l2+a2))/(F1+F2) +1;\t\t#feet\n", + "x_p1 = (0.5*l1-a1) - a1*2/b2;\t\t# feet\n", + "\n", + "def f2(y): \n", + "\t return (62.4/2)*(36-y**2)*(y+1)\n", + "\n", + "M = quad(f2,0,6)[0]\n", + "\n", + "x_p2 = M/F2;\t\t# feet\n", + "x_p = (x_p2*F2 - F1*x_p1)/(F1+F2);\t\t# feet\n", + "\n", + "# results \n", + "print 'Total force on composite area is %d lb'%(F1+F2); \n", + "print ' Vertical location of resultant force is %.2f ft below the water surface'%(l_p);\n", + "print ' Horizontal location of resultant force is %.3f ft right of the water surface'%(x_p);\n", + "\n", + "#incorrect answer given in textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Total force on composite area is 8509 lb\n", + " Vertical location of resultant force is 4.38 ft below the water surface\n", + " Horizontal location of resultant force is 1.423 ft right of the water surface\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.6 Page No : 45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "import numpy\n", + "\t\t\n", + "# variables\n", + "l = 8.; \t\t#feet\n", + "b = 10.;\t \t# feet\n", + "\n", + "# calculations \n", + "F_h = 0.5*l*b*62.4*(b+2.5);\t\t# lb\n", + "x = 83.2/(40*(b+2.5));\t\t# feet\n", + "F_v = (b+5)*62.4*40-(l*62.4*(25 - 0.25*math.pi*25));\t\t# lb\n", + "F = math.sqrt(F_h**2 + F_v**2);\t\t# lb\n", + "e = (2680*3.91 + 37440*(0.25*b))/F_v ;\t\t# feet\n", + "theta = 180*numpy.arctan(F_v/F_h) /math.pi;\t\t# degrees\n", + "x_p = 0.25*b-x;\t\t# feet\n", + "\n", + "# results \n", + "print 'Magnitude of resultant force is %d lb'%(F);\n", + "print 'Theta = %d degrees'%(theta);\n", + "print 'Location is %.3f feet above and %.2f feet to the right of B'%(x_p,e);\n", + "\n", + "#there are errors in the answer given in textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Magnitude of resultant force is 46709 lb\n", + "Theta = 48 degrees\n", + "Location is 2.334 feet above and 2.99 feet to the right of B\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.7 Page No : 48" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables\n", + "A = 4000.;\t\t# sq.ft\n", + "d1 = 10.;\t\t# feet\n", + "d2 = 2.;\t\t# inches\n", + "rho = 64.;\t\t# lb/cuft\n", + "\n", + "# calculations \n", + "W = A*(d2/12)*rho;\t\t# lb\n", + "\n", + "# results \n", + "print 'Weight of cargo = %d lb'%(round(W,-2));\n", + "\n", + "\t\t" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Weight of cargo = 42700 lb\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.8 Page No : 49" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# variables\n", + "gam = 53.0;\t\t# lb/cuft\n", + "D = 17.;\t\t# inches\n", + "d = 12.;\t\t# inches\n", + "\n", + "# calculations \n", + "V = (math.pi/6)*(D/d)**3;\n", + "V1 = 0.584;\t\t#cuft\n", + "V2 = 0.711;\t\t#cuft\n", + "W = V*gam;\n", + "F_B = V1*62.4;\n", + "F_ACA = (V2)*62.4;\n", + "F = W+F_ACA-F_B;\n", + "\n", + "# results \n", + "print 'The force exerted between sphere and orfice plate = %.1f lb'%(F);\n", + "\n", + "#incorrect answer for W in textbook. Hence the answer differs" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The force exerted between sphere and orfice plate = 86.8 lb\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.9 Page No : 51" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from scipy.integrate import quad \n", + "\n", + "# variables\n", + "v = 15;\t\t# ft/sec**2\n", + "d = 5;\t\t# ft\n", + "\n", + "# calculations \n", + "def f3(z): \n", + "\t return -62.4*(v+32.2)/32.2\n", + "\n", + "p = quad(f3,0,-5)[0]\n", + "\n", + "# results \n", + "print 'p = %d psf'%(p);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "p = 457 psf\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.10 Page No : 52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "from scipy.integrate import quad \n", + "\n", + "# variables\n", + "m = -0.229;\t\t#slope\n", + "a_z = 1.96;\t\t# ft/sec**2\n", + "a_x = 4*a_z;\t\t# ft/sec**2\n", + "a = math.sqrt(a_x**2 + a_z**2);\t\t# ft/sec**2\n", + "\n", + "def f1(z): \n", + "\t return -(32.2 + a_z)*(62.4/32.2)\n", + "\n", + "p = quad(f1,0,-2.75)[0]\n", + "\n", + "# results \n", + "print 'p = %.1f psf'%(p);\n", + "\n", + "#there is an error in the answer given in textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "p = 182.0 psf\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.11 Page No : 54" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "l1 = 2.;\t\t# feet\n", + "l2 = 3.;\t\t# feet\n", + "rpm = 100;\n", + "\n", + "# calculations \n", + "p_A = (l1+l2)-(2./3)*(2*math.pi*rpm/60)**2 /(2*32.2);\n", + "p_B = (l1+l2)+(1./3)*(2*math.pi*rpm/60)**2 /(2*32.2);\n", + "\n", + "# results \n", + "print 'Pressure heads at point A and point B ae %.2f ft and %.2ft ft respectively'%(p_A,p_B);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Pressure heads at point A and point B ae 3.86 ft and 5.57t ft respectively\n" + ] + } + ], + "prompt_number": 9 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_Fluid_Mechanics/ch3.ipynb b/Elementary_Fluid_Mechanics/ch3.ipynb new file mode 100755 index 00000000..eeb66d6e --- /dev/null +++ b/Elementary_Fluid_Mechanics/ch3.ipynb @@ -0,0 +1,250 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:78fe20b770d35c2d23108a58f50bb1579f3cddf4dfa3efb233fab2fa5b1b32cd" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 3 : Kinematics of Fluid Motion" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.1 Page No : 83" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# variables\n", + "v_mag = 3.;\n", + "x = 8.;\n", + "y = 6.;\n", + "\n", + "# calculations \n", + "s = math.sqrt(x**2 + y**2);\n", + "v = v_mag*s;\t\t# fps\n", + "a_t = v_mag*s*v_mag;\t\t# ft/sec**2\n", + "a_r = 0;\n", + "a = math.sqrt(a_r**2 + a_t**2);\n", + "\n", + "# results \n", + "print 'v = %d fps a = %d ft/sec**2'%(v,a);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "v = 30 fps a = 90 ft/sec**2\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.2 Page No : 83" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# variables\n", + "v = 5;\t\t# fps\n", + "a_t = 0;\n", + "\n", + "# calculations \n", + "a_r = v**2 /2.;\t\t# ft/sec**2\n", + "\n", + "# results \n", + "print 'Radial component of acceleration = %.1f ft/sec**2 \\nTangential component of acceleration = %d '%(a_r,a_t);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Radial component of acceleration = 12.5 ft/sec**2 \n", + "Tangential component of acceleration = 0 \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3 Page No : 85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# variables\t\t\n", + "v = 5.;\t\t# fps\n", + "r = 2.;\n", + "theta = 60.;\t\t# degrees\n", + "x = 1.;\n", + "y = math.sqrt(3);\n", + "v_t = v;\n", + "v_r = 0;\n", + "\n", + "# calculations \n", + "u = -v*y/(math.sqrt(x**2 + y**2));\n", + "v = v*x/(math.sqrt(x**2 + y**2));\n", + "a_x = -50*x/8;\n", + "a_y = -50*y/8;\n", + "a_r = -v_t**2 /r;\n", + "a_t = v_r*v_t/r;\n", + "\n", + "# results \n", + "print 'u = %.2f fps, v = %.2f fps'%(u,v);\n", + "print ' v_r = %d, v_t = %d fps'%(v_r,v_t);\n", + "print ' a_x = %.2f ft/sec**2, a_y = %.2f ft/sec**2'%(a_x,a_y);\n", + "print ' a_r = %.1f ft/sec**2, a_t = %d'%(a_r,a_t);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "u = -4.33 fps, v = 2.50 fps\n", + " v_r = 0, v_t = 5 fps\n", + " a_x = -6.25 ft/sec**2, a_y = -10.83 ft/sec**2\n", + " a_r = -12.5 ft/sec**2, a_t = 0\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.4 Page No : 88" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# variables\n", + "w = 600.;\t\t# pounds\n", + "l1 = 12.;\t\t#inches\n", + "l2 = 8.;\t\t#inches\n", + "\n", + "# calculations \n", + "Q = w/(62.4);\n", + "V_12 = Q/(0.25*math.pi*(l1/12)**2);\n", + "V_8 = Q/(0.25*math.pi*(l2/12)**2);\n", + "\n", + "# results \n", + "print 'Q = %.2f cfs'%(Q);\n", + "print 'V_12 = %.2f fps and V_8 = %.2f fps'%(V_12,V_8);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Q = 9.62 cfs\n", + "V_12 = 12.24 fps and V_8 = 27.55 fps\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.5 Page No : 89" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "l = 12.;\t\t# inches\n", + "W = 6.;\t\t# pounds\n", + "w = 0.0624\t\t# lb/cuft\n", + "l1 = 8.;\t\t# inches\n", + "rho = 0.050;\t\t# lb/cuft\n", + "\n", + "# calculations \n", + "Q_12 = W/w ;\n", + "Q_8 = W/rho ;\n", + "V_12 = Q_12/(0.25*math.pi*(l/12)**2);\n", + "V_8 = Q_8/(0.25*math.pi*(l1/12)**2);\n", + "\n", + "# results \n", + "print 'Q_12 = %.1f cfs, Q_8 = %d cfs'%(Q_12,Q_8);\n", + "print ' V_12 = %.1f fps, V_8 = %.f fps'%(V_12,V_8);\n", + "\n", + "\t\t" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Q_12 = 96.2 cfs, Q_8 = 120 cfs\n", + " V_12 = 122.4 fps, V_8 = 344 fps\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_Fluid_Mechanics/ch4.ipynb b/Elementary_Fluid_Mechanics/ch4.ipynb new file mode 100755 index 00000000..16e36170 --- /dev/null +++ b/Elementary_Fluid_Mechanics/ch4.ipynb @@ -0,0 +1,491 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a6ceb1d5d0407522d73c0769aac7029af109171f45086c52eac4f70c895f07f9" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 4 : Flow of an Incompressible Ideal Fluid" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.1 Page No : 103" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "d = 4.;\t\t #feet\n", + "theta = 30.;\t\t# degrees\n", + "p_C = 5. \t\t# psi\n", + "\n", + "# calculations \n", + "p_A = p_C-(62.4/144)*math.cos(theta*math.pi/180) *2;\n", + "p_B = p_C+(62.4/144)*math.cos(theta*math.pi/180) *2;\n", + "h = p_C*144/62.4;\n", + "\n", + "# results \n", + "print 'The static pressures at A and B are %.2f psi and %.2f psi respectively.'%(p_A,p_B);\n", + "print 'The hydraulic grade line is %.2f ft vertically above C'%(h);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The static pressures at A and B are 4.25 psi and 5.75 psi respectively.\n", + "The hydraulic grade line is 11.54 ft vertically above C\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.2 Page No : 105" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables \n", + "h = 100.;\t\t#ft\n", + "d1 = 5.;\t\t#in\n", + "d2 = 8.;\t\t#in\n", + "h1 = 60.;\t\t# ft\n", + "h2 = 10.;\t\t#ft\n", + "h3 = 40.;\t\t#ft\n", + "h4 = 102.;\t\t#ft\n", + "H = 300.;\t\t#ft\n", + "theta = 30.;\t\t#degrees\n", + "gam = 0.43;\n", + "\n", + "# calculations \n", + "V5 = math.sqrt(h*2*32.2);\n", + "Q = V5*0.25*math.pi*(d1/12)**2;\n", + "V1 = (d1/12)**4 *h;\n", + "V2 = h*(d1/d2)**4;\n", + "p1 = (h1-V1)*gam;\n", + "p2 = -(h2-V2)*2.04*gam;\n", + "p3 = (h3-V1)*gam;\n", + "p4 = (h4-V1)*gam;\n", + "V6 = V5*math.cos(theta*math.pi/180);\n", + "e = H - (V6**2)/(2*32.2);\n", + "\n", + "# results \n", + "print 'p1 = %.1f psi, p2 = %.1f in. of Hg vacuum, p3 = %.f psi and p4 = %.1f psi'%(p1,p2,p3,p4);\n", + "print 'elevation = %.1f ft'%(e);\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "p1 = 24.5 psi, p2 = 4.6 in. of Hg vacuum, p3 = 16 psi and p4 = 42.6 psi\n", + "elevation = 225.0 ft\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.3 Page No : 107" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables \n", + "p = 14.;\t\t#psia\n", + "gam = 62.;\t\t#lb/cuft\n", + "l1 = 35.;\t\t# ft\n", + "l2 = 10.;\t\t# ft\n", + "d = 6.; \t\t#in\n", + "\n", + "# calculations \n", + "p_v = 2.2*gam;\n", + "p_B = p*144;\n", + "k_c = l1-l2+(p_B/gam)-(p_v/gam);\n", + "K6 = l1;\n", + "d_c = d*(K6/k_c)**0.25;\n", + "\n", + "# results \n", + "print 'd = %.2f in'%(d_c);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "d = 5.35 in\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.4 Page No : 108" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# variables\n", + "rho = 0.00238;\t\t#slug/cuft\n", + "h = 6. \t\t#in\n", + "\n", + "# calculations \n", + "V_0 = math.sqrt(2*(h/12)*(62.4 - rho*32.2)/rho);\n", + "\n", + "# results \n", + "print 'The velocity of the air stream = %.f fps'%(V_0);\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The velocity of the air stream = 162 fps\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.5 Page No : 110" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "sg = 0.82;\n", + "p1 = 20.;\t\t#psia\n", + "p2 = 10.;\t\t#psia\n", + "d1 = 6.;\t\t#in\n", + "d2 = 12.;\t\t#in\n", + "del_z = 4.;\t\t#ft\n", + "d = 18.7;\t\t#in\n", + "\n", + "# calculations \n", + "h1 = (p1-p2)*144/(sg*62.4) - del_z;\n", + "A1 = 0.25*math.pi*(d1/12)**2;\n", + "A2 = 0.25*math.pi*(d2/12)**2;\n", + "V2 = math.sqrt(-2*h1*32.2/(1-(A2/A1)**2));\n", + "V1 = (A2/A1)*V2;\n", + "Q = A1*V1;\n", + "\n", + "# results \n", + "print 'Flow rate = %.2f cfs'%(Q);\n", + "\n", + "#there is a small error in the answer given in textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Flow rate = 8.00 cfs\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.6 Page No : 112" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# variables\n", + "e1 = 100.; \t\t#ft\n", + "theta = 60.;\t\t#degrees\n", + "e2 = 98.5;\t \t#ft\n", + "V_s2 = 20.;\t \t#fps\n", + "e3 = 95.;\t \t#ft\n", + "\n", + "# calculations \n", + "t2 = (e1-e2)/math.cos(theta*math.pi/180);\n", + "p2 = 3*62.4*math.cos(theta*math.pi/180);\n", + "V_F2 = math.sqrt((e1 + (V_s2**2 /(2*32.2)) - p2/62.4 -e2)*2*32.2);\n", + "q = 3*1*V_s2;\n", + "y = 11.22;\t\t#ft\n", + "y1 = 10.74;\t\t#ft\n", + "V1 = math.sqrt((y-y1)*2*32.2);\n", + "\n", + "# results \n", + "print 'On spillway: Pressure = %.1f psf , velocity = %d fps' %(p2,V_F2);\n", + "print 'In the approach channel: Depth = %.2f ft, V1 = %.1f fps'%(y1,V1);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "On spillway: Pressure = 93.6 psf , velocity = 20 fps\n", + "In the approach channel: Depth = 10.74 ft, V1 = 5.6 fps\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.7 Page No : 113" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "d = 10.;\t\t# in\n", + "p = 40.;\t\t#psi\n", + "G = 5.;\t\t#cfs\n", + "y1 = 92.4;\t\t#ft\n", + "k1 = -11.3;\t\t#ft\n", + "k2 = 92.4;\t\t#ft\n", + "k3 = 3.2;\t\t#ft\n", + "k4 = 10.1;\t\t#ft\n", + "\n", + "# calculations \n", + "E_p = k4+y1+d-k1-k3;\n", + "hp = G*62.4*E_p/550;\n", + "\n", + "# results \n", + "print 'Pump horsepower = %.1f hp'%(hp);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Pump horsepower = 68.4 hp\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.9 Page No : 122" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# variables\n", + "sw = 20.;\t\t# specific weight in lb/cuft\n", + "p_B = 6.;\t\t#psi\n", + "p_A = 2.;\t\t#psi\n", + "L = 17.28;\t\t#ft\n", + "l = 10.;\t\t#ft\n", + "\n", + "# calculations \n", + "V_A = math.sqrt(2*32.2*((p_B-p_A)*144/50 - l));\n", + "\n", + "# results \n", + "print 'The mean velocity = %.2f fps'%(V_A);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The mean velocity = 9.89 fps\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.11 Page No : 126" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "D = 6.;\t\t#in\n", + "v = 100.;\t#fps\n", + "p = 0.;\t\t#psi\n", + "gam = 0.08;\t#specific weight in lb/cuft\n", + "R = 6.;\t\t#in\n", + "theta = 60.;\t\t#degrees\n", + "\n", + "# calculations \n", + "v_r = v*(1-(0.5*D/R)**2)*math.cos(theta*math.pi/180);\n", + "v_t = -v*(1+(0.5*D/R)**2)*math.sin(theta*math.pi/180);\n", + "V = math.sqrt(v_r**2 + v_t**2);\n", + "p = ((v**2 /(2*32.2)) - (V**2 /(2*32.2)) - (math.cos(theta*math.pi/180)*math.sin(theta*math.pi/180)))*gam;\n", + "\n", + "# results \n", + "print 'Velocity = %.1f fps Pressure = %.2f psf'%(V,p);\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Velocity = 114.6 fps Pressure = -3.92 psf\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.12 Page No : 127" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "from scipy.integrate import quad \n", + "\t\t\n", + "# variables\n", + "p_A = 0;\n", + "p_B = 0;\n", + "p_C = 0;\n", + "p_D = 0;\n", + "#velocity heads\n", + "V1 = 15.28;\t\t#fps\n", + "V2 = 16.78;\t\t#fps\n", + "V3 = 15.50;\t\t#fps\n", + "V4 = 16.50;\t\t#fps\n", + "\n", + "# calculations \n", + "def f0(h): \n", + "\t return h**(1./2)\n", + "\n", + "\n", + "q = math.sqrt(2*32.2)* quad(f0,3.771,4.229)[0]\n", + "\n", + "# results \n", + "print 'V_A = %.2f fps, V_B = %.2f fps, V_C = %.2f fps, V_D = %.2f fps'%(V1,V2,V3,V4);\n", + "print 'Flow rate = %.2f cfs/ft'%(q);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "V_A = 15.28 fps, V_B = 16.78 fps, V_C = 15.50 fps, V_D = 16.50 fps\n", + "Flow rate = 7.35 cfs/ft\n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_Fluid_Mechanics/ch5.ipynb b/Elementary_Fluid_Mechanics/ch5.ipynb new file mode 100755 index 00000000..ecbe1153 --- /dev/null +++ b/Elementary_Fluid_Mechanics/ch5.ipynb @@ -0,0 +1,364 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:f32ce99f1a0136b5cca3224fda443fb508c92875ffc981f9eb147519b1c553ca" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 5 : Flow of a Compressible Ideal Fluid" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.1 Page No : 152" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\t\n", + "\n", + "# variables\n", + "v1 = 100.;\t\t# fps\n", + "p1 = 50.;\t\t# psia\n", + "T1 = 300.;\t\t# degreeF\n", + "v2 = 500.;\t\t# fps\n", + "Cp = 186.5;\n", + "gam = 1.4;\n", + "\n", + "# calculations \n", + "T2 = T1 - (v2**2 - v1**2)/(2*36.2*Cp);\n", + "p2 = p1*(1 - (v2**2 - v1**2)/(2*36.2*53.3*(T1+460)*(gam/(gam-1))))**(1/0.286);\n", + "\n", + "# results \n", + "print 'T2 = %d degreeF'%(round(T2,-1));\n", + "print 'p2 = %.1f psia'%(p2);\n", + "\n", + "#there is an error in the answer given in textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "T2 = 280 degreeF\n", + "p2 = 46.0 psia\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.2 Page No : 153" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "p1 = 300.;\t\t# psia\n", + "T1 = 900.;\t\t# degreeF\n", + "p2 = 200.;\t\t# psia\n", + "T2 = 780.;\t\t# degreeF\n", + "H2 = 1414.;\t\t#Btu/lb\n", + "H1 = 1471.;\t\t# Btu/lb\n", + "\n", + "# calculations \n", + "V2 = math.sqrt(2*31.1*778*(H1-H2));\n", + "\n", + "# results \n", + "print 'T2 = %d degreeF and V2 = %d fps'%(T2,V2);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "T2 = 780 degreeF and V2 = 1660 fps\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.3 Page No : 155" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "v = 586.;\t\t# fps\n", + "p = 13.;\t\t# psia\n", + "T = 0.;\t\t# degreeF\n", + "gam = 1.4;\n", + "\n", + "# calculations \n", + "rho_0 = p*144/(32.2*53.3*(460+T));\n", + "a_0 = math.sqrt(gam*32.2*53.3*(T+460));\n", + "M_0 = v/a_0;\n", + "p_8_approx = p+(0.5/144)*rho_0*v**2 *(1+0.25*M_0**2);\n", + "p_8_exact = p*(1+M_0**2 *(gam-1)/2)**(gam/(gam-1));\n", + "T_8 = v**2 /(2*32.2*186.5) +460;\n", + "rho_8 = p_8_exact*144/(T_8*32.2*53.3);\n", + "\n", + "# results \n", + "print 'At stagnetion point, p = %.2f psia, T = %.1f degreeR and density = %.5f slug/cuft'%(p_8_exact,T_8,rho_8);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "At stagnetion point, p = 16.05 psia, T = 488.6 degreeR and density = 0.00276 slug/cuft\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.4 pageno : 159" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# variables\n", + "g = 32.2\n", + "k = 1.40\n", + "R = 53.3\n", + "A2 = 0.005454\n", + "p1 = 114.7 * 144\n", + "T1 = 560\n", + "\n", + "# calculations\n", + "y1 = p1 /(R * T1)\n", + "p2 = 0.528 * 114.7\n", + "G1 = math.sqrt(g*k/R * (2/(k+1))**((k+1)/(k-1))) * A2*p1/math.sqrt(T1)\n", + "\n", + "p2 = 94.7 # psia\n", + "p2byp1 = 0.825\n", + "G = A2*math.sqrt( (2*g*k)/(k-1) * p1*y1 * ((p2byp1)**(2/k) - (p2byp1)**((k+1)/k)))\n", + "# result\n", + "print \"part a\"\n", + "print \"G = %.2f lb/sec\"%G1\n", + "print \"\\npart b\"\n", + "print \"G = %.2f lb/sec\"%G" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "part a\n", + "G = 2.03 lb/sec\n", + "\n", + "part b\n", + "G = 1.58 lb/sec\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.5 Page No : 161" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "d = 1.;\t\t # in\n", + "p = 100.;\t\t# psi\n", + "T = 10.;\t\t# degreeF\n", + "p_i = 80.;\t\t#psi\n", + "p_b = 14.7;\t\t#psi\n", + "p1 = 16520.;\t# psfa\n", + "gam1 = 0.553;\t# lb/cuft\n", + "k = 0.874;\n", + "\n", + "# calculations \n", + "G = (0.5*k*0.25*math.pi*(d/12)**2 /(1-(2./3)**4)) *math.sqrt(2*32.2*(p-p_i)*144/gam1);\n", + "\n", + "# results \n", + "print 'flow rate = %.2f lb/sec'%(G);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "flow rate = 1.72 lb/sec\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.6 Page No : 163" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# variables\t\t\n", + "d = 1.;\t\t# in\n", + "p_r = 100.;\t\t#psi\n", + "T_r = 100.;\t\t# degreeF\n", + "p_b = 14.7;\t\t# psi\n", + "p3 = 14.7;\t\t#psi\n", + "G = 2.03;\t\t# lb/sec\n", + "gam1 = 0.553;\n", + "gam = 1.4;\n", + "\n", + "# calculations \n", + "V3 = math.sqrt(2*32.2*(gam/(gam-1))*(p_r+p_b)*144/gam1 *(1-(p3/(p_r+p_b))**((gam-1)/gam)));\n", + "T3 = (T_r+460) - V3**2 /(2*32.2*186.5);\n", + "a3 = math.sqrt(gam*32.2*53.3*T3);\n", + "M3 = V3/a3;\n", + "A3 = G/(gam1*V3);\n", + "d3 = (A3/(0.25*math.pi))**(1./2);\n", + "p3_dash = 103.3;\t\t# psia\n", + "p_B = p3*(1+ (2*gam/(gam+1))*(M3**2 -1));\n", + "\n", + "# results \n", + "print 'V3 = %d fps, a3 = %d fps, M3 = %.2f '%(V3,a3,M3);\n", + "print 'p3_dash = %.1f psia, p_B = %.1f psia'%(p3_dash,p_B);\n", + "\n", + "#there are rounding-off errors in the answer given in textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "V3 = 1728 fps, a3 = 864 fps, M3 = 2.00 \n", + "p3_dash = 103.3 psia, p_B = 66.1 psia\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.7 Page No : 166" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "V_0 = 586.;\t\t# fps\n", + "t_0 = 0;\t\t# degreeF\n", + "P_0 = 13.;\t\t# psia\n", + "a_0 = 1052.;\t\t# fps\n", + "M_0 = 0.557;\n", + "V_A = 800.;\t\t#fps\n", + "V_B = 900.;\t\t#fps\n", + "gam = 1.4;\n", + "\n", + "# calculations \n", + "T_A = 488.5- V_A**2 /(2*32.2*186.5);\n", + "T_B = 488.5- V_B**2 /(2*32.2*186.5); \n", + "p_A = 16.18*(T_A/488.5)**(gam/(gam-1));\n", + "p_B = 16.18*(T_B/488.5)**(gam/(gam-1));\n", + "a_A = math.sqrt(gam*32.2*53.3*T_A);\n", + "a_B = math.sqrt(gam*32.2*53.3*T_B);\n", + "M_A = V_A/a_A;\n", + "M_B = V_B/a_B;\n", + "\n", + "# results \n", + "print 'At point A, p = %.2f psia, T = %.1f degreeR, a = %d fps, M = %.3f'%(p_A,T_A,a_A,M_A);\n", + "print 'At point B, p = %.2f psia, T = %.1f degreeR, a = %d fps, M = %.1f'%(p_B,T_B,a_B,M_B);\n", + "\n", + "#answer differs due to rounding-off errors in textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "At point A, p = 10.80 psia, T = 435.2 degreeR, a = 1022 fps, M = 0.782\n", + "At point B, p = 9.62 psia, T = 421.1 degreeR, a = 1005 fps, M = 0.9\n" + ] + } + ], + "prompt_number": 12 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_Fluid_Mechanics/ch6.ipynb b/Elementary_Fluid_Mechanics/ch6.ipynb new file mode 100755 index 00000000..9b793271 --- /dev/null +++ b/Elementary_Fluid_Mechanics/ch6.ipynb @@ -0,0 +1,622 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:8c235771f8d92166381a73459d7317269e7af4580e65930e8ce9765659a9e720" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 6 : The Impulse Momentum Principle" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.1 Page No : 176" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "G = 10.;\t\t#cfs\n", + "d1 = 12.;\t\t#in\n", + "d2 = 8.;\t\t#in\n", + "p1 = 10.;\t\t#psi\n", + "V = 3.;\t\t #cuft\n", + "theta = 60.;\t#degrees\n", + "p2 = 3.43;\t\t# psi\n", + "w = 187.;\t\t#lb\n", + "\n", + "# calculations \n", + "V1 = G/(0.25*math.pi*(d1/12)**2);\n", + "V2 = G/(0.25*math.pi*(d2/12)**2);\n", + "F1 = 0.25*math.pi*(d1**2)*p1;\n", + "F2 = 0.25*math.pi*d2**2 *p2;\n", + "Fx = F1+F2*math.cos(theta*math.pi/180) - G*1.935*(-V2*math.cos(theta*math.pi/180) - V1);\n", + "Fz = F2*math.sin(theta*math.pi/180) + w + G*1.935*(V2*math.sin(theta*math.pi/180));\n", + "F = round(math.sqrt(Fx**2 + Fz**2),-1);\n", + "alpha = (180/math.pi)*math.atan(Fz/Fx);\n", + "\n", + "# results \n", + "print 'The force on the bend = %d lb at %d degrees with the horizontal'%(F,alpha);\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The force on the bend = 1920 lb at 25 degrees with the horizontal\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.2 Page No : 178" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "l1 = 5.;\t\t#ft\n", + "l2 = 2.;\t\t#ft\n", + "\n", + "# calculations \n", + "V1 = round(math.sqrt(2*32.2*(l2-l1)/(1-(l1/l2)**2)),2);\n", + "V2 = round((l1/l2)*V1,1);\n", + "q = round(l1*V1,1);\n", + "F1 = 62.4*(l1**2)/2;\n", + "F2 = 62.4*(l2**2)/2;\n", + "Fx = F1-F2-q*1.935*(V2-V1);\n", + "\n", + "# results \n", + "print 'Force = %d lb and direction is in downstream direction'%(Fx);\n", + "\n", + "#answer differs due to incorrect value of F2 in textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Force = 118 lb and direction is in downstream direction\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.3 Page No : 182" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# variables\n", + "d = 2.;\t\t#ft\n", + "Q = 40.;\t\t#cfs/ft\n", + "\n", + "# calculations \n", + "V1 = Q/d;\n", + "y1 = d;\n", + "K1 = V1**2 /(32.2*y1);\n", + "y2 = (-1 +math.sqrt(1+8*K1));\n", + "V2 = Q/y2 ;\n", + "delta = d + (V1**2 /(2*32.2)) - y2 - (V2**2 /(2*32.2));\n", + "hp = Q*62.4*delta/550.;\n", + "\n", + "# results \n", + "print 'y2 = %.2f ft, delta = %.2f ft, Horsepower dissipated = %.1f hp'%(y2,delta,hp);\n", + "\n", + "#incorrect answer given in textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "y2 = 6.12 ft, delta = 1.43 ft, Horsepower dissipated = 6.5 hp\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.4 Page No : 184" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# variables\n", + "y1 = 2.;\t\t#ft\n", + "V1 = 20.;\t\t#fps\n", + "beta = 40.;\t\t#degrees\n", + "\n", + "# calculations \n", + "K1 = (V1**2)/(32.2*y1);\n", + "y2 = (-1 + math.sqrt(1+8*K1*(math.sin(beta*math.pi/180))**2));\n", + "k = (y1/y2)*V1*math.sin(beta*math.pi/180);\n", + "del_angle = (180/math.pi)*math.tan(math.sqrt((math.tan(beta*math.pi/180)) *(1+2*k**2 /(32.2*y2))/(1+2*K1*(math.sin(beta*math.pi/180))**2)));\n", + "theta = beta-del_angle;\n", + "\n", + "# results \n", + "print 'The required wedge angle = %.d degrees'%(2*theta);\n", + " \n", + "#rounding-off errors in the answer given in textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The required wedge angle = 16 degrees\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.5 Page No : 186" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "p1 = 14.7;\t\t#psia\n", + "V1 = 1732.;\t\t#pfs\n", + "a1 = 862.;\t\t#fps\n", + "\n", + "# calculations \n", + "M1 = round(V1/a1,2);\n", + "M2 = (math.sqrt((1+(0.4*M1**2)/2)/(1.4*M1**2 - (0.4/2))));\n", + "p2 = p1*(1+2*(1.4/2.4)*(M1**2 -1));\n", + "V2 = V1*((2+0.4*M1**2)/(2.4*M1**2));\n", + "a2 = V2/M2;\n", + "T2 = a2**2/(1.4*32.2*53.3);\n", + "T1 = a1**2/(1.4*32.2*53.3);\n", + "del_T = T2-T1;\n", + "\n", + "# results \n", + "print 'p2 = %.1f psia, V2 = %d fps, a2 = %d fps, T2 = %d degreeR'%(p2,V2,a2,T2);\n", + "print 'Rise of temperature = %d degreeF'%(del_T);\n", + "\n", + "#Answer differs due to rounding-off errors in M2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "p2 = 66.8 psia, V2 = 645 fps, a2 = 1122 fps, T2 = 523 degreeR\n", + "Rise of temperature = 214 degreeF\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6 Page No : 188" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# variables\n", + "p1 = 14.7;\t\t# psia\n", + "v1 = 1732.;\t\t# fps\n", + "a1 = 862.;\t\t# fps\n", + "beta = 40.;\t\t# degrees\n", + "\n", + "# calculations \n", + "M1 = round(v1/a1,2);\n", + "T1 = round(a1**2 /(1.4*32.2*53.3),-1);\n", + "p2 = round(p1*(1 + 2*(1.4/2.4)*(M1**2 *(math.sin(beta*math.pi/180))**2 -1)));\n", + "beta_theta = math.degrees(math.atan(math.tan(math.radians(beta)) * (0.4*(M1*math.sin(math.radians(beta)))**2 +2)/(2.4*(M1*math.sin(math.radians(beta)))**2)))\n", + "theta = round(beta - beta_theta,1)\n", + "M2 = round(math.sqrt((1/math.sin((beta-theta)*math.pi/180)**2) *(1 + (0.4/2)*((M1*math.sin(beta*math.pi/180))**2) )/((1.4*(M1*math.sin(beta*math.pi/180))**2) -(0.4/2))),2)\n", + "v2 = v1*math.cos(beta*math.pi/180)/math.cos((29.3)*math.pi/180);\n", + "a2 = v2/M2;\n", + "T2 = a2**2 /(1.4*32.2*53.3);\n", + "\n", + "# results \n", + "print 'Angle required = %.1f degrees, p2 = %.1f psia, v2 = %d fps, a2 = %.f fps, T2 = %.1f degreeR'%(theta,p2,v2,round(a2,-1),T2);\n", + "\n", + "# note : answers differ due to rounding off error\t\t" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Angle required = 10.8 degrees, p2 = 26.0 psia, v2 = 1521 fps, a2 = 940 fps, T2 = 367.1 degreeR\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.7 Page No : 190" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# variables\n", + "F = 1000;\t\t#lb\n", + "H = 30000;\t\t#ft\n", + "v1 = 500.;\t\t#fps\n", + "v2 = 4000.;\t\t#fps\n", + "p2 = 5;\t\t #psia\n", + "A2 = 1;\t\t #sqft\n", + "p1 = 4.37;\t\t#psia\n", + "\n", + "# calculations \n", + "G_a = (F - (p2-p1)*A2*144)*32.2/(v2-v1);\n", + "\n", + "# results \n", + "print 'Ga = %.1f lb/sec'%(G_a);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ga = 8.4 lb/sec\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.8 Page No : 194" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "gam = 0.0765;\t# lb/cuft\n", + "V1 = 293.;\t\t#fps\n", + "hp = 1500.;\n", + "h = 10.;\t\t#ft\n", + "V4 = 338.;\t\t#fps\n", + "\n", + "# calculations \n", + "V = 0.5*(V1+V4);\n", + "Q = hp*550/((V4-V1)*V*gam/32.2);\n", + "d1 = math.sqrt(Q/(V1*0.25*math.pi));\n", + "d4 = math.sqrt(Q/(V4*0.25*math.pi));\n", + "F = Q*(gam/32.2)*(V4-V1);\n", + "eta = V1/V;\n", + "\n", + "# results \n", + "print 'V4 = %d fps, V = %.1f fps, d1 = %.1f ft, d4 = %.2f ft, F = %d lb, efficiency = %.1f percentage'%(V4,V,d1,d4,F,eta*100);\n", + "\n", + "#answer differs due to rounding-off errors" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "V4 = 338 fps, V = 315.5 fps, d1 = 10.3 ft, d4 = 9.60 ft, F = 2614 lb, efficiency = 92.9 percentage\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.9 Page No : 198" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "D = 6.;\t\t#ft\n", + "d = 2.;\t\t#in\n", + "V1 = 200.;\t\t#fps\n", + "rpm = 250.;\n", + "theta = 150.;\t\t#degrees\n", + "\n", + "# calculations \n", + "u = (rpm/60)*2*math.pi*0.5*D;\n", + "v1 = V1-u;\n", + "v2 = v1;\n", + "V_2x = v1*math.cos(theta*math.pi/180) + u;\n", + "V_2y = v2*math.sin(theta*math.pi/180);\n", + "V2 = math.sqrt(V_2x**2 + V_2y**2);\n", + "Q = 0.25*math.pi*(d/12)**2 *V1;\n", + "F_x = Q*1.935*(V_2x-V1);\n", + "P = F_x*u/550;\n", + "\n", + "# results \n", + "print 'The working component of force on fluid = %d lb, P = %.f hp'%(F_x,-P);\n", + "\n", + "#answer differs due to rounding-off errors" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The working component of force on fluid = -1913 lb, P = 273 hp\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.10 Page No : 199" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "P = 100.;\t\t#hp\n", + "V = 75.;\t\t#fps\n", + "V1 = 150.;\t\t#fps\n", + "d = 2.;\t\t#in\n", + "alpha1 = 60.;\t\t#degrees\n", + "\n", + "# calculations \n", + "Q = round(0.25*math.pi*(d/12)**2 *V1,2);\n", + "F_y = round(550*P/V);\n", + "V2 = math.sqrt(V1**2 - P*550/(Q*1.935/2));\n", + "alpha2 = (180/math.pi)*math.asin((V1*math.sin(alpha1*math.pi/180) - (F_y/(Q*1.935)))/V2);\n", + "beta1 = 90 - (180/math.pi)*math.atan((V1*math.sin(alpha1*math.pi/180) - V)/(V1*math.cos(alpha1*math.pi/180)));\n", + "beta2 = 90 + (180/math.pi)*math.atan((V-V2*math.sin(alpha2*math.pi/180))/(V1*math.cos(alpha1*math.pi/180)));\n", + "\n", + "\n", + "# results \n", + "print 'Beta1 = %.f degrees, Beta2 = %.f degrees'%(beta1,round(beta2,-1));\n", + "\n", + "\n", + "# note : answer differs due to rounding off error\t\t" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Beta1 = 54 degrees, Beta2 = 130 degrees\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.11 Page No : 203" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "r1 = 5.;\t\t#ft\n", + "r2 = 3.5;\t\t#ft\n", + "beta1 = 60.;\t\t#degrees\n", + "beta2 = 150.;\t\t#degrees\n", + "t = 1.;\t\t#ft\n", + "alpha1 = 15.;\t\t#degree\n", + "Q = 333.;\t\t#cfs\n", + "gam = 0.434;\n", + "\n", + "# calculations \n", + "V_r1 = round(Q/(2*math.pi*r1),1);\n", + "V_r2 = round(Q/(2*math.pi*r2),2);\n", + "V_t1 = round(V_r1*(1/math.tan(alpha1*math.pi/180)),1);\n", + "u1 = round(V_t1 - V_r1*math.tan((90-beta1)*math.pi/180),1);\n", + "omega = round(u1/r1,1);\n", + "u2 = round(omega*r2,1);\n", + "V_t2 = round(u2 - V_r2*(1/math.tan((90-beta1)*math.pi/180)),1);\n", + "T = Q*1.935*(V_t1*r1 - (V_t2*r2));\n", + "hp = T*omega/550;\n", + "E_T = hp*550/(Q*62.4);\n", + "V1 = math.sqrt(V_r1**2 + V_t1**2);\n", + "V2 = math.sqrt(V_r2**2 + V_t2**2);\n", + "del_p = E_T*gam + (gam/(2*32.2))*(V2**2 - V1**2);\n", + "\n", + "# results \n", + "print 'V1 = %.1f fps, V2 = %.1f fps, T = %d ft-lb, hp = %d lb, E_T = %.1f ft-lb/lb, p1-p2 = %.1f psi'%(V1,V2,T,hp,E_T,del_p);\n", + "\n", + "#incorrect answer given in textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "V1 = 41.0 fps, V2 = 15.4 fps, T = 133896 ft-lb, hp = 1631 lb, E_T = 43.2 ft-lb/lb, p1-p2 = 9.0 psi\n" + ] + } + ], + "prompt_number": 49 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.12 Page No : 204" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "r1 = 3.;\t\t#in\n", + "r2 = 10.;\t\t#in\n", + "beta1 = 120.;\t\t#degrees\n", + "beta2 = 135.;\t\t#degrees\n", + "t = 1.;\t\t#in\n", + "Q = 4.;\t\t#cfs\n", + "gam = 0.434;\n", + "\n", + "# calculations \n", + "V1 = Q*144/(2*math.pi*r1);\n", + "V_r1 = V1;\n", + "V_r2 = Q*144/(2*math.pi*r2);\n", + "u1 = V1*math.tan((beta1-90)*math.pi/180);\n", + "omega = u1/(r1/12);\n", + "u2 = omega*(r2/12);\n", + "V_t2 = u2 - V_r2/math.tan((180-beta2)*math.pi/180);\n", + "T = Q*1.935*(V_t2*(r2/12));\n", + "P = T*omega/547.561;\t\t#hp\n", + "E_P = P*550/(Q*62.4);\n", + "V2 = math.sqrt(V_r2**2 + V_t2**2);\n", + "del_p = E_P*gam + (gam/(2*32.2))*(V1**2 - V2**2);\n", + "\n", + "# results \n", + "print 'Rotational speed = %.1f rad/sec '%(omega)\n", + "print ' T = %d ft-lb, P = %.1f hp'%(T,P);\n", + "print ' The energy given to each pound of water = %d ft'%(E_P);\n", + "print ' The pressure rise = %.1f psi'%(del_p);\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rotational speed = 70.6 rad/sec \n", + " T = 320 ft-lb, P = 41.3 hp\n", + " The energy given to each pound of water = 90 ft\n", + " The pressure rise = 28.6 psi\n" + ] + } + ], + "prompt_number": 15 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_Fluid_Mechanics/ch7.ipynb b/Elementary_Fluid_Mechanics/ch7.ipynb new file mode 100755 index 00000000..219a93a5 --- /dev/null +++ b/Elementary_Fluid_Mechanics/ch7.ipynb @@ -0,0 +1,292 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:46e45d2f6c46e25d719ba327e4ef8cabd3b2b67788136ad3286b5847149f6378" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 7 : Flow of a Real Fluid" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.1 Page No : 225" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "nu = 0.00001;\t\t# sqft/sec\n", + "d = 1.;\t\t#in\n", + "R_c = 2100.;\n", + "\n", + "# calculations \n", + "V = R_c*nu/(d/12);\n", + "Q = V*0.25*math.pi*(d/12)**2;\n", + "\n", + "# results \n", + "print 'Q = %.6f cfs'%(Q);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Q = 0.001374 cfs\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.2 pageno : 230" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# calculations\n", + "e = 0.2/(1./0.33)\n", + "l2 = 0.2/(1.94 * (1/0.33)**2)\n", + "l = math.sqrt(l2)\n", + "k2 = 0.2/(1.94 * (1/0.33)**2 / (-1/(0.33**2)**0.5)**2)\n", + "k = math.sqrt(k2)\n", + "\n", + "# results\n", + "print \"E = %.3f lb-sec/sqft\"%e\n", + "print \"l = %.3f ft\"%l\n", + "print \"k = %.2f\"%k\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "E = 0.066 lb-sec/sqft\n", + "l = 0.106 ft\n", + "k = 0.32\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.4 Page No : 240" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "G = 240.;\t\t#lb/sec\n", + "A1 = 4.;\t\t#sqft\n", + "A2 = 2.;\t\t#sqft\n", + "z1 = 30.;\t\t#ft\n", + "z2 = 80.;\t\t#ft\n", + "V1 = 600.;\t\t# fps\n", + "V2 = 800.;\t\t#fps\n", + "p1 = 20.;\t\t#psia\n", + "p2 = 35.;\t\t# psia\n", + "\n", + "# calculations \n", + "gam1 = G/(A1*V1);\n", + "gam2 = G/(A2*V2);\n", + "T1 = p1*144/(53.3*gam1);\n", + "T2 = p2*144/(53.3*gam2);\n", + "del_H = 186.5*(T2-T1);\n", + "E_H1 = (V2**2)/(2*32.2) - (V1**2)/(2*32.2) +del_H+z2-z1;\n", + "E_H2 = (V2**2)/(2*32.2) - (V1**2)/(2*32.2) +del_H;\n", + "Q = G*E_H2/550.;\n", + "\n", + "# results \n", + "print 'T1 = %d degreeR, T2 = %d degreeR'%(T1,T2);\n", + "print ' The net heat energy added = %d hp'%(round(Q,-1));\n", + "\n", + "#answer differs due to rounding-off errors" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "T1 = 540 degreeR, T2 = 630 degreeR\n", + " The net heat energy added = 9230 hp\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.5 Page No : 240" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "G = 50.;\t\t#cfs\n", + "Q = 400.;\t\t#hp\n", + "A1 = 4.;\t\t#sqft\n", + "A2 = 2.;\t\t#sqft\n", + "z1 = 30.;\t\t#ft\n", + "z2 = 80.;\t\t#ft\n", + "p1 = 20.;\t\t#psi\n", + "p2 = 10.;\t\t#psi\n", + "\n", + "# calculations \n", + "V1 = G/A1;\n", + "V2 = G/A2;\n", + "E_p = Q*(550/62.4)/G;\n", + "h_L = (p1-p2)*144/62.4 + (V1**2 - V2**2)/(2*32.2) +(z1-z2)+E_p;\n", + "\n", + "# results \n", + "print 'Head lost = %.1f ft'%(h_L);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Head lost = 36.3 ft\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.6 Page No : 243" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "b = 3.;\t\t#ft\n", + "d = 2.;\t\t#ft\n", + "l = 200.;\t\t#ft\n", + "h_L = 30.;\t\t#ft\n", + "\n", + "# calculations \n", + "tau_0 = h_L*62.4*b*d/(10*l);\t\t#0.00694\n", + "\n", + "# results \n", + "print 'The resistance stress exerted between fluid and conduit walls = %.2f psf = %.3f psi'%(tau_0,tau_0*0.00694);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The resistance stress exerted between fluid and conduit walls = 5.62 psf = 0.039 psi\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.7 Page No : 244" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables \n", + "h_L = 30.;\t\t#ft\n", + "l = 200.;\t\t#ft\n", + "d = 2.;\t\t#ft\n", + "r = 8.;\t\t#in\n", + "\n", + "# calculations \n", + "#part (a)\n", + "tau_0 = h_L*62.4/(d*l);\n", + "#part(b)\n", + "tau = (0.5*r/12)*(tau_0*0.00694);\n", + "\n", + "# results \n", + "print 'Parta): Shear stress = %.2f psf = %.4f psi '%(tau_0,tau_0*0.00694);\n", + "print 'Partb): Shear stress = %.4f psi '%(tau);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Parta): Shear stress = 4.68 psf = 0.0325 psi \n", + "Partb): Shear stress = 0.0108 psi \n" + ] + } + ], + "prompt_number": 10 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_Fluid_Mechanics/ch8.ipynb b/Elementary_Fluid_Mechanics/ch8.ipynb new file mode 100755 index 00000000..7725321e --- /dev/null +++ b/Elementary_Fluid_Mechanics/ch8.ipynb @@ -0,0 +1,193 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:736a6a7d4118b4fdd4663c0d1027cf5bfb979230c26b70e1555cf0d37da46444" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 8 : Similitude and Dimensional Analysis" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.1 Page No : 266" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\t\n", + "\n", + "# variables\n", + "Tw = 32.;\t\t# degreeF\n", + "d1 = 3.;\t\t# in\n", + "v = 10.;\t\t#fps\n", + "delp = 2.;\t\t#psi\n", + "h1 = 30.;\t\t# ft\n", + "Tb = 68.;\t\t#degreeF\n", + "d2 = 1.;\t\t#in\n", + "h2 = 10.;\t\t#ft\n", + "\n", + "# calculations \n", + "V = v*(d1/12)*0.0000137/((d2/12)*0.88*0.0000375);\n", + "del_p = delp/h2**2 *0.88*V**2;\n", + "\n", + "# results \n", + "print 'V = %.2f fps del_p = %.2f psi'%(V,del_p);\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "V = 12.45 fps del_p = 2.73 psi\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.2 Page No : 266" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "l = 400.;\t\t# ft\n", + "h = 10.;\t\t#ft\n", + "v = 30.;\t\t# fps\n", + "D = 2.;\t\t#lb\n", + "\n", + "# calculations \n", + "V = math.sqrt((v**2 /l)*h);\n", + "D_p = (D/V**2) *(v**2)*(l**2)/h**2;\n", + "\n", + "# results \n", + "print 'V = %.2f fps Prototype drag = %d lb'%(V,D_p);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "V = 4.74 fps Prototype drag = 128000 lb\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.3 Page No : 266" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\t\t\n", + "# variables\n", + "G = 20000.;\t\t#cfs\n", + "k = 1./15;\n", + "\n", + "# calculations \n", + "Q_m = G*(k)**(2+ 1./2);\n", + "\n", + "# results \n", + "print 'Qm = %.f cfs'%(Q_m);\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Qm = 23 cfs\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.4 Page No : 266" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "k = 1./10;\n", + "v = 3000.;\t\t#fps\n", + "h = 15000.;\t\t#altitude\n", + "T = 68.;\t\t# degreeF\n", + "am = 870.;\t\t#fps\n", + "ap = 1057.;\t\t#fps\n", + "\n", + "# calculations \n", + "Vm = v*(am/ap);\n", + "rho_m = v*(1/k)*0.001495*0.031/(0.033*Vm);\n", + "p_m = 32.2*rho_m*34.9*(T+460)/(144);\n", + "\n", + "# results \n", + "print 'Vm = %d fps p_m = %d psia'%(Vm,p_m);\n", + "\n", + "#rounding-off errors" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Vm = 2469 fps p_m = 70 psia\n" + ] + } + ], + "prompt_number": 6 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_Fluid_Mechanics/ch9.ipynb b/Elementary_Fluid_Mechanics/ch9.ipynb new file mode 100755 index 00000000..bbe5b2d4 --- /dev/null +++ b/Elementary_Fluid_Mechanics/ch9.ipynb @@ -0,0 +1,780 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:0b5a0b6f6c6e071c339201ca91727b19aa4afe9cff0ebf23f723c14d23de7be3" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 9 : Fluid Flow in Pipes" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.1 Page No : 281" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# variables\t\t\n", + "d = 6.;\t \t#inches\n", + "v = 15.;\t\t#fps\n", + "l = 100.;\t\t#ft\n", + "h_L = 17.5;\t\t#ft\n", + "\n", + "# calculations \n", + "f = round(h_L*(d/(12*l))*(2*32.2/v**2),3);\n", + "V_f = v*math.sqrt(f/8.);\n", + "\n", + "# results \n", + "print 'The friction velocity = %.2f fps'%(V_f);\n", + "\n", + "#incorrect answer in textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "0.025\n", + "The friction velocity = 0.84 fps\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.2 Page No : 285" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "T = 100.;\t\t# degreeF\n", + "d = 3.;\t\t # inches\n", + "Re = 80000.;\t# Reynolds number\n", + "e = 0.006;\t\t# inches\n", + "l = 1000.;\t\t#feet\n", + "f1 = 0.021;\t\t#friction factor\n", + "nu = 0.729*10**-5;\t\t# sqft/sec\n", + "\n", + "# calculations \n", + "V = Re*nu/0.25;\n", + "h_L1 = f1*(l/0.25)*(V**2 /(2*32.2));\n", + "f = 0.316/Re**0.25;\n", + "h_L = (f/f1)*h_L1;\n", + "\n", + "# results \n", + "print 'Head loss expected = %.1f ft and head loss expected if the pipe were smooth = %.2f ft'%(h_L1,h_L);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Head loss expected = 7.1 ft and head loss expected if the pipe were smooth = 6.35 ft\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.3 Page No : 288" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "T = 100.;\t\t#degreeF\n", + "d = 3.;\t\t # inches\n", + "Re = 80000.;\t# Reynolds number\n", + "e = 0.006;\t\t#inches\n", + "l = 1000.;\t\t#ft\n", + "f = 0.0255;\t\t#friction factor\n", + "V = 2.33;\t\t#fps\n", + "\n", + "# calculations \n", + "h_L = f*(l/0.25)*(V**2 /(2*32.2));\n", + "\n", + "# results \n", + "print 'Head loss expected = %.1f ft'%(h_L);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Head loss expected = 8.6 ft\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.4 Page No : 290" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "Q = 100.;\t\t#gallons per minute\n", + "sg = 0.90;\n", + "nu = 0.0012;\t# lb-sec/sqft\n", + "d = 3.;\t\t # in\n", + "l = 1000.;\t\t#ft\n", + "r = 1.;\t\t #in\n", + "V = 4.53;\t\t#fps\n", + "\n", + "# calculations \n", + "Re = V*(d/12)*sg*1.935/nu;\n", + "h_L = (64/Re)*(12*l/d)*(V**2 /(2*32.2));\n", + "v = 2*V*(1 - (2/d)**2);\n", + "tau = 62.4*sg*h_L/(2*l*12);\n", + "\n", + "# results \n", + "print 'v = %.2f fps, h_L = %.1f ft of oil and tau = %.3f psf'%(v,h_L,tau);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "v = 5.03 fps, h_L = 49.6 ft of oil and tau = 0.116 psf\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.5 pageno : 293" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# variables\n", + "# assume\n", + "v_vc = 0.80 \n", + "V = 8.35 #fps\n", + "R = 737000\n", + "f = 0.019 # from fig.\n", + "\n", + "# calculations\n", + "V_Vc = 1./(1+ 4.07* math.sqrt(f/8))\n", + "Q = math.pi * V/4\n", + "# results\n", + "print \"V/Vc = %.3f\"%V_Vc\n", + "print \"Q = %.2f cfs\"%Q" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "V/Vc = 0.834\n", + "Q = 6.56 cfs\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.6 Page No : 295" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "Q = 90.;\t\t# gallons per minute\n", + "T = 68.;\t\t#degreeF\n", + "d = 3.;\t\t # in\n", + "l = 3000.;\t\t#ft\n", + "r = 1.;\t\t # in\n", + "f = 0.018;\n", + "\n", + "# calculations \n", + "V = Q/(60*7.48*0.25*math.pi*(d/12)**2);\n", + "Re = V*(d/12)*1.935/(0.000021);\n", + "h_L = f*(l/0.25)*(V**2 /(2*32.2));\n", + "tau_0 = f*1.935*V**2 /8;\n", + "tau1 = 2*tau_0/d;\n", + "v_c = V*(1+4.07*math.sqrt(f/8));\n", + "v_ = math.sqrt(tau_0/1.935);\n", + "v1 = v_*(5.50+5.75*math.log10(v_*(r/(2*12))/0.00001085));\n", + "v1_ = v_c-v_*5.75*math.log10(0.5*d/(r/2));\n", + "delta = d*32.8/(Re*math.sqrt(f));\n", + "\n", + "# results \n", + "print 'Head lost = %.1f ft of water \\\n", + "\\nWall shear stress = %.3f psf \\\n", + "\\nthe center velocity = %.2f fps \\\n", + "\\nshearing stress = %.3f psf \\\n", + "\\nv1 = %.2f fps \\\n", + "\\ndelta = %.4f in.'%(h_L,tau_0,v_c,tau1,v1_,delta);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Head lost = 56.0 ft of water \n", + "Wall shear stress = 0.073 psf \n", + "the center velocity = 4.87 fps \n", + "shearing stress = 0.048 psf \n", + "v1 = 4.34 fps \n", + "delta = 0.0078 in.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.7 Page No : 298" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "d = 12.;\t\t# in\n", + "v = 10.;\t\t#fps\n", + "e = 2.;\t\t #in\n", + "k = 0.002;\t\t#relative roughness\n", + "l = 1000.;\t\t#ft\n", + "\n", + "# calculations \n", + "f = (1/(1.14+2*math.log10(1/k)))**2;\n", + "v_c = v*(1+4.07*math.sqrt(f/8));\n", + "tau_0 = f*1.935*v**2 /8;\n", + "v2 = v_c - tau_0*5.75*math.log10(0.5*d/e);\n", + "v2_ = 8.48*tau_0 + tau_0*5.75*math.log10(e/(12*k));\n", + "h_L = f*(l)*v**2 /(2*32.2); \n", + "\n", + "# results \n", + "print 'f = %.4f, v_c = %.2f fps, v2 = %.1f fps and h_L = %.1f ft of water'%(f,v_c,v2_,h_L);\n", + "\n", + "#there are small errors in the answer given in textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "f = 0.0234, v_c = 12.20 fps, v2 = 11.0 fps and h_L = 36.3 ft of water\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.8 Page No : 300" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "V = 4.08;\t\t # fps\n", + "Re = 93800.;\t\t#Reynolds number\n", + "r = 1.;\t\t#in\n", + "m = 1./7;\n", + "R = 3.;\t\t#in\n", + "\n", + "# calculations \n", + "f = 0.316/(Re**0.25);\n", + "v_c = V/(2/((m+1)*(m+2)));\n", + "v1 = v_c*(r/R)**(1./7);\n", + "tau_0 = f*1.935*V**2 /8;\n", + "\n", + "# results \n", + "print 'f = %.3f, v_c = %.2f fps, v1 = %.2f fps and wall shear = %.3f ps'%(f,v_c,v1,tau_0);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "f = 0.018, v_c = 5.00 fps, v1 = 4.27 fps and wall shear = 0.073 ps\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.9 Page No : 302" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "p = 14.7;\t\t#psia\n", + "T = 60.;\t\t# degreeF\n", + "l = 2000.;\t\t#ft\n", + "b = 18.;\t\t#in\n", + "h = 12.;\t\t# in\n", + "v = 10.;\t\t# fps\n", + "\n", + "# calculations \n", + "R_h = (b*h)/(2*12*(b+h));\n", + "Re = v*4*R_h*0.0763/(32.2*0.000000375);\n", + "f = 0.019;\n", + "h_L = f*(l/(4*R_h))*v**2 /(2*32.2);\n", + "del_p = 0.0763*h_L;\n", + "\n", + "# results \n", + "print 'loss of head = %.1f ft of air and the pressure drop = %.2f psf = %.3f psi'%(h_L,del_p,del_p*0.0069);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "loss of head = 49.2 ft of air and the pressure drop = 3.75 psf = 0.026 psi\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.10 Page No : 305" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# variables\n", + "Q = 90.;\t\t#gpm\n", + "d = 3.;\t\t#in\n", + "l = 3000.;\t\t#ft\n", + "\n", + "# calculations \n", + "V = Q/(60*7.48*0.25*math.pi*(d/12)**2);\n", + "R_h = (d/12)/4;\n", + "C_hw = 140;\n", + "S = (V/(1.318*140*R_h**0.63))**(1/0.54);\n", + "h_L = S*l;\n", + "\n", + "# results \n", + "print 'The loss of head = %.1f ft of water'%(h_L);\n", + "\n", + "\t\t" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The loss of head = 65.7 ft of water\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.11 Page No : 307" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import *\n", + "import math \n", + "\n", + "# variables\n", + "G = 40.;\t\t# lb/min\n", + "d = 3.;\t\t# in\n", + "T = 100.;\t\t# degreeF\n", + "p = 50.;\t\t# psia\n", + "l = 2000.;\t\t#ft\n", + "\n", + "# calculations \n", + "Re = ((G/60)*(d/12))/(0.0491*32.2*4*10**-7);\n", + "f = 0.015;\n", + "gam1 = p*(144/(53.3*(T+460)));\n", + "V1 = (G/60)/(gam1*0.0491);\n", + "a = math.sqrt(1.4*32.2*53.3*(T+460));\n", + "M1 = V1/a;\n", + "M2_limit = math.sqrt(1/1.4);\n", + "l = (((1-(M1/M2_limit)**2)/(1.4*M1**2)) - 2*math.log(M2_limit/M1))*(0.25/0.015);\n", + "p2 = 38.9;\t\t#psia, from trial and error method \n", + "#p2 = Symbol('p2')\n", + "#ans = solve((G/60)**2 * 53.3*560/(32.2 * 0.0491**2) * (2*log(p/p2) + gam1*l/0.25) - (144**2 * (p**2 - p2**2)))\n", + "\n", + "# results \n", + "print 'p2 = %.1f psia'%(p2);\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "p2 = 38.9 psia\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.12 Page No : 312" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "d = 12.;\t\t# in\n", + "D = 24.;\t\t#in\n", + "theta = 20.;\t\t#degrees\n", + "G = 10.;\t\t#cfs\n", + "p = 20.;\t\t#psi\n", + "\n", + "# calculations \n", + "V12 = G/(0.25*math.pi);\n", + "V24 = V12/4;\n", + "K_L = 0.43;\n", + "p24 = ((p*144/62.4) + (V12**2 /(2*32.2)) - ((V24**2)/(2*32.2)) - K_L*(V12-V24)**2 /(2*32.2))/2.314;\n", + "\n", + "# results \n", + "print 'Pressure in the larger pipe = %.1f psi'%(p24);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Pressure in the larger pipe = 20.7 psi\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.13 Page No : 322" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# variables\n", + "d = 12.;\t\t# in\n", + "l = 1000.;\t\t#ft\n", + "h1 = 200.;\t\t#elevation\n", + "h2 = 250.;\t\t#elevation\n", + "T = 50.;\t\t#degreeF\n", + "f1 = 0.030;\n", + "\n", + "# calculations \n", + "V1 = math.sqrt((h2-h1)*2*32.2/(0.5+f1*l +1));\n", + "R1 = V1/0.00000141;\n", + "f2 = 0.019;\n", + "V2 = math.sqrt((h2-h1)*2*32.2/(0.5+f2*l +1));\n", + "R2 = V1/0.00000141;\n", + "Q = 0.25*math.pi*(d/12)**2 *V2; \n", + "\n", + "# results \n", + "print 'Velocity = %.1f fps \\\n", + "\\nflow rate = %.1f cfs'%(V2,Q);\n", + "\n", + "\t\t#there is a minute error in the answer given in textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Velocity = 12.5 fps \n", + "flow rate = 9.8 cfs\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.14 Page No : 322" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "l = 200.;\t\t#ft\n", + "Q = 0.1;\t\t#cfs\n", + "del_h = 5.;\t\t#ft\n", + "T = 50.;\t\t#degreeF\n", + "d = 0.187;\t\t#ft\n", + "\n", + "# calculations \n", + "V = Q/(0.25*math.pi*d**2);\n", + "R = V*d/0.0000141;\n", + "f = (del_h*2*32.2/V**2 -(1+0.5))*(d/l);\n", + "\n", + "# results \n", + "print 'Required diameter of the pipe = %.2f in.'%(d*12);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Required diameter of the pipe = 2.24 in.\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.15 Page No : 324" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "Q = 2.5;\t\t#cfs\n", + "T = 50.;\t\t#degreeF\n", + "d1 = 8.;\t\t#in\n", + "d2 = 6.;\t\t#in\n", + "l1 = 1000.;\t\t#ft\n", + "l2 = 2000.;\t\t#ft\n", + "\n", + "# calculations \n", + "V8 = Q/(0.25*math.pi*(d1/12)**2);\n", + "V6 = Q/(0.25*math.pi*(d2/12)**2);\n", + "R8 = V8*0.667/0.0000141;\n", + "f8 = 0.020;\n", + "R6 = V6*0.5/0.0000141;\n", + "f6 = 0.019;\n", + "h_L8 = f8*(l1/0.667)*(V8**2 /(2*32.2));\n", + "h_L6 = f6*(l2/0.5)*(V6**2 /(2*32.2));\n", + "Ep = 100+h_L8+h_L6;\n", + "n = Q*62.4*(Ep)/550;\n", + "V8 = math.sqrt((30/f8)*2*32.2/(l1/0.667));\n", + "Q_max = V8*0.25*math.pi*(d1/12)**2;\n", + "\n", + "# results \n", + "print 'Maximum reliable flow that can be pumped = %.1f cfs'%(Q_max);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum reliable flow that can be pumped = 2.8 cfs\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.16 Page No : 327" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\t\t\n", + "# variables\n", + "Q = 5.;\t\t#cfs\n", + "d = 12.;\t\t#in\n", + "l = 5000.;\t\t#ft\n", + "h = 70.;\t\t#ft\n", + "L = 2000.;\t\t#ft\n", + "\n", + "# calculations \n", + "K = (h/Q**1.85);\n", + "a = (L/l)*K;\n", + "b = ((l-L)/l)*K;\n", + "Q_ = (h/((b+a*(0.5**(1.85)))))**(1/1.85);\n", + "Q_A = Q_/2;\n", + "Q_B = Q_/2;\n", + "del1 = Q_-Q;\t\t#gain capcaity\n", + "percent = (del1/Q)*100;\t\t#gain percentage\n", + "\n", + "# results \n", + "print 'The gain of capacity by looping the pipe is %.1f cfs or %d percentage'%(del1,percent);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The gain of capacity by looping the pipe is 1.0 cfs or 20 percentage\n" + ] + } + ], + "prompt_number": 19 + } + ], + "metadata": {} + } + ] +}
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--- /dev/null +++ b/Elementary_heat_Power/Chapter1.ipynb @@ -0,0 +1,272 @@ +{ + "metadata": { + "name": "EHP-1" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Matter and Energy" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1.1, Page 17" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\ng=32.2 #ft/s^2\nm=500.0 #lb\nrate=10.0 #ft/s^2\n\n#calculations\nF1=m/g *rate\nms=m/g\nF2=ms*rate\n\n#results\nprint \"Force in case 1 in lbf\",round(F1,3)\nprint \"Force in case 2 in lbf\",round(F2,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Force in case 1 in lbf 155.28\nForce in case 2 in lbf 155.28\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1.2,Page 18" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\ng=32.2 #ft/s^2\ng2=32.0 #ft/s^2\nrate=10 #ft/s^2\nw1=500 #lbf\n\n#calculations\nfd1=w1*g2/g\nF=fd1/g2 *rate\nms=w1/g\nF2=ms*rate\n#results\nprint \"Net weight of body in case 1 in lbf\",round(F,3)\nprint \"Force in case 2 in lbf\",round(F2,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Net weight of body in case 1 in lbf 155.28\nForce in case 2 in lbf 155.28\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1.3, Page 19" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\ng=32.174 #ft/s^2\nm=500.0 #lbm\nrate=10.0 #ft/s^2\n\n#calculations\nF=1/g *m*rate\n\n#results\nprint \"Force required in lbf\",round(F,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Force required in lbf 155.405\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1.4, Page 21" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\ng1=32.174 #ft/s^2\ngc=g1\ng2=30 #ft/s^2\nm=100 #lbm\n\n#calculations\nw1=g1/gc *m\nw2=g2/gc *m\n\n#results\nprint \"Weight in case 1 in lbf\",round(w1,3)\nprint \" Weight in case 2 in lbf\",round(w2,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Weight in case 1 in lbf 100.0\n Weight in case 2 in lbf 93.243\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1.5,Page 22" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nge=32.174 #ft/s^2\ngm=5.47 #ft/s^2\nwe=50 #lbm\n\n#calculations\nwm=we*gm/ge\n\n#results\nprint \"In case a, it will weigh the same, weight in lbm\",round(we,2)\nprint \" In case b, weight in lbf\",round(wm,2)\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "In case a, it will weigh the same, weight in lbm 50.0\n In case b, weight in lbf 8.5\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1.6,Page 31" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\ng=32.2 #ft/s^2\np1=100 #psig\np2=29.0 #in of Hg\n\n#calculations\nBP=p2*0.491\nAP=BP+p1\n\n#results\nprint \"Absolute pressure in psia\",round(AP,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Absolute pressure in psia 114.239\n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1.7,Page 32" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\ng=32.2 #ft/s^2\nPb=29.5 #in of Hg\nPv=10 #in of Hg\n\n#calculations\nAP=(Pb-Pv)*0.491\n\n#results\nprint \"Absoulte pressure in psia\",round(AP,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Absoulte pressure in psia 9.575\n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1.8,Page 38" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\ng=32.2 #ft/s^2\nv1=1 #cu ft\np1=100 #psia\n\n#calculations\nv2=2*v1\nW=144*p1*(v2-v1)\n\n#results\nprint \"Work done in ft-lb\",round(W,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Work done in ft-lb 14400.0\n" + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1.9,Page 39" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\ng=32.2 #ft/s^2\nv1=1 #cu ft\np1= 100.0 #psia\np2=50.0 #psia\nv2=3.0 #cu ft\n\n#calculations\npa=(p1+p2)/2\nW=pa*(v2-v1)*144\n\n#results\nprint \"Work done in ft-lb\",round(W,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Work done in ft-lb 21600.0\n" + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1.10,Page 40" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\ng=32.2 #ft/s^2\np1=100.0 #psia\np2=25.0 #psia\nv2=2.0 #cu ft\n\n#calculations\nW=p1*144*v2 - p2*144*v2\n\n#results\nprint \"Work done in ft-lb\",round(W,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Work done in ft-lb 21600.0\n" + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1.11,Page 41" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\ng=32.2 #ft/s^2\nn=100 #rpm\np1=100 #psia\np2=25 #psia\nv2=2 #cu ft\n\n#calculations\nW=p1*144*v2 - p2*144*v2\nHp=W*n/33000\n\n#results\nprint round(Hp,3),\"Horsepower developed in hp\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "65.0 Horsepower developed in hp\n" + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1.12,Page 46" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#initialisation of variable\nP=50.0 #hp\nm=30.0 #lb\nE=19000.0 #Btu/lb\n\n#calculations\neta= P*2545/(m*E) *100 #efficiency\n\n#results\nprint \"Efficiency in percent\",round(eta,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Efficiency in percent 22.325\n" + } + ], + "prompt_number": 12 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_heat_Power/Chapter10.ipynb b/Elementary_heat_Power/Chapter10.ipynb new file mode 100755 index 00000000..3662d932 --- /dev/null +++ b/Elementary_heat_Power/Chapter10.ipynb @@ -0,0 +1,167 @@ +{ + "metadata": { + "name": "EHP-10" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Drafts Fans Blowers and Compressors" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 10.1,Page 515" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan, cos\nfrom numpy import *\nhb=29.0 #in of Hg\nsg=0.491 #specific gravity\nRa=53.3\nTa=460+40.0 #R\nTg=540+460.0 #R\nH=300 #ft\ngam=62.4 #lb/cu ft\n\n#calculations\npb=hb*sg*144\nrhoa=pb/(Ra*Ta) #density\nrhog=pb/(Ra*Tg) #density\ndp=H*(rhoa-rhog)\nD=dp/(gam)\n\n#results\nprint \"Theoretical draft in psf\",round(dp,3)\nprint \" Draft in ft H2O\",round(D,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Theoretical draft in psf 11.541\n Draft in ft H2O 0.185\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 10.2,Page 516" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan, cos\nfrom numpy import *\nmd=15.0 #lb per lb of coal\nx=0.1\nmss=1.0 #basis\nrea=29.0 #in of Hg\nsg=0.491\nR=53.3\nT=540+460.0 #R\nV=25.0 #fps\ngam=0.038 #lb/ft^3\n\n#calculations\nm=mss-mss*x+md\nms=m\nrhog=rea*0.491*144/(R*T) #density\nA=ms/(gam*V)\n\n#results\nprint \"stack area in sq ft\",round(A,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "stack area in sq ft 16.737\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 10.3,Page 526" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan, cos\nfrom numpy import *\np=144*29*0.491 #psf\nR=53.3\nT=70+460.0 #R\ngamw=62.4 #lb/ft^3\ngama=0.073 #lb/ft^3\nhw=3.0/12 #ft\nhw2=3.5/12 #ft\nhv=32.2 #ft/s^2\nms=9.0 #lb\ng=32.2 #ft/s^2\n\n#calculations\nrhoa=p/(R*T)\nhs=hw*gamw/gama\nht=hw2*gamw/gama\nhv=ht-hs\nV=sqrt(2*g*hv)\nmsv=ms*V*60.0\nmm=msv*gama\nairhp= ht*mm/33000.0\n\n#results\nprint \"Velocity head in ft of air\",round(hv,3)\nprint \" velocity of air in the duct in fps\",round(V,3)\nprint \" volume in cu ft per min\",round(msv,3)\nprint \" Mass flow rate in lb/min\",round(mm,3)\nprint \" Air horsepower in hp\",round(airhp,3)\nprint \"The answers in the textbook are a bit different due to rounding off error in the textbook Please use a calculator\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Velocity head in ft of air 35.616\n velocity of air in the duct in fps 47.893\n volume in cu ft per min 25861.99\n Mass flow rate in lb/min 1887.925\n Air hp in hp 14.263\nThe answers in the textbook are a bit different due to rounding off error in the textbook Please use a calculator\n" + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 10.4,Page 528" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan, cos\nfrom numpy import *\nA2=9.0 #sq ft\np2=3.0/12 *62.4 #psf\np1=-1.0/12 *62.4 #psf\nms=20000.0 #cfm\nA1=16.0 #sq ft\ngam=0.075 #lb/ft^3\ng=32.2 #ft/s^2\ninp=17.0 #hp\n\n#calculations\nV2=ms/60 *1/A2\nV1=ms/60 *1/A1\nht=(p2-p1)/gam +(V2**2 -V1**2)/(2*g)\nairhp=ht*ms*gam/33000\neta=airhp/inp *100\n\n#results\nprint \"Total head in ft of air\",round(ht,3)\nprint \" Air horsepower in hp\",round(airhp,3)\nprint \" Effifciency in percent\",round(eta,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Total head in ft of air 291.894\n Air hp in hp 13.268\n Effifciency in percent 78.047\n" + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 10.5,Page 532" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "\n# Initialization of Variable\nfrom math import pi\nfrom math import atan, cos\nfrom numpy import *\nn1=400.0 #rpm\nmv1=10000.0 #lb\nmv2=15000.0 #lb\nh1=2.0 #in of water\nhp1=4.0 #hp\n\n#calculations\nn2=mv2/mv1 *n1 #speed\nh2=h1*(n2/n1)**2 #pressure height\nhp2=hp1 *(n2/n1)**3 #power\n\n#results\nprint \"The speed in rpm\",round(n2,3)\nprint \" The pressure in inch of water\",round(h2,3)\nprint \" Power in hp\",round(hp2,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The speed in rpm 600.0\n The pressure in in of water 4.5\n Power in hp 13.5\n" + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 10.6,Page 546" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan, cos\nfrom numpy import *\nm=100000.0 #lb/hr\np1=1.0 #psia\nx=0.8\np2=14.7 #psia\nt2=300.0 #F\n\n#calculations\n#\"from table A3 and A2\"\nh2=1192.8 #Btu/lb\nhf=69.7 #Btu/lb\nhfg=1036.3 #Btu/lb\nh1=hf+x*hfg\nW=h2-h1\npower=m*W\nhp=power/2545.0\n\n#results\nprint \"Power required in hp\",round(hp,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "11554.42 Power required in hp\n" + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 10.7,Page 546" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan, cos\nfrom numpy import *\np1=14.7 #psia\nt1=60.0 #F\np2=60.0 #psia\nt2=440.0 #F\nm=10.0 #lb/sec\n\n#calculations\n#\"From mollier charts\"\nh2=216.3 #Btu/lb\nh1=124.3 #Btu/lb\nW21=h2-h1\npower=W21*m\nhp=power*3600/2545\ncp=0.237\nW212=cp*(t2-t1)\npower2=W212*m\nhp2=power2*3600.0/2545.0\n\n#results\nprint \"Power required in hp\",round(hp,3)\nprint \" Power required in hp\",round(hp2,3)\nprint \" Work done in Btu/lb\",round(W212,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Power required in hp 1301.375\n Power required in hp 1273.933\n Work done in Btu/lb 90.06\n" + } + ], + "prompt_number": 14 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_heat_Power/Chapter11.ipynb b/Elementary_heat_Power/Chapter11.ipynb new file mode 100755 index 00000000..138cf155 --- /dev/null +++ b/Elementary_heat_Power/Chapter11.ipynb @@ -0,0 +1,104 @@ +{ + "metadata": { + "name": "EHP-11" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Feed Water Heaters and Condensers" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 11.1,Page 554" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan, cos\nfrom numpy import *\nm1=1000.0 #lb/hr\nm2=5000.0 #lb/hr\nm3=3000.0 #lb/hr\n#From mollier charts\nh5=196.16 #Btu/lb\nh1=38.04 #Btu/lb\nh2=67.97 #Btu/lb\nh3=117.89 #Btu/lb\nh4=1156.3 #Btu/lb\n\n#calculations\nm4=(m1*h1+m2*h2+m3*h3-(m1+m2+m3)*h5)/(h5-h4) #mass rate\n\n#results\nprint \"Pounds of steam entering the heater in lb/hr\", round(m4,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Pounds of steam entering the heater in lb/hr 1076.801\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 11.2,Page 558" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan, cos\nfrom numpy import *\nP1=100.0 #psia\nT1=400.0 #F\nT2=70.0 #F\n#\"From mollier charts\"\nh1=1227.6 #Btu/lb\nh2=298.4 #Btu/lb\nh3=279.9 #Btu/lb\nh4=38.04 #Btu/lb\n\n#calculations\nm1=(h3-h4)/(h1-h2)\n#results\nprint \"Mass of steam required in lb steam per lb water\",round(m1,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Mass of steam required in lb steam per lb water 0.26\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 11.3,Page 567" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan, cos\nfrom numpy import *\nh0=1260.0 #Btu/lb\nmsr=15.0 #lb\nm4=15.0 #lb per hr per kw\nt2=80.0 #F\nt3=60.0 #F\n\n#calculations\nh1=h0-3413/msr \n#\"from mollier charts\"\nh4=58 #Btu/lb\ndt=t2-t3\nm3=m4*(h1-h4)/dt\n\n#results\nprint \"enthalpy of steam entering the condenser in Btu/lb\",round(h1,3)\nprint \" mass of cooling water in lb per hr per kw\",round(m3,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "enthalpy of steam entering the condenser in Btu/lb 1032.467\n mass of cooling water in lb per hr per kw 730.85\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 11.4,Page 568" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan, cos\nfrom numpy import *\nm4=8*1000000 #lb per hr\ndt=12 #F\n\n#calculations\n#from mollier charts\")\ndh4=950 #Btu/lb\nm3=m4*(dh4)/dt #mass rate\n\n#results\nprint \" mass of cooling water in lb per hr\",round(m3,4)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": " mass of cooling water in lb per hr 633333333.0\n" + } + ], + "prompt_number": 6 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_heat_Power/Chapter12.ipynb b/Elementary_heat_Power/Chapter12.ipynb new file mode 100755 index 00000000..c40e0935 --- /dev/null +++ b/Elementary_heat_Power/Chapter12.ipynb @@ -0,0 +1,62 @@ +{ + "metadata": { + "name": "EHP-12" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "The Gas Turbine Power Plant" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 12.1,Page 577" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nT1=80.0 #F\nT2=460.0 #F\nT3=1300.0 #F\nT4=780.0 #F\n\n#calculations\n#\"from mollier charts\"\nh1=129.1 #Btu/lb\nh2 = 221.2 #Btu/lb\nh3= 438.8 #Btu/lb\nh4 = 301.5 #Btu/lb\nwcom=h2-h1\nwcob=h3-h2\nwtur=h3-h4\neta=(wtur-wcom)/wcob *100\n\n#results\nprint \" work done by compressor in btu input as work per lb of air compressed\",round(wcom,3)\nprint \" Heat supplied in the combustor in Btu supplied per lb of air \",round(wcob,3)\nprint \" work done in the turbine in Btu output as work per lb of air\",round(wtur,3)\nprint \" Cycle efficiency in percent\",round(eta,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": " work done by compressor in btu input as work per lb of air compressed 92.1\n Heat supplied in the combustor in Btu supplied per lb of air 217.6\n work done in the turbine in Btu output as work per lb of air 137.3\n Cycle efficiency in percent 20.772\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 12.2,Page 579 " + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nT1=80 #F\nT2=460 #F\nT=700 #F\nT3=1300 #F\nT4=780 #F\n#\"from mollier charts\"\nh1=129.1 #Btu/lb\nh2 = 221.2 #Btu/lb\nh3= 438.8 #Btu/lb\nh4 = 301.5 #Btu/lb\nh=281.1 #Btu/lb\n\n#calculations\nwcom=h2-h1\nwcob=h3-h2\nwtur=h3-h4\noutput=-wcom+wtur\nQ=h3-h\neff=output/Q *100\n\n#results\nprint \" Heat supplied in the combustor in Btu supplied per lb of air \",round(Q,3)\nprint \" Cycle efficiency in percent\",round(eff,3)\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": " Heat supplied in the combustor in Btu supplied per lb of air 157.7\n Cycle efficiency in percent 28.662\n" + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_heat_Power/Chapter13.ipynb b/Elementary_heat_Power/Chapter13.ipynb new file mode 100755 index 00000000..9d4ff984 --- /dev/null +++ b/Elementary_heat_Power/Chapter13.ipynb @@ -0,0 +1,62 @@ +{ + "metadata": { + "name": "Chapter 13" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Mechanical Refrigeration" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 13.2,Page 596" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nmr=3.0 #lb\nmj=5.0 #lb\nt2=67.0 #F\nt1=60.0 #lb\nihp=7.25\n#\"From mollier charts\"\nh4=709 #Btu/b\nh3=618 #Btu/lb\n\n#calculations\nenergyin=ihp*2545/60\nenergyout=mr*(h4-h3) + mj*(t2-t1)\n\n#results\nprint \"Energy in Btu/min\",round(energyin,3)\nprint \"Energy out Btu/min\",round(energyout,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Energy in Btu/min 307.521\nEnergy out Btu/min 308.0\n" + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 13.3,Page 602" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nr=3.0 #lb\nhp=10.0 #hp\n\n#calculations\nh3=618 #Btu/lb\nh1=131 #Btu/lb\nQe=mr*(h3-h1)\nwork=hp*2545/60\ncop=Qe/work\n\n#results\nprint \"Coefficient of performance \",round(cop,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Coefficient of performance 3.444\n" + } + ], + "prompt_number": 11 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_heat_Power/Chapter2.ipynb b/Elementary_heat_Power/Chapter2.ipynb new file mode 100755 index 00000000..1b4fab13 --- /dev/null +++ b/Elementary_heat_Power/Chapter2.ipynb @@ -0,0 +1,340 @@ +{ + "metadata": { + "name": "EHP-2" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Fuels and Combustion" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2.1,Page 76" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nx1=0.135 #moisture \nx2=0.056 #ash content\nveca=array([32.5, 48.4, 5.6, 13.5]) #materials as received\nB1=11788.0 #Btu/lb\n\n#calculations\nvecb=veca/(1-x1) #moisture free\nvecc=veca/(1-x1-x2) #moisture and ash free\nB2=B1/(1-x1)\nB3=B1/(1-x1-x2)\nvecb[3]=0\nvecc[3]=0\nvecc[2]=0\n\n#results\nprint \"In Moisture free case\",vecb\nprint \"In Moisture and Ash free case \",vecc\nprint \"Energy in Moisture free case Btu per lb\",round(B2,3)\nprint \"Energy in Moisture and ash free case Btu per lb\",round(B3,2)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "In Moisture free case [ 37.57225434 55.95375723 6.47398844 0. ]" + }, + { + "output_type": "stream", + "stream": "stdout", + "text": "\nIn Moisture and Ash free case [ 40.17305315 59.82694685 0. 0. ]\nEnergy in Moisture free case Btu per lb 13627.746\nEnergy in Moisture and ash free case Btu per lb 14571.08\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2.2,Page 77" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\ny1=13.5\nx1=0.135 #moisture \nx2=0.056 #ash \nveca=array([66.0, 1.5, 1.1, 5.6, 5.9, 19.9]) # fram table 2.3\nvecb=array([66.0, 1.5, 1.1, 5.6, 5.9, 19.9,13.5]) #with moisture \n\n#calculations\nvecb[4] = vecb[4] - 1.0/9*y1 ;\nvecb[5] = vecb[5] - 8.0/9*y1 ;\nvecc=vecb/(1-x1)\nvecd=vecb/(1-x1-x2) # moisture and ash free\nvecd[3]=0\nvecd[6]=0\nvecc[6]=0\ns1=sum(vecc)\ns2=sum(vecd)\n\n#results\nprint \"With moisture as a separate item\",vecb\nprint \"In Moisture free case, \",vecc\nprint \"In Moisture and Ash free case\",vecd\nprint \"Total Mositure free content percent\",round(s1,3)\nprint \" Total Mositure and ash free content percent\",round(s2,3)\nprint \"slight error due to rounding off error\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "With moisture as a separate item [ 66. 1.5 1.1 5.6 4.4 7.9 13.5]\nIn Moisture free case, [ 76.30057803 1.73410405 1.2716763 6.47398844 5.0867052\n 9.13294798 0. ]\nIn Moisture and Ash free case [ 81.58220025 1.85414091 1.35970334 6.92212608 0. 9.76514215\n 0. ]\nTotal Mositure free content percent 100.0\n Total Mositure and ash free content percent 101.483\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2.3,Page 78" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nH=5.9 \nO=19.9 \nH2=4.4 \nO2=7.9 \n\n#calculations\nHa1=H-O/8\nHa2=H2-O2/8\n\n#results\nprint \"Available hydrogen in case 1 percent by weight\",round(Ha1,3)\nprint \" Available hydrogen in case 2 percent by weight\",round(Ha2,3)\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Available hydrogen in case 1 percent by weight 3.413\n Available hydrogen in case 2 percent by weight 3.413\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2.4,Page 93" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nH1=0.059\nO1=0.199\nH2=0.044\nO2=0.079\nC=0.66\nS=0.011\n\n#calculations\nQh1= 14600*C+62000*(H1-O1/8)+4050*S #heating value\nQh2=14600*C+62000*(H2-O2/8)+4050*S #heating value\n\n#results\nprint \"Heating value in case 1 Btu/lb \",round(Qh1,3)\nprint \" Heating value in case 2 Btu/lb \",round(Qh2,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Heating value in case 1 Btu/lb 11796.3\n Heating value in case 2 Btu/lb 11796.3\n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2.5,Page 110" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nH1=0.059\nO1=0.199\nC=0.66\nS=0.011\n\n#calculations\nQh1= 11.52*C+34.56*(H1-O1/8)+4.32*S\n\n#results\nprint \"Theoretical air required in lb of air per lb of coal \",round(Qh1,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Theoretical air required in lb of air per lb of coal 8.83\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2.6,Page 110" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nmf=10000 #lb\nmr=700 #lb\nCr=0.20\nCco2=14.1 #% CO2\nCo2=5.1 #% O2\nCco=0.1 #% CO\nCf=0.66 \n\n#calculations\nCn2=100-(Cco2+Co2+Cco) #% N2\nCi=mf*Cf #carbon input\nCa=mr*Cr #carbon in ashpit\nCb=(Ci-Ca)/mf\nCb2=((mf*Cf)-mr*Cr)/(mf)\nMco2=Cco2 *44 #mass of CO2\nMo2=Co2 *32 #mass of O2\nMco=Cco *28 #mass of CO\nMn2=Cn2 *28 #mass of N2\nsumvec=Mo2+Mco2+Mco+Mn2;\nLbc=Cco2*12 + Cco*12\nGc=sumvec/Lbc\nGf=Gc*Cb\n\n#results\nprint \"Carbon in the dry products combustion in lb per lb of fuel\",round(Cb,3)\nprint \" In case 2, Carbon in the dry products combustion in lb per lb of fuel\",round(Cb2,3)\nprint \" Dry gaseous products of combstion per lb of coal in lb \",round(Gf,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Carbon in the dry products combustion in lb per lb of fuel 0.646\n In case 2, Carbon in the dry products combustion in lb per lb of fuel 0.646\n Dry gaseous products of combstion per lb of coal in lb 11.548\n" + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2.7,Page 111" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nmf=10000.0 #lb\nmr=700.0 #lb\nCr=0.20\nCco2=14.1\nCo2=5.1\nCco=0.1\nCf=0.66\n\n#calculations\nCn2=100-(Cco2+Co2+Cco)\nCi=mf*Cf\nCa=mr*Cr\nCb=(Ci-Ca)/mf\nCb2=((mf*Cf)-mr*Cr)/(mf)\nMco2=Cco2 *44 #mass of CO2\nMo2=Co2 *32 #mass of O2\nMco=Cco *28 #mass of CO\nMn2=Cn2 *28 #mass of N2\nsumvec=Mn2+Mco2+Mco+Mn2;\nCbb1=Cb*Cco*12/(Cco2*12 + Cco*12)\nCbb2= Cb*(Cco /(Cco + Cco2))\n\n#results\nprint \"In case 1, Carbon burned per lb of fuel in lb per lb of fuel\",round(Cbb1,6)\nprint \" In case 2, Carbon burned per lb of fuel in lb per lb of fuel\",round(Cbb2,6)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "In case 1, Carbon burned per lb of fuel in lb per lb of fuel 0.004549\n In case 2, Carbon burned per lb of fuel in lb per lb of fuel 0.004549\n" + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2.8,Page 111" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nH=4.4/100 #hydrogen\nM=13.5/100 #moisture\nH2=0.059 #hydrogen\n\n#calculations\npro=M+9*H\npro2=9*H2\n\n#results\nprint \"In case 1, watervapor present in products in lb\",round(pro,3)\nprint \"In case 2, watervapor present in products in lb\",round(pro2,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "In case 1, watervapor present in products in lb 0.531\nIn case 2, watervapor present in products in lb 0.531\n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2.9,Page 113" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nGf=11.57;#lb\nf=11.57 #lb per lb of fuel\nH=4.4/100\nM=13.5/100\nmr=700.0 #lb\nmf=10000.0 #lb\nmc=1 #lb\n\n#calculations\npro=M+9*H\nmrf=mr/mf\nAa=Gf+pro+mrf-mc\n\n#results\nprint \"Actual air supplied in lb of air supplied per lb of fuel\",round(Aa,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Actual air supplied in lb of air supplied per lb of fuel 11.171\n" + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2.10,Page 114" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nGf=11.57 #lb per lb of fuel\nH=4.4/100\nM=13.5/100\nmr=700.0\nmf=10000.0\nmc=1.0 #lb\n\n#calculations\npro=M+9*H\nmrf=mr/mf\nAa=Gf+pro+mrf-mc\nAt=8.83\nea=(Aa-At)/At *100\n\n#results\nprint \"Excess air in percent\",round(ea,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "26.512 Excess air in percent\n" + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2.11a,Page 114" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nGf=11.57 #lb per lb of fuel\ntg=500 #F\nta=70 #F\n#calculations\n\nQ1=0.24*Gf*(tg-ta)\n\n#results\nprint \"Heat loss in Btu per lb of fuel\",round(Q1,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Heat loss in Btu per lb of fuel 1194.024\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2.11b,Page 115" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nCo=0.1\nCo2=14.1\nCb=0.646\n\n#calculations\nQ2=Co/(Co+Co2) *Cb*10160\n\n#results\nprint \"Heat loss in Btu per lb of fuel\",round(Q2,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "46.221 Heat loss in Btu per lb of fuel\n" + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2.11c ,Page 115" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nmf=10000.0 #lb \nmr=700.0 #lb\nCr=0.2\n\n#calculations\nQ3=mr*Cr/mf *14600\n\n#results\nprint \"Heat loss in Btu per lb of fuel\",round(Q3,2)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "204.4 Heat loss in Btu per lb of fuel\n" + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2.11d,Page 115" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nM=0.135\ntg=500.0 #F\nta=70.0 #F\n\n#calculations\nQ4=M*(1089+0.46*tg-ta) #heat loss\n\n#results\nprint \"Heat loss in Btu per lb of fuel\",round(Q4,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "168.615 Heat loss in Btu per lb of fuel\n" + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2.11e,Page 115" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nPer=0.044 #percentage\ntg=500 #F\nta=70 #F\n\n#calculations\nQ5=9*Per*(1089+0.46*tg-ta) #heat loss\n\n#results\nprint \"Heat loss in Btu per lb of fuel\",round(Q5,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Heat loss in Btu per lb of fuel 494.604\n" + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_heat_Power/Chapter3.ipynb b/Elementary_heat_Power/Chapter3.ipynb new file mode 100755 index 00000000..cc592814 --- /dev/null +++ b/Elementary_heat_Power/Chapter3.ipynb @@ -0,0 +1,251 @@ +{ + "metadata": { + "name": "EHP-3" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Internal Combustion Engines" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 3.1 , Page 131" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nre=6.0\nk=1.4\n\n#calculations\nnt=1-1/re**(k-1)\nntt=nt*100 #efficiency\n\n#results\nprint \"Thermal efficiency in percent\",round(ntt,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Thermal efficiency in percent 51.164\n" + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 3.2,Page 132" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nd=3.25 #in\nstroke=4.0 #in\nv=6.0 #cu in\n\n#calculations\nDp=d**2 *pi*stroke/4\nr=(Dp+v)/v #compression ratio\n\n#results\nprint \"Compression ratio \",round(r,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Compression ratio 6.531\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 3.3,Page 133" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nper=20.0\nDp=100.0\n\n#calculations\nr=Dp/per +1\n\n#results\nprint \"Compression ratio \",round(r,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Compression ratio 6.0\n" + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 3.4,Page 136" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nr=16.0\nrc=4.0\nk=1.4\n\n#calculations\netat=1-1/r**(k-1) *((rc**k -1)/(k*(rc-1)))\neta=etat*100; #efficiency\n\n#results\nprint \"Thermal efficiency in percent\",round(eta,3)\nprint \"The answer is a bit different due to rounding off error in the textbook\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Thermal efficiency in percent 53.154\nThe answer is a bit different due to rounding off error in the textbook\n" + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 3.5,Page 142" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nF=200 #lb\narea=1.65 #sq. in\nlength=3.5 #in\n\n#calculations\nord=area/length\nmep=ord*F\n\n#results\nprint \"MEP of an engine in psi\",round(mep,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "MEP of an engine in psi 94.286\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 3.6,Page 143" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nPi=90 #psi\nL=5.0/12.0 #ft\nr=5.0 #in\nx=1.5 #ft\nrpm=1500.0 #rpm\n\n#calculations\nA=pi*x*x\nN=rpm*4/2\nIhp=Pi*L*A*N/33000.0\n\n#results\nprint \"IHP of cylinder \",round(Ihp,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "IHP of cylinder 24.097\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 3.7,Page 144" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nr=4.0 #ft\nn=300.0 #rpm\nF=60.0 #lb\n\n#calculations\nBhp=2*pi*r*F*n/33000\n\n#results\nprint \"Bhp of the engine \",round(Bhp,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Bhp of the engine 13.709\n" + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 3.8,Page 147" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nC=1.0/4000.0\nF=125.0 #lb\nn=3500 #rpm\n\n#calculations\nBhp=F*n*C\n\n#results\nprint \"Bhp of the engine \",round(Bhp,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Bhp of the engine 109.375\n" + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 3.9,Page 151" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nr=1.75 #ft\nF1=72.0 #lb\nF2=24.0 #lb\nn=500.0 #rpm\nm=1.8 #lb\nmi=15.0 #min\nQh=20000.0 #Btu/lb\n\n#calculations\nBhp=2*pi*r*F1*n/33000.0\nFhp=2*pi*r*F2*n/33000.0\nIhp=Bhp+Fhp\nFc=m*60/mi\nBsfc=Fc/Bhp\nIsfc=Fc/Ihp\netam=Bhp/Ihp *100\netabt=Bhp*2545/(Fc*Qh)\netait=Ihp*2545/(Fc*Qh)\n\n#results\nprint \"Thermal efficiency in percent\",round(etam,3)\nprint \" Brake thermal effficiency in percent\",round(etabt*100,3)\nprint \" Indicated thermal effficiency in percent\",round(etait*100,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Thermal efficiency in percent 75.0\n Brake thermal effficiency in percent 21.2\n Indicated thermal effficiency in percent 28.266\n" + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 3.10,Page 191" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nbore=3.0 #in\nstroke=4.0 #in\nrpm=3000.0 #rpm\nair=110.0 #cu ft per min\n\n#calculations\npdv=bore*bore*pi*stroke*2*bore/4\npde=pdv*rpm /2\nreq=air*1728\neff=req/pde *100; #efficiency\n\n#results\nprint \"Volumetric efficiency in percent\",round(eff,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Volumetric efficiency in percent 74.697\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 3.11, Page 244" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nx1=11.5 # CO2\nx2=4.1 #CO\nx3=0.4 #O2\nx4=2.3 # H2\nx5=0.2 #CH4\nx6=81.5 #N2\nyc=0.842 #lb of C \nyh=0.158 #lb of H2\nbasis=1\nbhp=100\nburn=8.9 #gal/hr\nsg=0.731\nQh=20750.0 #Btu/lbm\nrate=66.0 #gpm\nex=1100.0 #F\nair=70.0 #F\ncp=0.254\nh2=4330.0 #btu/lb\nh4=62000.0 #btu/lb \nh5=23700.0 #btu/lb\n\n#calculations\nc1=x1*44 #mass CO2\nc2=x2*28 #mass CO\nc3=x3*32 #mass O2\nc4=x4*2 #mass H2\nc5=x5*16 #mass CH4\nc6=x6*28 #mass N2\nsumm=c1+c2+c3+c4+c5+c6\ncarbon=x1*12 + x2*12+x5*12\nhydrogen=x4*2+x5*4\nlbdrygas=summ/carbon *yc\nlbfuel=carbon/yc\nlbH=lbfuel*yh\nlbH2=lbH-hydrogen\nlb3=lbH2*9\nlbwater=lb3/lbfuel\nlbair=lbdrygas+lbwater-basis\nbsfc=burn*sg*8.33/bhp\nfuelmin=bsfc*bhp/60\nenergy=2545/bsfc\nper1=energy/Qh #energy supplied\nEc=rate*8.33*10\nEclb=Ec/fuelmin\nper2=Eclb/Qh #energy absorbed\ndryloss=(ex-air)*cp*lbdrygas\nper3=dryloss/Qh #energy lost\nhv2=h2*c2/lbfuel\nhv4=h4*c4/lbfuel\nhv5=h5*c5/lbfuel\nhv=hv2+hv4+hv5\nper4=hv/Qh #energy supplied\neh2=lbwater*(1066+0.5*ex-air)\nper5=eh2/Qh #energy supplied\nrad=1017 \nper6=rad/Qh; #energy supplied\n\n#results\nprint \"Air supplied per lb of fuel in lb air per lb fuel\",round(lbair,3)\nprint \"Percentage of energy supplied utilized in Btu in percent\",round(per1*100,3)\nprint \"Percentage of energy absorbed by coolant in percent\",round(per2*100,3)\nprint \"Energy lost in sensible heat in percent\",round(per3*100,3)\nprint \"Energy supplied in combustiles in exhaust in percent\",round(per4*100,3)\nprint \"Energy supplied in water formed by combustion in percent\",round(per5*100,3)\nprint \"Energy supplied unaccounted for in percent\",round(per6*100,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Air supplied per lb of fuel in lb air per lb fuel 13.189\nPercentage of energy supplied utilized in Btu in percent 22.632\nPercentage of energy absorbed by coolant in percent 29.334\nEnergy lost in sensible heat in percent 16.369\nEnergy supplied in combustiles in exhaust in percent 18.366\nEnergy supplied in water formed by combustion in percent 8.987\nEnergy supplied unaccounted for in percent 4.901\n" + } + ], + "prompt_number": 27 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_heat_Power/Chapter5.ipynb b/Elementary_heat_Power/Chapter5.ipynb new file mode 100755 index 00000000..89ee8fb7 --- /dev/null +++ b/Elementary_heat_Power/Chapter5.ipynb @@ -0,0 +1,251 @@ +{ + "metadata": { + "name": "EHP-5" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Steam Generation" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 5.1,Page 296" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nx=0.98\nvg=26.80\nvf=0.01672\n\n#calculations\nvx=x*vg+(1-x)*vf #specific wet volume\n\n#results\nprint \"Specific volume of wet steam in cu ft per lb\",round(vx,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Specific volume of wet steam in cu ft per lb 26.264\n" + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 5.2,Page 298" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nhf=167.99 #Btu/lb\nhg=4.5 #Btu/lb\n\n#calculations\nhc=hf+hg\n\n#results\nprint \"Enthalpy of water in Btu/lb\",round(hc,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Enthalpy of water in Btu/lb 172.49\n" + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 5.3,Page 300" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nx=0.97\nhg=1187.2 #Btu/lb\nhf=298.40 #Btu/lb\nhfg=888.8 #Btu/lb\n\n#calculations\nhx1=x*hg+(1-x)*hf\nhx2=hf+x*hfg\nhx3=hg-(1-x)*hfg\n\n#results\nprint \" In case 1, enthalpy in Btu/lb\",round(hx1,3)\nprint \" In case 2, enthalpy in Btu/lb\",round(hx2,3)\nprint \" In case 3, enthalpy in Btu/lb\",round(hx3,3)\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": " In case 1, enthalpy in Btu/lb 1160.536\n In case 2, enthalpy in Btu/lb 1160.536\n In case 3, enthalpy in Btu/lb 1160.536\n" + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 5.4,Page 302" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nh1=1172.0 #Btu/lb\nhf1=355.36 #Btu/lb\nhfg1=843.0 #Btu/lb\n\n#calculations\nh2=h1\nx1= (h2-hf1)/hfg1\n\n#results\nprint \"Quality of steam in percent\",round(x1*100,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "96.873 Quality of steam in percent\n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 5.7,Page 306" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nP=200 #psia\nT=260 #F\n\n#calculations\n#\"From mollier chart\"\nhx=1174 #Btu/lb\nx1=2.8 #percent\ny1=100-x1 #percent\n\n#results\nprint \"Quality in percent\",round(y1,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Quality in percent 97.2\n" + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 5.8,Page 306" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nP=200 #psia\nT=500 #F\n\n#calculations\n#\"From mollier chart\"\nhi=1269 #Btu/lb\nhf=1063 #Btu/lb\ndh=hi-hf\ny1=91\n\n#results\nprint \"Quality in percent\",round(y1,3)\nprint \" Change in enthalpy in Btu/lb\",round(dh,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Quality in percent 91.0\n Change in enthalpy in Btu/lb 206.0\n" + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 5.9,Page 307" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nP=200 #psia\nTs=260 #F\nTf=220 #F\nm=10000 #lb\nPc=20 #psia\n\n#calculations\n#\"From mollier charts\"\nhf=188.0 #Btu/lb\nh2=1172.0 #Btu/lb\nQ=m*(h2-hf)\n\n#results\nprint \"Heat absorption in Btu/hr\",round(Q,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "9840000.0 Heat absorption in Btu/hr\n" + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 5.10,Page 307" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nm=150000.0 #lb\nP1=1000.0 #psia\nTs=900.0 #F\nTf=200.0 #F\n\n#calculations\n#\"From mollier charts\"\nh2=1448.2 #Btu/lb\nhf=167.99 #Btu/lb\ncorrec=2.2 #Btu/lb\nhc=hf+correc\nQ=m*(h2-hc)\n\n#results\nprint \"Heat absorption in Btu/hr\",round(Q,3)\nprint\"The answer is a bit different due to rounding off error in textbook\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Heat absorption in Btu/hr 191701500.0\nThe answer is a bit different due to rounding off error in textbook\n" + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 5.11,Page 308" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nm=150000.0 #lb\nP1=1000.0 #psia\nTs=900.0 #F\nTf=200.0 #F\n\n#calculations\n#\"From mollier charts\"\nh2=1448.2 #Btu/lb\nhf=167.99 #Btu/lb\ncorrec=2.2 #Btu/lb\nhc=hf+correc\nQ=m*(h2-hc)\noutput=Q/1000\n\n#results\nprint \"Output of the steam generating unit kB/hr\",round(output,3)\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Output of the steam generating unit kB/hr 191701.5\n" + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 5.12,Page 308" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nm=150000 #lb\nP1=1000 #psia\nTs=900 #F\nTf=200 #F\nm2=21000 #lb\nHV=12000 #Btu/lb\n\n#calculations\n#\"From mollier charts\"\nh2=1448.2 #Btu/lb\nhf=167.99 #Btu/lb\ncorrec=2.2 #Btu/lb\nhc=hf+correc\nQ=m*(h2-hc)\noutput=Q\ninpu=m2*HV\neta=output/inpu\n\n#results\nprint \"Efficiency of the steam generating unit in percent\",round(eta*100,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Efficiency of the steam generating unit in percent 76.072\n" + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 5.13,Page 309" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nhv=11780 #Btu/lb\nsteam=55000 #lb/hr\ncoal=6480 #lb\nx1=0.66 # %Carbon\nx2=0.044 #% Hydrogen\nx3=0.079 #% Oxygen\nx4=0.015 #% Nitrogen\nx5=0.11 # % Sulphur\nz1=14.5 # % CO2\nz2=0.2 # CO\nz3=4.4 # O2\nz4=80.9 # N2\nxash=0.076 #% ash\nxmois=0.115 # % moisture\nyc=0.21 # % Carbon in refuse\nrefuse=622 #lb/hr\ncp=0.24 #coeff of heat\ntg=400 #F\nta=70 #F\nQco=10160 #Btu/lb\nQc=14600 #Btu/lb\n\n#calculations\n#\"From steam tables\"\nhf=269.6 #Btu/lbm\nhfg=1.5 #Btu/lbm\nh1=hf+hfg #Btu/lbm\nh2=1196.5\nQb=h2-h1\nh3=1407.7 #Btu/lbm\nQs=h3-h2\nh4=h3-h1\nout=steam*h4/1000\neff=steam*h4/(coal*hv)\n#Energy balance\nCi=coal*x1\nCr=refuse*yc\nCb=(Ci-Cr)/coal\nlbt= z1*44+z2*28+z3*32+z4*28\nlbC=z1*12+z2*12\ndry=lbt/lbC *Cb\nloss1=dry*cp*(tg-ta)\nloss2=z2*12/(lbC) *Cb*Qco\nloss3=Cr*Qc/coal\nloss4=xmois*(1089+0.46*tg-ta)\nloss5=x2*9*(1089+0.46*tg-ta)\nloss6=steam*h4/coal\n\n#results\nprint \"Heat absorbed in the boiler in Btu per lb of steam generated\",round(Qb,3)\nprint \" Heat absorbed in the superheater in Btu/lb of steam\",round(Qs,3)\nprint \" Heat absorbed in steam generating in Btu/lb of steam generated\",round(h4,3)\nprint \" Output of steam generating unit in kB\",round(out,3)\nprint \" Efficiency of steam generating unit in percent\",round(eff*100,3)\nprint \" Carbon burned to CO and CO2 in lb of C per lb of fuel\",round(Cb,3)\nprint \" Dry products of combustion in lb per lb of fuel\",round(dry,3)\nprint \" Loss due to sensible heat in dry gaseous products of combustion in Btu/lb of fuel\",round(loss1,3)\nprint \" Loss due to CO in dry products of combustion in Btu/lb of fuel\",round(loss2,3)\nprint \" Loss due to C in refuse in Btu/lb of fuel\",round(loss3,3)\nprint \" Loss due to evaporating moisture in fuel in Btu/lb of fuel\",round(loss4,3)\nprint \" Loss due to water vapor formed from H in Btu/lb of fuel\",round(loss5,3)\nprint \" Energy absorbed in generating steam in Btu/lb of fuel\",round(loss6,3)\nprint \"The answers are a bit different due to rounding off error in the textbook\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Heat absorbed in the boiler in Btu per lb of steam generated 925.4\n Heat absorbed in the superheater in Btu/lb of steam 211.2\n Heat absorbed in steam generating in Btu/lb of steam generated 1136.6\n Output of steam generating unit in kB 62513.0\n Efficiency of steam generating unit in percent 81.894\n Carbon burned to CO and CO2 in lb of C per lb of fuel 0.64\n Dry products of combustion in lb per lb of fuel 11.062\n Loss due to sensible heat in dry gaseous products of combustion in Btu/lb of fuel 876.078\n Loss due to CO in dry products of combustion in Btu/lb of fuel 88.446\n Loss due to C in refuse in Btu/lb of fuel 294.298\n Loss due to evaporating moisture in fuel in Btu/lb of fuel 138.345\n Loss due to water vapor formed from H in Btu/lb of fuel 476.388\n Energy absorbed in generating steam in Btu/lb of fuel 9647.068\nThe answers are a bit different due to rounding off error in the textbook\n" + } + ], + "prompt_number": 31 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_heat_Power/Chapter6.ipynb b/Elementary_heat_Power/Chapter6.ipynb new file mode 100755 index 00000000..b2d3e81f --- /dev/null +++ b/Elementary_heat_Power/Chapter6.ipynb @@ -0,0 +1,104 @@ +{ + "metadata": { + "name": "EHP-6" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Steam Power Plant Cycles" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 6.1,Page 366" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nP1=200.0 #psia\nT1=600.0 #F\nP2=2.0 #psia\nJ=778.0\n\n#calculations\n#\"from mollier charts,\"\nh1=1322 #Btu/lb\nh2=974 #Btu/lb\nvf2=0.01623 #cu ft per lb\nhf2=94.0 #Btu/lb\nt2=126.0 #F\nWtj=h1-h2\nQout=h2-hf2\nWp=(P1-P2)*vf2\nWpj=Wp/J\nh3=hf2+Wpj\nQin=h1-h3\netat=((h1-h2)-Wpj)/(h1-(hf2+Wpj)) #efficiency\neta=((h1-h2))/(h1-(hf2)) #efficiency\n\n#results\nprint round(etat*100,3),\"Efficiency of rankine cycle in percent\"\nprint round(eta*100,3),\" Efficiency of rankine cycle neglecting boiler feed pump in percent\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "28.339 Efficiency of rankine cycle in percent\n28.339 Efficiency of rankine cycle neglecting boiler feed pump in percent\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 6.2,Page 366" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nB=70 #F\nP1=140 #psia\nx=0.986\nP2=14.7 #psia\nms=2000.0 #lb/hr\nIhp=80.0\n\n#calculations\n#\"From mollier charts\"\nhc=38 #Btu/lb\nhf=324.82 #Btu/lb\nhfg=868.2 #Btu/lb\nh1=hf+x*hfg\nQin=ms*(h1-hc)\neta=Ihp*2545*100/(Qin) #efficiency\nQw=Ihp*2545 \nQr=Qin-Qw\nper=Qr/Qin *100 #percent\n\n#results\nprint \"Heat input to the boiler in Btu/hr\",round(Qin,3)\nprint \" Cycle efficiency in percent\",round(eta,3)\nprint \" Heat rejected to waste in Btu/hr is\",round(Qr,3),\" or in percent of Qin\",round(per,3)\nprint \"The answer is a bit different due to rounding off error in textbook\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Heat input to the boiler in Btu/hr 2285730.4\n Cycle efficiency in percent 8.907\n Heat rejected to waste in Btu/hr 2082130.4 or in percent of Qin 91.093\nThe answer is a bit different due to rounding off error in textbook\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 6.3,Page 372" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nB=70.0 #F\nP1=140.0 #psia\nx=0.986\nP2=14.7 #psia\nms=2000 #lb/hr\nIhp=80\n\n#calculations\n#\"From mollier charts\"\nhc=180.0 #Btu/lb\nhf=324.82 #Btu/lb\nhfg=868.2 #Btu/lb\nh1=hf+x*hfg\nQin=ms*(h1-hc)\neta=Ihp*2545*100/(Qin)\nQw=Ihp*2545 #Btu/hr\nQr=Qin-Qw #Btu/hr\nper=Qr/Qin *100 #percent\n\n#results\nprint \"Heat input to the boiler in Btu/hr\",round(Qin,3)\nprint \" Cycle efficiency in percent\",round(eta,3)\nprint \" Heat rejected to waste in Btu/hr is\",round(Qr,3),\" or in percent of Qin\",round(per,3)\nprint \"The answer is a bit different due to rounding off error in textbook\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Heat input to the boiler in Btu/hr 2001730.4\n Cycle efficiency in percent 10.171\n Heat rejected to waste in Btu/hr 1798130.4 in percent of Qin 89.829\nThe answer is a bit different due to rounding off error in textbook\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 6.4,Page 380" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nm=1.24 #lb\nHV=11300 #Btu/lb\n\n#calculations\nHR=m*HV\neff=3413/HR #efficiency\n\n#results\nprint \"Plant heat rate in Btu/kw hr\",round(HR,3)\nprint \" Overall efficiency in percent\",round(eff*100,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Plant heat rate in Btu/kw hr 14012.0\n Overall efficiency in percent 24.358\n" + } + ], + "prompt_number": 10 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_heat_Power/Chapter7.ipynb b/Elementary_heat_Power/Chapter7.ipynb new file mode 100755 index 00000000..b3a3267c --- /dev/null +++ b/Elementary_heat_Power/Chapter7.ipynb @@ -0,0 +1,188 @@ +{ + "metadata": { + "name": "EHP-7" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Steam Turbines" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 7.1,Page 390" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nP1=200.0 #psia\nT1=500.0 #psia\nm=1.0 #lb /s\nP4=140.0 #psia\nP11=1.0 #psia\nx=0.808\n\n#calculations\n#\"From mollier charts\"\nh1=1268.9 #Btu/lb\nh4=1234.7 #Btu/lb\nV4=223.8*sqrt(h1-h4)\nv4=3.584 #cu ft/lb\nA4=m*v4/V4\nh11=907.4 #Btu/lb\nV11=223.8*sqrt(h1-h11)\nvf=0.01614 #cu ft/lb\nvg=333.6 #cu ft/lb\nvfg=vg-vf\nv11=x*vg\nA11=m*v11/V11 #area\n\n#results\nprint \"Area of nozzle in sq ft\",round(A4,6)\nprint \" Area of nozzle in sq ft\",round(A11,6)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Area of nozzle in sq ft 0.002738\n Area of nozzle in sq ft 0.063347\n" + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 7.4,Page 410" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan, cos\nfrom numpy import *\nP1=200.0 #psia\nT1=500.0 #F\nP2=1.0 #psia\nalpha=20.0 #degrees\nn=3600.0\ng=32.2 #ft/s^2\n\n#calculations\n#\"From mollier charts\"\nV1=4240.0 #fps\nVb=V1*cos(alpha*pi/180) /2\nR=Vb*60/(n*2*pi)\nwork=1/32.2 *(V1*cos(alpha*pi/180))*Vb\neff=work/(V1**2 /(2*g)) *100 #efficiency\n\n#results\nprint \"Blade velocity in fps\",round(Vb,3)\nprint \" Blade radius in ft\",round(R,3)\nprint \" Work done in per lb of steam\",round(work,3)\nprint \" Blade efficiency in percent\",round(eff,3)\nprint \"The answers are a bit different due to rounding off error in textbook\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Blade velocity in fps 1992.148\n Blade radius in ft 5.284\n Work done in per lb of steam 246500.315\n Blade efficiency in percent 88.302\nThe answers are a bit different due to rounding off error in textbook\n" + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 7.5,Page 411" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nP1=200.0 #psia\nT1=500.0 #F\nP2=1.0 #psia\nalpha=20.0 #degrees\nn=3600.0\ng=32.2 #ft/s^2\nVb=1200.0 #fps\n\n#calculations\n#\"From mollier charts\"\nV1=4240.0 #fps\nV1x=3980.0 #fps\nV2x=-1580.0 #fps\nwork=1/32.2 *(V1x - V2x)*Vb\neff=work/(V1**2 /(2*g)) *100 #efficiency\n\n#results\nprint \" Work done in per lb of steam\",round(work,3)\nprint \" Blade efficiency in percent\",round(eff,3)\nprint \"The answers are a bit different due to rounding off error in textbook\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": " Work done in per lb of steam 207204.969\n Blade efficiency in percent 74.226\nThe answers are a bit different due to rounding off error in textbook\n" + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 7.6,Page 413" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan, cos, tan\nfrom numpy import *\nP1=200.0 #psia\nT1=500.0 #F\nP2=1.0 #psia\nalpha=20.0 #degrees\nn=3600.0\ng=32.2 #ft/s^2\n\n#calculations\n#\"From mollier charts\")\nV1=2450 #fps\nVb=V1*cos(alpha*pi/180) /2\nR=Vb*60/(n*2*pi)\nwork=1/32.2 *(V1*cos(alpha*pi/180))*Vb\nw3=3*work #workdone\n\n#results\nprint \"Blade velocity in fps\",round(Vb,3)\nprint \"Blade radius in ft\",round(R,3)\nprint \" Work done in per lb of steam\",round(w3,3)\nprint \"The answers are a bit different due to rounding off error in textbook\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Blade velocity in fps 1151.123\nBlade radius in ft 3.053\n Work done in per lb of steam 246910.29\nThe answers are a bit different due to rounding off error in textbook\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 7.7, Page 416" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nP1=200.0 #psia\nT1=500.0 #F\nP2=1.0 #psia\nalpha=20.0 #degrees\nn=3600.0\ng=32.2 #ft/s^2\nstage=2.0\n\n#calculations\n#\"From mollier charts\"\nV1=4240.0 #fps\nVb=V1*cos(alpha*pi/180) /(2*stage)\nR=Vb*60/(n*2*pi)\nV1x=3980 #fps\nV2x=-1990 #fps\nwork1=1/g *(V1x-V2x)*Vb #work\nwork2=1/g *(-V2x)*Vb #work\nwt=work1+work2 #work done\n\n#results\nprint \"Blade velocity in fps\",round(Vb,3)\nprint \" Blade radius in ft\",round(R,3)\nprint \" Work done in per lb of steam\",round(wt,3)\nprint \"The answers are a bit different due to rounding off error in textbook\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Blade velocity in fps 996.074\n Blade radius in ft 2.642\n Work done in per lb of steam 246234.486\nThe answers are a bit different due to rounding off error in textbook\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 7.8,Page 422" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan,cos, tan\nfrom numpy import *\nalpha=20.0 #degrees\nn=3600.0\ng=32.2 #ft/s^2\nV1=500 #fps\n\n#calculations\nVb=V1*cos(alpha*pi/180) #blade velocity\nV1x=Vb\nwork=1/32.2 *(V1x)*Vb #work done\n\n#results\nprint \"Blade velocity in fps\",round(Vb,3)\nprint \" Work done in per lb of steam\",round(work,3)\nprint \"The answers are a bit different due to rounding off error in textbook.\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Blade velocity in fps 469.846\n Work done in per lb of steam 6855.763\nThe answers are a bit different due to rounding off error in textbook.\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 7.9,Page 436" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\npower=1000.0 #kw\nms=16000.0 #lb/hr\nP=200.0 #psia\nT=540 #F\n\n#calculations\n#\"From mollier charts\"\nh1=1290.0 #Btu/hr\nh2=940.0 #Btu/hr\ndh=h1-h2\nrate=3413.0/dh\nact=ms/power #steam rate\n\n#results\nprint \"Ideal steam rate in per kw hr\",round(rate,3)\nprint \" Actual steam rate in per kw hr\",round(act,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Ideal steam rate in per kw hr 9.751\n Actual steam rate in per kw hr 16.0\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 7.10,Page 439" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nP=200 #psia\nT=540 #F\npower=1000 #kw\nms=16000 #lb/hr\n\n#calculations\n#\"From mollier charts\")\nh1=1290.0 #Btu/hr\nh2=940.0 #Btu/hr\ndh=h1-h2\nhf2=83.0 #Btu/lb\netat=(h1-h2)/(h1-hf2)\nact=power*3413/(ms*(h1-hf2)) #thermal efficiency\netae=act/etat #efficiency of engine\n\n#results\nprint \"Ideal thermal efficiency in percent\",round(etat*100,3)\nprint \"Actual thermal efficiency in percent\",round(act*100,3)\nprint \" Engine efficiency in percent\",round( etae*100,3)\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Ideal thermal efficiency in percent 28.998\nActual thermal efficiency in percent 17.673\n Engine efficiency in percent 60.946\n" + } + ], + "prompt_number": 7 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_heat_Power/Chapter8.ipynb b/Elementary_heat_Power/Chapter8.ipynb new file mode 100755 index 00000000..94c8be36 --- /dev/null +++ b/Elementary_heat_Power/Chapter8.ipynb @@ -0,0 +1,62 @@ +{ + "metadata": { + "name": "EHP-8" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Steam Engines" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 8.1,Page 461" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\narea1=2.7\nlength=3.4\nscale=60\narea2=2.75\ndia=12 #ft\nd2=2.5 #ft\nL=15.0/12 #ft\nn=250.0 #rpm\nF=600.0 #lb\nr=3 #ft\n\n#calculations\nAh=dia**2 *pi/4\nAc=(dia**2 -d2**2)*pi/4\nPih=area1/length *scale\nPic=area2/length *scale\nHihp=Pih*L*Ah*n/33000.0\nCihp=Pic*L*Ac*n/33000.0\nTihp=Hihp+Cihp\nBhp=2*pi*r*F*n/33000.0\nFhp=Tihp-Bhp\neff=Bhp/Tihp *100 #efficiency\n\n#results\nprint \"Ihp in ihp\",round(Tihp,3)\nprint \"Bhp in bhp\",round(Bhp,3)\nprint \"Fhp in fhp\",round(Fhp,3)\nprint \"Efficiency in percent\",round(eff,3)\nprint \"The answer is a bit different due to rounding off error in the textbook\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Ihp in ihp 100.749\nBhp in bhp 85.68\nFhp in fhp 15.069\nEfficiency in percent 85.043\nThe answer is a bit different due to rounding off error in the textbook\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 8.2,Page 468" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nIhp=101.1\nBhp=85.7\nmd=3000.0 #lb/hr\nh1=1172.0 #Btu/hr\nh2=180.0 #Btu/hr\nh3=1025.0 #Btu/hr\n\n#calculations\neta1=Ihp*2545/(md*(h1-h2)) *100 #efficiency\neta2=Bhp*2545/(md*(h1-h2)) *100 #efficiency\netat=(h1-h3)/(h1-h2) *100 #efficiency\nengeff=eta1/etat *100 #engine efficiency\nrate1= md/Ihp\nrate2=md/Bhp\nh22=h1-2545/rate1 #enthalpy\n\n#results\nprint \"Actual thermal efficiency based upon Ihp in lb per ihp hr\",round(eta1,3)\nprint \"Actual thermal efficiency based upon Bhp in lb per ihp hr\",round(eta2,3)\nprint \"Ideal thermal efficiency in percent \",round(etat,3)\nprint \"Engine efficiency in percent\",round(engeff,3)\nprint \"Steam rate in lb per ihp hr\",round(rate1,3)\nprint \"Steam rate in lb per bhp hr\",round(rate2,3)\nprint \"Enthalpy of exhaust steam in Btu per lb of steam\",round(h22,3)\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Actual thermal efficiency based upon Ihp in lb per ihp hr 8.646\nActual thermal efficiency based upon Bhp in lb per ihp hr 7.329\nIdeal thermal efficiency in percent 14.819\nEngine efficiency in percent 58.345\nSteam rate in lb per ihp hr 29.674\nSteam rate in lb per bhp hr 35.006\nEnthalpy of exhaust steam in of steam 1086.234\n" + } + ], + "prompt_number": 5 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_heat_Power/Chapter9.ipynb b/Elementary_heat_Power/Chapter9.ipynb new file mode 100755 index 00000000..aa870a60 --- /dev/null +++ b/Elementary_heat_Power/Chapter9.ipynb @@ -0,0 +1,230 @@ +{ + "metadata": { + "name": "EHP-9" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Pumps" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 9.1,Page 472" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan, cos\nfrom numpy import *\nh=200.0 #ft\ngam=64.0 #lb per cu ft\n\n#calculations\nP=h*gam/144.0 #pressure\n\n#results\nprint \"Pressure in psi\",round(P,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "88.889 Pressure in psi\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 9.2,Page 472" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nP=20.0 #psi\ngam=62.4 #lb per cu ft\n\n#calculations\nh=P*144/gam #height\n\n#results\nprint \"height in ft\",round(h,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "46.154 height in ft\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 9.3,Page 472" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nh=3.0/12.0 #ft\ngam=62.4 #lb per cu ft\ngam2=0.075 #lb per cu ft\n\n#calculations\nP=h*gam #pressure\nh2=P/gam2 #height\n\n#results\nprint \"Air height required in ft of air\",round(h2,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Air height required in ft of air 208.0\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 9.4,Page 477" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\ndif=4.0 #in\ngam=62.4 #lb per cu ft\ndensity=13.6 #g/cc\n\n#calculations\npv=dif*1/12 *density*gam/144 - dif/12 *gam/144 #pressure\nhv=pv*144/gam #height\n\n#results\nprint \"velocity pressure in psi\",round(pv,3)\nprint \" velocity head in ft of water \",round(hv,3)\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "velocity pressure in psi 1.82\n velocity head in ft of water 4.2\n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 9.5,Page 479" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\ndif=4.0 #in\ngam=62.4 #lb per cu ft\ngam2=0.08 #lb per cu ft\n\n#calculations\npv=dif*1/12 *gam/144 #pressure\nhv=pv*144/gam2 #height\n\n#results\nprint \"velocity pressure in psi\",round(pv,3)\nprint \" velocity head in ft of water \",round(hv,3)\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "velocity pressure in psi 0.144\n velocity head in ft of water 260.0\n" + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 9.6,Page 482" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan,sqrt\nfrom numpy import *\nhw=3.0/12 #ft\ngam1=62.4 #lb/ft^3\ngam2=0.07 #lb/ft^3\ng=32.2 #ft/s^2\n\n#calculations\np=hw*gam1 #pressure\nhg=p/gam2 #height\nV=sqrt(2*g*hg) #velocity\n\n#results\nprint \"velocity of gas in fps\",round(V,3)\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "velocity of gas in fps 119.8\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 9.7,Page 485" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nh=4.0 #in\nden=13.6 #g/cc\nAr=1.0/9\nA1=12 #sq in\ngam=62.4 #lb/ft^3\ng=32.2 #ft/s^2\n\n#calculations\ndh=(h*den-h)/12.0\nVr=1/Ar\nV22=2*g*dh/(1-Ar**2)\nV2=sqrt(V22)\nA2=A1*Ar\nv2=1/gam\nms=A2*V2/(v2*144) #mass flow rate\n\n#results\nprint \"Flow rate of water in lb/sec\",round(ms,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Flow rate of water in lb/sec 9.561\n" + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 9.8,Page 488" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nmdot=8000.0 #lb/min\nA1=1.0 #sq ft\nA2=3.0/4 #sq ft\nP2=50.0 #psi\nP1=10.0 #psi\ngam=62.4 #lb/ft^3\ny2=-2.0 #ft\ny1=-4.0 #ft\ng=32.2 #ft/s^2\neff=0.7\n\n#calculations\nv=1/gam\ncap=mdot/8.33\nV1=mdot*v/A1 /60.0 #velocity\nV2=mdot*v/A2 /60.0 #velocity\nht= (y2-y1) + (V2**2 -V1**2)/(2*g) + (P2-P1)*144/gam #height\nHhp=mdot*ht/33000.0 #horsepower\nPhp=Hhp/eff #horsepower\n\n#results\nprint \"Capacity in gpm\",round(cap,3)\nprint \" Total dynamic head in ft\",round(ht,3)\nprint \" Hydraulic horsepower in hp\",round(Hhp,3)\nprint \" pump horsepower in hp\",round(Php,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Capacity in gpm 960.384\n Total dynamic head in ft 94.363\n Hydraulic hp in hp 22.876\n pump horsepower in hp 32.68\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 9.9,Page 491" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nz=12.0 #ft\ngam1=62.4 #lb/ft^3\nsg=0.8\nP2=100.0 #psia\nP1=-10.0 #psia\nmm=10000.0 #lb/min\n\n#calculations\ngam2=sg*gam1\np2g=P2*144/(gam2) +z #pressure\np1g=P1*144*0.491/(gam2) #pressure\nht=p2g-p1g #height\nHhp=mm*ht/33000 #horsepower\n\n#results\nprint \"Total dynamic head in ft of oil\",round(ht,3)\nprint \" Hydraulic horsepower in hp\",round(Hhp,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Total dynamic head in ft of oil 314.625\n Hydraulic hp in hp 95.341\n" + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 9.10,Page 510" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# Initialization of Variable\nfrom math import pi\nfrom math import atan\nfrom numpy import *\nsr=2\n\n#calculations\nhr=sr**2 #ratio\ncapr=sr #ratio\nhpr=sr**3 #ratio\n\n#results\nprint \"head is \",round(hr,3),\"times the original\"\nprint \" capacity is \",round(capr,3),\"times the original\"\nprint \" power is\",round(hpr,3),\"times the original\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "head is 4.0 times the original\n capacity is 2.0 times the original\n power is 8.0 times the original\n" + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_heat_Power/README.txt b/Elementary_heat_Power/README.txt new file mode 100755 index 00000000..c837b597 --- /dev/null +++ b/Elementary_heat_Power/README.txt @@ -0,0 +1,10 @@ +Contributed By: Avik Kumar Das +Course: btech +College/Institute/Organization: Indian Institute of Technology Bombay +Department/Designation: Civil Engineering +Book Title: Elementary heat Power +Author: H L Solberg +Publisher: John Wiley and Sons, London +Year of publication: 1986 +Isbn: 0131106724 +Edition: 2
\ No newline at end of file diff --git a/Elementary_heat_Power/screenshots/chapter10.png b/Elementary_heat_Power/screenshots/chapter10.png Binary files differnew file mode 100755 index 00000000..45b6135c --- /dev/null +++ b/Elementary_heat_Power/screenshots/chapter10.png diff --git a/Elementary_heat_Power/screenshots/chapter2.png b/Elementary_heat_Power/screenshots/chapter2.png Binary files differnew file mode 100755 index 00000000..c8f462da --- /dev/null +++ b/Elementary_heat_Power/screenshots/chapter2.png diff --git a/Elementary_heat_Power/screenshots/chapter3.png b/Elementary_heat_Power/screenshots/chapter3.png Binary files differnew file mode 100755 index 00000000..0e9459d3 --- /dev/null +++ b/Elementary_heat_Power/screenshots/chapter3.png diff --git a/Engineering_&_Chemical_Thermodynamics/README.txt b/Engineering_&_Chemical_Thermodynamics/README.txt new file mode 100755 index 00000000..86222029 --- /dev/null +++ b/Engineering_&_Chemical_Thermodynamics/README.txt @@ -0,0 +1,10 @@ +Contributed By: Vaibhav Shah +Course: mca +College/Institute/Organization: IIT +Department/Designation: Developer +Book Title: Engineering & Chemical Thermodynamics +Author: M. D. Koretsky +Publisher: Wiley India Pvt. Ltd., New Delhi +Year of publication: 2010 +Isbn: 978-81-265-2449-5 +Edition: 2
\ No newline at end of file diff --git a/Engineering_&_Chemical_Thermodynamics/ch1.ipynb b/Engineering_&_Chemical_Thermodynamics/ch1.ipynb new file mode 100755 index 00000000..b3ff4f91 --- /dev/null +++ b/Engineering_&_Chemical_Thermodynamics/ch1.ipynb @@ -0,0 +1,164 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 1 : Measured thermodynamic Properties and Other Basic Concepts" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.3 Page No : 28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "P = 1.4 ; \t\t\t# [MPa]\n", + "T = 333. \t\t\t#[K]\n", + "T1 = 320. \t\t\t#[K]\n", + "T2 = 360. \t\t\t#[K]\n", + "P_low = 1. \t\t\t#[MPa]\n", + "P_high = 1.5\n", + "V_cap_T1_P1 = 0.2678\n", + "V_cap_T2_P1 = 0.2873\n", + "V_cap_T1_P1_5 = 0.1765\n", + "V_cap_T2_P1_5 = 0.1899\n", + "\n", + "# Calculations\n", + "#At P = 1 MPa\n", + "V_cap_T333_P1 = V_cap_T1_P1 + (V_cap_T2_P1 - V_cap_T1_P1)*((T - T1)/(T2- T1))\n", + "\n", + "#Similarly at P=1.5 MPa\n", + "V_cap_T333_P1_5 = V_cap_T1_P1_5 + (V_cap_T2_P1_5 - V_cap_T1_P1_5)*((T - T1)/(T2 - T1))\n", + "\n", + "#At T=333*C\n", + "V_cap_P1_5 = V_cap_T333_P1_5 ;\n", + "V_cap_P1 = V_cap_T333_P1 ;\n", + "V_cap_P1_4 = V_cap_P1 + (V_cap_P1_5 - V_cap_P1)*((P - P_low)/(P_high - P_low)) ; \t\t\t#[m**3/kg]\n", + "\n", + "# Results\n", + "print 'Required specific volume = %.5f m**3/kg'%(V_cap_P1_4);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Required specific volume = 0.19951 m**3/kg\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.4 Page No : 29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "P = 1.4 ; \t\t\t#[MPa]\n", + "P_low = 1 ;\t\t\t#[MPa]\n", + "P_high = 1.5;\t\t\t#[MPa]\n", + "\n", + "#At T=333*C from interpolation we have\n", + "v_cap_P1_5 = 0.18086 ;\t\t\t#[m**3/kg]\n", + "v_cap_P1 = 0.27414 ;\t\t\t#[m**3/kg]\n", + "\n", + "# Calculations\n", + "#Molar volume is inversely proportional to pressure\n", + "v_cap_P1_4 = v_cap_P1 +(v_cap_P1_5 - v_cap_P1)*((1/P - 1/P_low)/(1/P_high - 1/P_low));\n", + "x=(0.19951-v_cap_P1_4)/v_cap_P1_4*100 ;\n", + "\n", + "# Results\n", + "print 'Specific volume m**3/kg) = %g m**3/kg'%(v_cap_P1_4);\n", + "print 'Percentage difference = %.1f %%'%(x);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Specific volume m**3/kg) = 0.194186 m**3/kg\n", + "Percentage difference = 2.7 %\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.5 page no : 29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# variables\n", + "V1 = 1 # volume\n", + "m1 = 2.5 # water\n", + "Vv = 5.042 # m**3/kg\n", + "Vt = .001023 # m**3/kg\n", + "T_44632 = 145 # C\n", + "T_39278 = 150 # C\n", + "\n", + "# Calculations\n", + "v1 = V1/m1 \n", + "x1 = (v1 - Vt)/(Vv - Vt)\n", + "\n", + "T2 = T_44632 + ( T_39278 - T_44632) * (( 0.4 - 0.44632)/(0.39278 - 0.44632))\n", + "\n", + "# Restuls\n", + "print \"Approximately %.f %% of the mass of water in the vapor\"%(round(x1,2)*100)\n", + "print \"Temperature T2 = %.1f C\"%T2\n", + "\n", + "# Note : Answer in book is wrong. Please calculate manually.\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Approximately 8 % of the mass of water in the vapor\n", + "Temperature T2 = 149.3 C\n" + ] + } + ], + "prompt_number": 8 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_&_Chemical_Thermodynamics/ch2.ipynb b/Engineering_&_Chemical_Thermodynamics/ch2.ipynb new file mode 100755 index 00000000..e1006ddc --- /dev/null +++ b/Engineering_&_Chemical_Thermodynamics/ch2.ipynb @@ -0,0 +1,1120 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 2 : The First law of Thermodynamics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.1 Page No : 38" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "z1 = 10. \t\t\t#[m]\n", + "z2 = 0. \t\t\t#[m],Taking ground as state 2,reference\n", + "v1 = 0.\n", + "g = 9.81 #m/s\n", + "\n", + "#From conservation of total energy we get\n", + "# (1/2*m*v2**2-1 / 2*m*v1**2)+(m*g*z2 - m*g*z1) = 0\n", + "# 1/2*m*v2**2 - m*g*z1 = 0\n", + "v2 = math.sqrt(2 * g * z1) ; \t\t\t#[m/s]\n", + "\n", + "# Results\n", + "print 'Final velocity = %.f m/s'%(v2);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Final velocity = 14 m/s\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.2 Page No : 41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "V2 = 14. \t\t\t# [m/s]\n", + "u_cap_l1 = 104.86 ; \t\t\t#[kJ/kg],at 25*C internal energy of saturated water\n", + "u_cap_l_t25 = 104.86 ; \t\t\t#[kJ/kg], From steam table \n", + "u_cap_l_t30 = 125.77 ; \t\t\t#[kJ/kg], From steam table\n", + "T1 = 25. \t\t \t#[*C]\n", + "T2 = 30. \t\t \t #[*C]\n", + "\n", + "# Calculations\n", + "#For unit mass change in kinetic energy\n", + "Delta_e_cap_k = 1./2 * V2**2 * 10**-3 ; \t\t\t#[kJ/kg]\n", + "Delta_u_cap = Delta_e_cap_k ;\n", + "#For final state of water:\n", + "u_cap_l2 = Delta_u_cap + u_cap_l1 ;\n", + "\n", + "x = (u_cap_l2 - u_cap_l_t25) / (u_cap_l_t30 - u_cap_l_t25) ;\n", + "T_unknown = T1 + x*(T2 - T1) ;\n", + "\n", + "# Results\n", + "print 'Final temperature of water = %.2f degreeC'%(T_unknown);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Final temperature of water = 25.02 degreeC\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.3 Page No : 43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "import scipy.integrate\n", + "\n", + "# Variables\n", + "P_ex = 1*10**5 ; \t\t\t#[Pa]}\n", + "\n", + "# Calculations\n", + "#To calculte work done\n", + "def f(x):\n", + " return 1\n", + " \n", + "I = scipy.integrate.quadrature(f,10,15.2)[0]\n", + "W = -P_ex * I * 10**-3 ; \t\t\t#[J]\n", + "\n", + "# Results\n", + "print 'Work done = %g J'%(W);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Work done = -520 J\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.4 page no : 57" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from scipy.integrate import quad\n", + "\n", + "# Variables\n", + "V = 1.0 # m**3\n", + "m = 10 # kg\n", + "p1p2 = .2\n", + "u1 = 2600.3 #kJ/kg\n", + "P1 = 20\n", + "v1 = 0.1\n", + "\n", + "# Calculations\n", + "v = V/m\n", + "v2 = (p1p2*v**1.5)**(1./1.5)\n", + "\n", + "def fun(v):\n", + " return P1*v1**1.5/v**1.5\n", + "\n", + "w = -quad(fun,0.1,.0342)[0]\n", + " \n", + "u2 = 3045.8 + (3144.5 - 3045.8) * (v2 - .03279)/(.03564 - .03279)\n", + "T2 = 500 + (550 - 500) * (v2 - .03279)/(.03564 - .03279)\n", + "\n", + "# results\n", + "print \"The specific volume of initial state = %.2f m**3/kg\"%v\n", + "print \"V2 = %.4f m**3/kg\"%v2\n", + "print \"Work done = %.f kJ/kg\"%(w*100)\n", + "print \"U2 = %.1f kJ/kg\"%u2\n", + "print \"T2 = %.f C\"%T2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The specific volume of initial state = 0.10 m**3/kg\n", + "V2 = 0.0342 m**3/kg\n", + "Work done = 284 kJ/kg\n", + "U2 = 3094.6 kJ/kg\n", + "T2 = 525 C\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.5 Page No : 63" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# variables\n", + "#From steam table specific enthalpy at state1 and state2 are\n", + "h_cap_1 = 3373.6 \t\t\t#[kJ/kg]\n", + "h_cap_2 = 2675.5 \t\t\t#[kJ/kg]\n", + "m_dot1 = 10. \t\t\t#[kg/s],As we are dealing with steady state\n", + "m_dot2 = 10. \t \t\t#[kg/s]\n", + "\n", + "# Calculations\n", + "Ws_dot = m_dot1 * (h_cap_2 - h_cap_1) ; \t\t\t#[kW]\n", + "\n", + "# Results\n", + "print 'Power generated = %g kW'%(Ws_dot);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Power generated = -6981 kW\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.6 Page No : 64" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "h_cap_in = 3241.\t\t#[kJ/kg] , From steam table\n", + "P_final = 10. \t\t\t#[MPa]\n", + "\n", + "# Calculations and Results\n", + "u_cap_2 = h_cap_in ;\n", + "T2 = 600. ; \t\t\t# From steam table .No calculation is involved .\n", + "print 'a) The final temperature of the system = %g *C'%(T2);\n", + "\n", + "u_cap_2 = h_cap_in ;\n", + "# So temperature is T2 = 600*C (From table).\n", + "#Solution(b)\n", + "print \"(b) The temperature of the fluid increases in the system due to the receipent of flow work .\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a) The final temperature of the system = 600 *C\n", + "(b) The temperature of the fluid increases in the system due to the receipent of flow work .\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.7 Page No : 72" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import scipy.integrate.quadrature \n", + "\n", + "# Variables\n", + "n = 2. \t\t\t#[mol]\n", + "A = 3.470 ;\n", + "B = 1.450*10**-3 ;\n", + "D = 0.121*10**5 ;\n", + "T1 = 473. ; \t\t\t#[K]\n", + "T2 = 773. ; \t\t\t#[K]\n", + "\n", + "# Calculations and Results\n", + "def f(T):\n", + " return 8.314*(A + B*T + D*T**-2)\n", + " \n", + "Delta_h = scipy.integrate.quadrature(f,T1,T2)[0]\n", + "\n", + "Q = n * Delta_h ;\n", + "\n", + "print 'a)Heat required = %d J'%(Q);\n", + "\n", + "#From steam table\n", + "h_cap_1 = 2827.9 ; \t\t\t#[kJ/kg]\n", + "h_cap_2 = 3478.4 ; \t\t\t#[kJ/kg]\n", + "m = 2*0.018 ; \t\t\t#[kg]\n", + "\n", + "Delta_h_cap = (h_cap_2 - h_cap_1) * 10**3 ; \t\t\t#[J/kg]\n", + "Q = m * Delta_h_cap;\n", + "print 'b)Heat required = %g J'%(Q);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a)Heat required = 21981 J\n", + "b)Heat required = 23418 J\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.8 Page No : 73" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "T1 = 298.;\n", + "T2_start = 300.;\n", + "A = 3.355;\n", + "B = 0.575*10**-3;\n", + "D = -0.016*10**5;\n", + "\n", + "# Calculations\n", + "def f(T):\n", + " return 8.314*(A*T + B/2*T**2 - D/T)\n", + "\n", + "for T2_start in range(300,1000+1,100):\n", + " del_h = f(T2_start) - f(T1);\n", + " Cp = del_h /(T2_start - 298);\n", + " print 'At temperatureK %g, Molar heat capacity J/molK, %.2f'%(T2_start,Cp); \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "At temperatureK 300, Molar heat capacity J/molK, 29.17\n", + "At temperatureK 400, Molar heat capacity J/molK, 29.45\n", + "At temperatureK 500, Molar heat capacity J/molK, 29.71\n", + "At temperatureK 600, Molar heat capacity J/molK, 29.97\n", + "At temperatureK 700, Molar heat capacity J/molK, 30.22\n", + "At temperatureK 800, Molar heat capacity J/molK, 30.46\n", + "At temperatureK 900, Molar heat capacity J/molK, 30.71\n", + "At temperatureK 1000, Molar heat capacity J/molK, 30.95\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.9 Page No : 73" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "n_dot_air = 10. ; \t\t\t#[mol/min]\n", + "C_bar_P_900 = 30.71 ; \t\t\t#[J/molK]\n", + "C_bar_P_600 = 29.97 ; \t\t\t#[J/molK]\n", + "T1 = 600. ; \t\t\t#[K]\n", + "T2 = 900. ; \t\t\t#[K]\n", + "T_ref = 298. ; \t\t\t#[K]\n", + "\n", + "# Calculations\n", + "# Q_dot = n_dot_air * (h_900 - h_600)...........Eqn E2.8A\n", + "Q_dot = n_dot_air * (C_bar_P_900 * (T2 - T_ref) - C_bar_P_600 * (T1 - T_ref)); \n", + "\n", + "# Results\n", + "print 'Heat rate required = %.3f J/min'%(Q_dot/1000);\n", + "\n", + "# note: answer may vary because of rounding error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat rate required = 94.365 J/min\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.10 Page No : 74" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "P1 = 100000. ; \t\t\t # [N/m**2]\n", + "T1 = 298. ; \t\t \t#[K]\n", + "V1 = 0.1 * 0.1 ; \t\t\t# [m**3]\n", + "T2 = 373. ; \t\t\t # [N]\n", + "P_ext = 100000. ; \t\t\t#[N/m**2]\n", + "k = 50000. \t\t\t#[N/m]\n", + "A = 0.1 ; \t\t \t#[m**2]\n", + "\n", + "# Calculations and Results\n", + "a = k / (T2 * A**2) ;\n", + "b = (P_ext / T2) - k * V1 / (A**2 * T2) ;\n", + "c = -P1 * V1 / T1 ;\n", + "V2 = (-b + math.sqrt ( b**2 - (4*a*c))) / (2 * a) ;\n", + "W = -P_ext * (V2 - V1) - ( k * (V2 - V1)**2)/(2 * A**2);\t\t\t#From eqn E2.9C\n", + "\n", + "print 'a) Work required = %.f J '%(W);\n", + "\n", + "\n", + "A = 3.355 ;\n", + "B = 0.575 * 10**-3 ;\n", + "D = -0.016 * 10**5 ;\n", + "P1 = 10**5 ; \t\t\t#[N/m**2]\n", + "V1 = 0.01 ; \t\t\t#[m**3]\n", + "R = 8.314 ;\n", + "T1 = 298 ;\n", + "\n", + "n = (P1 * V1) / (R * T1) ;\n", + "def f(T):\n", + " return R*((A - 1) * T + B/2 * T**2 -D/T)\n", + "\n", + "del_u = f(373) - f(298) ;\n", + "del_U = n * del_u ;\n", + "Q = del_U - W;\n", + "print 'b).Heat transfered = %.f J'%(Q);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a) Work required = -166 J \n", + "b).Heat transfered = 803 J\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.11 Page No : 78" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "n_dot = 10. \t \t\t#[mol/s]\n", + "T1 = 298.2 \t\t\t#[K]\n", + "T2 = 342. \t\t\t #[K]\n", + "T3 = 373.2 \t\t \t#[K]\n", + "Cp_298_342 = 216.3 \t\t\t#[J/molK]\n", + "A = 3.025 \n", + "B = 53.722 * 10**-3\n", + "C = -16.791 * 10**-6\n", + "del_h_vap = 28.88 \t\t\t#[kJ/mol]\n", + "\n", + "# Calculations\n", + "del_h_1 = Cp_298_342 * (T2 - T1) * 10**-3 ; \t\t\t#[kJ/mol]\n", + "del_h_2 = del_h_vap ;\n", + "def f(T):\n", + " return 8.314*(A*T + (B/2)*(T**2) + (C/3)*(T**3))* 10**-3 ;\n", + "\n", + "del_h_3 = f(T3) - f(T2) ;\n", + "Q = n_dot * (del_h_1 + del_h_2 + del_h_3) ;\n", + "\n", + "print 'Rate of heat supplied = %d kJ/s'%(Q );\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rate of heat supplied = 435 kJ/s\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.12 Page No : 79" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "m_1_v = 4.3 \t\t\t#[kg]\n", + "m_1_l = 50. \t\t\t#[kg]\n", + "u_cap_1_v = 2437.9 \t\t\t#[kJ/kg],From steam table\n", + "u_cap_1_l = 191.8 \t\t\t#[kJ/kg],From steam table\n", + "v_cap_1_v = 14.67 \t \t\t#[m**3],From steam table\n", + "v_cap_1_l = 0.001 \t\t \t#[m**3],From steam table\n", + "\n", + "# Calculations\n", + "V2 = m_1_l * v_cap_1_l + m_1_v * v_cap_1_v ;\n", + "m_2_v = m_1_l + m_1_v ;\n", + "v_cap_2_v = V2 / m_2_v ; \t\t\t#[m**3/kg]\n", + "P2= 0.15 \t\t\t#[MPa]\n", + "u_cap_2_v = 2519.6 \t\t\t#(kJ/kg)\n", + "Q = ((m_2_v * u_cap_2_v) -(m_1_l * u_cap_1_l + m_1_v * u_cap_1_v))*1000;\n", + "\n", + "# Results\n", + "print 'Minimum amount of heat required = %.2e J'%(Q);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum amount of heat required = 1.17e+08 J\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.13 Page No : 83" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "del_h0_f_CO2 = -393.51 ; \t\t\t# [kJ/mol]\n", + "del_h0_f_H2 = 0 ; \t\t\t# [kJ/mol]\n", + "del_h0_f_H2O = -241.82 ; \t\t\t# [kJ/mol]\n", + "del_h0_f_CH3OH = -200.66 ; \t\t\t# [kJ/mol]\n", + "\n", + "# Calculations\n", + "del_h0 = del_h0_f_CO2 + 3 * del_h0_f_H2 - del_h0_f_H2O - del_h0_f_CH3OH ;\n", + "\n", + "# Results\n", + "print 'Enthalpy of reaction = %.f kJ/mol'%(del_h0);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Enthalpy of reaction = 49 kJ/mol\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.14 page no : 84" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# variables\n", + "Vp = 3*10**6 *0.7*10**6\n", + "MW = 114.2 # g/mol\n", + "\n", + "\n", + "#Calculations\n", + "n = Vp/MW\n", + "H = -9.3 * 10**16 # J\n", + "Ep = 200 * 0.10*24*3600 # energy density\n", + "A = -H/Ep\n", + "\n", + "# Results\n", + "print \"a) Area A = %.1e m**2\"%A\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a) Area A = 5.4e+10 m**2\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.15 Page No : 85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "from numpy import *\n", + "from scipy.integrate import *\n", + "\n", + "# Variables\n", + "del_h0_f_CO2 = -393.51 \t\t\t#[kJ/mol], From Appendix A.3 \n", + "del_h0_f_CO = -110.53 \t \t\t#[kJ/mol], From Appendix A.3\n", + "del_h0_f_H2O = -241.82 \t\t\t#[kJ/mol], From Appendix A.3\n", + "del_h0_f_C3H8 = -103.85 \t\t\t#[kJ/mol], From Appendix A.3\n", + "del_h0_f_O2 = 0. \t\t\t #[kJ/mol], From Appendix A.3\n", + "A_CO2 = 5.457 \t\t\t # From table E2.13\n", + "B_CO2 = 1.05 * 10**-3 \n", + "D_CO2 = -1.16 * 10**5 \n", + "A_CO = 3.379 \n", + "B_CO = 5.57 * 10**-4\n", + "D_CO = -3.1 * 10**3 \n", + "A_H2O = 3.470 \n", + "B_H2O = 1.45 * 10**-3\n", + "D_H2O = 1.21 * 10**4 \n", + "A_N2 = 3.280 \n", + "B_N2 = 5.93 * 10**-4 \n", + "D_N2 = 4.00 * 10**3 \n", + " \n", + "# Calculations \n", + "n_C3H8 = 10. \t\t\t#[mol]\n", + "n_N2 = (0.79/0.21) * (9.7/2) * n_C3H8 ; \t\t\t#[mol]\n", + "n_CO2 = 2.7 * n_C3H8 ; \t\t\t#[mol]\n", + "n_CO = 0.3 * n_C3H8 ; \t\t\t#[mol]\n", + "n_H2O = 4 * n_C3H8 ; \t\t\t#[mol]\n", + "n_O2 = (9.7 / 2)* n_C3H8 \t\t#[mol]\n", + "T_reff = 298. \t\t\t#[K]\n", + "del_H_rxn_298 = n_CO2 * del_h0_f_CO2 + n_CO * del_h0_f_CO + n_H2O * del_h0_f_H2O - n_C3H8 * del_h0_f_C3H8 - n_O2 * del_h0_f_O2 ; \t\t\t#[kJ]\n", + "\n", + "#The co-efficients of T2 in the equation of degree 3 are\n", + "a = 8.314*(n_CO2 * (B_CO2/2) + n_CO * (B_CO/2) + n_H2O * (B_H2O/2) + n_N2 * (B_N2/2));\n", + "b = 8.314*(n_CO2 * A_CO2 + n_CO * A_CO + n_H2O * A_H2O + n_N2 * A_N2) ;\n", + "d =8.314*(- n_CO2 * D_CO2 - n_CO * D_CO - n_H2O * D_H2O -n_N2 * D_N2) ;\n", + "c = (del_H_rxn_298 *1000) + 8.314 * (n_CO2 * (- T_reff * A_CO2 - B_CO2/2 * T_reff**2 + D_CO2/T_reff) + n_CO * (- T_reff * A_CO - B_CO/2 * T_reff**2 + D_CO/T_reff) + n_H2O * (- T_reff * A_H2O - B_H2O/2 * T_reff**2 + D_H2O/T_reff) + n_N2 * (-T_reff * A_N2 - B_N2/2 * T_reff**2 + D_N2/T_reff));\n", + "\n", + "T2=poly1d([a,b,c,d])\n", + "M = roots(T2);\n", + "\n", + "\n", + "# Results\n", + "print \"T2 = %.f [K]\"%(round(round(M[1]),-1))\n", + "\n", + "# Note: python has only 1 method to find roots so part 1-2 can be calculated same way here." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "T2 = 2350 [K]\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.16 pageno : 89" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# variables\n", + "\n", + "# Calculations\n", + "deltaH1rxn298 = (-393.51) - 1*(-110.53) -0 # Reaction 1\n", + "deltaH2rxn298 = 1*(-110.53) -0 - 0 # Reaction 2\n", + "E1 = 3\n", + "E2 = 1\n", + "\n", + "deltaHrxn = E1*deltaH1rxn298*1000 + E2*deltaH2rxn298*1000\n", + "nCO2 = 4 - E1 + E2\n", + "nO22 = 4 - (1./2*E1) - 1./2*E2\n", + "nCO22 = 0 + E1\n", + "nC2 = 2 - E2\n", + "s = 52000.\n", + "Q = deltaHrxn + s\n", + "\n", + "# Results\n", + "print \"extensive enthalpy of reaction %.2e J\"%deltaHrxn\n", + "print \"amount of energy transferred by heat %.1e J\"%Q" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "extensive enthalpy of reaction -9.59e+05 J\n", + "amount of energy transferred by heat -9.1e+05 J\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.18 Page No : 96" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import *\n", + "\n", + "# Variables\n", + "V1 = 350. \t\t \t#[m/s]\n", + "A = 3.355 \n", + "B = 0.575*10**-3\n", + "D = -0.016*10**5\n", + "Tin = 283. \t \t\t#[K]\n", + "MW = 29.*10**-3 ; \t\t\t#[kg/mol]\n", + "\n", + "ek = 1./2 * MW * V1**2 \n", + "a = B/2.\n", + "b = A ;\n", + "c = -(Tin * A + Tin**2*B/2 - (D/Tin) + ek/8.314)\n", + "d=-D \n", + "\n", + "T2=poly1d([a,b,c,d])#'T2',0);\n", + "M = roots(T2);\n", + "\n", + "\n", + "# Results\n", + "print \"Temperature T2 = %.f [K]\"%M[1]\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temperature T2 = 344 [K]\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.19 Page No : 97" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "V_dot_2 = 0.001 ; \t\t\t#[m**3/kg]\n", + "v_cap_2 = 0.001 ; \t\t\t#[m**3/kg], Specific volume of water\n", + "z2 = 250. \t\t\t#[m] ; Taking ground as the reference level\n", + "e_cap_2 = 9.8 * z2 \t\t\t#[kg*m**2/s**2]\n", + "\n", + "# Calculations\n", + "m_dot_2 = V_dot_2 / v_cap_2 \t\t\t#[kg/s]\n", + "W_dot_s = m_dot_2 * e_cap_2 * 10**-3 ;\n", + "\n", + "# Results\n", + "print 'Minimum power required is = %.1f kW'%(W_dot_s);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum power required is = 2.5 kW\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.20 Page No : 98" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "n_dot = 10. \t\t\t#[mol/min]\n", + "del_h_vap_CO2 = 10400. ; \t\t\t#[J/mol]\n", + "A_CO2 = 5.457 ; \t\t \t#From appendix A.3\n", + "B_CO2 = 1.045 * 10**-3 ;\n", + "D_CO2 = -1.157 * 10**5 ;\n", + "A_air = 3.355 ; \n", + "B_air = 0.575 * 10**-3 ;\n", + "D_air = -0.016 * 10**5 ;\n", + "T1 = 273. ; \t\t\t#[K]\n", + "T2 = 283. ; \t\t\t#[K]\n", + "T3 = 323. ; \t\t\t#[K]\n", + "T4 = 293. ; \t\t\t#{k}\n", + "\n", + "def f1(T):\n", + " return 8.314 * (A_CO2 * T + (B_CO2/2) * T**2 - D_CO2/T)\n", + "\n", + "sen_heat_CO2 = f1(T2) - f1(T1) ;\n", + "Q_dot = n_dot * (del_h_vap_CO2 + sen_heat_CO2) ; \t\t\t#[J/min]\n", + "\n", + "def f2(T):\n", + " return 8.314 * (A_air * T + B_air/2*T**2 - D_air /T)\n", + "\n", + "sen_heat_air = f2(T4) - f2(T3);\n", + "n_dot_air = - Q_dot / sen_heat_air ;\n", + "print 'Air required = %.F mol/min'%(n_dot_air); \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Air required = 123 mol/min\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.21 Page No : 100" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "m_dot_1 = 10. \t\t\t#[kg/s]\n", + "h_cap_1 = 3238.2 ;\t\t\t#[kJ/kg], Super heated steam at 500*C & 200bar\n", + "h_cap_2 = 93.3 ;\t\t\t#[kL/kg], subcooled liquid at 20*C & 100bar\n", + "h_cap_3 = 2724.7 ;\t\t\t#{kJ/kg}, Super heated vapour at 100bar \n", + "\n", + "# Calculations\n", + "m_dot_2 = m_dot_1 * (h_cap_1 - h_cap_3) / (h_cap_3 - h_cap_2);\n", + "\n", + "# Results\n", + "print 'Flow of liquid stream = %.2f kg/s'%(m_dot_2);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Flow of liquid stream = 1.95 kg/s\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.22 Page No : 101" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "h_cap_st_1 = 2923.4 ; \t\t\t# [kJ/kg]\n", + "h_cap_200 = 2875.3 ; \t\t\t# {kJ/kg} , At 100kPa\n", + "h_cap_250 = 2974.3 ; \t\t\t# {kJ/kg} , At 100 kPa\n", + "del_T = 250.-200 ;\n", + "T1 = 200. \t\t\t#[K]\n", + "\n", + "# Calculations\n", + "h_cap_st_2 = h_cap_st_1 ;\t\t\t#Assumimg bulk kinetic energy of the stream and heat transfered is negligible\n", + "T2 = T1 + del_T * (h_cap_st_2 - h_cap_200) / (h_cap_250 - h_cap_200) ;\n", + "\n", + "# Results\n", + "print 'The exit temperature is = %d *C'%(T2) ;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The exit temperature is = 224 *C\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.23 Page No : 105" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from numpy import zeros\n", + "\n", + "# Variables\n", + "Cv = 3./2 * 8.314 ;\n", + "Cp = 5./2 * 8.314 ;\n", + "n = 1.; \n", + "R = 8.314 ; \n", + "T1 = 1000. ; \t\t\t#[K]\n", + "P1 = 10. ; \t \t\t#[bar]\n", + "T2 = 1000. ; \t\t\t#[K]\n", + "P2 = 0.1 ; \t\t\t#[bar]\n", + "T3 = 300. ; \t\t\t#[K]\n", + "T4 = 300. ; \t\t\t#[K]\n", + "\n", + "# Calculations and Results\n", + "k = Cp / Cv ;\n", + "P3 = P2 * (T3 / T2)**(k/(k-1)); \t\t\t#[bar]\n", + "P4 = P1 * (T4 / T1)**(k/(k-1)) ; \t\t\t#[bar]\n", + "\n", + "del_U_12 = 0 ; \t\t\t# As process 1-2 is isothermal \n", + "W_12 = n * R * T1 * math.log(P2 / P1);\n", + "Q_h_12 = W_12 ;\n", + "print 'a) 1) del_U = %d J'%(del_U_12) ;\n", + "print ' Work = %d J'%(W_12) ;\n", + "print ' Heat = %d J'%(Q_h_12) ;\n", + "\n", + "Q_23 = 0 ; \t\t\t# As adiabatic process\n", + "del_U_23 = n * Cv *(T3 - T2) ;\n", + "W_23 = del_U_23 ;\n", + "print ' 2) del_U = %g J'%(round(del_U_23,-1)) ;\n", + "print ' Work J = %d J'%(round(W_23,-1)) ;\n", + "print ' Heat J = %d J'%(Q_23) ;\n", + "\n", + "del_U_34 = 0 ; \t\t\t# As isothermal process\n", + "W_34 = n * R * T3 * math.log(P4 / P3) ; \t\t\t# Eqn E2.20.A\n", + "Q_c_34 = del_U_34 - W_34 ;\n", + "print ' 3) del_U = %g J'%(del_U_34) ;\n", + "print ' Work = %d J'%(W_34) ;\n", + "print ' Heat = %d J'%(Q_c_34) ;\n", + "\n", + "Q_41 = 0 ; \t\t\t# As adiabatic process\n", + "del_U_41 = n * Cv * (T1 - T4) ;\n", + "W_41 = del_U_41 ;\n", + "print ' 4) del_U = %g J'%(round(del_U_41,-1)) ;\n", + "print ' Work = %d J'%(round(W_41,-1)) ;\n", + "print ' Heat = %d J'%(Q_41) ;\n", + "\n", + "#Solution (c)\n", + "W_total = W_12 + W_23 + W_34 + W_41 ;\n", + "Q_absor = Q_h_12 ;\n", + "effi = W_total / Q_absor ;\n", + "print 'c) efficiency = %g'%(effi)\n", + "\n", + "#Solution (d)\n", + "x = 1 - T3 / T1 ;\n", + "print 'd) 1 - Tc/Th = %g'%(x);\n", + "print \" i.e Efficiency = 1 - Tc/Th\"\n", + "\n", + "#Solution (e)\n", + "print \"(e) The process can be made more efficient by raimath.sing Th or by lowering Tc .\"\n", + "print \"Table E2.20B\"\n", + "print \" T(K) P(bar) v(m**3/mol)\"\n", + "P = [P1 , P2 , P3 , P4 ] ;\n", + "T = [T1 , T2 , T3 , T4 ] ;\n", + "v = zeros(4)\n", + "for i in range(4):\n", + " v[i] = R * T[i] * 10**-5/ P[i] ;\n", + " print \" %6d %8.4f %.4f \"%(T[i],P[i],v[i]) ;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a) 1) del_U = 0 J\n", + " Work = -38287 J\n", + " Heat = -38287 J\n", + " 2) del_U = -8730 J\n", + " Work J = -8730 J\n", + " Heat J = 0 J\n", + " 3) del_U = 0 J\n", + " Work = 11486 J\n", + " Heat = -11486 J\n", + " 4) del_U = 8730 J\n", + " Work = 8730 J\n", + " Heat = 0 J\n", + "c) efficiency = 0.7\n", + "d) 1 - Tc/Th = 0.7\n", + " i.e Efficiency = 1 - Tc/Th\n", + "(e) The process can be made more efficient by raimath.sing Th or by lowering Tc .\n", + "Table E2.20B\n", + " T(K) P(bar) v(m**3/mol)\n", + " 1000 10.0000 0.0083 \n", + " 1000 0.1000 0.8314 \n", + " 300 0.0049 5.0597 \n", + " 300 0.4930 0.0506 \n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_&_Chemical_Thermodynamics/ch3.ipynb b/Engineering_&_Chemical_Thermodynamics/ch3.ipynb new file mode 100755 index 00000000..d04254c8 --- /dev/null +++ b/Engineering_&_Chemical_Thermodynamics/ch3.ipynb @@ -0,0 +1,661 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 3 : Entropy and the Second law of Thermodynamics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.2 Page No : 144" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "\n", + "# Variables\n", + "del_U = 0. ; \t\t\t# As no work or heat transfered across its boundaries during the process \n", + "T_1 = 500. ; \t\t\t# [K]\n", + "\n", + "# Calculations and Results\n", + "V1 = 1.6682 / 2 * 10**-3; \t\t\t# [m**3]\n", + "V2 = 2. * V1 ; \n", + "del_S_sur = 0 ; \t\t\t# As no heat transfered across its boundaries during the process \n", + "\n", + "\n", + "print \"(a)For an ideal gas u = u(T only)\"\n", + "print ' Final temperature = %g K '%(T_1);\n", + "\n", + "q_rev = 8.314 * T_1 * math.log(V2/V1) ;\n", + "del_S_sys = q_rev / T_1 ;\n", + "del_S_univ = del_S_sys + del_S_sur ;\n", + "\n", + "print 'b)Entropy change for universe = %.2f J/molK)'%(del_S_univ);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)For an ideal gas u = u(T only)\n", + " Final temperature = 500 K \n", + "b)Entropy change for universe = 5.76 J/molK)\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3 Page No : 145" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "T_1_1 = 273. ;\t\t\t# {K}\n", + "T_1_2 = 373. ; \t\t\t#[K]\n", + "Cp = 24.5 ; \t\t\t# [J/molK]\n", + "del_S_sur = 0. ; \t\t\t#Since the system is isolated \n", + "\n", + "# Calculations\n", + "T2 = (T_1_1 + T_1_2)/2 ;\n", + "del_S = Cp / 2 * math.log(T2**2 / (T_1_1 * T_1_2)) ;\n", + "\n", + "# Results\n", + "print \"Entropy change for the system = %.2f J/mol K)\"%( del_S);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Entropy change for the system = 0.30 J/mol K)\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.4 Page No : 147" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "del_h_vap = 38.56 * 10**3 ; \t\t\t#[J/mol] , From Table \n", + "Tb = 78.2 + 273. \t\t\t#[K] ,From table\n", + "\n", + "# Calculations\n", + "del_S = - del_h_vap / Tb * 10**-3 ;\n", + "\n", + "# Results\n", + "print \"Change in entropy = %.4f kJ/mol K\"%(del_S);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Change in entropy = -0.1098 kJ/mol K\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.5 Page No : 149" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "P_1 = 300. * 10**3 \t\t#[N/m**2]\n", + "T_1 = 700. \t\t \t# [*C]\n", + "V_bar_1 = 20. ; \t\t\t #[m/s]\n", + "P_2 = 200. * 10**3 ; \t\t\t# [N/m**2]\n", + "h_cap_1 = 3927.1 * 10**3 ; \t\t# [J/kg] , From table\n", + "S_cap_1 = 8.8319 \t\t\t # [kJ/kgK] , From table\n", + "\n", + "# Calculations\n", + "S_cap_2 = S_cap_1 \t\t\t# Reverssible adiabatic process \n", + "T2 = 623. \t\t\t# [*C] ,From table by interpolation\n", + "h_cap_2 = 3754.7 * 10**3\t\t# [J/kgK] ,From table by interpolation\n", + "V_bar_2 = math.sqrt(2 * (h_cap_1 - h_cap_2) + V_bar_1**2) ; \n", + "\n", + "# Results\n", + "print 'The final temperature is %g C and the exit velocity is %.1f m/s'%(T2,V_bar_2);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The final temperature is 623 C and the exit velocity is 587.5 m/s\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.6 page no : 150" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "m1 = 10 # kg/s\n", + "m2 = 1.95 # kg/s\n", + "s1 = 6.14 # kJ / kg K\n", + "s2 = .2945 # kJ / kg K\n", + "s3 = 5.6140 # kJ / kg K\n", + "\n", + "# Calculations\n", + "m3 = m1 + m2\n", + "dSdT = (m3 * s3) - (m1 *s1) + (m2*s2)\n", + "\n", + "#results\n", + "print \"dS/dTuniv = %.2f kW/k\"%dSdT\n", + "\n", + "# note : answer in book is wrong. " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "dS/dTuniv = 6.26 kW/k\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.7 Page No : 153" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "V_1 = 0.5 ; \t\t\t#[m**3]\n", + "P_1 = 150. ; \t\t\t#[kPa]\n", + "T_1 = 20. + 273.\t\t#[K]\n", + "P_2 = 400. ; \t\t\t# [kPa]\n", + "Cp = 2.5 * 8.314 ;\n", + "\n", + "\n", + "# Calculations and Results\n", + "Q = V_1 * (P_1 - P_2)\n", + "print \"a)Heat transferd = %g kJ\"%(Q);\n", + "del_S_sys = (P_1 * V_1) / T_1 * -math.log(P_2 / P_1) ;\n", + "print 'b)Entropy change of system = %.2f kJ/K '%(del_S_sys);\n", + "Q_surr = - Q ;\n", + "del_S_surr = Q_surr / T_1 ;\n", + "print ' Entropy change of surrounding = %.2f kJ/K '%(del_S_surr) ; \n", + "del_S_univ = del_S_sys + del_S_surr ;\n", + "print ' Entropy change of universe =%.2f kJ/K '%(del_S_univ) ;\n", + "\n", + "print \"c)Since entropy of the universe increases , the process is irreverssible .\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a)Heat transferd = -125 kJ\n", + "b)Entropy change of system = -0.25 kJ/K \n", + " Entropy change of surrounding = 0.43 kJ/K \n", + " Entropy change of universe =0.18 kJ/K \n", + "c)Since entropy of the universe increases , the process is irreverssible .\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.8 Page No : 154" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "import math\n", + "A = 3.355 \t\t\t# from table\n", + "B = 0.575 * 10**-3 ; \t\t\t# from table\n", + "D = -0.016 * 10**5 ; \t\t\t# from table\n", + "R = 8.314 ;\n", + "P1 = 1. \t\t\t#[bar]\n", + "P2 = 0.5 ; \t\t\t #[bar]\n", + "\n", + "# Calculations\n", + "def f(T):\n", + " return R * (A * math.log(T) + B * T + D / (2 * T**2))\n", + "\n", + "S1 = f(373.) - f(298.) ;\n", + "S2 = R * math.log(P1 / P2) ; \n", + "del_S = S1 - S2 ; \n", + "\n", + "# Results\n", + "print 'Entropy change = %.2f J/mol K'%(del_S);\n", + "\n", + "# Note : Answer may be different because of rouding error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Entropy change = 0.88 J/mol K\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.9 Page No : 155" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# Variables\n", + "P = 1. \t\t\t#[bar]\n", + "p_O2 = 0.5 ; \t\t\t#[bar]\n", + "p_N2 = 0.5 ; \t\t\t# [bar]\n", + "n_O2 = 1. \t\t\t#[mol]\n", + "n_N2 = 1. \t\t\t#[mol]\n", + "R = 8.314 \t\t\t# J/mol K\n", + "\n", + "# Calculations\n", + "del_S_1_O2 = -n_O2 * R * math.log(p_O2 / P) ;\n", + "del_S_1_N2 = -n_N2 * R * math.log(p_N2 / P) ; \n", + "del_S_2 = 0.\n", + "del_S = del_S_2 + del_S_1_O2 + del_S_1_N2 ;\n", + "\n", + "# Results\n", + "print \" Entropy of mixing = %.2f J/K\"%(del_S);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Entropy of mixing = 11.53 J/K\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.11 Page No : 157" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#Variables\n", + "P_1 = 10. ; \t\t\t#[bar]\n", + "T_1 = 298. ; \t\t\t# [K]\n", + "P_2 = 1. ; \t\t\t#[bar]\n", + "T_2 = 298. ; \t\t\t# [K]\n", + "P_3 = 1. ; \t\t\t#[bar]\n", + "R = 8.314 ; \t\t\t# [J/mol K]\n", + "n = 4. ; \t\t\t#[mol]\n", + "X = 0.01 ;\n", + "\n", + "# Calculations\n", + "del_S_sys = - R * math.log(P_2 / P_1);\n", + "del_S_surr = - R * (1 - P_2 / P_1) ;\n", + "del_s_univ_1 = del_S_sys + del_S_surr ; \n", + "Del_S_univ_1 = n * del_s_univ_1 ;\n", + "\n", + "Del_S_univ_2 = 0 ;\n", + "n_3 = n * P_3 / P_1 ;\n", + "\n", + "n_out = n - n_3 ;\n", + "del_S_sys_3 = - n_out * R * math.log(X) ;\n", + "Del_S_univ_3 = del_S_sys_3\n", + "Del_S_univ = Del_S_univ_1 + Del_S_univ_2 + Del_S_univ_3 ;\n", + "\n", + "# Results\n", + "print \"Total entropy change of universe = %.2f J/K \"%(Del_S_univ) ;\n", + "print \"No matter how slow the leak , the driving force for the expansion is finite . \\\n", + "So the process canot be reverssible .\"\n", + "\n", + "# Note : answer is different becuase of rounding error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Total entropy change of universe = 184.48 J/K \n", + "No matter how slow the leak , the driving force for the expansion is finite . So the process canot be reverssible .\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.12 Page No : 162" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "n_dot = 250. \t\t\t# [mol/s]\n", + "P_1 = 125. * 10**5 ; \t\t\t# [N/m**2]\n", + "V_cap_1 = 5. * 10**-4 \t\t\t# [m**3/mol]\n", + "P_2 = 8 * 10.**5 ; \t\t \t# [N/m**2]\n", + "\n", + "# Calculations\n", + "X = 3 * P_1**0.6667 * V_cap_1 * ( P_2**(1./3) - P_1**(1./3)) ;\n", + "W_dot_s = n_dot * X * 10**-6 \n", + "\n", + "# Results\n", + "print 'Power generated = %.1f MW'%(W_dot_s) ;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Power generated = -2.8 MW\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.13 Page No : 163" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Ws_real = -2.1 ; \t\t\t#[MW]\n", + "Ws_rev = -2.8 ; \t\t\t# [MW]\n", + "\n", + "# Calculations\n", + "n_tur = Ws_real / Ws_rev ;\n", + "\n", + "# Results\n", + "print \"Isentropic efficiency of turbine = %.2f %%\"%( n_tur * 100);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Isentropic efficiency of turbine = 75.00 %\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.14 Page No : 166" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "P_1 = 10. * 10**6 ; \t\t\t# [N/m**2]\n", + "T_1 = 600. + 273 ; \t\t\t #[K]\n", + "T_H = T_1 ;\n", + "T_C = 100. + 273 ; \t\t\t #[K]\n", + "P_3 = 10. * 10**4 ; \t\t\t#[N/m**2]\n", + "P_4 = P_1 ;\n", + "h_cap_1 = 3625.3 ; \t\t\t# [kJ/kg],From steam table\n", + "S_cap_1 = 6.9028 ; \t\t\t#[kJ/kgK],From steam table\n", + "S_cap_2 = S_cap_1 ; \t\t\t#[kJ/kgK],From steam table\n", + "S_cap_v = 7.3593 ; \t\t\t#[kJ/kgK],From steam table\n", + "S_cap_l = 1.3025 ; \t\t\t#[kJ/kgK],From steam table\n", + "h_cap_l = 417.44 ;\t\t\t# [kJ/kg],From steam table\n", + "h_cap_v = 2675.5 ;\t\t\t# [kJ/kg],From steam table\n", + "V_cap_l = 10**-3 ; \t\t\t# [m**3/kg],From steam table\n", + "\n", + "# Calculations\n", + "X = (S_cap_2 - S_cap_l) / (S_cap_v - S_cap_l);\n", + "h_cap_2 = (1 - X) * h_cap_l + X * h_cap_v ;\n", + "W_cap_s = h_cap_2 - h_cap_1 ;\n", + "h_cap_3 = h_cap_l ;\n", + "\n", + "W_cap_c = V_cap_l * (P_4 - P_3) * 10**-3 ;\n", + "h_cap_4 = h_cap_3 + W_cap_c ;\n", + "W_net = W_cap_s + W_cap_c ; \t\t\t# [kJ/kg]\n", + "n_turb = ( -W_cap_s - W_cap_c) / (h_cap_1 - h_cap_4) ;\n", + "\n", + "# Results\n", + "print \"Efficiency of the Rankine cycle = %.1f %% \"%(n_turb * 100 );\n", + "n_carnot = 1 - T_C / T_H ;\n", + "print \"Efficiency of the Carnot cycle = %.1f %%\"%(n_carnot * 100);\n", + "print \"The Rankine efficiecy is lower than Carnot efficiency .\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Efficiency of the Rankine cycle = 34.7 % \n", + "Efficiency of the Carnot cycle = 57.3 %\n", + "The Rankine efficiecy is lower than Carnot efficiency .\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.15 Page No : 167" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "n_turb = 0.85 ;\n", + "n_comp = 0.85 ;\n", + "W_cap_s_rev = -1120. ; \t\t\t#[kJ/kg]\n", + "h_cap_1 = 3625.3 ; \t\t\t#[kJ/kg]\n", + "h_cap_l = 417.44 ; \t \t\t#[kJ/kg]\n", + "W_cap_c_rev = 9.9 ; \t\t\t#[kJ/kg]\n", + "\n", + "# Calculations\n", + "W_cap_s_act = n_turb * W_cap_s_rev ;\n", + "h_cap_2_act = W_cap_s_act + h_cap_1 ;\n", + "h_cap_3 = h_cap_l ;\n", + "W_cap_c_act = W_cap_c_rev / n_comp ;\n", + "h_cap_4_act = W_cap_c_act + h_cap_3 ;\n", + "W_cap_net = W_cap_s_act + W_cap_c_act ;\n", + "n_rank_act = (-W_cap_s_act - W_cap_c_act) / (h_cap_1 - h_cap_4_act) ;\n", + "\n", + "# Results\n", + "print \"W_cap_net = %.1f kJ/kg\"%(W_cap_net) ;\n", + "print \"Efficiency of Rankine cycle = %.1f %%\"%(n_rank_act*100) \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "W_cap_net = -940.4 kJ/kg\n", + "Efficiency of Rankine cycle = 29.4 %\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.16 Page No : 171" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "P_1 = 120. * 10**3 ; \t\t\t#[N]\n", + "P_2 = 900. * 10**3 ; \t\t\t#[N]\n", + "h_4 = 25.486 \t\t\t#[kJ/mol], From table\n", + "h_1 = h_4 ;\n", + "h_2 = 39.295 ; \t\t \t#[kJ/mol], From table\n", + "S_2 = 177.89 ; \t\t \t#[kJ/molK], From table\n", + "S_3 = S_2 ; \t\t\t #[kJ/mol]\n", + "h_3 = 43.578 \t\t\t#[kJ/mol] , Enthalpy corresponding to S3 value which equales to S2\n", + "Q_dot_c_des = 10. \t \t\t#[kW]\n", + "\n", + "# Calculations\n", + "q_c = h_2 - h_1 ;\n", + "Q_dot_c = h_2 - h_1 ;\n", + "W_dot_c = h_3 - h_2 ;\n", + "COP = Q_dot_c / W_dot_c ;\n", + "n_dot = Q_dot_c_des / q_c ;\n", + "\n", + "# Results\n", + "print \"COP of the refrigerator is = %.2f \\nMass flow rate needed = %.3f mol/s\"%(COP,n_dot)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "COP of the refrigerator is = 3.22 \n", + "Mass flow rate needed = 0.724 mol/s\n" + ] + } + ], + "prompt_number": 11 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_&_Chemical_Thermodynamics/ch4.ipynb b/Engineering_&_Chemical_Thermodynamics/ch4.ipynb new file mode 100755 index 00000000..dd2a21ec --- /dev/null +++ b/Engineering_&_Chemical_Thermodynamics/ch4.ipynb @@ -0,0 +1,434 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 4 : Equation of states and intermolecular forces" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.1 Page No : 220" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "from numpy import *\n", + "\n", + "# Variables\n", + "H2O = 1. ;\n", + "NH3 = 2. ;\n", + "CH4 = 3. ;\n", + "CH3Cl = 4. ;\n", + "CCl4 = 5. ;\n", + "\n", + "M_11 = 1.85\n", + "alp_12 = 14.80\n", + "I_13 = 12.62 ;\n", + "M_12 = 1.47\n", + "alp_22 = 22.20\n", + "I_23 = 10.07 ;\n", + "M_31 = 0.00\n", + "alp_32 = 26.00\n", + "I_33 = 12.61 ;\n", + "M_41 = 1.87\n", + "alp_42 = 45.30\n", + "I_43 = 11.26 ; \n", + "M_51 = 0.00\n", + "alp_52 = 105.0\n", + "I_53 = 11.47 ;\n", + "\n", + "# Calculations\n", + "k = 1.38 * 10**-16 ; \t\t\t#[ J/K]\n", + "T = 298. \t\t\t#[K]\n", + "A =[[M_11 , alp_12 , I_13, 0,0,0,0],[M_12 , alp_22 , I_23,0,0,0,0 ],[M_31 , alp_32 , I_33,0,0,0,0],[M_41 , alp_42 , I_43,0,0,0,0],[M_51 , alp_52 , I_53,0,0,0,0 ]]\n", + "print A\n", + "\n", + "print (\" Molecule M alp*10**25 I C*10**60 Cd_d Cind Cdis\") ;\n", + "for i in range(5):\n", + " A[i][4] = ceil( 2./3 * A[i][0]**4 / (k * T) * 10**-12) ;\n", + " A[i][5] = ceil(2 * A[i][1] * A[i][0]**2 * 10**-1) ; \n", + " A[i][6] = ceil(3./4 * A[i][1]**2 * A[i][2] * 1.6 * 10**-2) ;\n", + " A[i][3] = ceil(A[i][4] + A[i][5] + A[i][6]) ; \t\t\t# ....E4.1D\n", + "\n", + "\n", + "print \" H2O %8.2f %5.1f %5.2f %7d %7d %5d %d \"%(A[0][0],A[0][1],A[0][2],A[0][3],A[0][4],A[0][5],A[0][6]) ;\n", + "print \" NH3 %8.2f %5.1f %5.2f %7d %7d %5d %d \"%(A[1][0],A[1][1],A[1][2],A[1][3],A[1][4],A[1][5],A[1][6]) ;\n", + "print \" CH4 %8.2f %5.1f %5.2f %7d %7d %5d %d \"%(A[2][0],A[2][1],A[2][2],A[2][3],A[2][4],A[2][5],A[2][6]) ;\n", + "print \" CH3Cl %8.2f %5.1f %5.2f %7d %7d %5d %d \"%(A[3][0],A[3][1],A[3][2],A[3][3],A[3][4],A[3][5],A[3][6]) ;\n", + "print \" CCl4 %8.2f %5.1f %5.2f %7d %7d %5d %d \"%(A[4][0],A[4][1],A[4][2],A[4][3],A[4][4],A[4][5],A[4][6]) ;\n", + "\n", + "print \"Even though it is non polar , CCl4 exhibit the largest intermolecular forces . It is due to the large polarizability accociated with the four Cl atom in CCl4 .\"\n", + "\n", + "# Note : Answer are slightly different because of rounding off error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "[[1.85, 14.8, 12.62, 0, 0, 0, 0], [1.47, 22.2, 10.07, 0, 0, 0, 0], [0.0, 26.0, 12.61, 0, 0, 0, 0], [1.87, 45.3, 11.26, 0, 0, 0, 0], [0.0, 105.0, 11.47, 0, 0, 0, 0]]\n", + " Molecule M alp*10**25 I C*10**60 Cd_d Cind Cdis\n", + " H2O 1.85 14.8 12.62 235 190 11 34 \n", + " NH3 1.47 22.2 10.07 146 76 10 60 \n", + " CH4 0.00 26.0 12.61 103 0 0 103 \n", + " CH3Cl 1.87 45.3 11.26 509 199 32 278 \n", + " CCl4 0.00 105.0 11.47 1518 0 0 1518 \n", + "Even though it is non polar , CCl4 exhibit the largest intermolecular forces . It is due to the large polarizability accociated with the four Cl atom in CCl4 .\n" + ] + } + ], + "prompt_number": 41 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.2 Page No : 222" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "import math\n", + "C6_Ar_HCl_tab = 76 * 10**-60 ;\t\t\t#From table E4.2\n", + "C6_Ar_Ar_tab = 52 * 10**-60 ;\t\t\t#From table E4.2\n", + "C6_HCl_HCl_tab = 134 * 10**-60 ;\t\t\t#From table E4.2\n", + "\n", + "# Calculations\n", + "C6_Ar_HCl_gmean = math.sqrt(C6_Ar_Ar_tab * C6_HCl_HCl_tab) ; \t\t\t#[erg/cm**6]\n", + "x = (C6_Ar_HCl_gmean - C6_Ar_HCl_tab) / C6_Ar_HCl_tab * 100 ;\n", + "\n", + "# Results\n", + "print \"The geometric mean is different from that in table E4.2 by %d %%\"%(x)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The geometric mean is different from that in table E4.2 by 9 %\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.4 Page No : 230" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "import math\n", + "Psat_wat_25 = 3.169 * 10**3 ;\t\t\t# From steam table\n", + "Psat_wat_50 = 1.235 * 10**4 ;\t\t\t# From steam table\n", + "Psat_wat_100 = 1.014 * 10**5 ;\t\t\t# From steam table\n", + "A =11.9673 ;\n", + "B = 3626.55 ;\n", + "C = -34.29 ;\n", + "T1 = 25 ; \t\t\t#[*C]\n", + "T2 = 50 ; \t\t\t#[*C]\n", + "T3 = 100 ; \t\t\t#[*C]\n", + "\n", + "# Calculations\n", + "M = [[T1, Psat_wat_25 , 0],[T2 , Psat_wat_50, 0],[T3 , Psat_wat_100, 0]]\n", + "#M = array(M)\n", + "\n", + "print (\" T(*C) Water(Pa) Methanol(Pa)\")\n", + "for i in range(3):\n", + " M[i][2] = math.exp(A - B / (M[i][0] + 273 + C)) * 10**5 ;\n", + " print \"%5d %7.3e %7.2e\"%(M[i][0],M[i][1],M[i][2])\n", + " #t.append(math.exp(A - B / (M[i][0] + 273 + C)) * 10**5)\n", + "#M.append(t)\n", + "\n", + "# Results\n", + "\n", + "print \"1) Water can form two hydrogen bonds . While CH4Oh can form only one . Thus at\\\n", + " a given temperature, water has stronger attractive forces in the liquid and a lower vapour pressure .\"\n", + "print \"2) Since the Maxwell-Boltzmann distribution depends exponentially on temperature,\\\n", + " Psat also increses exponentially with temperature .\"\n", + "\n", + "# Note: Answers may vary because of rounding off error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " T(*C) Water(Pa) Methanol(Pa)\n", + " 25 3.169e+03 1.68e+04\n", + " 50 1.235e+04 5.52e+04\n", + " 100 1.014e+05 3.53e+05\n", + "1) Water can form two hydrogen bonds . While CH4Oh can form only one . Thus at a given temperature, water has stronger attractive forces in the liquid and a lower vapour pressure .\n", + "2) Since the Maxwell-Boltzmann distribution depends exponentially on temperature, Psat also increses exponentially with temperature .\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.6 Page No : 236" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Pc_B = 49.1 ; \t\t\t# [bar] , From table\n", + "Pc_T = 42.0 ; \t\t\t# [bar] , From table\n", + "Pc_C = 40.4 ; \t\t\t# [bar] , From table\n", + "Tc_B = 562 ; \t\t\t# [K] , From table\n", + "Tc_T = 594 ; \t\t\t# [K] , From table \n", + "Tc_C = 553 ; \t\t\t# [K] , From table\n", + "R = 8.314 ;\n", + "\n", + "# Calculations\n", + "A = [[Pc_B , Tc_B, 0,0],[Pc_T , Tc_T,0,0],[Pc_C , Tc_C,0,0]]\n", + "\n", + "# Results\n", + "print \" P_c T_c a b \"\n", + "for i in range(3):\n", + " A[i][2] = 27./64 * (R * A[i][1])**2 /( A[i][0] * 10**5) ;\n", + " A[i][3] = R * A[i][1] / (8 * A[i][0] * 10**5) ;\n", + " print \" %5.1f %5d %7.2f %7.2e\"%(A[i][0],A[i][1],A[i][2],A[i][3])\n", + "\n", + "print \"The attractive interactions of all three compounds are dominated by\\\n", + " print ersion interactions ( parameter a) , while size affects parameter b .\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " P_c T_c a b \n", + " 49.1 562 1.88 1.19e-04\n", + " 42.0 594 2.45 1.47e-04\n", + " 40.4 553 2.21 1.42e-04\n", + "The attractive interactions of all three compounds are dominated by print ersion interactions ( parameter a) , while size affects parameter b .\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.11 Page No : 246" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "import math\n", + "B = 0.0486 * 10**-3 ;\n", + "T1 = 20 + 273 ; \t\t\t#[K]\n", + "T2 = 500 + 273 ; \t\t\t#[K]\n", + "v1 = 7.11 ; \t\t\t# [cm**3/mol]\n", + "\n", + "# Calculations\n", + "v2 = v1 * math.exp( B * (T2 - T1)) ;\n", + "\n", + "# Results\n", + "print \" Molar volume of solid state 2 = %.2f cm**3/mol\"%( v2);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Molar volume of solid state 2 = 7.28 cm**3/mol\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.12 Page No : 248" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "P_c = 37.9 * 10**5 ;\t\t\t#[N/m**2] , From compressibility chart\n", + "T_c = 425.2 \t\t\t# [K , From compressibility chart\n", + "P = 50. * 10**5 ; \t \t\t#N/m**2]\n", + "T = 333.2 ;\t\t\t #[K]\n", + "R = 8.314 ;\n", + "z_0 = 0.2148 ; \t\t\t\n", + "z_1 = -0.0855 ; \t\t\n", + "w = 0.199 ;\n", + "m = 10. ;\n", + "MW = 0.05812 ;\n", + "\n", + "# Calculations\n", + "a = (0.42748 * R**2 * T_c**2.5) / P_c ;\n", + "b = 0.08664 * R * T_c / P_c ;\n", + "A = P * T**(1./2) ;\n", + "B = -R * T**(3./2) ;\n", + "C = (a - P * T**(1./2) * b**2 - R * T**(3./2)*b) ;\n", + "D = - a * b;\n", + "\n", + "#mycoeff = [ D , C , B , A] ;\n", + "mycoeff = [ A,B,C,D]\n", + "#p = poly1d(mycoeff , \"v\" , \"coeff\" ); \n", + "M = roots(mycoeff);\n", + "\n", + "# Results\n", + "for i in range(3):\n", + " ans = sign(M[i])\n", + " if ans == 1:\n", + " V = m / MW *(M[i]) ;\n", + " #print \"Using Redlich Kwong equation the volume is = %.3f m**3\"%(V)\n", + "\n", + "z = z_0 + w * z_1 ; \n", + "v = z * R * T / P ;\n", + "V = m / MW * v ;\n", + "print \"Using compressibility chart the volume is = %.3f m**3\"%(V)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Using compressibility chart the volume is = 0.019 m**3\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.13 Page No : 253" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "import math\n", + "T = 100. + 273 ; \t\t\t#[K]\n", + "P = 70. * 10**5 ; \t\t\t#[N/m**2]\n", + "P_c = 42.2 * 10 ** 5 ;\n", + "T_c = 370. ; \t\t\t#[K]\n", + "w = 0.153 \t \t\t# Interpolating from table C.1 and C.2\n", + "z_0 = 0.2822 ;\t\t \t# Interpolating from table C.1 and C.2\n", + "z_1 = - 0.0670 ;\t\t\t# Interpolating from table C.1 and C.2\n", + "m = 20. * 10**3 ;\t\t\t#[g]\n", + "MW = 44. ; \t\t \t#[g/mol]\n", + "R = 8.314 ;\n", + "\n", + "# Calculations and Results\n", + "P_r = P / P_c ;\n", + "T_r = T / T_c ;\n", + "z = z_0 + w * z_1 ;\n", + "V = m / MW *z * R * T / P ;\n", + "\n", + "print \"1) Volume = %.4f m**3 \"%( V )\n", + "\n", + "\n", + "T = 295. ;\t\t\t#[K]\n", + "n = 50. ; \t\t\t# [mol]\n", + "a = 0.42748 * R**2 * T_c**2.5 / P_c ;\n", + "b = 0.08664 * R * T_c / P_c ;\n", + "v = 0.1 ;\n", + "P = R * T / (v - b) - a / (T**0.5 * v * (v + b)) ;\n", + "x = P * n * 10**-6 ;\n", + "\n", + "print \"2) Pressure = %.1f MPa \"%( x )\n", + "\n", + "y1 = 0.4 ;\n", + "y2 = 1 - y1 ;\n", + "n = 50. ;\n", + "P_c = 48.7 * 10**5 ;\t\t\t#[N/m**2]\n", + "T_c = 305.5 ; \t\t\t#[K]\n", + "a1 = a ;\n", + "b1 = b ;\n", + "a2 = 0.42748 * R**2 * T_c**2.5 / P_c ;\n", + "b2 = 0.08664 * R * T_c / P_c ;\n", + "\n", + "a_mix = y1**2 * a1 + 2 * y1 * y2 * math.sqrt(a1 * a2) + y2**2 * a2 ;\n", + "b_mix = y1 * b1 + y2 * b2 ;\n", + "P = R * T / (v - b_mix) - a_mix /(T**0.5 * v * (v + b_mix));\n", + "x = P * n * 10**-6 ;\n", + "\n", + "print \"3) Pressure = %.2f MPa \"%( x )\n", + "\n", + "# Note : Answers are slightly different because of rounding off error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1) Volume = 0.0548 m**3 \n", + "2) Pressure = 1.2 MPa \n", + "3) Pressure = 1.22 MPa \n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_&_Chemical_Thermodynamics/ch5.ipynb b/Engineering_&_Chemical_Thermodynamics/ch5.ipynb new file mode 100755 index 00000000..be3b6851 --- /dev/null +++ b/Engineering_&_Chemical_Thermodynamics/ch5.ipynb @@ -0,0 +1,313 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 5 : The thermodynamic web" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.2 Page No : 278" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "T_c = 370. ; \t\t\t#[K]\n", + "P_c = 41.58 * 10**5 ;\t\t\t#[N/m**2]\n", + "R = 8.314 ;\n", + "V1 = 0.001 ;\t\t\t#[m**3]\n", + "V2 = 0.04 ; \t\t\t#[m**3]\n", + "q = 10400. ; \t\t\t#[J]\n", + "\n", + "# Calculations\n", + "a = 27./64 * (R **2)*(T_c)**2 / P_c ;\n", + "del_U = -0.96 * (1 / V2 - 1 / V1) ; \n", + "W = round(del_U - q,-1) ;\n", + "\n", + "# Results\n", + "print \" Work done for the expansion = %.f J/mol\"%(W) ;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Work done for the expansion = -9460 J/mol\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.4 Page No : 282" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from scipy.optimize import fsolve \n", + "from scipy.integrate import quad\n", + "import math \n", + "\n", + "# Variables\n", + "P_1 = 9.43 * 10**5 ; \t\t\t#[N/m**2]\n", + "P_2 = 18.9 * 10**5 ; \t\t\t#[N/m**2]\n", + "T_1 = 80 + 273 ; \t\t\t#[K]\n", + "T_2 = 120 + 273 ; \t\t\t#[K]\n", + "A = 1.935 ;\n", + "B = 36.915 * 10**-3 ;\n", + "C = -11.402 * 10**-6 ;\n", + "T_c = 425.2 ; \t\t\t# [K]\n", + "P_c = 37.9 * 10**5 ;\t\t\t#[N/m**2]\n", + "R = 8.314 ;\n", + "\n", + "\n", + "del_h_1 = 1368 ; \t\t\t#[J/mol]\n", + "del_h_3 = -2542 ; \t\t\t#[J/mol]\n", + "Ws = 2100. ; \t\t\t#[J/mol]\n", + "\n", + "a = 0.42748 * R**2 * T_c**2.5 / P_c ;\n", + "b = 0.08664 * R * T_c / P_c ;\n", + " \n", + "\n", + "def f1(v):\n", + " return R * T_1 / (v - b) - a / (math.sqrt(T_1) * v *(v + b)) - P_1;\n", + "\n", + "za= fsolve( f1,0.001)\n", + "\n", + "def f2(v):\n", + " return R * T_2 / (v - b) - a / (math.sqrt(T_2) * v *(v + b)) - P_2;\n", + "\n", + "zb= fsolve( f2,0.001)\n", + "v1 = zb\n", + "v2 = zb\n", + "def f3(v):\n", + " T = 353.15\n", + " return b*R*T/(v-b) + a/T**0.5 * (3./(2*b) * math.log(v/(v+b)) - 1./(v+b))\n", + " \n", + "delta_h1 = quad(f3,za,zb)\n", + "\n", + " \n", + "def f(T):\n", + " return R * ( A * T + B/2 * T**2 + C/3 * T**3) ;\n", + "\n", + "del_h_2 = f(T_2) - f(T_1) ;\n", + "del_h_total = del_h_1 + del_h_2 + del_h_3 ;\n", + "q = del_h_total - Ws ;\n", + "\n", + "# Results\n", + "print \"v1 = %.2e m**3/mol, v2 = %.2e m**3/mol\"%(za,zb ) ;\n", + "print \"The heat input = %.f J/mol\"%(q);\n", + "\n", + "# Note : answer is slightly different because of rounding error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "v1 = 2.60e-03 m**3/mol, v2 = 1.27e-03 m**3/mol\n", + "The heat input = 1421 J/mol\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.6 Page No : 289" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import log\n", + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "\n", + "#Variables\n", + "A = 2.104 ; \n", + "B = 2.98 * 10**-3 ;\n", + "R = 8.314 ;\n", + "Beta = 3.5 * 10**-5 ;\n", + "v = 7.1 * 10**-6 # m**3/mol\n", + "P2 = 10000*10**5 ; \t\t#[N/m**2]\n", + "P1 =1 ;\t\t\t #[N/m**2] \n", + "\n", + "#Calculations\n", + "T2 = Symbol('T2')\n", + "q = solve(A*R*log(T2/1000)+B*R*(T2-1000)-(Beta*v*(P2-P1)),T2)\n", + "\n", + "#Result \n", + "print \"Heat input = \",q,\"J/mol\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat input = [1005.88624122037] J/mol\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.7 page no : 296" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from scipy.integrate import quad\n", + "\n", + "# Variables ( same as example 5.4)\n", + "R = 8.314\n", + "Tc = 425.2 # K \n", + "Pc = 37.9 # bar \n", + "w = 0.199\n", + "T1 = 353.15 # K\n", + "Tc = 425.2 # K\n", + "P1 = 9.47 # bar\n", + "Pc = 38. # bar\n", + "T2 = 393.15 # K\n", + "P2 = 18.9 # bar\n", + "R = 8.314\n", + "\n", + "# calculations\n", + "T1r = T1/Tc\n", + "P1r = P1/Pc\n", + "T2r = T2/Tc\n", + "P2r = P2/Pc\n", + "\n", + "def f(T):\n", + " return 1.935 + 36.915 * 10**-3 * T - 11.402 * 10**-6 * T**2\n", + "\n", + "cpDt = R * quad(f,353,393)[0]\n", + "h2h1 = .536 * R*Tc +cpDt - .969*R*Tc\n", + "\n", + "# Results\n", + "print \"the heat capacity = %.f J/mol\"%cpDt\n", + "print \"h2 - h1 = %.f J/mol\"%h2h1\n", + "\n", + "# Note : answer is slightly differnt becuase of rounding off error. and this problem is same as 5.4" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the heat capacity = 4695 J/mol\n", + "h2 - h1 = 3164 J/mol\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.10 Page No : 302" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from scipy.integrate import quad\n", + "\n", + "# Variables\n", + "T_c = 126.2; \t \t\t#[K] , From appendix A.1\n", + "P_c = 33.8 * 10**5 ; \t\t\t#[N/m**2] , From appendix A.1\n", + "w = 0.039 ; \t\t \t# From appendix A.1\n", + "enth_dep_1 = -2.81 ; \t\t\t# From table C.1 Appendix C\n", + "A = 3.28 ;\t\t\t # From Appendix A.2\n", + "B = 0.593 * 10**-3 ;\t\t\t# From Appendix A.2\n", + "del_h_dep_l = -5.1 ; \n", + "del_h_dep_v = -0.1 ;\n", + "T1 = 151. \t\t\t#[K]\n", + "P1 = 100. * 10**5 ; \t \t\t#[N/m**2]\n", + "P2 = 1. * 10**5 ; \t\t \t#[N/m**2]\n", + "T2_r = 0.61 ; \t\t\t # From figure 5.4\n", + "\n", + "# Calculations\n", + "T1_r = T1 / T_c ;\n", + "P1_r = round(P1 / P_c) ;\n", + "P2_r = P2 / P_c ;\n", + "T2 = round(T2_r * T_c) ; \t\t\t#[K]\n", + "\n", + "def f(T):\n", + " #return A * T + B/2 * T**2 \n", + " return A + B*T\n", + "\n", + "ans = quad(f,151,T2)[0]\n", + "\n", + "x = 1. / T_c *ans\n", + "y = enth_dep_1 - x ;\n", + "\n", + "# Results\n", + "X = ( y - del_h_dep_l) / (del_h_dep_v - del_h_dep_l);\n", + "print \"Quality = %.2f\"%(X) ;\n", + "\n", + "# Note: answer is different because of quad is giving different value -1.9 instead of -1.28. " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Quality = 0.85\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_&_Chemical_Thermodynamics/ch6.ipynb b/Engineering_&_Chemical_Thermodynamics/ch6.ipynb new file mode 100755 index 00000000..f7f5fdcf --- /dev/null +++ b/Engineering_&_Chemical_Thermodynamics/ch6.ipynb @@ -0,0 +1,519 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 6 : Multi component Phase Equillibrium" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.3 Page No : 330" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "from numpy import *\n", + "from matplotlib.pyplot import *\n", + "\n", + "\n", + "# Variables\n", + "slop = -4222.1 ;\n", + "R = 8.314 ;\n", + "\n", + "# Calculations\n", + "del_h_vap = -R * slop * 10**-3 ;\n", + "%pylab inline\n", + "\n", + "def f(x):\n", + " return -4222.1 * x + 17.556\n", + "\n", + "xdata = linspace(0.0032,0.004,8) ;\n", + "ydata = f(xdata) ;\n", + "plot(xdata,ydata) ;\n", + "xlabel(\"1/T[k**-1]\")\n", + "ylabel(\"Ln(Psat[kPa])\")\n", + "suptitle(\"least-squares linear fit\")\n", + "show()\n", + "\n", + "# Results\n", + "print \"Enthalpy of vapourisation of GaCH3)3 = %.1f kJ/mol\"%(del_h_vap) ;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Populating the interactive namespace from numpy and matplotlib\n" + ] + }, + { + "output_type": "stream", + "stream": "stderr", + "text": [ + "WARNING: pylab import has clobbered these variables: ['draw_if_interactive', 'new_figure_manager']\n", + "`%pylab --no-import-all` prevents importing * from pylab and numpy\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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XXcb27dupVasWR44coXPnzmzfvr288wEaeUjl8Msv1ha4ixfD44/DvfeC+m3F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\n8PO0adPMpEmTzvqaVq1amUO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+ "text": [ + "<matplotlib.figure.Figure at 0x3453bd0>" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Enthalpy of vapourisation of GaCH3)3 = 35.1 kJ/mol\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.5 pageno : 332\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# variables\n", + "MWni = 58.69 #g/mol\n", + "Psni = 8900 # kg/m**3 \n", + "Sni = 1.75 # J/m**2 surface tension of solid Ni\n", + "deltaHfus = -17.48 # kJ/mol\n", + "Tm = 1728 #K\n", + "R = 2*10**-9\n", + "\n", + "\n", + "# calculations\n", + "Uni = MWni/(Psni*1000)\n", + "T = Tm*(1+((2*Uni*Sni)/(deltaHfus*R)))\n", + "\n", + "# results\n", + "print \"molar density of the solid %.1e m**3/mol\" %Uni\n", + "print \"T = %.f K\"%T\n", + "# incorrect answer in textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "molar density of the solid 6.6e-06 m**3/mol\n", + "T = -1139085 K\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.8 Page No : 347" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%pylab inline\n", + "\n", + "from numpy import zeros\n", + "from matplotlib.pylab import *\n", + "\n", + "# Variables\n", + "MW1 = 119.5 ;\n", + "MW2 = 58 ;\n", + "A = [0,4.77,9.83,14.31,19.38,23.27,25.53,25.07,21.55,13.56,0] ;\n", + "B = [0,.1 , .2 , .3 ,.4 ,.5 ,.6 ,.7 ,.8 ,.9,1] ;\n", + "\n", + "C = zeros(11)\n", + "D = zeros(11)\n", + "# Calculations\n", + "print \" For weight percent %.3f del_h_mix = %.1f J/mol\"%(C[0],D[0])\n", + "for i in range(1,11):\n", + " x1 = (B[i] / MW1) / (B[i]/MW1 + (1 - B[i]) / MW2) ;\n", + " x2 = 1 - x1 ;\n", + " MW = x1 * MW1 + x2 * MW2 ;\n", + " del_h_mix = - 1*(A[i]) * MW ;\n", + " C[i] = del_h_mix ;\n", + " D[i] = x1 ;\n", + " print \" For weight percent %.3f del_h_mix = %.1f J/mol\"%(x1,del_h_mix)\n", + "\n", + "xdata = D ;\n", + "ydata = C ;\n", + "plot(xdata ,ydata) ;\n", + "xlabel(\"xCHCI3\")\n", + "ylabel(\"delta Hmix[J/mol]\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Populating the interactive namespace from numpy and matplotlib\n", + " For weight percent 0.000 del_h_mix = 0.0 J/mol\n", + " For weight percent 0.051 del_h_mix = -291.7 J/mol\n", + " For weight percent 0.108 del_h_mix = -635.6 J/mol\n", + " For weight percent 0.172 del_h_mix = -981.5 J/mol\n", + " For weight percent 0.244 del_h_mix = -1415.4 J/mol\n", + " For weight percent 0.327 del_h_mix = -1817.3 J/mol\n", + " For weight percent 0.421 del_h_mix = -2142.2 J/mol\n", + " For weight percent 0.531 del_h_mix = -2272.9 J/mol\n", + " For weight percent 0.660 del_h_mix = -2124.7 J/mol\n", + " For weight percent 0.814 del_h_mix = -1465.1 J/mol\n", + " For weight percent 1.000 del_h_mix = 0.0 J/mol\n" + ] + }, + { + "output_type": "stream", + "stream": "stderr", + "text": [ + "WARNING: pylab import has clobbered these variables: ['f']\n", + "`%pylab --no-import-all` prevents importing * from pylab and numpy\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 2, + "text": [ + "<matplotlib.text.Text at 0x361ce50>" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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kOHI99eoBn38u1wbodEDLlqojIrqz336Tv6/vvAN066Y6GvdhsVts06ZNGDRo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\nto/p3r27+P7770VWVpbQarX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+ "text": [ + "<matplotlib.figure.Figure at 0x3453b90>" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.9, Page 349" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy.solvers import solve\n", + "from sympy import Symbol\n", + "\n", + "#Variables\n", + "cp1 = 27.5 #J/mol-K\n", + "cp2 = 25 #J/mol-K\n", + "cp3 = 20 #J/mol-K\n", + "T1 = 4+273 #K\n", + "\n", + "#Calculations\n", + "#Since only species 2 and 3 are involved,\n", + "deltaH1_unmix = -160 #J\n", + "deltaH11_mix = 1100 #J\n", + "\n", + "deltaH_mix = deltaH1_unmix+deltaH11_mix #J\n", + "T2 = Symbol('T2')\n", + "Tf = solve(deltaH_mix+235*(T2-T1))\n", + "\n", + "#Result\n", + "print \"The final temperature is\",Tf,\"K\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The final temperature is [273] K\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.10 Page No : 353" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "A = [-32669,-31840,-28727,-26978,-24301,-20083,-13113] ;\n", + "B = [20 ,10 ,5 ,4 ,3 ,2 ,1] ;\n", + "\n", + "# Calculations and Results\n", + "C = zeros(7)\n", + "D = zeros(7)\n", + "\n", + "for i in range(7):\n", + " del_h_mix = A[i] / (1. + B[i]) ;\n", + " C[i] = del_h_mix ;\n", + " D[i] = 1. / (1 + B[i]) ;\n", + " print \"For mole fraction %.3f the entropy of mixing is %.0f J/mol\"%(D[i],C[i])\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For mole fraction 0.048 the entropy of mixing is -1556 J/mol\n", + "For mole fraction 0.091 the entropy of mixing is -2895 J/mol\n", + "For mole fraction 0.167 the entropy of mixing is -4788 J/mol\n", + "For mole fraction 0.200 the entropy of mixing is -5396 J/mol\n", + "For mole fraction 0.250 the entropy of mixing is -6075 J/mol\n", + "For mole fraction 0.333 the entropy of mixing is -6694 J/mol\n", + "For mole fraction 0.500 the entropy of mixing is -6556 J/mol\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.12 Page No : 359" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "x1 = 0.3 ;\n", + "x2 = 1- x1 ;\n", + "B11 = -910. ;\n", + "B22 = -1330. ;\n", + "B12 = -2005. ;\n", + "T = 333. ; \t\t\t#[K]\n", + "P = 10. * 10**5 ;\n", + "R = 8.314 ;\n", + "\n", + "# Calculations and Results\n", + "v1 = round(R * T /P * 10**6 + B11,-1) ;\t\t\t#....E6.9A\n", + "print \" v1 = %g cm**3/mol\"%(v1)\n", + "\n", + "V_bar_1 = (R * T / P) * 10**6+ (x1**2 + 2 * x1 * x2) *3.12 * B11 + x2**2 * B12 - x2**2 * B22 ;\t\t\t#.....E6.9B\n", + "print \" V_bar_1 = %.f cm**3/mol\"%(V_bar_1) ;\n", + "\n", + "del_v_mix = x1 * x2 * (2 * B12 - B11 - B22) ;\t\t\t#.....E6.9C\n", + "print \" del_v = %.f cm**3/mol\"%(del_v_mix);\n", + "\n", + "# Note : answer are slightly different because of rounding error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " v1 = 1860 cm**3/mol\n", + " V_bar_1 = 990 cm**3/mol\n", + " del_v = -372 cm**3/mol\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.13 Page No : 360" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import zeros\n", + "from matplotlib.pyplot import *\n", + "\n", + "# Variables\n", + "h_H2SO4 = 1.596 ; \t\t\t#[kJ/mol]\n", + "h_H2O = 1.591 ; \t\t\t#[kJ/mol]\n", + "C1 = -74.40 ;\n", + "C2 = 0.561 ;\n", + "A = [0 ,0.1 , 0.2 ,0.3 ,0.4 ,0.5 ,0.6 ,0.7 ,0.8 ,0.9 ,1] ;\n", + "B = [1 ,0.9 ,0.8 ,0.7 ,0.6 ,0.5 ,0.4 ,0.3 ,0.2 ,0.1 ,0] ;\n", + "\n", + "# Calculations\n", + "y_data_1 = zeros(11)\n", + "y_data_2 = zeros(11)\n", + "x_data = zeros(11)\n", + "for i in range(11):\n", + " H_bar_H2SO4 = h_H2SO4 + C1 * B[i]**2 - 2 * C2 * C1 * A[i] * B[i]**2 ;\n", + " H_bar_H2O = h_H2O + C1 * A[i]**2 -C2 * C1 * A[i]**2 * (1 - 2 * B[i]) ;\n", + " y_data_1[i] = H_bar_H2SO4 ;\n", + " y_data_2[i] = H_bar_H2O ;\n", + " x_data[i] = A[i] ;\n", + "\n", + "plot(x_data,y_data_1) ;\n", + "plot(x_data,y_data_2) ;\n", + "xlabel(\"xH2SO4\")\n", + "ylabel(\"Partial molar enthalpy (J/mol)\")\n", + "#suptitle(\"Partial molar enthalpies of water and sulfuric acid at 21\u00b0C.\")\n", + "m = y_data_1[5] ;\n", + "s = y_data_2[5] ;\n", + "print \"For equimolar mixture del_H_H2SO4 = %.1f kJ/mol del_H_H2O = %.1f kJ/mol\"%(m,s);\n", + "show()\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For equimolar mixture del_H_H2SO4 = -6.6 kJ/mol del_H_H2O = -17.0 kJ/mol\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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wwsPDi72dx7YoHvzVn5qayvHjx2ndujUVKlRQP2Rnx+bNm4u804sXL+Lm5gao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\ndnZGr9cXa18FceIEzJ+vtia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MgoKCAGjbti3ffvstsbGxjBgxAjs7O9LT0+natSvBwcHo\ndDq2bdvGmDFjmDJlCkajkb/97W/Mnj3bqr+rEEUh04wLIYTIk1x6EkIIkScJCiGEEHmSoBBCCJEn\nCQohhBB5kqAQQgiRJwkKIYQQeZKgEEIIkScJCiGEEHn6f5bpWFZTbdHmAAAAAElFTkSuQmCC\n", + "text": [ + "<matplotlib.figure.Figure at 0x3613750>" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.16 Page No : 365" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%pylab inline\n", + "\n", + "from matplotlib.pyplot import *\n", + "\n", + "# Variables\n", + "C1 = 1.596 ;\n", + "C2 = 1.591 ;\n", + "C3 = -74.40 ;\n", + "C4 = -0.561 ;\n", + "A = [ 0 ,0.1 ,0.2 ,0.3 ,0.4 ,0.5 ,0.6 ,0.7 ,0.8 ,0.9 ,1] ;\n", + "m = (-C1 + C2 + C3 * ( C4 * 0.25)) * 1000 ;\n", + "\n", + "# Calculations\n", + "C = zeros(11)\n", + "for i in range(11):\n", + " x_H2O = A[i] ;\n", + " x_H2SO4 = 1- x_H2O ;\n", + " h = C1 * x_H2SO4 + C2 * x_H2O + C3 * x_H2SO4 * x_H2O *(1 + C4 * x_H2SO4) ;\n", + " C[i] = h * 10**3;\n", + "\n", + "y1 = C[5]\n", + "\n", + "def f613(x):\n", + " return -m * (x - 0.5 ) + y1 ;\n", + "\n", + "F = zeros(11)\n", + "for i in range(11):\n", + " F[i] = f613(A[i]) ;\n", + "\n", + "# Results\n", + "plot(A,C);\n", + "plot(A,F)\n", + "plot(0.5,-12000,\"go\")\n", + "xlabel(\"XH2O\")\n", + "ylabel(\"h(J/mol)\")\n", + "suptitle(\"Graphical determination of values for the partial molar enthalpies of sulfuric acid and water\")\n", + "\n", + "print \"The partial molar property can be obtained by drawing tangent at mole fraction 0.5 .\"\n", + "\n", + "show()\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Populating the interactive namespace from numpy and matplotlib\n", + "The partial molar property can be obtained by drawing tangent at mole fraction 0.5 .\n" + ] + }, + { + "output_type": "stream", + "stream": "stderr", + "text": [ + "WARNING: pylab import has clobbered these variables: ['draw_if_interactive']\n", + "`%pylab --no-import-all` prevents importing * from pylab and numpy\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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XfvON4rapanVnY6M+WjXWpg1w6hQglQIDBgCJiUJHRAhRZ6GhQHAwNyMdURxK\ntGpOS4ubOerNN7nl9m7fFjoiQog6KivjbuVZt46ujikaJdomQCLhBkX98APg7Q2cOyd0RIQQdfPr\nr0DbtsCkSUJH0vRQH60c1KGf4do1YNw47r7bmTOFjoYQog7S0wELC+6OBhsbxW9fHepOZaJEKwd1\nKSwxMdzqP6NHA2vWcIvLE0JIbWbNApo3B9avV8721aXuVBZKtHJQp8KSns4l2q5dgV276KZzQkjN\nIiK4KV4fPgR0dJSzD3WqO5WB2jpNlK4udxlISwvw9ATS0oSOiBCiamQybgDU6tXKS7KEEm2T1qIF\n8PvvwODB3IhkWpCAEFLR7t1csvXzEzqSpo0uHctBnS9/bNsG/N//AYcPc/fcEkLELSsL6N2buxe/\nTx/l7kud605FoEQrB3UvLEFB3P2269cDkycLHQ0hREgffgjk53O39SibutedDUWJVg5NobDcu8fN\nkTxzJvDpp7QgASFidO8eN3bjwQOgY0fl768p1J0NQYlWDk2lsCQnc6v/2NsDW7dyi8sTQsSBMcDD\nA5gwAZgzp3H22VTqzldFg6FESF8fuHSJW6Vj+HAgO1voiAghjeXgQe47TxPaNB5KtCLVujVw4gRg\nZgb07w88eSJ0RIQQZcvL46Zr3bQJ0NQUOhrxoEQrYlpawMaNwDvvAP36AbduCR0RIUSZvv6au2zs\n5iZ0JOJCfbRyaMr9DMeOcZeSduzgBksRQpqWR4+4H9T37nEzxjWmplx31ge1aAkAYOxY4I8/gBkz\nuGX3CCFNB2PA/PncnQaNnWQJJVpSgYsLt/Dzhg3AwoXcjDGEEPV36hQ3DmP+fKEjESe6dCwHsVz+\nyMjgWri6usCePUCrVkJHRAh5VQUF3BJ4v/0GDBkiTAxiqTtrQy1aUk2HDkBAAJdgBw8Gnj0TOiJC\nyKv69lvAyUm4JEuoRSsXsf0qYwxYvpxbmODsWcDcXOiICCHyiI3lkmxkJNC9u3BxiK3urEptW7Qr\nVqyAoaEh7O3tYW9vj3PnzvHPrVmzBqampjA3N0dgYCD/eEREBKytrWFqaooFCxbwjxcVFWHixIkw\nNTVF3759ER8f36jHoqokEmDVKuCLL4BBg7hJLggh6uPjj7l/QiZZosaJViKR4OOPP0ZkZCQiIyMx\ndOhQAMCDBw9w8OBBPHjwAP7+/pgzZw7/S2r27NnYtm0boqKiEBUVBX9/fwDAtm3boKuri6ioKHz0\n0Uf45JN0WV3OAAAgAElEQVRPBDsuVeTnB+zbB4wfz7VuCSGqb8cO4P59bmAjEZbaJloANV6KOHny\nJCZPngxtbW1IpVKYmJggLCwMKSkpyM3NhbOzMwBg2rRpOHHiBADg1KlTmD59OgBg3LhxCA4ObryD\nUBOensDFi8CyZcCXX3KXlQkhqikggLuV5/Rpbl1qIiy1TrQbN26Era0t3n33XWRlZQEAkpOTYWho\nyL/G0NAQSUlJ1R43MDBAUlISACApKQndunUDAGhpaaFdu3bIyMhoxCNRD5aWwPXr3K0Cb70FvHgh\ndESEkKoiI4GpU4GjR4FevYSOhgCAltAB1MXLywupqanVHv/6668xe/ZsfPHFFwCAzz//HAsXLsS2\nbduUHtOKFSv4/3d3d4e7u7vS96lKunTh+mpnzwZcXbkZpUxMhI6KEAIA8fHcylxbtwo7zWJISAhC\nQkKEC0DFqHSiDQoKqtfr3nvvPYwcORIA11JNSEjgn0tMTIShoSEMDAyQmJhY7fHy9zx58gT6+voo\nLS1FdnY2OnToUOO+KiZasWrVCti5E/jlF25Kt//9Dxg1SuioCBG3jAxg6FBgyRJg3DhhY6naCFm5\ncqVwwagAtb10nJKSwv//8ePHYW1tDQAYNWoUDhw4gOLiYsTGxiIqKgrOzs7o0qUL2rZti7CwMDDG\nsGfPHrz++uv8e3bt2gUAOHLkCDw9PRv/gNSMRALMmsX1Ac2dy/XdlpUJHRUh4lRYCIweDQwbRrM/\nqSK1vY922rRpuH37NiQSCYyMjPDLL79AT08PALB69Wps374dWlpaWL9+PXx8fABwt/f4+fmhoKAA\nw4YNw4YNGwBwt/dMnToVkZGR0NXVxYEDByCVSqvtU+z3gtXm2TNg0iRuNaB9+4COHYWOiBDxkMm4\n759EAuzfD2ioYPNJ7HWnUhJtYWEhJBIJmjdvruhNC0rshaUupaVcq3b/fuDIEaBPH6EjIkQcFi4E\nbt7kRhqr6ghjsdedCvntI5PJcOzYMYwfPx4GBgYwMjJCjx49YGBggDfeeAPHjx8X9UkWAy0tYO1a\n4McfgeHDuXlV6SMnRLl++gk4dw44cUJ1kyxRUIt24MCBGDBgAEaNGgU7Ozu+JVtUVITIyEicOnUK\nV69exeXLlxscsJDE/qusvv75h1uUoG9fYNMmoGVLoSMipOk5cgT48ENuxa0ePYSOpm5irzsVkmiL\niopeepm4Pq9RdWIvLPLIywPee49bbProUcDISOiICGk6rl7lfswGBgJ2dkJH83JirzsVkmhfNrlD\nbbfKqBuxFxZ5Mcatbbt6NXc70L+zZBJCGuDhQ8DdHdi9G/D2Fjqa+hF73amQRCuVSiGRSGregUSC\nx48fN3QXKkHsheVVXbnCjYqcOZMbMKWKoyIJUQepqdxEMStWAP/OGqsWxF53qu3tPUIQe2FpiJQU\nYMIEoG1bbmGC9u2FjogQ9ZKby7Vkx4zhfrCqE7HXnQpPtCdPnsTly5chkUgwaNAgfsampkDshaWh\nSkqATz4BTp7k+m3VoW+JEFVQUsLNvtatGzcjWy0XEFWW2OtOhSbapUuXIjw8HG+++SYYYzhw4ACc\nnJywZs0aRe1CUGIvLIpy8CA3m9T33wPTpgkdDSGqjTFuYGFqKvcjVUulJ86tmdjrToUmWmtra9y+\nfRuampoAgLKyMtjZ2eHevXuK2oWgJBIJ7j29B8tOlrX2SZP6uX+fGzXp6cnde6vmA9IJUZqVK4E/\n/uCWqWzdWuhoXo3YE61Ch6VIJBJ+uToAyMrKanIJaeT+kej2Yze8c/IdHLp/CBkFtJzeq7C0BMLD\nuV/pAwcCFdaBIIT8a/t2bnTxH3+ob5IlCm7R7t+/H0uXLuVXbbh06RLWrl2LSZMmKWoXgpJIJJDJ\nZIjKiEJAdAACYgJwOf4yLDpZwMfYBz4mPnA2cIaWhhpe2xEIY8C6dVyr9vffuRYuIQTw9wf8/IDL\nlwEzM6GjaRixt2gVPhgqOTkZ4eHhkEgk/Ko5TUVNhaWotAhXn1xFQAyXeBOyE+DZ05NLvMY+6Nau\nm0DRqpcLF4A33+RmulmyRP0GexCiSLduAb6+XJ+sq6vQ0TQcJVoFH/2dO3cQFxeH0tJS/rLx2LFj\nFbkLwdSnsCTnJiMwJhABMQEIiglC59c6w9fEFz7GPhjYYyBaatN8hLVJSADGjwf09YEdO4B27YSO\niJDGFxfHLdq+eTO39F1TQIlWgUf/9ttv4969e7C0tIRGhVkJduzYoahdCErewlImK8OtlFsIiAmA\nf7Q/7jy9g37d+sHH2Ae+Jr7o3bF3k+vDbqiiIuDjj4GgIODYMcDKSuiICGk8GRlAv37AvHnABx8I\nHY3iUKJV4NFbWFjg/v37TTZ5NLSwZBVm4ULsBb5/t4yV8ZeYh/QcgvYtaRaHcrt3c8t/bdgATJ4s\ndDSEKF9hITBkCNea/eYboaNRLEq0Cjz66dOnY8mSJbC0tFTUJlWKIgsLYwyP0h/BP9ofATEBuPrk\nKqw6W/GDqvro94GmhqZC9qWubt8Gxo0DRo7kBkxpawsdESHKIZNxM6c1a8YNCmxq05RSolXg0YeE\nhGDUqFHo0qULv1KPRCLB3bt3FbULQSmzsBSWFnKDqqID4B/jj+TcZHgaecLXxBfext4wbGuolP2q\nusxMYOpUICsLOHwY6NpV6IgIUbyPPuJ+WPr7N817yinRKvDojY2N8eOPP8LKyqpSH61UKlXULgTV\nmIUlKSfpv0FVj4PQtXVXvrU7sMdAtNASzyrPMhm3AtDWrcCBA8CAAUJHRIji/PgjsG0bt/Sdjo7Q\n0SgHJVoFHr2rqyv+/PNPRW1O5QhVWMpkZbiZfJO/heje03tw6+7G9++adzRvsv3iFQUEcCuWLF0K\nLFhAtwAR9XfoEDcW4do1bh7jpooSrQKPfs6cOcjKysLIkSPRrFkzbgcSiahu72kMWYVZCH4czPfv\nAuBvIfLs6QmdFk30ZzG4Wx/GjQNMTYH//Y9myyHq6/Jl4I03gPPnARsboaNRLlWpO4Wi0ETr5+fH\nbbRKU0Ost/c0BsYYHj5/yN9CFJoQChs9G/4WIseujk1uUFVhIXfrw7Vr3BR1TeGGfiIuDx4AHh7A\nvn3imA1NFevOxqSQRLtv3z74+PhAV1dXETGpLHUoLAUlBbjy5Ap/C1FqXiqG9BzC9+/qt9EXOkSF\nOXSIm0lq9GhgzRqa4IKoh+Rk7l7ZL7/kBvqJgTrUncqkkES7du1aBAYGori4GEOGDMHQoUPh7Ozc\n5PoN1bGwJOYk8kn3/OPzMGxryCfd/t37q/2gqsxM4NNPuUnXf/qJu6zcxIodaUJyc7lFNCZM4Mqt\nWKhj3alICr10nJOTg/PnzyMgIAA3btyAubk5hg4dCh8fH+jp6SlqN4JR98JSJitDeHI437d7/9l9\nDOgxgB9UZaZrprY/jkJDgfffB3r25Kau695d6IgIqaykBBgxgiujW7aI6wehutedDaXwuY4run//\nPs6dO4fAwEAEBgYqazeNpqkVloyCDAQ/Dub7d7U0tPjWrqeRJ9q1UK9rscXF/60E9NlnwPz56rlI\nNml6GAPeeQdIT+emFhVbuWxqdae8FJJoIyIiam0JNWvWDN27d0fbtm0buhvBNeXCwhjDg7QH/C1E\n1xKuwa6LHd/addR3hIZEPaariYoCZs3iJrn49VfA0VHoiIjYLV/OTUZx4QLw2mtCR9P4mnLdWR8K\nSbTu7u61JtrS0lLEx8fjgw8+wCeffNLQXQlKTIUlvyQfl+Mv8/27aflp8OrpBR9jH3gbe6NrG9We\nookxYM8ebsm9yZO5gSd0KxARwv/+B6xdy42S79xZ6GiEIaa6syZKvXRcrrCwEA4ODnjw4IGyd6VU\nYi4sT7KfIDAmEP7R/giODUb3dt35W4jcurmhuZZqzhv3/DmwaBFw8SLXdztihNARETE5exZ4913g\nyhXAxEToaIQj5roTUFCivXTpEgYNGlTna9atW4fFixc3dFeCEnthKVcqK8WNpBt8a/dB2gMM7DGQ\nT7wmHUxUblDVhQvc5WRbW2D9em7NW0KU6eZNYNgw4PRpwMVF6GiEJfa6UyGJdtGiRbh8+TKGDBkC\nJycndOnSBYwxpKam4ubNmzh//jw8PDzw7bffKiJmwYi9sNQmPT8d5x+f5/t3m2s25wdVDTYajLbN\nVaN/vrAQ+Ppr4OefgZUrucTb1FZJIarh8WNuTu6tW4FRo4SORnhirzsVduk4NzcXJ0+eRGhoKOLj\n4wEAPXr0QP/+/fH666+jdRPoIBN7YakPxhjup93nbyG6nngdDl0d+EFV9l3tBR9U9eABdytQWRk3\nWMraWtBwSBPC2H8TqSxfzv2YI1R3KrSPtrCwEEePHkVcXBxKS0v5x5cvX66oXQhK7IXlVeSX5ONS\n3CW+tZuenw5vY29+UJVea2Hur5bJuEEqy5ZxfWhffAG0bClIKKSJSEwE5szhWrP/+x/Qt6/QEakO\nsdedCk20Pj4+0NHRgaOjIzQ1/5tfd+HChYrahaDEXlgUIT4rnk+6F2IvQKoj5Vu7bt3d0EyzWaPG\nk5rKtT7Cw7lLyl5ejbp70gTIZMBvv3E/2ubO5WZ8ata4xVjlib3uVGiitbKywl9//aWozakcsRcW\nRSuVlSIsMYxPvA+fP8SgHoP4/l2TDo03TPPcOa414uYG/PCDeG/DIPKJigJmzOD6/7dtAywthY5I\nNYm97lRoZ1m/fv1w9+5dRW6SNGFaGlpw6+6GVR6rEPZeGGLmx2CK9RTcTLmJgTsGwniDMT448wFO\n/XMKuUW5So1l6FDgr7+Arl0BKytuVSAR1wvkJUpLgW+/5VaOGj2amwKUkiypjUISrbW1NaytrXH1\n6lU4OjrCzMyMf8ymgQstHj58GJaWltDU1MStW7cqPbdmzRqYmprC3Ny80hSPERERsLa2hqmpKRYs\nWMA/XlRUhIkTJ8LU1BR9+/blB20BwK5du2BmZgYzMzPs3r27QTGTV9OxVUdMspqEHa/vQNLHSTg+\n8TikOlJsCNsA/R/04b7THWuurEFkSiRkTKbw/b/2GjeFY2AgdxnZwwP45x+F74aoudu3udt1goK4\nLocPPwQ0m9ZKlETRmALExsbW+a8h/v77b/bPP/8wd3d3FhERwT9+//59Zmtry4qLi1lsbCwzNjZm\nMpmMMcZYnz59WFhYGGOMsaFDh7Jz584xxhjbvHkzmz17NmOMsQMHDrCJEycyxhhLT09nPXv2ZJmZ\nmSwzM5P//6oUdLrIK8grymNnHp1h88/OZ7029mKd13Vmbx17i+25s4c9zXuq8P2VljK2YQNjHTsy\ntmIFY4WFCt8FUTMFBYx99hljnToxtn07Y/9WN6QexF53KmRqa6lUqojN1Mjc3LzGx0+ePInJkydD\nW1sbUqkUJiYmCAsLQ48ePZCbmwtnZ2cAwLRp03DixAn4+vri1KlTWLlyJQBg3LhxmDt3LgAgICAA\n3t7e0NHRAQB4eXnB398fkyZNUtpxEfm81uw1DDMdhmGmwwAAcVlxCIgOwLG/j2Hu2bkw7mDMD6py\n7eba4EFVmprAvHnAmDHcf+3sgF9+4ZY4I+Jz9Srw3ntct8Ldu0CXLkJHRNSJ2q4hkZycjL4Vxs8b\nGhoiKSkJ2traMDQ05B83MDBAUlISACApKQndunUDAGhpaaFdu3ZIT09HcnJypfeUb4uoLqmOFDOd\nZmKm00yUlJXgeuJ1BMQEYFHQIkSlR2GQdBB8jX3hY+KDnu17vvJ+DA2B48e5f2++Cfj4cH1zHToo\n8GCIysrN5UYRHz8ObNwIjB0rdEREHanEvDheXl58n27Ff6dPnxY6NKIGtDW1MaDHAHw1+CuEzwhH\n9PxoTLKchLCkMLhtd4PpRlPMPTsXp/85jbzivFfax5gxwP373L22lpbA3r00WKqpO3uWa8EWFnID\n5SjJklelEi3aoKAgud9jYGCAhIQE/u/ExEQYGhrCwMAAiYmJ1R4vf8+TJ0+gr6+P0tJSZGdnQ1dX\nFwYGBggJCeHfk5CQgMGDB9e43xUrVvD/7+7uDnd3d7ljJ8rVsVVHTLaejMnWk8EYw92ndxEQE4Cf\nwn7ClGNT0Ee/D38Lka2ebb3nZW7blmvVvPUWN7PU7t3cAt7Gxko+INKonj/nBjhdu8bdsjNkiNAR\nqZ+QkJBKdaroCd1JXF/u7u7s5s2b/N/lg6GKiorY48ePWc+ePfnBUM7Ozuz69etMJpNVGww1a9Ys\nxhhj+/fvrzQYysjIiGVmZrKMjAz+/6tSo9NFapFblMtO/3OazT0zl5luMGV66/TY1GNT2d67e9mz\nvGf13k5xMWPffsuYri5jCxcyFhWlxKBJo5DJGNu3jzE9PcY+/pixvDyhI2o6xF53qvzRHzt2jBka\nGrIWLVowPT095uvryz/39ddfM2NjY9arVy/m7+/PP37z5k1mZWXFjI2N2bx58/jHCwsL2fjx45mJ\niQlzcXGpNCJ6+/btzMTEhJmYmLCdO3fWGIvYC0tTFJMRw7bc2MJe3/86a7emHXP8xZF9dv4zdinu\nEisuLX7p++PiGFu0iBud7O3N2PHjjJWUNELgRKESEhgbMYIxKyvG/r1hgSiQ2OvORlmPtqkQ++wm\nTV1JWQn+TPwTAdEB8I/xR0xGDDyMPPjRzEbtjWp9b2EhcPgwt1pLQgI3W9B779FyfKpOJuMWlvj8\nc250+dKlNH2iMoi97qREKwexFxaxefbiGYJigvgpIts1b8f37bpL3dG6Wc0rUt25w014cfAgMHgw\nMHs2918VW6JX9B494n4QFRdziwDQzE7KI/a6kxKtHMReWMRMxmS4k3qHT7rhSeFwMXThW7s2ejbV\nBlXl5HCjk7duBYqKuCXTpk+nW4OEVloKfP89NwvY559zCwHQzE7KJfa6kxKtHMReWMh/cotycTHu\nIgKiucT7ouQFn3S9jL3QsVVH/rWMcSNYt24Fzpzh5sadPRvo04dauY0tMpJbFrFjR24CEqPaewOI\nAom97qREKwexFxZSu5iMGATEBMA/2h+X4i/BTNcMPsY+8DXxhYuBC7Q1tQEAaWnAjh3cpeX27bmE\nO3kyN88yUZ7CQmDVKu4S8bp1wLRp9COnMYm97qREKwexFxZSP8VlxbiWcI0fVBWbGYvBRoP5/l2p\njhQyGbd4wdat3PR+U6ZwSdfCQujom54rV7iBaTY23H3QNH1i4xN73UmJVg5iLyzk1aTmpfKDqgJj\nAtGhZQc+6Q7qMQjpqa/ht9+41paZGZdwx46l0a8NlZPDTZ944gSwaRM3uxcRhtjrTkq0chB7YSEN\nJ2MyRKZE8oOqbqXcgouBC3xNfDG4hw+iQ63w888SPHjA9SW+/z7Qo4fQUaufM2e4Hyze3tyl4vbt\nhY5I3MRed1KilYPYCwtRvJyiHFyMvQj/aH8ExASgqKwI3sbesGrpg0fnvHBkty5cXbmk4etLo2Pr\nkp0NXL8O7NoFhIVx98d6egodFQGo7qREKwexFxaiXIwxRGdE863dS3GX0Eu3N7q+8MGjcz4ojHbB\nzBlaePddoHNnoaMVFmNATAw3mvvaNSA0FIiLA5ycuLmJP/oIaNVK6ChJObHXnZRo5SD2wkIaV1Fp\nEUITQvlbiB6nx0M3xxNPr/lgiNQHi2Z0x4AB4hg9W1gIRET8l1ivXeP6sN3cgH79uH+2toC2ttCR\nkpqIve6kRCsHsRcWIqyU3BQExgTi9N8B8I8KQklWRzT7yxTtS5Ng0LUV2jZrhflT5mO413ChQ22w\n1NTKSfXOHaB378qJ9d+lpYkaEHvdSYlWDmIvLER1yJgM6w9uwPJfv0TuoAz+cS1/XTh0XAVfx9kw\nkkrQowcglXIL2Ktqa6+sjFvrt/wS8LVrQGYm4OrKJVQ3N25yD7rXWH2Jve6kRCsHsRcWolp83vZB\noDSw2uNaF1qgpYcu9HJ9oP3EBzm3hyDtSQd06cIl3fJ/5UlYKuVah42ViHNyuMFK5a3VsDDu3tby\nlmq/foC5OaCh0TjxEOUTe91JiVYOYi8sRLW4+7njktGlao8PfDwQv37/Kz+o6kr8FVh0skTfjj4w\nlfigRUYfJMRrIS4O/L+UFEBPr+5E/Cr39TLGbb+8pXrtGhAdDTg4/HcZ2NWVmxKRNF1irzsp0cpB\n7IWFqJbaWrQ+8T7w3+7P/11YWoirT67yg6oScxLh2dMTvsa+8DHxgWFbQ5SUAElJQHw8KiXg8n/J\nydxI57oScfPm3OIJkZGVLwNLJFxSLU+sdnY0GYfYiL3upEQrB7EXFqJazgSdwYLNCxBjH8M/ZnzL\nGOvnrq9zQFRybjICYwLhH+2P84/PQ6+1Hr8gwsAeA9FSu2W195SWcom4YvKtmJSTkrhWaXY2N7tV\n+SVgNzege3dxjIwmtRN73UmJVg5iLyxE9ZwJOoON+zeiUFaIFhotMG/yPLlGHZfJyhCREsG3du88\nvQO3bm78FJG9O/autvxfTUpLuVZv+/ZAmzYNOSLSFIm97qREKwexFxbS9GUVZiH4cTDfv8sY45Ou\np5En2rekuQyJ/MRed1KilYPYCwsRF8YYHj5/yCfdq0+uwrqzNb/8n5O+EzQ1aE5I8nJirzsp0cpB\n7IWFiFthaSGuxF/h191NyUvBkJ5D+P5dg7YGQodIVJTY605KtHIQe2EhpKLEnEQExgQiICYAQTFB\nMGhrwCfdAT0GoIVWC6FDJCpC7HUnJVo5iL2wEFKbMlkZwpPD+UFV957dw4DuA/j+3V66veo1qIo0\nTWKvOynRykHshYWQ+sosyERwbDC//J+GRIPv2/U08kS7Fu2EDpE0IrHXnZRo5SD2wkLIq2CM4e/n\nf/Ot3dCEUNjq2fKtXceujjSoqokTe91JiVYOYi8shChCQUkBLsdf5kczP817Ci9jL/gY+8Db2Bv6\nbfSFDpEomNjrTkq0chB7YSFEGRKyE7iZqmL8Efw4GIZtDeFr4gsfYx/0794fzbWaCx0iaSCx152U\naOUg9sJCiLKVykoRnhTOt3bvP7uPAT0G8P27ph1MaVCVGhJ73UmJVg5iLyyENLaMggycf3ye79/V\n1tTmbyHy7OmJts3bCh0iqQex152UaOUg9sJCiJAYY3iQ9oBv7V5LuAb7Lvb8oCqHrg7QkNAitqpI\n7HUnJVo5iL2wEKJK8kvyuUFV/7Z20/LT4NXTC74mvvA29kaX1l2EDpH8S+x1JyVaOYi9sBCiyp5k\nP+GX/wuODUaPdj341q5bNzcaVCUgsdedlGjlIPbCQoi6KJWV4kbSDQREB8A/xh9/p/2NQdJBfP+u\nSQcTGlTViMRed1KilYPYCwsh6io9P50bVPVv/25zzeb8SObBRoPRpjktoqtMYq87KdHKQeyFhZCm\ngDGG+2n3+ekhrydeh0NXB761a9/VngZVKZjY605KtHIQe2EhpCnKL8nHpbhLfOLNKMiAt7E3P1OV\nXms9oUNUe2KvOynRykHshYUQMYjPiucvMQc/DoZReyO+tevW3Q3NNJsJHaLaEXvdqfLXRw4fPgxL\nS0toamri1q1b/ONxcXFo2bIl7O3tYW9vjzlz5vDPRUREwNraGqampliwYAH/eFFRESZOnAhTU1P0\n7dsX8fHx/HO7du2CmZkZzMzMsHv37sY5OEKIyumh0wPvO76PoxOOIm1xGjYO3QhtDW18cv4TdFrX\nCaP2j8LmG5sRnREtdKhETah8i/bhw4fQ0NDAzJkz8f3338PBwQEAl2hHjhyJe/fuVXuPs7MzNm3a\nBGdnZwwbNgzz58+Hr68vtmzZgr/++gtbtmzBwYMHcfz4cRw4cAAZGRno06cPIiIiAACOjo6IiIiA\njo5Ope2K/VcZIWL3PP85gmKC+BZvK+1W8DX2hY+JDzykHjSoqhZirztVvkVrbm4OMzOzer8+JSUF\nubm5cHZ2BgBMmzYNJ06cAACcOnUK06dPBwCMGzcOwcHBAICAgAB4e3tDR0cHOjo68PLygr+/v4KP\nhBCi7jq26ojJ1pOxc/ROJH+cjOMTj0OqI8X6sPXQ/0Ef7jvdsebKGtxKuQUZkwkdLlERWkIH0BCx\nsbGwt7dHu3bt8NVXX6F///5ISkqCoaEh/xoDAwMkJSUBAJKSktCtWzcAgJaWFtq1a4f09HQkJydX\neo+hoSH/HkIIqYlEIoGNng1s9Gyw2G0xXhS/QEhcCAJiAjDpyCRkF2VXGlTV+bXOQodMBKISidbL\nywupqanVHl+9ejVGjhxZ43v09fWRkJCA9u3b49atWxg9ejTu37+v7FCxYsUK/v/d3d3h7u6u9H0S\nQlTfa81ew3Cz4RhuNhwAEJsZi4CYABz9+yjmnp0L4w7G/KAq126uTXpQVUhICEJCQoQOQ2WoRKIN\nCgqS+z3NmjVDs2ZcQXVwcICxsTGioqJgYGCAxMRE/nWJiYl8a9XAwABPnjyBvr4+SktLkZ2dDV1d\nXRgYGFQqFAkJCRg8eHCN+62YaAkhpDZG7Y0wy2kWZjnNQklZCa4nXod/tD8WBi5EVEYU3KXufP9u\nz/Y9hQ5Xoao2QlauXClcMCpA5ftoK6rYmf78+XOUlZUBAB4/foyoqCj07NkTXbt2Rdu2bREWFgbG\nGPbs2YPXX38dADBq1Cjs2rULAHDkyBF4enoCALy9vREYGIisrCxkZmYiKCgIPj4+jXx0hJCmSltT\nGwN6DMDXnl/j5vs3ET0vGhMtJ+LPxD/Rb1s/mG40xdyzc3H6n9PIK84TOlyiYCo/6vj48eOYP38+\nnj9/jnbt2sHe3h7nzp3D0aNHsXz5cmhra0NDQwOrVq3C8OHcJZuIiAj4+fmhoKAAw4YNw4YNGwBw\nt/dMnToVkZGR0NXVxYEDByCVSgEAO3bswOrVqwEAy5Yt4wdNVST2kXOEEMWTMRnuPr3Lr0IUnhyO\nPvp9+AURbPVs1X5eZrHXnSqfaFWJ2AsLIUT58orzcDH2In8LUV5xHj+oyqunFzq91knoEOUm9rqT\nElrF0jEAAAy2SURBVK0cxF5YCCGNLyYjhk+6IXEhMO1gCl8TX/gY+6CvYV9oa2oLHeJLib3upEQr\nB7EXFkKIsIrLivFnwp8IiAmAf7Q/Hmc+hoeRBz+a2ai9kdAh1kjsdSclWjmIvbAQQlTL07ynCHrM\nzVQVGBMInRY6fNJ1l7rjtWavCR0iAKo7KdHKQeyFhRCiumRMhtupt/lBVREpEXAxcOEHVVl3thZs\nUJXY605KtHIQe2EhhKiP3KJcXIi9wPfvFpQUwMfEhx9UpdtKt9FiEXvdSYlWDmIvLIQQ9RWdEc23\ndkPiQmDe0Rw+xj7wNfGFi6ELtDSUN3+R2OtOSrRyEHthIYQ0DcVlxQh9Esq3duOy4jDYaDDfv9tD\np4dC9yf2upMSrRzEXlgIIU1Tal4qgmKC4B/jj6CYIOi20uWT7iDpILTSbtWg7Yu97qREKwexFxZC\nSNMnYzJEpkTyrd1bKbfgaujKD6qy7GQp96AqsdedlGjlIPbCQggRn5yiHG5Q1b/9u8Vlxf/NVGXs\nhQ4tO7x0G2KvOynRykHshYUQIm6MMW5Q1b8TZlyOvwyLThZ8a9fZwLnGQVVirzsp0cpB7IWFEEIq\nKiotQmhCKN/afZL9BJ49Pfn+3W7tugGgupMSrRzEXlgIIaQuKbkpCIwJREBMAIIeB6FTq07wNfHF\nj74/irrupEQrB0q0hBBSPzImw62UWwiIDsCyQctEXXdSopUDJVpCCJGf2OtODaEDIIQQQpoySrSE\nEEKIElGiJYQQQpSIEi0hhBCiRJRoCSGEECWiREsIIYQoESVaQgghRIko0RJCCCFKRImWEEIIUSJK\ntIQQQogSUaIlhBBClIgSLSGEEKJElGgJIYQQJaJESwghhCgRJVpCCCFEiSjREkIIIUpEiZYQQghR\nIkq0hBBCiBJRoiWEEEKUiBItIYQQokQqn2gXL16M3r17w9bWFmPHjkV2djb/3Jo1a2Bqagpzc3ME\nBgbyj0dERMDa2hqmpqZYsGAB/3hRUREmTpwIU1NT9O3bF/Hx8fxzu3btgpmZGczMzLB79+7GOThC\nCCFNnsonWm9vb9y/fx937tyBmZkZ1qxZAwB48OABDh48iAcPHsDf3x9z5swBYwwAMHv2bGzbtg1R\nUVGIioqCv78/AGDbtm3Q1dVFVFQUPvroI3zyyScAgIyMDKxatQo3btzAjRs3sHLlSmRlZQlzwGoi\nJCRE6BBUBp2L/9C5+A+dC1JO5ROtl5cXNDS4MF1cXJCYmAgAOHnyJCZPngxtbW1IpVKYmJggLCwM\nKSkpyM3NhbOzMwBg2rRpOHHiBADg1KlTmD59OgBg3LhxCA4OBgAEBATA29sbOjo60NHRgZeXF5+c\nSc2oEvkPnYv/0Ln4D50LUk7lE21F27dvx7BhwwAAycnJMDQ05J8zNDREUlJStccNDAyQlJQEAEhK\nSkK3bt0AAFpaWmjXrh3S09Nr3RYhhBDSUFpCBwBwrdbU1NRqj69evRojR44EAHz99ddo1qwZpkyZ\n0tjhEUIIIa9MJRJtUFBQnc/v3LkTZ8+e5S/1AlxLNSEhgf87MTERhoaGMDAw4C8vV3y8/D1PnjyB\nvr4+SktLkZ2dDV1dXRgYGFS6zJOQkIDBgwdXi8PY2BgSieRVD7PJWblypdAhqAw6F/+hc/EfOhcc\nY2NjoUMQFlNx586dYxYWFiwtLa3S4/fv32e2trasqKiIPX78mPXs2ZPJZDLGGGPOzs7s+vXrTCaT\nsaFDh7Jz584xxhjbvHkzmzVrFmOMsf3797OJEycyxhhLT09nRkZGLDMzk2VkZPD/TwghhDSUSrRo\n6zJv3jwUFxfDy8sLAODq6ootW7bAwsICEyZMgIWFBbS0tLBlyxa+tbllyxb4+fmhoKAAw4YNg6+v\nLwDg3XffxdSpU2FqagpdXV0cOHAAANChQwd8/vnn6NOnDwBg+fLl0NHREeBoCSGENDUSxv69J4YQ\nQgghCqdWo44bi7+/P8zNzWFqaopvvvmmxtfMnz8fpqamsLW1RWRkZCNH2Hhedi727t0LW1tb2NjY\nwM3NDXfv3hUgysZRn3IBAOHh4dDS0sKxY8caMbrGVZ9zERISAnt7e1hZWcHd3b1xA2xELzsXz58/\nh6+vL+zs7GBlZYWdO3c2fpCN4J133oGenh6sra1rfY1Y6s1qhL52rWpKS0uZsbExi42NZcXFxczW\n1pY9ePCg0mvOnDnDhg4dyhhj7Pr168zFxUWIUJWuPufi2rVrLCsrizHG9aeL+VyUv87Dw4MNHz6c\nHTlyRIBIla8+5yIzM5NZWFiwhIQExhirNsaiqajPuVi+fDlbunQpY4w7Dx06dGAlJSVChKtUly9f\nZrdu3WJWVlY1Pi+WerMm1KKt4saNGzAxMYFUKoW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+ "text": [ + "<matplotlib.figure.Figure at 0x7f36783f0250>" + ] + } + ], + "prompt_number": 5 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_&_Chemical_Thermodynamics/ch7.ipynb b/Engineering_&_Chemical_Thermodynamics/ch7.ipynb new file mode 100755 index 00000000..16e8e4f7 --- /dev/null +++ b/Engineering_&_Chemical_Thermodynamics/ch7.ipynb @@ -0,0 +1,197 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 7 : Phase Equilibia 2 Fugacity" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.1 Page No : 397" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# Variables\n", + "h_cap_H2O = 2676.0 ; \t\t\t#[kJ/kg],From steam table\n", + "S_cap_H2O = 7.3548 ; \t\t\t#[kJ/kgK],From steam table\n", + "h_cap_0_H2O = 2687.5 ; \t\t\t#[kJ/kg],From Appendix B\n", + "S_cap_0_H2O = 8.4479 ; \t\t\t#[kJ/kgK],From Appendix B\n", + "P_0_H2O = 10. ;\t\t\t#[kPa]\n", + "T = 373.15 ;\t\t\t#[K]\n", + "R = 8.314 / 18 ;\n", + "P_sys = 101.35 ;\t\t\t#[kPa]\n", + "\n", + "# Calculations\n", + "g_cap_H2O = h_cap_H2O - T * S_cap_H2O ;\n", + "g_cap_0_H2O = h_cap_0_H2O - T * S_cap_0_H2O ; \n", + "\n", + "f_H2O = P_0_H2O * math.exp((g_cap_H2O - g_cap_0_H2O ) / (R * T)) ;\n", + "\n", + "Sai_H2O = f_H2O / P_sys ;\n", + "\n", + "\n", + "# Results\n", + "print \"The fugacity = %.2f kPa The fugacity coefficient = %.3f\"%(f_H2O,Sai_H2O);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The fugacity = 99.73 kPa The fugacity coefficient = 0.984\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.3 Page No : 402" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "P = 50. \t\t\t#[bar]\n", + "T = 25. + 273.2\t\t\t#[K]\n", + "P_c = 48.7 ; \t\t\t#[bar] , From Appendix A.1 Table C.7 & C.8\n", + "T_c = 303.5 ; \t\t\t#[K] , From Appendix A.1 Table C.7 & C.8\n", + "w = 0.099 ; \t\t\t# From Appendix A.1 Table C.7 & C.8\n", + "log_w_0 = -0.216 ;\t\t\t# By interpolation\n", + "log_w_1 = -0.060 ;\t\t\t# By interpolation\n", + "\n", + "# Calculations\n", + "X = log_w_0 + w * log_w_1 ;\n", + "sai_eth = 10**(X) ;\n", + "f_eth = sai_eth * P ;\n", + "\n", + "# Results\n", + "print \"Fugacity = %.f bar\"%(f_eth);\n", + " \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Fugacity = 30 bar\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.10 Page No : 435" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "import math\n", + "gama_a_inf = 0.88 ;\n", + "gama_b_inf = 0.86 ;\n", + "R = 8.314 ;\n", + "T = 39.33 + 273 ;\n", + "\n", + "# Calculations\n", + "A_1 = R * T * math.log(gama_a_inf) ;\n", + "A_2 = R * T * math.log(gama_b_inf) ;\n", + "A = (A_1 + A_2) / 2 ;\n", + "\n", + "# Results\n", + "print \"The average value of two-suffix Margules parameter A = %.f J/mol\"%(A);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The average value of two-suffix Margules parameter A = -362 J/mol\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.13 Page No : 448" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "A_T1 = 1401. ; \t\t\t#[J/mol]\n", + "T1 = 10 + 273. ;\t\t\t#[K]\n", + "T2 = 60 + 273. ; \t\t\t#[K]\n", + "C = 3250. ;\n", + "A_T2_prev = 1143. ;\t\t\t#[J/mol]\n", + "\n", + "# Calculations and Results\n", + "A_T2 = T2 * (C *(1./T2 - 1./T1) + A_T1 / T1);\n", + "print \"Value of A at 60*C = %.f J/mol\"%(A_T2) ;\n", + "\n", + "x = (A_T2_prev - A_T2) / A_T2_prev* 100 ;\n", + "print \"The values differ by = %.f %%\"%(x)\n", + "\n", + "# Note: Answer may vary because of rounding off error. " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Value of A at 60*C = 1074 J/mol\n", + "The values differ by = 6 %\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_&_Chemical_Thermodynamics/ch8.ipynb b/Engineering_&_Chemical_Thermodynamics/ch8.ipynb new file mode 100755 index 00000000..c1f9c289 --- /dev/null +++ b/Engineering_&_Chemical_Thermodynamics/ch8.ipynb @@ -0,0 +1,942 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 8 : Phase Equilibria III Phase Diagrams" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.1 Page No : 470" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from scipy.optimize import fsolve \n", + "import math \n", + "\n", + "# Variables\n", + "A_C5H12 = 9.2131 ; \t\t\t#From table E8.2A\n", + "B_C5H12 = 2477.07 ; \t\t\t#From table E8.2A\n", + "C_C5H12 = -39.94 ; \t\t\t#From table E8.2A\n", + "A_C6H12 = 9.1325 ; \t\t\t#From table E8.2A\n", + "B_C6H12 = 2766.63 ; \t\t\t#From table E8.2A\n", + "C_C6H12 = -50.50 ; \t\t\t#From table E8.2A\n", + "A_C6H14 = 9.2164 ; \t\t\t#From table E8.2A\n", + "B_C6H14 = 2697.55 ; \t\t\t#From table E8.2A\n", + "C_C6H14 = -48.78 ; \t\t\t#From table E8.2A\n", + "A_C7H16 = 9.2535 ; \t\t\t#From table E8.2A\n", + "B_C7H16 = 2911.32 ; \t\t\t#From table E8.2A\n", + "C_C7H16 = -56.51 ; \t\t\t#From table E8.2A\n", + "x_C5H12 = 0.3 ;\n", + "x_C6H12 = 0.3 ;\n", + "x_C6H14 = 0.2 ;\n", + "x_C7H16 = 0.2 ;\n", + "\n", + "# Calculations\n", + "def f82(T):\n", + " return -1 + (x_C5H12 * math.exp(A_C5H12 - B_C5H12 / (T + C_C5H12)) + x_C6H12 *\\\n", + " math.exp(A_C6H12 - B_C6H12 / (T + C_C6H12)) + x_C6H14 * \\\n", + " math.exp(A_C6H14 - B_C6H14 / (T + C_C6H14)) + x_C5H12 * math.exp(A_C5H12 - B_C5H12 / (T \\\n", + " + C_C5H12)) + x_C7H16 * math.exp(A_C7H16 - B_C7H16 / (T + C_C7H16)));\n", + "\n", + "y =fsolve(f82,300)\n", + "\n", + "# Results\n", + "print \" The temperature at which the liquid develops the first bubble of vapour = %d K\"%(y); \n", + "\n", + "# Note : answer may vary because of fsolve function of python." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The temperature at which the liquid develops the first bubble of vapour = 317 K\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.2 Page No : 471" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from scipy.optimize import fsolve \n", + "import math \n", + "\n", + "# Variables\n", + "A_C5H12 = 9.2131 ; \t\t\t#From table E8.2A\n", + "B_C5H12 = 2477.07 ; \t\t\t#From table E8.2A\n", + "C_C5H12 = -39.94 ; \t\t\t#From table E8.2A\n", + "A_C6H12 = 9.1325 ; \t\t\t#From table E8.2A\n", + "B_C6H12 = 2766.63 ; \t\t\t#From table E8.2A\n", + "C_C6H12 = -50.50 ; \t\t\t#From table E8.2A\n", + "A_C6H14 = 9.2164 ; \t\t\t#From table E8.2A\n", + "B_C6H14 = 2697.55 ; \t\t\t#From table E8.2A\n", + "C_C6H14 = -48.78 ; \t\t\t#From table E8.2A\n", + "A_C7H16 = 9.2535 ; \t\t\t#From table E8.2A\n", + "B_C7H16 = 2911.32 ; \t\t\t#From table E8.2A\n", + "C_C7H16 = -56.51 ; \t\t\t#From table E8.2A\n", + "\n", + "y_C5H12 = 0.3 ;\n", + "y_C6H12 = 0.3 ;\n", + "y_C6H14 = 0.2 ;\n", + "y_C7H16 = 0.2 ;\n", + "P = 1 ; \t\t\t#[bar]\n", + "\n", + "# Calculations and Results\n", + "def f83(T):\n", + " return -1 + P * ( y_C5H12 / math.exp(A_C5H12 - B_C5H12 / (T + \n", + " C_C5H12)) + y_C6H12 / math.exp(A_C6H12 - B_C6H12 / (T + C_C6H12)) + \\\n", + " y_C6H14 / math.exp(A_C6H14 - B_C6H14 / (T + C_C6H14)) + y_C7H16 / math.exp(A_C7H16 - B_C7H16 / (T + C_C7H16)));\n", + "\n", + "y =fsolve(f83,300)\n", + "\n", + "print \"The temperature at which vapour develops the first drop of liquid = %.f K\"%(y) ;\n", + "\n", + "T = y ;\n", + "P_sat_C5H12 = math.exp(A_C5H12 - B_C5H12 / (T + C_C5H12)) ;\n", + "p_sat_C6H12 = math.exp(A_C6H12 - B_C6H12 / (T + C_C6H12)) ;\n", + "P_sat_C6H14 = math.exp(A_C6H14 - B_C6H14 / (T + C_C6H14)) ;\n", + "P_sat_C7H16 = math.exp(A_C7H16 - B_C7H16 / (T + C_C7H16)) ;\n", + "\n", + "x_C5H12 = y_C5H12 * P / P_sat_C5H12 ;\n", + "x_C6H12 = y_C6H12 * P / p_sat_C6H12 ;\n", + "x_C6H14 = y_C6H14 * P / P_sat_C6H14 ;\n", + "x_C7H16 = y_C7H16 * P / P_sat_C7H16 ;\n", + "\n", + "print \"x_C5H12 = %.3f x_C6H12 = %.3f x_C6H14 = %.3f x_C7H16 = %.3f\"%(x_C5H12,x_C6H12,x_C6H14,x_C7H16) ;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The temperature at which vapour develops the first drop of liquid = 349 K\n", + "x_C5H12 = 0.091 x_C6H12 = 0.345 x_C6H14 = 0.159 x_C7H16 = 0.405\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.3 Page No : 471" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from scipy.integrate import *\n", + "from numpy import *\n", + "import math \n", + "\n", + "# Variables\n", + "P_a_sat = 0.53 ; \t\t\t#[bar]\n", + "P_b_sat = 0.16 ; \t\t\t#[bar]\n", + "X = 1./3 ;\n", + "Y = 1- X ;\n", + "x_a_feed = 0.5 ;\n", + "x_b_feed = 0.5 ;\n", + "\n", + "# Calculations\n", + "a = Y * -(x_a_feed + x_b_feed) + Y**2 ;\n", + "b = X * Y *(P_a_sat + P_b_sat) - (x_a_feed * P_b_sat + x_b_feed * P_a_sat)*X ;\n", + "c = P_a_sat * P_b_sat * X**2;\n", + "\n", + "k=poly1d([a,b,c])#'k',0);\n", + "#P = c + b*k**1 + a*k**2 ;\n", + "M = roots(k);\n", + "\n", + "# Results\n", + "for i in range(2):\n", + " ans = sign(M[0]) ;\n", + " if ans == 1:\n", + " print \" Pressure = %.2f bar\"%(M[i]) ; \n", + " Xa = x_a_feed / (P_a_sat / M[0] * X + Y) ;\t\t\t#....E8.4D\n", + " Ya = Xa * P_a_sat / M[0] \t\t\t#.....E8.4B\n", + " print \" Xa = %.2f Ya = %.1f\"%(Xa,Ya);\n", + " break\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Pressure = 0.31 bar\n", + " Xa = 0.40 Ya = 0.7\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.5 Page No : 476" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from scipy.optimize import fsolve \n", + "import math \n", + "\n", + "# Variables\n", + "P1_sat = 0.72 ; \t\t\t#[bar]\n", + "P2_sat = 0.31 ; \t\t\t#[bar]\n", + "A = 3590. ;\n", + "B = -1180. ;\n", + "R = 8.314 ;\n", + "T = 70. + 273 ;\t\t\t#[K]\n", + "\n", + "# Calculations\n", + "def f85(x1):\n", + " return -.48 + ( x1 * math.exp((A + 3*B) * (1 - x1)**2 / (R * T) - \n", + " 4 * B * (1 - x1)**3 / (R * T)) * P1_sat) / ( x1 * math.exp((A + 3*B) * \n", + " (1 - x1)**2 / (R * T) -4 * B * (1 - x1)**3 / (R * T)) * P1_sat +(1 - x1) * \n", + " math.exp((A - 3*B) * x1**2 / (R * T) -4 * B * x1**3 / (R * T)) * P2_sat ) ; \n", + "\n", + "y = fsolve(f85,0.1)[0]\n", + "x1 = y ;\n", + "P = ( x1 * math.exp((A + 3*B) * (1 - x1)**2 / (R * T) - 4 * B * \n", + "(1 - x1)**3 / (R * T)) * P1_sat) + (1 - x1) * math.exp((A - 3*B) * x1**2 / \n", + "(R * T) -4 * B * x1**3 / (R * T)) * P2_sat ;\n", + "\n", + "# Results\n", + "print \"The value of x1 = %.3f\"%(y) ;\n", + "print \"Pressure = %.2f bar\"%(P) ;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of x1 = 0.114\n", + "Pressure = 0.55 bar\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.7 Page No : 488" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "import math\n", + "P = 0.223 ; \t\t\t#[bar]\n", + "P_a_sat = 0.156 ; \t\t\t# [bar]\n", + "P_b_sat = 0.124 ; \t\t\t#[bar]\n", + "R = 8.314 ;\n", + "T = 50 + 273 ;\n", + "Xa = 0.554 ;\n", + "Xb = 1 - Xa ;\n", + "\n", + "# Calculations\n", + "gama_a = P / P_a_sat ;\n", + "A1 = R * T * math.log(gama_a) / (Xb**2) * 10**-3 ;\n", + "gama_b = P / P_b_sat ;\n", + "A2 = R * T * math.log(gama_b) / (Xa**2) * 10**-3 ; \n", + "\n", + "A = math.ceil((A1 + A2) / 2) ;\n", + "\n", + "# Results\n", + "print \"Value of two suffix Marguels parameter = %.1f kJ/mol\"%(A);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Value of two suffix Marguels parameter = 5.0 kJ/mol\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.9 Page No : 491" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "import math\n", + "R = 8.314 ;\n", + "T = 10 + 273 ; \t\t\t#[K]\n", + "A_B = 9.2806 ; \t\t\t# From Appendix A , Table A1.1\n", + "B_B = 2788.5 ; \t\t\t# From Appendix A , Table A1.1\n", + "C_B = -52.36 ; \t\t\t# From Appendix A , Table A1.1\n", + "A_C = 9.1325 ; \t\t\t# From Appendix A , Table A1.1\n", + "B_C = 2766.63 ; \t\t\t# From Appendix A , Table A1.1\n", + "C_C = -50.50 ; \t\t\t# From Appendix A , Table A1.1\n", + "\n", + "x1 = [0 ,0.0610 ,0.2149 ,0.3187 ,0.4320 ,0.5246 ,0.6117 ,0.7265 ,0.8040 ,0.8830 ,0.8999 ,1] ; \t\t\t#From table E8.9A\n", + "P_exp = [6344 ,6590 ,6980 ,7140 ,7171 ,7216 ,7140 ,6974 ,6845 ,6617 ,6557 ,6073] ; \t\t\t#From table E8.9A\n", + "\n", + "P_1_sat = 6072.15 ; \t\t\t#[Pa]\n", + "P_2_sat = 6344 ; \t\t\t#[Pa] \n", + "\n", + "A = [1390 ,1391 ,1392 ,1393 ,1394 ,1395 ,1396 ,1397 ,1398 ,1399 ,1400 ,1401 ,1402 ,1403 ,1404 ,1405 ,1406 ,1407 ,1408 ,1409 ,1410 ] ;\n", + "\n", + "C = zeros([21,12])\n", + "for k in range(21):\n", + " y = A[k] ;\n", + " P = zeros(12)\n", + " for i in range(12):\n", + " P[i] = x1[i] * math.exp( y / (R * T ) * (1 - x1[i])**2) * P_1_sat+(1 - x1[i]) * math.exp(y / (R * T ) * x1[i]**2) * P_2_sat ;\n", + " C[k,i] = (P[i] - P_exp[i])**2 ;\n", + "\n", + "R = zeros(21)\n", + "for k in range(21):\n", + " y = 0 ;\n", + " for i in range(12):\n", + " y = y + C[k,i] ; \n", + " R[k] = y ;\n", + "\n", + "\n", + "k = 100000. ;\n", + "for i in range(21):\n", + " K = R[i] ; \n", + " if K < k:\n", + " k = K ;\n", + "\n", + "for i in range(21):\n", + " if R[i] == k :\n", + " print \"The two suffix Margules co-efficient is = %g J/mol\" %(A[i]) ; \n", + "\n", + "#Note : answer may vary because of rounding off error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The two suffix Margules co-efficient is = 1401 J/mol\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.10 Page No : 494" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "import math\n", + "R = 8.314 ;\n", + "T = 10 + 273.15\t\t\t#[K]\n", + "A_B = 9.2806 ; \t\t\t# From Appendix A , Table A1.1\n", + "B_B = 2788.5 ; \t\t\t# From Appendix A , Table A1.1\n", + "C_B = -52.36 ; \t\t\t# From Appendix A , Table A1.1\n", + "A_C = 9.1325 ; \t\t\t# From Appendix A , Table A1.1\n", + "B_C = 2766.63 ; \t\t\t# From Appendix A , Table A1.1\n", + "C_C = -50.50 ; \t\t\t# From Appendix A , Table A1.1\n", + "\n", + "x1 = [0,0.0610 ,0.2149 ,0.3187 ,0.4320 ,0.5246 ,0.6117 ,0.7265 ,0.8040 ,0.8830 ,0.8999 ,1] ; \t\t\t#From table E8.9A\n", + "P_exp = [6344 ,6590 ,6980 ,7140 ,7171 ,7216 ,7140 ,6974 ,6845 ,6617 ,6557 ,6073] ; \t\t\t#From table E8.9A\n", + "\n", + "P_1_sat = 6073. ; \t\t\t#[Pa]\n", + "P_2_sat = 6344. ; \t\t\t#[Pa] \n", + "A = linspace(1390,1410,21) ;\n", + "B = linspace(60,80,21)\n", + "w = 1 / (R * T) ;\n", + "S = zeros([21,21])\n", + "for k in range(21):\n", + " y = A[k] ;\n", + " for i in range(21):\n", + " z = B[i] ;\n", + " P = zeros(12)\n", + " R = zeros(12)\n", + " for j in range(12):\n", + " P[j] = x1[j] * math.exp((y + 3 * z) * (1 - (x1[j]))**2 *w-4*z*(1-x1[j])**3* w )* P_1_sat + (1-x1[j])*exp((y -3*z)*(x1[j])**2 * w + 4 * z * (x1[j]**3) * w )*P_2_sat ;\n", + " R[j] =(P[j] - P_exp[j])**2 ;\n", + " m = 0 ;\n", + " for l in range(12):\n", + " m = m + R[l]\n", + " S[k,i] = m ;\n", + "\n", + "D = zeros(21)\n", + "for i in range(21):\n", + " k = S[i,0]\n", + " for l in range(1,21):\n", + " if S[i,l] < k:\n", + " k = S[i,l]\n", + " D[i] = k ;\n", + "\n", + "a = D[0] ;\n", + "for i in range(2,21):\n", + " if D[i] < a:\n", + " a = D[i] ;\n", + "\n", + "# Results\n", + "for i in range(21):\n", + " if D[i] == a :\n", + " for l in range(21):\n", + " if S[i,l] == a:\n", + " print \" A = %g J/mol\"%(A[i]);\n", + " print \" B = %g J/mol\"%(B[l]) ;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " A = 1397 J/mol\n", + " B = 69 J/mol\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.11 Page No : 494" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%pylab inline\n", + "\n", + "from matplotlib.pyplot import *\n", + "\n", + "# Given\n", + "R = 8.314 ;\n", + "T = 10 + 273.15 ; \t\t\t#[K]\n", + "x1 = [0 ,0.0610 ,0.2149 ,0.3187 ,0.4320 ,0.5246 ,0.6117 ,0.7265 ,0.8040 ,0.8830 ,0.8999 ,1] ; \t\t\t#From table E8.9A\n", + "P_exp = [6344 ,6590 ,6980 ,7140 ,7171 ,7216 ,7140 ,6974 ,6845 ,6617 ,6557 ,6073 ,6073] ; \t\t\t#From table E8.9A\n", + "y1 = [ 1 ,0.0953 ,0.2710 ,0.3600 ,0.4453,0.5106 ,0.5735 ,0.6626 ,0.7312 ,0.8200 ,0.8382, 0 ] ;\t\t\t#From table E8.9A\n", + "P_1_sat = 6073. ; \t\t\t#[Pa]\n", + "P_2_sat = 6344. ; \t\t\t#[Pa]\n", + "\n", + "n = 0 ;\n", + "x2 = zeros(11)\n", + "y2 = zeros(11)\n", + "g_E = zeros(11)\n", + "ydata = zeros(11)\n", + "xdata = zeros(11)\n", + "for i in range(1,11):\n", + " x2[i] = 1 - x1[i] ;\n", + " y2[i] = 1 - y1[i] ;\n", + " g_E[i] = R * T *( x1[i] * math.log (( y1[i] * P_exp[i]) / (x1[i]* P_1_sat)) + x2[i] * math.log((y2[i] * P_exp[i]) / (x2[i] * P_2_sat)) ) ;\n", + " n = n + g_E[i] / ((x1[i] * x2[i]) * 10) ;\n", + " ydata[i-1] = (g_E[i]/(x1[i]*x2[i]));\n", + " xdata[i-1] = x1[i] - x2[i] ;\n", + "\n", + "m= 0\n", + "n=0\n", + "o = 0\n", + "p= 0\n", + "N = 10\n", + "for i in range(1,11):\n", + " m = m + g_E[i] * (2 * x1[i] - 1) / ( x1[i] * x2[i]) ;\n", + " n = n + g_E[i] / ( x1[i] * x2[i]) ;\n", + " o = o + (2 * x1[i] - 1) ;\n", + " p = p + (2 * x1[i] - 1)**2 ;\n", + "\n", + "x_bar = o / N ;\n", + "y_bar = n / N ;\n", + "a1 = (N * m - n * o)/(N * p - o**2) ;\n", + "a0 = y_bar - a1 * x_bar ;\n", + "\n", + "ydata2 = zeros(11)\n", + "for i in range(11):\n", + " ydata2[i] = a0 + a1*xdata[i] ;\n", + "\n", + "# Results\n", + "plot(xdata,ydata,\"+\") ;\n", + "plot(xdata,ydata2) ;\n", + "xlabel(\"x1-x2\")\n", + "ylabel(\"g**E/x1x2 [J/mol]\")\n", + "#suptitle(\"Fit of g E /x 1 x 2 vs. x 1 2 x 2 for benzene (1)\u2013cyclohexane (2) at 10\u00b0C to straight line\")\n", + "ylim(1320,1500)\n", + "\n", + "print \" From average, the value of A = %d J/mol\"%(n/10) ;\n", + "print \" From linear regression best fit line the values of A and B are \\\n", + " %.1f J/mol & %.1f J/mol respectively .\"%(a0, a1) ;\n", + "\t\t\t#Readers can refer figure E8.11 .\n", + "show()\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Populating the interactive namespace from numpy and matplotlib\n", + " From average, the value of A = 1408 J/mol" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " From linear regression best fit line the values of A and B are 1401.3 J/mol & 76.1 J/mol respectively .\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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EuCOLnZL9BhcXFyZOnMjEiRPJz89nz549Zs9MiOos+XQyEfoIkk4n8ab/m6wa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\nhNmkPIQQQphNykMIIYTZpDyEEEKY7f8BfWbvvZfeqaEAAAAASUVORK5CYII=\n", + "text": [ + "<matplotlib.figure.Figure at 0x3713c10>" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.12 Page No : 500" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "H_O2 = 44253.9\t\t\t#[bar] , From table 8.1\n", + "p_O2 = 0.21 \t\t\t#[bar]\n", + "\n", + "# Calculations\n", + "x_O2 = p_O2 / H_O2 ;\n", + "v_H2O = 1/(1/0.001 * 1/0.018 * 0.001 );\n", + "_O2_ = x_O2 / v_H2O ; \t\t\t#[M]\n", + "\n", + "# Results\n", + "print \"Mole fraction of O2 = %.2e\"%(x_O2 ) ;\n", + "print \"Concentration of O2 = %.2e M \"%(_O2_) ;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mole fraction of O2 = 4.75e-06\n", + "Concentration of O2 = 2.64e-04 M \n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.13 Page No : 500" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "import math\n", + "P = 300. ; \t\t\t#[bar]\n", + "V_bar_inf_N2 = 3.3 * 10**-5 ;\n", + "R = 8.314 ;\n", + "T = 298. ; \t\t\t#[K]\n", + "y_N2 = 1. ; \t\t\t# At 25*C vapour pressure of water is small\n", + "H_N2_1 = 87365. ; \t\t\t#[bar]\n", + "P_c = 33.8 ; \t\t\t#[bar]\n", + "T_c = 126.2 ;\t\t\t# [K]\n", + "w = 0.039 ; \t\t\t# From Appendix A.1 \n", + "log_w_0 = 0.013 ;\n", + "log_w_1 = 0.210 ;\n", + "\n", + "# Calculations\n", + "H_N2_300 = H_N2_1 * math.exp((V_bar_inf_N2 * (P -1) * 10**5 )/ (R * T)) ;\n", + "k = log_w_0 + w * log_w_1 ;\n", + "sai_N2 = 10**k ;\n", + "x_N2 = y_N2 * sai_N2 * P / H_N2_300 ;\n", + "\n", + "# Results\n", + "print \"Solubility of N2 in water = %.5f\"%(x_N2) ;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Solubility of N2 in water = 0.00242\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.20 Page No : 518" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%pylab inline\n", + "\n", + "import math \n", + "from matplotlib.pyplot import *\n", + "from numpy import *\n", + "\n", + "\n", + "# Variables\n", + "R = 8.314 ;\n", + "T = 20. + 273 ;\t\t\t#[K]\n", + "A = 6000. ; \t\t\t#[J/mol]\n", + "B = -384. ; \t\t\t#[J/mol]\n", + "x_a = [0.001 ,0.03 ,0.05 ,0.06 ,0.075 ,0.1 ,0.12 , 0.13 ,0.15 ,0.2 ,0.25 ,0.3 ,0.35 ,0.4 ,0.45,0.475 ,0.5 ,0.55 ,0.6 ,0.65 ,0.7 ,0.75 ,0.8 ,0.8475 ,0.85 ,0.9 ,0.925 ,0.95 ,0.975 ,0.999] ;\n", + "\n", + "y_data = zeros(30)\n", + "y_data2 = zeros(30)\n", + "\n", + "# Calculations\n", + "for i in range(30):\n", + " y_data[i] = R * T * ( x_a[i] * math.log(x_a[i]) + (1 - x_a[i]) * math.log(1- x_a[i])) + x_a[i] * (1 - x_a[i]) * (A + B * (2*x_a[i] - 1 )) ;\n", + " y_data2[i] = - 82 * x_a[i]- 185.6 ;\n", + "\n", + "m = min(y_data) ;\n", + "for i in range(30):\n", + " if y_data[i] == m:\n", + " a = x_a[i] ;\n", + "\n", + "for i in range(30):\n", + " y_data2[i] = -(R * T *( math.log(a) - math.log(1 - a)) + A * (1 - 2*a) + B * (6 * a - 1 - 6 * a**2)) * (x_a[i] - a) + m ;\n", + "\n", + "y_data3 = zeros(20)\n", + "for i in range(20):\n", + " y_data3[i] = y_data[i] - y_data2[i] ;\n", + "n = min(y_data3) ;\n", + "\n", + "for i in range(20):\n", + " if y_data3[i] == n:\n", + " b = x_a[i] ;\n", + "\n", + "\n", + "# Results\n", + "plot(x_a ,y_data) ;\n", + "plot(x_a ,y_data2) ;\n", + "xlabel(\"Xa\")\n", + "ylabel(\"g-XaGa-XbGb\")\n", + "suptitle(\"the binary system described \")\n", + "print \"The equilibrium composition can be found by drawing a line tangent to\\\n", + " the minima . In this case the answer is %.2f and %.1f .\" %(a ,b)\n", + "show()\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Populating the interactive namespace from numpy and matplotlib\n", + "The equilibrium composition can be found by drawing a line tangent to the minima . In this case the answer is 0.85 and 0.1 ." + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n" + ] + }, + { + "output_type": "stream", + "stream": "stderr", + "text": [ + "WARNING: pylab import has clobbered these variables: ['draw_if_interactive']\n", + "`%pylab --no-import-all` prevents importing * from pylab and numpy\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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4UPj4+Ag/Pz/xyy+/GB4/evSoCAwMFL6+vmL69OmlHruMj1Ru+flCeHoKceKE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\nAABJRU5ErkJggg==\n", + "text": [ + "<matplotlib.figure.Figure at 0x35de510>" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.21 Page No : 519" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from scipy.optimize import fsolve \n", + "import math \n", + "\n", + "\n", + "# Variables\n", + "A = 6349 ; \t\t\t#[J/mol]\n", + "B = -384. ; \t\t\t#[J/mol]\n", + "R = 8.314 ;\n", + "T = 20 + 273 ; \t\t\t#[K]\n", + "k = 0.000001 ;\n", + "\n", + "# Calculations\n", + "def f816(x_a):\n", + " return R * T * (1/x_a + 1/(1 - x_a)) - 2 * A +6 * B * (1 - 2 * x_a) + k\n", + "\n", + "ans1 = fsolve(f816,0.1)[0]\n", + "ans2 = fsolve(f816,0.5)[0]\n", + "\n", + "print \"%.3f < x_a < %.3f \"%(ans1,ans2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "0.225 < x_a < 0.706 \n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.22 Page No : 522" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from scipy.optimize import fsolve\n", + "\n", + "# Variables\n", + "import math\n", + "T = 300. \t\t\t#[K]\n", + "A = 6235. \t \t\t#[J/mol] \n", + "P_a_sat = 100. * 10**3 ; \t\t\t#[Pa]\n", + "P_b_sat = 50. * 10**3 ; \t\t\t#{Pa}\n", + "R = 8.314 ;\n", + "w = 1./(R * T) \n", + "\n", + "def f817(R):\n", + " x_a_a = R[0]\n", + " x_a_b = R[1]\n", + " Z817 = [0,0]\n", + " Z817[0] = x_a_b * math.exp(A * (1 - x_a_b) ** 2 * w) - x_a_a * math.exp(A*(1-x_a_a)**2*w)\n", + " Z817[1] = (1 - x_a_b) * math.exp(A * ( x_a_b) ** 2 * w) - (1 - x_a_a) * math.exp(A * (x_a_a) ** 2 * w )\n", + " return Z817\n", + "x0 = [0.75 , 0.1] ;\n", + "z = fsolve(f817,x0)\n", + "\n", + "# Results\n", + "print \"The compositions are : x_a_a = %.3f and x_a_b = %.3f\"%(z[0],z[1]) ;\n", + "P = z[0] * math.exp(A * z[1] ** 2 * w) * P_a_sat + z[1] * math.exp(A * z[0] ** 2 * w) * P_b_sat ;\n", + "print \"Total pressure = %d kPa\"%(P * 10**-3) ;\n", + "y_a = z[0] * math.exp(A * z[1] ** 2 * w) * P_a_sat / P ;\n", + "print \"y_a = %.3f\" %( y_a ) ;\n", + "\n", + "# Note : incorrect answer in textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The compositions are : x_a_a = 0.272 and x_a_b = 0.272\n", + "Total pressure = 49 kPa\n", + "y_a = 0.667\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.23 Page No : 535" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "T_b = 373.15 ; \t \t\t#[K]\n", + "del_h_vap = 2257. ; \t\t#[J/g]\n", + "MW_salt = 58.5 ; \t\t\t#[g/mol]\n", + "MW_water = 18. ; \t\t\t#[g/mol]\n", + "w_salt = 3.5 ;\n", + "w_water = 100 - w_salt ;\n", + "R = 8.314 ;\n", + "\n", + "# Calculations\n", + "x_salt = (w_salt / MW_salt) / (w_salt / MW_salt + w_water / MW_water) ;\n", + "x_b = 2 * x_salt\n", + "del_T = R * T_b**2 / (del_h_vap * MW_water)* x_b ;\n", + "\n", + "# Results\n", + "print \"The temperature that sea water boils is = %.2f degreeC\"%(100 + del_T);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The temperature that sea water boils is = 100.63 degreeC\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.24 Page No : 538" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "rho_w = 1000. \t \t\t# [kg/m**3]\n", + "g = 9.8 ; \t \t \t# [m/s**2]\n", + "h = 0.0071 ;\t\t\t #[m]\n", + "m_b = 1.93 * 10**-3 ; \t\t\t# [kg]\n", + "V = 520. * 10**-6 ; \t\t\t#[m**3]\n", + "R = 8.314\n", + "T = 298.\n", + "\n", + "# Calculations\n", + "PI = rho_w * g * h ;\n", + "C_b = m_b / V ;\n", + "MW_b = R * T * C_b / PI ;\n", + "\n", + "# Results\n", + "print \"The molecular weight of the protein = %d kg/mol\"%( MW_b );\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The molecular weight of the protein = 132 kg/mol\n" + ] + } + ], + "prompt_number": 18 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_&_Chemical_Thermodynamics/ch9.ipynb b/Engineering_&_Chemical_Thermodynamics/ch9.ipynb new file mode 100755 index 00000000..ba9e9ffc --- /dev/null +++ b/Engineering_&_Chemical_Thermodynamics/ch9.ipynb @@ -0,0 +1,1019 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 9 : Chemical reaction Equilibria" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.1 Page No : 570" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "n_o_CH3OH = 1. ; \t\t\t#[mol]\n", + "n_o_H2O = 3. \t\t\t#[mol]\n", + "S = 0.87 ;\n", + "\n", + "# Calculations\n", + "n_CH3OH = 1 - S ;\n", + "n_H2O = 2 - S ;\n", + "n_CO2 = S ;\n", + "n_H2 = 3 * S ;\n", + "n_v = n_CH3OH + n_CO2 + n_H2O + n_H2 ; \n", + "y_H2 = n_H2 / n_v ;\n", + "\n", + "# Results\n", + "print \"No of moles of H2 produced for 1mol of CH3OH = %.3f mol\" %( n_H2)\n", + "print \"Mole fraction of H2 = %.2f\" %( y_H2) ;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "No of moles of H2 produced for 1mol of CH3OH = 2.610 mol\n", + "Mole fraction of H2 = 0.55\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.2 Page No : 574" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "del_gf_0_CO2 = -394.36 ; \t\t\t#[kJ/mol],From Appendix A.3\n", + "del_gf_0_H2 = 0 ; \t\t\t#[kJ/mol],From Appendix A.3\n", + "del_gf_0_H2O = -228.57 ; \t\t\t#[kJ/mol],From Appendix A.3\n", + "del_gf_0_CH3OH = -161.96 ; \t\t\t#[kJ/mol],From Appendix A.3\n", + "n_CO2 = 1 ;\n", + "n_H2 = 3 ;\n", + "n_CH3OH = 1 ;\n", + "n_H2O = 1 ;\n", + "T = 298.15 ;\t\t\t#[K]\n", + "R = 8.314 ; \t\t\t#[J/molK]\n", + "\n", + "# Calculations\n", + "del_g0_rxn = (n_CO2 * del_gf_0_CO2 + n_H2 * del_gf_0_H2 - n_H2O * del_gf_0_H2O - n_CH3OH * del_gf_0_CH3OH) * 10**3 ; \t\t\t# [J/mol]\n", + "K_298 = math.exp( - del_g0_rxn / (R * T)) ;\n", + "\n", + "# Results\n", + "print \"The equillibrium constant K298 = %.2f \"%(K_298) ;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The equillibrium constant K298 = 4.69 \n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.3 page no : 575" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "K_298 = 4.69\n", + "\n", + "# calculations\n", + "deltaH = (-393.51) + 3 * 0 - (-241.82) - (-200.66) \n", + "lnK333_469 = round((-deltaH*1000/8.314 )*(1./333 - 1./298),2)\n", + "lnK333 = lnK333_469 + math.log(K_298)\n", + "K333 = math.e**lnK333\n", + "\n", + "# Results\n", + "print \"Equilibrium constant = %.2f \"%K333\n", + "\n", + "# Note: answer is slightly different because of rounding off error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equilibrium constant = 37.54 \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.4 Page No : 577" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# Variables\n", + "del_gf_0_CH2O = -110.0 ; \t\t\t#[kJ/mol],From Appendix A.2 & A.3\n", + "del_gf_0_H2 = 0 ; \t\t\t#[kJ/mol],From Appendix A.2 & A.3\n", + "del_gf_0_CH4O = -162.0 ; \t\t\t#[kJ/mol],From Appendix A.2 & A.3\n", + "del_hf_0_CH2O = -116.0 ; \t\t\t#[kJ/mol],From Appendix A.2 & A.3\n", + "del_hf_0_H2 = 0 ; \t\t\t#[kJ/mol],From Appendix A.2 & A.3\n", + "del_hf_0_CH4O = -200.7 ; \t\t\t#[kJ/mol],From Appendix A.2 & A.3n_CH20 = 1 ;\n", + "n_H2 = 1. ;\n", + "n_CH4O = 1. ;\n", + "n_CH2O = 1. ;\n", + "T1 = 298. \t\t\t#[K]\n", + "T2 = 873. ; \t\t\t# [K]\n", + "R = 8.314 ; \t\t\t#[J/molK]\n", + "Del_A = 3.302 ;\n", + "Del_B = -4.776 * 10**-3 ;\n", + "Del_C = 1.57 * 10**-6 ;\n", + "Del_D = 0.083 * 10**5 ; \n", + "\n", + "# Calculations and Results\n", + "del_g_rxn_298 = n_CH2O * del_gf_0_CH2O + n_H2 * del_gf_0_H2 - n_CH4O * del_gf_0_CH4O ;\n", + "K_298 = math.exp( - del_g_rxn_298 * 10**3 / (R * T1)) ;\n", + "print \"a) K_298 = %.2e As the equilibrium constant is very small very little amount of\\\n", + " formaldehyde will be formed .\"%(K_298) ;\n", + "\n", + "#Solution(b)\n", + "del_h_rxn_298 = (n_CH2O * del_hf_0_CH2O + n_H2 * del_hf_0_H2 - n_CH4O * del_hf_0_CH4O) * 10**3 ;\t\t\t#[J/mol]\n", + "K_873 = K_298 * math.exp((-del_h_rxn_298 * (1/T2 - 1/T1)) / R) ;\n", + "print \"b) i) K_873 = %.2f \"%(K_873) ;\n", + "\n", + "#Solution(c)\n", + "x = ( -del_h_rxn_298 / R + Del_A * T1 + Del_B / 2 * T1**2 + Del_C /3 * T1**3 - Del_D / T1 ) *(1/T2 - 1/T1) + Del_A * math.log(T2 / T1) + Del_B / 2 * (T2 -T1) + Del_C / 6 * (T2**2 -T1**2) + Del_D / 2 * (1/(T2**2) -1/(T1**2)) ;\n", + "K_873 = K_298 * math.exp(x) ;\n", + "print \" ii) K_873 = %.2f \"%(K_873) ;\n", + "\n", + "# Note : answer are slightly different because of rounding off error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a) K_298 = 7.67e-10 As the equilibrium constant is very small very little amount of formaldehyde will be formed .\n", + "b) i) K_873 = 4.61 \n", + " ii) K_873 = 8.71 \n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.6 Page No : 581" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "import math\n", + "del_g0_f_C6H6 = -32.84 ; \t\t\t#[kJ/mol] , From Table E9.6\n", + "del_g0_f_C2H4 = 68.15 ; \t\t\t#[kJ/mol] , From Table E9.6\n", + "del_g0_f_H2 = 0 ; \t\t\t #[kJ/mol] , From Table E9.6\n", + "del_h0_f_C6H6 = -84.68 ; \t\t\t#[kJ/mol] , From Table E9.6\n", + "del_h0_f_C2H4 = 52.26 ; \t\t\t#[kJ/mol] , From Table E9.6\n", + "del_h0_f_H2 = 0 ; \t\t\t #[kJ/mol] , From Table E9.6\n", + "T1 = 298.2 ;\t\t\t #[K]\n", + "P = 1. \t\t\t#[bar]\n", + "R = 8.31 ;\n", + "T2 = 1273. \t\t\t# [K]\n", + "\n", + "# Calculations\n", + "del_g0_f_rxn = del_g0_f_C2H4 + del_g0_f_H2 - del_g0_f_C6H6 ;\n", + "K_298 = math.exp ( - (del_g0_f_rxn * 10**3) / (R * T1)) ;\n", + "del_h0_f_rxn = (del_h0_f_C2H4 + del_h0_f_H2 - del_h0_f_C6H6) * 10**3 ;\n", + "K_1273 = K_298 * math.exp( - del_h0_f_rxn / R * (1/T2 - 1/T1)) ;\n", + "x = math.sqrt( K_1273 / ( K_1273 + P)) ;\n", + "\n", + "# Results\n", + "print \"n_C2H6 = %.2f mol n_C2H4 = %.2f mol n_H2 = %.2f mol\"%(1-x,x,x) ;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "n_C2H6 = 0.09 mol n_C2H4 = 0.91 mol n_H2 = 0.91 mol\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.7 Page No : 583" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from matplotlib.pyplot import *\n", + "from numpy import *\n", + "from scipy.integrate import *\n", + "import math \n", + "\n", + "\n", + "# Variables\n", + "del_h0_f_NH3 = -46.11 ; \t\t\t# [kJ/mol],From table E9.7\n", + "del_h0_f_N2 = 0 \t\t\t# [kJ/mol],From table E9.7\n", + "del_h0_f_H2 = 0 ; \t\t \t# [kJ/mol],From table E9.7\n", + "del_g0_f_NH3 = -16.45 ; \t\t\t# [kJ/mol],From table E9.7\n", + "del_g0_f_N2 = 0 ; \t\t \t# [kJ/mol],From table E9.7\n", + "del_g0_f_H2 = 0 ; \t\t\t # [kJ/mol],From table E9.7\n", + "n_NH3 = 2. ;\n", + "n_N2 = -1. ;\n", + "n_H2 = -3. ;\n", + "A_NH3 = 3.578 \n", + "B_NH3 = 3.02 * 10**-3\n", + "D_NH3 = -0.186 * 10**5 ; \n", + "A_N2 = 3.280\n", + "B_N2 = 0.593 * 10**-3\n", + "D_N2 = 0.040 * 10**5 ;\n", + "A_H2 = 3.249\n", + "B_H2 = 0.422 * 10**-3\n", + "D_H2 = 0.083 * 10**5 ;\n", + "R = 8.314 ;\n", + "T = 298. ;\n", + "T2 = 773. ;\n", + "P = 1. ; \t\t\t#[bat]\n", + "\n", + "# Calculations\n", + "Del_h0_rxn = (n_NH3 * del_h0_f_NH3 + n_N2 * del_h0_f_N2 + n_H2 * del_h0_f_H2) * 10**3 ;\n", + "Del_g0_rxn = (n_NH3 * del_g0_f_NH3 + n_N2 * del_g0_f_N2 + n_H2 * del_g0_f_H2) * 10**3 ;\n", + "del_A = n_NH3 * A_NH3 + n_N2 * A_N2 + n_H2 * A_H2 ;\n", + "del_B = n_NH3 * B_NH3 + n_N2 * B_N2 + n_H2 * B_H2 ;\n", + "del_D = n_NH3 * D_NH3 + n_N2 * D_N2 + n_H2 * D_H2 ;\n", + "\n", + "K_298 = math.exp( - Del_g0_rxn / ( R * T)) ;\n", + "K_T = K_298 * math.exp( - Del_h0_rxn / R * (1 / T2 - 1 / T)) ;\n", + "A = K_T * P**2 *27 -16 ;\n", + "B = 64 - K_T * P**2 * 108 ;\n", + "C = -64 + K_T * P**2 * 162 ;\n", + "D = -108 * K_T * P**2 ;\n", + "E = 27 * K_T * P**2 ;\n", + "\n", + "\n", + "mycoeff =[E , D ,C , B ,A];\n", + "M = roots([A,B,C,D,E]);\n", + "\n", + "for i in range(3):\n", + " ans = isreal(M[i]) ;\n", + " if ans == True:\n", + " y = M[i] / M[i+1] - 1 ;\n", + " ans = sign(y) ;\n", + " \n", + " if ans == 1 :\n", + " x = M[i]\n", + " else:\n", + " x = M[i+1]\n", + " \n", + "# Results\n", + "print \"a)Extent of reaction = %.3f \"%(x);\n", + "\n", + "#(b)\n", + "X = (-Del_h0_rxn / R + del_A * T + del_B / 2 * T**2 - del_D / T) * (1/T2 - 1/T) + del_A * math.log(T2 / T) + del_B / 2 * (T2 - T) + del_D / 2 * (1/(T2**2) - 1/(T**2) );\n", + "K_T = K_298 * math.exp(X) ;\n", + "\n", + "A = K_T * P**2 *27 -16 ;\n", + "B = 64 - K_T * P**2 * 108 ;\n", + "C = -64 + K_T * P**2 * 162 ;\n", + "D = -108 * K_T * P**2 ;\n", + "E = 27 * K_T * P**2 ;\n", + "\n", + "mycoeff =[E , D ,C , B ,A];\n", + "M1 = roots([A,B,C,D,E]);\n", + "\n", + "for i in range(3):\n", + " ans = isreal(M1[i])\n", + " if ans == True :\n", + " y = M1[i] / M1[i+1] - 1 ;\n", + " ans = sign(y) ;\n", + " if ans == True:\n", + " x1 = M1[i]\n", + " else:\n", + " x1 = M1[i+1]\n", + "\n", + "print \"b) Extent of reaction = %.3f\"%(x1);\n", + "print \"Under these conditions we do not expect to produce an appreciable amount of ammonia .\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a)Extent of reaction = 0.005 \n", + "b) Extent of reaction = 0.003\n", + "Under these conditions we do not expect to produce an appreciable amount of ammonia .\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.9 Page No : 585" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from scipy.optimize import fsolve \n", + "import math\n", + "\n", + "# Variables\n", + "K_T = 1.51 * 10**-5 ;\n", + "P = 300. ; \t\t\t#[bar]\n", + "T = 500 + 273.2 ; \t\t\t#[K]\n", + "R = 8.314 ;\n", + "\n", + "# Calculations and Results\n", + "def f991(k):\n", + " return ((2 * k)**2 * (4 - 2 * k)**2 / ((1 - k) * (3 - 3*k)**3)) * P**-2 - K_T \n", + "\n", + "z1 = fsolve(f991,0.3)[0]\n", + "\n", + "print \"a) Extent of reaction = %.2f \"%(z1);\n", + "\n", + "P_c = [111.3 * 101325 , 33.5 * 101325 , 12.8 * 101325] ;\n", + "T_c = [405.5 , 126.2 , 33.3] ;\n", + "\n", + "a = zeros(3)\n", + "b = zeros(3)\n", + "V = zeros(3)\n", + "sai = zeros(3)\n", + "for i in range(3):\n", + " a[i] = 27./ 64 * (R * T_c[i])**2 / P_c[i] ;\n", + " b[i] = (R * T_c[i]) / (8 * P_c[i]) ;\n", + " \n", + " def f992(v):\n", + " return (R * T) / (v - b[i]) - a[i] / (v**2) - P * 100000 ;\n", + " V[i] = fsolve(f992,.0002)\n", + " sai[i] = math.exp( - math.log((V[i] - b[i]) * P * 10**5/ ( R * T)) + b[i] / (V[i] - b[i]) - 2 * a[i] / (R * T * V[i])) ;\n", + "\n", + "def f993(k):\n", + " return ((2 * k)**2 * sai[0]**2 * (4 - 2 * k)**2 * 3 / ((1 - k) * sai[1]* (3 - 3*k)**3 * sai[2]**3 ))* P**-2 - K_T \n", + "\n", + "z2 = fsolve(f993,0.3)\n", + "\n", + "x = (z1 - z2) / z1 * 100 ;\n", + "\n", + "print \"b) Extent of reaction = %.2f \"%(z2);\n", + "print \" A correction of about %d%% is observed from accounting for nonideal behaviour . \"%(x)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a) Extent of reaction = 0.37 \n", + "b) Extent of reaction = 0.33 \n", + " A correction of about 11% is observed from accounting for nonideal behaviour . \n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.10 Page No : 586" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "import math\n", + "del_g0_f_1 = 31.72 ; \t\t\t#[kJ/mol]\n", + "del_g0_f_2 = 26.89 ; \t\t\t#[kJ/mol]\n", + "R = 8.314 ;\n", + "T = 298 ;\t\t\t#[K]\n", + "\n", + "# Calculations\n", + "del_g0_rxn = del_g0_f_2 - del_g0_f_1 ;\n", + "K = math.exp( - del_g0_rxn * 10**3 / (R * T) ) ;\n", + "x = K / (1 + K) ;\n", + "\n", + "# Results\n", + "print \"x = %.3f \\nAt equilibrium %.1f %% of the liquid exists as cyclohexane.\"%(x ,x * 100) ;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "x = 0.875 \n", + "At equilibrium 87.5 % of the liquid exists as cyclohexane.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.11 Page No : 587" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "import math\n", + "del_g0_f_CaCO3 = -951.25 ;\n", + "del_g0_f_CaO = -531.09 ;\n", + "del_g0_f_CO2 = -395.81 ;\n", + "R = 8.314 ;\n", + "T = 1000. ;\t\t\t# [K]\n", + "\n", + "# Calculations\n", + "del_g0_rxn = del_g0_f_CaO + del_g0_f_CO2 - del_g0_f_CaCO3 ;\n", + "K = math.exp(-del_g0_rxn * 10**3 / (R * T)) ;\n", + "p_CO2 = K ;\n", + "\n", + "# Results\n", + "print \"Equilibrium pressure = %.3f bar \"%(p_CO2) ;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equilibrium pressure = 0.053 bar \n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.12 Page No : 588" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "import math\n", + "del_g0_f_B = 124.3 ; \t\t\t#[kJ/mol] , From Appendix A.3\n", + "del_g0_f_Ac = 209.2 ; \t\t\t#[kJ/mol] , From Appendix A.3\n", + "R = 8.314 ;\n", + "T = 298. ; \t\t\t# [K]\n", + "A = 9.2806 ;\n", + "B = 2788.51 ;\n", + "C = -52.36 ;\n", + "\n", + "# Calculations\n", + "del_g0_rxn = del_g0_f_B - 3 * del_g0_f_Ac ;\n", + "K = math.exp( - del_g0_rxn * 10**3 / (R * T)) ;\n", + "P = 1 / K**(1./3) ;\n", + "X = A - B / (T + C) ;\n", + "P_b = math.exp(X) ;\n", + "\n", + "# Results\n", + "print (\"At equilibrium , the cylinder is almost completely filled with Benzene .\")\n", + "print \"System pressure = %.3f bar \"%(P_b)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "At equilibrium , the cylinder is almost completely filled with Benzene .\n", + "System pressure = 0.126 bar \n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.13 Page No : 596" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "import math\n", + "E_0_c = 0.153 ; \t\t\t#[V]\n", + "E_0_a = -0.521 ; \t\t\t# [v]\n", + "T = 298. ; \t\t\t#[K]\n", + "z = 1. ;\n", + "F =96485 ; \t\t\t#[C/mol e-]\n", + "R =8.314 ; \t\t\t#[J/mol K ]\n", + "\n", + "# Calculations\n", + "E_0_rxn = E_0_c + E_0_a ;\n", + "del_g_0_rxn = - z * F * E_0_rxn ;\n", + "\n", + "K = math.exp( - del_g_0_rxn / ( R * T )) ;\n", + "\n", + "# Results\n", + "print \"The equilibrium constant = %.3g \"%(K)\n", + "print \"The equilibrium constant is small . So the etching will not proceed\\\n", + " spontaneously . However if we apply work through application of an electrical potential\\\n", + " , we can etch the copper .\"\n", + "\n", + "# Note: answer is slightly different because of rounding off error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The equilibrium constant = 5.97e-07 \n", + "The equilibrium constant is small . So the etching will not proceed spontaneously . However if we apply work through application of an electrical potential , we can etch the copper .\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.14 Page No : 596" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "import math\n", + "E_0_c = 0.34 ; \t\t\t#[V]\n", + "E_0_a = -1.23 \t\t\t#[V]\n", + "T = 298. ; \t\t\t# [K]\n", + "pH = 1. ;\n", + "z = 2. ;\n", + "Cu2 = 0.07\n", + "F = 96485. \t \t\t#[C/mol e-]\n", + "R = 8.314 ;\n", + "\n", + "# Calculations\n", + "E_0_rxn = E_0_c + E_0_a ;\n", + "E = E_0_rxn + 2.303 * R * T * 2 * pH / (z * F) + R * T * math.log(Cu2) / (z * F) ;\n", + "\n", + "# Results\n", + "print \"Del_E_0_rxn = %.2f \"%(E_0_rxn ) ;\n", + "print \"We have to apply potential greater than %.1f V\"%(-E) ;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Del_E_0_rxn = -0.89 \n", + "We have to apply potential greater than 0.9 V\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.17 Page No : 602" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "m = 4. ;\n", + "T = 2. ;\n", + "Pai = 1. ;\n", + "S = 1. ;\n", + "\n", + "# Calculations\n", + "R = m - T + 2 - Pai - S ;\n", + "\n", + "# Results\n", + "print \"We must specify %g independent equations .\"%(R)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "We must specify 2 independent equations .\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.19 Page No : 603" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%pylab inline\n", + "\n", + "from numpy import *\n", + "from scipy.optimize import fsolve \n", + "from matplotlib.pyplot import *\n", + "import math \n", + "\n", + "# Variables\n", + "del_g_f_CH4 = -50.72 ;\n", + "del_g_f_H2 = 0 ;\n", + "del_g_f_H2O = -228.57 ;\n", + "del_g_f_CO = -137.17 ;\n", + "del_g_f_CO2 = -394.36 ;\n", + "del_h_f_CH4 = -74.81 ;\n", + "del_h_f_H2 = 0 ;\n", + "del_h_f_H2O = -241.82 ;\n", + "del_h_f_CO = -110.53 ;\n", + "del_h_f_CO2 = -393.51 ;\n", + "\n", + "v1_CH4 = -1. ;\n", + "v1_H2 = 3. ;\n", + "v1_H2O = -1. ;\n", + "v1_CO = 1. ;\n", + "v1_CO2 = 0. ;\n", + "v2_CH4 = -1. ;\n", + "v2_H2 = 4. ;\n", + "v2_H2O = -2. ;\n", + "v2_CO = 0. ;\n", + "v2_CO2 = 1. ;\n", + "\n", + "A_CH4 = 1.702 ;\n", + "B_CH4 = 9.08 * 10**-3 ;\n", + "C_CH4 = -2.16 * 10**-6 ;\n", + "D_CH4 = 0 ;\n", + "A_H2 = 3.249 ;\n", + "B_H2 = 4.22 * 10**-4 ;\n", + "C_H2 = 0 ;\n", + "D_H2 = 8.30 * 10**3 ;\n", + "A_H2O = 3.47 ;\n", + "B_H2O = 1.45 * 10**-3 ;\n", + "C_H2O = 0 ;\n", + "D_H2O = 1.21 * 10**4 ;\n", + "A_CO = 3.376 ;\n", + "B_CO = 5.57 * 10**-4 ;\n", + "C_CO = 0 ;\n", + "D_CO = -3.10 * 10**3 ;\n", + "A_CO2 = 5.457 ;\n", + "B_CO2 = 1.05 * 10**-3 ;\n", + "C_CO2 = 0 ;\n", + "D_CO2 = -1.16 * 10**5 ;\n", + "M = zeros((12,10))\n", + "M[:,0] = linspace(600,1150,12)\n", + "\n", + "R = 8.314 ;\n", + "P = 1 ; \t\t\t#[bar]\n", + "T_ref = 298.15 ; \t\t\t#[K]\n", + "\n", + "# Calculations\n", + "del_g_f_1 = (v1_CO * del_g_f_CO + v1_H2 *del_g_f_H2 + v1_CH4 * del_g_f_CH4 + v1_H2O * del_g_f_H2O) * 1000 ;\n", + "del_h_f_1 = (v1_CO * del_h_f_CO + v1_H2 *del_h_f_H2 + v1_CH4 * del_h_f_CH4 + v1_H2O * del_h_f_H2O) * 1000 ;\n", + "del_g_f_2 = (v2_CO2 * del_g_f_CO2 + v2_H2 *del_g_f_H2 + v2_CH4 * del_g_f_CH4 + v2_H2O * del_g_f_H2O) * 1000 ;\n", + "del_h_f_2 = (v2_CO2 * del_h_f_CO2 + v2_H2 *del_h_f_H2 + v2_CH4 * del_h_f_CH4 + v2_H2O * del_h_f_H2O) * 1000;\n", + "Del_A_1 = v1_CO * A_CO + v1_H2 * A_H2 + v1_CH4 * A_CH4 + v1_H2O * A_H2O ;\n", + "Del_B_1 = v1_CO * B_CO + v1_H2 * B_H2 + v1_CH4 * B_CH4 + v1_H2O * B_H2O ;\n", + "Del_C_1 = v1_CO * C_CO + v1_H2 * C_H2 + v1_CH4 * C_CH4 + v1_H2O * C_H2O ;\n", + "Del_D_1 = v1_CO * D_CO + v1_H2 * D_H2 + v1_CH4 * D_CH4 + v1_H2O * D_H2O ;\n", + "Del_A_2 = v2_CO2 * A_CO2 + v2_H2 * A_H2 + v2_CH4 * A_CH4 + v2_H2O * A_H2O ;\n", + "Del_B_2 = v2_CO2 * B_CO2 + v2_H2 * B_H2 + v2_CH4 * B_CH4 + v2_H2O * B_H2O ;\n", + "Del_C_2 = v2_CO2 * C_CO2 + v2_H2 * C_H2 + v2_CH4 * C_CH4 + v2_H2O * C_H2O ;\n", + "Del_D_2 = v2_CO2 * D_CO2 + v2_H2 * D_H2 + v2_CH4 * D_CH4 + v2_H2O * D_H2O ;\n", + "\n", + "\n", + "K_298_1 = math.exp( - del_g_f_1 / (R * T_ref)) ;\n", + "K_298_2 = math.exp( - del_g_f_2 / (R * T_ref)) ;\n", + "\n", + "for i in range(12):\n", + " X = (-del_h_f_1 / R + Del_A_1 * T_ref + Del_B_1 / 2 * T_ref**2 + \\\n", + " Del_C_1 /3* T_ref**3- Del_D_1 / T_ref) * (1./M[i,0] - 1./T_ref) + \\\n", + " Del_A_1*math.log(M[i,0] / T_ref)+ Del_B_1 / 2 * (M[i,0] - T_ref) + \\\n", + " Del_C_1 / 6 *(M[i,0]**2 - T_ref**2) + Del_D_1 / 2 * (1./(M[i,0]**2) - 1./(T_ref**2))\n", + " \n", + " M[i,1] = K_298_1 * math.exp(X) ;\n", + " \n", + " Y = (-del_h_f_2 / R + Del_A_2 * T_ref + Del_B_2 / 2 * T_ref**2 + Del_C_2/3* T_ref**3- Del_D_2 / T_ref) * \\\n", + " (1/M[i,0] - 1/T_ref) + Del_A_2 * math.log(M[i,0] / T_ref)+ Del_B_2 / 2 * (M[i,0] - T_ref) + \\\n", + " Del_C_2 / 6 *(M[i,0]**2 - T_ref**2) + Del_D_2 / 2* (1/(M[i,0]**2) - 1/(T_ref**2));\n", + " \n", + " M[i,2] = K_298_2 * math.exp(Y) ;\n", + " def f918(R):\n", + " s1 = R[0] ;\n", + " s2 = R[1] ;\n", + " y = [0,0]\n", + " y[0] = (s1 * (3 * s1 + 4 * s2)**3) / ((5 + 2 * s1 + 2 * s2)**2 * (1 - s1 -s2) * (4 - s1 - 2 * s2)) * P**2 - M[i,1] ;\n", + " y[1] = (s2 * (3 * s1 + 4 * s2)**4) / ((5 + 2 * s1 + 2 * s2)**2 * (1 - s1 -s2) * (4 - s1 - 2 * s2)**2) * P**2 - M[i,2] ;\n", + " return y\n", + " z = fsolve(f918,[0.0001,0.0001])\n", + " M[i,3] = z[0] ;\n", + " M[i,4] = z[1] ; \n", + " M[i,5] = (1 - M[i,3] - M[i,4]) / (5 + 2 * M[i,3] + 2 * M[i,4]) ;\n", + " M[i,6] = (4 - M[i,3] - 2 * M[i,4]) / (5 + 2 * M[i,3] + 2 * M[i,4]) ;\n", + " M[i,7] = (3 * M[i,3] + 4 * M[i,4]) / (5 + 2 * M[i,3] + 2 * M[i,4]) ;\n", + " M[i,8] = M[i,3] / (5 + 2 * M[i,3] + 2 * M[i,4]) ; \n", + " M[i,9] = M[i,4] / (5 + 2 * M[i,3] + 2 * M[i,4]) ; \n", + "\n", + "n1 = zeros([12,7])\n", + "for i in range(12):\n", + " for j in range(7):\n", + " n1[i,j] = M[i,j] ;\n", + "n2 = zeros([12,3])\n", + "for i in range(12):\n", + " for j in range(3):\n", + " n2[i,j] = M[i,j+7]\n", + "print \" T K1 K2 S1 S2 y_CH4 y_H2\"\n", + "for row in n1:\n", + " print \" %5d %7.2e %7.2e %7.3f %7.3f %7.3f %7.3f \"%(row[0],row[1],row[2],row[3],row[4],row[5],row[6])\n", + "#print (n1) ;\n", + "print (\" y_H20 y_CO y_CO2 \") ;\n", + "for row in n2:\n", + " print \" %7.3f %7.3f %7.3f \"%(row[0],row[1],row[2])\n", + "\n", + "\n", + "#print (n2) ;\n", + "N = zeros([10,10])\n", + "for i in range(10):\n", + " for j in range(10):\n", + " N[i,j] = M[i,j]\n", + "'''\n", + "plot(N[3],N[0],\"+\") ;\n", + "plot(N[4], N[0],\".\") ;\n", + "plot(N[5] , N[0], \"o-\") ; \n", + "plot(N[6] , N[0], \"s-\");\n", + "plot(N[7] , N[0], \"*-\") ;\n", + "plot(N[8] , N[0], \"x-\") ;\n", + "plot(N[9] , N[0], \".-\") ;\n", + "show()\n", + "'''\n", + "\n", + "suptitle(\"Figure E9.18 Extent of reaxn vs temp\")\n", + "xlabel(\"Temperature(K)\")\n", + "ylabel(\"S\") ;\n", + "#legend(\"S1\",\"S2\") ;\n", + "plot(N[:,0] , N[:,5], \"o-\") ; \n", + "plot(N[:,0] , N[:,6], \"s-\");\n", + "plot(N[:,0] , N[:,7], \"^-\") ;\n", + "plot(N[:,0] , N[:,8], \"x-\") ;\n", + "plot(N[:,0] , N[:,9], \".-\") ;\n", + "\n", + "# Note : some answers are different because of rounding off error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Populating the interactive namespace from numpy and matplotlib\n", + " T K1 K2 S1 S2 y_CH4 y_H2" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " 600 4.91e-07 1.40e-05 0.000 0.113 0.170 0.722 \n", + " 650 1.42e-05 2.24e-04 0.003 0.191 0.150 0.671 \n", + " 700 2.59e-04 2.47e-03 0.011 0.294 0.124 0.606 \n", + " 750 3.24e-03 2.01e-02 0.037 0.410 0.094 0.534 \n", + " 800 3.00e-02 1.29e-01 0.099 0.515 0.062 0.461 \n", + " 850 2.15e-01 6.69e-01 0.207 0.578 0.033 0.401 \n", + " 900 1.25e+00 2.93e+00 0.332 0.584 0.012 0.366 \n", + " 950 6.06e+00 1.11e+01 0.424 0.551 0.004 0.356 \n", + " 1000 2.52e+01 3.70e+01 0.485 0.509 0.001 0.358 \n", + " 1050 9.20e+01 1.11e+02 0.531 0.468 0.000 0.362 \n", + " 1100 2.99e+02 3.02e+02 0.319 0.322 0.057 0.484 \n", + " 1150 8.78e+02 7.56e+02 0.343 0.295 0.058 0.488 \n", + " y_H20 y_CO y_CO2 \n", + " 0.087 0.000 0.022 \n", + " 0.144 0.000 0.036 \n", + " 0.215 0.002 0.052 \n", + " 0.297 0.006 0.070 \n", + " 0.378 0.016 0.083 \n", + " 0.447 0.032 0.088 \n", + " 0.488 0.049 0.085 \n", + " 0.500 0.061 0.079 \n", + " 0.499 0.069 0.073 \n", + " 0.495 0.076 0.067 \n", + " 0.357 0.051 0.051 \n", + " 0.352 0.055 0.047 \n" + ] + }, + { + "output_type": "stream", + "stream": "stderr", + "text": [ + "/home/jovina/virtualenvs/scipy/local/lib/python2.7/site-packages/scipy/optimize/minpack.py:236: RuntimeWarning: The iteration is not making good progress, as measured by the \n", + " improvement from the last ten iterations.\n", + " warnings.warn(msg, RuntimeWarning)\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 2, + "text": [ + "[<matplotlib.lines.Line2D at 0x34b06d0>]" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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S3ntPfqx5CGNqKms6diS5e/dye01pUYhKY/vx7by4/EV2/b2L1299nQFtB6DX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\nvLwcmZmZWLhwITQaDTQaDc6fP29TTfX19cK9NRo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+ "text": [ + "<matplotlib.figure.Figure at 0x2619950>" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.21 Page No : 607" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "import math\n", + "del_g_0_f_SiCl2 = - 216012. ; \n", + "del_g_0_f_SiCl4 = - 492536. ;\n", + "del_g_0_f_SiCl3H = -356537. ;\n", + "del_g_0_f_SiCl2H2 = -199368. ;\n", + "del_g_0_f_SiClH3 = -28482. ;\n", + "del_g_0_f_SiH4 = -176152. ;\n", + "del_g_0_f_HCl = -102644. ;\n", + "del_g_0_f_H2 = 0. ;\n", + "del_g_0_f_Si = 0. ;\n", + "R = 8.314 ;\n", + "T = 1300. ; \t\t\t#[K]\n", + "\n", + "# Calculations\n", + "Del_g_rxn_1 = del_g_0_f_SiCl2 + 2 * del_g_0_f_HCl - del_g_0_f_SiCl4 - del_g_0_f_H2 ;\n", + "Del_g_rxn_2 = del_g_0_f_SiCl3H + del_g_0_f_HCl - del_g_0_f_SiCl4 - del_g_0_f_H2 ;\n", + "Del_g_rxn_3 = del_g_0_f_SiCl2H2 + del_g_0_f_HCl - del_g_0_f_SiCl3H - del_g_0_f_H2 ;\n", + "Del_g_rxn_4 = del_g_0_f_SiClH3 + del_g_0_f_HCl - del_g_0_f_SiCl2H2 - del_g_0_f_H2 ;\n", + "Del_g_rxn_5 = del_g_0_f_SiH4 + del_g_0_f_HCl - del_g_0_f_SiCl3H - del_g_0_f_H2 ;\n", + "Del_g_rxn_6 = del_g_0_f_Si + 4 * del_g_0_f_HCl - del_g_0_f_SiCl4 - 2 * del_g_0_f_H2 ;\n", + "\n", + "M = zeros([6,4])\n", + "\n", + "M[0,0] = math.exp( - Del_g_rxn_1 / (R * T)) ;\n", + "M[1,0] = math.exp( - Del_g_rxn_2 / (R * T)) ;\n", + "M[2,0] = math.exp( - Del_g_rxn_3 / (R * T)) ;\n", + "M[3,0] = math.exp( - Del_g_rxn_4 / (R * T)) ;\n", + "M[4,0] = math.exp( - Del_g_rxn_5 / (R * T)) ;\n", + "M[5,0] = math.exp( - Del_g_rxn_6 / (R * T)) ;\n", + "\n", + "S = [0.0763,0.1979,0.0067,0.0001,0.0000,-0.0512] ;\n", + "K_cal = [.00137,0.0457,0.00644,0.00181,0.000752,0.000509] ;\n", + "\n", + "for i in range(6):\n", + " M[i,1] = S[i] ;\n", + " M[i,2] = K_cal[i] ;\n", + " M[i,3] = M[i,0] - M[i,2] ;\n", + "# Results\n", + "print (\" K_i S K_i_cal K_i - K_i_cal\") ;\n", + "for row in M:\n", + " print \" %5.2e %7.4f %7.2e %7.2e\"%(row[0],row[1],row[2],row[3])\n", + "\n", + "# Readers can refer figure E9.19 .\n", + "# Note : answers are different because of rounding off error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " K_i S K_i_cal K_i - K_i_cal\n", + " 1.37e-03 0.0763 1.37e-03 2.77e-06\n", + " 4.57e-02 0.1979 4.57e-02 -1.95e-05\n", + " 6.44e-03 0.0067 6.44e-03 2.87e-06\n", + " 1.81e-03 0.0001 1.81e-03 9.39e-07\n", + " 7.52e-04 0.0000 7.52e-04 -4.00e-09\n", + " 5.09e-04 -0.0512 5.09e-04 -3.50e-08\n" + ] + } + ], + "prompt_number": 69 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_&_Chemical_Thermodynamics/screenshots/graphical_determination.png b/Engineering_&_Chemical_Thermodynamics/screenshots/graphical_determination.png Binary files differnew file mode 100755 index 00000000..9e79c2e5 --- /dev/null +++ b/Engineering_&_Chemical_Thermodynamics/screenshots/graphical_determination.png diff --git a/Engineering_&_Chemical_Thermodynamics/screenshots/least-squares-linear-fit.png b/Engineering_&_Chemical_Thermodynamics/screenshots/least-squares-linear-fit.png Binary files differnew file mode 100755 index 00000000..f3276fd3 --- /dev/null +++ b/Engineering_&_Chemical_Thermodynamics/screenshots/least-squares-linear-fit.png diff --git a/Engineering_&_Chemical_Thermodynamics/screenshots/partial.png b/Engineering_&_Chemical_Thermodynamics/screenshots/partial.png Binary files differnew file mode 100755 index 00000000..25753278 --- /dev/null +++ b/Engineering_&_Chemical_Thermodynamics/screenshots/partial.png diff --git a/Engineering_Physics/Chapter_1.ipynb b/Engineering_Physics/Chapter_1.ipynb new file mode 100755 index 00000000..4dd3b7b8 --- /dev/null +++ b/Engineering_Physics/Chapter_1.ipynb @@ -0,0 +1,2094 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 1: Interference" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.1, Page 1.40" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " # Given \n", + "l = 6.6e-7 # wavelength of light in meter\n", + "L = 1.32e-5 # coherence length in meter\n", + "\n", + "#Calculation\n", + "coherence_time = L / (3 * 10 ** 8)#calculation for coherence time\n", + "\n", + "#Result\n", + "print \"Coherence time = %.1e sec\"%coherence_time" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Coherence time = 4.4e-14 sec\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.2, Page 1.40" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "l = 5.896e-7 # wavelength of light in meter\n", + "L = 2.945e-2 # coherence length in meter\n", + "\n", + "#Calculations\n", + "coherence_time = L / (3 * 10**8) # calculation for coherence time\n", + "n = L / l # calculation for number of oscillations \n", + "\n", + "#Results\n", + "print \"Coherence time = %.3e sec.\"%coherence_time\n", + "print \"No. of oscillations = %.2e\"%n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Coherence time = 9.817e-11 sec.\n", + "No. of oscillations = 4.99e+04\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.3, Page 1.40" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "l = 6.058e-7 # wavelength of light in meter\n", + "L = 0.2 # coherence length in meter\n", + "\n", + "#Calculations\n", + "line_width = (l**2) / L#calculation for line width\n", + "f_spread = (3 * 10**8) / L# calculation for frequency spread\n", + "f = (3 * 10**8) / l # calculation for frequency\n", + "f_stability = f_spread / f # calculation for frequency stability\n", + "coherence_time = L / (3 * 10 ** 8) # calculation for coherence time\n", + "\n", + "#Results\n", + "print(\"Coherence time = %.3e sec\"%coherence_time)\n", + "print(\"Line width = %.3e meter\"%line_width)\n", + "print(\"Frequency stability = %.1e\"%f_stability) #incorrect answer in the textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Coherence time = 6.667e-10 sec\n", + "Line width = 1.835e-12 meter\n", + "Frequency stability = 3.0e-06\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.4, Page 1.41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "lambda_D = 5.5e-13 # Doppler width of orange light in meter\n", + "l = 6.058e-7 # wavelength of light in meter\n", + "\n", + "#Calculation\n", + "coherence_length = (l ** 2) / lambda_D# calculation for coherence light\n", + "\n", + "#Result\n", + "print(\"Coherence length = %.4f meter\"%coherence_length)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Coherence length = 0.6673 meter\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.5, Page 1.41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "lambda1 = 5.461e-7 # wavelength of light emitted by mercury vapour lamp in meter\n", + "band_width1 = 6e8 # band width for mercury vapour lamp in Hz\n", + "lambda2 = 6.328e-7 # the operating wavelength of light for He Ne laser \n", + "band_width2 = 1e6 # band width for laser in Hz\n", + "\n", + "#Calculations\n", + "delta_lambda1 = (lambda1**2 * band_width1) / 3e8 # calculation for difference between two wavelength for mercury vapour\n", + "delta_L1 = lambda1**2 / delta_lambda1 # calculation for coherence length for mercury vapour lamp\n", + "delta_lambda2 = (lambda2**2 * band_width2) / 3e8 # calculation for difference between two wavelength for He Ne laser\n", + "delta_L2 = lambda2**2 / delta_lambda2 # calculation for coherence length for He Ne laser\n", + "R = delta_L1/delta_L2 # calculation for ratio of coherence length of mercury vapour lamp to the coherence length of He Ne laser\n", + "\n", + "#Result\n", + "print(\"The ratio of coherence length of mercury vapour lamp to the coherence length of He Ne laser = 1:%d\"%(1./R))\n", + "#Answer differes due to rounding-off values" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The ratio of coherence length of mercury vapour lamp to the coherence length of He Ne laser = 1:600\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.6, Page 1.42" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "band_width = 3000 # band width of laser in hertz\n", + " \n", + "#Calculation\n", + "coherence_length = (3 * 10 ** 8) / band_width#calculation for coherence length \n", + "\n", + "#Result\n", + "print(\"Coherence length of laser = %.f meter\"%(coherence_length))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Coherence length of laser = 100000 meter\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.7, Page 1.42" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "l = 6.328e-7 # wavelength of monochromatic light in meter\n", + "t = 10**-10 # chopping time in sec\n", + "\n", + "#Calculations\n", + "coherence_length = (3 * 10 ** 8) * t # calculation for coherence length of monochromatic light \n", + "band_width = 1 / t # calculation for band width \n", + "line_width = ((l ** 2) * band_width) / (3 * 10 ** 8) # calculation for line width \n", + "\n", + "#Result\n", + "print(\"Coherence length of monochromatic light = %.e meter. \\nband width = %.f Hz. \\nline width = %.4f A.\"%(coherence_length, band_width, line_width*1e10))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Coherence length of monochromatic light = 3e-02 meter. \n", + "band width = 10000000000 Hz. \n", + "line width = 0.1335 A.\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.8, Page 1.42" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "l = 6.438e-7 # wavelength of red cadmium line in meter\n", + "L = 3.8e-1 # coherence length in meter\n", + "\n", + "#Calculations\n", + "coherence_time = L / (3 * 10 ** 8)# calculation for coherence time\n", + "spectral_line_width = (l**2) / L # calculation for spectral line width\n", + "\n", + "#Result\n", + "print(\"Coherence time of red cadmium line = %.3e sec. \\nSpectral line width = %.2e meter.\"%(coherence_time,spectral_line_width))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Coherence time of red cadmium line = 1.267e-09 sec. \n", + "Spectral line width = 1.09e-12 meter.\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.9, Page 1.43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "from sympy import *\n", + "\n", + "# Given \n", + "ratio = 16 # ratio of intensities of two waves\n", + "\n", + "#Calculation\n", + "a1 = sqrt(ratio) # by the formula amplitude = sqrt(intensity)\n", + "a2 = 1\n", + "R = ((a1 + a2) ** 2) / ((a1 - a2) ** 2)# calculation for ratio of maximum intensity with minimum intensity\n", + "R = nsimplify(R)\n", + "\n", + "#Result\n", + "print \"Ratio of maximum intensity with minimum intensity =\",R" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ratio of maximum intensity with minimum intensity = 25/9\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.10, Page 1.43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "d = 0.0001 # distance between two slits in meter\n", + "Beta = 0.005 # width of the fringes formed in meter\n", + "D = 1 # distance between slit and screen in meter\n", + "\n", + "#Calculation\n", + "l = (Beta * d) / D # calculation for wavelength of light = %e meter\n", + "\n", + "#Result\n", + "print(\"Wavelength of light = %.f A. \"%(l*1e10))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of light = 5000 A. \n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.11, Page 1.43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "# Given \n", + "alpha = pi / 180 # angle of bi prism in radian\n", + "mu = 1.5 # refractive index of biprism \n", + "a = 0.4 # distance of bi prism from slit in meter\n", + "b = 0.6 # distance of bi prism from screen in meter\n", + "l = 5.893e-7 # wavelength of light in meter\n", + "\n", + "#Calculation\n", + "D = a + b # calculation for distance between slits and screen\n", + "fringe_width = (l * D) / (2 * a * (mu - 1) * alpha) # calculation for fringe width\n", + "\n", + "#Result\n", + "print(\"Fringe width = %.3e meter.\"%(fringe_width))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Fringe width = 8.441e-05 meter.\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.12, Page 1.44" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "# Given \n", + "d1 = 4.05e-3 # distance between slits in first position in meter\n", + "d2 = 2.90e-3 # distance between slits in second position in meter\n", + "l = 5.893e-7 # wavelength of light in meter\n", + "D = 1 # distance between slit and screen\n", + "\n", + "#Calculations\n", + "d = sqrt(d1 * d2)# calculation for distance between fringe\n", + "fringe_width = (l * D) / d # calculation for fringe width\n", + "\n", + "#Result\n", + "print(\"Fringe width = %.3f mm\"%(fringe_width*1000))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Fringe width = 0.172 mm\n" + ] + } + ], + "prompt_number": 48 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.13, Page 1.44" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import floor\n", + "\n", + "# Given \n", + "fringe_width = 3.42e-4 # fringe width in meter\n", + "mu = 1.542 # refractive index of glass\n", + "Xn = 2.143e-3 # shift of central fringe in meter\n", + "l = 5.89e-7 # wavelength of light in meter\n", + "\n", + "#Calculations\n", + "n = Xn / fringe_width # calculation for order of the fringe\n", + "t = (floor(n) * l) / (mu - 1) # calculation for thickness of the glass\n", + "\n", + "#Result\n", + "print(\"Thickness of glass sheet = %.2e meter. \"%t)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thickness of glass sheet = 6.52e-06 meter. \n" + ] + } + ], + "prompt_number": 49 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.14, Page 1.45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "fringe_width = 9e-4 # fringe width in meter\n", + "a = 0.1 # distance of bi prism from slit in meter\n", + "b = 0.9 # distance of bi prism from screen in meter\n", + "l = 5.896e-7 # wavelength of light in meter\n", + "\n", + "#Calculation\n", + "D = a + b # calculation for distance between slits and screen\n", + "d = (l * D) / fringe_width # calculation for distance between coherent sources\n", + "\n", + "#Result\n", + "print(\"Distance between coherent sources = %.2e meter. \"%d)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Distance between coherent sources = 6.55e-04 meter. \n" + ] + } + ], + "prompt_number": 50 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.15, Page 1.45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi,degrees\n", + "\n", + "# Given \n", + "fringe_width = 0.0135 # fringe width in meter\n", + "a = 0.5 # distance of bi prism from slits in meter\n", + "b = 0.5 # distance of bi prism from screen in meter\n", + "mu = 1.5 # refractive index of bi prism \n", + "alpha = pi / 360 # angle of bi prism in radian \n", + "\n", + "#Calculations\n", + "D = a + b # calculation for distance between slits and screen \n", + "l = (2. * a * (mu - 1) * alpha * fringe_width) / D # calculation for wavelength of light = %e meter\n", + "\n", + "\n", + "#Result\n", + "print(\"Wavelength of light = %.f A \"%(l*1e8))\n", + "#Answer differs due to rounding-off errors\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of light = 5890 A \n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.16, Page 1.45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "# Given \n", + "a = 45 # distance between slit and bi prism in cm\n", + "alpha = pi / 180 # angle of bi prism in radian\n", + "Mu = 1.5 # refractive index of bi prism\n", + "fringe_width = 15.6e-3 # fringe width in meter \n", + "D = 90 #cm\n", + "\n", + "#Calculations\n", + "d2 = (2*a*(Mu-1)*alpha) # calculation for distance between screen and slit\n", + "l = (fringe_width * d2) / D # calculation for wavelength\n", + "\n", + "#Result\n", + "print(\"Wavelength of light = %.f A.\"%(l/1e-8))\n", + "#Answer differs due to rounding-off errors" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of light = 13614 A.\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.17, Page 1.45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "D = 1.20 # distance between source and eye piece in meter\n", + "Xn = 1.9e-2 # distance move by eye piece for 20 fringe in meter\n", + "n = 20 # no. of fringes\n", + "d = 6e-4 # distance between slits in meter \n", + "\n", + "#Calculation\n", + "l = (Xn * d) / (D * n)# calculation for wavelength\n", + "\n", + "#Result\n", + "print(\"Wavelength of light = %.f A.\"%(l*1e10))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of light = 4750 A.\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.18, Page 1.46" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import floor\n", + "\n", + "# Given \n", + "lambda1 = 5.890e-7 # wavelength of first light in meter\n", + "lambda2 = 4.358e-7 # wavelength of second light in meter\n", + "n1 = 40 # no. of fringes observed in the field of in first case \n", + "\n", + "#Calculation\n", + "n2 = (n1 * lambda1) / lambda2 # by using formula n1*lambda1=n2*lambda2\n", + "\n", + "#Result\n", + "print(\"No. of fringes observed in field of view in second case = %d. \"%(floor(n2)))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "No. of fringes observed in field of view in second case = 54. \n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.19, Page 1.46" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import cos\n", + "\n", + "# Given \n", + "l = 5.893e-7 # wavelength of light in meter\n", + "Mu = 1.42 # refractive index of soap film \n", + "i = 0 # incidence angle in radian \n", + "r = 0 # refracted angle in radian\n", + "n = 1 # for smallest thickness\n", + "\n", + "#Calculations\n", + "t1 = ((2 * n - 1) * l) / (4 * Mu * cos(r)) # calculation for east thickness of soap film for bright fringe\n", + "t2 = (n * l) / (2 * Mu * cos(r)) # calculation for east thickness of soap film for dark fringe\n", + "\n", + "#Result\n", + "print(\"Least thickness of soap film -\\n (a) For bright fringe = %.3e mm. \\n (b) For dark fringe = %.3e mm.\"%(t1*1000,t2*1000))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Least thickness of soap film -\n", + " (a) For bright fringe = 1.038e-04 mm. \n", + " (b) For dark fringe = 2.075e-04 mm.\n" + ] + } + ], + "prompt_number": 97 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.20, Page 1.46" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import cos,pi,sin,asin\n", + "\n", + "# Given \n", + "l = 5.89e-7 # wavelength of light in meter\n", + "Mu_o = 1.4 # refractive index of oil film \n", + "Mu_w = 1.33 # refractive index of water\n", + "i = pi / 6 # incidence angle in radian \n", + "n = 6 # no. of fringes seen\n", + "\n", + "#Calculation\n", + "r = asin(sin(i) / Mu_o) # calculation for angle of refraction\n", + "t = (n * l) / (2 * Mu_o * cos(r)) # calculation for thickness of film\n", + "\n", + "#Result\n", + "print(\"Thickness of oil film = %.3e mm.\"%(t*1000))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thickness of oil film = 1.351e-03 mm.\n" + ] + } + ], + "prompt_number": 98 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.21, Page 1.47" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import cos\n", + "\n", + "# Given \n", + "l = 6e-7 # wavelength of light in meter\n", + "Mu = 1.463 # refractive index of soap film \n", + "i = 0 # incidence angle in radian \n", + "r = 0 # refracted angle in radian\n", + "n = 1 # for smallest thickness\n", + "\n", + "#Calculation\n", + "t = ((2 * n - 1) * l) / (4 * Mu * cos(r)) # calculation for least thickness of soap film for bright fringe\n", + "\n", + "#Result\n", + "print(\"Least thickness of soap film for bright fringe = %.3e mm. \"%(t*1000))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Least thickness of soap film for bright fringe = 1.025e-04 mm. \n" + ] + } + ], + "prompt_number": 99 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.22, Page 1.47" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import cos,pi,sin\n", + "\n", + "# Given That\n", + "l = 5.89e-7 # wavelength of light \n", + "Mu_o = 1.46 # refractive index of oil film \n", + "i = pi / 6 # incidence angle in radian \n", + "n = 8 # no. of fringe is seen\n", + "\n", + "#Calculations\n", + "r = asin(sin(i) / Mu_o) # calculation for angle of refraction\n", + "t = (n * l) / (2 * Mu_o * cos(r)) # calculation for thickness of oil film\n", + "\n", + "#Result\n", + "print(\"Thickness of oil film = %.2e mm. \"%(t*1000))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thickness of oil film = 1.72e-03 mm. \n" + ] + } + ], + "prompt_number": 100 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.23, Page 1.47" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import cos,pi,floor\n", + "\n", + "# Given That\n", + "lambda1 = 4e-7 # max. wavelength of light in meter\n", + "lambda2 = 5e-7 # min. wavelength of light in meter\n", + "Mu = 1.4 # refractive index of soap film \n", + "i = pi / 4 # incidence angle in radian \n", + "t = 1e-5 # thickness of oil film in meter\n", + "\n", + "#Calculations\n", + "r = asin(sin(i) / Mu) # calculation for angle of refraction\n", + "n1 = (2 * Mu * t * cos(r)) / lambda1 # calculation for no. of dark bands seen in the case of max. wavelength \n", + "n2 = (2 * t * Mu * cos(r)) / lambda2 # calculation for no. of dark seen in the case of min. wavelength \n", + "n = floor(n1) - floor(n2) # claculation for no. of dark bands seen between wavelengths\n", + "\n", + "#Result\n", + "print(\"No. of dark bands seen between wavelengths = %d\"%n)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "No. of dark bands seen between wavelengths = 12\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.24, Page 1.48" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import cos,pi\n", + "\n", + "# Given \n", + "l = 5.89e-7 # wavelength of light in meter\n", + "Mu = 1.5 # refractive index of soap film \n", + "r = pi / 3 # refracted angle in radian\n", + "n = 1 # for smallest thickness\n", + "\n", + "#Calculation\n", + "t = (n * l) / (2 * Mu * cos(r)) # calculation for least thickness of soap film for bright fringe\n", + "\n", + "#Result\n", + "print(\"Least thickness of soap film for bright fringe = %.3e meter \"%t)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Least thickness of soap film for bright fringe = 3.927e-07 meter \n" + ] + } + ], + "prompt_number": 104 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.25, Page 1.48" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import cos,pi,sin\n", + "\n", + "# Given That\n", + "lambda1 = 6.1e-7 # max. wavelength of light in meter\n", + "lambda2 = 6e-7 # min. wavelength of light in meter\n", + "Mu = 1.333 # refractive index of film \n", + "i = pi / 4 # incidence angle in radian \n", + "\n", + "#Calculation\n", + "r = asin(sin(i) / Mu) # calculation for angle of refraction\n", + "n = lambda2 / (lambda1 - lambda2) # calculation for no. of bright band\n", + "t = (n * lambda1) / (2 * Mu * cos(r)) # calculation for thickness of the film\n", + "\n", + "#Result\n", + "print(\"Thickness of the film = %.2e meter \"%t)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thickness of the film = 1.62e-05 meter \n" + ] + } + ], + "prompt_number": 105 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.26, Page 1.49" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import cos\n", + "\n", + "# Given \n", + "l = 6e-7 # wavelength of light in meter\n", + "Mu = 1.463 # refractive index of soap film \n", + "i = 0 # incidence angle in radian \n", + "r = 0 # refracted angle in radian\n", + "n = 1 # for smallest thickness\n", + "\n", + "#Calculation\n", + "t = ((2 * n - 1) * l) / (4 * Mu * cos(r)) # calculation for thickness of soap film\n", + "\n", + "#Result\n", + "print(\"\\n Least thickness of soap film for bright fringe = %.3e meter. \"%t)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " Least thickness of soap film for bright fringe = 1.025e-07 meter. \n" + ] + } + ], + "prompt_number": 106 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.27, Page 1.49" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import cos,sin,asin,pi,degrees,radians\n", + "\n", + "# Given \n", + "lambda1 = 6.1e-7 # max. wavelength of light in meter\n", + "lambda2 = 6e-7 # min. wavelength of light in meter\n", + "Mu = 4. / 3 # refractive index of film \n", + "i = asin(4. / 5) # incidence angle in radian \n", + "\n", + "#Calculation\n", + "r = asin(sin(i) / Mu)*180/pi # calculation for angle of refraction\n", + "n = lambda2 / (lambda1 - lambda2) # calculation for order of fringe\n", + "t = (n * lambda1) / (2 * Mu * cos(r*pi/180)) # calculation for thickness of film\n", + "\n", + "#Result\n", + "print(\"Thickness of the film = %.3e mm. \"%(t*1000))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thickness of the film = 1.716e-02 mm. \n" + ] + } + ], + "prompt_number": 127 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.28, Page 1.50" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "l = 5.893e-7 # wavelenth of light in meter\n", + "n = 20 # no. of interference fringes are observed \n", + "Mu = 1 # refractive index of air\n", + "i = 0 # incidence angle in radian \n", + "r = 0 # refracted angle in radian \n", + "\n", + "#Calculation\n", + "t = (n * l) / (2 * Mu) # calculation for thickness of fringe\n", + "\n", + "#Result\n", + "print(\"Thickness of wire = %.3e mm. \"%(t*1000))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thickness of wire = 5.893e-03 mm. \n" + ] + } + ], + "prompt_number": 128 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.29, Page 1.50" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "lamda = 6e-7 # wavelength of light in meter\n", + "Mu = 1 # refractive index of air film\n", + "l = 0.06e-3 # diameter of wire in meter\n", + "L = 0.15 # distance of wire from edge in meter\n", + "\n", + "#Calculation\n", + "theta = l / L #calculation for theta\n", + "fringe_width = (lamda * L)/ (2 * Mu * l) # calculation for fringe width\n", + "\n", + "#Result\n", + "print(\"Fringe width = %.1e mm.\"%fringe_width)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Fringe width = 7.5e-04 mm.\n" + ] + } + ], + "prompt_number": 140 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.30, Page 1.51" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "l = 4.56e-7 # wavelength of light in meter\n", + "theta = 1.9e-4 # angle of wedge in radian \n", + "Mu = 1 # refractive index of air\n", + "\n", + "#Calculation\n", + "fringe_width = l / (2 * Mu * theta)# calculation for fringe width \n", + "\n", + "#Result\n", + "print(\"Fringe width = %.1f mm.\"%(fringe_width*1000))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Fringe width = 1.2 mm.\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.31, Page 1.51" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "lamda = 6e-7 # wavelength of light in meter\n", + "Mu = 1 # refractive index of air film\n", + "l = 0.03*10**-3 # diameter of wire in meter\n", + "L = 0.15 # distance of wire from edge in meter\n", + "i = 0 # incidence angle in radian \n", + "r = 0 # refracted angle in radian \n", + "\n", + "#Calculation\n", + "theta = l / L # calculation for theta\n", + "fringe_width = lamda / (2 * Mu * theta) # calculation for fringe width \n", + "\n", + "#Result\n", + "print(\"Fringe width = %.1f mm.\"%(fringe_width*1000))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Fringe width = 1.5 mm.\n" + ] + } + ], + "prompt_number": 141 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.32, Page 1.51" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "l = 5.890e-7 # wavelength of light in meter\n", + "theta = 1e-2 # angle of wedge in radian \n", + "n = 12 # no. of dark fringe\n", + "Mu = 1 # refractive index of air\n", + "i = 0 # incidence angle in radian\n", + "r = 0 # refracted angle in radian\n", + "\n", + "#Calculatiom\n", + "x = ( n * l) / (2 * theta) # calculation for distance\n", + "\n", + "#Result\n", + "print(\"Distance = %.2f mm. \"%(x*1000))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Distance = 0.35 mm. \n" + ] + } + ], + "prompt_number": 142 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.33, Page 1.52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "# Given \n", + "l = 5.5e-7 # wavelength of light in meter\n", + "w = 2e-5 # fringe width in meter\n", + "Mu = 1.5 # refractive index of film\n", + "i = 0 # incidence angle in radian\n", + "r = 0 # refracted angle in radian\n", + "\n", + "#Calculation\n", + "theta = l / (2 * Mu * w)# calculation for the angle of the film\n", + "\n", + "#Result\n", + "print(\"Angle of wedge = %.3f degree. \"%(theta * 180/pi))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Angle of wedge = 0.525 degree. \n" + ] + } + ], + "prompt_number": 143 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + " Example 1.34, Page 1.52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "d1 = 5.9e-3 # diameter of 15th ring in meter\n", + "d2 = 3.36e-3 # diameter of 5th ring in meter\n", + "R = 1 # radius of the plano-convex lens in meter\n", + "\n", + "#Calculations\n", + "p = 15 - 5\n", + "l = ((d1**2) - (d2**2)) / (4 * p * R) # calculation for wavelength of light\n", + "\n", + "#Result\n", + "print(\"Wavelength of light = %.f A.\"%(l*1e10))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of light = 5880 A.\n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.35, Page 1.52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "d1 = 2e-3 # diameter of 10th ring in meter\n", + "d2 = 3e-3 # diameter of 20th ring in meter\n", + "f = 0.9 # focal length of the plano-convex lens in meter\n", + "mu = 1.5 # refractive index of lens\n", + "\n", + "#Calculations\n", + "p = 20 - 10\n", + "R = (f * (mu - 1)) # calculation for radius of convex surface of lens\n", + "l = ((d2**2) - (d1**2)) / (4 * p * R)\n", + "\n", + "#Result\n", + "print(\"Wavelength of light = %.f nm.\"%(l*1e9))\n", + "#Incorrect answer in the textbook\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of light = 2778 nm.\n" + ] + } + ], + "prompt_number": 145 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.36, Page 1.53" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt \n", + "\n", + "# Given \n", + "l = 5.896e-7 # wavelength of light in meter\n", + "f = 1 # focal length of the plano-convex lens in meter\n", + "mu = 1.5 # refractive index of lens \n", + "n = 7 # no. of bright ring\n", + "\n", + "#Calculations\n", + "p = 20 - 10\n", + "R = (f * (mu - 1)) * 2 # calculation for radius of lens\n", + "D = sqrt(4 * n * l * R) # calculation for diameter of 7th ring \n", + "\n", + "#Result\n", + "print(\"Diameter of 7th bright ring = %.3e meter.\"%(D))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diameter of 7th bright ring = 4.063e-03 meter.\n" + ] + } + ], + "prompt_number": 146 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.37, Page 1.53" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "# Given That\n", + "lambda1 = 6e-7 # wavelength of first light in meter\n", + "lambda2 = 4.8e-7 # wavelength of second light in meter\n", + "r = 0.96 # radius of curvature of curved surface of lens in meter\n", + "\n", + "#Calculations\n", + "n = lambda2 / (lambda1 - lambda2) # calculation for order of fringe\n", + "D = sqrt(4 * (n + 1) * lambda2 * r) # calculation for diameter of ring\n", + "\n", + "#Result\n", + "print(\"Diameter of (n +1)th dark ring of lambda2. = %.2e meter.\"%D)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diameter of (n +1)th dark ring of lambda2. = 3.04e-03 meter.\n" + ] + } + ], + "prompt_number": 147 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.38, Page 1.54" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "# Given \n", + "lambda1 = 6e-7 # wavelength of first light in meter\n", + "lambda2 = 5.9e-7 # wavelength of second light in meter\n", + "r = 0.9 # radius of curvature of curved surface of lens in meter\n", + "\n", + "#Calculations\n", + "n = lambda2 / (lambda1 - lambda2) # calculation for order of ring\n", + "D = sqrt(4 * (n + 1) * lambda1 * r) # calculation for diameter of ring\n", + "\n", + "#Result\n", + "print(\"Diameter of nth dark ring of lambda1 = %.4f meter.\"%D)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diameter of nth dark ring of lambda1 = 0.0114 meter.\n" + ] + } + ], + "prompt_number": 148 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.39, Page 1.54" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "l = 5.896e-7 # wavelength of light in meter\n", + "D = 4e-3 # diameter of 7th brighter fringe in m\n", + "R = 1 # radius of curvature in m\n", + "n = 7 # for seventh brighter fringe\n", + "\n", + "#Calculation\n", + "mu = 2*(2*n-1)*l*R / D**2 # calculation for refractive index of liquid\n", + "\n", + "#Result\n", + "print(\"Refractive index of liquid = %.2f.\"%mu)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Refractive index of liquid = 0.96.\n" + ] + } + ], + "prompt_number": 149 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.40, Page 1.54" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "D1 = 3e-3 # diameter of nth dark fringe when liquid is absent between the lens and the plate in m\n", + "D2 = 2.5e-3 # diameter of nth dark fringe when liquid is introduced between the lens and the plate in m\n", + "c = 3e8 # velocity of light in vacuum in m/sed\n", + "\n", + "#Calculations\n", + "mu = D1**2 / D2**2# calculation for refractive index\n", + "v = 3e8 / mu # calculation for velocity of light \n", + "\n", + "#Result\n", + "print(\"Refractive index of liquid = %.2f.\\n velocity of light in the liquid = %.2e m/sec.\"%(mu,v))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Refractive index of liquid = 1.44.\n", + " velocity of light in the liquid = 2.08e+08 m/sec.\n" + ] + } + ], + "prompt_number": 150 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.41, Page 1.55" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "l = 5.896e-7 # wavelength of light in meter\n", + "D = 5.1e-3 # diameter of 16th brighter fringe in m\n", + "R = 1 # radius of curvature in m\n", + "n = 16 # for sixteenth brighter fringe\n", + "\n", + "#Calculation\n", + "mu = 4*n*l*R / D**2 # calculation for refractive index of liquid\n", + "\n", + "#Result\n", + "print(\"Refractive index of liquid = %.2f\"%mu)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Refractive index of liquid = 1.45\n" + ] + } + ], + "prompt_number": 170 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.42, Page 1.55" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "# Given \n", + "l = 6.3e-7 # wavelength of light in meter\n", + "mu = 1.63 # refractive index of liquid \n", + "R = 0.9 # the radius of curvature of convex lens in meter\n", + "\n", + "#Calculation\n", + "r = sqrt(l*R/mu) # calculation for the radius of smallest dark ring\n", + "\n", + "#Result\n", + "print(\"The radius of smallest dark ring = %.2f mm.\"%(r*1000))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The radius of smallest dark ring = 0.59 mm.\n" + ] + } + ], + "prompt_number": 171 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.43, Page 1.55" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "r = 10./7 # ratio of nth ring diameter for two media\n", + "\n", + "#Calculation\n", + "R = (1/r)**2 # calculation for the ratio of refractive index of media\n", + "\n", + "#Result\n", + "print(\"The ratio refractive index of media = %.f:100.\"%(R*100))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The ratio refractive index of media = 49:100.\n" + ] + } + ], + "prompt_number": 173 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.44, Page 1.56" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "R = 0.9 # radius of curvature of the lower face of the lens in meter\n", + "D = 4.8e-3 # diameter of the 10th dark ring in meter\n", + "n = 10 # for 10th dark ring\n", + "\n", + "#Calculation\n", + "l = D**2 / (4 * n * R) # calculation for wavelength of light\n", + "\n", + "#Result\n", + "print(\"Wavelength of light = %.f A.\"%(l * 1e10))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of light = 6400 A.\n" + ] + } + ], + "prompt_number": 174 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.45, Page 1.56" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "r = 1./2 # ratio of 5th ring diameter \n", + "\n", + "#Calculatio\n", + "R = (1/r)**2 # calculation for refractive index of liquid\n", + "\n", + "#Result\n", + "print(\"Refractive index of liquid = %.f. \"%R)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Refractive index of liquid = 4. \n" + ] + } + ], + "prompt_number": 175 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.46, Page 1.56" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "# Given That\n", + "R = 1 # radius of curvature of lens of both side in meter\n", + "l = 5.4e-7 # wavelength of monochromatic light in meter\n", + "n1 = 5 # for 5th dark ring\n", + "n2 = 15 # for 10th dark ring\n", + "\n", + "#Calculation\n", + "r1 = sqrt((n1*l)/(1/R + 1/R)) # calculation for radius of 5th dark ring\n", + "r2 = sqrt((n2*l)/(1/R + 1/R)) # calculation for radius of 15th dark ring\n", + "d = r2 - r1 # calculation for distance between 5th and 15th dark ring\n", + "\n", + "#Result\n", + "print(\"Distance between 5th and 15th dark ring = %.3f cm.\"%(d * 100))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Distance between 5th and 15th dark ring = 0.085 cm.\n" + ] + } + ], + "prompt_number": 177 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.47, Page 1.57" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "x = 2.5e-5 # distance moved by movable mirror in meter\n", + "t = 5e-5 # thickness of mica sheet in meter\n", + "\n", + "#Calculation\n", + "mu = x / t + 1 # calculation for refractive index of mica\n", + "\n", + "#Result\n", + "print(\"Refractive index of mica = %.1f\"%mu)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Refractive index of mica = 1.5\n" + ] + } + ], + "prompt_number": 179 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.48, Page 1.57" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "x = 6e-5 # distance moved by movable mirror in meter\n", + "N = 200 # no. of fringes crossed the field of view \n", + "\n", + "#Calculation\n", + "l = (2 * x) / N # calculation for wavelength of light\n", + "\n", + "#Result\n", + "print(\"Wavelength of light = %.f A.\"%(l * 1e10))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of light = 6000 A.\n" + ] + } + ], + "prompt_number": 180 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.49, Page 1.57" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "n = 50 # no. of bands crosses the line of observation \n", + "l = 5.896e-7 # wavelength of light in meter\n", + "mu = 1.4 # refractive index \n", + "\n", + "#Calculation\n", + "t = n*l / (2*(mu-1)) # calculation for thickness of the plate\n", + "\n", + "#Result\n", + "print(\"Thickness of the plate = %.2e m.\"%t)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thickness of the plate = 3.69e-05 m.\n" + ] + } + ], + "prompt_number": 181 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.50, Page 1.57" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "n = 50 # no. of bands crosses the line of observation \n", + "lambda1 = 5.896e-7 # max. wavelength of light in meter\n", + "lambda2 = 5.89e-7 # min. wavelength of light in meter\n", + "\n", + "#Calculation\n", + "x = lambda1 * lambda2 /(lambda1 - lambda2) # calculation for the path difference\n", + "\n", + "#Result\n", + "print(\"The path difference = %.4f mm.\"%(x*10**3))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The path difference = 0.5788 mm.\n" + ] + } + ], + "prompt_number": 182 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.51, Page 1.58" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "x = 2.948e-5 # distance moved by movable mirror in meter\n", + "n = 100 # no. of fringes cross the field of view \n", + "\n", + "#Calculation\n", + "l = 2*x/n # calculation for wavelength of monochromatic light\n", + "\n", + "#Result\n", + "print(\"Wavelength of monochromatic light = %.f A.\"%(l * 1e10))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of monochromatic light = 5896 A.\n" + ] + } + ], + "prompt_number": 183 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.52, Page 1.58" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "lambda1 = 5.896e-7 # max. wavelength of light in meter\n", + "lambda2 = 5.89e-7 # min. wavelength of light in meter\n", + "\n", + "#Calculation\n", + "x = lambda1 * lambda2 /(2*(lambda1 - lambda2)) # calculation for the path difference\n", + "\n", + "#Result\n", + "print(\"The distance through which the movable mirror is move = %.3f mm.\"%(x*10**3))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The distance through which the movable mirror is move = 0.289 mm.\n" + ] + } + ], + "prompt_number": 184 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.53, Page 1.58" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "x = 2.945e-4 # distance moved by movable mirror in meter\n", + "l = 5.893e-7 # mean wavelength of light in meter\n", + "\n", + "#Calculation\n", + "delta_lambda = l**2 / (2*x) # calculation for difference between two wavelengths\n", + "\n", + "#Result\n", + "print(\"Difference between two wavelengths = %.3f A.\"%(delta_lambda*1e10))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Difference between two wavelengths = 5.896 A.\n" + ] + } + ], + "prompt_number": 185 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.54, Page 1.58" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "n = 140 # no. of shift in fringe\n", + "l = 5.46e-7 # wavelength of light in meter\n", + "t = 0.2 # length of tube in meter\n", + "\n", + "#Calculation\n", + "mu = (n*l)/(2*t) + 1 # calculation for refractive index of gas\n", + "\n", + "#Result\n", + "print(\"Refractive index of gas = %.5f\"%mu)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Refractive index of gas = 1.00019\n" + ] + } + ], + "prompt_number": 188 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter_10.ipynb b/Engineering_Physics/Chapter_10.ipynb new file mode 100755 index 00000000..39a354b1 --- /dev/null +++ b/Engineering_Physics/Chapter_10.ipynb @@ -0,0 +1,613 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 10: Electromagnetism" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.15, Page 10.42" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "n = 2000 # flux lines enter in given volume in Vm\n", + "n_ = 4000 # flux lines diverge from given volume in Vm\n", + "e0 = 8.85e-12 # permittivity of space\n", + "\n", + "#Calculations\n", + "fi = n_ - n\n", + "q = e0 * fi\n", + "\n", + "#Result\n", + "print 'The total charge within volume(in C) = ',q" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The total charge within volume(in C) = 1.77e-08\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.16, Page 10.42" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "n = 20000 # flux lines entering in given volume in Vm\n", + "n_ = 45000 # flux lines entering out from given volume in Vm\n", + "e0 = 8.85e-12 # permittivity of space\n", + "\n", + "#Calculations\n", + "fi = n_ - n\n", + "q = e0 * fi\n", + "\n", + "#Result\n", + "print \"The total charge enclosed by closed surface is %.3e C\"%q" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The total charge enclosed by closed surface is 2.212e-07 C\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.17, Page 10.43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "q = 13.5e-6 # charge enclosed at the centre of cube in C\n", + "l = 6 # length of the side of cube in cm\n", + "e0 = 8.85e-12 # permittivity of space\n", + "\n", + "#Calculations\n", + "fi = q / e0\n", + "fi_ = fi / 6\n", + "q = e0 * fi\n", + "\n", + "#Results\n", + "print \"Electric flux through the whole volume of the cube is %.3e Nm^2/C\\nElectric flux through one face of the cube is %.2e Nm^2/C\"%(fi,fi_)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Electric flux through the whole volume of the cube is 1.525e+06 Nm^2/C\n", + "Electric flux through one face of the cube is 2.54e+05 Nm^2/C\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.18, Page 10.43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "q = 11 # charge enclosed at the centre of cube in C\n", + "l = 5 # length of the side of cube in cm\n", + "e0 = 8.85e-12 # permittivity of space\n", + "\n", + "#Calculation\n", + "fi_ = (q / e0) / 6\n", + "\n", + "#Result\n", + "print \" Electric flux through each surface of the cube = %.2e Nm^2/C\"%fi_" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Electric flux through each surface of the cube = 2.07e+11 Nm^2/C\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.19, Page 10.43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "# Given \n", + "q = 1e-8 # charge uniformly spread over metallic sphere in C\n", + "r = .1 #radius of sphere in m\n", + "d = 7 # distance of a point from centre of the sphere in cm\n", + "d_ = .5 # distance of another point from centre of the sphere in m\n", + "e0 = 8.85e-12 # permittivity of space\n", + "\n", + "#calculations\n", + "E1 = (1 / (4 * pi * e0) * (q / r**2))\n", + "E2 = 0 #because sphere is metallic\n", + "E3 = (1 / (4 * pi * e0) * (q / d_**2))\n", + "\n", + "#Result\n", + "print \"Electric field intensity-\\n(1) On the surface of the sphere = %.e N/C\\n(2) At first point = %d N/C\\n(3)At second point = %.2e N/C\"%(E1,E2,E3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Electric field intensity-\n", + "(1) On the surface of the sphere = 9e+03 N/C\n", + "(2) At first point = 0 N/C\n", + "(3)At second point = 3.60e+02 N/C\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.20, Page 10.44" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "# Given \n", + "q = 1.6e-19 # charge on a proton in C\n", + "d = 1e-10 # distance of a point from proton in m\n", + "e0 = 8.85e-12 # permittivity of space\n", + "\n", + "#Calculation\n", + "E = (1 / (4 * pi * e0)) * (q / d**2)\n", + "\n", + "#Result\n", + "print \"Electric field = %.2e V/m\"%E" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Electric field = 1.44e+11 V/m\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.21, Page 10.44" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "v = 1000 # potential through which alpha particle accelerated in V\n", + "q = 3.2e-19 # charge on an alpha particle in C\n", + "e0 = 8.85e-12 # electric permittivity of space\n", + "\n", + "#calculation\n", + "E = q * v\n", + "\n", + "#Result\n", + "print \"Energy gained by alpha particle = %.1e J\"%E" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy gained by alpha particle = 3.2e-16 J\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.22, Page 10.44" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "# Given \n", + "q = 1.6e-19 # charge on a proton in C\n", + "d = 1e-10 # distance of a point from proton in m\n", + "d_ = 2e-11 # distance of another point from proton in m\n", + "e0 = 8.85e-12 # permittivity of space\n", + "\n", + "#calculations\n", + "v = (1 / (4 * pi * e0)) * (q / d)#calculation for potential at first point\n", + "E = -q * v#calculation for energy at first point in J\n", + "delta_v = (1 / (4 * pi * e0)) * q * ((1 / d_) - (1 / d))#calculation for potential difference between points\n", + "\n", + "#Result\n", + "print \"Potential energy at first point = %.1f eV\\nPotential difference between points = %.1f V\"%(E/q,delta_v)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Potential energy at first point = -14.4 eV\n", + "Potential difference between points = 57.5 V\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.23, Page 10.45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi, ceil\n", + "\n", + "# Given \n", + "q = 1.5e-6 # charge in C\n", + "v = 30 # potential of a surface in V\n", + "e0 = 8.85e-12 # permittivity of space\n", + "\n", + "#Calculation\n", + "r = (1 / (4 * pi * e0)) * (q / v)\n", + "\n", + "#Result\n", + "print \"Radius of equipotential surface = %d m\"%ceil(r)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Radius of equipotential surface = 450 m\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.24, Page 10.45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "# Given \n", + "p = 3.8e26 # power radiated by sun in W\n", + "r = 7e8 # radius of sun in m\n", + "e0 = 8.85e-12 # permittivity of space\n", + "\n", + "#Calculation\n", + "s = p / (4 * pi * r**2)\n", + "\n", + "#Result\n", + "print \"The value of poynting vector at the surface of the sun = %.3e W/m^2\"%s" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of poynting vector at the surface of the sun = 6.171e+07 W/m^2\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.25, Page 10.45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "Se = (2*4.2)/(60*10**-4) #J/m^2-sec\n", + "c = 3*10**8 #m/s\n", + "res = 1.4*10**11 #m\n", + "rs = 7*10**7 #W/m^2\n", + "\n", + "#Calculations\n", + "Prad_e = Se/c\n", + "Ss = Se*((res/rs)**2)\n", + "Prad_s = Ss/c\n", + "\n", + "#Results\n", + "print \"Radiation pressure at the surface of the earth =%.2e N/m^2\"%Prad_e\n", + "print \"Radiation pressure at the surface of the sun =%.3e N/m^2\"%Prad_s" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Radiation pressure at the surface of the earth =4.67e-06 N/m^2\n", + "Radiation pressure at the surface of the sun =1.867e+01 N/m^2\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.28, Page 10.47" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "# Given \n", + "s = 2 # energy received by the earth in cal/cm^2.min\n", + "e0 = 8.85e-12 # electric permittivity of space\n", + "mu0 = 1.2567e-6 # magnetic permittivity of space\n", + "c = 3e8 # speed of light in meter/sec\n", + "\n", + "#calculations\n", + "r = sqrt(mu0 / e0)\n", + "P = s*4.2/(60*1e-4)\n", + "E = sqrt(P*r)\n", + "H = E/r\n", + "\n", + "#Result\n", + "print \"Magnitude of electric field vector = %.1f v/m\\nMagnitude of magnetic field vector = %.3f A/m\"%(E * sqrt(2),H*sqrt(2))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Magnitude of electric field vector = 1027.2 v/m\n", + "Magnitude of magnetic field vector = 2.726 A/m\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.29, Page 10.48" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "# Given \n", + "H = 1 # magnitude of magnetic field vector A/m\n", + "e0 = 8.85e-12 # electric permittivity of space\n", + "mu0 = 1.2567e-6 # magnetic permittivity of space\n", + "c = 3e8 # speed of light in meter/sec\n", + "\n", + "#Calculations\n", + "r = sqrt(mu0 / e0) # ratio of E,H\n", + "E = H * r\n", + "\n", + "#Result\n", + "print \"Magnitude of electric field vector = %.2f v/m.\"%E" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Magnitude of electric field vector = 376.83 v/m.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.31, Page 10.48" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt, pi\n", + "\n", + "# Given \n", + "p = 1000 # power of lamp in W\n", + "d = 2 # distance of a point from lamp in meter\n", + "e0 = 8.85e-12 # electric permittivity of space\n", + "mu0 = 1.2567e-6 # magnetic permittivity of space\n", + "c = 3e8 # speed of light in meter/sec\n", + "\n", + "#Calculations\n", + "s = p / (4 * pi * d**2) #calculation for \n", + "r = sqrt(mu0 / e0) # ratio of E,H\n", + "E = sqrt(s * r)#calculation for average value of intensity of electric field of radiation\n", + "\n", + "#Result\n", + "print \"Average value of the intensity of electric field of radiation = %.2f v/m.\"%E" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Average value of the intensity of electric field of radiation = 86.58 v/m.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.32, Page 10.49" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "# Given \n", + "k = 81 # relative permittivity of water \n", + "c = 3e8 # speed of light in meter/sec\n", + "\n", + "#Calculations\n", + "mu = sqrt(k)\n", + "v = c / mu\n", + "\n", + "#Result\n", + "print \"Refractive index of distilled water is %d \\nSpeed of light in water is %.2e m/sec\"%(mu,v)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Refractive index of distilled water is 9 \n", + "Speed of light in water is 3.33e+07 m/sec\n" + ] + } + ], + "prompt_number": 5 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter_11.ipynb b/Engineering_Physics/Chapter_11.ipynb new file mode 100755 index 00000000..0cd9f3f6 --- /dev/null +++ b/Engineering_Physics/Chapter_11.ipynb @@ -0,0 +1,1203 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 11: Theory of Relativity" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.5, Page 11.19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "# Given \n", + "x = 100 # in meter\n", + "y = 10 # in meter\n", + "z = 5 # in meter\n", + "t = 1e-4 # in sec\n", + "# coordinates of point in frame F\n", + "v = 2.7e8 # velocity of frame F_ w.r.t. frame F in m/sed\n", + "c=3e8 # speed of light in m/sec\n", + "\n", + "#Calculations\n", + "# according to Galilean transformation\n", + "x__ = x-v*t\n", + "y__=y\n", + "z__=z\n", + "t__=t\n", + "# according to Lorentz transformation\n", + "x_ = (x-v*t)/sqrt(1-(v/c)**2)\n", + "y_=y\n", + "z_=z\n", + "t_=(t-(v*x/c**2))/sqrt(1-(v/c)**2)\n", + "\n", + "#Results\n", + "print \"Coordinate of the event in reference frame F_ using \\n(a)Galilean transformation-x=%.f m, y=%.f m, z = %.f m, t = %.e sec\"%(x__,y__,z__,t__) \n", + "print \"(b)Lorentz transformation-x=%.f m,y =%.f m, z = %.f m, t=%.e sec \"%(x_,y_,z_,t_)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Coordinate of the event in reference frame F_ using \n", + "(a)Galilean transformation-x=-26900 m, y=10 m, z = 5 m, t = 1e-04 sec\n", + "(b)Lorentz transformation-x=-61713 m,y =10 m, z = 5 m, t=2e-04 sec \n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.6, Page 11.19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "r = 4 # ratio of mass of particle to the rest mass\n", + "\n", + "#Calculations\n", + "v = 3e8 * sqrt(1 - (1 / r)**2)\n", + "\n", + "#Result\n", + "print \"Speed of particle is %.e meter/sec\"%v" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Speed of particle is 3e+08 meter/sec\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.7, Page 11.20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "r = 1.2 # ratio of mass of particle to the rest mass\n", + "\n", + "#Calculations\n", + "v = 3e8 * sqrt(1 - (1 / r)**2)\n", + "\n", + "#Results\n", + "print \"Speed of particle is %.3f c\"%(v/3e8)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Speed of particle is 0.553 c\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.9, Page 11.20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "E = 2 # kinetic energy of electron in Mev\n", + "\n", + "#Calculations\n", + "v = 3e8 * sqrt(1 - (1 / (1 + (1.6e-19 * 2e6) / (9.1e-31 * 3e8**2)))**2)\n", + "m = (9.1e-31) / sqrt(1 - (v / 3e8)**2)\n", + "\n", + "#Results\n", + "print \"Speed of electron is %.2f c\\nMass of electron is %.2e kg\"%(v/3e8,m)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Speed of electron is 0.98 c\n", + "Mass of electron is 4.47e-30 kg\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.11, Page 11.21" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "r = 2.25 # ratio of mass of particle to the rest mass\n", + "\n", + "#Calculations\n", + "v = 3e8 * sqrt(1 - (1 / r)**2)\n", + "\n", + "#Result\n", + "print \"Speed of particle is %.2e meter/sec\"%v" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Speed of particle is 2.69e+08 meter/sec\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.12, Page 11.22" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "r = 2. # ratio of kinetic energy of body to its rest mass\n", + "\n", + "#Calculations\n", + "v = 3e8 * sqrt(1 - (1 / (r + 1))**2)\n", + "\n", + "#Result\n", + "print \"Velocity of particle is %.2f c\"%(v/3e8)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Velocity of particle is 0.94 c\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.13, Page 11.22" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "r = 11. # ratio of mass of particle to the rest mass of electron\n", + "\n", + "#Calculations\n", + "KE = (r - 1) * 9.1e-31 * 3e8**2 / (1.6e-19)\n", + "m = 3e8 * sqrt(1 - (1 / r)**2) * 9.1e-31 * 11\n", + "\n", + "#Result\n", + "print \"Kinetic energy of electron is %.1f MeV\\nMomentum of particle is %.2e N-sec\"%(KE/1e6,m)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Kinetic energy of electron is 5.1 MeV\n", + "Momentum of particle is 2.99e-21 N-sec\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.14, Page 11.23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "m = 1.67e-27 # mass of electron in kg\n", + "m_ = 9.1e-31 # rest mass of electron in kg\n", + "\n", + "#Calculations\n", + "v = 3e8 * sqrt(1 - (m_ / m)**2)\n", + "\n", + "#Result\n", + "print \"Velocity of electron is %.3e meter/sec\"%v" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Velocity of electron is 3.000e+08 meter/sec\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.15, Page 11.23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "E = 0.1 # kinetic speed of electron in Mev\n", + "\n", + "#Calculations\n", + "v_ = sqrt(2 * (E * 1.6e-13) / 9.1e-31)\n", + "v = 3e8 * sqrt(1 - (1 / (1 + (1.6e-13 * E) / (9.1e-31 * 3e8**2)))**2)\n", + "\n", + "#Results\n", + "print \"Speed according to classical mechanics is %.2e meter/sec\\nSpeed according to relativistic mechanics is %.2e meter/sec\"%(v_,v)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Speed according to classical mechanics is 1.88e+08 meter/sec\n", + "Speed according to relativistic mechanics is 1.64e+08 meter/sec\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.17, Page 11.24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "r1 = 3. # ratio of kinetic energy of body to its rest mass in first case\n", + "r2 = 1. # ratio of kinetic energy of body to its rest mass in second case\n", + "\n", + "#Calculations\n", + "v1 = 3e8 * sqrt(1 - (1 / (r1 + 1))**2)\n", + "v2 = 3e8 * sqrt(1 - (1 / (r2 + 1))**2)\n", + "\n", + "#Results\n", + "print \"Velocity of particle if kinetic energy is %d times of rest mass energy is %.2e meter/sec\\nVelocity of particle if kinetic energy is %.1e times of rest mass energy is %e meter/sec\"%(r1,v1,r2,v2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Velocity of particle if kinetic energy is 3 times of rest mass energy is 2.90e+08 meter/sec\n", + "Velocity of particle if kinetic energy is 1.0e+00 times of rest mass energy is 2.598076e+08 meter/sec\n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.19, Page 11.25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "E = 1.5 # kinetic energy of electron in Mev\n", + "\n", + "#Calculations\n", + "v = 3e8 * sqrt(1 - (1 / (1 + (1.6e-19 * 2e6) / (9.1e-31 * 3e8**2)))**2)\n", + "m = (E * 1.6e-13 / (3e8)**2) + 9.1e-31\n", + "\n", + "#Results\n", + "print \"Speed of electron is %.2e meter/sec\\nMass of electron is %.1e kg\"%(v,m)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Speed of electron is 2.94e+08 meter/sec\n", + "Mass of electron is 3.6e-30 kg\n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.20, Page 11.26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "r = 3. / 2 # ratio of mass of metre stick to the rest mass of metre stick\n", + "\n", + "#Calculations\n", + "l = 1 * (1. / r)\n", + "\n", + "#Result\n", + "print \"Length of meter stick is %.2f meter if mass is %.2f times of its rest mass\"%(l,r)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Length of meter stick is 0.67 meter if mass is 1.50 times of its rest mass\n" + ] + } + ], + "prompt_number": 43 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.21, Page 11.26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "# Given \n", + "r = 1. / 2 # ratio of area of circular lamina in frame S_ to the ratio of area of circular lamina in frame S\n", + "\n", + "#Calculations\n", + "v = 3e8 * sqrt(1 - r**2)\n", + "\n", + "#Result\n", + "print \"Velocity of frame S_ w.r.t. frame S is %.1e meter/sec\"%v" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Velocity of frame S_ w.r.t. frame S is 2.6e+08 meter/sec\n" + ] + } + ], + "prompt_number": 45 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.22, Page 11.27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import * \n", + "# Given \n", + "t = 1 # lose in time in an hour in minute\n", + "\n", + "#Calculations\n", + "v = 3e8 * sqrt(1 - ((60. - t) / 60)**2)\n", + "\n", + "#Result\n", + "print \"Speed of clock is %.2e meter/sec\"%v" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Speed of clock is 5.45e+07 meter/sec\n" + ] + } + ], + "prompt_number": 47 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.23, Page 11.27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "# Given \n", + "t_ = 2.5e-8 # proper life of pi+ mesons in sec\n", + "v = 2.4e8 # velocity of beam of mesons in m/sec\n", + "r = 1. / exp(2) # ratio of final flux to initial flux of the meson beam\n", + "\n", + "#Calculations\n", + "t = t_ / sqrt(1 - (v / 3e8)**2)\n", + "T = t * log(1. / r)\n", + "d = T * v\n", + "\n", + "#Result\n", + "print \"Distance travel by the beam is %.2f meter\"%d" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Distance travel by the beam is 20.00 meter\n" + ] + } + ], + "prompt_number": 48 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.24, Page 11.27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "v = 1.8e8 # velocity of space ship away from the earth in m/sec\n", + "v1 = 2.1e8 # velocity of rocket w.r.t. space ship away from the earth in first case in m/sec\n", + "v2 = -2.1e8 # velocity of rocket w.r.t. space ship away from the earth in second case in m/sec\n", + "\n", + "#Calculations\n", + "u1 = (v1 + v) / (1 + ((v1 * v) / (3e8)**2))\n", + "u2 = (v2 + v) / (1 + ((v2 * v) / (3e8)**2))\n", + "\n", + "#Results\n", + "print \"Velocity of rocket w.r.t. earth in first case = %.2f c away from the earth\\nVelocity of rocket w.r.t. earth in second case = %.2f c away from the earth \"%(u1/3e8,u2/3e8)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Velocity of rocket w.r.t. earth in first case = 0.92 c away from the earth\n", + "Velocity of rocket w.r.t. earth in second case = -0.17 c away from the earth \n" + ] + } + ], + "prompt_number": 49 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.25, Page 11.28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "l = 1 # length of the rod in meter\n", + "v = 1.8e8 # speed of rod along its length in meter/sec\n", + "\n", + "#Calculations\n", + "L = l * sqrt(1- (v / 3e8)**2)\n", + "\n", + "#Result\n", + "print \"Length as it appear to the observer is %.1f meter\"%L" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Length as it appear to the observer is 0.8 meter\n" + ] + } + ], + "prompt_number": 50 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.26, Page 11.28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "l = 2 # length of the rod in meter\n", + "v = 2.7e8 # speed of rod along its length in meter/sec\n", + "\n", + "#Calculations\n", + "L = l * sqrt(1- (v / 3e8)**2)\n", + "\n", + "#Result\n", + "print \"Length as it appear to the observer is %.3f meter\"%L" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Length as it appear to the observer is 0.872 meter\n" + ] + } + ], + "prompt_number": 51 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.27, Page 11.28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "l = 100. # consider the length of the rod in meter\n", + "v = 2.4e8 # speed of rod in meter/sec\n", + "theta = pi / 3 # direction of velocity of rod along its length in radian\n", + "\n", + "#Calculations\n", + "Lx = l * cos(theta)\n", + "Ly = l * sin(theta)\n", + "L_x = Lx * sqrt(1 - (v / 3e8)**2)\n", + "L_y = Ly \n", + "L = sqrt(L_x**2 + L_y**2)\n", + "p_l = ((l - L) / l) * 100 \n", + "\n", + "#Result\n", + "print \"Percentage length contraction is %.1f percent\"%p_l" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Percentage length contraction is 8.3 percent\n" + ] + } + ], + "prompt_number": 52 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.28, Page 11.29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "# Given \n", + "r = 0.5 # ratio of length of rod when it is in motion to the length of the rod when it is in rest \n", + "\n", + "#Calculations\n", + "v = 3e8 * sqrt(1 - r**2)\n", + "\n", + "#Result\n", + "print \"Speed of the rod relative to observer is %.3f c\"%(v/3e8)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Speed of the rod relative to observer is 0.866 c\n" + ] + } + ], + "prompt_number": 53 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.29, Page 11.29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "l = 5 # length of the rod in meter\n", + "v = 1.8e8 # speed of rod in meter/sec\n", + "theta = pi / 6 # direction of velocity of rod along its length in radian\n", + "\n", + "#Calculations\n", + "Lx = l * cos(theta)\n", + "Ly = l * sin(theta)\n", + "L_x = Lx * sqrt(1 - (v / 3e8)**2)\n", + "L_y = Ly \n", + "L = sqrt(L_x**2 + L_y**2)\n", + "orientation = atan(L_y / L_x) * (180 / pi)\n", + "\n", + "#Results\n", + "print \"Length of the rod in moving frame is %.2f meter\\nOrientation of the rod is %.2f degree\"%(L,orientation)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Length of the rod in moving frame is 4.27 meter\n", + "Orientation of the rod is 35.82 degree\n" + ] + } + ], + "prompt_number": 55 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.30, Page 11.30" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "T = 17.8e-9 # half-life of prticle at rest in sec\n", + "v = 2.4e8 # speed of particle in meter/sec\n", + "\n", + "#Calculations\n", + "t = T / (sqrt(1 - (v / 3e8)**2))\n", + "\n", + "#Result\n", + "print \"New half-life of particle is %.2f nanosec\"%(t/1e-9)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "New half-life of particle is 29.67 nanosec\n" + ] + } + ], + "prompt_number": 56 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.31, Page 11.30" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "T = 24 # no. of hours in a day\n", + "v = 1e8 # speed of spaceship in meter/sec\n", + "\n", + "#Calculations\n", + "t = T * (sqrt(1 - (v / 3e8)**2))\n", + "T_ = T - t\n", + "m=(T_-1)*60\n", + "s=(m-22)*60\n", + "\n", + "#Result\n", + "print \"Time lost per day is %d hours %d minute %d sec\"%(T_,m,s)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time lost per day is 1 hours 22 minute 21 sec\n" + ] + } + ], + "prompt_number": 58 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.32, Page 11.30" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "T = 4. # no. of year when rocket is moving corresponding to one year \n", + "\n", + "#Calculations\n", + "v = 3e8 * sqrt(1 - (1 / T)**2)\n", + "\n", + "#Result\n", + "print \"Speed of rocket is %.2f c\"%(v/3e8)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Speed of rocket is 0.97 c\n" + ] + } + ], + "prompt_number": 60 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.33, Page 11.31" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "d = 4 # distance of star from the earth in light years\n", + "v = 3e8 * sqrt(0.9999) # speed of rocket in meter/sec\n", + "\n", + "#Calculations\n", + "t = (2 * d * 3e8) / v\n", + "T_ = t * sqrt(1 - (v / 3e8)**2)\n", + "\n", + "#Result\n", + "print \"Time taken by the rocket is %.2f year\"%T_" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time taken by the rocket is 0.08 year\n" + ] + } + ], + "prompt_number": 61 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.34, Page 11.31" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "t = 2e-7 # life time of particle when it is moving in sec\n", + "v = 2.8e8 # speed of particle in meter/sec\n", + "\n", + "#Calculations\n", + "T_ = t * sqrt(1 - (v / 3e8)**2)\n", + "\n", + "#Result\n", + "print \"Proper life time of particle is %.2e sec\"%T_" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Proper life time of particle is 7.18e-08 sec\n" + ] + } + ], + "prompt_number": 62 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.35, Page 11.31" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "v1 = 2.7e8 # velocity of first electron beam in meter/sec\n", + "v2 = -2.7e8 # velocity of second electron beam in meter/sec\n", + "\n", + "#Calculations\n", + "u = v1 - v2\n", + "u_ = (v1 - v2) / (1 - (v1 * v2) / (3e8)**2)\n", + "\n", + "#Results\n", + "print \"Velocity of electrons beam w.r.t. another electron beam according to Newtonian mechanics is %.1f c\\nVelocity of electrons beam measured by the observer moving with other electron beam = %.2f c\"%(u/3e8,u_/3e8)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Velocity of electrons beam w.r.t. another electron beam according to Newtonian mechanics is 1.8 c\n", + "Velocity of electrons beam measured by the observer moving with other electron beam = 0.99 c\n" + ] + } + ], + "prompt_number": 63 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.37, Page 11.32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "E = 900 # total relativistic energy of proton in Mev\n", + "m = 1.63-27 # rest mass of proton in kg\n", + "c = 3e8 # velocity of photon in meter/sec\n", + "\n", + "#Calculations\n", + "m_ = (E * 1.6e-13) / (c)**2\n", + "\n", + "#Results\n", + "print \"Relativistic mass of proton is %.2e kg\\nHere relativistic mass is same as rest mass\\n hence proton is at rest and speed and kinetic energy of proton will be zero\"%(m_)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Relativistic mass of proton is 1.60e-27 kg\n", + "Here relativistic mass is same as rest mass\n", + " hence proton is at rest and speed and kinetic energy of proton will be zero\n" + ] + } + ], + "prompt_number": 64 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.38, Page 11.32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "E = 5.4e6 # energy liberates during dynamite explosion in J/kg\n", + "c = 3e8 # velocity of photon in meter/sec\n", + "\n", + "#Calculations\n", + "E_ = 1 * c**2 # energy liberated by 1 kg content in J\n", + "f = E / E_\n", + "\n", + "#Result\n", + "print \"Fraction of total energy content in it is %.1e per kg\"%f" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Fraction of total energy content in it is 6.0e-11 per kg\n" + ] + } + ], + "prompt_number": 65 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.39, Page 11.32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import * \n", + "# Given \n", + "k = 1.02 # kinetic energy of electron in Mev\n", + "E_ = 0.51 # rest mass energy of electron in Mev\n", + "c = 3e8 # velocity of photon in meter/sec\n", + "\n", + "#Calculations\n", + "E = k + E_\n", + "v = c * sqrt(1 - (E_ / E)**2)\n", + "\n", + "#Result\n", + "print \"Speed of the electron is %.2e meter/sec\"%v" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Speed of the electron is 2.83e+08 meter/sec\n" + ] + } + ], + "prompt_number": 66 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.40, Page 11.33" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "E = 1400 # solar energy receives by the earth in W/square meter\n", + "d = 1.5e11 # distance between earth and the sun in meter\n", + "c = 3e8 # velocity of photon in meter/sec\n", + "\n", + "#Calculations\n", + "E_ = 4 * pi * d**2 * E\n", + "m = E_ / c**2\n", + "\n", + "#Result\n", + "print \"Rate of decrement of mass of the sun is %.2e kg/sec\"%m" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rate of decrement of mass of the sun is 4.40e+09 kg/sec\n" + ] + } + ], + "prompt_number": 67 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter_12.ipynb b/Engineering_Physics/Chapter_12.ipynb new file mode 100755 index 00000000..1b8c5610 --- /dev/null +++ b/Engineering_Physics/Chapter_12.ipynb @@ -0,0 +1,1613 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 12: Applied Nuclear Physics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.1, Page 12.31" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "\n", + "# Given \n", + "E = 1.14 # energy of gamma radiation in Mev\n", + "l = 0.2 # length of aluminium in meter\n", + "p = 0.03 # reduce in intensity in beam\n", + "d = 2700 # density of aluminium in kg/m^3 \n", + "\n", + "#Calculations\n", + "mu = (1 / l) * log(1 / p)\n", + "k = mu / d\n", + "x = 0.693 / mu\n", + "\n", + "#Result\n", + "print \"Mass absorption coeffiecient of Al for this radiation is %f m^2/kg\\nHalf value thickness is %.4f meter\"%(k,x)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mass absorption coeffiecient of Al for this radiation is 0.006494 m^2/kg\n", + "Half value thickness is 0.0395 meter\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.2, Page 12.32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given that\n", + "E = 1.1 # energy of gamma radiation in Mev\n", + "l = 0.25 # length of aluminium in meter\n", + "p = 0.02 # reduce in intensity in beam\n", + "d = 2700 # density of aluminium in kg/m^3 \n", + "\n", + "#Calculations\n", + "mu = (1 / l) * log(1 / p)\n", + "k = mu / d\n", + "x = 0.693 / mu\n", + "\n", + "#Result\n", + "print \"Mass attenuation coefficient of Al for this radiation is %.1e m^2/kg\\nHalf value thickness is %.4f meter\"%(k,x)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mass attenuation coefficient of Al for this radiation is 5.8e-03 m^2/kg\n", + "Half value thickness is 0.0443 meter\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.3, Page 12.32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "t = 15 # half-life for Na(23) in hours\n", + "r = 93.75 # percentage fraction of sample which decayed \n", + "\n", + "#Calculations\n", + "lamda = 0.693 / t\n", + "T = (1 / lamda) * (log(100 / (100 - r)))\n", + "\n", + "#Result\n", + "print \"Time taken for 93.75 per decay of sample is %d hours\"%T" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time taken for 93.75 per decay of sample is 60 hours\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.4, Page 12.33" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "t = 4 # half-life of radioactive element in years\n", + "r = 1. / 64 # ratio of mass of element present in specimen to the initial mass of element \n", + "\n", + "#Calculations\n", + "lamda = 0.693 / t\n", + "T = (1 / lamda) * log(1 / r)\n", + "\n", + "#Result\n", + "print \"Time after which element present in specimen reduce to 1/64 of its original value is %d years\"%T" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time after which element present in specimen reduce to 1/64 of its original value is 24 years\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.5, Page 12.33" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "t = 15 # half-life of radioactive element in years\n", + "r = 0.025 # ratio of mass of element present in specimen to the intial mass of element \n", + "\n", + "#Calculations\n", + "lamda = 0.693 / t\n", + "T = (1 / lamda) * log(1 / r)\n", + "\n", + "#result\n", + "print \"Period in which 2.5 percent of the initial quantity left over is %.2f years\"%T" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Period in which 2.5 percent of the initial quantity left over is 79.85 years\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.6, Page 12.33" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "t = 3.8 # half-life for radon in days\n", + "r = 60. # percentage fraction of sample which decayed \n", + "\n", + "#Calculations\n", + "lamda = 0.693 / t\n", + "T = (1 / lamda) * (log(100 / (100 - r)))\n", + "\n", + "#Result\n", + "print \"Time taken for 60 percent decay of sample is %.3f days\"%T" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time taken for 60 percent decay of sample is 5.024 days\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.7, Page 12.34" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "lamda = 4.28e-4 # decay constant in per year\n", + "\n", + "#Calculations\n", + "T = 0.693 / lamda\n", + "t = 1 / lamda\n", + "\n", + "#Result\n", + "print \"Half life time is %.2f years\\nMean life time is %.2f years\"%(T,t)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Half life time is 1619.16 years\n", + "Mean life time is 2336.45 years\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.8, Page 12.34" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "t = 30 # time in years\n", + "r = 1. / 64 # ratio of final mass of element to the intial mass of element \n", + "\n", + "#Calculations\n", + "lamda = log(1 / r) / t\n", + "T = 0.693 / lamda\n", + "\n", + "#Result\n", + "print \"Half life of radioactive material is %d years\"%(ceil(T))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Half life of radioactive material is 5 years\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.9, Page 12.34" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "t = 2.1 # half life in minute\n", + "r = 60 # percentage fraction of sample which decayed \n", + "\n", + "#Calculation\n", + "lamda = 0.693 / t\n", + "\n", + "#Result\n", + "print \"Decay constant is %.2f per minute\"%lamda" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Decay constant is 0.33 per minute\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.10, Page 12.35" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "t = 2.7 # half-life of Au(198) in days\n", + "m = 1e-6 # mass of sample in gm\n", + "T = 8 * 86400 # time in seconds\n", + "\n", + "#Calculations\n", + "lamda = 0.693 / (t * 86400)\n", + "N = (m * 6.023e23) / 198 # by the formula (N = mass*Avogadro number/molar mass)\n", + "A_ = lamda * N\n", + "A = A_ * (1 / exp(lamda * T))\n", + "\n", + "#Result\n", + "print \"Activity of sample is %.4e decays/sec\"%A" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Activity of sample is 1.1595e+09 decays/sec\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.11, Page 12.35" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "n = 3 # no. of half lives\n", + "\n", + "#Calculation\n", + "f = (1. / 2)**n\n", + "\n", + "#Result\n", + "print \"Fraction of sample left after %d half lives is %.3f \"%(n,f)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Fraction of sample left after 3 half lives is 0.125 \n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.12, Page 12.35" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "t = 2 # life period of radioactive substance in years\n", + "T = 4 # time in years\n", + "m = 10. # mass of substance in mg\n", + "\n", + "#Calculation\n", + "N = m / T # in mg\n", + "\n", + "#Result\n", + "print \"Substance remained unchanged after 4 years is %.1f mg\"%N" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Substance remained unchanged after 4 years is 2.5 mg\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.13, Page 12.36" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given that\n", + "m = 1. # initial mass of radium in gm\n", + "m_ = 0.0021 # final mass of radium in gm\n", + "t = 5. # time for decay from m to m_ in years\n", + "\n", + "#Calculations\n", + "lamda = log(m / (1 - m_)) / t\n", + "T = 0.693 / lamda\n", + "T_ = 1. / lamda\n", + "\n", + "#Result\n", + "print \"Decay constant is %.4f per year\\nHalf life of sample is %.1f years\\nAverage life of sample is %.3f years\"%(lamda,T,T_)\n", + "#Incorrect answer in the textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Decay constant is 0.0004 per year\n", + "Half life of sample is 1648.3 years\n", + "Average life of sample is 2378.452 years\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.14, Page 12.36" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "t = 10. # time in days\n", + "r = 15. # percentage fraction of sample which remain \n", + "\n", + "#Calculations\n", + "lamda = log(100. / 15) / t\n", + "T = 0.693 / lamda\n", + "\n", + "#Result\n", + "print \"Half life of sample is %.3f days\"%T" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Half life of sample is 3.653 days\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.15, Page 12.36" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "t = 12.3 # half life in year\n", + "T = 50 # time in year \n", + "\n", + "#Calculations\n", + "lamda = 0.693 / t\n", + "f = 1 / exp(lamda * T)\n", + "\n", + "#Result\n", + "print \"Fraction of radioactive isotope remained is %.2f \"%f" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Fraction of radioactive isotope remained is 0.06 \n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.16, Page 12.37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "R = 1 # radioactivity of Pb(214) in curie\n", + "t = 26.8 # half life in minute \n", + "\n", + "#calculations\n", + "lamda = 0.693 / (t * 60)\n", + "R = 1 * 3.7e10 # in disintegration per sec\n", + "m = (R * 214) / (6.023e23 * lamda)\n", + "\n", + "#Result\n", + "print \"Mass of Pb(214)is %.2e gm\"%m" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mass of Pb(214)is 3.05e-08 gm\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.17, Page 12.37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "R = 1e6 # radioactivity of Pb(214) in disintegrations per sec\n", + "t = 26.8 # half life in minute \n", + "\n", + "#Calculations\n", + "lamda = 0.693 / (t * 60)\n", + "m = (R * 214) / (6.023e23 * lamda)\n", + "\n", + "#Result\n", + "print \"Mass of Pb(214) is %.2e gm\"%m" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mass of Pb(214) is 8.24e-13 gm\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.18, Page 12.37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "m = 1. # mass of Ra(226) in gm \n", + "R = 1. # radioactivity of Ra(226) in curie\n", + "\n", + "#Calculations\n", + "r = R * 3.7e10 # in disintegrations per sec\n", + "N = 6.023e23 * m / 226\n", + "lamda = r * 226 / 6.023e23\n", + "T = 1. / lamda\n", + "T_ = 0.693 / lamda\n", + "\n", + "#Result\n", + "print \"Mean life of radium is %.2e year\\nHalf life of radium is %.2e year\"%(T,T_)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mean life of radium is 7.20e+10 year\n", + "Half life of radium is 4.99e+10 year\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.19, Page 12.38" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "m = 0.0001 # mass of Sr(90) in gm\n", + "t = 28. # half life of Sr(90) in year\n", + "t_ = 9. # time in sec\n", + "\n", + "#Calculations\n", + "lamda = 0.693 / (t * 86400 * 365)\n", + "N_ = 6.023e23 * m / 90\n", + "n = N_ * lamda * t_\n", + "\n", + "#Result\n", + "print \"Activity of Sr is %.2e disintegration/sec.\"%n\n", + "#Incorrect answer in the textbook\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Activity of Sr is 4.73e+09 disintegration/sec.\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.20, Page 12.38" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "t = 1600 # the half life of radium(226) in year\n", + "t1 = 3.8 # the half life of radon(222) in days\n", + "m = 1 # mass of Ra(226) in gm\n", + "\n", + "#Calculation\n", + "m_ = (222 * t1 * m) / (226 * 365 * t) # by the formula N1*t = N2*t1\n", + "\n", + "#Result\n", + "print \"Mass of radon is %.2e gm\"%m_" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mass of radon is 6.39e-06 gm\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.21, Page 12.39" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "m1 = 4.002603 # mass of He(4) in a.m.u.\n", + "m2 = 3.016056 # mass of H(3) in a.m.u.\n", + "m3 = 1.007276 # mass of H(1) in a.m.u.\n", + "\n", + "#Calculations\n", + "k = m2 + m3 - m1\n", + "E = k * 931\n", + "\n", + "#Result\n", + "print \"Energy of gamma ray photon is %.3f MeV\"%E" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy of gamma ray photon is 19.299 MeV\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.22, Page 12.39" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "E = 3 # kinetic energy of proton in Mev\n", + "m1 = 1.007276 # mass of H(1) in a.m.u.\n", + "m2 = 3.016056 # mass of H(3) in a.m.u.\n", + "m3 = 1.008665 # mass of neutron in a.m.u.\n", + "m4 = 3.016036 # mass of He(3) in a.m.u.\n", + "\n", + "#calculations\n", + "k = m1 + m2 - m3 - m4\n", + "E = k * 931.5\n", + "\n", + "#Result\n", + "print \"Q value of reaction is %.4f MeV\"%E" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Q value of reaction is -1.2752 MeV\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.23, Page 12.40" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "E = 200 # energy released per fission in Mev\n", + "m = 0.01 # mass of U(235) in gm\n", + "n = 235 # atomic no of sample\n", + "N_0=6.023e23 # Avogadro constant\n", + "\n", + "#calculations\n", + "E_ = E * 1.6e-13\n", + "k = E_ * N_0 * m / n\n", + "H = k / 4.168\n", + "\n", + "#Result\n", + "print \"Heat produce by complete disintegration is %.3e cal\\nEnergy released is %.1e J\"%(H,E_)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat produce by complete disintegration is 1.968e+08 cal\n", + "Energy released is 3.2e-11 J\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.24, Page 12.40" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "E = 200 # energy released per fission in Mev\n", + "m = 1 # mass of U(235) in kg\n", + "\n", + "#Calculations\n", + "E_ = E * 1.6e-13\n", + "k = E_ * 6.023e26 * m / 235\n", + "\n", + "#Result\n", + "print \"Energy released by fission of 1 kg of U(235)is %.2e J\"%k" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy released by fission of 1 kg of U(235)is 8.20e+13 J\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.25, Page 12.40" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "P = 1e9 # power required for enlighten the city in watt\n", + "e = 30 # percentage efficiency of nuclear reactor\n", + "E = 3.2e-11 # energy released per fission in J\n", + "\n", + "#Calculations\n", + "E_ = E * 30 / 100\n", + "N = P / E_\n", + "N_ = N * 24 * 3600\n", + "m = N_ * 235 / 6.023e26 \n", + "\n", + "#Result\n", + "print \"Amount of fuel required per day is %.2f kg\"%m" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Amount of fuel required per day is 3.51 kg\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.26, Page 12.41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "E = 200 # energy released per fission of U(235)in Mev\n", + "m = 3.7 # mass of U(235) consumed in a day in kg\n", + "e = 20 # percentage efficiency of reactor\n", + "\n", + "#Calculations\n", + "E_ = E * 1.6e-13\n", + "N = m * 6.023e26 / 235\n", + "H = E_ * e / 100\n", + "k = H * N / (24 * 3600)\n", + "\n", + "#Result\n", + "print \"Power output of reactor is %.3f GW\"%(k * 1e-9)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Power output of reactor is 0.702 GW\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.27, Page 12.42" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "m1 = 4.00260 # mass of He(4) in a.m.u.\n", + "m2 = 0.00055 # mass of electron in a.m.u.\n", + "m3 = 12 # mass of C(12) in a.m.u.\n", + "\n", + "#Calculations\n", + "delta_m = 3 * m1 - m3\n", + "E = delta_m * 931\n", + "\n", + "#Result\n", + "print \"Energy produce by each reaction is %.4f MeV\"%E" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy produce by each reaction is 7.2618 MeV\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.28, Page 12.42" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "P = 5e7 # power in watt\n", + "e = 33. # percentage efficiency of nuclear reactor\n", + "m1 = 2.01478 # mass of H(2) in a.m.u.\n", + "m2 = 4.00388 # mass of He(4) in a.m.u.\n", + "\n", + "#calculations\n", + "m = (2 * m1) - m2\n", + "E = m * 931 * 1.6e-13\n", + "E_O = (E * e) /(2 * 100)\n", + "N = P / E_O\n", + "k = N * m1 / 6.023e26\n", + "M = k * 24 * 3600 \n", + "\n", + "#Result\n", + "print \"Mass of deuterium consumed per day is %.4f kg\"%M" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mass of deuterium consumed per day is 0.0229 kg\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.29, Page 12.43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given\n", + "d = 1.8 # diameter in meter\n", + "B = 0.8 # magnetic field in tesla\n", + "m = 6.68e-27 # mass of He(4) in kg\n", + "e = 1.6e-19 # charge on an electron in Coulomb\n", + "\n", + "#Calculations\n", + "r = d / 2\n", + "E = (B**2 * (2 * e)**2 * r**2) / (2 * m * 1.6e-19 * 10**6) \n", + "f = B * 2 * e / (2 * pi * m)\n", + "N = f / 2\n", + "\n", + "#Result\n", + "print \"Energy is %.2f MeV\\nNumber of revolution made by particle to obtain above energy is %.2e per sec\"%(E,N)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy is 24.83 MeV\n", + "Number of revolution made by particle to obtain above energy is 3.05e+06 per sec\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.30, Page 12.43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given\n", + "f = 12e6 # oscillator frequency of cyclotron in Hz\n", + "r = 0.53 # radius of dee in meter\n", + "e = 1.6e-19 # charge on an electron in Coulomb\n", + "\n", + "#Calculations\n", + "B = (2 * pi * f * 2 * 1.67e-27) / e\n", + "\n", + "#Result\n", + "print \"Value of magnetic induction needed to accelerate deuteron in it is %.3f T\"%B" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Value of magnetic induction needed to accelerate deuteron in it is 1.574 T\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.31, Page 12.44" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "f = 10e6 # frequency of applied Emf in Hz\n", + "r = 0.32 # radius in meter\n", + "m = 3.32e-27 # mass of deuteron in kg\n", + "e = 1.6e-19 # charge on an electron in Coulomb\n", + "\n", + "#Calculations\n", + "B = (2 * pi * f * m) / e\n", + "v = (e * B * r) / m\n", + "\n", + "#Result\n", + "print \"Flux density of the magnetic field is %.3f T\\nVelocity of the deuterons emerging out of the cyclotron is %.2e m/sec \"%(B, v)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Flux density of the magnetic field is 1.304 T\n", + "Velocity of the deuterons emerging out of the cyclotron is 2.01e+07 m/sec \n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.32, Page 12.44" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "f = 60 # operating frequency in Hz\n", + "d = 1.6 # diameter in meter\n", + "B = 0.5 # magnetic field at the orbit in tesla\n", + "e = 1.6e-19 # charge on an electron in Coulomb\n", + "\n", + "#Calculations\n", + "r = d / 2\n", + "w = 2 * pi * f\n", + "E = 4 * e * w * r**2 * B\n", + "E_ = 3e8 * r * B / 1e6\n", + "\n", + "#Result\n", + "print \"Energy gained per turn is %.1f eV\\nFinal energy is %d MeV\"%(E / e,E_)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy gained per turn is 482.5 eV\n", + "Final energy is 120 MeV\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.33, Page 12.45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "E = 70 # energy of betatron synchrotron in Mev\n", + "r = 0.28 # radius in meter\n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#Calculations\n", + "E_ = E * 1.6e-13\n", + "B = E_ / (3e8 * e * r)\n", + "\n", + "#Result\n", + "print \"Magnitude of magnetic field is %.2f T\"%B" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Magnitude of magnetic field is 0.83 T\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.34, Page 12.45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "E = 4.18 # energy of alpha particle in Mev\n", + "n = 12 # no. of particle enter the chamber per sec\n", + "E_ = 40 # required energy of an ion pair in ev\n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#Calculations\n", + "R = n * E * 10**6 # in eV\n", + "N = R / E_\n", + "i = N * e\n", + "\n", + "#Result\n", + "print \"The current produced = %.e Amp\"%i" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current produced = 2e-13 Amp\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.35, Page 12.46" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "n = 10**8 # no. of electron per discharge counted by GM counter\n", + "r = 500. # counting rate in counts per minutes\n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#Calculations\n", + "N = r / 60\n", + "i = N * n * e \n", + "\n", + "#Result\n", + "print \"Average current in the circuit = %.2e Amp\"%i" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Average current in the circuit = 1.33e-10 Amp\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.36, Page 12.46" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "E = 10 # energy of electron in kev\n", + "B = 5e-5 # magnetic field of earth in tesla\n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#Calculations\n", + "f = e * B / 9.1e-31\n", + "E_ = E * 1.6e-16\n", + "v = sqrt((2 * E_) / 9.1e-31)\n", + "r = v / f\n", + "\n", + "#Result\n", + "print \"Frequency of cyclotron = %.3e per sec\\nLarmour radius = %.3f meter\"%(f,r)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Frequency of cyclotron = 8.791e+06 per sec\n", + "Larmour radius = 6.745 meter\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.37, Page 12.46" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "B = 5e-9 # magnetic field in tesla\n", + "v = 3e5 # velocity of proton stream in m/sec\n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#calculations\n", + "r = (1.67e-27 * v) / (e * B)\n", + "\n", + "#Result\n", + "print \"Larmour radius is %.2e meter\"%r" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Larmour radius is 6.26e+05 meter\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.38, Page 12.46" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "E = 1. # energy of He+ in kev\n", + "r = 0.188 # Larmour radius in meter\n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#Calculations\n", + "E_ = E * 1.6e-16\n", + "v = sqrt((2 * E_) / (4 * 1.67e-27))\n", + "B = (4 * 1.67e-27 * v) / (e * r)\n", + "\n", + "#Result\n", + "print \"Magnetic field is %.3e tesla\"%B" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Magnetic field is 4.861e-02 tesla\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.39, Page 12.47" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "E = 3.5 # energy of He++ ash particle in Mev\n", + "B = 8 # magnetic field in tesla\n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#Calculations\n", + "E_ = E * 1.6e-13\n", + "v = sqrt(2 * E_ / (4 * 1.67e-27))\n", + "r = (4 * 1.67e-27 * v) / (2 * e * B)\n", + "\n", + "#Result\n", + "print \"Larmour radius is %.2e meter\"%r" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Larmour radius is 3.38e-02 meter\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.40, Page 12.47" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "d = 1e12 # electron density in number per m^3\n", + "E = 0.1 # thermal energy in eV\n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#Calculations\n", + "lamda = sqrt((8.85e-12 * E * e) / (d * e * e))\n", + "omega = sqrt(d * e**2 / (9.1e-31 * 8.85e-12))\n", + "f = omega / (2 * pi)\n", + "\n", + "#Results\n", + "print \"Debye length is %.2e meter\\nPlasma frequency is %.2f MHz\"%(lamda,f / 1e6)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Debye length is 2.35e-03 meter\n", + "Plasma frequency is 8.97 MHz\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.41, Page 12.48" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "d = 1e16 # density in per m^3\n", + "E = 2 # thermal energy in eV\n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#Calculations\n", + "lamda = sqrt((8.85e-12 * E * e) / (d * e * e))\n", + "omega = sqrt(d * e**2 / (9.1e-31 * 8.85e-12))\n", + "f = omega / (2 * pi)\n", + "\n", + "#Result\n", + "print \"Debye length is %.4e meter\\nPlasma frequency is %.3e Hz\"%(lamda,f)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Debye length is 1.0518e-04 meter\n", + "Plasma frequency is 8.973e+08 Hz\n" + ] + } + ], + "prompt_number": 25 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter_13.ipynb b/Engineering_Physics/Chapter_13.ipynb new file mode 100755 index 00000000..db97b36d --- /dev/null +++ b/Engineering_Physics/Chapter_13.ipynb @@ -0,0 +1,995 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 13: Crystal Structure\n" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.1, Page 13.24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Given \n", + "x = 2 # intercepts cut by the plane along vector a of crystallographic axes\n", + "y = 3 # intercepts cut by the plane along vector b of crystallographic axes\n", + "z = 1 # intercepts cut by the plane along vector c of crystallographic axes\n", + "\n", + "#Calculations\n", + "x_ = 6 / x\n", + "y_ = 6 / y\n", + "z_ = 6 / z\n", + "\n", + "#Result\n", + "print \"Miller indices of the plane are (%d %d %d)\"%(x_,y_,z_)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Miller indices of the plane are (3 2 6)\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.2, Page 13.24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Given \n", + "x = 1 # intercepts cut by the plane along vector a of crystallographic axes\n", + "y = 2 # intercepts cut by the plane along vector b of crystallographic axes\n", + "z = -3. / 2 # intercepts cut by the plane along vector c of crystallographic axes\n", + "\n", + "#Calculations\n", + "x_ = 6 / x\n", + "y_ = 6 / y\n", + "z_ = 6 / z\n", + "\n", + "print \"Miller indices of the plane are (%d %d %d)\"%(x_,y_,z_)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Miller indices of the plane are (6 3 -4)\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.3, Page 13.25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "x1 = 3. # intercepts cut by the plane along vector a of crystallographic axes in first case\n", + "y1 = 3. # intercepts cut by the plane along vector b of crystallographic axes in first case \n", + "z1 = 2. # intercepts cut by the plane along vector c of crystallographic axes in first case\n", + "x2 = 1. # intercepts cut by the plane along vector a of crystallographic axes in second case\n", + "y2 = 2. # intercepts cut by the plane along vector b of crystallographic axes in second case\n", + "k2 = 0. # raciprocal of intercepts cut by the plane along vector c of crystallographic axes in second case\n", + "x3 = 1. # intercepts cut by the plane along vector a of crystallographic axes in third case\n", + "y3 = 1./2 # intercepts cut by the plane along vector b of crystallographic axes in third case\n", + "z3 = 1. # intercepts cut by the plane along vector c of crystallographic axes in third case\n", + "\n", + "#Calculations\n", + "x_1 = 6. / x1\n", + "y_1 = 6. / y1\n", + "z_1 = 6. / z1\n", + "x_2 = 2. / x2\n", + "y_2 = 2. / y2\n", + "z_2 = 2*k2\n", + "x_3 = 2 * x3\n", + "y_3 = 2 * y3\n", + "z_3 = 2 * z3\n", + "\n", + "#Result\n", + "print \"Miller indices of the plane (i) In first case are (%d %d %d)\\n(ii) In second case are (%d %d %d)\\n(iii)In the third case are (%d %d %d).\"%(x_1,y_1,z_1,x_2,y_2,z_2,x_3,y_3,z_3)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Miller indices of the plane (i) In first case are (2 2 3)\n", + "(ii) In second case are (2 1 0)\n", + "(iii)In the third case are (2 1 2).\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.4, Page 13.25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "# Given \n", + "x1 = 1 # coordinate on x axis for first plane\n", + "y1 = 0 # coordinate on y axis for first plane\n", + "z1 = 0 # coordinate on z axis for first plane\n", + "x2 = 1 # coordinate on x axis for second plane\n", + "y2 = 1 # coordinate on y axis for second plane\n", + "z2 = 1 # coordinate on z axis for second plane\n", + "\n", + "#Calculations\n", + "d1 = 1 / sqrt(x1**2 + y1**2 + z1**2)\n", + "d2 = 1 / sqrt(x2**2 + y2**2 + z2**2)\n", + "\n", + "#Result\n", + "print \"Spacing between the plane in first case is a / %d\\nSpacing between the plane in second case is a / %f\"%(d1,d2)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Spacing between the plane in first case is a / 1\n", + "Spacing between the plane in second case is a / 0.577350\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.5, Page 13.26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "# Given that\n", + "x = 1. # intercepts cut by the plane along vector a of crystallographic axes\n", + "y = 2. # intercepts cut by the plane along vector b of crystallographic axes\n", + "k = 0. # raciprocal of intercepts cut by the plane along vector c of crystallographic axes\n", + "a = 5. # length of vector a of crystallographic axes in angstrom\n", + "b = 5. # length of vector b of crystallographic axes in angstrom \n", + "c = 5. # length of vector c of crystallographic axes in angstrom\n", + "\n", + "#Calculations\n", + "x_ = 2. / x\n", + "y_ = 2. / y\n", + "z_ = 2 * k\n", + "d = a / sqrt(x_**2 + y_**2 + z_**2)\n", + "D=d**2\n", + "\n", + "#Result\n", + "print \"Miller indices of the plane are (%d %d %d)\\nInter planar distance is sqrt(%d) angstrom\"%(x_,y_,z_,D)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Miller indices of the plane are (2 1 0)\n", + "Inter planar distance is sqrt(0) angstrom\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.6, Page 13.26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "x = 2 # intercepts cut by the plane along vector a of crystallographic axes\n", + "y = 2. / 3 # intercepts cut by the plane along vector b of crystallographic axes\n", + "k = 0 # raciprocal of intercepts cut by the plane along vector c of crystallographic axes\n", + "\n", + "#Calculations\n", + "x_ = 2. / x\n", + "y_ = 2. / y\n", + "z_ = 2 * k\n", + "\n", + "#Result\n", + "print \"Miller indices of the plane are (%d %d %d)\"%(x_,y_,z_)\n", + "\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Miller indices of the plane are (1 3 0)\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.7, Page 13.27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "x1 = 2 # coordinate on x axis \n", + "y1 = 3 # coordinate on y axis \n", + "z1 = 1 # coordinate on z axis \n", + "r = 0.175 # atomic radius of fcc structure in nm\n", + "\n", + "#Calculations\n", + "a = (4 * r) / sqrt(2)\n", + "d = a / sqrt(x1**2 + y1**2 + z1**2)\n", + "\n", + "#Result\n", + "print \"Inter planar spacing is %.3f nm\"%d\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Inter planar spacing is 0.132 nm\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.8, Page 13.27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "x1 = 1 # coordinate on x axis in first case\n", + "y1 = 2 # coordinate on y axis in first case \n", + "z1 = 3 # coordinate on z axis in first case\n", + "x2 = 1\n", + "y2 = 1\n", + "z2 = 0\n", + "# coordinate of first plane in second case\n", + "x3 = 1\n", + "y3= 1\n", + "z3 = 1\n", + "# coordinate of second plane in second case\n", + "\n", + "#Calculations\n", + "x_=6/x1\n", + "y_=6/y1\n", + "z_=6/z1\n", + "d1 = 1 / sqrt(x2^2 + y2^2 + z2^2)\n", + "d2= 1/ sqrt(x3^2 + y3^2 + z3^2)\n", + "d = d1/d2\n", + "\n", + "#Results\n", + "print \"The ratio of intercepts of three axes by the point are %d : %d : %d\\nThe ratio of spacing between two planes is %.3f\"%(x_,y_,z_,d)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The ratio of intercepts of three axes by the point are 6 : 3 : 2\n", + "The ratio of spacing between two planes is 1.225\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.9, Page 13.28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "a = 5 # the lattice constant of the structure in angstrom\n", + "\n", + "#Calculations\n", + "d = (sqrt(3) / 4) * a\n", + "\n", + "#Result\n", + "print \"Distance between two atoms is %.2f A\"%d\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Distance between two atoms is 2.17 A\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.10, Page 13.28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given that\n", + "a = 3.56 # the length of cube edge in angstrom\n", + "\n", + "#Calculations\n", + "d = a / sqrt(2)\n", + "\n", + "#Result\n", + "print \"Permitive translation vector is %.2f A\"%d\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Permitive translation vector is 2.52 A\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.11, Page 13.28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "w = 207.2 # atomic weight of Pb\n", + "d = 11.36e3 # density of Pb in kg/m^3\n", + "a = 3.2e-10 # length of cube edge in meter\n", + "N = 6.023e26 # Avogadro's no. in per kg mole\n", + "\n", + "#Calculations\n", + "n = (a**3 * d * N) / w\n", + "\n", + "#Result\n", + "print \"Number of atom per unit cell is %d\"%n\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of atom per unit cell is 1\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.12, Page 13.28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "M = 60.2 # molecular weight\n", + "d = 6250 # density in kg/m^3\n", + "N = 6.023e+26 # Avogadro's no. in per kg mole\n", + "n = 4. # for fcc lattice\n", + "\n", + "#Calculations\n", + "a = (((4 * M) / (N * d))**(1. / 3)) * 1e10\n", + "\n", + "#Result\n", + "print \"Lattice constant is %.f A\"%a\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Lattice constant is 4 A\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.13, Page 1329" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "x1 = 1 # coordinate on x axis of plane\n", + "y1 = 0 # coordinate on y axis of plane\n", + "z1 = 0 # coordinate on z axis of plane\n", + "d = 2.82 # the space between successive plane in angstrom\n", + "theta = 8.8 # glancing angle in degree\n", + "\n", + "#Calculations\n", + "n = 1\n", + "lamda = 2 * d * sin(theta*pi/180) / n\n", + "\n", + "#Result\n", + "print \"Wavelength of x-ray is %.3f A\"%lamda\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of x-ray is 0.863 A\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.14, Page 13.29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "d = 2.51 # the space between adjacent plane in angstrom\n", + "theta = 9 # glancing angle in degree\n", + "\n", + "#Calculations\n", + "n = 1 # for n=1\n", + "lamda = 2 * d * sin(theta*pi/180) / n\n", + "n = 2 # for n=2\n", + "theta = degrees(asin(lamda / d))\n", + "\n", + "#Result\n", + "print \"Wavelength of x-ray is %.4f A\\nGlancing angle for second order diffraction is %.1f degree\"%(lamda,theta)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of x-ray is 0.7853 A\n", + "Glancing angle for second order diffraction is 18.2 degree\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.15, Page 13.29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "lamda = 1.5 # wavelength of x-ray in angstrom\n", + "theta = 60 # glancing angle in degree\n", + "\n", + "#Calculations\n", + "n = 1 # for first order\n", + "d = ( n * lamda) / (2 * sin(theta*pi/180))\n", + "\n", + "#Result\n", + "print \"Lattice constant of NaCl is %.2f A\"%d\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Lattice constant of NaCl is 0.87 A\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.16, Page " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "lamda = 1.4 # wavelength of x-ray in angstrom\n", + "x1 = 1 # coordinate on x axis of plane\n", + "y1 = 1 # coordinate on y axis of plane\n", + "z1 = 1 # coordinate on z axis of plane\n", + "a = 5. # lattice parameter of of crystal in angstrom\n", + "\n", + "#Calculations\n", + "n = 1 # for first order\n", + "d = a / sqrt(x1**2 + y1**2 + z1**2)\n", + "theta = degrees(asin((n * lamda) / (2 * d)))\n", + "\n", + "#Result\n", + "print \"Angle of incidence of x-ray on the plane is %.f degree\"%theta\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Angle of incidence of x-ray on the plane is 14 degree\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.17, Page 13.30" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "lamda = 0.710 # wavelength of x-ray in angstrom\n", + "x1 = 1 # coordinate on x axis of plane\n", + "y1 = 0 # coordinate on y axis of plane\n", + "z1 = 0 # coordinate on z axis of plane\n", + "a = 2.814 # lattice parameter of of crystal in angstrom\n", + "\n", + "#Calculations\n", + "n = 2 # for second order\n", + "d = a / sqrt(x1**2 + y1**2 + z1**2)\n", + "theta = degrees(asin((n * lamda) / (2 * d)))\n", + "\n", + "#Result\n", + "print \"Glancing angle is %.1f degree\"%theta\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Glancing angle is 14.6 degree\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.18, Page 13.30" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "n = 1 # order of brag reflection \n", + "d = 3.84e-10 # the space between successive plane in m\n", + "theta = 30. # glancing angle in degree\n", + "\n", + "#Calculations\n", + "lamda = 2 * d * sin(theta*pi/180) / n\n", + "v = 6.62e-34 / (1.67e-27 * lamda)\n", + "\n", + "#Result\n", + "print \"Wavelength of neutron beam is %.2f A\\nSpeed of neutron beam is %.2e meter/sec\"%(lamda * 10**10,v)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of neutron beam is 3.84 A\n", + "Speed of neutron beam is 1.03e+03 meter/sec\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.19, Page 13.31" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "v = 120 # voltage at which electron is accelerated in v\n", + "n = 1 # order of Bragg reflection \n", + "x1 = 1 # coordinate on x axis of plane\n", + "y1 = 1 # coordinate on y axis of plane\n", + "z1 = 1 # coordinate on z axis of plane\n", + "theta = 22 # angle at which maximum reflection is obtain in degree\n", + "n = 1 # order of reflection\n", + "\n", + "#Calculations\n", + "lamda = 6.62e-34 / sqrt(2 * 9.1e-31 * 1.6e-19 * v)\n", + "d = (n * lamda) / (2 * sin(theta*pi/180))\n", + "a = d * sqrt(3) \n", + "\n", + "#Result\n", + "print \"Lattice parameter is %.3f A\"%(a * 10**10) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Lattice parameter is 2.589 A\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.20, Page 13.31" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "lamda = 1.24e-10 # wavelength of X-ray in A\n", + "x1 = 1 # coordinate on x axis of first plane\n", + "y1 = 0 # coordinate on y axis of first plane\n", + "z1 = 0 # coordinate on z axis of first plane\n", + "x2 = 1 # coordinate on x axis of second plane\n", + "y2 = 1 # coordinate on y axis of second plane\n", + "z2 = 0 # coordinate on z axis of second plane\n", + "x3 = 1 # coordinate on x axis of third plane\n", + "y3 = 1 # coordinate on y axis of third plane\n", + "z3 = 1 # coordinate on z axis of third plane\n", + "M = 74.5 # molecular weight of KCl\n", + "d = 1980 # density of KCl in kg/m^3\n", + "N = 6.023e+26 # Avogadro's No per Kg mole\n", + "\n", + "#Calculations\n", + "a = (4*M / (N*d))**(1./3)\n", + "D1 = a/sqrt(x1**2 + y1**2 + z1**2)\n", + "D2 = a/sqrt(x2**2 + y2**2 + z2**2)\n", + "D3 = a/sqrt(x3**2 + y3**2 + z3**2)\n", + "\n", + "#Result\n", + "print \"Inter planner distances are - \\n(1) in first case %.1f A \\n(2) in second case %.2f A \\n(3) in third case %.2f A\"%(D1*10**10,D2*10**10,D3*10**10)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Inter planner distances are - \n", + "(1) in first case 6.3 A \n", + "(2) in second case 4.45 A \n", + "(3) in third case 3.64 A\n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.21, Page 13.32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "d = 0.15e-9 # distance between K(+) and Cl(-) in m\n", + "\n", + "#Calculations\n", + "v = -1.6e-19 / (4 * pi * 8.85e-12 * d)\n", + "\n", + "#Result\n", + "print \"Potential energy of molecule is %.1f eV\"%v\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Potential energy of molecule is -9.6 eV\n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.22, Page 13.32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "d = 0.32e-9 # equilibrium separation in m\n", + "alpha = 1.748 \n", + "n = 9\n", + "e = 4 # ionization energy in eV\n", + "a = -2.16 # electron affinity in eV\n", + "\n", + "#Calculations\n", + "E = -((alpha * 1.6e-19) / (4 * pi * 8.85e-12 * d)) * (1 - (1. / n))\n", + "\n", + "#Result\n", + "print \"Cohesive energy of Nacl is %f eV\"%E\n", + "#Incorrect answer in the textbook " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Cohesive energy of Nacl is -6.985633 eV\n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.23, Page 13.33" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "E = 2.02 # average energy required to produce a Schottky defect at room temperature in eV\n", + "k = 1.38e-23 # Boltzmann constant in J/k\n", + "T = 300 # room temperature in K\n", + "\n", + "#Calculations\n", + "r = exp(-(E * 1.6e-19) / (2 * k * T))\n", + "\n", + "#Result\n", + "print \"Ratio of number of Schottky defects to total number of cation-anion pairs is %.2e\"%r\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ratio of number of Schottky defects to total number of cation-anion pairs is 1.12e-17\n" + ] + } + ], + "prompt_number": 44 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter_14.ipynb b/Engineering_Physics/Chapter_14.ipynb new file mode 100755 index 00000000..498814f0 --- /dev/null +++ b/Engineering_Physics/Chapter_14.ipynb @@ -0,0 +1,1951 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 14: Development of Quantum Mechanics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.1, Page 14.20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "E = 75 # energy of photon in eV\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in J \n", + "\n", + "#Calculations\n", + "f = E * e / h\n", + "lamda = c / f\n", + " \n", + "#Result\n", + "print \"Frequency is %.2e Hz\\nWavelength is %.1f A\"%(f,lamda * 10**10)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Frequency is 1.81e+16 Hz\n", + "Wavelength is 165.5 A\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.2, Page 14.20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "P = 2e5 # radiated power in W\n", + "f = 98e6 # frequency in Hz \n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "\n", + "#Calculations\n", + "E = h * f\n", + "n = P / E\n", + "\n", + "#Result\n", + "print \"Number of quanta emitted per sec is %.2e\"%n\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of quanta emitted per sec is 3.08e+30\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.3, Page 14.20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "lamda = 4e-7 # wavelength of spectral line in meter\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "\n", + "#Calculations\n", + "E = (h * c) / lamda\n", + "\n", + "#Result\n", + "print \"Energy of photon is %.3e J\"%E\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy of photon is 4.965e-19 J\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.4, Page 14.21" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Given \n", + "lamda = 5e-7 # wavelength of green light in meter\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "P = 1. # energy in erg\n", + "\n", + "#Calculations\n", + "E = ((h * c) / lamda) * (10**7)\n", + "n = P / E\n", + "\n", + "#Result\n", + "print \"Number of photons of green light emitted is %.2e\"%n\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of photons of green light emitted is 2.52e+11\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.5, Page 14.21" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Given \n", + "E = 5e-19 # energy of photon in J\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#Calculations\n", + "lamda = (c * h) / E\n", + "\n", + "#Result\n", + "print \"Wavelength is %.f A\"%(lamda * 10**10)\n", + "#Incorrect answer in the textbook " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength is 3972 A\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.6, Page 14.21" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "lamda = 4.35e-7 # wavelength of green light in meter\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "P = 1 # energy in erg\n", + "\n", + "#Calculations\n", + "E = ((h * c) / lamda)\n", + "\n", + "#Result\n", + "print \"Energy of an electron is %.3e J\"%E\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy of an electron is 4.566e-19 J\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.7, Page 14.22" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "lamda = 5.6e-7 # wavelength of light in meter\n", + "n = 120 # no. of photons per second\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "\n", + "#Calculations\n", + "E = ((h * c) / lamda)\n", + "p = E * n\n", + "\n", + "#Result\n", + "print \"Energy received by the eye per second is %.3e W\"%p\n", + "#Incorrect answer in the textbook " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy received by the eye per second is 4.256e-17 W\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.8, Page 14.22" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "lamda = 5.5e-7 # wavelength of light in meter\n", + "E = 1.5 # energy in J\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "\n", + "#Calculations\n", + "E_ = ((h * c) / lamda)\n", + "n = E / E_\n", + "\n", + "#Result\n", + "print \"Number of photons of yellow light = %.3e\"%n\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of photons of yellow light = 4.154e+18\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.9, Page 14.22" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "lamda = 4.35e-7 # wavelength of light in meter\n", + "lambda_ = 5.42e-7 # threshold wavelength of photoelectron in meter\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of an electron in kg \n", + "\n", + "#Calculations\n", + "w = ((h * c) / lambda_)\n", + "v = sqrt(((2 * h * c) / m) * (1 / lamda - 1 / lambda_))\n", + "V = m * v**2 / (2 * e)\n", + "\n", + "#Result\n", + "print \"Work function is %.3e J\\nStopping potential is %.2f V\\nMaximum velocity is %.3e m/sec\"%(w,V,v)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Work function is 3.664e-19 J\n", + "Stopping potential is 0.56 V\n", + "Maximum velocity is 4.451e+05 m/sec\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.10, Page 14.23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "f = 1.2e15 # frequency of light in Hz\n", + "f_ = 1.1e+15 # threshold frequency of photoelectron emission in copper in Hz\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#Calculations\n", + "E = h * (f - f_) / e\n", + "\n", + "#Result\n", + "print \"Maximum energy of photoelectron is %.3f eV\"%E\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum energy of photoelectron is 0.414 eV\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.11, Page 14.23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "lamda = 6.2e-7 # threshold wavelength of photoelectron in first case in meter\n", + "lambda_ = 5e-7 # threshold wavelength of photoelectron in second case in meter\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#Calculations\n", + "w = ((h * c) / lamda) * (1 / e)\n", + "w_ = ((h * c) / lambda_) * (1 / e)\n", + "\n", + "#Result\n", + "print \"Work function for wavelength %.e A is %.f eV\\nWork function for wavelength %.e A is %.2f eV\"%(lamda,w,lambda_,w_)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Work function for wavelength 6e-07 A is 2 eV\n", + "Work function for wavelength 5e-07 A is 2.48 eV\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.12, Page 14.24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Given \n", + "lamda = 3.132e-7 # wavelength of light in meter\n", + "V = 1.98 # stopping potential in V\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of an electron in kg \n", + "\n", + "#Calculations\n", + "E = e * V\n", + "lambda_ = 1. / ((1. / lamda) - (E / (h * c)))\n", + "f = c / lambda_\n", + "w = ((h * c) / lambda_)\n", + "\n", + "#Result\n", + "print \"Work function is %.3e J\\nMaximum energy is %.3e J\\nThreshold frequency is %.3e Hz\"%(w,E,f)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Work function is 3.173e-19 J\n", + "Maximum energy is 3.168e-19 J\n", + "Threshold frequency is 4.793e+14 Hz\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.13, Page 14.24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Given \n", + "w = 4.8 # work function in eV\n", + "lambda1 = 5e-7 # wavelength of incident radiation in first case in meter\n", + "lambda2 = 2e-7 # wavelength of incident radiation in second case in meter\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#Calculations\n", + "E_k1 = h*c/lambda1\n", + "E_k2 = h*c / lambda2\n", + "\n", + "#Result\n", + "print \"The energy corresponding to wavelength 5000 A is %.2f which is found to be less than the work function 4.8 eV\"%(E_k1/e) \n", + "print \"The energy corresponding to wavelength 2000 A %.2f is found to be greater than the work function\"%(E_k2/e)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The energy corresponding to wavelength 5000 A is 2.48 which is found to be less than the work function 4.8 eV\n", + "The energy corresponding to wavelength 2000 A 6.21 is found to be greater than the work function\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.14, Page 14.25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "lamda = 5.893e-7 # wavelength of light in meter\n", + "V = 0.36 # stopping potential for emitted electron in eV\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of an electron in kg \n", + "\n", + "#Calculations\n", + "E = h * c / lamda\n", + "w = ((h * c) / lamda) * (1 / e) - V\n", + "f = w * e / h\n", + "\n", + "#Result\n", + "print \"Maximum energy is %.2f eV\\nWork function is %.2f eV\\nThreshold frequency is %.2e cycles/sec\"%(E/e,w,f)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum energy is 2.11 eV\n", + "Work function is 1.75 eV\n", + "Threshold frequency is 4.22e+14 cycles/sec\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.15, Page 14.25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Given \n", + "lamda = 5.89e-7 # wavelength of light in meter\n", + "lambda_ = 7.32e-7 # threshold wavelength of photoelectron in meter\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of an electron in kg \n", + "\n", + "#Calculations\n", + "E = (h * c) * (1 / lamda - 1 / lambda_)\n", + "V = E / e\n", + "\n", + "#Result\n", + "print \"Stopping potential is %.3f V\\nMaximum kinetic energy is %.3e J\"%(V,E)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Stopping potential is 0.412 V\n", + "Maximum kinetic energy is 6.587e-20 J\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.16, Page 14.26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "E = 1.5 # maximum energy in eV\n", + "lambda_ = 2.3e-7 # threshold wavelength of photoelectron in meter\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of an electron in kg \n", + "\n", + "#Calculations\n", + "lamda = 1 / ((E * e / (h * c)) + (1 / lambda_))\n", + "\n", + "#Resuult\n", + "print \"Wavelength of light is %.1f A\"%(lamda * 1e10)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of light is 1799.8 A\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.17, Page 14.26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "lamda = 1.5e-7 # wavelength of light in in meter\n", + "w = 4.53 # work function of tungsten in eV\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#Calculations\n", + "E = ((h * c) / lamda) * (1 / e)\n", + "k = E - w\n", + "\n", + "#Result\n", + "print \"Energy of incident photon is %.2f eV,which is greater than the work function \\nSo it causes photoelectric emission.\\nKinetic energy of the emitted electron is %.2f eV\"%(E,k)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy of incident photon is 8.28 eV,which is greater than the work function \n", + "So it causes photoelectric emission.\n", + "Kinetic energy of the emitted electron is 3.75 eV\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.18, Page 14.26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "w = 2.3 # work function of sodium in eV\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#Calculations\n", + "lamda = ((h * c) / w) * (1 / e)\n", + "\n", + "#Result\n", + "print \"Longest wavelength required for photoemission is %.2f A\"%(lamda * 1e10)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Longest wavelength required for photoemission is 5396.74 A\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.19, Page 14.27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "w = 2 # work function of sodium in eV\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#Calculations\n", + "lamda = ((h * c) / w) * (1 / e)\n", + "\n", + "#Result\n", + "print \"Threshold wavelength for photo emission is %d A\"%(lamda * 1e10)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Threshold wavelength for photo emission is 6206 A\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.20, Page 14.27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "k = 4 # maximum kinetic energy of electron in eV\n", + "w = 2.2 # work function of sodium in eV\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#Calculations\n", + "lambda_ = ((h * c) / (w * e)) \n", + "lamda = (1 / ((((k * e) / (h * c))) + (1 / lambda_)))\n", + "\n", + "#Result\n", + "print \"Threshold wavelength is %d A\\nIncident electromagnetic wavelength is %.f A\"%(lambda_ * 1e10,lamda * 1e10)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Threshold wavelength is 5642 A\n", + "Incident electromagnetic wavelength is 2002 A\n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.21, Page 14.28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "lamda = 3.5e-7 # wavelength of light in meter\n", + "i = 1 # intensity in W/m^2\n", + "p = 0.5 # percent of incident photon produce electron\n", + "a = 1 # surface area of potassium in cm^2\n", + "w = 2.1 # work function of potassium in eV\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of an electron in kg \n", + "\n", + "#Calculations\n", + "E = (((h * c) / lamda) * (1 / e) - w) * e\n", + "E_ = (p * a * 1e-4) / 100 # in W/cm^2\n", + "n = E_ / E\n", + "\n", + "#Result\n", + "print \"Maximum kinetic energy is %.3e J\\nNumber of electrons emitted per sec from 1cm^2 area is %.2e\"%(E,n)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum kinetic energy is 2.314e-19 J\n", + "Number of electrons emitted per sec from 1cm^2 area is 2.16e+12\n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.22, Page 14.28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "lamda = 5.896e-7 # wavelength of first light in meter\n", + "lambda_ = 2.83e-7 # wavelength of second light in meter\n", + "V1 = 0.12 # stopping potential for emitted electrons for first light in V\n", + "V2 = 2.2 # stopping potential for emitted electrons for second light in V\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "\n", + "#Calculations\n", + "h = (e * (V2 - V1) / c) / (1 / lambda_ - 1 / lamda)\n", + "\n", + "#Result\n", + "print \"Value of Planck constant is %.2e J-sec\"%h\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Value of Planck constant is 6.04e-34 J-sec\n" + ] + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.23, Page 14.29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "lamda = 1e-10 # wavelength of light in meter\n", + "theta = 90 # angle at which scattered radiation is viewed in degree\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "m = 9.1e-31 # mass of an electron in kg\n", + "\n", + "#Calculations\n", + "delta_lambda = (h * (1 - cos(theta*pi/180))) / (m * c)\n", + "\n", + "#Result\n", + "print \"Compton shift is %.3f A\"%(delta_lambda * 1e10)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Compton shift is 0.024 A\n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.24, Page 14.29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import * \n", + "# Given \n", + "lamda = 1e-10 # wavelength of light in meter\n", + "theta = 90 # angle in degree\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "m = 9.1e-31 # mass of an electron in kg\n", + "\n", + "#Calculations\n", + "delta_lambda = (h * (1 - cos(theta*pi/180))) / (m * c)\n", + "E = (h * c) / delta_lambda\n", + "\n", + "#Result\n", + "print \"Compton shift is %.3f A\\nEnergy of incident beam is %.3f MeV\"%(delta_lambda * 1e10,E / 1.6e-13)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Compton shift is 0.024 A\n", + "Energy of incident beam is 0.512 MeV\n" + ] + } + ], + "prompt_number": 41 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.25, Page 14.30" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + " \n", + "# Given \n", + "E = 4 # enrgy of recoil electron in KeV\n", + "theta = 180 # scattered angle of photon in degree\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "m = 9.1e-31 # mass of an electron in kg\n", + "\n", + "#Calculations\n", + "p = sqrt(2 * E * 10**3 * e * m)\n", + "lamda = (2 * h * c) / (p * c + E * 10**3 * e)\n", + "\n", + "#Result\n", + "print \"Wavelength of incident beam is %.3f A\"%(lamda * 1e10)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of incident beam is 0.365 A\n" + ] + } + ], + "prompt_number": 42 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.26, Page 14.31" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "lamda = 1e-10 # wavelength of light in meter\n", + "theta = 90 # angle in degree\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "m = 9.1e-31 # mass of an electron in kg\n", + "\n", + "#Calculations\n", + "delta_lambda = (h * (1 - cos(theta*pi/180))) / (m * c)\n", + "E = (h * c) * ((1 / lamda) - (1 / (lamda + delta_lambda)))\n", + "\n", + "#Result\n", + "print \"Compton shift is %.3e m\\nKinetic energy is %.f eV\"%(delta_lambda,E / 1.6e-19)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Compton shift is 2.425e-12 m\n", + "Kinetic energy is 294 eV\n" + ] + } + ], + "prompt_number": 44 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.27, Page 14.31" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "lamda = 0.144e-10 # wavelength of x-ray in meter\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "m = 9.1e-31 # mass of an electron in kg\n", + "\n", + "#Calculations\n", + "theta = 180 # for maximum shift\n", + "d_lamda = (h * (1 - cos(theta*pi/180))) / (m * c)\n", + "E = (h * c) * ((1. / lamda) - (1. / (d_lamda+lamda)))\n", + "\n", + "#Result\n", + "print \"Maximum Compton shift is %.4f A\\nKinetic energy is %.2f KeV\"%(delta_lambda * 1e10,E / 1.6e-16)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum Compton shift is 0.0485 A\n", + "Kinetic energy is 21.72 KeV\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.28, Page 14.32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "lamda = 0.2e-10 # wavelength of x-ray in meter\n", + "theta = 45 # scattered angle in degree\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "m = 9.1e-31 # mass of an electron in kg\n", + "\n", + "#Calculations\n", + "delta_lambda = (h * (1 - cos(theta*pi/180))) / (m * c)\n", + "E = (h * c) * ((1 / lamda) - (1 / (lamda + delta_lambda)))\n", + "theta_ = 180 # for maximum\n", + "delta_lambda_ = (h * (1 - cos(theta_*pi/180))) / (m * c)\n", + "lambda_ = lamda + delta_lambda_\n", + "E_k = h*c*(1/lamda - 1/lambda_)\n", + "\n", + "#Result\n", + "print \"Wavelength of x-ray is %.4f A\\nMaximum kinetic energy %.2e J\"%(lambda_ * 1e10,E_k)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of x-ray is 0.2485 A\n", + "Maximum kinetic energy 1.94e-15 J\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.29, Page 14.33" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "v = 96 # speed of automobile in km/hr\n", + "e = 1.6e-19 # charge on an electron in C \n", + "m = 2e3 # mass of automobile in kg\n", + "\n", + "#Calculations\n", + "v_ = v * (5. / 18)\n", + "lamda = h / (m * v_)\n", + "\n", + "#Result\n", + "print \"de-Broglie wavelength is %.2e m\"%lamda\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "de-Broglie wavelength is 1.24e-38 m\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.30, Page 14.33" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "v = 50 # potential differece in volt\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "m = 9.1e-31 # mass of an electron in kg\n", + "\n", + "#Calculations\n", + "lamda = h / sqrt(2 * m * v * e)\n", + "\n", + "#Result\n", + "print \"de-Broglie wavelength is %.2f A\"%(lamda * 1e10)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "de-Broglie wavelength is 1.73 A\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.31, Page 14.33" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "t = 300 # temperature in K\n", + "k = 1.37e-23 # Boltzmann's constant in J/K\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "m = 1.67e-27 # mass of neutron in kg\n", + "\n", + "#Calculations\n", + "lamda = h / sqrt(3 * m * k * t)\n", + "\n", + "#Result\n", + "print \"Wavelength of thermal neutron is %.3f A\"%(lamda * 1e10)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of thermal neutron is 1.459 A\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.32, Page 14.34" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "v = 2e8 # speed of proton in m/sec\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "m = 1.67e-27 # mass of proton in kg\n", + "\n", + "#Calculations\n", + "lamda = h / (m * v)\n", + "\n", + "#Result\n", + "print \"Wavelength of matter wave associated with proton is %.2e m\"%lamda\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of matter wave associated with proton is 1.98e-15 m\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.33, Page 14.34" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "lamda = 0.1e-10 # DE Broglie wavelength associated with electron in M\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "m = 9.1e-31 # mass of electron in kg\n", + "\n", + "#Calculations\n", + "V = h**2 / (2 * m* e * lamda**2)\n", + "\n", + "#Result\n", + "print \"Potential difference is %.2f KV\"%(V * 10**-3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Potential difference is 15.05 KV\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.34, Page 14.34" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "v = 200 # potential differece in volt\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "q = 3.2e-19 # charge on an alpha particle in C \n", + "m = 4 * 1.67e-27 # mass of alpha particle in kg\n", + "\n", + "#Calculations\n", + "lamda = h / sqrt(2 * m * v * q)\n", + "\n", + "#Result\n", + "print \"de-Broglie wavelength = %.2e m\"%lamda\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "de-Broglie wavelength = 7.16e-13 m\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.35, Page 14.34" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "t = 400 # temperature in K\n", + "k = 1.38e-23 # Boltzmann's constant in J/K\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "m = 4 * 1.67e-27 # mass of helium atom in kg\n", + "\n", + "#Calculations\n", + "lamda = h / sqrt(3 * m * k * t)\n", + "\n", + "#Result\n", + "print \"de-Broglie wavelength = %.4f A\"%(lamda * 1e10)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "de-Broglie wavelength = 0.6294 A\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.36, Page 14.35" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "v = 2000 # velocity of neutron in m/sec\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "m = 1.67e-27 # mass of neutron in kg\n", + "\n", + "#Calculations\n", + "lamda = h / (m * v)\n", + "\n", + "#Result\n", + "print \"de-Broglie wavelength is %.2f A\"%(lamda * 1e10)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "de-Broglie wavelength is 1.98 A\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.37, Page 14.35" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "lamda = 1e-10 # wavelength in m\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "m = 9.1e-31 # mass of electron in kg\n", + "m_ = 1.7e-27 # mass of neutron in kg\n", + "\n", + "#Calculations\n", + "v = h / (m_ * lamda)\n", + "E = h**2 / (2 * m * lamda**2)\n", + "E_ = h**2 / (2 * m_ * lamda**2)\n", + "\n", + "#Result\n", + "print \"Energy for electron is %.f eV\\nEnergy for neutron is %.3f eV\"%(E / e,E_ / e)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy for electron is 150 eV\n", + "Energy for neutron is 0.081 eV\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.38, Page 14.36" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "E1 = 500 # kinetic energy of electron in first case in eV\n", + "E2 = 50 # kinetic energy of electron in second case in eV\n", + "E3 = 1 # kinetic energy of electron in third case in eV\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "m = 9.1e-31 # mass of electron in kg\n", + "\n", + "#Calculations\n", + "lambda1 = h / sqrt(2 * m * E1 * e)\n", + "lambda2 = h / sqrt(2 * m * E2 * e)\n", + "lambda3 = h / sqrt(2 * m * E3 * e)\n", + "\n", + "#Result\n", + "print \"de-Broglie wavelength of electron - \\n(1) In first case is %.4f A \\n(2) In second case is %.3f A \\n(3) In third is %.3f A\"%(lambda1*1e10,lambda2*1e10,lambda3*1e10)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "de-Broglie wavelength of electron - \n", + "(1) In first case is 0.5486 A \n", + "(2) In second case is 1.735 A \n", + "(3) In third is 12.268 A\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.39, Page 14.36" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "E1 = 1 # kinetic energy of neutron in first case in eV\n", + "E2 = 510 # kinetic energy of neutron in second case in eV\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "m = 1.67e-27 # mass of neutron in kg\n", + "\n", + "#Calculations\n", + "lambda1 = h / sqrt(2 * m * E1 * e)\n", + "lambda2 = h / sqrt(2 * m * E2 * e)\n", + "r = lambda1 / lambda2\n", + "\n", + "#Result\n", + "print \"Ratio of de-Broglie wavelengths is %.2f:1\"%r\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ratio of de-Broglie wavelengths is 22.58:1\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.40, Page 14.37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "E = 20 # kinetic energy of proton in MeV\n", + "E2 = 510 # kinetic energy of neutron in second case in eV\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "m = 1.67e-27 # mass of proton in kg\n", + "m_ = 9.1e-31 # mass of electron in kg\n", + "\n", + "#Calculations\n", + "lambda1 = h / sqrt(2 * m * 10**6 * E * e)\n", + "lambda2 = h / sqrt(2 * m_ * E * 10**6 * e)\n", + "r = lambda2 / lambda1\n", + "\n", + "#Result\n", + "print \"Ratio of de-Broglie wavelengths is 1:%.f\"%r\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ratio of de-Broglie wavelengths is 1:43\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.41, Page 14.37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "E = 1 # kinetic energy of proton in MeV\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "m = 1.67e-27 # mass of proton in kg\n", + "\n", + "#Calculations\n", + "v = sqrt(2 * E * 1.6e-13 / m)\n", + "\n", + "#Result\n", + "print \"Velocity is %.2e m/sec\"%v\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Velocity is 1.38e+07 m/sec\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.42, Page 14.38" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "r = 1. / 20 # ratio of velocity of proton to the velocity of light \n", + "c = 3e8 # velocity of light in m/sec\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "m = 1.67e-27 # mass of proton in kg\n", + "\n", + "#Calculations\n", + "v = r * c\n", + "lamda = h / (m * v)\n", + "\n", + "#Result\n", + "print \"de-Broglie wavelength is %.3e m\"%lamda\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "de-Broglie wavelength is 2.643e-14 m\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.43, Page 14.38" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "lamda = 5.0e-7 # wavelength in m\n", + "c = 3.e8 # velocity of light in m/sec\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "m = 1.67e-27 # mass of proton in kg\n", + "m_ = 9.1e-31 # mass of electron in kg\n", + "\n", + "#Calculations\n", + "E1 = h**2 / (2 * m * lamda**2)\n", + "E2 = h**2 / (2 * m_ * lamda**2)\n", + "\n", + "#Results\n", + "print 'kinetic energy of proton(in J) =%.3e'%E1\n", + "print 'kinetic energy of electron(in J) =%.2e'%E2\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "kinetic energy of proton(in J) =5.248e-28\n", + "kinetic energy of electron(in J) =9.63e-25\n" + ] + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.44, Page 14.38" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "n = 1 # no. of Bohr's orbit of hydrogen atom\n", + "c = 3e8 # velocity of light in m/sec\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "m = 9.1e-31 # mass of electron in kg\n", + "\n", + "#Calculations\n", + "E = (13.6 / n**2) * e\n", + "lamda = h / sqrt(2 * m * E)\n", + "\n", + "#Result\n", + "print \"de-Broglie wavelength is %.1f A\"%(lamda*1e10)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "de-Broglie wavelength is 3.3 A\n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.45, Page 14.38" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "t = 300 # temperature in K\n", + "k = 1.376e-23 # Boltzmann's constant in J/K\n", + "c = 3e8 # velocity of light in m/sec\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "m_ = 4 * 1.67e-27 # mass of helium atom in kg\n", + "m = 1.67e-27 # mass of hydrogen atom in kg\n", + "\n", + "#Calculations\n", + "lambda1 = h / sqrt(3 * m * k * t)\n", + "lambda2 = h / sqrt(3 * m_ * k * t)\n", + "r = lambda1 / lambda2\n", + "\n", + "#Result\n", + "print \"Ratio of de-Broglie wavelengths is %d:1\"%r\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ratio of de-Broglie wavelengths is 2:1\n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.47, Page 14.40" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "lamda = 1.2e-10 # DE Broglie wavelength in m\n", + "c = 3e8 # velocity of light in m/sec\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "e = 1.6e-19 # charge on an electron in C \n", + "m = 9.1e-31 # mass of electron in kg\n", + "\n", + "#Calculations\n", + "v1 = h / (m * lamda)\n", + "v2 = h / (2 * m * lamda)\n", + "\n", + "#Result\n", + "print \"Group velocity is %.2e m/sec\\nPhase velocity is %.2e m/sec\"%(v1,v2)\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Group velocity is 6.06e+06 m/sec\n", + "Phase velocity is 3.03e+06 m/sec\n" + ] + } + ], + "prompt_number": 41 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter_15.ipynb b/Engineering_Physics/Chapter_15.ipynb new file mode 100755 index 00000000..2b3eeafd --- /dev/null +++ b/Engineering_Physics/Chapter_15.ipynb @@ -0,0 +1,983 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 15: Quantum Mechanics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.1, Page 15.24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "E = 1000 # energy of electron in eV\n", + "delta_x = 1e-10 # error in position in m\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of electron in kg\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "\n", + "#Calculations\n", + "p = sqrt(2 * m * E * e)\n", + "delta_p = h / (4 * pi * delta_x)\n", + "P = (delta_p / p) * 100\n", + "\n", + "#Result\n", + "print \"Percentage of uncertainty in momentum is %.1f%%\"%P" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Percentage of uncertainty in momentum is 3.1%\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.3, Page 15.25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "E = 500 # energy of electron in eV\n", + "delta_x = 2e-10 # error in position in m\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of electron in kg\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "\n", + "#Calculations\n", + "p = sqrt(2 * m * E * e)\n", + "delta_p = h / (4 * pi * delta_x)\n", + "P = (delta_p / p) * 100\n", + "\n", + "#Result\n", + "print \"Percentage of uncertainty in momentum is %.2f%%\"%P" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Percentage of uncertainty in momentum is 2.18%\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.4, Page 15.25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "delta_lambda = 1e-6 # accuracy in wavelength of its one part\n", + "lamda = 1e-10 # wavelength of x-ray in m\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "\n", + "#Calculations\n", + "delta_x = lamda / (4 * pi * delta_lambda)\n", + "\n", + "#Result\n", + "print \"Uncertainty in position is %.2f micrometer\"%(delta_x*10**6)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Uncertainty in position is 7.96 micrometer\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.5, Page 15.26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "delta_x = 1e-10 # error in position in m\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of electron in kg\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "\n", + "#Calculations\n", + "delta_p = h / (4 * pi * delta_x)\n", + "\n", + "#Result\n", + "print \"Uncertainty in momentum is %.2e kg m/sec\"%delta_p" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Uncertainty in momentum is 5.27e-25 kg m/sec\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.6, Page 15.26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "M = 5.4e-26 # momentum of electron in kg-m/sec\n", + "p = 0.05 # percentage accuracy in momentum\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of electron in kg\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "\n", + "#Calculations\n", + "delta_m = p * M / 100\n", + "delta_x = h / (4 * pi * delta_m)\n", + "\n", + "#Result\n", + "print \"Uncertainty in position is %.3f micrometer\"%(delta_x * 10**6)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Uncertainty in position is 1.951 micrometer\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.7, Page 15.27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "r = 0.53e-10 # radius of hydrogen atom in m\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of electron in kg\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "\n", + "#Calculations\n", + "delta_M = h / (4 * pi * r)\n", + "delta_k = delta_M**2 / (2 * m)\n", + "\n", + "#Result\n", + "print \"Minimum energy of electron is %.3e J\"%delta_k" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum energy of electron is 5.428e-19 J\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.8, Page 15.27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "v = 5e3 # speed of electron in m/sec\n", + "a = 0.003 # percentage accuracy in measurement of speed \n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of electron in kg\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "\n", + "#Calculations\n", + "delta_v = v * a / 100\n", + "delta_p = m * delta_v\n", + "delta_x = h / (4 * pi * delta_p)\n", + "\n", + "#Result\n", + "print \"Uncertainty in determining the position of electron is %.3e m\"%delta_x" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Uncertainty in determining the position of electron is 3.859e-04 m\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.9, Page 15.27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "v = 6.6e4 # speed of electron in m/sec\n", + "a = 0.01 # percentage accuracy in measurement of speed \n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of electron in kg\n", + "h = 6.6e-34 # Planck constant in J-sec\n", + "\n", + "#Calculations\n", + "delta_v = v * a / 100\n", + "delta_p = m * delta_v\n", + "delta_x = h / (4 * pi * delta_p)\n", + "\n", + "#Result\n", + "print \"Uncertainty in determining the position is %.2e m\"%delta_x" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Uncertainty in determining the position is 8.74e-06 m\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.10, Page 15.28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "v = 3e7 # speed of electron in m/sec \n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of electron in kg\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "c = 3e8 # speed of light in m/sec\n", + "\n", + "#Calculations\n", + "delta_p = m * v / sqrt(1 - (v/c)**2)\n", + "delta_x = h / (4 * pi * delta_p)\n", + "\n", + "#Result\n", + "print \"Uncertainty in determining the position is %.2e m\"%delta_x" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Uncertainty in determining the position is 1.92e-12 m\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.11, Page 15.28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "# Given \n", + "t = 2.5e-14 # life time of hydrogen atom in exited state in sec\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "\n", + "#Calculations\n", + "delta_E = h / (4 * pi * t)\n", + "\n", + "#Result\n", + "print \"Minimum error in measurement of the energy is %.2e J\"%delta_E" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum error in measurement of the energy is 2.11e-21 J\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.12, Page 15.28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "t = 10**-8 # life time of atom in exited state in sec\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "\n", + "#Calculations\n", + "delta_f = 1 / (4 * pi * t)\n", + "\n", + "#Result\n", + "print \"Minimum uncertainty in frequency is %.2e sec\"%delta_f" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum uncertainty in frequency is 7.96e+06 sec\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.13, Page 15.29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "# Given \n", + "delta_x = 20e-10 # uncertainty in position in m\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of electron in kg\n", + "m_ = 1.67e-27 # mass of proton in kg\n", + "c = 3e8 # speed of light in m/sec\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "\n", + "#Calculations\n", + "delta_v1 = h / (4 * pi * m * delta_x)\n", + "delta_v2 = h / (4 * pi * m_ * delta_x)\n", + "r = delta_v2 / delta_v1\n", + "\n", + "#Result\n", + "print \"Ratio of uncertainty in velocity of a proton and an electron is %.2e\"%r" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ratio of uncertainty in velocity of a proton and an electron is 5.45e-04\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.14, Page 15.29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "delta_x = 1e-10 # width of box in m\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of electron in kg\n", + "c = 3e8 # speed of light in m/sec\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "n = 1 # for n=1\n", + "\n", + "#Calculations\n", + "E = (n**2 * h**2) / (8 * m * delta_x**2)\n", + "n = 2 # for n=2\n", + "E_ = (n**2 * h**2) / (8 * m * delta_x**2)\n", + "\n", + "#Result\n", + "print \"Energy of electron - \\nFor (n=1) energy is %.2e J\\nFor (n=2) energy is %.2e J\"%(E,E_)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy of electron - \n", + "For (n=1) energy is 6.02e-18 J\n", + "For (n=2) energy is 2.41e-17 J\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.15, Page 15.30" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Given \n", + "l = 1e-10 # width of box in m\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of electron in kg\n", + "c = 3e8 # speed of light in m/sec\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "n = 1 # for n=1\n", + "\n", + "#Calculations\n", + "E = (n**2 * h**2) / (8 * m * l**2)\n", + "n = 2 # for n=2\n", + "E_ = (n**2 * h**2) / (8 * m * l**2)\n", + "d = E_ - E\n", + "\n", + "#Result\n", + "print \"Energy difference is %.2e J\"%d" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy difference is 1.81e-17 J\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.16, Page 15.30" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "l = 3e-10 # width of box in m\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of electron in kg\n", + "c = 3e8 # speed of light in m/sec\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "n = 1 # For n=1\n", + "\n", + "#Calculations\n", + "E = (n**2 * h**2) / (8 * m * l**2)\n", + "n = 2 # For n=2\n", + "E_ = (n**2 * h**2) / (8 * m * l**2)\n", + "n = 3 # For n=3\n", + "E__ = (n**2 * h**2) / (8 * m * l**2)\n", + "\n", + "#Result\n", + "print \"Energy of electron -\\nFor (n=1) is %.1e J\\nFor (n=2) is %.2e J\\nFor (n=3) is %.2e J\"%(E,E_,E__)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy of electron -\n", + "For (n=1) is 6.7e-19 J\n", + "For (n=2) is 2.68e-18 J\n", + "For (n=3) is 6.02e-18 J\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.17, Page 15.30" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "l = 2.5e-10 # width of box in m\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of electron in kg\n", + "c = 3e8 # speed of light in m/sec\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "\n", + "#Calculations\n", + "n = 1 # for n=1\n", + "E = (n**2 * h**2) / (8 * m * l**2)\n", + "n = 2 # for n=2\n", + "E_ = (n**2 * h**2) / (8 * m * l**2)\n", + "\n", + "#Result\n", + "print \"Energy of electron -\\nFor (n=1) is %.2e J\\nFor (n=2) is %.3e J\"%(E,E_)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy of electron -\n", + "For (n=1) is 9.63e-19 J\n", + "For (n=2) is 3.853e-18 J\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.18, Page 15.31" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "l = 1e-14 # width of box in m\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 1.67e-27 # mass of neutron in kg\n", + "c = 3e8 # speed of light in m/sec\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "\n", + "#Calculations\n", + "n = 1 # for n=1\n", + "E = (h**2) / (8 * m * l**2)\n", + "\n", + "#Result\n", + "print \"Lowest energy of neutron confined in the nucleus is %.2e J\"%E" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Lowest energy of neutron confined in the nucleus is 3.28e-13 J\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.19, Page 15.31" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "l = 1e-10 # width of box in m\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of electron in kg\n", + "c = 3e8 # speed of light in m/sec\n", + "h = 6.63e-34 # Planck constant in J-sec\n", + "\n", + "#Calculations\n", + "n = 1 # for n=1\n", + "p1 = (n * h) / (2 * l)\n", + "E = (n**2 * h**2) / (8 * m * l**2)\n", + "n = 2 # for n=2\n", + "p2 = (n * h) / (2 * l)\n", + "E_ = (n**2 * h**2) / (8 * m * l**2)\n", + "\n", + "#Result\n", + "print \"Energy of electron -\\nFor (n=1) is %.2e J\\nFor (n=2) is %.2e J\"%(E,E_)\n", + "print \"\\nMomentum of electron -\\nFor (n=1) is %.3e kg-m/sec\\nFor (n=2) is %.2e kg-m/sec\"%(p1,p2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy of electron -\n", + "For (n=1) is 6.04e-18 J\n", + "For (n=2) is 2.42e-17 J\n", + "\n", + "Momentum of electron -\n", + "For (n=1) is 3.315e-24 kg-m/sec\n", + "For (n=2) is 6.63e-24 kg-m/sec\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.20, Page 15.32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "l = 1e-10 # length of box in m\n", + "m = 9.1e-31 # mass of electron in kg\n", + "c = 3e8 # speed of light in m/sec\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "\n", + "#Calculations\n", + "n = 1 # for n=1\n", + "E1 = (n**2 * h**2) / (8 * m * l**2)\n", + "lambda1 =2*l\n", + "n = 2 # for n=2\n", + "E2 = (n**2 * h**2) / (8 * m * l**2)\n", + "lambda2 =2*l/2\n", + "n = 3 # for n=3\n", + "E3 = (n**2 * h**2) / (8 * m * l**2)\n", + "lambda3 =2*l/3\n", + "\n", + "#Results\n", + "print \"Energy Eigen value of electron -\\nFor (n=1) is %.2e J\\nFor (n=2) is %.2e J\\nFor (n=3) is %.2e J\"%(E1,E2,E3)\n", + "print \"\\nde-Broglie wavelength of electron -\\nFor (n=1) is %.f A\\nFor (n=2) is %.f A \\nFor (n=3) is %.3f A\"%(lambda1*1e10,lambda2*1e10,lambda3*1e10)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy Eigen value of electron -\n", + "For (n=1) is 6.02e-18 J\n", + "For (n=2) is 2.41e-17 J\n", + "For (n=3) is 5.42e-17 J\n", + "\n", + "de-Broglie wavelength of electron -\n", + "For (n=1) is 2 A\n", + "For (n=2) is 1 A \n", + "For (n=3) is 0.667 A\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.21, Page 15.32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "E1 = 3.2e-18 # minimum energy possible for a particle entrapped in a one dimensional box in J\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of electron in kg\n", + "c = 3e8 # speed of light in m/sec\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "\n", + "#Calculations\n", + "E1 = E1 / e # in eV\n", + "n = 2 # for n=2\n", + "E2 = n**2 * E1\n", + "n = 3 # for n=3\n", + "E3 = n**2 * E1\n", + "n = 4 # for n=4\n", + "E4 = n**2 * E1\n", + "\n", + "#Result\n", + "print \"Energy Eigen values -\\nFor (n=2) for %.f eV\\nFor (n=3) is %.f eV\\nFor (n=4) is %.f eV\"%(E2,E3,E4)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy Eigen values -\n", + "For (n=2) for 80 eV\n", + "For (n=3) is 180 eV\n", + "For (n=4) is 320 eV\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.22, Page 15.33" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "l = 4e-10 # width of box in m\n", + "E = 9.664e-17 # energy of electron in J\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of electron in kg\n", + "c = 3e8 # speed of light in m/sec\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "\n", + "#Calculations\n", + "n = 1 # for n=1\n", + "E1 = (n**2 * h**2) / (8 * m * l**2)\n", + "N = sqrt(E / E1)\n", + "p = ((N) * h) / (2 * l)\n", + "\n", + "#Result\n", + "print \"Order of exited state is %d\\nMomentum of electron is %.2e kg-m/sec\"%(N,p)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Order of exited state is 16\n", + "Momentum of electron is 1.33e-23 kg-m/sec\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.23, Page 15.33" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "l = 10e-10 # width of box containing electron in m\n", + "E = 9.664e-17 # energy of electron in J\n", + "M = 0.001 # mass of glass marble in kg\n", + "l_ = 0.2 # width of box containing marble in m\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of electron in kg\n", + "c = 3e8 # speed of light in m/sec\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "\n", + "#Calculations\n", + "# For electron\n", + "n = 1 # for n=1\n", + "E1 = (n**2 * h**2) / (8 * m * l**2)\n", + "E2 = 2**2* E1\n", + "E3 = 3**2 * E1\n", + "# For glass marble\n", + "E1_ = h**2/(8*M*l_**2)\n", + "E2_ = 2**2 * E1_\n", + "E3_ = 3**2 *E1_\n", + "\n", + "#Result\n", + "print \"\\nEnergy levels of electron- \\nFor (n=1) is %.2e J\\nFor (n=2) is %.2e J\\n For (n=3) is %.2e J\"%(E1,E2,E3)\n", + "print \"\\nEnergy levels of marble- \\nFor (n=1) is %.2e J\\nFor (n=2) is %.2e J\\nFor (n=3) is %.2e J\"%(E1_,E2_,E3_)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "Energy levels of electron- \n", + "For (n=1) is 6.02e-20 J\n", + "For (n=2) is 2.41e-19 J\n", + " For (n=3) is 5.42e-19 J\n", + "\n", + "Energy levels of marble- \n", + "For (n=1) is 1.37e-63 J\n", + "For (n=2) is 5.48e-63 J\n", + "For (n=3) is 1.23e-62 J\n" + ] + } + ], + "prompt_number": 25 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter_16.ipynb b/Engineering_Physics/Chapter_16.ipynb new file mode 100755 index 00000000..0fab1922 --- /dev/null +++ b/Engineering_Physics/Chapter_16.ipynb @@ -0,0 +1,467 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 16: Free Electron Theory" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.1, Page 16.14" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "t = 0 # temperature in K\n", + "E = 10 # Fermi energy of electron in eV\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of electron in kg\n", + "\n", + "#Calculations\n", + "E_ = E * 3 / 5\n", + "v = sqrt(2 * E_ * e / m)\n", + "\n", + "#Result\n", + "print \"\\nAverage energy of electron is %.f eV\\nSpeed of electron is %.2e m/sec\"%(E_,v)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "Average energy of electron is 6 eV\n", + "Speed of electron is 1.45e+06 m/sec\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.2, Page 16.14" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "t = 0 # temperature in K\n", + "E = 7.9 # Fermi energy in eV\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of electron in kg\n", + "\n", + "#Calculations\n", + "E_ = E * 3 / 5\n", + "v = sqrt(2 * E_ * e / m)\n", + "\n", + "#Result\n", + "print \"Average energy of electron is %.2f eV\\nSpeed of electron is %.2e m/sec\"%(E_,v)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Average energy of electron is 4.74 eV\n", + "Speed of electron is 1.29e+06 m/sec\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.3, Page 16.15" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "n = 2.5e28 # no. of free electron in per meter cube\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of electron in kg\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "\n", + "#Calculations\n", + "E = (h**2 / (8 * pi**2 * m)) * (3 * pi**2 * n)**(2./3) * (1. / e)\n", + "v = (h / (2 * pi * m)) * (3 * pi**2 * n)**(1./3)\n", + "\n", + "#Results\n", + "print \"Fermi energy is %.2f eV\\nSpeed of electron is %.2e m/sec\"%(E,v)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Fermi energy is 3.12 eV\n", + "Speed of electron is 1.05e+06 m/sec\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.4, Page 16.15" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "d = 8940 # density of copper in kg/m^3\n", + "w = 63.55 # atomic weight of copper\n", + "t = 0 # temperature in K\n", + "N = 6.02e26 # Avogadro no. in per kg\n", + "m = 9.1e-31 # mass of electron in kg\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#Calculations\n", + "V = w / d\n", + "n = N / V \n", + "E = (h**2 / (8 * pi**2 * m)) * (3 * pi**2 * n)**(2./3) * (1. / e)\n", + "E_ = 3 * E / 5\n", + "\n", + "#Results\n", + "print \"Fermi energy is %.3f eV\\nAverage energy is %.2f eV\"%(E,E_)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Fermi energy is 7.036 eV\n", + "Average energy is 4.22 eV\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.5, Page 16.16\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "d = 10.5e6 # density of silver in g/m^3\n", + "w = 108 # atomic weight of silver\n", + "t = 0 # temperature in K\n", + "N = 6.02e23 # Avogadro no. in per kg\n", + "m = 9.1e-31 # mass of electron in kg\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#Calculations\n", + "V = w / d\n", + "n = N / V \n", + "E = (h**2 / (8 * pi**2 * m)) * (3 * pi**2 * n)**(2./3) * (1 / e)\n", + "\n", + "#Result\n", + "print \"Fermi energy is %.1f eV\"%E" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Fermi energy is 5.5 eV\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.6, Page 16.16" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "a = 4e-10 # lattice constant in mr\n", + "t = 0 # temperature in K\n", + "N = 6.02e23 # Avogadro no. in per kg\n", + "m = 9.1e-31 # mass of electron in kg\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#Calculations\n", + "V = a**3\n", + "n = 4 / V \n", + "E = (h**2 / (8 * pi**2 * m)) * (3 * pi**2 * n)**(2./3) * (1 / e)\n", + "k = (3 * pi**2 *n)**(1./3)\n", + "KE = (3 * E / 5) * (n)\n", + "\n", + "#Results\n", + "print \"Fermi energy is %.2f eV\\nFermi vector is %.2e per m\\nTotal kinetic energy is %.2e eV\"%(E,k,KE)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Fermi energy is 5.75 eV\n", + "Fermi vector is 1.23e+10 per m\n", + "Total kinetic energy is 2.15e+29 eV\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.7, Page 16.17" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "# Given \n", + "d = 0.9e-3 # diameter of aluminium in m\n", + "i = 6 # current in amp\n", + "n = 4.5e28 # no. of electron available for conduction per meter^3 \n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#Calculations\n", + "J = i * 4 / (pi * (d)**2)\n", + "v = J / (n * e)\n", + "\n", + "#Result\n", + "print \"Drift velocity of electron is %.3e m/sec\"%v" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Drift velocity of electron is 1.310e-03 m/sec\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.8, Page 16.17" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "d = 8.92e3 # density of copper in kg/m^3\n", + "i = 5 # current in amp\n", + "w = 63.5 # atomic weight of copper\n", + "r = 0.7e-3 # radius in meter\n", + "N = 6.02e28 # Avogadro no.\n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#Calculations\n", + "V = (w / d)\n", + "n = N / V \n", + "J = i / (pi * r**2)\n", + "v = J / (n * e)\n", + "\n", + "#Result\n", + "print \"Current density = %.2e amp/m^2\\nDrift velocity is %.1e m/sec\"%(J,v)\n", + "#Incorrect units in the textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current density = 3.25e+06 amp/m^2\n", + "Drift velocity is 2.4e-06 m/sec\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.9, Page 16.17" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given\n", + "d1= 0.534*10**3 # densiy of Li in kg/m^3\n", + "d2= 0.971*10**3 # densiy of Na in kg/m^3\n", + "d3= 0.86*10**3 # densiy of K in kg/m^3\n", + "w1 = 6.939 # atomic weight of Li \n", + "w2 = 22.99 # atomic weight of Na\n", + "w3 = 39.202 # atomic weight of K\n", + "h = 6.62e-34 # Planck constant in J sec\n", + "m = 9.1e-31 # mass of an electron in kg\n", + "NA = 6.023e26 # Avogadro no.\n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#Calculations\n", + "# For Li\n", + "n1 = NA * d1/w1\n", + "E1 = h**2/(8*pi**2*m)*(3*pi**2*n1)**(2./3)\n", + "# For Na\n", + "n2 = NA * d2/w2\n", + "E2 = h**2/(8*pi**2*m)*(3*pi**2*n2)**(2./3)\n", + "# For K\n", + "n3 = NA * d3/w3\n", + "E3 = h**2/(8*pi**2*m)*(3*pi**2*n3)**(2./3)\n", + "\n", + "#Results\n", + "print \"Fermi Energy \\nFor Li is %.2f eV\\nFor Na is %.3f eV \\nFor K is %.3f eV\"%(E1/e,E2/e,E3/e)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Fermi Energy \n", + "For Li is 4.71 eV\n", + "For Na is 3.156 eV \n", + "For K is 2.039 eV\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.10, Page 16.18\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "l = 1e-10 #length of box in m\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of electron in kg\n", + "h = 6.62e-34 # Planck constant in J-sec\n", + "\n", + "#Calculations\n", + "n = 1 # for n=1\n", + "E = (n**2 * h**2) / (8 * m * l**2)\n", + "n = 2 # for n=2\n", + "E_ = (n**2 * h**2) / (8 * m * l**2)\n", + "d = (E_ - E) * (1 / e)\n", + "\n", + "#Result\n", + "print \"Energy difference is %.2f eV\"%d" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy difference is 112.87 eV\n" + ] + } + ], + "prompt_number": 10 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter_17.ipynb b/Engineering_Physics/Chapter_17.ipynb new file mode 100755 index 00000000..01028f54 --- /dev/null +++ b/Engineering_Physics/Chapter_17.ipynb @@ -0,0 +1,258 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 17: Band Theory of Solids" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.1, Page 17.18" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "a = 3e-10 # side of square lattice in m\n", + "h = 6.62e-34 # Planck constant in J sec\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of electron in kg\n", + "\n", + "#Calculations\n", + "p = (h / (2 * a))\n", + "E = (p**2 / (2 * m)) * (1. / e)\n", + "\n", + "#Result\n", + "print \"Electron momentum value at the sides of first Brilloin zone is %.1e kg-m/sec\\nEnergy of free electron is %.1f eV\"%(p,E)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Electron momentum value at the sides of first Brilloin zone is 1.1e-24 kg-m/sec\n", + "Energy of free electron is 4.2 eV\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.2, Page 17.19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log,pi\n", + "\n", + "# Given \n", + "n = 5e22 # no. of atoms per m^3\n", + "t = 300 # room temperature in K\n", + "k = 1.37e-23 # Boltzmann's constant in J/K\n", + "h = 6.62e-34 # Planck constant in J sec\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of electron in kg\n", + "\n", + "#Calculations\n", + "d = (k * t) * log(n * h**3 / (2 * (2 * pi * m * k * t)**(3./2)))\n", + "\n", + "#Result\n", + "print \"Position of fermi level is %.3f eV\"%(-d/e)\n", + "#Answer varies due to round-off errors" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Position of fermi level is 0.159 eV\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.3, Page 17.19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "E = 0.3 # Fermi energy in eV\n", + "T = 330 # temperature in K\n", + "t = 300 # room temperature in K\n", + "k = 1.37e-23 # Boltzmann's constant in J/K\n", + "h = 6.62e-34 # Planck constant in J sec\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of electron in kg\n", + "\n", + "#Calculations\n", + "d = (T / t) * (E)\n", + "\n", + "print \"New position of fermi level is %.2f eV\"%d" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "New position of fermi level is 0.30 eV\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.4, Page 17.20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given\n", + "E = 0.7 # band gap for semiconductor in eV\n", + "t = 300 # room temperature in K\n", + "k = 1.38e-23 # Boltzmann's constant in J/K\n", + "h = 6.62e-34 # Planck constant in J sec\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of electron in kg\n", + "\n", + "#Calculations\n", + "n = 2 * ((2 * pi * k * t * m) / h**2)**(3./2) * exp(-(E * e / (2 * k * t)))\n", + "\n", + "#Result\n", + "print \"Density of holes and electron is %.1e per m^3\"%n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Density of holes and electron is 3.4e+19 per m^3\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.5, Page 17.20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "n = 5e28 # no. of atoms in per m^3\n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#Calculations\n", + "R = -(1 / (n * e))\n", + "\n", + "#Resilt\n", + "print \"Hall coefficient is %.3e m^3/C\"%R" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Hall coefficient is -1.250e-10 m^3/C\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.6, Page 17.20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "a = 4.28e-10 # cell side of Na in m\n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#Calculations\n", + "n = (2 / a**3)\n", + "R = -(1 / (n * e))\n", + "\n", + "#Result\n", + "print \"Hall coefficient is %.3e m^3/C\"%R" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Hall coefficient is -2.450e-10 m^3/C\n" + ] + } + ], + "prompt_number": 7 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter_18.ipynb b/Engineering_Physics/Chapter_18.ipynb new file mode 100755 index 00000000..de6fffda --- /dev/null +++ b/Engineering_Physics/Chapter_18.ipynb @@ -0,0 +1,876 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 18: Magnetic Properties of Solids" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 18.1, Page 18.21" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "r = 0.53e-10 # radius of orbit in m\n", + "f = 6.6e15 # frequency of revolution in Hz\n", + "h = 6.6e-34 # Planck constant in J sec\n", + "e = 1.6e-19 # charge on an electron in C\n", + "m = 9.1e-31 # mass of electron in kg\n", + "\n", + "#Calculations\n", + "M = e * f * pi * r**2\n", + "mu = (e * h) / (4 * pi * m) \n", + "\n", + "#Result\n", + "print \"Magnetic moment is %.3e Am^2\\nBohr magneton is %.2e J/T\"%(M,mu)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Magnetic moment is 9.319e-24 Am^2\n", + "Bohr magneton is 9.23e-24 J/T\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 18.2, Page 18.21" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "X = -4.2e-6 # magnetic susceptibility\n", + "H = 1.19e5 # magnetic field in A/m\n", + "mu_ = 4 * pi * 1e-7 # magnetic permeability of space\n", + "\n", + "#Calculations\n", + "I = X * H\n", + "B = mu_ * (H + I)\n", + "mur = (1 + I/H)\n", + "\n", + "#Results\n", + "print \"Magnetisation is %.2f A/m\\nFlux density is %.3f T\\nRelative permeability is %.2f\"%(I,B,mur)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Magnetisation is -0.50 A/m\n", + "Flux density is 0.150 T\n", + "Relative permeability is 1.00\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 18.3, Page 18.22" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "X = 1.2e-5 # magnetic susceptibility of magnesium\n", + "\n", + "#Calculations\n", + "p = 100 * X\n", + "\n", + "#Result\n", + "print \"Percentage increase in magnetic induction is %.4f percent\"%p" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Percentage increase in magnetic induction is 0.0012 percent\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 18.4, Page 18.22" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "X = -0.4e-5 # magnetic susceptibility of material\n", + "H = 1e4 # magnetic field in A/m\n", + "mu_ = 4 * pi * 1e-7 # magnetic permittivity of space\n", + "\n", + "#Calculations\n", + "I = X * H\n", + "B = mu_ * (H + I)\n", + "\n", + "#Result\n", + "print \"Magnetisation is %.2f A/m\\nMagnetic flux density is %f T\"%(I,B)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Magnetisation is -0.04 A/m\n", + "Magnetic flux density is 0.012566 T\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 18.5, Page 18.23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "X = 2.3e-5 # magnetic susceptibility of aluminium\n", + "mu_ = 4 * pi * 1e-7 # magnetic permeability of space\n", + "\n", + "#Calculations\n", + "mur = 1 + X\n", + "mu = mu_ * mur\n", + "\n", + "#Result\n", + "print \"Permeability of aluminium is %.2e N/A^2\"%mu" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Permeability of aluminium is 1.26e-06 N/A^2\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 18.6, Page 18.23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "# Given \n", + "X = 9.4e-2 # magnetic susceptibility\n", + "mu_ = 4 * pi * 1e-7 # magnetic permeability of space\n", + "\n", + "#Calculations\n", + "mu_r = 1 + X\n", + "mu = mu_ * mu_r\n", + "\n", + "#Results\n", + "print \"Absolute permeability is %.2e N/A^2\\nRelative permeability is %.3f\"%(mu,mu_r)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Absolute permeability is 1.37e-06 N/A^2\n", + "Relative permeability is 1.094\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 18.7, Page 18.23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "mu = 0.126 # maximum value of the permeability in N/A^2\n", + "mu_ = 4 * pi * 1e-7 # magnetic permeability of space\n", + "\n", + "#Calculations\n", + "mu_r = mu / mu_\n", + "X = mu_r - 1\n", + "\n", + "#Results\n", + "print \"Magnetic susceptibility is %.f\\nRelative permeability is %e\"%(X,mu_r)\n", + "#Answers differ due to rounding off values" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Magnetic susceptibility is 100267\n", + "Relative permeability is 1.002676e+05\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 18.8, Page 18.24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "# Given \n", + "r = 0.6e-10 # radius of the atom\n", + "N = 28e26 # no. of electron in per m^3\n", + "mu_ = 4 * pi * 1e-7 # magnetic permeability of space\n", + "Z = 2 # atomic no. of helium\n", + "m = 9.1e-31 # mass of an electron in kg\n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#Calculations\n", + "Chi = -(mu_ * Z * N * r**2 * e**2) / (6 * m)\n", + "\n", + "#Result\n", + "print \"Diamagnetic susceptibility is %.3e\"%Chi" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diamagnetic susceptibility is -1.188e-07\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 18.9, Page 18.24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "# Given \n", + "H = 1e3 # magnetisation field in A/m\n", + "phi = 2e-5 # magnetic flux in Weber\n", + "a = 0.2e-4 # area of cross section in m^2\n", + "mu_ = 4 * pi * 1e-7 # magnetic permeability of space\n", + "\n", + "#Calculations\n", + "B = phi / a\n", + "mu = B / H\n", + "X = mu / mu_ - 1\n", + "\n", + "#Results\n", + "print \"Permeability is %.e N/A^2\\nSusceptibility is %.3f\"%(mu,X)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Permeability is 1e-03 N/A^2\n", + "Susceptibility is 794.775\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 18.10, Page 18.24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "l = 1 # length of iron rod in m\n", + "a = 4e-4 # area in m^2\n", + "mu = 50e-4 # permeability of iron in H/m\n", + "Phi = 4e-4 # magnetic flux in Weber\n", + "\n", + "#Calculations\n", + "B = Phi / a\n", + "NI = B / mu\n", + "\n", + "#Result\n", + "print \"Number of ampere turns is %d A/m\"%NI" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of ampere turns is 200 A/m\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 18.11, Page 18.25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "n = 200 # no. of turns \n", + "l = 0.5 # the mean length of iron wire in m\n", + "phi = 4e-4 # magnetic flux in Weber\n", + "a = 4e-4 # area of cross section in m^2\n", + "mu = 6.5e-4 # permeability of iron in wb/Am\n", + "mu_ = 4 * pi * 1e-7 # magnetic permeability of space\n", + "\n", + "#Calculations\n", + "B = phi / a\n", + "N = n / l\n", + "I = B / (mu * N)\n", + "\n", + "#Result\n", + "print \"Current through the winding is %.2f A\"%I" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current through the winding is 3.85 A\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 18.12, Page 18.25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "# Given \n", + "X = -5.6e-6 # magnetic susceptibility of material\n", + "a = 2.55e-10 # lattice constant in m\n", + "H = 1e4 # magnetic field in A/m\n", + "mu_ = 4 * pi * 1e-7 # magnetic permittivity of space\n", + "m = 9.1e-31 # mass of electron in kg\n", + "e = 1.6e-19 # charge in an electron in C\n", + "\n", + "#Calculations\n", + "N = 2 / a**3\n", + "z = 1 \n", + "R = ((-X * 6 * m) / (mu_ * z * e**2 * N))**(1./2)\n", + "\n", + "#Result\n", + "print \"Radius of atom is %.2f A\"%(R * 1e10)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Radius of atom is 0.89 A\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 18.13, Page 18.25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "# Given \n", + "N = 6.5e25 # no. of atom per m^3\n", + "T = 300 # room temperature in K\n", + "mu_ = 4 * pi * 1e-7 # magnetic permittivity of space\n", + "k = 1.38e-23 # Boltzmann's constant in J/K\n", + "m = 9.1e-31 # mass of electron in kg\n", + "e = 1.6e-19 # charge in an electron in C\n", + "h = 6.62e-34 # Planck constant in J sec\n", + "\n", + "#Calculations\n", + "M = (e * h) / (4 * pi * m)\n", + "X = (mu_ * N * M**2) / (3 * k * T)\n", + "\n", + "#Result\n", + "print \"Susceptibility is %.3e\"%X" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Susceptibility is 5.642e-07\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 18.14, Page 18.26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "w = 168.5 # molecular weight \n", + "d = 4370 # density of material in kg/m^3\n", + "H = 2e5 # magnetic field in A/m\n", + "T = 300 # room temperature in K\n", + "mu_ = 4 * pi * 1e-7 # magnetic permittivity of space]\n", + "NA = 6.02e26 # Avogadro no. in per kg\n", + "mu_b = 9.24e-24 # Bohr magnetons in Am^2\n", + "k = 1.38e-23 # Boltzmann's constant in J/K\n", + "\n", + "#Calculations\n", + "N = d * NA / w\n", + "X = (mu_ * N * (2 * mu_b)**2) / (3 * k * T)\n", + "I = X * H\n", + "\n", + "#Result\n", + "print \"Magnetisation is %.2f A/m\"%I" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Magnetisation is 107.89 A/m\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 18.15, Page 18.26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given that\n", + "A = 2500 # area of hysteresis loop \n", + "m = 10000 # weight in kg\n", + "d = 7.5 # density of material in g/cm^3\n", + "f = 50 # frequency in Hz\n", + "\n", + "#Calculations\n", + "E = f * A * 3600\n", + "V = m / d\n", + "L = E * V\n", + "\n", + "#Result\n", + "print \"Total loss of energy per hour is %.e ergs\"%L" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Total loss of energy per hour is 6e+11 ergs\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 18.16, Page 18.27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "H = 5e3 # coercivity in A/m\n", + "l = 0.10 # length of solenoid in m\n", + "n = 50 # no. of turns \n", + "\n", + "#Calculations\n", + "N = n / l\n", + "i = H / N\n", + "\n", + "#Result\n", + "print \"Current in solenoid should be %d A\"%i" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current in solenoid should be 10 A\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 18.17, Page 18.27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "l = 0.50 # length of iron rod in m\n", + "a = 4e-4 # area of cross section of rod in m^2\n", + "mu = 65e-4 # permeability of iron in H/m\n", + "fi = 4e-5 # flux in weber \n", + "\n", + "#Calculations\n", + "B = fi / a\n", + "H = B / mu\n", + "N = H * l\n", + "\n", + "#Result\n", + "print \"Number of turns are %.2f\"%N" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of turns are 7.69\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 18.18, Page 18.27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "H = 600 # magnetic flux in A/m\n", + "a = 0.2e-4 # area of cross section of rod in m^2\n", + "phi = 2.4e-5 # flux in weber \n", + "mu_ = 4*pi * 1e-7 # permeability of space in N/A^2\n", + "\n", + "#Calculations\n", + "B = phi / a\n", + "mu = B / H\n", + "X = mu / mu_ - 1\n", + "\n", + "#Result\n", + "print \"Permeability is %.3f N/A^2\\nSusceptibility is %.f\"%(mu,X)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Permeability is 0.002 N/A^2\n", + "Susceptibility is 1591\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 18.19, Page 18.28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "X = 9.5e-9 # susceptibility of medium \n", + "mu_ = 4*pi * 1e-7 # permeability of space in N/A^2\n", + "\n", + "#Calculations\n", + "mu = mu_ * (1 + X)\n", + "mu_r = mu / mu_\n", + "\n", + "#Result\n", + "print \"Relative permeability is 1 + %.3e\"%(mu_r -1)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Relative permeability is 1 + 9.500e-09\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 18.20, Page 18.28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "a = 250. # area of the B-H loop in J/m^3\n", + "f = 50. # frequency in Hz\n", + "d = 7.5e3 # density of iron in kg/m^3\n", + "m = 100. # mass of core in kg\n", + "\n", + "#Calculations\n", + "V = m / d\n", + "n = 3600 * f\n", + "A = a * V * n\n", + "\n", + "\n", + "#Result\n", + "print \"Energy loss per hour is %.3e J\"%A\n", + "#Answer varies due to rounding-off values" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy loss per hour is 6.000e+05 J\n" + ] + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 18.21, Page 18.28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "B_max = 1.375 # maximum value of B in Wb/m^2\n", + "a = 0.513 # area of the loop in cm^2\n", + "k = 1000 # value of 1 cm on x axis in A/m\n", + "k_ = 1 # value of 1 cm on y axis in Wb/m^2\n", + "B = 1.375 # alternating magnetic flux density in Wb/m^2\n", + "v = 1e-3 # volume of specimen in m^3\n", + "f = 50 # frequency in Hz\n", + "\n", + "#Calculations\n", + "K = a * k * k_\n", + "L = K * v * f\n", + "\n", + "#Result\n", + "print \"Hysteresis loss per sec is %.2f W\"%L" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Hysteresis loss per sec is 25.65 W\n" + ] + } + ], + "prompt_number": 38 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter_19.ipynb b/Engineering_Physics/Chapter_19.ipynb new file mode 100755 index 00000000..8e043325 --- /dev/null +++ b/Engineering_Physics/Chapter_19.ipynb @@ -0,0 +1,333 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 19: Superconductivity" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 19.1, Page 19.13" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Given \n", + "T_c = 7.2 # critical temperature in K\n", + "T = 5.1 # temperature in K\n", + "lambda_ = 380 # penetration depth at 0 K in A\n", + "\n", + "#Calculations\n", + "lamda = lambda_ * (1 - (T / T_c)**4)**(-1./2)\n", + "\n", + "#Result\n", + "print \"Penetration depth is %.2f A\"%lamda" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Penetration depth is 439.30 A\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 19.2, Page 19.13" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "Hc1 = 1.41e5 # first critical field at 14.1K\n", + "Hc2 = 4.205e5 # second critical field at 12.9K \n", + "T1 = 14.11 # temperature in K\n", + "T2 = 12.9 # temperature in K \n", + "T = 4.2 # temperature in K\n", + "lambda_ = 380 # penetration depth at 0 K in A\n", + "\n", + "#Calculations\n", + "Tc = sqrt((Hc2*T1**2 - Hc1*T2**2) / (Hc2 - Hc1))\n", + "H_ = Hc1 / (1 - (T1 / Tc)**2)\n", + "Hc = H_ * (1 - (T/Tc)**2)\n", + "\n", + "#Result\n", + "print \"Transition temperature is %.2f K\\nCritical field at temperate at 4.2 k is %.2e A/m\"%(Tc,Hc)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Transition temperature is 14.68 K\n", + "Critical field at temperate at 4.2 k is 1.69e+06 A/m\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 19.3, Page 19.14" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "d = 1e-3 # diameter of wire in m\n", + "T1 = 4.2 # temperature in K\n", + "T2 = 7.18 # temperature in K\n", + "H_ = 6.51e4 # critical magnetic field at 0 K\n", + "\n", + "#Calculations\n", + "r = d / 2\n", + "Hc = H_ * (1 - (T1 / T2)**2)\n", + "Jc = (2 * pi * r * Hc) / (pi * r**2)\n", + "\n", + "#Result\n", + "print \"Critical current density is %.3e A/m^2\"%Jc" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Critical current density is 1.713e+08 A/m^2\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 19.4, Page 19.15" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "w = 199.5 # isotopic mass of Hg\n", + "Tc = 4.186 # critical temperature in K \n", + "w_ = 203.4 # increased isotope mass of Hg\n", + "\n", + "#Calculations\n", + "Tc_ = Tc * (w / w_)**(1./2)\n", + "\n", + "#Result\n", + "print \"Critical temperature is %.4f K\"%Tc_" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Critical temperature is 4.1457 K\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 19.5, Page 19.15" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "T_c = 4.2 # critical temperature in K\n", + "T = 2.9 # temperature in K\n", + "lamda = 57 # penetration depth at 2.9 K in nm\n", + "\n", + "#Calculations\n", + "lambda_ = lamda * (1 - (T / T_c)**4)**(1./2)\n", + "\n", + "#Result\n", + "print \"Penetration depth at 0 K is %.2f nm\"%lambda_" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Penetration depth at 0 K is 50.10 nm\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 19.6, Page 19.15" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given\n", + "T1 = 2.18 # temperature in first case in K\n", + "lambda1 = 16 # penetration depth at 2.18 K in nm\n", + "T2 = 8.1 # temperature in second case in K\n", + "lambda2 = 96 # penetration depth at 8.1 K in nm\n", + "\n", + "#Calculations\n", + "Tc = (((lambda2**2 * T2**4) - (T1**4 * lambda1**2)) / (lambda2**2 - lambda1**2))**(1./4)\n", + "\n", + "#Result\n", + "print \"Critical temperature is %.2f K\"%Tc" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Critical temperature is 8.16 K\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 19.7, Page 19.16" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "w = 26.91 # isotopic mass of superconducting sample\n", + "Tc = 1.19 # first critical temperature in K \n", + "w_ = 32.13 # increased isotope mass of superconducting sample\n", + "\n", + "#Calculations\n", + "Tc_ = Tc * (w / w_)**(1./2)\n", + "\n", + "#Result\n", + "print \"Critical temperature is %.3f K\"%Tc_" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Critical temperature is 1.089 K\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 19.8, Page 19.16" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "k = 1.38e-23 # Boltzmann's constant in J/K\n", + "h = 6.62e-34 # Planck constant in J sec\n", + "Tc = 4.2 # critical temperature of Hg in K\n", + "c = 3e8 # speed of light in m/sec \n", + "\n", + "#Calculations\n", + "E = 3 * k * Tc\n", + "lamda = h * c / E\n", + "\n", + "#Result\n", + "print \"Energy gap is %.2e eV\\nWavelength of photon is %.2e m\"%(E/1.6e-19,lamda)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy gap is 1.09e-03 eV\n", + "Wavelength of photon is 1.14e-03 m\n" + ] + } + ], + "prompt_number": 8 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter_2.ipynb b/Engineering_Physics/Chapter_2.ipynb new file mode 100755 index 00000000..dccc71e2 --- /dev/null +++ b/Engineering_Physics/Chapter_2.ipynb @@ -0,0 +1,1712 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 2: Diffraction" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.1, Page 2.38" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt,pi\n", + "\n", + "# Given \n", + "l = 5e-7 # wavelength of light in meter\n", + "d = 1 # distance of wavefront received on the screen from the opening in meter\n", + "n = 80 # no. of half period zone\n", + "\n", + "#Calculations\n", + "Rn = sqrt(n * l * d)# calculation for radius of nth half period zone\n", + "A = pi * d * l# calculation for area of half period zone\n", + "\n", + "#Result\n", + "print(\"Radius of 80th half period zone = %.3f cm. \\nArea of half period zone = %.4f square cm.\"%(Rn*100,A*10000))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Radius of 80th half period zone = 0.632 cm. \n", + "Area of half period zone = 0.0157 square cm.\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.2, Page 2.38" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "# Given \n", + "l = 6e-7 # wavelength of light in meter\n", + "f = 0.6 # focal length of convex lens in meter\n", + "n = 1 # no. of half period zone\n", + "\n", + "#Calculation\n", + "Rn = sqrt(n * l * f)# calculation for radius of half period zone\n", + "\n", + "print(\"Radius of half period zone = %.1f mm \"%(Rn*1000))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Radius of half period zone = 0.6 mm \n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.3, Page 2.38" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "# Given \n", + "l = 6e-7 # wavelength of light in meter\n", + "f = 0.60 #focal length in m\n", + "n = 1 # no. of half period zone\n", + "\n", + "#Calculation\n", + "r1 = sqrt(f* l ) # because at maxima intensity is four time the individual intensity of light\n", + "\n", + "#Result\n", + "print(\"Radius of 80th half period zone = %.4f mm. \"%(r1)) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Radius of 80th half period zone = 0.0006 mm. \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.4, Page 2.39" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "l = 6e-7 # wavelength of light in meter\n", + "d = 0.5 # distance of observation point from circular opening in meter\n", + "r1 = 2e-3 # radius of circular opening in first case in meter\n", + "r2 = 2e-2 # radius of circular opening in second case in meter \n", + "\n", + "#Calculation\n", + "n1 = (r1**2) / (d * l) # calculation for no. of half period zone in first case \n", + "n2 = (r2**2) / (d * l) # calculation for no. of half period zone in second case\n", + "\n", + "print(\"No. of half period zone in first case = %d \\nNo. of half period zone in second case = %d \"%(n1,n2))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "No. of half period zone in first case = 13 \n", + "No. of half period zone in second case = 1333 \n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.5, Page 2.39" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "l = 5e-7 # wavelength of light in meter\n", + "d = 1e-3 # diameter of the first ring of zone plate in meter\n", + "n = 1 # no. of half period zone\n", + "\n", + "#Calculation\n", + "D = (d**2) / (4 * l * n) # calculation for distance of screen from opening\n", + "\n", + "#Result\n", + "print(\"Distance of screen from opening = %.1f meter \"%D)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Distance of screen from opening = 0.5 meter \n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.6, Page 2.40" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "# Given \n", + "l = 5.893e-7 # wavelength of light in meter\n", + "f = 1 # focal-length of convex lens in meter\n", + "n1 = 1 # no. of first half period zone\n", + "n2 = 3 # no. of second half period zone\n", + "n3 = 5 # no. of third half period zone\n", + "\n", + "#Calculations\n", + "R1 = sqrt(n1 * l * f) # calculation for Radius of first half period zone\n", + "R2 = sqrt(n2 * l * f) # calculation for Radius of second half period zone\n", + "R3 = sqrt(n3 * l * f) # calculation for Radius of third half period zone\n", + "\n", + "#Result\n", + "print(\"Radius of first ,second and third half period zone = %.3e, %.3e and %.3e meter respectively. \"%(R1,R2,R3))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Radius of first ,second and third half period zone = 7.677e-04, 1.330e-03 and 1.717e-03 meter respectively. \n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.7, Page 2.40" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "# Given \n", + "l = 5e-7 # wavelength of light in meter\n", + "f = 0.2 # focal length of convex lens in meter\n", + "n = 10 # no. of half period zone\n", + "\n", + "#Calculation\n", + "Rn = sqrt(n * l * f) # calculation for radius of 10th half period zone\n", + "\n", + "#Result\n", + "print(\"Radius of 10th half period zone = %.1f mm. \"%(Rn*1000))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Radius of 10th half period zone = 1.0 mm. \n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.8, Page 2.40" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "# Given \n", + "l = 5.89e-7 # wavelength of light in meter\n", + "d1 = 1. # distance of wavefront recieved on the screen from the opening in first side in meter\n", + "d2 = 2. # distance of wavefront recieved on the screen from the opening in other side in meter\n", + "\n", + "#Calculations\n", + "f = (d1 * d2) / (d1 + d2)\n", + "p = 1. / f # beacause zone plate act as a convex lens\n", + "n = 1 # for first zone\n", + "Rn = sqrt(n * l * f) # calculation for radius of first zone\n", + "Dn = 2 * Rn # calculation for diameter of first zone\n", + "\n", + "#Result\n", + "print(\"Focal length = %.2f meter. \\n Power = %.1f D. \\n Diameter of first zone = %.3f mm. \"%(f,p,Dn*1000))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Focal length = 0.67 meter. \n", + " Power = 1.5 D. \n", + " Diameter of first zone = 1.253 mm. \n" + ] + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.9, Page 2.41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "lambda1 = 6e-7 # wavelength of first light in meter\n", + "lambda2 = 5e-7 # wavelength of second light in meter\n", + "f1 = 1 # focal length in first case in meter \n", + "\n", + "#Calculation\n", + "f2 = (lambda1 * f1) / lambda2 # calculation for focal length in second case\n", + "\n", + "#Result\n", + "print(\"Focal length in second case = %.1f meter\"%f2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Focal length in second case = 1.2 meter\n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.10, Page 2.41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "l = 4e-7 # wavelength of light in meter\n", + "u = 0.2 # distance of object from zone plate in meter\n", + "v = 0.2 # distance of brightest image from from zone plate in meter \n", + "r = 0.01 # radius in meter\n", + "\n", + "#Calculations\n", + "f = (u * v) / (u + v) # calculation for focal length\n", + "n = (r**2) / (f * l) # calculation for no. of zone of Fresnel\n", + "\n", + "#Result\n", + "print(\"No. of zone of Fresnel = %.f\"%n)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "No. of zone of Fresnel = 2500\n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.11, Page 2.42" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "l = 5.893e-7 # wavelength of light in meter\n", + "d = 2.3e-3 # diameter of the central zone of zone plate in meter\n", + "u = 6 # distance between point source from zone plate in meter\n", + "n = 1 # for central zone\n", + "\n", + "#Calculations\n", + "r = d/2\n", + "f = (r**2) / (l) # calculation for focal length\n", + "v = (f * u) / (u - f) # calculation for distance of first image from zone plate\n", + "\n", + "#Result\n", + "print(\"Distance of first image from zone plate = %.2f meter \"%v) #answer differs due to rounding-off values" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Distance of first image from zone plate = 3.59 meter \n" + ] + } + ], + "prompt_number": 41 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.12, Page 2.42" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "R = 2 # radius of curvature in meter\n", + "\n", + "#Calculation\n", + "f = R # calculation for principal focal length of zone plate\n", + "\n", + "#Result\n", + "print(\"Principal focal length of zone plate = %.1f meter \"%f)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Principal focal length of zone plate = 2.0 meter \n" + ] + } + ], + "prompt_number": 43 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.13, Page 2.42" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import asin, pi\n", + "\n", + "# Given \n", + "l = 5.89e-7 # wavelength of light in meter\n", + "b = 1e-3 # slit-width in meter\n", + "m = 1 # for first minima\n", + "\n", + "#Calculation\n", + "theta = asin((m * l) / b) # calculation for angular spread of the central maxima in radian\n", + "theta_ = theta * (180 / pi) # calculation for angular spread of the central maxima in degree\n", + "\n", + "#Result\n", + "print(\"Angular spread of the central maxima = %.4f degree \"%(2 * theta_))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Angular spread of the central maxima = 0.0675 degree \n" + ] + } + ], + "prompt_number": 44 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.14, Page 2.43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "d = 1.2 # distance of screen from slit in meter\n", + "x = 3.7e-3 # distance between first maxima to central maxima in meter\n", + "b = 2e-4 # slit-width in meter\n", + "\n", + "#Calculation\n", + "l = (x * b) / d # calculation for wavelength of light\n", + "\n", + "#Result\n", + "print \"Wavelength of light = \",round(l/1e-10),\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of light = 6167.0 A\n" + ] + } + ], + "prompt_number": 70 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.15, Page 2.43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import asin, pi\n", + "\n", + "# Given \n", + "l = 5.5e-7 # wavelength of light in meter\n", + "b = 2.2e-6 # slit-width in meter\n", + "\n", + "#Calculations\n", + "m2 = 2 # for second minima\n", + "theta2 = asin((m2 * l) / b) * (180 / pi) # calculation for angular position of second minima\n", + "m3 = 3 # for third minima\n", + "theta3 = asin((m3 * l) / b) * (180 / pi) # calculation for angular position of third minima\n", + "\n", + "#Result\n", + "print(\"Angular position of second and third minima = %.f degrees and %.2f degrees respectively \"%(theta2 ,theta3))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Angular position of second and third minima = 30 degrees and 48.59 degrees respectively \n" + ] + } + ], + "prompt_number": 72 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.16, Page 2.44" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import asin, pi\n", + "\n", + "# Given \n", + "l = 5.89e-7 # wavelength of light in meter\n", + "b = 1.2e-6 # slit-width in meter\n", + "\n", + "#Calculation\n", + "m = 1 # for first minima\n", + "theta = asin((m * l) / b) # calculation for half angular width of the central bright maxima in radian\n", + "theta_ = theta * (180 / pi) # calculation for half angular width of the central bright maxima in degree\n", + "\n", + "#Result\n", + "print(\"Half angular width of the central bright maxima = %.2f degrees \"%theta_)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Half angular width of the central bright maxima = 29.40 degrees \n" + ] + } + ], + "prompt_number": 74 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.17, Page 2.44" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sin, pi \n", + "\n", + "# Given \n", + "l = 5e-7 # wavelength of light in meter\n", + "theta = pi / 6 # half angular width of central maximum in first case in radian\n", + "theta_ = pi / 2 # half angular width of central maximum in second case in radian\n", + "\n", + "#Calculation\n", + "m = 1 # for first minima\n", + "b1 = (l * m) / sin(theta) # calculation for slit width in first case\n", + "b2 = (l * m) / sin(theta_) # calculation for slit width in second case\n", + "\n", + "#Result\n", + "print(\"Slit width in first case = %.f micro-meter \\nSlit width in second case = %.1f micro-meter\"%(b1*1e6,b2*1e6))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Slit width in first case = 1 micro-meter \n", + "Slit width in second case = 0.5 micro-meter\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.18, Page 2.44" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import asin, pi\n", + "\n", + "# Given \n", + "l = 5.89e-7 # wavelength of light in meter\n", + "d = 1 # distance of screen from slit in meter\n", + "b = 1e-4 # slit-width in meter\n", + "\n", + "#Calculations\n", + "theta = (asin(l / b)) * (180 / pi) # calculation for angular spread\n", + "x = (2 * d * l) / b# calculation for linear width\n", + "\n", + "#Result\n", + "print(\"Angular spread = %.3f degree\\nLinear width = %.3f cm \"%(2*theta,x*1e2))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Angular spread = 0.675 degree\n", + "Linear width = 1.178 cm \n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.20, Page 2.45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import asin, pi\n", + "\n", + "# Given \n", + "l = 6e-7 # wavelength of light in meter\n", + "b = 1.2e-6 # slit-width in meter\n", + "\n", + "#Calculations\n", + "m = 1 # for first minima\n", + "theta = asin((m * l) / b) # calculation for angular width of the central maxima in radian\n", + "theta_ = theta * (180 / pi) # calculation for angular width of the central maxima in degree\n", + "\n", + "#Result\n", + "print(\"Angular width of the central maxima = %.f degree \"%(2 * theta_))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Angular width of the central maxima = 60 degree \n" + ] + } + ], + "prompt_number": 104 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.21, Page 2.46" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "l = 4.890e-7 # wavelength of light in meter\n", + "b = 5e-3 # slit-width in meter\n", + "f = 0.4 # focal-length of convex lens in meter\n", + "\n", + "#Calculation\n", + "m = 1 # for first dark fringe\n", + "x = (f * m * l) / b \n", + "n = 1 # for first secondary maxima\n", + "x_ = ((2 * n + 1) * l * f) / (2 * b) \n", + "delta_x = x_ - x # calculation for separation of dark band \n", + "\n", + "#Result\n", + "print(\"Separation of dark band = %.3e meter.\"%(delta_x))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Separation of dark band = 1.956e-05 meter.\n" + ] + } + ], + "prompt_number": 105 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.22, Page 2.47" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "l = 5.893e-7 # wavelength of light in meter\n", + "b = 5e-4 # slit-width in meter\n", + "f = 1 # focal length of convex lens in meter\n", + "\n", + "#Calculation\n", + "x = (2 * l * f) / b # calculation for Separation of dark band on either side of the cenral maximum\n", + "\n", + "#Result\n", + "print(\"Separation of dark band on either side of the central maximum = %.3e meter\"%x)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Separation of dark band on either side of the central maximum = 2.357e-03 meter\n" + ] + } + ], + "prompt_number": 106 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.23, Page 2.47" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "d = 4e-4 # separation between slits in meter\n", + "b = 8e-5 # slit-width in meter\n", + "\n", + "#Calculations\n", + "r = (b + d) / b # calculation for ratio of n with m\n", + "m1 = 1\n", + "n1 = r * m1 # calculation for Missing orders \n", + "m2 = 2\n", + "n2 = r * m2 # calculation for Missing orders \n", + "m3 = 3\n", + "n3 = r * m3 # calculation for Missing orders \n", + "\n", + "#Result\n", + "print(\"Missing orders = %d,%d,%d,......etc.\"%(n1,n2,n3))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Missing orders = 6,12,18,......etc.\n" + ] + } + ], + "prompt_number": 107 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.24, Page 2.47" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "d = 4e-4 # separation between slits in meter\n", + "b = 2e-4 # slit-width in meter\n", + "fringe_width = 2.5e-3 # fringe width in meter\n", + "D = 1.6 # distance between screen and slits\n", + "\n", + "#Calculations\n", + "l = (fringe_width * d) / D # calculation for wavelength of light\n", + "r = (b + d) / b # calculation for ratio of n with m\n", + "m1 = 1\n", + "n1 = r * m1 # calculation for missing order\n", + "m2 = 2\n", + "n2 = r * m2 # calculation for missing order\n", + "m3 = 3\n", + "n3 = r * m3 # calculation for missing order\n", + "\n", + "#Result\n", + "print(\"Wavelength of light = %.3e meter. \\nMissing order = %d,%d,%d....etc.\"%(l,n1,n2,n3))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of light = 6.250e-07 meter. \n", + "Missing order = 3,6,9....etc.\n" + ] + } + ], + "prompt_number": 109 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.25, Page 2.48" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi, sin\n", + "\n", + "# Given \n", + "N = 425000 # no. of lines in plane transmission grating per meter\n", + "theta = pi / 6 # angle at which second order spectral line is observed in radian\n", + "n = 2 # order of spectral line\n", + "\n", + "#Calculation\n", + "l = sin(theta) / (2 * N) # calculation for wavelength of light\n", + "\n", + "#Result\n", + "print \"Wavelength of light = \",round(l/1e-10),\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of light = 5882.0 A\n" + ] + } + ], + "prompt_number": 115 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.26, Page 2.48" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi, sin\n", + " \n", + "# Given \n", + "N = 500000 # no. of lines in plane transmission grating per meter\n", + "theta = pi / 6 # angle at which second order spectral line is observed in radian\n", + "n = 2 # order of spectral line\n", + "\n", + "#Calculation\n", + "l = sin(theta) / (2 * N) # calculation for wavelength of light\n", + "\n", + "#Result\n", + "print \"wavelength of light = \",l/1e-10,\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "wavelength of light = 5000.0 A\n" + ] + } + ], + "prompt_number": 116 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.27, Page 2.48" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import ceil\n", + "\n", + "# Given \n", + "lambda2 = 5.461e-7 # wavelength of light in second case in meter\n", + "n1 = 4 # no. of order in first case\n", + "n2 = 3 # no. of order in second case \n", + "\n", + "#Calculation\n", + "lambda1 = (n2 * lambda2) / n1 # calculation for Wavelength of light in first case\n", + "\n", + "#Result\n", + "print(\"Wavelength of light in first case = %d A\"%(ceil(lambda1*1e10)))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of light in first case = 4096 A\n" + ] + } + ], + "prompt_number": 90 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.28, Page 2.49" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi, sin\n", + "\n", + "# Given \n", + "l = 5e-7 # wavelength of light in meter\n", + "theta = pi / 6 # angle at which second order spectral line is observed in radian\n", + "n = 2 # order of spectral line\n", + "\n", + "#Calculations\n", + "k = (n * l) / sin(theta) # calculation for (b+d)\n", + "N = 1 / k # calculation for no. of lines in per cm\n", + "\n", + "#Result\n", + "print(\"No. of lines per cm = %.f \"%(N / 100))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "No. of lines per cm = 5000 \n" + ] + } + ], + "prompt_number": 117 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.29, Page 2.49" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import asin,pi \n", + "\n", + "# Given \n", + "lambda1 = 5.048e-7 # wavelength of light in first case in meter\n", + "lambda2 = 5.016e-7 # wavelength of light in second case in meter\n", + "n = 2 # no. of order in first case\n", + "N = 15000 # no. of lines in grating per inch \n", + "\n", + "#Calculations\n", + "k = 2.54 / 1500000 # in meter\n", + "theta1 = asin(n * lambda1 / k) * (180 / pi) # calculation for angle in first case\n", + "theta2 = asin(n * lambda2 / k) * (180 / pi) # calculation for angle in second case\n", + "delta_theta = theta1 - theta2 # calculation for angle of separation\n", + "\n", + "#Result\n", + "print(\"Angle of separation = %.2f degree\"%delta_theta)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Angle of separation = 0.27 degree\n" + ] + } + ], + "prompt_number": 118 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.30, Page 2.50" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import asin,pi \n", + "\n", + "# Given \n", + "lambda1 = 5.89e-7 # wavelength of light in first case in meter\n", + "lambda2 = 5.896e-7 # wavelength of light in second case in meter\n", + "n = 2 # no. of order in first case\n", + "N = 600000 # no. of lines in grating per meter \n", + "\n", + "#Calculations\n", + "k = 1. / N # in meter\n", + "theta1 = asin(n * lambda1 / k) * (180 / pi) # calculation for angle in first case\n", + "theta2 = asin(n * lambda2 / k) * (180 / pi) # calculation for angle in second case\n", + "delta_theta = theta2 - theta1 # calculation for angle of separation\n", + "\n", + "#Result\n", + "print(\"Angle of separation = %.2f degree\"%delta_theta)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Angle of separation = 0.06 degree\n" + ] + } + ], + "prompt_number": 119 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.31, Page 2.50" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "# Given \n", + "lambda1 = 5.4e-7 # wavelength of light for nth order in meter\n", + "lambda2 = 4.05e-7 # wavelength of light for (n+1)th order in meter \n", + "theta = pi / 6 # angle of diffraction in radian \n", + "\n", + "#Calculations\n", + "k = (lambda1 * lambda2) / ((lambda1 - lambda2) * sin(theta)) # calculation for b+d\n", + "N = (1 / k) * (0.01) # calculation for no. of lines per cm\n", + "\n", + "#Result\n", + "print(\"No. of lines per cm = %d \"%N)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "No. of lines per cm = 3086 \n" + ] + } + ], + "prompt_number": 97 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.32, Page 2.51" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi, cos, sin\n", + "\n", + "# Given \n", + "d_theta = 0.01 # angular separation between two wavelengths in radian \n", + "theta = pi / 6 # angle of diffraction in radian \n", + "l = 5e-7 # wavelength of light in meter\n", + "\n", + "#Calculation\n", + "d_lambda = (l * cos(theta) * d_theta) / sin(theta) # calculation for difference in two waveligth\n", + "\n", + "#Result\n", + "print \"Difference in two wavelength = \",round(d_lambda/1e-10,1),\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Difference in two wavelength = 86.6 A\n" + ] + } + ], + "prompt_number": 127 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.33, Page 2.51" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "N = 2620 # no. of lines in plane transmission grating per inch\n", + "l = 5e-7 # wavelength of incident radiation in meter\n", + "\n", + "#Calculations\n", + "k = 2.54 / N * 1 / 100 # calculation for b+d in meter\n", + "n = k / l # calculation for order of spectrum\n", + "\n", + "#Result\n", + "print(\"Order of spectrum = %d\"%n)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Order of spectrum = 19\n" + ] + } + ], + "prompt_number": 99 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.34, Page 2.51" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "N = 500000. # no. of lines in plane transmission grating per meter\n", + "l = 5e-7 # wavelength of incident radiation in meter\n", + "\n", + "#Calculations\n", + "k = 1 / N # calculation for b+d in meter\n", + "n = k / l # calculation for order of spectrum \n", + "\n", + "#Result\n", + "print \"Order of spectrum = %d\"%n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Order of spectrum = 4\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.35, Page 2.52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "N = 4000. # no. of lines in plane transmission grating per meter\n", + "lambda1 = 4.e-7 # wavelength of light in first case in meter\n", + "lambda2 = 7.e-7 # wavelength of light in second case in meter\n", + "\n", + "#Calculations\n", + "b_plus_d = (1/N)*10**-2\n", + "n1 = b_plus_d / lambda1 # calculation for Observed order in first case\n", + "n2 = b_plus_d / lambda2 # calculation for Observed order in second case\n", + "\n", + "#Result\n", + "print \"Observed order = %.2f,%.2f\"%(n1,n2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Observed order = 6.25,3.57\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.36, Page 2.52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "# Given \n", + "N = 4000 # no. of lines in grating per meter\n", + "l = 5e-5 # wavelength of incident radiation in cm\n", + "n = 3 # no. of order\n", + "\n", + "#Calculation\n", + "p = (n * N) / (sqrt(1 - (N * n * l)))# dispersive power (p) = d(theta)/d(lambda)\n", + "\n", + "#Result\n", + "print \"Dispersive power = %.3e rad/m\"%p" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Dispersive power = 1.897e+04 rad/m\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.37, Page 2.52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "n = 2 # no. of order\n", + "lambda1 = 5.89e-7 # wavelength of light in first case in meter\n", + "lambda2 = 5.896e-7 # wavelength of light in second case in meter\n", + "\n", + "#Calculation\n", + "N = lambda1 / (n * (lambda2 - lambda1)) # calculation for minimum no. of lines in grating \n", + "\n", + "#Result\n", + "print \"Minimum no. of lines in grating = %.1f\"%N" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum no. of lines in grating = 490.8\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.38, Page 2.53" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "n = 1 # no. of order\n", + "lambda1 = 5.89e-7 # wavelength of light in first case in meter\n", + "lambda2 = 5.896e-7 # wavelength of light in second case in meter\n", + "\n", + "#Calculation\n", + "N = lambda1 / (n * (lambda2 - lambda1)) # calculation for minimum no. of lines in grating\n", + "\n", + "#Result\n", + "print \"Minimum no. of lines in grating = %.2f\"%N" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum no. of lines in grating = 981.67\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.39, Page 2.53" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi,sin\n", + "\n", + "# Given \n", + "n = 3 # no. of order\n", + "theta = pi / 6 # view angle of third order in radian\n", + "lambda1 = 5.89e-7 # min. wavelength of light in meter\n", + "lambda2 = 5.896e-7 # max.wavelength of light in meter\n", + "\n", + "#Calculations\n", + "mean_lambda = (lambda1 + lambda2) / 2 # calculation for mean wavelength\n", + "s = (n * mean_lambda) / sin(theta) # calculation for grating space b+d\n", + "N = lambda1 / (n * (lambda2 - lambda1)) # calculation for minimum no. of lines in grating\n", + "\n", + "#Result\n", + "print \"Grating space = %.3e meter. \\nTotal width of ruled surface = %.3e meter. \"%(s,s * N)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Grating space = 3.536e-06 meter. \n", + "Total width of ruled surface = 1.157e-03 meter. \n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.40, Page 2.53" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "l = 5.5e-7 # wavelength of light in meter\n", + "a = 5 # diameter of objective lens of telescope in meter\n", + "R = 3.8e8 # distance of moon in meter\n", + "\n", + "#Calculations\n", + "theta = (1.22 * l) / a # calculation for angle \n", + "x = (R * theta) # calculation for the separation of two points on moon\n", + "\n", + "#Result\n", + "print \"The separation of two points on moon = %.3f meter\"%x" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The separation of two points on moon = 50.996 meter\n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.41, Page 2.54" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "# Given \n", + "l = 5e-7 # wavelength of light in meter\n", + "theta = (1e-3) * (pi / 180) # separation angle of stars in radian\n", + "\n", + "#Calculation\n", + "a = (1.22 * l) / theta # calculation for diameter of telescope objective\n", + "\n", + "#Result\n", + "print \"Diameter of telescope objective = %.5f meter\"%a" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diameter of telescope objective = 0.03495 meter\n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.42, Page 42" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "l = 6e-7 # wavelength of light in meter\n", + "theta = 2.44e-6 # separation angle of stars in radian\n", + "\n", + "#Calculation\n", + "a = (1.22 * l) / theta # calculation for diameter of telescope objective\n", + "\n", + "#Result\n", + "print \"Diameter of telescope objective = %.2f meter\"%a" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diameter of telescope objective = 0.30 meter\n" + ] + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.43, Page 2.54" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "l = 5.5e-7 # wavelength of light in meter\n", + "a = 0.004 # diameter of objective lens of telescope in meter\n", + "x = 1.5e-3 # distance between two pin holes in meter\n", + "\n", + "#Calculations\n", + "theta = (1.22 * l) / a # calculation for angle \n", + "R = x / theta # calculation for max. distance of pin holes from microscope\n", + "\n", + "#Result\n", + "print \"Max. distance of pin holes from microscope = %.4f meter\"%R" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Max. distance of pin holes from microscope = 8.9419 meter\n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.44, Page 2.55" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi, sin\n", + "\n", + "# Given \n", + "l = 5.5e-7 # wavelength of light in meter\n", + "theta = pi / 6 # semi-angle of cone in radian\n", + "\n", + "#Calculation\n", + "d = (1.22 * l) / (2 * sin(theta)) # calculation for the resolving limit of microscope \n", + "\n", + "#Result\n", + "print \"The resolving limit of microscope = %.1e meter\"%d" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The resolving limit of microscope = 6.7e-07 meter\n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.45, Page 2.55" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "l = 5.461e-7 # wavelength of light in meter\n", + "d = 4e-7 # separation between objects in meter\n", + "\n", + "#Calculation\n", + "NA = (1.22 * l) / (2 * d) # calculation for numerical aperture of objective \n", + "\n", + "#Result\n", + "print \"Numerical aperture of objective = %.3f\"%NA" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Numerical aperture of objective = 0.833\n" + ] + } + ], + "prompt_number": 41 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter_20.ipynb b/Engineering_Physics/Chapter_20.ipynb new file mode 100755 index 00000000..6252eb1b --- /dev/null +++ b/Engineering_Physics/Chapter_20.ipynb @@ -0,0 +1,755 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 20: X-Ray" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.1, Page 20.7" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "V1 = 40e3 # voltage in first case in V\n", + "V2 = 20e3 # voltage in second case in V\n", + "V3 = 100e3 # voltage in second in V\n", + "\n", + "#Calculations\n", + "v1 = 0.593e6 * sqrt(V1)\n", + "lambda1 = 12400 / V1\n", + "v2 = 0.593e6 * sqrt(V2)\n", + "lambda2 = 12400 / V2\n", + "v3 = 0.593e6 * sqrt(V3)\n", + "lambda3 = 12400 / V3\n", + "\n", + "#Results\n", + "print \"Max. speed of electrons at %d Volts is %.3e m/sec\\nMax. speed of electrons at %d Volts is %.2e m/sec/sec\\nMax. speed of electrons at %d Volts is %.3e m/sec\\nShortest wavelength of x-ray = %.2f A\\nShortest wavelength of x-ray = %.2f A\\nShortest wavelength of x-ray = %.3f A\"%(V1,v1,V2,v2,V3,v3,lambda1,lambda2,lambda3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Max. speed of electrons at 40000 Volts is 1.186e+08 m/sec\n", + "Max. speed of electrons at 20000 Volts is 8.39e+07 m/sec/sec\n", + "Max. speed of electrons at 100000 Volts is 1.875e+08 m/sec\n", + "Shortest wavelength of x-ray = 0.31 A\n", + "Shortest wavelength of x-ray = 0.62 A\n", + "Shortest wavelength of x-ray = 0.124 A\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.2, Page 20.7\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "V = 30e3 # voltage in V\n", + "lambda_min = 0.414e-10 # shortest wavelength in m\n", + "e = 1.6e-19 # charge on an electron in C\n", + "c = 3e8 # speed of light in m/sec\n", + "\n", + "#Calculations\n", + "h = (e * V * lambda_min) / c\n", + "\n", + "#Result\n", + "print \"Planck constant is %.3e J sec\"%h" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Planck constant is 6.624e-34 J sec\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.3, Page 20.8" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "V = 25e3 # voltage in V\n", + "\n", + "#Calculations\n", + "lambda_min = 12400 / V\n", + "\n", + "#Result\n", + "print \"Minimum wavelength of x-ray is %.3f A\"%lambda_min" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum wavelength of x-ray is 0.496 A\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.4, Page 20.8" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "V = 13.6e3 # voltage in V\n", + "\n", + "#Calculations\n", + "v = 0.593e6*sqrt(V)\n", + "\n", + "#Result\n", + "print \"Maximum speed of electron is %.2e m/sec\"%v" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum speed of electron is 6.92e+07 m/sec\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.5, Page 20.8" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "V = 10e3 # voltage in V\n", + "i = 2e-3 # current in amp\n", + "\n", + "#Calculations\n", + "v = 0.593e6*sqrt(V)\n", + "\n", + "#Result\n", + "print \"Velocity of electron is %.2e m/sec\"%v" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Velocity of electron is 5.93e+07 m/sec\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.6, Page 20.8" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given\n", + "V = 9.8e3 # voltage in V\n", + "i = 2e-3 # current in amp\n", + "c = 3e8 # speed of light in m/sec\n", + "\n", + "#Calculations\n", + "lamda = 12400 / V\n", + "f = c / (lamda*10**-10)\n", + "\n", + "#Results\n", + "print \"Highest frequency is %.2e Hz\\nMinimum wavelength is %.2f A\"%(f,lamda)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Highest frequency is 2.37e+18 Hz\n", + "Minimum wavelength is 1.27 A\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.7, Page 20.9" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "# Given \n", + "V = 12.4e3 # voltage in V\n", + "i = 2e-3 # current in amp\n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#Calculations\n", + "n = i / e\n", + "v = 0.593e6*sqrt(V)\n", + "\n", + "#Result\n", + "print \"Number of electrons striking the target per sec is %.2e\\nSpeed of electrons is %.1e m/sec\"%(n,v)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of electrons striking the target per sec is 1.25e+16\n", + "Speed of electrons is 6.6e+07 m/sec\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.8, Page 20.9" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "V = 10e3 # voltage in V\n", + "i = 15e-3 # current in amp\n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#Calculations\n", + "n = i / e\n", + "lamda = 12400 / V \n", + "\n", + "#Results\n", + "print \"Number of electrons striking the anode per sec is %.2e\\nMinimum wavelength produced is %.2f A\"%(n,lamda)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of electrons striking the anode per sec is 9.38e+16\n", + "Minimum wavelength produced is 1.24 A\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.9, Page 20.9" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "V = 50e3 # voltage in V\n", + "i = 1e-3 # current in amp\n", + "e = 1.6e-19 # charge on an electron in C\n", + "\n", + "#Calculations\n", + "n = i / e\n", + "\n", + "#Result\n", + "print \"Number of electrons striking the anode per sec is %.2e\"%n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of electrons striking the anode per sec is 6.25e+15\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.10, Page 20.10" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "lambda1 = 40e-12 # minimum wavelength in first case in m\n", + "lambda2 = 1e-10 # minimum wavelength in second case in m\n", + "\n", + "#Calculations\n", + "V1 = 12400e-10 / lambda1\n", + "V2 = 12400e-10 / lambda2\n", + "\n", + "#Results\n", + "print \"Applied voltage to get wavelength of %.e meter is %.f KV\\nApplied voltage to get wavelength of %.e meter is %.1f KV\"%(lambda1,V1/10**3,lambda2,V2/10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Applied voltage to get wavelength of 4e-11 meter is 31 KV\n", + "Applied voltage to get wavelength of 1e-10 meter is 12.4 KV\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.11, Page 20.10" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "V1 = 44e3 # voltage in first case in V\n", + "V2 = 50e3 # voltage in second case in V\n", + "lambda1 = 0.284e-10 # shortest wavelength in first case in m\n", + "lambda2 = 0.248e-10 # shortest wavelength in second case in m\n", + "e = 1.6e-19 # charge on an electron in C\n", + "c = 3e8 # speed of light in m/sec\n", + "\n", + "#Calculations\n", + "h1 = e * V1 * lambda1 / c\n", + "h2 = e * V2 * lambda2 / c\n", + "\n", + "#Results\n", + "print \"Planck constant is %.2e J sec if shortest wavelength is %.3e m \\nPlanck constant is %.3e Jsec if shortest wavelength is %.3e m \"%(h1,lambda1,h2,lambda2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Planck constant is 6.66e-34 J sec if shortest wavelength is 2.840e-11 m \n", + "Planck constant is 6.613e-34 Jsec if shortest wavelength is 2.480e-11 m \n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.12, Page 20.10" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "lamda = 1e-11 # K-absorption limit for uranium in m\n", + "\n", + "#Calculations\n", + "V = 12400e-10 / lamda\n", + "\n", + "#Result\n", + "print \"Excitation potential is %d kV\"%(V/10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Excitation potential is 124 kV\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.13, Page 20.11" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "lamda = 1.4e-11 # K-absorption edge for lead in m\n", + "V = 88.6e3 # minimum voltage required for producing k-lines in V\n", + "c = 3e8 # speed of light in m/sec\n", + "\n", + "#Calculations\n", + "r = V * lamda / c\n", + "\n", + "#Result\n", + "print \"The value of the ratio of h/e = %.3e Jsec/C\"%r" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of the ratio of h/e = 4.135e-15 Jsec/C\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.14, Page 20.11" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "Z = 92 # atomic no. of atom\n", + "Rh = 1.1e5 # Rydberg constant in cm^-1\n", + "c = 3e8 # speed of light in m/sec\n", + "\n", + "#Calculations\n", + "lamda = 1 / (Rh *(Z-1)**2 * (1 - (1 / 2**2)))\n", + "\n", + "#Result\n", + "print \"Wavelength of K line = %.2f A\"%(lamda*1e8)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of K line = 0.11 A\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.15, Page 20.11" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "Z = 42 # atomic no. of Mo\n", + "lamda = 0.71e-10 # wavelength in m\n", + "Z_ = 29 # atomic no. of Cu\n", + "\n", + "#Calculations\n", + "lambda_ = (Z-1)**2 * lamda / (Z_-1)**2\n", + "\n", + "#Result\n", + "print \"Wavelength of the corresponding radiation of Cu is %.2f A\"%(lambda_*1e10)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of the corresponding radiation of Cu is 1.52 A\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.16, Page 20.12" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "Z = 79 # atomic no. of element\n", + "b = 1 # a constant\n", + "a = 2.468e15 # a constant in per sec\n", + "c = 3e8 # speed of light in m/sec\n", + "\n", + "#Calculations\n", + "f = a * (Z - b)**2\n", + "lamda = c / f\n", + "\n", + "#Result\n", + "print \"Wavelength of x-ray is %.4f A\"%(lamda*1e10)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of x-ray is 0.1998 A\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.17, Page 20.12" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "Z = 29 # atomic no. of Cu\n", + "R = 1.097e7 # Rydberg constant in m^-1\n", + "c = 3e8 # speed of light in m/sec\n", + "h = 6.62e-34 # Planck constant in J sec\n", + "\n", + "#Calculations\n", + "f = 3./4 * (R * c) * (Z-1)**2\n", + "E = h * f / 1.6e-16\n", + "E_L = 0.931 # let E_L = 0.931 KeV\n", + "E_ = E + E_L\n", + "\n", + "#Result\n", + "print \"Ionization potential of K-shell electron of Cu is %.3f keV\"%E_" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ionization potential of K-shell electron of Cu is 8.938 keV\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.18, Page 20.13" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "Z = 79 # atomic no. of anticathode\n", + "R = 1.097e7 # Rydberg constant in m^-1\n", + "c = 3e8 # speed of light in m/sec\n", + "\n", + "#Calculations\n", + "f = 3./4 * (R * c) * (Z-1)**2\n", + "\n", + "#Result\n", + "print \"Frequency of k line is %.3e Hz\"%f" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Frequency of k line is 1.502e+19 Hz\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.19, Page 20.13" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "Z = 27 # atomic no. of Co\n", + "R = 1.097e7 # Rydberg constant in m^-1\n", + "c = 3e8 # speed of light in m/sec\n", + "h = 6.62e-34 # Planck constant in J sec\n", + "\n", + "#Calculations\n", + "f = 3./4 * (R * c) * (Z-1)**2\n", + "E = h * f\n", + "lamda = c / f\n", + "\n", + "#Results\n", + "print \"Energy is %.2f keV\\nWavelength of x-ray is %.2f A\"%(E / 1.6e-16,lamda*1e10)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy is 6.90 keV\n", + "Wavelength of x-ray is 1.80 A\n" + ] + } + ], + "prompt_number": 24 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter_22.ipynb b/Engineering_Physics/Chapter_22.ipynb new file mode 100755 index 00000000..a955c292 --- /dev/null +++ b/Engineering_Physics/Chapter_22.ipynb @@ -0,0 +1,62 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 22: Nanopyhsics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.1, Page 22.13" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Given \n", + "d = 12e-6 # diameter in m\n", + "d_ = 90e-9 # diameter of nanoparticle in m\n", + "\n", + "#Calculations\n", + "r = d / 2\n", + "r_ = d_ / 2\n", + "k = r / 3\n", + "k_ = r_ / 3\n", + "R = k_ / k\n", + "\n", + "#Result\n", + "print \"The ratio of the value of Nb/Ns of spherical particle and nanoparticle = %.1e\"%R" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The ratio of the value of Nb/Ns of spherical particle and nanoparticle = 7.5e-03\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter_3.ipynb b/Engineering_Physics/Chapter_3.ipynb new file mode 100755 index 00000000..1e5fdb51 --- /dev/null +++ b/Engineering_Physics/Chapter_3.ipynb @@ -0,0 +1,996 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 3: Polarisation" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.1, Page 3.23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi, atan\n", + "\n", + "# Given \n", + "mu = 1.5 # refractive index of glass\n", + "\n", + "#Calculations\n", + "Ip = atan(mu) * (180 / pi) # by brewster's law\n", + "r = 90 - Ip # calculation for angle of refraction\n", + "\n", + "#Result\n", + "print \"Brewster angle = %.f degree\\nAngle of refraction = %.f degree\"%(Ip,r)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Brewster angle = 56 degree\n", + "Angle of refraction = 34 degree\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.2, Page 3.24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi, atan\n", + "\n", + "# Given \n", + "mu = 1.33 # refractive index of glass\n", + "\n", + "#Calculations\n", + "Ip = atan(mu) * (180 / pi) # by Brewster's law\n", + "\n", + "#Result\n", + "print \"Angle of brewster = %.2f degree\"%Ip" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Angle of brewster = 53.06 degree\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3, Page 3.24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi, atan\n", + "\n", + "# Given \n", + "mu_w = 1.33 # refractive index of water\n", + "mu_g = 1.54 # refractive index of glass\n", + "\n", + "#Calculations\n", + "Ip_1 = atan(mu_g / mu_w) * (180 / pi)#calculation for polarizing angle for water\n", + "Ip_2 = atan(mu_w / mu_g) * (180 / pi) # calculation for polarizing angle for glass\n", + "\n", + "#Result\n", + "print \"Polarizing angle for water to glass = %.2f degree,\\n Polarizing angle for glass to water = %.2f degree\"%(Ip_1,Ip_2)\n", + "print \"So polarizing angle is greater for a beam incident from water to glass\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Polarizing angle for water to glass = 49.18 degree,\n", + " Polarizing angle for glass to water = 40.82 degree\n", + "So polarizing angle is greater for a beam incident from water to glass\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.4, Page 24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi, asin, tan, sin\n", + "\n", + "# Given \n", + "Ip = pi / 3 # polarizing angle of piece of glass for green light in radian\n", + "a = pi / 3 # angle of prism in radian \n", + "\n", + "#Calculations\n", + "mu = tan(Ip) # calculation for refractive index\n", + "delta_m = 2 * (asin(mu * sin(a / 2)) - (a / 2)) * (180 / pi) # calculation for angle of minimum deviation\n", + "\n", + "#Result\n", + "print \"Angle of minimum deviation = %.f degree\"%delta_m" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Angle of minimum deviation = 60 degree\n" + ] + } + ], + "prompt_number": 47 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.5, Page 3.25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi, atan\n", + "\n", + "# Given \n", + "mu_w = 1.33 # refractive index of water\n", + "mu_g = 1.5 # refractive index of glass\n", + "\n", + "#Calculations\n", + "Ip = atan(mu_g / mu_w) * (180 / pi) # calculation for Brewster angle\n", + "\n", + "#Result\n", + "print \"Brewster angle = %.1f degree\"%Ip\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Brewster angle = 48.4 degree\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.6, Page 3.25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi, atan\n", + "\n", + "# Given \n", + "mu = 1.732 # refractive index of glass\n", + "\n", + "#Calculations\n", + "Ip = atan(mu) * (180 / pi) # by Brewster's law\n", + "r = 90 - Ip# calculation for angle of refraction\n", + "\n", + "#Result\n", + "print \"Angle of incidence = %.f degree\\nAngle of refraction = %.f degree\"%(Ip,r)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Angle of incidence = 60 degree\n", + "Angle of refraction = 30 degree\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.7, Page 3.25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi, cos\n", + "\n", + "# Given \n", + "alpha = pi / 3 # angle between polarizer and analyzer\n", + "\n", + "#Calculation\n", + "r = (cos(alpha))**2 # where r = transmitted intensity / incident intensity\n", + "\n", + "#Result\n", + "print \"Ratio between transmitted intensity to incident intensity = %.2f \"%r" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ratio between transmitted intensity to incident intensity = 0.25 \n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.8, Page 3.25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt,acos,degrees\n", + "\n", + "#Given \n", + "r1 = 1./3 #ratio of intensity of transmitted light to the intensity of transmitted beam in first case\n", + "r2 = 1./3 #ratio of intensity of transmitted light to the intensity of incident beam in second case\n", + "p = 50 #percentage reduction in intensity of unpolarized light by the sheet \n", + "\n", + "#Calculations\n", + "theta1 = degrees(acos(sqrt(r1))) #calculation for the angle between characteristics directions of the sheet in first case\n", + "theta2 = degrees(acos(sqrt(2*r2))) #calculation for the angle between characteristics directions of the sheet in second case\n", + "\n", + "#Result\n", + "print \"The angle between characteristics directions of the sheet in 1st case = %.2f degrees.\"%(theta1)\n", + "print \"The angle between characteristics directions of the sheet in 2nd case = %.2f degrees.\"%(theta2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The angle between characteristics directions of the sheet in 1st case = 54.74 degrees.\n", + "The angle between characteristics directions of the sheet in 2nd case = 35.26 degrees.\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.9, Page 3.26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import acos, sqrt, pi\n", + "\n", + "# Given \n", + "r = 3. / 4 # ratio of intensity of transmitted light to the intensity of incident light\n", + "\n", + "#Calculation\n", + "theta = acos(sqrt(r)) * (180 / pi) # calculation for angle between the nicol prisms\n", + "\n", + "#Result\n", + "print \"Angle between the nicol prisms = %.f degree\"%theta" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Angle between the nicol prisms = 30 degree\n" + ] + } + ], + "prompt_number": 45 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.10, Page 3.26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi, cos\n", + "\n", + "# Given \n", + "theta1 = pi / 6 # angle between Nicole prisms in first case in radian\n", + "theta2 = pi / 4 # angle between Nicole prisms in second case in radian\n", + "theta3 = pi / 3 # angle between Nicole prisms in third case in radian\n", + "theta4 = pi / 2 # angle between Nicole prisms in fourth case in radian\n", + "\n", + "#Calculations\n", + "I1 = (1 - (cos(theta1))**2) * 100\n", + "I2 = (1 - (cos(theta2))**2) * 100\n", + "I3 = (1 - (cos(theta3))**2) * 100\n", + "I4 = (1 - (cos(theta4))**2) * 100\n", + "\n", + "#Result\n", + "print \"Percentage reduction in intensity of light-\\n(i)%.f %%\\n(ii)%.f %%\\n(iii)%.f %%\\n(iv)%.f %%\"%(I1,I2,I3,I4)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Percentage reduction in intensity of light-\n", + "(i)25 %\n", + "(ii)50 %\n", + "(iii)75 %\n", + "(iv)100 %\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.11, Page 3.27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi, acos, sqrt \n", + "\n", + "# Given \n", + "i1 = 1. / 2 # reduced intensity ratio in first case\n", + "i2 = 1. / 4 # reduced intensity ratio in second case\n", + "\n", + "#Calculations\n", + "theta1 = acos(sqrt(i1)) * (180 / pi)# calculation for angle between nicols in first case \n", + "theta2 = acos(sqrt(i2)) * (180 / pi)# calculation for angle between nicols in second case\n", + "\n", + "#Result\n", + "print \"Angle between the Nicols in first case = %.f degree\\nAnd in second case = %.f degree\"%(theta1,theta2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Angle between the Nicols in first case = 45 degree\n", + "And in second case = 60 degree\n" + ] + } + ], + "prompt_number": 44 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.12, Page 3.27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "l = 5e-7 # wavelength of light in meter\n", + "mu_e = 1.553 # refractive index for extraordinary light\n", + "mu_o = 1.544 # refractive index for ordinary light\n", + "\n", + "#Calculations\n", + "t = l / (2 * (mu_e - mu_o)) # calculation for thickness of half-wave plate of quartz\n", + "\n", + "#Result\n", + "print \"Thickness of half-wave plate of quartz = %.2e meter\"%t" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thickness of half-wave plate of quartz = 2.78e-05 meter\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.13, Page 3.27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "l = 5.893e-7 # wavelength of light in meter\n", + "mu_e = 1.533 # refractive index for extraordinary light\n", + "mu_o = 1.554 # refractive index for ordinary light\n", + "\n", + "#Calculation\n", + "t = l / (4 * (mu_o - mu_e)) # calculation for thickness of quartz plate\n", + "\n", + "#Result\n", + "print \"Thickness of quartz plate = %.2e meter\"%t" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thickness of quartz plate = 7.02e-06 meter\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.14, Page 3.28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "l = 5.89e-7 # wavelength of light in meter\n", + "mu_e1 = 1.5 # refractive index for extraordinary light in first case\n", + "mu_o1 = 1.55 # refractive index for ordinary light in first case\n", + "mu_e2 = 1.57 # refractive index for extraordinary light in second case\n", + "mu_o2 = 1.55 # refractive index for ordinary light in second case\n", + "\n", + "#Calculations\n", + "t1 = l / (4 * (mu_o1 - mu_e1))\n", + "t2 = l / (4 * (mu_e2 - mu_o2))\n", + " # calculation for thickness of plate of quartz\n", + "\n", + "#Result\n", + "print \"Thickness of plate of quartz in first case = %.3e meter,\\nAnd thickness of plate of quartz in second case = %.2e meter\"%(t1,t2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thickness of plate of quartz in first case = 2.945e-06 meter,\n", + "And thickness of plate of quartz in second case = 7.36e-06 meter\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.15, Page 3.28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "l = 5.89e-7 # wavelength of light in meter\n", + "mu_e = 1.486 # refractive index for extraordinary light\n", + "mu_o = 1.658 # refractive index for ordinary light\n", + "\n", + "#Calculation\n", + "t = l / (4 * (mu_o - mu_e)) # calculation for thickness of calcite plate \n", + "\n", + "#Result\n", + "print \"Thickness of calcite plate = %.2e meter\"%t" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thickness of calcite plate = 8.56e-07 meter\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.16, Page 3.28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "l = 5e-7 # wavelength of light in meter\n", + "mu_e = 1.5533 # refractive index for extraordinary light\n", + "mu_o = 1.5442 # refractive index for ordinary light\n", + "\n", + "#Calculation\n", + "t = l / (4 * (mu_e - mu_o)) # calculation for thickness of quartz plate\n", + "\n", + "#Result\n", + "print \"Thickness of quartz plate = %.2e meter\"%t" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thickness of quartz plate = 1.37e-05 meter\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.17, Page 3.28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "l = 5.89e-7 # wavelength of light in meter\n", + "mu_e = 1.54 # refractive index for extraordinary light\n", + "mu_o = 1.55 # refractive index for ordinary light\n", + "\n", + "#Calculation\n", + "t = l / (4 * (mu_o - mu_e)) # calculation for thickness of quartz plate\n", + "\n", + "#Result\n", + "print \"Thickness of quartz plate = %.2e meter\"%t" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thickness of quartz plate = 1.47e-05 meter\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.18, Page 3.28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "l = 5.89e-7 # wavelength of light in meter\n", + "mu_e = 1.553 # refractive index for extraordinary light\n", + "mu_o = 1.544 # refractive index for ordinary light\n", + "\n", + "#Calculation\n", + "t = l / (4 * (mu_e - mu_o)) # calculation for thickness of quartz plate\n", + "\n", + "#Result\n", + "print \"Thickness of quartz plate = %.2e meter\"%t" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thickness of quartz plate = 1.64e-05 meter\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.19, Page 3.29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "mu_e = 1.5442 # refractive index for extraordinary light\n", + "mu_o = 1.5533 # refractive index for ordinary light\n", + "l = 5e-7 # wavelength of plane polarized light in meter\n", + "\n", + "#Calculation\n", + "t = l / (2 * (mu_o - mu_e))# calculation for thickness of quartz plate\n", + "\n", + "#Result\n", + "print \"Thickness of quartz plate = %.2e meter\"%t" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thickness of quartz plate = 2.75e-05 meter\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.20, Page 3.29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "theta = 10 # rotation of plane of polarization in degree\n", + "s = 60 # specific rotation of sugar solution in degree per decimeter per unit concentration\n", + "l = 2.5 # length of Polari meter in decimeter\n", + "\n", + "#Calculation\n", + "c = theta / (s * l) # calculation for concentration of sugar solution\n", + "\n", + "#Result\n", + "print \"Concentration of sugar solution = %.3f gm/cc\"%c" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Concentration of sugar solution = 0.067 gm/cc\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.21, Page 3.29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "theta = 26.4 # rotation of plane of polarization in degree\n", + "c = 0.2 # concentration of sugar solution in gm/cc\n", + "l = 2 # length of polarizing tube in decimeter\n", + "\n", + "#Calculation\n", + "s = theta / (l * c)# calculation for specific rotation of sugar solution\n", + "\n", + "#Result\n", + "print \"Specific rotation of sugar solution = %.f degree/(dm-cc)\"%s" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Specific rotation of sugar solution = 66 degree/(dm-cc)\n" + ] + } + ], + "prompt_number": 43 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.22, Page 3.29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "theta = 6.5 # rotation of plane of polarization in degree\n", + "c = 0.05 # concentration of sugar solution in gm/cc\n", + "l = 2 # length of polarizing tube in decimeter\n", + "\n", + "#Calculation\n", + "s = theta / (l * c) # calculation for specific rotation of sugar solution\n", + "\n", + "#Result\n", + "print \"Specific rotation of sugar solution = %.f degree/(dm-cc)\"%s" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Specific rotation of sugar solution = 65 degree/(dm-cc)\n" + ] + } + ], + "prompt_number": 42 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.23, Page 3.30" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "w = 80 # weight of impure sugar in gm\n", + "theta = 9.9 # rotation of plane of polarization in degree\n", + "s = 66 # specific rotation of sugar solution in degree per decimeter per unit concentration\n", + "l = 2 # length of Polari meter in decimeter\n", + "\n", + "#Calculations\n", + "c = theta / (s * l) * (1000) # in gm/l\n", + "per_c = (c * 100) / w # calculation for concentration of sugar solution\n", + "\n", + "#Result\n", + "print \"Concentration of sugar solution = %.2f percent\"%per_c" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Concentration of sugar solution = 93.75 percent\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.24, Page 3.30" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "theta = 11. # rotation of plane of polarization in degree\n", + "s = 66 # specific rotation of sugar solution in degree per decimeter per unit concentration\n", + "l = 2 # length of Polari meter in decimeter\n", + "\n", + "#Calculation\n", + "c = theta / (s * l) # calculation for concentration of sugar solution\n", + "\n", + "#Result\n", + "print \"Concentration of sugar solution = %.4f gm/cc\"%c" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Concentration of sugar solution = 0.0833 gm/cc\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.25, Page 3.30" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "theta = 26.4 # rotation of plane of polarization in degree\n", + "c = 0.2 # concentration of sugar solution in gm/cc\n", + "l = 2 # length of polarizing tube in decimeter\n", + "\n", + "#calculation\n", + "s = theta / (l * c) # calculation for specific rotation of sugar solution\n", + "\n", + "#Result\n", + "print \"Specific rotation of sugar solution = %.f degree/(dm-cc)\"%s" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Specific rotation of sugar solution = 66 degree/(dm-cc)\n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.26, Page 3.30" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "theta = 13 # rotation of plane of polarization in degree\n", + "r = (1. / 3) # ratio of the final concentration to the initial solution\n", + "l = 2 # length of Polari meter in decimeter\n", + "l_ = 3 # length of second polarizing tube in decimeter \n", + "\n", + "#Calculation\n", + "theta_ = (l_ * r * theta) / l# calculation for optical rotation of diluted solution\n", + "\n", + "#Result\n", + "print \"Optical rotation of diluted solution = %.1f degree\"%theta_" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Optical rotation of diluted solution = 6.5 degree\n" + ] + } + ], + "prompt_number": 39 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter_4.ipynb b/Engineering_Physics/Chapter_4.ipynb new file mode 100755 index 00000000..981a3d82 --- /dev/null +++ b/Engineering_Physics/Chapter_4.ipynb @@ -0,0 +1,311 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 4: Lasers and Holography" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.1, Page 4.23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import exp\n", + "\n", + "# Given \n", + "l = 5.5e-7 # wavelength of light in meter\n", + "c = 3e+8 # speed of light in m/sec\n", + "h = 6.63e-34 # Planck constant in j/sec\n", + "e = 1.6e-19 # charge on electron in coulomb \n", + "k = 8.62e-5 # Boltzmann constant in eV/K\n", + "T = 300 # temperature in kelvin\n", + "\n", + "#Calculations\n", + "delta_E = (h * c) / (l * e) # calculation for energy difference \n", + "r = exp(-delta_E / (k * T)) # calculation for ratio of population of upper level to the lower energy level\n", + "T_ = (delta_E / (k * 0.693)) # calculation for temperature for the second condition\n", + "\n", + "#Result\n", + "print \"Ratio of population of upper level to the lower energy level = %.1e. \\nTemperature for the second condition = %.f K. \"%(r,T_)\n", + "#Incorrect answer in the textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ratio of population of upper level to the lower energy level = 1.1e-38. \n", + "Temperature for the second condition = 37837 K. \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.2, Page 4.24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "# Given \n", + "lambda1 = 6.328e-7 # wavelength of light in first case in meter\n", + "lambda2 =2e-7 # wavelength of light in second case in meter\n", + "r1 = 2.3e-4 # the radius of internal beam of laser in first case in meter\n", + "r2 = 2.4e-3 # the radius of internal beam of laser in second case in meter\n", + "\n", + "#Calculations\n", + "theta1 = lambda1 / (pi * r1) # calculation for beam divergence angle in first case\n", + "theta2 = lambda2 / (pi * r2) # calculation for beam divergence angle in second case\n", + "\n", + "#Result\n", + "print \"Beam divergence angle in first case = %.2e radian. \\nBeam divergence angle in second case = %.2e radian. \"%(theta1,theta2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Beam divergence angle in first case = 8.76e-04 radian. \n", + "Beam divergence angle in second case = 2.65e-05 radian. \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.3, Page 4.25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi,ceil\n", + "\n", + "# Given \n", + "l = 6.0*10**-2 # length of laser in meter\n", + "D = 1.0*10**-2 # diameter of laser in meter\n", + "L = 6.944e-7 # wavelength of light in meter\n", + "d = 3700 # density of aluminium oxide in kg/meter cube\n", + "Na = 6e+23 # Avogadro number\n", + "M = 0.102 # molar mass of aluminium oxide in kg/meter cube\n", + "h = 4.1e-15 # Planck constant in eV-sec\n", + "c = 3e+8 # speed of light in meter/sec\n", + "\n", + "#Calculations\n", + "v = (pi * (D**2) * l) / 4 # calculation for volume \n", + "N = (2 * Na * d * v) / M # calculation for no. of aluminium ions\n", + "N_ = N / 3500 # calculation for the no. of chromium ions\n", + "E = (h * c) / L # calculation for the energy of stimulated emission photon \n", + "Et = N_ * E * (1.6e-19) # calculation for total energy\n", + "\n", + "#Result\n", + "print \"Total energy = %.f J\"%(ceil(Et))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Total energy = 17 J\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.4, Page 4.26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "# Given \n", + "p = 4e-3 # energy of laser pulse in meter\n", + "r = 1.5e-5 # radius of spot in meter\n", + "t = 1e-9 # pulse length in time in sec\n", + "\n", + "#Calculations\n", + "p_ = p / t# calculation for power in watt\n", + "I = p_ / (pi * r**2)# calculation for power per unit area delivered by the laser\n", + "\n", + "#Result\n", + "print \"Power per unit area delivered by the laser = %.1e watt/square meter\"%I" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Power per unit area delivered by the laser = 5.7e+15 watt/square meter\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.5, Page 4.26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "D = 5e-3 # diameter of laser in meter\n", + "l = 7.2e-7 # wavelength of light in meter\n", + "d = 4e8 # distance at moon from earth in meter\n", + "\n", + "#Calculations\n", + "r = (D / 2) # calculation for radius\n", + "theta = (0.637 * l) / r # calculation for angular spread\n", + "areal_spread = (d * theta)**2 # calculation for areal spread\n", + "\n", + "#Result\n", + "print \"Angular spread = %.3e radian ,\\nAreal spread = %.2e square meter\"%(theta,areal_spread)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Angular spread = 1.835e-04 radian ,\n", + "Areal spread = 5.38e+09 square meter\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.6, Page 4.27\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "D = 5.0e-3 # diameter of laser in meter\n", + "l = 6.943e-7 # wavelength of light in meter\n", + "f =0.1 # focal length in meter\n", + "P = 0.1 # power of laser in watt\n", + "\n", + "#Calculations\n", + "r = (D / 2)# calculation for \n", + "theta = (0.637 * l) / r# calculation for angular spread\n", + "areal_spread = (f * theta)**2# calculation for areal spread\n", + "I = P / areal_spread# calculation for intensity\n", + "\n", + "#Result\n", + "print \"Areal spread = %.3e square meter,\\nIntensity = %.3e watt/square meter\"%(areal_spread,I)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Areal spread = 3.130e-10 square meter,\n", + "Intensity = 3.195e+08 watt/square meter\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.7, Page 4.28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "tou = 1e-10 # coherence time in sec\n", + "l = 5.4e-7 # wavelength of light in meter\n", + "\n", + "#Calculations\n", + "delta_v = 1 / tou \n", + "v_ = (3e+8) / l # calculation for frequency\n", + "d = delta_v / v_ # calculation for degree of non-monochromaticity\n", + "\n", + "#Result\n", + "print \"Degree of non-monochromaticity = %f \"%d" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Degree of non-monochromaticity = 0.000018 \n" + ] + } + ], + "prompt_number": 9 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter_5.ipynb b/Engineering_Physics/Chapter_5.ipynb new file mode 100755 index 00000000..e9ebdd41 --- /dev/null +++ b/Engineering_Physics/Chapter_5.ipynb @@ -0,0 +1,549 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 5: Fibre Optics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.1, Page 5.15" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi, sqrt, asin\n", + "\n", + "# Given \n", + "mu1 = 1.52 # refractive index for core\n", + "mu2 = 1.41 # refractive index for cladding\n", + "\n", + "#Calculations\n", + "theta_c = asin(mu2 / mu1) * (180 / pi)\n", + "NA = sqrt(mu1**2 - mu2**2)\n", + "theta_0 = asin(NA) * (180 / pi)\n", + "\n", + "#Result\n", + "print \"Critical angle = %.2f degree \\nNumerical aperture = %.3f\\nMaximum incidence angle = %.1f degree\"%(theta_c,NA,theta_0)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Critical angle = 68.07 degree \n", + "Numerical aperture = 0.568\n", + "Maximum incidence angle = 34.6 degree\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.2, Page 5.16" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi, sqrt, asin\n", + "\n", + "# Given \n", + "mu1 = 1.6 # refractive index for core\n", + "mu2 = 1.5 # refractive index for cladding\n", + "\n", + "#Calculations\n", + "NA = sqrt(mu1**2 - mu2**2)#calculation for numerical aperture\n", + "theta_0 = asin(NA) * (180 / pi)#calculation for maximum incidence angle\n", + "\n", + "#Result\n", + "print \"Numerical aperture = %.3f\\nMaximum incidence angle = %.2f degree\"%(NA,theta_0)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Numerical aperture = 0.557\n", + "Maximum incidence angle = 33.83 degree\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.3, Page 5.16" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi, sqrt, asin\n", + " \n", + "# Given \n", + "mu_0 = 1 # refractive index of air\n", + "mu1 = 1.5 # refractive index for core\n", + "mu2 = 1.48 # refractive index for cladding\n", + "\n", + "#Calculations\n", + "theta_c = asin(mu2 / mu1) * (180 / pi)\n", + "delta_mu = (mu1 - mu2) / mu1\n", + "NA = sqrt(mu1**2 - mu2**2)\n", + "theta_0 = asin(NA) * (180 / pi)\n", + "\n", + "#Result\n", + "print \"Critical angle = %.2f degrees \\nNumerical aperture = %.3f \\nAcceptance angle = %.2f degrees\\nFractional refractive index = %.2f %%\"%(theta_c,NA,theta_0,delta_mu*100)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Critical angle = 80.63 degrees \n", + "Numerical aperture = 0.244 \n", + "Acceptance angle = 14.13 degrees\n", + "Fractional refractive index = 1.33 %\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.4, Page 5.17" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import * \n", + "# Given \n", + "mu1 = 1.62 # refractive index for core\n", + "mu2 = 1.52 # refractive index for cladding\n", + "\n", + "#Calculations\n", + "NA = sqrt(mu1**2 - mu2**2)\n", + "theta_0 = asin(NA) * (180 / pi)\n", + "\n", + "#Results\n", + "print \"Numerical aperture = %.2f \\nMaximum incidence angle = %.1f degrees\"%(NA,theta_0)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Numerical aperture = 0.56 \n", + "Maximum incidence angle = 34.1 degrees\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.5, Page 5.17" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "# Given \n", + "NA = 0.22 # numerical aperture\n", + "delta_mu = 0.012 # fractional refractive index\n", + "\n", + "#Calculations\n", + "mu1 = sqrt(NA**2 / (1 - (1 - delta_mu)**2))\n", + "mu2 = (1 - delta_mu) * mu1\n", + "\n", + "#Result\n", + "print \"Refractive index for core = %.3f\\nRefractive index for cladding = %.2f\"%(mu1,mu2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Refractive index for core = 1.424\n", + "Refractive index for cladding = 1.41\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.6, Page 5.17" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi, sqrt, asin, sin\n", + "\n", + "# Given \n", + "d = 0.0064 # diameter of fiber in cm\n", + "mu1 = 1.53 # refractive index for core\n", + "mu2 = 1.39 # refractive index for clad\n", + "L = 90 # length of fiber in cm\n", + "mu_0 = 1 # refractive index of air\n", + "\n", + "#calculations\n", + "NA = sqrt(mu1**2 - mu2**2)\n", + "theta_0 = asin(NA) * (180 / pi)\n", + "N1 = L / (d * sqrt((mu1 / (mu_0 * sin(theta_0 * (pi / 180))))**2 - 1))\n", + "N2 = L / (d * sqrt((mu1 / (mu_0 * sin(theta_0 * (pi / 360))))**2 - 1))\n", + "\n", + "#Result\n", + "print \"Numerical aperture = %.2f\\nAcceptance angle = %.1f degrees \\nNumber of reflections at maximum incidence = %.f \\nNumber of reflections in second case = %.f \"%(NA,theta_0,N1,N2)\n", + "#Incorrect answer in the textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Numerical aperture = 0.64\n", + "Acceptance angle = 39.7 degrees \n", + "Number of reflections at maximum incidence = 6468 \n", + "Number of reflections in second case = 3204 \n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.7, Page 5.18" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "# Given \n", + "d = 0.05 # diameter of fiber in mm\n", + "NA = 0.22 # numerical aperture\n", + "l = 8.5e-4 # wavelength of light in mm\n", + "\n", + "#calculations\n", + "Vn = (pi * d * NA) / l\n", + "Mm = 0.5 * (Vn)**2\n", + "\n", + "#Result\n", + "print \"The normalized frequency = %.2f\\nNumber of guided in the core = %d\"%(Vn,Mm)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The normalized frequency = 40.66\n", + "Number of guided in the core = 826\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.8, Page 5.18" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt, pi\n", + "\n", + "# Given \n", + "l = 1.25e-6 #wavelength of light in meter\n", + "mu1 = 1.465 # refractive index for core\n", + "mu2 = 1.460 # refractive index for cladding\n", + "\n", + "#Calculations\n", + "NA = sqrt(mu1**2 - mu2**2)\n", + "k = (2.4 * l) / ( pi * NA)\n", + "Mm = 0.5 * ((pi * 50e-6 * NA) / l)**2\n", + "\n", + "#Result\n", + "print \"Diameter of core < %.1e meter,\\n number of modes = %d\"%(k,Mm)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diameter of core < 7.9e-06 meter,\n", + " number of modes = 115\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.9, Page 5.19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt, pi\n", + "\n", + "# Given \n", + "l = 0.85e-6 #wavelength of light in meter\n", + "mu1 = 1.461 # refractive index for core\n", + "mu2 = 1.456 # refractive index for clad\n", + "d = 4e-5 # diameter of core in meter\n", + "\n", + "#Calculations\n", + "NA = sqrt(mu1**2 - mu2**2)\n", + "Mm = 0.5 * ((pi * d * NA) / l)**2\n", + "\n", + "#Result\n", + "print \"Numerical aperture = %.3f\\n Number of modes = %d \"%(NA,Mm)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Numerical aperture = 0.121\n", + " Number of modes = 159 \n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.10, Page 5.19\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt, pi\n", + "\n", + "# Given\n", + "mu1 = 3.6 # refractive index for core\n", + "mu2 = 3.55 # refractive index for cladding\n", + "\n", + "#Calculations\n", + "NA = sqrt(mu1**2 - mu2**2)#calculation for numerical aperture\n", + "Mm1 = 0.5 * (pi * 5 * NA)**2#calculation for no. of modes in first case\n", + "Mm2 = 0.5 * (pi * 50 * NA)**2#calculation for no. of modes in second case\n", + "\n", + "#Result\n", + "print \"Number of modes in first case = %d \\nNumber of modes in second case = %d\"%(Mm1,Mm2)\n", + "#Incorrect answer in the textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of modes in first case = 44 \n", + "Number of modes in second case = 4410\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.11, Page 5.20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt, pi\n", + "\n", + "# Given \n", + "l = 1.25e-6 #wavelength of light in meter\n", + "mu1 = 1.46 # refractive index for core\n", + "mu2 = 1.457 # refractive index for cladding\n", + "\n", + "#Calculations\n", + "NA = sqrt(mu1**2 - mu2**2)#calculation for numerical aperture\n", + "k = (2.4 * l) / ( pi * NA)\n", + "\n", + "#Result\n", + "print \"Maximum diameter of core = %.2f micro meter\"%(k*1e6)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum diameter of core = 10.21 micro meter\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.12, Page 5.20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "# Given \n", + "L = 0.1 # length of fiber in km\n", + "p = 5e-6 # power of signal in watt\n", + "p_ = 1e-6 # power of signal inside the fiber in watt\n", + "\n", + "#Calculation\n", + "alpha = (10 * log10(p / p_)) / L#calculation for absorption coefficient\n", + "\n", + "#Result\n", + "print \"Absorption coefficient = %d dB/km \"%alpha" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Absorption coefficient = 69 dB/km \n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.13, Page 5.20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import exp\n", + "\n", + "# Given \n", + "L = 3 # length of optical fiber in km\n", + "l = 6 # losses in dB\n", + "p = 5e-3 # input power in watt \n", + "\n", + "#calculations\n", + "alpha = (l * 3) / L\n", + "p_ = p / (exp((2.303 * alpha * L) / 10))\n", + "\n", + "#Result\n", + "print \"Output power = %.3f mW \"%(p_*1e3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Output power = 0.079 mW \n" + ] + } + ], + "prompt_number": 19 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter_6.ipynb b/Engineering_Physics/Chapter_6.ipynb new file mode 100755 index 00000000..76ec5aef --- /dev/null +++ b/Engineering_Physics/Chapter_6.ipynb @@ -0,0 +1,399 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 6: Electron Optics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.1, Page 6.20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "# Given \n", + "V = 500 # voltage across the electrode in eV\n", + "m = 9e-31 # mass of electron in kg\n", + "e = 1.6e-19 # charge on an electron in coulomb\n", + "\n", + "#Calculations\n", + "E = e * V\n", + "v = sqrt((2 * e * V) / m)\n", + "p = m * v\n", + "\n", + "#Result\n", + "print \"Energy gained by electron = %.e J\\nSpeed of electron = %.2e meter/sec\\nMomentum of electron = %.2e kg-meter/sec\"%(E,v,p)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy gained by electron = 8e-17 J\n", + "Speed of electron = 1.33e+07 meter/sec\n", + "Momentum of electron = 1.20e-23 kg-meter/sec\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.2, Page 6.20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "v = 2.5e6 # speed of electron in meter/sec\n", + "B = 2e-4 # magnetic field in tesla\n", + "r = 1.76e11 # ratio of charge on electron to the mass of electron in C/kg\n", + "\n", + "#Calculations\n", + "a = (B * r * v)\n", + "\n", + "#Result\n", + "print \"Momentum of acceleration = %.2e meter/square sec.\"%a" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Momentum of acceleration = 8.80e+13 meter/square sec.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.4, Page 6.21" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given that\n", + "v = 5.2e6 # speed of electron in meter/sec\n", + "B = 1.3e-4 # magnetic field in tesla\n", + "r = 1.76e11 # ratio of charge on electron to the mass of electron in C/kg\n", + "E = 3.2e-12 # energy of the electron beam in J\n", + "M = 9e-31 # mass of an electron in kg\n", + "\n", + "R = v / (r * B)\n", + "v_ = sqrt((2 * E) / M )\n", + "\n", + "print \"Radius of circle traced by the beam = %.1f cm. \\nSpeed of beam in second case = %.2e meter/sec\"%(R*100,v_)\n", + "print \"Speed of beam in second case is greater than speed of light so we cannot use above formula.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Radius of circle traced by the beam = 22.7 cm. \n", + "Speed of beam in second case = 2.67e+09 meter/sec\n", + "Speed of beam in second case is greater than speed of light so we cannot use above formula.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.5, Page 6.22" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "V = 2.500e3 # voltage across the electrode in V\n", + "E = 3.6e4 # strength of electric field in V/m\n", + "B = 1.2e-3 # magnetic field in tesla\n", + "\n", + "#Calculation\n", + "r = (E / B)**2 / (2 * V)#calculation for ratio of the charge on an electron to the mass of an electron\n", + "\n", + "#Result\n", + "print \"Ratio of the charge on an electron to the mass of an electron = %.1e C/kg.\"%r" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ratio of the charge on an electron to the mass of an electron = 1.8e+11 C/kg.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6, Page 6.23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "# Given \n", + "M = 9.1e-31 # mass of electron in kg\n", + "E = 1.6e-15 # energy of electron in J\n", + "B = 5e-5 # magnetic field in tesla\n", + "e = 1.6e-19 # charge on an electron in coulomb\n", + "\n", + "#Calculations\n", + "v = sqrt((2 * E) / M)\n", + "r = (M * v) / (e * B)\n", + "\n", + "#Result\n", + "print \"Larmoure radius = %.2f meter\"%r" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Larmoure radius = 6.75 meter\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.7, Page 6.23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "Mp = 1.67e-27 # mass of proton in kg\n", + "v = 3e5 # speed of proton in meter/sec\n", + "B = 5e-9 # magnetic field in tesla\n", + "e = 1.6e-19 # charge on a proton in coulomb\n", + "\n", + "#Calculation\n", + "r = (Mp * v) / (e * B)#calculation for Larmour radius\n", + "\n", + "#Result\n", + "print \"Larmour radius = %.2e meter\"%r" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Larmour radius = 6.26e+05 meter\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.8, Page 6.23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt, pi\n", + "\n", + "# Given \n", + "M = 6.68e-27 # mass of helium ion in kg\n", + "E = 1.6e-16 # energy of helium ion in J\n", + "B = 5e-2 # magnetic field in tesla\n", + "e = 1.6e-19 # charge on helium ion in coulomb\n", + "\n", + "#calculations\n", + "v = sqrt((2 * E) / M)#calculation for velocity\n", + "r = (M * v) / (e * B)#calculation for Larmour radius\n", + "A = pi * r**2#calculation for area traced by the trajectory of helium ion\n", + "\n", + "#Result\n", + "print \"Area traced by the trajectory of helium ion = %.3f square meter\"%A" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Area traced by the trajectory of helium ion = 0.105 square meter\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.9, Page 6.23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "E = 100 # strength of electric field in V/m\n", + "B = 1e-3 # magnetic field in tesla\n", + "\n", + "#Calculation\n", + "v = E / B\n", + "\n", + "#Result\n", + "print \"The drift of the guiding center = %.e m/sec\"%v" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drift of the guiding center = 1e+05 m/sec\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.10, Page 6.24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "v = 1e6 # velocity of ion beam in m/sec\n", + "B = 1 # magnetic field in tesla\n", + "\n", + "#Calculation\n", + "E = B * v\n", + "\n", + "#Result\n", + "print \"Internal electric field = %.e V/m\"%E" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Internal electric field = 1e+06 V/m\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.12, Page 6.24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "r = 1.1 # ratio of new number of turns to the initial number of turns\n", + "\n", + "#Calculation\n", + "r_ = (1 / r)**2\n", + "\n", + "#Result\n", + "print \"Ratio of the new focus length to the initial focus length = %.3f \"%r_" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ratio of the new focus length to the initial focus length = 0.826 \n" + ] + } + ], + "prompt_number": 11 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter_7.ipynb b/Engineering_Physics/Chapter_7.ipynb new file mode 100755 index 00000000..6762cbc5 --- /dev/null +++ b/Engineering_Physics/Chapter_7.ipynb @@ -0,0 +1,569 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 7: Waves and Oscillations" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.1, Page 7.22" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi, sin, sqrt\n", + "\n", + "# Given \n", + "E = 1.024e-3 # total energy of particle in J\n", + "T = 2 * pi # time period of S.H.M. in sec\n", + "x = 0.08 * sqrt(2) # distance of partile in meter\n", + "t = pi / 4 # time in second\n", + "\n", + "#Calculations\n", + "A = x / sin((2 * pi * t) / T)\n", + "M = (E * T**2) / (2 * pi**2 * A**2)\n", + "\n", + "#Result\n", + "print \"Amplitude = %.2f meter\\nMass of particle = %.f g\"%(A,M/1e-3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Amplitude = 0.16 meter\n", + "Mass of particle = 80 g\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.2, Page 7.22" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "# Given \n", + "A = 0.05 # amplitude in meter\n", + "T = 10 # time period of S.H.M. in sec\n", + "\n", + "#Calculations\n", + "v = (A * 2 * pi) / T\n", + "\n", + "#Result\n", + "print \"Maximum amplitude of velocity = %.4f meter/sec\"%v" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum amplitude of velocity = 0.0314 meter/sec\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.3, Page 7.23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi, sqrt\n", + "\n", + "# Given \n", + "E = 9 # total energy of particle in J\n", + "U = 5 # potential energy in J\n", + "A = 1 # amplitude in meter\n", + "m = 2. # mass of harmonic oscillator in kg\n", + "\n", + "#Calculations\n", + "kE = E - U# calculation for kinetic energy\n", + "k = (2 * kE) / A**2# calculation for force constant\n", + "T = (2 * pi) * sqrt(m / k)# calculation for time period\n", + "\n", + "#Result\n", + "print \"Force constant = %.f J/m\\nTime period = %.2f sec\"%(k,T)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Force constant = 8 J/m\n", + "Time period = 3.14 sec\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.4, Page 7.23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import asin, sin, pi\n", + "\n", + "# Given \n", + "A = 0.06 # amplitude in meter\n", + "T = 6 # time period of S.H.M. in sec\n", + "x = 0.03 # position of particle in meter\n", + "\n", + "#Calculations\n", + "delta = asin(1) # by the formula x=Asin(wt+delta) and (at t = 0,x=A) \n", + "t = x / (A * sin(((2 * pi) / T) + delta))\n", + "\n", + "#Result\n", + "print \"Time taken by the particle = %.f sec\"%t" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time taken by the particle = 1 sec\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.5, Page 7.24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "# Given \n", + "A = 0.05# amplitude in meter\n", + "T = 10 * pi # time period of s.h.m. in sec\n", + "\n", + "#Calculations\n", + "v = A * (2 * pi / T)\n", + "a = A * (2 * pi / T)**2\n", + "\n", + "#Result\n", + "print \"Maximum velocity = %.e meter/sec\\nacceleration = %.e m/sec^2\"%(v,a)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum velocity = 1e-02 meter/sec\n", + "acceleration = 2e-03 m/sec^2\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.6, Page 7.24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "# Given \n", + "A = 0.06# amplitude in meter\n", + "T = 10 * pi # time period of s.h.m. in sec\n", + "\n", + "#Calculation\n", + "v = A * (2 * pi / T)\n", + "\n", + "#Result\n", + "print \"Maximum velocity = %.1e meter/sec\"%v" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum velocity = 1.2e-02 meter/sec\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.7, Page 7.24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt, pi\n", + "\n", + "# Given \n", + "k = 16 # stiffness constant of spring n/m\n", + "m = 1 # mass of particle in kg\n", + "\n", + "#Calculations\n", + "n = sqrt(k / m) / (2 * pi)\n", + "\n", + "#Result\n", + "print \"natural frequency = %.2f Hz\"%n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "natural frequency = 0.64 Hz\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.8, Page 7.25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt, pi\n", + "\n", + "# Given \n", + "l = 1 # length of pendulum in meter\n", + "m = 2 # mass of particle in kg\n", + "g = 9.8 # acceleration due to gravity in m/sec^2\n", + "\n", + "#Calculation\n", + "T = 2 * pi * sqrt(l / g)\n", + "\n", + "#Result\n", + "print \"The time period of pendulum = %.f sec\"%T" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The time period of pendulum = 2 sec\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.9, Page 7.25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt, pi\n", + "\n", + "# Given \n", + "m = 100. # mass of particle in gm\n", + "\n", + "#Calculation\n", + "n = (1 / (2 * pi)) * sqrt(10 / m) # by using given formula \n", + "\n", + "#Result\n", + "print \"Frequency = %.2f Hz\"%n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Frequency = 0.05 Hz\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.10, Page 7.25" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt, pi\n", + "\n", + "# Given \n", + "f = 3 # acceleration of pendulum in m/sec^2\n", + "l = 1 # length of pendulum in meter\n", + "g = 9.8 # acceleration due to gravity in m/sec^2\n", + "\n", + "#calculatiom\n", + "T = 2 * pi * sqrt(l / (g + f))\n", + "\n", + "#Result\n", + "print \"Time period of pendulum = %.2f sec\"%T" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time period of pendulum = 1.76 sec\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.11, Page 7.26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt, pi\n", + "\n", + "# Given \n", + "x = 0.3 # stretch in spring in meter\n", + "m1 = 6 # mass of first body in kg\n", + "m2 = 1 # mass of second body in kg\n", + "g = 9.8 # gravitational acceleration of earth in m/sec^2\n", + "\n", + "#Calculations\n", + "k = (m1 * g) / x\n", + "T = (2 * pi) * sqrt(m2 / k)\n", + "\n", + "#Result\n", + "print \"Time period of motion = %.2f sec \"%T" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time period of motion = 0.45 sec \n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.12, Page 7.26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt, pi\n", + "\n", + "# Given \n", + "x = 0.1 # compression in spring in m\n", + "F = 10 # restoring force in N\n", + "m = 4 # mass of body in kg\n", + "g = 9.8 # acceleration due to gravity in m/sec^2\n", + "\n", + "#Calculations\n", + "k = F / x\n", + "F_ = m * g\n", + "x_ = F_ / k\n", + "T = (2 * pi) * sqrt(m / k)\n", + "\n", + "#Result\n", + "print \"Time period of motion = %.2f sec \\nCompression of the spring due to the weight of the body = %.3f m \"%(T,x_)\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time period of motion = 1.26 sec \n", + "Compression of the spring due to the weight of the body = 0.392 m \n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.13, Page 7.26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import exp\n", + "\n", + "# Given \n", + "t = 50. # relaxation time in sec\n", + "r = 1 / exp(1) # falls in amplitude and energy\n", + "\n", + "#Calculations\n", + "s = 1 / (2 * t)\n", + "T = 1 / s # by using formula A=A_exp(-st) and using r=A/A_\n", + "\n", + "#Result\n", + "print \"Time = %.f sec\"%T\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time = 100 sec\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.14, Page 7.27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi, exp\n", + "\n", + "# Given \n", + "n = 260 # frequency in Hz\n", + "Q = 2000 # quality factor\n", + "r = 1 / (exp(1)**2) # decrease in amplitude \n", + "\n", + "#Calculations\n", + "tou = Q / (2 * pi * n)\n", + "t = 2 * tou # by using formula A=A_exp(-st) and using r=A/A_\n", + "\n", + "#Result\n", + "print \"Time = %.3f sec\"%t\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time = 2.449 sec\n" + ] + } + ], + "prompt_number": 17 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter_8.ipynb b/Engineering_Physics/Chapter_8.ipynb new file mode 100755 index 00000000..60f3ed3a --- /dev/null +++ b/Engineering_Physics/Chapter_8.ipynb @@ -0,0 +1,413 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 8: Sound Waves and Acoustics of Buildings \n" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.1, Page 8.17" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "v = 34500 # speed of sound in cm/sec\n", + "f = 20 # lower limit of frequency for human hearing ear in Hz\n", + "f_ = 20000 # upper limit of frequency for human hearing ear in Hz\n", + "\n", + "#Clculations\n", + "l1 = v / f\n", + "l2 = v / f_\n", + "\n", + "#Results\n", + "print \"Wavelength range of the sound wave is %.f cm to %.f cm.\"%(l1,l2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength range of the sound wave is 1725 cm to 1 cm.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.2, Page 8.18" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "# Given \n", + "T = 373. # temperature in kelvin\n", + "d = 1.293e-3 # density of air at S.T.P. in gm/cm^3\n", + "d_ = 13.6 # density of mercury in gm/cm^3\n", + "Cp = 0.2417 # specific heat of air at constant pressure \n", + "Cv = 0.1715 # specific heat of air at constant volume\n", + "g = 980 # gravitational constant i dynes/cm^3\n", + "\n", + "#calculations\n", + "p = 76 * d_ * g\n", + "gama = Cp / Cv\n", + "v = sqrt(gama * (p / d))\n", + "v_ = v * sqrt(T / 273)\n", + "\n", + "#Result\n", + "print \"Velocity of sound in the air in %.f cm/sec.\"%v_\n", + "#Incorrect answer in textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Velocity of sound in the air in 38839 cm/sec.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.3, Page 8.18" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + " \n", + "# Given that\n", + "n = 512. # frequency of tuning fork in Hz \n", + "T = 290. # temperature in kelvin\n", + "l = 66.5 # wavelength of the gas emitted by tuning fork in cm\n", + "d = 1.293e-3 # density of air at S.T.P. in gm/cm^3\n", + "d_ = 13.6 # density of mercury in gm/cm^3\n", + "g = 980 # gravitational constant i dynes/cm^3\n", + "\n", + "#Calculations\n", + "p = 76 * d_ * g# calculation for pressure\n", + "v_ = n * l# calculation for velocity of sound in air at temperature 17 c \n", + "v = v_ * sqrt(273 / T)# calculation for velocity of sound in air at temp 0 c\n", + "gama = v**2 * (d / p)# calculation for ratio of two specific heat\n", + "\n", + "#Result\n", + "print \"Ratio of two principal specific heats of air is %.2f\"%gama" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ratio of two principal specific heats of air is 1.39\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.4, Page 8.19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "A = 15 * 30 # area of the floor in square meter\n", + "h = 6 # height of hall in meter\n", + "N = 500 # no. of people\n", + "t = 1.36 # optimum time for orchestral music in sec\n", + "k = 0.44 # absorption coefficient per person\n", + "\n", + "#Calculations\n", + "aS = 0.161 * ((A * h) / t)\n", + "a = N * k\n", + "a_ = aS - a\n", + "w = a_ + (N / 2) * k + (N / 2) * 0.02\n", + "t = (0.161 * (A * h)) / w\n", + "\n", + "#Results\n", + "print \"Coefficient of absorption to be provided by the walls, floor and ceiling when the hall is fully occupied is %.f SI unit.\"%a_\n", + "print \"Reverberation time if only half upholstered seats are occupied is %.2f sec.\"%t\n", + "#Answer varies due to rounding-off " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "319.632352941\n", + "Coefficient of absorption to be provided by the walls, floor and ceiling when the hall is fully occupied is 100 SI unit.\n", + "Reverberation time if only half upholstered seats are occupied is 2.03 sec.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.5, Page 8.19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given\n", + "V = 8000 # volume of hall in meter^3\n", + "t = 1.8 # reverberation time in sec\n", + "\n", + "#Calculation\n", + "aS = (0.161 * V) / t# calculation for the total absorption constant\n", + "\n", + "#Result\n", + "print \"The total absorption constant = %.3f O.W.U.\"%aS\n", + "#Incorrect answer in the textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The total absorption constant = 715.556 O.W.U.\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.6, Page 8.20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "V = 1700 # volume in meter^3\n", + "a1 = 98 # area of plastered wall in m^2\n", + "a2 = 144 # area of plastered ceiling in m^2\n", + "a3 = 15 # area of wooden door in m^2\n", + "a4 = 88 # area of cushioned chairs in m^2\n", + "a5 = 150 # area of audience (each person) in m^2\n", + "k1 = 0.03 # coefficient of absorption for plastered wall in O.W.U.\n", + "k2 = 0.04 # coefficient of absorption for plastered ceiling in O.W.U.\n", + "k3 = 0.06 # coefficient of absorption for wooden door in O.W.U.\n", + "k4 = 1 # coefficient of absorption for cushioned chair in O.W.U.\n", + "k5 = 4.7 # coefficient of absorption for audience (each person) in O.W.U.\n", + "\n", + "#Calculations\n", + "A1 = a1 * k1# calculation for the absorption by the plaster wall\n", + "A2 = a2 * k2# calculation for the absorption by the plastered ceiling\n", + "A3 = a3 * k3# calculation for wooden door\n", + "A4 = a4 * k4# calculation for cushioned chairs \n", + "A = A1 + A2 + A3 + A4# calculation for total absorption\n", + "T = 0.161 * (V / A)# calculation for reverberation time\n", + "\n", + "#Result\n", + "print \"Reverberation time is %.2f sec\"%T" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Reverberation time is 2.80 sec\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.7, Page 8.20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "V = 1400 # volume of hall in meter^3\n", + "C = 110 # seating capacity of hall \n", + "a1 = 98 # area of plastered wall in m^2\n", + "a2 = 144 # area of plastered ceiling in m^2\n", + "a3 = 15 # area of wooden door in m^2\n", + "a4 = 88 # area of cushioned chairs in m^2\n", + "a5 = 150 # area of audience (each person) in m^2\n", + "k1 = 0.03 # coefficient of absorption for plastered wall in O.W.U.\n", + "k2 = 0.04 # coefficient of absorption for plastered ceiling in O.W.U.\n", + "k3 = 0.06 # coefficient of absorption for wooden door in O.W.U.\n", + "k4 = 1 # coefficient of absorption for cushioned chair in O.W.U.\n", + "k5 = 4.7 # coefficient of absorption for audience (each person) in O.W.U.\n", + "\n", + "#Calculations\n", + "A1 = a1 * k1# calculation for the absorption by the plaster wall\n", + "A2 = a2 * k2# calculation for the absorption by the plastered ceiling\n", + "A3 = a3 * k3# calculation for wooden door\n", + "A4 = a4 * k4# calculation for cushioned chairs \n", + "A5 = C*k5 # the absorption due to persons\n", + "A = A1 + A2 + A3 + A4 + A5 # calculation for total absorption\n", + "T = (0.161 * V) / A# calculation for the reverberation time\n", + "\n", + "#Result\n", + "print \"Reverberation time is %.3f sec\"%T" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Reverberation time is 0.367 sec\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.8, Page 8.21" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "V = 980 # volume in meter^3\n", + "a1 = 150 # area of wall in m^2\n", + "a2 = 95 # area of ceiling in m^2\n", + "a3 = 90 # area of floor in m^2\n", + "k1 = 0.03 # coefficient of absorption for wall in O.W.U.\n", + "k2 = 0.80 # coefficient of absorption for ceiling in O.W.U.\n", + "k3 = 0.06 # coefficient of absorption for floor in O.W.U.\n", + "\n", + "#calculations\n", + "A1 = a1 * k1\n", + "A2 = a2 * k2\n", + "A3 = a3 * k3\n", + "A = A1 + A2 + A3\n", + "T = 0.161 * (V / A)\n", + "\n", + "#Result\n", + "print \"Reverberation time = %.2f sec\"%T" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Reverberation time = 1.84 sec\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.9, Page 8.21" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "V = 980 # volume in meter^3\n", + "a = 1.58 # area of window in m^2\n", + "I_ = 1e-12 # standard intensity level of sound wave in W/m^2\n", + "l = 60 # intensity level in dB\n", + "\n", + "#calculations\n", + "I = I_ * 10**(l / 10)# calculation for intensity\n", + "AP = I * a# calculation for acoustic power\n", + "\n", + "#Result\n", + "print \"Acoustic power = %.2e watt\"%AP" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Acoustic power = 1.58e-06 watt\n" + ] + } + ], + "prompt_number": 14 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_Physics/Chapter_9.ipynb b/Engineering_Physics/Chapter_9.ipynb new file mode 100755 index 00000000..a43c0fed --- /dev/null +++ b/Engineering_Physics/Chapter_9.ipynb @@ -0,0 +1,254 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 9: Dielectrics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.1, Page 9.11" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "E = 10**6 # electric field inside the plates in V/m\n", + "d = 0.02 # distance between the plates in meter\n", + "k = 3 # dielectric constant of slab\n", + "e_ = 8.85e-12 # electric permittivity of air in C^2/Nm^2\n", + "\n", + "#Calculations\n", + "D = e_*k*E\n", + "P = D-e_*E\n", + "\n", + "#Result\n", + "print \"Polarization vector is %.2e C/m^2 \\nDisplacement vector is %.3e C/m^2\"%(P,D)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Polarization vector is 1.77e-05 C/m^2 \n", + "Displacement vector is 2.655e-05 C/m^2\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.2, Page 9.11" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "# Given \n", + "E1 = 3*10**5 # electric intensity when space between plates evacuated in V/m\n", + "E2 = 1*10**5 # electric intensity when space between plates is filled with dielectric in V/m\n", + "e_ = 8.85e-12 # electric permittivity of air in C^2/Nm^2\n", + "\n", + "#calculation\n", + "sigma = e_*(E1 - E2)\n", + "\n", + "#result\n", + "print \"The included charge density on the surface of the dielectric is %.2e C/m^2\"%sigma" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The included charge density on the surface of the dielectric is 1.77e-06 C/m^2\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.3, Page 9.11" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "E = 1*10**5 # electric field strength inside the plates in V/m\n", + "d = 6. # distance between the plates in mm\n", + "k = 2.8 # dielectric constant of slab\n", + "e_ = 8.85e-12 # electric permittivity of air in C^2/Nm^2\n", + "\n", + "#Calculations\n", + "P = e_*(k-1)*E\n", + "D = e_*k*E\n", + "energy_density = 1./2 * k*e_*E**2\n", + "\n", + "#Result\n", + "print \"Polarization vector is %.1e C/m^2 \\nDisplacement vector is %.1eC/m^2 \\nEnergy density is %.3f J/m^3\"%(P,D,energy_density )" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Polarization vector is 1.6e-06 C/m^2 \n", + "Displacement vector is 2.5e-06C/m^2 \n", + "Energy density is 0.124 J/m^3\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.4, Page 9.12" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "D = 5e-4 # electric displacement vector in C/m^2\n", + "P = 4e-4 # electric polarization vector in C/m^2\n", + "v = 0.5 # volume of the slab in m^3\n", + "e_ = 8.85e-12 # electric permittivity of air in C^2/Nm^2\n", + "\n", + "#Calculations\n", + "E= (D-P)/ e_\n", + "k = D/(e_*E)\n", + "p = P*v\n", + "energy_density = 1./2 * k*e_*E**2\n", + "\n", + "#Result\n", + "print \"Value of relative permittivity is %d \\nTotal dipole moment of the slab is %.e C-m\"%(k,p) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Value of relative permittivity is 5 \n", + "Total dipole moment of the slab is 2e-04 C-m\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.5, Page .12" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given that\n", + "E = 3e4 # external field in V/m\n", + "k = 1.00074 # dielectric constant of gas at N.T.P.\n", + "e_ = 8.85e-12 # electric permittivity of air in C^2/Nm^2\n", + "\n", + "#Calculations\n", + "x = k-1\n", + "P = x*e_*E\n", + "N = 6.023e23/22.4e-3\n", + "p = P/N\n", + "\n", + "#Result\n", + "print \"Induce dipole moment of each is %.2e C-m\"%p" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Induce dipole moment of each is 7.31e-36 C-m\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.6, Page .13" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given \n", + "E = 3e4 # external field in V/m\n", + "k = 1.000041 # dielectric constant of gas at 0 degree centigrate\n", + "\n", + "#calculation\n", + "x = k-1\n", + "\n", + "#Result\n", + "print \"Electric susceptibility at 0 degree centigrate is %.1e\"%x" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Electric susceptibility at 0 degree centigrate is 4.1e-05\n" + ] + } + ], + "prompt_number": 8 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_Physics/README.txt b/Engineering_Physics/README.txt index bd7c4bfa..85c741ca 100755 --- a/Engineering_Physics/README.txt +++ b/Engineering_Physics/README.txt @@ -1,10 +1,10 @@ -Contributed By: KRISHNA CHAITANYA -Course: btech -College/Institute/Organization: JNTUH -Department/Designation: Computer Science +Contributed By: Muktesh Chaudhary +Course: be +College/Institute/Organization: Anglo Eastern ship management india Pvt. Ltd +Department/Designation: Electrical & Electronics Officer Book Title: Engineering Physics -Author: D. K. Bhattacharya & A. Bhaskaran -Publisher: Oxford University Press, New Delhi -Year of publication: 2013 -Isbn: 9780198065425 -Edition: 1
\ No newline at end of file +Author: H. K. Malik & A. Singh +Publisher: Tata McGraw Hill, New Delhi +Year of publication: 2010 +Isbn: 9780070671539 +Edition: 1st
\ No newline at end of file diff --git a/Engineering_Physics/screenshots/noofphotons.png b/Engineering_Physics/screenshots/noofphotons.png Binary files differnew file mode 100755 index 00000000..bb15b8cf --- /dev/null +++ b/Engineering_Physics/screenshots/noofphotons.png diff --git a/Engineering_Physics/screenshots/ref_indices.png b/Engineering_Physics/screenshots/ref_indices.png Binary files differnew file mode 100755 index 00000000..300447f7 --- /dev/null +++ b/Engineering_Physics/screenshots/ref_indices.png diff --git a/Engineering_Physics/screenshots/wavelenK.png b/Engineering_Physics/screenshots/wavelenK.png Binary files differnew file mode 100755 index 00000000..da79e117 --- /dev/null +++ b/Engineering_Physics/screenshots/wavelenK.png diff --git a/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/README.txt b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/README.txt new file mode 100755 index 00000000..c62d8c2b --- /dev/null +++ b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/README.txt @@ -0,0 +1,10 @@ +Contributed By: Hiren Shah +Course: mca +College/Institute/Organization: Financial Technology +Department/Designation: Developer +Book Title: Fundamentals Of Engineering Heat And Mass Transfer +Author: R. C. Sachdeva +Publisher: New Age Science Ltd., New Delhi +Year of publication: 2009 +Isbn: 9781906574123 +Edition: 4
\ No newline at end of file diff --git a/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch1.ipynb b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch1.ipynb new file mode 100755 index 00000000..8a321701 --- /dev/null +++ b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch1.ipynb @@ -0,0 +1,260 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:756d82bd4f6b1c40b9796c02104f4ba9dd307f7a3eb5083116664c3ed159b9c9" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 1 : Basic Concepts" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.1 Page No : 9" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "L = 0.02;\t\t \t#Thicness of stainless steel plate in m\n", + "T = [550,50];\t\t\t#Temperatures at both the faces in degree C\n", + "k = 19.1;\t\t \t#Thermal Conductivity of stainless steel at 300 degree C in W/m.K\n", + " \n", + "# Calculations\n", + "q = ((k*(T[0]-T[1]))/(L*1000));\t\t\t#Heat transfered per uni area in kW/m**2\n", + "\n", + "# Results\n", + "print 'The heat transfered through the material per unit area is %3.1f kW/m**2'%(q)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The heat transfered through the material per unit area is 477.5 kW/m**2\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.2 Page No : 11" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "L = 1.;\t\t\t#Length of the flat plate in m\n", + "w = 0.5;\t\t\t#Width of the flat plate in m\n", + "T = 30.;\t\t\t#Air stream temperature in degree C\n", + "h = 30.;\t\t\t#Convective heat transfer coefficient in W/m**2.K\n", + "Ts = 300.;\t\t\t#Temperature of the plate in degree C\n", + "\n", + "# Calculations\n", + "A = (L*w) \t\t\t#Area of the plate in m**2\n", + "Q = (h*A*(Ts-T)/(1000));\t\t\t#Heat transfer in kW\n", + "\n", + "# Results\n", + "print 'Heat transfer rate is %3.2f kW'%(Q)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer rate is 4.05 kW\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.3 Page No : 11" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "T = 55.\t\t\t#Surface temperature in degree C\n", + "\n", + "# Calculations\n", + "q = (5.6697*10**-8*(273+T)**4)/1000;\t\t\t#The rate at which the radiator emits radiant heat per unit area if it behaves as a black body in kW/m**2\n", + "\n", + "# Results\n", + "print 'The rate at which the radiator emits radiant heat per unit area is %3.2f kW/m**2'%(q)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The rate at which the radiator emits radiant heat per unit area is 0.66 kW/m**2\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.5 Page No : 20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "k = 0.145;\t\t\t#Thermal conductivity of Firebrick in W/m.K\n", + "e = 0.85;\t\t\t#Emissivity\n", + "L = 0.145;\t\t\t#Thickness of the wall in m\n", + "Tg = 800.;\t\t\t#Gas temperature in degree C\n", + "Twg = 798.;\t\t\t#Wall temperature ion gas side in degree C\n", + "hg = 40.;\t\t\t#Film conductance on gas side in W/m**2.K\n", + "hc = 10.;\t\t\t#Film conductance on coolant side in W/m**2.K\n", + "F = 1.; \t\t\t#Radiation Shape factor between wall and gas\n", + "\n", + "# Calculations\n", + "R1 = (((e*5.67*10**-8*F*((Tg+273)**4-(Twg+273)**4))/(Tg-Twg))+(1./hg));\t\t\t#Thermal resistance inverse\n", + "R2 = (L/k); \t \t\t#Thermal resistance\n", + "R3 = (1./hc);\t \t \t#Thermal resistance\n", + "U = 1./((1./R1)+R2+R3);\t\t\t#Overall heat transfer coefficient in W/m**2.K\n", + "\n", + "# Results\n", + "print 'Overall heat transfer coefficient is %3.3f W/m**2.K'%(U)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Overall heat transfer coefficient is 0.906 W/m**2.K\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.6 Page No : 21" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "D = 0.05;\t\t\t#Outside diameter of the pipe in m\n", + "e = 0.8;\t\t\t#Emmissivity\n", + "T = 30;\t\t\t#Room Temperature in degree C\n", + "Ts = 250;\t\t\t#Surface temperature in degree C\n", + "h = 10;\t\t\t#Convective heat transfer coefficient in W/m**2.K\n", + "\n", + "# Calculations\n", + "q = ((h*3.14*D*(Ts-T))+(e*3.14*D*5.67*10**-8*((Ts+473)**4-(T+273)**4)));\t\t\t#Heat loss per unit length of pipe in W/m\n", + "\n", + "# Results\n", + "print 'Heat loss per unit length of pipe is %3.1f W/m'%(q) \n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat loss per unit length of pipe is 2231.3 W/m\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.7 Page No : 21" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "A = 0.1;\t\t\t#Surface area of water heater in m**2\n", + "Q = 1000.;\t\t\t#Heat transfer rate in W\n", + "Twater = 40;\t\t\t#Temperature of water in degree C\n", + "h1 = 300;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "Tair = 40;\t\t\t#Temperature of air in degree C\n", + "h2 = 9;\t\t\t#Heat transfer coefficient in W/m**2.K \n", + "\n", + "# Calculations\n", + "Tsw = (Q/(h1*A))+Twater;\t\t\t#Temperature when used in water in degree C\n", + "Tsa = (Q/(h2*A))+Tair;\t\t\t#Temperature when used in air in degree C\n", + "\n", + "# Results\n", + "print 'Temperature when used in water is %3.1f degree C \\n \\\n", + "Temperature when used in air is %i degree C'%(Tsw,Tsa)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temperature when used in water is 73.3 degree C \n", + " Temperature when used in air is 1151 degree C\n" + ] + } + ], + "prompt_number": 7 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch10.ipynb b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch10.ipynb new file mode 100755 index 00000000..a0d3a1bb --- /dev/null +++ b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch10.ipynb @@ -0,0 +1,1000 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:4a3273e6ea6b2958563a9e1b4fdabb424a50f820382694523b368d9f9ad6e27a" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 10 : Radiative Heat exchange between surfaces" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.1 Page No : 403" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables\n", + "I = 1350;\t\t\t#Solar Irradiation in W/m**2\n", + "L = (1.5*10**8);\t\t\t#Approximate dismath.tance in km\n", + "D = (1.39*10**6);\t\t\t#Approximate diameter in km\n", + "\n", + "\n", + "# Calculations\n", + "E = (I*(L*1000)**2*3.14)/((3.14/4)*(D*1000)**2);\t\t\t#Emissive power of Earth \n", + "Ts = (E/(5.67*10**-8))**0.25;\t\t\t#Surface temperature of sun in K\n", + "\n", + "# Results\n", + "print 'Surface temperature of sun is %d K'%(Ts)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Surface temperature of sun is 5770 K\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.4 Page No : 409" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "S = 1;\t\t\t#Side of a square in m\n", + "L = 0.4;\t\t\t#Distance between the plates in m\n", + "T1 = 900;\t\t\t#Temperature of one plate in degree C\n", + "T2 = 400;\t\t\t#Temperature of the other plate in degree C\n", + "\n", + "# Calculations\n", + "R = (S/L);\t\t\t#Ratio of the side of the square to the distance between plates\n", + "F12 = 0.415;\t\t\t#From Fig.10.4 on page no.409\n", + "Q = (5.67*10**-8*S*S*F12*((T1+273)**4-(T2+273)**4))/1000;\t\t\t#The net heat transfer in kW\n", + "\n", + "# Results\n", + "print 'The net exchange of energy due to radiation between the plates is %3.1f kW'%(Q)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The net exchange of energy due to radiation between the plates is 39.7 kW\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.5 Page No : 411" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "A51 = 2;\t\t\t#Ratio of areas A5 and A1\n", + "A21 = 1;\t\t\t#Ratio of areas A2 and A1\n", + "F56 = 0.15;\t\t\t#Shape factor\n", + "F53 = 0.11;\t\t\t#Shape factor\n", + "F26 = 0.24;\t\t\t#Shape Factor\n", + "F23 = 0.2;\t\t\t#Shape Factor\n", + "\n", + "# Calculations\n", + "F14 = (A51*(F56-F53))-(A21*(F26-F23));\t\t\t#Shape factor\n", + "\n", + "# Results\n", + "print 'Shape factor F14 is %3.2f'%(F14)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Shape factor F14 is 0.04\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.8 Page No : 415" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Th = 40.;\t\t\t#Radiating heating panel in degree C\n", + "Tb = 5.;\t\t\t#Temperature of black plane in degree C\n", + "Tc = 31.;\t\t\t#Temperature of ceiling in degree C\n", + "A = (10.*12);\t\t#Area in m**2\n", + "\n", + "# Calculations\n", + "F56 = 0.075;\t\t\t#Using Fig.10.2 on page no. 408\n", + "F63 = 0.04;\t \t\t#Using Fig.10.2 on page no. 408\n", + "F12 = 0.052;\t\t\t#Shape factor\n", + "F1w = (1-F12);\t\t\t#Shape factor between the floor and all the walls but the window\n", + "Q12 = (A*F12*5.67*10**-8*((Th+273)**4-(Tb+273)**4));\t\t\t#Heat exchange between the floor and window in W\n", + "#Q1 = (5.67*10**-8*A*((Th+273.15)**4-((F12*(Th+273.15)**4)-(F1w*(Tb+273.15)**4))))/1000;\t\t\t#Net heat given up by the floor in kW\n", + "Q1 = (5.67*10**-8*A*((Th+273)**4-((F12*(Th+273)**4)-(F1w*(294)**4))))/1000;\t\t\t#Net heat given up by the floor in kW\n", + "\n", + "\n", + "# Results\n", + "print 'Heat exchange between the floor and window is %3.0f W \\n \\\n", + "Net heat given up by the floor is %3.1f kW'%(Q12,Q1)\n", + "\n", + "\n", + "# Note : rounding off error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat exchange between the floor and window is 1283 W \n", + " Net heat given up by the floor is 110.1 kW\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.14 Page No : 424" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "A2 = (6.*2);\t\t\t#Area of windows in m**2\n", + "A1 = (10.*12);\t\t\t#Area of floor in m**2\n", + "Th = 40.;\t\t\t#Radiating heating panel in degree C\n", + "Tb = 5.;\t\t\t#Temperature of black plane in degree C\n", + "F12 = 0.052;\t\t\t#Shape factor\n", + "\n", + "# Calculations\n", + "F12a = ((A2-(A1*F12**2))/(A1+A2-(2*A1*F12)));\t\t\t#Shape factor\n", + "Q12 = (A1*F12a*5.67*10**-8*((Th+273)**4-(Tb+273)**4));\t\t\t#Net heat exchange in W\n", + "X = (((A2/A1)-F12)/(1-F12));\t\t\t#X value for equilibrium temperature\n", + "T = (((Th+273)**4+(X*(Tb+273)**4))/(X+1))**0.25;\t\t\t#Equilibrium temperature in K\n", + "\n", + "# Results\n", + "print 'Net heat exchange is %3.0f W Equilibrium temperature is %3.2f K'%(Q12,T)\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Net heat exchange is 2409 W Equilibrium temperature is 311.57 K\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.15 Page No : 430" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "D = 0.2;\t\t\t #Diameter of each disc in m\n", + "L = 2;\t\t\t #Distance between the plates in m\n", + "T = 800+273,300+273;\t\t\t#Temperatures of the plates in K\n", + "e = [0.3,0.5] ;\t\t\t#Emissivities of plates\n", + "\n", + "# Calculations\n", + "e1 = (e[0]*e[1]);\t\t\t#Equivalent emissivity\n", + "R = (D/L);\t\t\t#Ratio between diameter and distance between the plates\n", + "F = 0.014;\t\t\t#F value from Fig.10.4 from page no. 409\n", + "Q = (e1*(3.14/4)*D**2*F*5.67*10**-8*((T[0]**4-(T[1]**4))));\t\t\t#Radiant heat exchange for the plates in W\n", + "\n", + "# Results\n", + "print 'Radiant heat exchange for the plates is %3.2f W'%(Q)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Radiant heat exchange for the plates is 4.55 W\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.16 Page No : 430" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "e = 0.8;\t\t\t#Emissivity of brick wall\n", + "D1 = [6,4];\t\t\t#Width and Height in m\n", + "L = 0.04;\t\t\t#Distance from the wall in m\n", + "D2 = [0.2,0.2];\t\t\t#Dimensions of the furnace wall in m\n", + "D3 = [1,1];\t\t\t#Dimensions at lower and left of the centre of the wall in m\n", + "T = [1523+273,37+273];\t\t\t#Furnace temperature and wall temperature in degree C\n", + "\n", + "# Calculations\n", + "F12 = 0.033;\t\t\t#Shape factor from Fig.10.3 on page no. 409\n", + "F13 = 0.05;\t\t\t#Shape factor from Fig.10.3 on page no. 409\n", + "F14 = 0.12;\t\t\t#Shape factor from Fig.10.3 on page no. 409\n", + "F15 = 0.08;\t\t\t#Shape factor from Fig.10.3 on page no. 409\n", + "Fow = (F12+F13+F14+F15);\t\t\t#Shape factor between opening and wall\n", + "Q = (e*L*Fow*5.67*10**-8*(T[0]**4-T[1]**4))/1000;\t\t\t#Net radiation exchange in kW\n", + "\n", + "# Results\n", + "print 'Net radiation exchange between the opening and the wall is %3.1f kW'%(Q)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Net radiation exchange between the opening and the wall is 5.3 kW\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.17 Page No : 431" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "D = [2,1,1];\t\t\t#Dimensions of the math.tank in m\n", + "A = 8;\t\t\t#Area of the tank in m**2\n", + "e = 0.9;\t\t\t#Surface emissivity \n", + "Ts = 25+273;\t\t\t#Surface temperature in K\n", + "Ta = 2+273;\t\t\t#Ambient temperature in K\n", + "e1 = 0.5;\t\t\t#Emissivity of aluminium \n", + "\n", + "# Calculations\n", + "Q = (e*A*5.67*10**-8*(Ts**4-Ta**4))/1000;\t\t\t#Heat lost by radiation in kW\n", + "r = ((e-e1)/e)*Q;\t\t\t#Reduction in heat loss if the tank is coated with an aluminium paint in kW\n", + "\n", + "# Results\n", + "print 'Heat lost by radiation is %3.2f kW \\n \\\n", + "Reduction in heat loss if the tank is coated with an aluminium paint is %3.3f kW'%(Q,r)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat lost by radiation is 0.88 kW \n", + " Reduction in heat loss if the tank is coated with an aluminium paint is 0.393 kW\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.18 Page No : 432" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "D = 0.2;\t\t\t#Outer diameter of the pipe in m\n", + "Ta = 30+273;\t\t\t#Temperature of the air in K\n", + "Ts = 400+273;\t\t\t#Surface temperature in K\n", + "e = 0.8;\t\t\t#Emissivity of the pipe surface\n", + "D1 = 0.4;\t\t\t#Diamter of brick in m\n", + "e1 = 0.91;\t\t\t#Emissivity of brick\n", + "\n", + "# Calculations\n", + "Q = (e*3.14*D*5.67*10**-8*(Ts**4-Ta**4))/1000;\t\t\t#Loss of heat by thermal radiation in kW/m\n", + "e2 = (1./((1./e)+((D/D1)*((1./e1)-1))));\t\t\t#Equivalent emissivity\n", + "Q1 = (e2*3.14*D*5.67*10**-8*(Ts**4-Ta**4))/1000;\t\t\t#Heat loss when brick is used in kW/m\n", + "r = (round(Q,2)-round(Q1,2))*1000;\t\t\t#Reduction in heat loss in W/m\n", + "\n", + "# Results\n", + "print 'Loss of heat by thermal radiation is %3.1f*10**3 W/m \\n \\\n", + "Reduction in heat loss is %3.0f W/m'%(Q,r)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Loss of heat by thermal radiation is 5.6*10**3 W/m \n", + " Reduction in heat loss is 210 W/m\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.19 Page No : 433" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "e = 0.03;\t\t\t#Emissivity of silver\n", + "T2 = -153.+273;\t\t\t#Temperature of the outer surface of the inner wall in K\n", + "T1 = 27.+273;\t\t\t#Temperature of the inner surface of the outer wall in K\n", + "D1 = 0.42;\t\t\t#Diamter of first sphere in m\n", + "D2 = 0.6;\t\t\t#Diamter of the second sphere in m\n", + "V = 220.;\t\t\t#Rate of vapourization in kJ/kg\n", + "\n", + "# Calculations\n", + "e1 = (1./((1./e)+((D1/D2)**2*((1./e)-1))));\t\t\t#Equivalent emissivity\n", + "A = (4*3.14*(D1/2)**2);\t\t\t#Area in m**2\n", + "Q = (e1*A*5.67*10**-8*(T1**4-T2**4))/(1000./3600);\t\t\t#Radiation heat transfer through walls into the vessel in kJ/h\n", + "R = (Q/V);\t\t\t#Rate of evaporation in kg/h\n", + "\n", + "# Results\n", + "print 'Radiation heat transfer through walls into the vessel is %3.3f kJ/h \\n \\\n", + "Rate of evaporation of liqiud oxygen is %3.4f kg/h'%(Q,R)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Radiation heat transfer through walls into the vessel is 18.146 kJ/h \n", + " Rate of evaporation of liqiud oxygen is 0.0825 kg/h\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.20 Page No : 433" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "T = [800+273,300+273];\t\t\t#Temperatures of the plates in K\n", + "e = [0.3,0.5];\t\t\t#Emissivities of the plates\n", + "\n", + "# Calculations\n", + "Q = ((5.67*10**-8*(T[0]**4-T[1]**4))/((1./e[0])+((1./e[1]))-1))/1000;\t\t\t#Net radiant heat exchange in kW/m**2\n", + "\n", + "# Results\n", + "print 'Net radiant heat exchange is %3.2f kW/m**2'%(Q)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Net radiant heat exchange is 15.93 kW/m**2\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.21 Page No : 434" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "T1 = 127+273;\t\t\t#Temperature of the outer side of the brick setting in K\n", + "T2 = 50+273;\t\t\t#Temperature of the inside of the steel plate in K\n", + "e1 = 0.6;\t\t\t#Emissivity of steel\n", + "e2 = 0.8;\t\t\t#Emissivity of fireclay\n", + "\n", + "# Calculations\n", + "Q = ((5.67*10**-8*(T1**4-T2**4))/((1./e1)+((1./e2))-1));\t\t\t#Net radiant heat exchange in W/m**2\n", + "\n", + "# Results\n", + "print 'Net radiant heat exchange is %3.0f W/m**2'%(Q)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Net radiant heat exchange is 435 W/m**2\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.22 Page No : 445" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "D = 1;\t\t\t#Dimension of the plate in m\n", + "L = 0.5;\t\t\t#Distance between the plates in m\n", + "Ts = 27+273;\t\t\t#Surface temperature of the walls in K\n", + "T = [900+273,400+273];\t\t\t#Temperature of the plates in K\n", + "e = [0.2,0.5];\t\t\t#Emissivities of the plates \n", + "\n", + "# Calculations\n", + "F12 = 0.415;\t\t\t#From Fig.10.4 on page no.409\n", + "F13 = (1-F12);\t\t\t#Shape factor\n", + "F23 = (1-F12);\t\t\t#Shape factor\n", + "R1 = (1-e[0])/(e[0]*D*D);\t\t\t#Resistance for 1\n", + "R2 = (1-e[1])/(e[1]*D*D);\t\t\t#Resistance for 2\n", + "R3 = 0;\t\t\t #Resistance for 3\n", + "A1F12I = (1./(D*D*F12));\t\t\t#Inverse of the product of area and Shape factor\n", + "A1F13I = (1./(D*D*F13));\t\t\t#Inverse of the product of area and Shape factor\n", + "A2F23I = (1./(D*D*F23));\t\t\t#Inverse of the product of area and Shape factor\n", + "Eb1 = (5.67*10**-8*T[0]**4)/1000;\t\t\t#Emissive power of 1 in kW/m**2\n", + "Eb2 = (5.67*10**-8*T[1]**4)/1000;\t\t\t#Emissive power of 2 in kW/m**2\n", + "Eb3 = (5.67*10**-8*Ts**4);\t\t\t#Emissive power of 3 in W/m**2\n", + "J1 = 25;\t\t\t#Radiosity at node 1 in kW/m**2\n", + "J2 = 11.53;\t\t\t#Radiosity at node 2 in kW/m**2\n", + "J3 = 0.46;\t\t\t#Radiosity at node 3 in kW/m**2\n", + "Q1 = ((Eb1-J1)/R1);\t\t\t#Total heat loss by plate 1 in kW\n", + "Q2 = ((Eb2-J2)/R2);\t\t\t#Total heat loss by plate 2 in kW\n", + "Q3 = ((J1-J3)/(A1F13I))+((J2-J3)/(A2F23I));\t\t\t#Total heat received by the room in kW\n", + "\n", + "# Results\n", + "print 'Total heat loss by plate 1 is %3.1f kW Total heat loss by plate 2 is %3.1f kW \\n \\\n", + "Total heat received by the room is %3.2f kW'%(Q1,Q2,Q3)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Total heat loss by plate 1 is 20.6 kW Total heat loss by plate 2 is 0.1 kW \n", + " Total heat received by the room is 20.83 kW\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.23 Page No : 447" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "T = [800+273,300+273];\t\t\t#Temperatures of the plates in K\n", + "e = [0.3,0.5];\t\t\t#Emissivities of the plates\n", + "e3 = 0.05;\t\t\t#Emissivity of aluminium\n", + "\n", + "# Calculations\n", + "q = ((5.67*10**-8*(T[0]**4-T[1]**4))/((1./e[0])+(1./e[1])-1))/1000;\t\t\t#Heat transfer without the shield in kW/m**2\n", + "R1 = (1-e[0])/e[0];\t\t\t#Resistance in 1\n", + "R2 = (1-e[1])/e[1];\t\t\t#Resistance in 2\n", + "R3 = (1-e3)/e3;\t\t\t#Resistance in 3\n", + "R = (R1+(2*R2)+(2*R3));\t\t\t#Total resismath.tance \n", + "q1 = ((5.67*10**-8*(T[0]**4-T[1]**4))/R)/1000;\t\t\t#Heat transfer with shield in kW/m**2\n", + "r = ((q-q1)*100)/q;\t\t\t#Reduction in heat transfer \n", + "X1 = ((1./e3)+(1./e[1])-1);\t\t\t#X1 for tempearture T3\n", + "X2 = ((1./e[0])+(1./e3)-1);\t\t\t#X1 for tempearture T3\n", + "T3 = (((X1*T[0]**4)+(X2*T[1]**4))/(X2+X1))**0.25;\t\t\t#Temperature of the sheild in K\n", + "T3c = T3-273;\t\t\t#Temperature of the sheild in degree C\n", + "\n", + "# Results\n", + "print 'Percentage reduction in heat transfer is %3.0f percent \\n \\\n", + "Temperature of the sheild is %3.2f degree C'%(r,T3c)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Percentage reduction in heat transfer is 90 percent \n", + " Temperature of the sheild is 641.02 degree C\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.24 Page No : 448" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Q = 79;\t\t\t#Reduction in net radiation from the surfaces \n", + "e1 = 0.05;\t\t\t#Emissivity of the screen\n", + "e2 = 0.8;\t\t\t#Emissivity of the surface\n", + "\n", + "# Calculations\n", + "n = (((Q*((2/e2)-1))-((2/e2)+1))/((2/e1)-1));\t\t\t#Number of screens to be placed between the two surfaces to achieve the reduction in heat exchange\n", + "\n", + "# Results\n", + "print 'Number of screens to be placed between the two surfaces to achieve the reduction in heat exchange is%3.0f'%(n)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of screens to be placed between the two surfaces to achieve the reduction in heat exchange is 3\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.25 Page No : 449" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "e = 0.8;\t\t\t#Emissivity of the pipe\n", + "D = 0.275;\t\t\t#Diameter of the pipe in m\n", + "Ts = 500+273;\t\t\t#Surface temperature in K\n", + "Te = 30+273;\t\t\t#Temperature of enclosure in K\n", + "D1 = 0.325;\t\t\t#Diamter of the steel screen in m\n", + "e1 = 0.7;\t\t\t#Emissivity of steel screen\n", + "Tsc = 240+273;\t\t\t#Temperature of screen in K\n", + "\n", + "#CALCUATIONS\n", + "Q = (e*5.67*10**-8*3.14*D*(Ts**4-Te**4))/1000;\t\t\t#Loss of heat per unit length by radiation in kW/m\n", + "e2 = (1./((1./e)+((D/D1)*((1./e1)-1))));\t\t\t#Equivalent emissivity\n", + "Q1 = (e2*5.67*10**-8*3.14*D*(Ts**4-Tsc**4))/1000;\t\t\t#Radiant heat exchange per unit length of header with screen in kW/m\n", + "R = (Q-Q1);\t\t\t#Reduction in heat by radiation due to the provision of the screen in kW/m\n", + "\n", + "# Results\n", + "print 'Loss of heat per unit length by radiation is %3.1f kW/m \\n \\\n", + "Reduction in heat by radiation due to the provision of the screen is %3.2f kW/m'%(Q,R)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Loss of heat per unit length by radiation is 13.7 kW/m \n", + " Reduction in heat by radiation due to the provision of the screen is 4.92 kW/m\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.26 Page No : 451" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "e = 0.6;\t\t \t#Emissivity of thermocouple\n", + "Ta = 20+273;\t\t\t#Ambient temperature in K\n", + "Tt = 500+273;\t\t\t#Temperature from the thermocouple in K\n", + "e = 0.3;\t \t\t#Emissivity of radiation shield\n", + "h = 200;\t\t \t#Convective heat transfer coefficient in W/m**2.K\n", + "Ts = 833;\t\t\t #Temperature in K\n", + "\n", + "# Calculations\n", + "T = ((5.67*10**-8*e*(Tt**4-Ta**4))/(h*1000))+Tt;\t\t\t#Temperature of the shield in K\n", + "T1 = (Ts-T);\t\t\t#Error between the thermocouple temperature and gas temperature in K\n", + "Ts = 825.; \t\t\t#Surface temperature with radiation shield in K\n", + "Tc = 829.;\t \t\t#Thermocouple temperature with radiation shield in K\n", + "e = (Tc-Ts);\t\t\t#Error between the thermocouple temperature and gas temperature with the shielded thermocouple arrangement in K\n", + "\n", + "# Results\n", + "print 'Error between the thermocouple temperature and gas temperature is%3.0f K \\n \\\n", + "Error between the thermocouple temperature and gas temperature with the shielded thermocouple arrangement is%3.0f K'%(T1,e)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Error between the thermocouple temperature and gas temperature is 60 K \n", + " Error between the thermocouple temperature and gas temperature with the shielded thermocouple arrangement is 4 K\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.27 Page No : 452" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "D = 0.2;\t\t\t#Diameter of pipe in m\n", + "Ta = 30+273;\t\t\t#Temperature of air in K\n", + "Ts = 200+273;\t\t\t#Temperature of surface in K\n", + "e = 0.8;\t\t\t#Emissivity of the pipe\n", + "\n", + "# Calculations\n", + "Q = (e*5.67*10**-8*3.14*D*(Ts**4-Ta**4));\t\t\t#Heat lost by thermal radiation in W/m\n", + "T = (Ta+Ts)/2;\t\t\t#Film temperature in degree C\n", + "k = 0.03306;\t\t\t#Thermal conductivity in W/m.K\n", + "v1 = (24.93*10**-6);\t\t\t#Kinematic viscosity in m**2/s\n", + "b = (1./388);\t\t\t#Coefficient of thermal expansion in 1./K\n", + "Pr = 0.687;\t\t\t#Prantl number\n", + "Ra = ((9.81*b*D**3*(Ts-Ta)*Pr)/(v1**2));\t\t\t#Rayleigh number\n", + "Nu = (0.53*(Ra)**0.25);\t\t\t#Nussults number\n", + "h = (k*Nu)/D;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "Q1 = (h*3.14*D*(Ts-Ta));\t\t\t#Heat lost by convection in W/m\n", + "Q2 = (Q+Q1);\t\t\t#Total heat lost per meter length in W/m\n", + "\n", + "# Results\n", + "print 'Heat lost by thermal radiation is %3.0f W/m \\n \\\n", + "Heat lost by convection is %3.1f W/m'%(Q,Q1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat lost by thermal radiation is 1186 W/m \n", + " Heat lost by convection is 734.4 W/m\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.28 Page No : 453" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Ts = 200+273;\t\t\t#Temperature of stream main in K\n", + "Ta = 30+273;\t\t\t#Rooom temperature in K\n", + "h = 17.98;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "e = 0.8;\t\t\t#Emissivity of the pipe surface\n", + "\n", + "# Calculations\n", + "q = (5.67*10**-8*e*(Ts**4-Ta**4));\t\t\t#Heat transfer by radiation in W/m**2\n", + "hr = (q/(Ts-Ta));\t\t\t #Heat transfer coefficient due to radiation in W/m**2.K\n", + "hc = (h-hr);\t\t\t #Heat transfer coefficient due to convection in W/m**2.K\n", + "\n", + "# Results\n", + "print 'Heat transfer coefficient due to radiation is %3.1f W/m**2.K \\n \\\n", + "Heat transfer coefficient due to convection is %3.2f W/m**2.K'%(hr,hc)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer coefficient due to radiation is 11.1 W/m**2.K \n", + " Heat transfer coefficient due to convection is 6.87 W/m**2.K\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.29 Page No : 461" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# Variables\n", + "t = 0.05;\t\t\t#Thickness of the gas layer in m\n", + "r = 0.1;\t\t\t#Remaining radiation intensity\n", + "\n", + "# Calculations\n", + "a = (-1./t)*2.3*(math.log(r)/math.log(10));\t\t\t#Extinction coefficient per m\n", + "\n", + "# Results\n", + "print 'Extinction coefficient is %3.2f/m'%(a)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Extinction coefficient is 46.00/m\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.30 Page No : 462" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "A = 30.;\t\t\t#Total surface area in m**2\n", + "V = 10.;\t\t\t#Volume in m**3\n", + "Ts = 1000.;\t\t\t#Temperature of the furnace in degree C\n", + "p = 2.; \t\t\t#Total pressure in atm\n", + "ph2o = 0.1;\t\t\t#Partial pressure of water vapour in atm\n", + "pco2 = 0.3;\t\t\t#Partial pressure of CO2\n", + "\n", + "# Calculations\n", + "lms = (3.6*V)/A;\t\t\t#Mean beam length in m\n", + "pco2lms = (pco2*lms);\t\t\t#pco2lms in m.atm\n", + "eco2 = 0.16;\t\t\t#From Fig.10.23 on page no. 458\n", + "cco2 = 1.11;\t\t\t#From Fig.10.23 on page no. 458\n", + "cco2eco2 = (cco2*eco2);\t\t\t#cco2eco2 value\n", + "ph2olms = (ph2o*lms);\t\t\t#ph2olms in m.atm\n", + "eh2o = 0.12;\t\t\t#From Fig.10.24 on page no. 459\n", + "P = (p+ph2o)/2;\t\t\t#P value in atm\n", + "ch2o = 1.43;\t\t\t#From Fig.10.26 on page no. 460\n", + "ch2oeh2o = (ch2o*eh2o);\t\t\t#ch2oeh2o value\n", + "P1 = (ph2o/(ph2o+pco2));\t\t\t#Ratio of pressures\n", + "X = (pco2lms+ph2olms);\t\t\t#X value in m.atm\n", + "e = 0.035;\t\t\t#Error value from Fig. 10.27 on page no.461\n", + "et = (cco2eco2+ch2oeh2o-e);\t\t\t#Total emissivity of the gaseous mixture \n", + "\n", + "# Results\n", + "print 'Emissivity of the gaseous mixture is %3.4f'%(et)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Emissivity of the gaseous mixture is 0.3142\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.31 Page No : 463" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Tg = 950+273;\t\t\t#Flue gas temperature in K\n", + "p = 1;\t\t\t#Total pressure in atm\n", + "pco2 = 0.1;\t\t\t#Percent of co2\n", + "ph2o = 0.04;\t\t\t#Percent of h2o\n", + "D = 0.044;\t\t\t#Diameter of the tube in m\n", + "e = 0.8;\t\t\t#Emissivity of grey surface \n", + "Tw = 500+273.;\t\t\t#Uniform temperature in K\n", + "\n", + "# Calculations\n", + "lms = (3*0.044);\t\t\t#lms value from Table 10.2 on page no. 457\n", + "pco2lms = (pco2*lms);\t\t\t#pco2lms in m.atm\n", + "ph2olms = (ph2o*lms);\t\t\t#ph2olms in m.atm\n", + "eco2 = 0.05;\t\t\t#From Fig.10.23 on page no. 458\n", + "eh2o = 0.005;\t\t\t#From Fig.10.24 on page no. 459\n", + "b = 1.05;\t\t\t#Correction factor from Fig. 10.28 on page no. 461\n", + "eg = 0.061;\t\t\t#Total emissivity of gaseous mixture\n", + "ag = ((0.056*(Tg/Tw)**0.65)+(b*0.021));\t\t\t#Absorbtivity of the gases \n", + "q = (0.5*(e+1)*5.67*10**-8*((eg*Tg**4)-(ag*Tw**4)));\t\t\t#Heat transfer rate by radiation in W/m**2\n", + "hr = (q/(Tg-Tw));\t\t\t#Radiation heat transfer coefficient in W/m**2.degree C\n", + "\n", + "# Results\n", + "print 'Net radiation exchange between the gas and the tube walls is %3.0f W/m**2 \\n \\\n", + "Radiation heat transfer coefficient is %3.2f W/m**2.degree C'%(q,hr)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Net radiation exchange between the gas and the tube walls is 5187 W/m**2 \n", + " Radiation heat transfer coefficient is 11.53 W/m**2.degree C\n" + ] + } + ], + "prompt_number": 6 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch11.ipynb b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch11.ipynb new file mode 100755 index 00000000..b7e9b3c7 --- /dev/null +++ b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch11.ipynb @@ -0,0 +1,486 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a7f87932a4eea8107911ad53ff496ae266946836f956d53145d3e7186e5c88d6" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 11 : Boiling and Condensation" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.1 Page No : 480" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables\n", + "Tsat = 100.;\t\t\t#Saturation temperature of water in degree C\n", + "p1 = 957.9;\t\t\t#Density of liquid in kg/m**3\n", + "Cpl = 4217.;\t\t\t#Specific heat in J/kg.K\n", + "u = (279.*10**-6);\t\t\t#Dynamic viscosity in N.s/m**2\n", + "Pr = 1.76;\t\t\t#Prantl number\n", + "hjg = 2257.;\t\t\t#Enthalpy in kJ/kg\n", + "s = (58.9*10**-3);\t\t\t#Surface tension in N/m\n", + "pv = 0.5955;\t\t\t#Density of vapour in kg/m**3\n", + "m = 30.;\t\t\t#Rate of water in kg/h\n", + "D = 0.3;\t\t\t#Diameter in m\n", + "\n", + "# Calculations\n", + "q = round((m*hjg*1000)/(3600*math.pi*D**2/4));\t\t\t#Heat transfer in W/m**2\n", + "Ts = Tsat+(((q/(u*hjg*1000))*math.sqrt(s/(9.81*(p1-pv)))))**0.33\n", + "\n", + "# Results\n", + "print 'Temperature of the bottom surface of the pan is %3.2f degree C'%(Ts)\n", + "\n", + "# NOte: answer in book is wrong." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temperature of the bottom surface of the pan is 101.02 degree C\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.2 Page No : 481" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Tsat = 100;\t\t\t#Saturation temperature of water in degree C\n", + "p1 = 957.9;\t\t\t#Density of liquid in kg/m**3\n", + "Cpl = 4217;\t\t\t#Specific heat in J/kg.K\n", + "u = (279*10**-6);\t\t\t#Dynamic vismath.cosity in N.s/m**2\n", + "Pr = 1.76;\t\t\t#Prantl number\n", + "hjg = 2257;\t\t\t#Enthalpy in kJ/kg\n", + "s = (58.9*10**-3);\t\t\t#Surface tension in N/m\n", + "pv = 0.5955;\t\t\t#Density of vapour in kg/m**3\n", + "m = 30;\t\t\t#Rate of water in kg/h\n", + "D = 0.3;\t\t\t#Diameter in m\n", + "\n", + "# Calculations\n", + "q = (0.18*hjg*1000*pv*((s*9.81*(p1-pv))/pv**2)**0.25)/10**6;\t\t\t#Burnout heat flux in MW/m**2\n", + "\n", + "# Results\n", + "print 'Burnout heat flux is %3.3f MW/m**2'%(q)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Burnout heat flux is 1.520 MW/m**2\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.3 Page No : 481" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "D = 0.0016;\t\t\t#Diameter of the wire in m\n", + "T = 255;\t\t\t#Temperature difference in degree C\n", + "p1 = 957.9;\t\t\t#Density of liquid in kg/m**3\n", + "Cpl = 4640;\t\t\t#Specific heat in J/kg.K\n", + "u = (18.6*10**-6);\t\t\t#Dynamic viscosity in N.s/m**2\n", + "hjg = 2257;\t\t\t#Enthalpy in kJ/kg\n", + "k = (58.3*10**-3);\t\t\t#Thermal conductivity in W/m.K\n", + "pv = 31.54;\t\t\t#Density of vapour in kg/m**3\n", + "Ts = 628;\t\t\t#Surface temperature in K\n", + "Tsat = 373;\t\t\t#Saturation temperature in K\n", + "\n", + "# Calculations\n", + "hc = (0.62*((k**3*pv*(p1-pv)*9.81*((hjg*1000)+(0.4*Cpl*T)))/(u*D*T))**0.25);\t\t\t#Convective heat transfer coefficient in W/m**2.K\n", + "hr = ((5.67*10**-8)*(Ts**4-Tsat**4))/(Ts-Tsat);\t\t\t#Radiative heat transfer coefficient in W/m**2.K\n", + "hm = (hc+(0.75*hr));\t\t\t#Mean heat transfer coefficient in W/m**2.K\n", + "Q = (hm*3.14*D*T)/1000;\t\t\t#Power dissipation rate per unit length of the heater in kW/m\n", + "\n", + "# Results\n", + "print 'Mean heat transfer coefficient is %3.1f W/m**2.K \\n \\\n", + "Power dissipation rate per unit length of the heater is %3.3f kW/m'%(hm,Q)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mean heat transfer coefficient is 1340.9 W/m**2.K \n", + " Power dissipation rate per unit length of the heater is 1.718 kW/m\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.4 Page No : 485" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Ts = 10.;\t\t\t#Surface temperature in degree C\n", + "p1 = 10.;\t\t\t#Pressure of water in atm\n", + "\n", + "# Calculations\n", + "hp = (5.56*Ts**0.4);\t\t\t#Heat transfer coefficient in kW/m**2.K\n", + "hp1 = (5.56*(2*Ts)**3*p1**0.4);\t\t\t#Heat transfer coefficient in kW/m**2.K\n", + "hp2 = (5.56*Ts**3*(2*p1)**0.4);\t\t\t#Heat transfer coefficient in kW/m**2.K\n", + "x1 = (hp1/hp)/1000;\t\t\t#Ratio of heat transfer coefficients\n", + "x2 = (hp2/hp)*100;\t\t\t#Ratio of heat transfer coefficients\n", + "\n", + "# Results\n", + "print 'Heat transfer coefficient becomes %.f times the original value in the first case\\n \\\n", + "Heat transfer coefficient is increased only by 32 percent in the second case'%(x1)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer coefficient becomes 8 times the original value in the first case\n", + " Heat transfer coefficient is increased only by 32 percent in the second case\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.5 Page No : 485" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "p = 6.;\t\t\t#Pressure of water in atm\n", + "D = 0.02;\t\t\t#Diameter of the tube in m\n", + "Ts = 10.;\t\t\t#Wall temperature in degree C\n", + "L = 1.;\t\t\t#Length of the tube in m\n", + "\n", + "# Calculations\n", + "p1 = (p*1.0132*10**5)/10**6;\t\t\t#Pressure in MN/m**2\n", + "h = (2.54*Ts**3*math.exp(p1/1.551));\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "Q = (h*math.pi*D*L*Ts);\t\t\t#Heat transfer rate in W/m\n", + "\n", + "\n", + "# Results\n", + "print 'Heat transfer rate is %3.1f W/m'%(Q)\n", + "\n", + "\n", + "# note : rounding off error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer rate is 2361.8 W/m\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.6 Page No : 489" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "p = 2.45;\t\t\t#Pressure of dry saturated steam in bar\n", + "h = 1;\t\t\t#Height of vertical tube in m\n", + "Ts = 117;\t\t\t#Tube surface temperature in degree C\n", + "d = 0.2;\t\t\t#Distance from upper end of the tube in m\n", + "\n", + "# Calculations\n", + "Tsat = 127;\t\t\t#Saturation temperature of water in degree C\n", + "p1 = 941.6;\t\t\t#Density of liquid in kg/m**3\n", + "k1 = 0.687;\t\t\t#Thermal conductivity in W/m.K\n", + "u = (227*10**-6);\t\t\t#Dynamic vismath.cosity in N.s/m**2\n", + "hfg = 2183;\t\t\t#Enthalpy in kJ/kg\n", + "pv = 1.368;\t\t\t#Density of vapour in kg/m**3\n", + "q = round((((4*k1*u*10*d)/(9.81*p1*(p1-pv)*hfg*1000))**0.25)*1000,2);\t\t\t#Thickness of condensate film in mm\n", + "h = (k1/(q/1000));\t\t\t#Local heat transfer coefficient at x = 0.2 in W/m**2.K\n", + "\n", + "# Results\n", + "print 'Thickness of condensate film is %3.2f mm \\n \\\n", + "Local heat transfer coefficient at x = 0.2 is %3.0f W/m**2.K'%(q,h)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thickness of condensate film is 0.09 mm \n", + " Local heat transfer coefficient at x = 0.2 is 7633 W/m**2.K\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.7 Page No : 491" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "D = 0.05;\t\t\t#Diameter of the tube in m\n", + "L = 2;\t\t\t#Length of the tube in m\n", + "Ts = 84;\t\t\t#Outer surface temperature in degree C\n", + "Tsat = 100;\t\t\t#Saturation temperature of water in degree C\n", + "Tf = (Tsat+Ts)/2;\t\t\t#Film temperature in degree C\n", + "p1 = 963.4;\t\t\t#Density of liquid in kg/m**3\n", + "u = (306*10**-6);\t\t\t#Dynamic vismath.cosity in N.s/m**2\n", + "hfg = 2257;\t\t\t#Enthalpy in kJ/kg\n", + "pv = 0.596;\t\t\t#Density of vapour in kg/m**3\n", + "k1 = 0.677;\t\t\t#Thermal conductivity in W/m.K\n", + "\n", + "# Calculations\n", + "hL = (1.13*((9.81*p1*(p1-pv)*k1**3*hfg*1000)/(u*16*L))**0.25);\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "Ref = ((4*hL*L*2)/(hfg*1000*u));\t\t\t#Reynolds nmber\n", + "Q = (hL*math.pi*D*L*10);\t\t\t#Heat transfer rate in W\n", + "m = (Q/(hfg*1000))*3600;\t\t\t#Condensate mass flow rate in kg/h\n", + "\n", + "# Results\n", + "print 'Heat transfer rate is %3.0f W \\n \\\n", + "Condensate mass flow rate is %3.1f kg/h'%(Q,m)\n", + "\n", + "# note : rounding off error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer rate is 17930 W \n", + " Condensate mass flow rate is 28.6 kg/h\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.8 Page No : 492" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "h = 2.8;\t\t\t#Height of the plate in m\n", + "T = 54.;\t\t\t#Temperature of the plate in degree C\n", + "Tsat = 100.;\t\t\t#Saturation temperature of water in degree C\n", + "Tf = (Tsat+T)/2;\t\t\t#Film temperature in degree C\n", + "p1 = 973.7;\t\t\t#Density of liquid in kg/m**3\n", + "u = (365.*10**-6);\t\t\t#Dynamic viscosity in N.s/m**2\n", + "hfg = 2257.;\t\t\t#Enthalpy in kJ/kg\n", + "pv = 0.596;\t\t\t#Density of vapour in kg/m**3\n", + "k1 = 0.668;\t\t\t#Thermal conductivity in W/m.K\n", + "\n", + "# Calculations\n", + "Re = (0.00296*((p1*9.81*(p1-pv)*k1**3*(Tsat-T)**3*h**3)/(u**5*(hfg*1000)**3))**(5./9));\t\t\t#Reynolds number\n", + "hL = (0.0077*((9.81*p1*(p1-pv)*k1**3)/u**2)**(1./3)*Re**0.4);\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "Q = (hL*h*(Tsat-T))/1000;\t\t\t#Heat transfer rate per unit width in kW/m\n", + "\n", + "# Results\n", + "print 'Heat transfer rate per unit width is %3.2f kW/m'%(Q)\n", + "\n", + "\n", + "# note : rounding error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer rate per unit width is 702.33 kW/m\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.9 Page No : 494" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "T = 100;\t\t\t#Temperature of dry steam in degree C\n", + "Do = 0.025;\t\t\t#Outer diameter of the pipe in m\n", + "Ts = 84;\t\t\t#Surface temmperature of pipe in degree C\n", + "Tf = (T+Ts)/2;\t\t#Film temperature in degree C\n", + "p1 = 963.4;\t\t\t#Density of liquid in kg/m**3\n", + "u = (306*10**-6);\t#Dynamic viscosity in N.s/m**2\n", + "hfg = 2257;\t\t\t#Enthalpy in kJ/kg\n", + "pv = 0.596;\t\t\t#Density of vapour in kg/m**3\n", + "k1 = 0.677;\t\t\t#Thermal conductivity in W/m.K\n", + "\n", + "# Calculations\n", + "h = (0.725*((9.81*p1*(p1-pv)*k1**3*hfg*1000)/(u*(T-Ts)*Do))**0.25);\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "q = (h*3.14*Do*(T-Ts))/1000;\t\t\t#Heat transfer per unit length in kW/m\n", + "m = (q/hfg)*3600;\t\t\t #Total mass flow of condensate per unit length in kg/h\n", + "\n", + "# Results\n", + "print 'Rate of formation of condensate per unit length is %3.2f kg/h'%(m)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rate of formation of condensate per unit length is 21.94 kg/h\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.10 Page No : 494" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "m = 50;\t\t\t#Mass of vapour per hour\n", + "n = 100;\t\t\t#Number of tubes\n", + "D = 0.01;\t\t\t#Diameter of the tube in m\n", + "L = 1;\t\t\t#Length of the tube in m\n", + "n1 = 10;\t\t\t#Array of 10*10\n", + "\n", + "# Calculations\n", + "mr = ((0.725/1.13)*(L/(n1*D))**0.25);\t\t\t#Ratio of horizontal and vertical position \n", + "mv = (m/mr);\t\t\t#Mass flow rate in the vertical position in kg/h\n", + "\n", + "# Results\n", + "print 'Mass flow rate in the vertical position is %3.2f kg/h'%(mv)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mass flow rate in the vertical position is 43.82 kg/h\n" + ] + } + ], + "prompt_number": 14 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch12.ipynb b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch12.ipynb new file mode 100755 index 00000000..eb722971 --- /dev/null +++ b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch12.ipynb @@ -0,0 +1,923 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:0803cdabadcb852bdbb9c79bc32fa04b5ddfc37dadc8bca5c47a13f22864faf2" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 12 : Heat Exchangers" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.1 Page No : 503" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#INPUT\n", + "T = 80.;\t\t \t#Bulk Temperature of water in degrees C\n", + "Di = 0.0254;\t\t\t#Inner diameter of steel pipe in m\n", + "Do = 0.0288;\t\t\t#Outer diameter of steel pipe in m\n", + "k = 50.;\t\t\t #Thermal conductivity of steel in W/m.K\n", + "ho = 30800.;\t\t \t#Average convection coefficient in W/m**2.K\n", + "v = 0.50;\t\t \t#Velocity of water in m/s\n", + "# Variables FROM HEAT AND MASS TRANSFER DATA BOOK FOR WATER AT BULK TEMPERATURE OF 80 degree C\n", + "d = 974.;\t\t \t#Density in kg/m**3\n", + "v1 = 0.000000364;\t\t#Kinematic viscosity in m**2/s\n", + "k1 = 0.6687;\t\t\t#Thermal conductivity in W/m.K\n", + "Pr = 2.2;\t\t\t #Prantl Number\n", + "\n", + "# Calculations\n", + "Re = (v*Di)/v1;\t\t\t#Reynold's number\n", + "Nu = (0.023*Re**0.8*Pr**0.4);\t\t\t#Nusselts number\n", + "hi = Nu*(k1/Di);\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "ri = (Di/2);\t\t\t#Inner radius of steel pipe in m\n", + "ro = (Do/2);\t\t\t#Outer radius of steel pipe in m\n", + "U = (1./((1./ho)+((ro/k)*math.log(ro/ri))+(ro/(ri*hi))));\t\t\t#Overall heat transfer coefficient in W/m**2.K\n", + "\n", + "# Results\n", + "print 'Overall heat transfer coefficient is %3.1f W/m**2.K'%(U)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Overall heat transfer coefficient is 2591.8 W/m**2.K\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.2 Page No : 504" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# Variables\n", + "Do = 0.0254;\t\t\t#Outer Diameter of heat exchanger tube in m\n", + "Di = 0.02286;\t\t\t#Inner Diameter of heat exchanger tube in m\n", + "k = 102;\t\t\t#Thermal conductivity of the tube in W/m.K\n", + "hi = 5500;\t\t\t#Heat transfer coefficients at the inner side of tube in W/m**2.K\n", + "ho = 3800;\t\t\t#Heat transfer coefficients at the outer side of tube in W/m**2.K\n", + "Rfi = 0.0002;\t\t\t#Fouling factor in m**2.W.K\n", + "Rfo = 0.0002;\t\t\t#Fouling factor in m**2.W.K\n", + "\n", + "# Calculations\n", + "ro = (Do/2);\t\t\t#Outer radius of heat exchanger tube in m\n", + "ri = (Di/2);\t\t\t#Inner radius of heat exchanger tube in m\n", + "U = (1./((1./ho)+Rfo+((ro/k)*math.log(ro/ri))+((ro*Rfi)/ri)+(ro/(ri*hi))));\t\t\t#Overall heat transfer coefficient in W/m**2.K\n", + "\n", + "# Results\n", + "print 'Overall heat transfer coefficient is %i W/m**2.K'%(U)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Overall heat transfer coefficient is 1110 W/m**2.K\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.3 Page No : 509" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# Variables\n", + "mh = 10000.;\t\t\t#Mass flow rate of oil in kg/h\n", + "ch = 2095.;\t\t\t#Specific heat of oil J/kg.K\n", + "Thi = 80.;\t\t\t#Inlet temperature of oil in degree C\n", + "Tho = 50.;\t\t\t#Outlet temperature of oil in degree C\n", + "mc = 8000.;\t\t\t#Mass flow rate of water in kg/h\n", + "Tci = 25.;\t\t\t#Inlet temperature of water in degree C\n", + "U = 300.;\t\t\t#Overall heat ransfer coefficient in W/m**2.K\n", + "cc = 4180.;\t\t\t#Specific heat of water in J/kg.K\n", + "\n", + "# Calculations\n", + "Q = (mh*ch*(Thi-Tho));\t\t\t#Heat transfer rate in W\n", + "Tco = ((Q/(mc*cc))+Tci);\t\t\t#Outlet temperature of water in degree C\n", + "T = (Thi-Tco);\t\t\t#Temperature difference between oil inlet temperature and water outlet temperature in degree C\n", + "t = (Tho-Tci);\t\t\t#Temperature difference between oil outlet temperature and water inlet temperature in degree C\n", + "A = (((Q/U)*math.log(t/T))/(3600*(t-T)));\t\t\t#Area of heat exchanger in m**2\n", + "\n", + "# Results\n", + "print 'Area of heat exchanger is %3.2f m**2'%(A)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Area of heat exchanger is 19.23 m**2\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.4 Page No : 510" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# Variables\n", + "Ch = 2500.;\t\t\t#Capacity rate of hot oil in W/K\n", + "Thi = 360.;\t\t\t#Temperature of hot fluid at the entrance of heat exchanger in degree C\n", + "Tho = 300.;\t\t\t#Temperature of hot fluid at the exit of heat exchanger in degree C\n", + "Tci = 30.;\t\t\t#Temperature of cold fluid at the entrance of heat exchanger in degree C\n", + "Tco = 200.;\t\t\t#Temperature of hot fluid at the exit of heat exchanger in degree C\n", + "U = 800.;\t\t\t#Overall heat transfer coefficient in W/m**2.K\n", + "\n", + "# Calculations\n", + "Q = (Ch*(Thi-Tho));\t\t\t#Heat transfer from the oil in W\n", + "#Parallel flow\n", + "T1 = Thi-Tci;\t\t\t#Temperature difference between hot fluid inlet temperature and cold fluid inlet temperature in degree C\n", + "T2 = Tho-Tco;\t\t\t#Temperature difference between hot fluid outlet temperature and cold fluid outlet temperature in degree C\n", + "Tlm1 = ((T1-T2)/math.log(T1/T2));\t\t\t#LMTD for parallel flow arrangement in degree C\n", + "A1 = (Q/(U*Tlm1));\t\t\t#Area of heat exchanger in m**2\n", + "#Counter flow\n", + "t1 = Thi-Tco;\t\t\t#Temperature difference between hot fluid inlet temperature and cold fluid outlet temperature in degree C\n", + "t2 = Tho-Tci;\t\t\t#Temperature difference between hot fluid outlet temperature and cold fluid inlet temperature in degree C\n", + "Tlm2 = ((t1-t2)/math.log(t1/t2));\t\t\t#LMTD for counter flow arrangement in degree C\n", + "A2 = (Q/(U*Tlm2));\t\t\t#Area of heat exchanger in m**2\n", + "\n", + "# Results\n", + "print 'Area of heat exchanger in parallel flow arrangement is %3.3f m**2 \\n \\\n", + "Area of heat exchanger in counter flow arrangement is %3.3f m**2'%(A1,A2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Area of heat exchanger in parallel flow arrangement is 0.973 m**2 \n", + " Area of heat exchanger in counter flow arrangement is 0.892 m**2\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.5 Page No : 511" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# Variables\n", + "ch = 2130.;\t\t\t#Specific heat of oil in J/kg.K\n", + "T1 = 160.;\t\t\t#Temperature of hot fluid (oil) at the entrance of heat exchanger in degree C\n", + "T2 = 60.;\t\t\t#Temperature of hot fluid (oil) at the exit of heat exchanger in degree C\n", + "t1 = 25.;\t\t\t#Temperature of cold fluid (water) at the entrance of heat exchanger in degree C\n", + "d = 0.5;\t\t\t#Inner diameter of the tube in m\n", + "mc = 2.;\t\t\t#Mass flow rate of cooling water in kg/s\n", + "D = 0.7;\t\t\t#outer annulus outer diameter in m\n", + "mh = 2.;\t\t\t#Mass flow rate of hot oil in kg/s\n", + "U = 250.;\t\t\t#Overall heat transfer coefficient in W/m**2.K\n", + "cc = 4186.;\t\t\t#Specific heat of water in J/kg.K\n", + "\n", + "# Calculations\n", + "Q = (mh*ch*(T1-T2));\t\t\t# Required heat transfer rate in W\n", + "t2 = ((Q/(mc*cc))+t1);\t\t\t#Outer water temperature in degree C\n", + "T = T1-t2; \t \t\t #Change in temperature between inlet tmperature of hot fluid and outlet temperature of cold fluid in degree C\n", + "t = T2-t1;\t \t \t#Change in temperature between outlet tmperature of hot fluid and inlet temperature of cold fluid in degree C\n", + "Tlm = ((T-t)/(math.log(T/t)));\t#Value of LMTD in degree C\n", + "L = (Q/(U*3.14*d*Tlm));\t\t\t#Length of the heat exchanger in m\n", + "\n", + "# Results\n", + "print 'Length of the heat exchanger is %3.2f m'%(L)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Length of the heat exchanger is 19.38 m\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.6 Page No : 512" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "T = 120.;\t\t\t#Saturated steam temperature in degree C\n", + "U = 1800.;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "m = 1000.;\t\t\t#mass flow rate of water in kg/h\n", + "t1 = 20.;\t\t\t#Inlet temperature of water in degree C\n", + "t2 = 90.;\t\t\t#Outlet tmperature of water in degree C\n", + "hfg = 2200.;\t\t#Enthalpy of steam in kJ/kg\n", + "c = 4186.;\t\t\t#Specific het of water in J/kg.K\n", + "\n", + "# Calculations\n", + "Tlm = (((T-t1)-(T-t2))/(math.log((T-t1)/(T-t2))));\t\t\t#LMTD in a condenser in degree C\n", + "Q = ((m/3600)*c*(t2-t1));\t \t\t#Rate of heat transfer in W\n", + "A = (Q/(U*Tlm));\t\t \t#Surface area of heat exchanger in m**2 \n", + "ms = ((Q*3600)/(hfg*1000));\t\t \t#Rate of condensation of steam in kg/h\n", + "\n", + "# Results\n", + "print 'Surface area of heat exchanger is %3.2f m**2 \\n \\\n", + "Rate of condensation of steam is %3.1f kg/h'%(A,ms)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Surface area of heat exchanger is 0.78 m**2 \n", + " Rate of condensation of steam is 133.2 kg/h\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.7 Page No : 516" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# Variables\n", + "T = 100.;\t\t\t#Temperature of saturated steam in degree C\n", + "t1 = 30.;\t\t\t#Inlet temperature of water in degree C\n", + "t2 = 70.;\t\t\t#Exit temperature of water in degree C\n", + "\n", + "# Calculations\n", + "#COUNTER FLOW\n", + "Tc = (T-t2);\t\t\t#Temperature difference between saturated steam and exit water temperature in degree C\n", + "tc = (T-t1);\t\t\t#Temperature difference between saturated steam and inlet water temperature in degree C\n", + "Tlmc = ((Tc-tc)/math.log(Tc/tc));\t\t\t#LMTD for counter flow in degree C\n", + "\n", + "#PARALLEL FLOW\n", + "Tp = (T-t1);\t\t\t#Temperature difference between saturated steam and inlet water temperature in degree C\n", + "tp = (T-t2);\t\t\t#Temperature difference between saturated steam and exit water temperature in degree C\n", + "Tlmp = ((Tp-tp)/math.log(Tp/tp));\t\t\t#LMTD for counter flow in degree C\n", + "#CROSS FLOW\n", + "R = ((T-T)/(t2-t1));\t\t\t#R value for Correction factor F\n", + "P = ((t2-t1)/(T-t1));\t\t\t#P value for Correction Factor F\n", + "F = 1;\t\t\t#Referring to Fig.12.12 in page no 515\n", + "Tlmx = (F*Tlmc);\t\t\t#LMTD for cross flow in degree C\n", + "\n", + "# Results\n", + "print 'The effective math.log mean temperature difference for: \\\n", + "\\ni)COUNTER FLOW is %3.1f degree C \\\n", + "\\nii)PARALLEL FLOW is %3.1f degree C \\niii)CROSS FLOW is %3.1f degree C'%(Tlmc,Tlmp,Tlmx)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The effective math.log mean temperature difference for: \n", + "i)COUNTER FLOW is 47.2 degree C \n", + "ii)PARALLEL FLOW is 47.2 degree C \n", + "iii)CROSS FLOW is 47.2 degree C\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.8 Page No : 516" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Ti = 18;\t\t\t#Inlet temperature of Shell fluid in degree C\n", + "To = 6.5;\t\t\t#Outlet temperature of Shell fluid in degree C\n", + "ti = -1.1;\t\t\t#Inlet temperature of Tube fluid in degree C\n", + "to = 2.9;\t\t\t#Outlet temperature of Tube fluid in degree C\n", + "U = 850;\t\t\t#Overall heat transfer coefficient in W/m**2.K\n", + "Q = 6000;\t\t\t#Design heat load in W\n", + "\n", + "# Calculations\n", + "T = (Ti-to);\t\t\t#Temperature difference between shell side inlet fluid and tube side outlet fluid in degree C\n", + "t = (To-ti);\t\t\t#Temperature difference between shell side outlet fluid and tube side inlet fluid in degree C\n", + "Tlm = ((T-t)/math.log(T/t));\t\t\t#LMTD for a counterflow arrangement in degree C\n", + "P = ((to-ti)/(Ti-ti));\t\t\t#P value to calculate correction factor\n", + "R = ((Ti-To)/(to-ti));\t\t\t#R value to calculate correction factor\n", + "F = 0.97\t\t\t#Taking correction factor from fig. 12.9 on page no.514\n", + "A = (Q/(U*F*Tlm));\t\t\t#Area of shell aand tube heat exchanger in m**2\n", + "\n", + "# Results\n", + "print 'Area of shell-and-tube heat exchanger is %3.2f m**2'%(A)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Area of shell-and-tube heat exchanger is 0.67 m**2\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.9 Page No : 517" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Q = 6000;\t\t\t#Taking design heat load value in W from Example no. 12.8 on page no.516\n", + "U = 850;\t\t\t#Taking overall heat transfer coefficient value in W/m**2.K from Example no. 12.8 on page no.516\n", + "Tlm = 10.92\t\t\t#Taking LMTD for a counterflow arrangement in degree C from Example no. 12.8 on page no.517\n", + "R = 2.875;\t\t\t#Taking R value from Example no. 12.8 on page no.517\n", + "P = 0.209;\t\t\t#Taking P value from Example no. 12.8 on page no.517\n", + "F = 0.985;\t\t\t#Taking correction factor from Fig. 12.10 on page no.514\n", + "\n", + "# Calculations\n", + "A = (Q/(U*F*Tlm));\t\t\t#Area of shell-and-tube heat exchanger in m**2\n", + "\n", + "# Results\n", + "print 'Area of shell aand tube heat exchanger is %3.3f m**2'%(A)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Area of shell aand tube heat exchanger is 0.656 m**2\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.10 Page No : 517" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Ti = 360.;\t\t\t#Inlet temperature of hot fluid in degree C taken from Example no. 12.4 on page no. 510\n", + "To = 300.;\t\t\t#Outlet temperature of hot fluid in degree C taken from Example no. 12.4 on page no. 510\n", + "ti = 30.;\t\t\t#Inlet temperature of cold fluid in degree C taken from Example no. 12.4 on page no. 510\n", + "to = 200.;\t\t\t#Outlet temperature of cold fluid in degree C taken from Example no. 12.4 on page no. 510\n", + "U = 800.;\t\t\t#Overall heat transfer coefficient in W/m**2.K taken from Example no. 12.4 on page no. 510\n", + "Q = 150000.;\t\t\t#Calculated heat transfer rate in W from Example no. 12.4 on page no. 510 \n", + "Tlm = 210.22\t\t\t#Calculated LMTD for counterflow arrangement in degree C taken from Example no. 12.4 on page no. 511\n", + "\n", + "# Calculations\n", + "P = ((to-ti)/(Ti-ti));\t\t\t#P value for calculation of correction factor\n", + "R = ((Ti-To)/(to-ti));\t\t\t#R value for calculation of correction factor\n", + "F = 0.98;\t\t\t#Correction Factor value taken from Fig.12.11 on page no.515\n", + "A = (Q/(U*F*Tlm));\t\t\t#Required surface area in a cross flow heat exchanger in m**2\n", + "\n", + "# Results\n", + "print 'The required surface area in a cross flow heat exchanger is %3.2f m**2'%(A)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The required surface area in a cross flow heat exchanger is 0.91 m**2\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.11 Page No : 518" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# Variables\n", + "mc = 4.;\t\t\t#Mass flow rate of cold water in kg/s\n", + "Tci = 38.;\t\t\t#Inlet Temperature of cold water in degree C\n", + "Tco = 55.;\t\t\t#Outlet Temperature of cold water in degree C\n", + "D = 0.02;\t\t\t#Diameter of the tube in m\n", + "v = 0.35;\t\t\t#Velocity of water in m/s\n", + "Thi = 95.;\t\t\t#Inlet Temperature of hot water in degree C\n", + "mh = 2.;\t\t\t#Mass flow rate of hot water in kg/s\n", + "L = 2.;\t\t\t #Length of the tube in m\n", + "U = 1500.;\t\t\t#Overall heat transfer coefficient in W/m**2.K\n", + "c = 4186.;\t\t\t#Specific heat of water in J/kg.K\n", + "d = 1000.;\t\t\t#Density of water in kg/m**3\n", + "\n", + "# Calculations\n", + "Q = (mc*c*(Tco-Tci));\t\t\t#Heat transfer rate for cold fluid in W\n", + "Tho = (Thi-(Q/(mh*c)));\t\t\t#Outlet temperature of hot fluid in degree C\n", + "T = Thi-Tco;\t\t\t#Difference of temperature between hot water inlet and cold water outlet in degree C\n", + "t = Tho-Tci;\t\t\t#Difference of temperature between hot water outlet and cold water inlet in degree C\n", + "Tlm = ((T-t)/math.log(T/t));\t\t\t#LMTD for counterflow heat exchanger\n", + "A = (Q/(U*Tlm));\t\t\t#Area of heat exchanger in m**2\n", + "A1 = (mc/(d*v));\t\t\t#Total flow area in m**2\n", + "n = ((A1*4)/(3.14*D**2));\t\t\t#Number of tubes\n", + "L = (A/(36*3.14*D));\t\t\t#Length of each tube taking n = 36 in m\n", + "N = 2;\t\t\t#Since this length is greater than the permitted length of 2m, we must use more than one tube pass. Let us try 2 tube passes\n", + "P = ((Tco-Tci)/(Thi-Tci));\t\t\t#P value for calculation of correction factor\n", + "R = ((Thi-Tho)/(Tco-Tci));\t\t\t#R value for calculation of correction factor\n", + "F = 0.9;\t\t\t#Corrction Factor from Fig.12.9 on page no. 514 \n", + "A2 = (Q/(U*F*Tlm));\t\t\t#Total area required for one shall pass,2 tube pass exchanger in m**2\n", + "L1 = (A2/(2*36*3.14*D));\t\t\t#Length of tube per pass taking n = 36 in m\n", + "\n", + "# Results\n", + "print 'Number of tubes per pass is %.f \\n \\\n", + "Number of passes is %i \\n \\\n", + "Length of tube per pass is %3.3f m'%(n,N,L1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of tubes per pass is 36 \n", + " Number of passes is 2 \n", + " Length of tube per pass is 1.518 m\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.12 Page No : 524" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# Variables\n", + "mh = 250.;\t\t\t#Mass flow rate of hot liquid in kg/h\n", + "ch = 3350.;\t\t\t#Specific heat of hot liquid in J/kg.K\n", + "Thi = 120.;\t\t\t#Inlet temperature of hot liquid in degree C\n", + "mc = 1000.;\t\t\t#Mass flow rate of cold liquid in kg/h\n", + "Tci = 10.;\t\t\t#Inlet temperature of cold liquid in degree C\n", + "U = 1160.;\t\t\t#Overall heat transfer coefficient in W/m**2.K\n", + "A = 0.25;\t\t\t#Surface area of heat exchanger in m**2\n", + "cc = 4186.;\t\t\t#Specific heat of cold liquid in J/kg.K\n", + "\n", + "# Calculations\n", + "Cc = ((mc*cc)/3600);\t\t\t#Heat capacity rate for cold liquid in W/K\n", + "Ch = ((mh*ch)/3600);\t\t\t#Heat capacity rate for hot liquid in W/K\n", + "Cmin = min(Cc,Ch);\t\t\t#Minimum heat capacity rate in W/K\n", + "Cmax = max(Cc,Ch);\t\t\t#Maximum heat capacity rate in W/K\n", + "r = (Cmin/Cmax);\t\t\t#Ratio of min amd max heat capacity rates\n", + "NTU = ((U*A)/Cmin);\t\t\t#Number of transfer units\n", + "e = ((1-math.exp(-NTU*(1+r)))/(1+r));\t\t\t#Effectiveness for a parallel flow heat exchanger\n", + "Qmax = (Cmin*(Thi-Tci));\t\t\t#Maximum possible heat transfer rate in W\n", + "Q = (e*Qmax);\t\t\t#Actual rate of heat transfer in W\n", + "Tco = ((Q/Cc)+Tci);\t\t\t#Outlet temperature of cold liquid in degree C\n", + "Tho = (Thi-(Q/Ch));\t\t\t#Outlet temperature of hot liquid in degree C\n", + "\n", + "\n", + "# Results\n", + "print 'Effectiveness for a parallel flow heat exchanger is %3.3f \\n \\\n", + "Outlet temperature of water is %3.2f degree C \\n \\\n", + "Outlet temperature of cooled liquid is %3.2f degree C'%(e,Tco,Tho)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Effectiveness for a parallel flow heat exchanger is 0.647 \n", + " Outlet temperature of water is 24.23 degree C \n", + " Outlet temperature of cooled liquid is 48.87 degree C\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.13 Page No : 527" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Tci = 15.;\t\t\t#Inlet temperature of water in degree C\n", + "mc = 1300.;\t\t\t#Mass flow rate of water in kg/h\n", + "ch = 2000.;\t\t\t#Specific heat of oil in J/kg.K\n", + "mh = 550.;\t\t\t#Mass flow rate of oil in kg/h\n", + "Thi = 94.;\t\t\t#Inlet temperature of oil in degree C\n", + "A = 1.;\t\t\t #Area of heat exchanger in m**2\n", + "U = 1075.;\t\t\t#Overall heat transfer coefficient in W/m**2.K\n", + "cc = 4186.;\t\t\t#Specific heat of water in J/kg.K\n", + "\n", + "# Calculations\n", + "Cc = ((mc*cc)/3600);\t\t\t#Heat capacity of water in W/K\n", + "Ch = ((mh*ch)/3600);\t\t\t#Heat capacity of oil in W/K\n", + "Cmin = min(Cc,Ch);\t\t\t#Minimum heat capacity in W/K\n", + "Cmax = max(Cc,Ch);\t\t\t#Maximum heat capacity in W/K\n", + "r = (Cmin/Cmax);\t\t\t#Ratio of min and max heat capacity\n", + "NTU = ((U*A)/Cmin);\t\t\t#Number of transfer Units\n", + "e = 0.94\t\t\t#Effectiveness of heat exchanger from Fig. 12.15 on page no.524\n", + "Qmax = (Cmin*(Thi-Tci));\t\t\t#Maximum possible heat transfer rate in W\n", + "Q = (e*Qmax);\t\t\t#Actual heat transfer rate in W\n", + "Tco = ((Q/Cc)+Tci);\t\t\t#Outlet Temperature of water in degree C\n", + "Tho = (Thi-(Q/Ch));\t\t\t#Outlet Temperature of oil in degree C\n", + "\n", + "# Results\n", + "print 'The total heat transfer is %3.1f W \\n \\\n", + "Outlet Temperature of water is %i degree C \\n \\\n", + "Outlet Temperature of oil is %3.2f degree C'%(Q,Tco,Tho)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The total heat transfer is 22690.6 W \n", + " Outlet Temperature of water is 30 degree C \n", + " Outlet Temperature of oil is 19.74 degree C\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.14 Page No : 528" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "N = 3000.;\t\t\t#Number of brass tubes\n", + "D = 0.02;\t\t\t#Diameter of brass tube in m\n", + "Tci = 20.;\t\t\t#Inlet temperature of cooling water in degree C\n", + "mc = 3000.;\t\t\t#Mass flow rate of cooling water in kg/s\n", + "ho = 15500.;\t\t\t#Heat transfer coefficient for condensation in W/m**2.K\n", + "Q = (2.3*10**8);\t\t\t#Heat load of the condenser in W\n", + "Thi = 50.;\t\t\t#Temperature at which steam condenses in degree C\n", + "hfg = 2380.\t\t\t#Enthalpy of liquid vapour mixture in kJ/kg\n", + "m = 1.;\t\t\t#Flow rate of each tube in kg/s\n", + "Cc = 4180.;\t\t\t#Specific heat of water in J/kg.K\n", + "#Properties of water at 300K from data book\n", + "Cc = 4186.;\t\t\t#Specific heat in J/kg.K\n", + "mu = (855.*10**-6);\t\t\t#Dynamic vismath.cosity in Ns/m**2\n", + "k = 0.613;\t\t\t#Thermal Conductivity in W/mK\n", + "Pr = 5.83\t\t\t#Prantl number\n", + "\n", + "# Calculations\n", + "Tco = ((Q/(mc*Cc))+Tci);\t\t\t#Outlet temperature of cooling water in degree C\n", + "Re = ((4*m)/(3.1415*D*mu));\t\t\t#Reynold's number\n", + "Nu = (0.023*Re**(4./5)*Pr**(2./5));\t\t\t#Nusselts number\n", + "hi = (Nu*(k/D));\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "U = (1./((1./ho)+(1./hi)));\t\t\t#Overall heat transfer coefficient in W/m**2.K\n", + "Cmin = (mc*Cc);\t\t\t#Minimum heat capacity in W/K\n", + "Qmax = (Cmin*(Thi-Tci));\t\t\t#Maximum heat transfer rate in W\n", + "e = (Q/Qmax);\t\t\t#Effectiveness of heat transfer\n", + "NTU = 0.8;\t\t\t#Number of transfer units from Fig. 12.16 on page no.525 \n", + "A = ((NTU*Cmin)/U);\t\t\t#Area of heat exchanger in m**2\n", + "L = (A/(2*N*3.1415*D));\t\t\t#Length of tube per pass in m\n", + "ms = (Q/(hfg*1000));\t\t\t#Amount of steam condensed in kg/s\n", + "\n", + "# Results\n", + "print 'The outlet temperature of the cooling water is %3.2f degree C \\n \\\n", + "The overall heat transfer coefficient is %3.1f W/m**2.K \\n \\\n", + "Tube length per pass using NTU method is %3.2f m \\n \\\n", + "The rate of condensation of steam is %3.0f kg/s'%(Tco,U,L,ms)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The outlet temperature of the cooling water is 38.32 degree C \n", + " The overall heat transfer coefficient is 6525.8 W/m**2.K \n", + " Tube length per pass using NTU method is 4.08 m \n", + " The rate of condensation of steam is 97 kg/s\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.15 Page No : 530" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Tci = 5.;\t\t\t#Inlet temperature of water in degree C\n", + "mc = 4600.;\t\t\t#Mass flow rate of water in kg/h\n", + "mh = 4000.;\t\t\t#Mass flow rate of air in kg/h\n", + "Thi = 40.;\t\t\t#Inlet temperature of air in degree C\n", + "U = 150.;\t\t\t#Overall heat transfer coefficient in W/m**2.K\n", + "A = 25.;\t\t\t#Area of heat exchanger in m**2\n", + "Cc = 4180.;\t\t\t#Specific heat of water in J/kg.K\n", + "Ch = 1010.;\t\t\t#Specific heat of air in J/kg.K\n", + "\n", + "# Calculations\n", + "C1 = ((mh*Ch)/3600);\t\t\t#Heat capacity of air in W/K\n", + "C2 = ((mc*Cc)/3600);\t\t\t#Heat capacity of water in W/K\n", + "Cmin = min(C1,C2);\t\t\t#Minimum value of heat capacity in W/K\n", + "Cmax = max(C1,C2);\t\t\t#Maximum value of heat capacity in W/K\n", + "r = (Cmin/Cmax);\t\t\t#Ratio of min and max heat capacity in W/K\n", + "NTU = ((U*A)/Cmin);\t\t\t#Number of heat transfer units\n", + "e = 0.92;\t\t\t#Effectiveness of heat exchanger from Fig. 12.18 on page no.526\n", + "Q = (e*Cmin*(Thi-Tci));\t\t\t#Heat transfer rate in W\n", + "Tco = ((Q/C2)+Tci);\t\t\t#Outlet temperature of water in degree C\n", + "Tho = (Thi-(Q/C1));\t\t\t#Outlet temperature of air in degree C\n", + "\n", + "# Results\n", + "print 'The exit temperature of water is %3.1f degree C \\n \\\n", + "The exit temperature of air is %3.1f degree C'%(Tco,Tho)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The exit temperature of water is 11.8 degree C \n", + " The exit temperature of air is 7.8 degree C\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.16 Page No : 532" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "A = 15.82;\t\t\t#Total outside area of heat exchanger in m**2\n", + "Thi = 110;\t\t\t#Inlet temperature of oil in degree C\n", + "Ch = 1900;\t\t\t#Specific heat of oil in J/kg.K\n", + "mh = 170.9;\t\t\t#Mass flow rate of oil in kg/min\n", + "mc = 68;\t\t\t#Mass flow rate of water in kg/min\n", + "Tci = 35;\t\t\t#Inlet temperature of water in degree C\n", + "U = 320;\t\t\t#Overall heat transfer coefficient in W/m**2.K\n", + "Cc = 4186;\t\t\t#Specific heat of water in J/kg.K\n", + "\n", + "# Calculations\n", + "C1 = ((mh*Ch)/60);\t\t\t#Heat capacity of oil in W/K\n", + "C2 = ((mc*Cc)/60);\t\t\t#Heat capacity of water in W/K\n", + "D = (U*A*((1./C1)-(1./C2)));\t\t\t#Constant\n", + "r = (C1/C2);\t\t\t#Ratio of heat capacity of oil and water\n", + "Tho = Thi-(((Thi-Tci)*(1-math.exp(D)))/(r-math.exp(D)));\t\t\t#Outlet temperature of oil in degree C\n", + "Tco = Tci+(r*(Thi-Tho));\t\t\t#Outlet temperature of water in degree C\n", + "\n", + "# Results\n", + "print 'The exit temperature of oil is %3.2f degree C \\n \\\n", + " The exit temperature of water is %3.1f degree C'%(Tho,Tco)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The exit temperature of oil is 74.97 degree C \n", + " The exit temperature of water is 75.0 degree C\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.17 Page No : 533" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Tci = 20.;\t\t\t#Inlet temperature of water in degree C\n", + "Tco = 50.;\t\t\t#Outlet temperature of water in degree C\n", + "Th = 120.;\t\t\t#Temperature at which steam condenses in degree C\n", + "newTci = 15.;\t\t\t#New Inlet temperature of water in degree C\n", + "\n", + "# Calculations\n", + "newTco = (((Tco-Tci)*(Th-newTci))/(Th-Tci))+newTci;\t\t\t#New outlet temperature of water in degree C\n", + "\n", + "# Results\n", + "print 'New outlet temperature of water is %3.1f degree C'%(newTco)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "New outlet temperature of water is 46.5 degree C\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.18 Page No : 534" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "T = 100;\t\t\t#Total length of tubes in m\n", + "\n", + "\n", + "# Calculations\n", + "D = ((3.14*4000)/(3.14*30000))**0.5;\t\t\t#Diameter of the exchanger in m\n", + "L = (2./(3.1415*D**2));\t\t\t#Length of the exchanger in m\n", + "Cost = (10000+(15000*D**3*L)+(2000*D*L));\t\t\t#Optimal math.cost in Rs\n", + "\n", + "# Results\n", + "print 'The diameter of the exchanger is %3.3f m \\n \\\n", + "The Length of the exchanger is %3.2f m \\n \\\n", + "Optimal cost is %3.0f Rs'%(D,L,Cost)\n", + "\n", + "# note : rounding error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The diameter of the exchanger is 0.365 m \n", + " The Length of the exchanger is 4.77 m \n", + " Optimal cost is 16974 Rs\n" + ] + } + ], + "prompt_number": 8 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch13.ipynb b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch13.ipynb new file mode 100755 index 00000000..bbf39026 --- /dev/null +++ b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch13.ipynb @@ -0,0 +1,444 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:154cdfdbffe4cd226ec00ca571625579dcd760fa5b25d6fc7f505d761a2beb76" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 13 : Diffusion Mass Transfer" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.1 Page No : 544" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "ro2 = 0.21;\t\t\t#Ratio of O2 in the mixture \n", + "rn2 = 0.79;\t\t\t#Ratio of N2 in the mixture \n", + "T = (25+273);\t\t\t#Temperature of container in degree C\n", + "p = 1;\t\t\t#Total pressure in atm\n", + "\n", + "# Calculations\n", + "Co2 = (ro2*10**5)/(8314*T);\t\t\t#Molar concentration of O2 in K.mol/m**3\n", + "Cn2 = (rn2*10**5)/(8314*T);\t\t\t#Molar concentration of N2 in K.mol/m**3\n", + "po2 = (32*Co2);\t\t\t#Mass density in kg/m**3\n", + "pn2 = (28*Cn2);\t\t\t#Mass density in kg/m**3\n", + "p = (po2+pn2);\t\t\t#Overall mass density in kg/m**3\n", + "mo2 = (po2/p);\t\t\t#Mass fraction of O2\n", + "mn2 = (pn2/p);\t\t\t#Mass fraction of N2\n", + "M = (ro2*32)+(rn2*28);\t\t\t#Average molecular weight \n", + "\n", + "# Results\n", + "print 'Molar concentration of O2 is %3.4f K.mol/m**3 \\n \\\n", + "Molar concentration of N2 is %3.3f K.mol/m**3 \\n \\\n", + "Mass density of O2 is %3.3f kg/m**3 \\n \\\n", + "Mass density of N2 is %3.3f kg/m**3 \\n \\\n", + "Mole fraction of O2 is %3.2f \\n \\\n", + "Mole fraction of N2 is %3.2f \\n \\\n", + "Mass fraction of O2 is %3.3f \\n \\\n", + "Mass fraction of N2 is %3.3f \\n \\\n", + "Average molecular weight is %3.2f'%(Co2,Cn2,po2,pn2,ro2,rn2,mo2,mn2,M)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Molar concentration of O2 is 0.0085 K.mol/m**3 \n", + " Molar concentration of N2 is 0.032 K.mol/m**3 \n", + " Mass density of O2 is 0.271 kg/m**3 \n", + " Mass density of N2 is 0.893 kg/m**3 \n", + " Mole fraction of O2 is 0.21 \n", + " Mole fraction of N2 is 0.79 \n", + " Mass fraction of O2 is 0.233 \n", + " Mass fraction of N2 is 0.767 \n", + " Average molecular weight is 28.84\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.2 Page No : 545" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "yh2 = 0.4;\t\t\t#Mole fraction og H2\n", + "yo2 = 0.6;\t\t\t#Mole fraction of O2\n", + "vh2 = 1;\t\t\t#velocity of H2 in m/s\n", + "vo2 = 0;\t\t\t#velocity of O2 in m/s\n", + "\n", + "# Calculations\n", + "V = (yh2*vh2)+(yo2*vo2);\t\t#Molar average velocity in m/s\n", + "M = (yh2*2)+(yo2*32);\t\t\t#Molecular weight of the mixture\n", + "mh2 = (yh2*2)/M;\t \t\t#Mass fraction of H2 \n", + "mo2 = (yo2*32)/M;\t\t \t#Mass fraction of O2\n", + "v = (mh2*vh2)+(mo2*vo2);\t\t#Mass average velocity in m/s\n", + "x1 = (mh2*vh2);\t\t\t #Mass flux \n", + "x2 = (mo2*vo2);\t\t\t #Mass flux\n", + "y1 = (v*vh2);\t\t \t #Molar flux\n", + "y2 = (yo2*vo2);\t \t\t #Molar flux\n", + "jh2 = (mh2*(vh2-v));\t\t\t#Mass diffusion flux\n", + "jo2 = (mo2*(vo2-v));\t\t\t#Mass diffusion flux\n", + "Jh2 = (yh2*(vh2-V));\t\t\t#Molar diffusion flux\n", + "Jo2 = (yo2*(vo2-V));\t\t\t#Molar diffusion flux\n", + "\n", + "# Results\n", + "print 'Molar average velocity is %3.1f m/s \\n \\\n", + "Mass average velocity is %3.2f m/s \\n \\\n", + "Mass flux of H2 when it is stationary is %3.2fp kg/m2.s3 \\n \\\n", + "Mass flux of O2 when it is stationary is %3.0f kg/m**2.s \\n \\\n", + "Molar flux of H2 when it is stationary is %3.2fC k.mol/m**2.s \\n \\\n", + "Molar flux of O2 when it is stationary is %3.0f k.mol/m**2.s \\n \\\n", + "Mass diffusion flux of H2 across a surface moving with mass average velocity is %3.4fp kg/m**2.s \\n \\\n", + "Mass diffusion flux of O2 across a surface moving with mass average velocity is %3.4fp kg/m**2.s \\n \\\n", + "Molar diffusion flux across a surface moving with molar average velociy for H2 is %3.2fC k.mol/m**2.s \\n \\\n", + "Molar diffusion flux across a surface moving with molar average velociy for O2\\\n", + " is %3.2fC k.mol/m**2.s'%(V,v,x1,x2,y1,y2,jh2,jo2,Jh2,Jo2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Molar average velocity is 0.4 m/s \n", + " Mass average velocity is 0.04 m/s \n", + " Mass flux of H2 when it is stationary is 0.04p kg/m2.s3 \n", + " Mass flux of O2 when it is stationary is 0 kg/m**2.s \n", + " Molar flux of H2 when it is stationary is 0.04C k.mol/m**2.s \n", + " Molar flux of O2 when it is stationary is 0 k.mol/m**2.s \n", + " Mass diffusion flux of H2 across a surface moving with mass average velocity is 0.0384p kg/m**2.s \n", + " Mass diffusion flux of O2 across a surface moving with mass average velocity is -0.0384p kg/m**2.s \n", + " Molar diffusion flux across a surface moving with molar average velociy for H2 is 0.24C k.mol/m**2.s \n", + " Molar diffusion flux across a surface moving with molar average velociy for O2 is -0.24C k.mol/m**2.s\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.3 Page No : 557" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "t = 0.001;\t\t\t#Thickness of the membrane in m\n", + "CA1 = 0.02;\t\t\t#Concentration of helium in the membrane at inner surface in k.mol/m**3\n", + "CA2 = 0.005;\t\t\t#Concentration of helium in the membrane at outer surface in k.mol/m**3\n", + "DAB = 10**-9;\t\t\t#Binary diffusion coefficient in m**2/s\n", + "\n", + "# Calculations\n", + "NAx = ((DAB*(CA1-CA2))/t)/10**-9;\t\t\t#Diffusion flux of helium through the plastic in k.mol/sm**2 *10**-9\n", + "\n", + "# Results\n", + "print 'Diffusion flux of helium through the plastic is %3.0f*10**-9 k.mol/sm**2'%(NAx)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diffusion flux of helium through the plastic is 15*10**-9 k.mol/sm**2\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.4 Page No : 557" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "T = 273+25;\t\t\t#Temperature of Helium gas in K\n", + "p = 4;\t\t\t#Pressure of helium gas in bar\n", + "Di = 0.1;\t\t\t#Inner diamter of wall in m\n", + "Do = 0.003;\t\t\t#Outer diamter of wall in m\n", + "DAB = (0.4*10**-13);\t\t\t#Binary diffusion coefficient in m**2/s\n", + "S = (0.45*10**-3);\t\t\t#S value for differentiation\n", + "\n", + "# Calculations\n", + "A = (3.14*Di**2);\t\t\t#Area in m**2\n", + "V = (3.14*Di**3)/6;\t\t\t#Volume in m**3\n", + "R = 0.08316\t\t\t#Gas consmath.tant in m**3 bar/kmol.K\n", + "d = ((-6*R*T*DAB*S*p)/(Do*Di))/10**-11;\t\t\t#Decrease of pressure with time in bar/s*10**-11\n", + "\n", + "# Results\n", + "print 'Initial rate of leakage for the system is provided by the decrease of pressure \\\n", + " with time which is %3.2f*10**-11 bar/s'%(d)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Initial rate of leakage for the system is provided by the decrease of pressure with time which is -3.57*10**-11 bar/s\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.5 Page No : 558" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "po2 = 2;\t\t\t#Pressure of O2 in bar\n", + "Di = 0.025;\t\t\t#inside diamter of the pipe in m\n", + "L = 0.0025;\t\t\t#Wall thickness in m\n", + "a = (0.21*10**-2);\t\t\t#Diffusivity of O2 in m**2/s\n", + "S = (3.12*10**-3);\t\t\t#Solubility of O2 in k.mol/m**3.bar\n", + "DAB = (0.21*10**-9);\t\t\t#Binary diffusion coefficient in m**2/s\n", + "\n", + "# Calculations\n", + "CAi = (S*po2);\t\t\t#Concentration of O2 on inside surface in kmol/m**3\n", + "RmA = ((math.log((Di+(2*L))/Di))/(2*3.14*DAB));\t\t\t#Diffusion resismath.tance in sm**2\n", + "Loss = (CAi/RmA)/10**-11;\t\t\t#Loss of O2 by diffusion per meter length of pipe *10**-11\n", + "\n", + "# Results\n", + "print 'Loss of O2 by diffusion per meter length of pipe is %3.2f*10**-11 kmol/s'%(Loss)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Loss of O2 by diffusion per meter length of pipe is 4.51*10**-11 kmol/s\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.6 Page No : 560" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "p = 1;\t\t\t#Pressure of system in atm\n", + "T = 25+273;\t\t\t#Temperature of system in K\n", + "pco2 = (190./760);\t\t\t#Partial pressure of CO2 at one end in atm\n", + "pco2o = (95./760);\t\t\t#Partial pressure of CO2 at other end in atm\n", + "DAB = (0.16*10**-4);\t\t\t#Binary diffusion coefficient in m**2/s from Table 13.3\n", + "R = 0.08205\t\t\t#Gas constant in m**3 atm/kmol.K\n", + "\n", + "# Calculations\n", + "NAx = (DAB*(pco2-pco2o))/(R*T*p);\t\t\t#Equimolar counter diffusion in kmol/m**2s\n", + "M = (NAx*3.14*(0.05**2/4)*3600);\t\t\t#Mass transfer rate in kmol/h\n", + "MCO2 = (M*44)/10**-5;\t\t\t#Mass flow rate of CO2 in kg/h *10**-5\n", + "Mair = (29*-M)/10**-5;\t\t\t#Mass flow rate of air in kg/h *10**-5\n", + "\n", + "# Results\n", + "print 'Mass transfer rate of CO2 is %3.2f*10**-5 kg/h Mass transfer rate of air is %3.2f*10**-5 kg/h'%(MCO2,Mair)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mass transfer rate of CO2 is 2.54*10**-5 kg/h Mass transfer rate of air is -1.68*10**-5 kg/h\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.7 Page No : 563" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# Variables\n", + "T = 27+273;\t\t\t#Temperature of water in K\n", + "D = 0.02;\t\t\t#Diameter of the tube in m\n", + "L = 0.4;\t\t\t#Length of the tube in m\n", + "DAB = (0.26*10**-4);\t\t\t#Diffusion coefficient in m**2/s\n", + "\n", + "# Calculations\n", + "p = 1.0132;\t\t\t#Atmospheric pressure in bar\n", + "pA1 = 0.03531;\t\t\t#Vapour pressure in bar\n", + "m = ((p*10**5*3.14*(D/2)**2*18*DAB)/(8316*T*L))*(1000*3600)*math.log(p/(p-pA1));\t\t\t#Diffusion rate of water in gram per hour\n", + "\n", + "# Results\n", + "print 'Diffusion rate of water is %3.4f gram per hour'%(m)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diffusion rate of water is 0.0019 gram per hour\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.8 Page No : 564" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# Variables\n", + "T = 25+273;\t\t\t#Temperature of water in K\n", + "D = 0.02;\t\t\t#Diameter of the tube in m\n", + "L = 0.08;\t\t\t#Length of the tube in m\n", + "m = (8.54*10**-4);\t\t\t#Diffusion coefficient in kg/h\n", + "\n", + "# Calculations\n", + "p = 1.0132;\t\t\t#Atmospheric pressure in bar\n", + "pA1 = 0.03165;\t\t\t#Vapour pressure in bar\n", + "DAB = (((m/3600)*8316*T*L)/(p*10**5*3.14*(D/2)**2*18*math.log(p/(p-pA1))*10**2))/10**-4;\t\t\t#Diffusion coefficient of water in m**2/s *10**-4\n", + "\n", + "# Calculations\n", + "print 'Diffusion coefficient of water is %3.3f*10**-4 m**2/s'%(DAB)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diffusion coefficient of water is 0.259*10**-4 m**2/s\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.9 Page No : 569" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "CAs = 0.02;\t\t\t#Carbon mole fraction\n", + "CAo = 0.004;\t\t#Content of steel\n", + "CA = 0.012;\t\t\t#Percet of depth\n", + "d = 0.001;\t\t\t#Depth in m\n", + "H = (6*10**-10);\t#Diffusivity of carbon in m**2/s\n", + "\n", + "# Calculations\n", + "X = (CA-CAs)/(CAo-CAs);\t\t\t#Calculation for erf function\n", + "n = 0.48; \t\t\t#erf(n) = 0.5; n = 0.48\n", + "t = ((d/(n*2.))**2/(3600.*H))*3600;\t\t\t#Time required to elevate the carbon content of steel in s\n", + "\n", + "\n", + "# Results\n", + "print 'Time required to elevate the carbon content of steel is %3.2f s'%(t)\n", + "\n", + "\n", + "# note : rounding error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time required to elevate the carbon content of steel is 1808.45 s\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch14.ipynb b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch14.ipynb new file mode 100755 index 00000000..45a0661c --- /dev/null +++ b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch14.ipynb @@ -0,0 +1,525 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:484a2df5b3762980a76b55ab94f7de4a4a7924fa2fa9ffe86e0603d655f531d8" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 14 : Convective Mass Transfer" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.1 Page No : 574" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables\n", + "D = 0.025;\t\t\t #Diameter of the cylinder in m\n", + "R = (2*10**-6);\t\t\t#Rate of sublime in kg/s\n", + "C = (6*10**-6);\t\t\t#Saturated vapour concentration in kmol/m**3\n", + "W = 128;\t\t\t #Molecular weight in kg/kmol\n", + "\n", + "# Calculations\n", + "q = (R/W);\t\t\t #Molar transfer rate in k.mol/sm\n", + "h = (q/(math.pi*D*C));\t\t\t#Convective mass transfer coefficient in m/s\n", + "\n", + "# Results\n", + "print 'Convective mass transfer coefficient is %3.3f m/s'%(h)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Convective mass transfer coefficient is 0.033 m/s\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.2 Page No : 576" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# Variables\n", + "pA = -0.9;\t\t\t#Partial pressure of water vapour in atm\n", + "t = 0.0025;\t\t\t#Boundary layer thickness in m\n", + "\n", + "# Calculations\n", + "y = 0;\n", + "pAs1 = math.exp(-33.35*y)-0.9;\t\t\t#Partial pressure in atm\n", + "y = t;\n", + "pAs2 = math.exp(-33.35*y)-0.9;\t\t\t#Partial pressure in atm\n", + "#partial derivative of pA wrt y is -33.35exp(y)-0.9\n", + "x = 0;\n", + "X = (-33.35*math.exp(x))-pA;\t\t\t#Partial derivative value at x = 0\n", + "DAB = (0.26*10**-4)\t\t\t#DAB value in m**2/s\n", + "h = (DAB*X)/(pAs2-pAs1);\t\t\t#Local mass transfer coefficient in m/s\n", + "\n", + "# Results\n", + "print 'Local mass transfer coefficient is %3.3f m/s'%(h)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Local mass transfer coefficient is 0.011 m/s\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.3 Page No : 583" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "T = 27;\t\t\t#Temperature of dry air in degree C\n", + "p = 1;\t\t\t#Pressure of dry air in atm\n", + "L = 0.5;\t\t\t#Length of the plate in m\n", + "v = 50;\t\t\t#Velocity in m/s\n", + "\n", + "# Calculations\n", + "DAB = (0.26*10**-4)\t\t\t#DAB value in m**2/s\n", + "p = 1.16;\t\t\t#Density in kg/m**3\n", + "u = (184.6*10**-7);\t\t\t#Dynamic viscosity in N.s/m**2\n", + "Pr = 0.707;\t\t\t#Prantl number\n", + "Sc = (u/(p*DAB));\t\t\t#Schmidt number\n", + "Re = (p*v*L)/u;\t\t\t#Reynolds number\n", + "jm = (0.0296*(Re**(-1./5)));\t\t\t#jm value\n", + "h = (jm*v)/Sc**(2./3);\t\t\t#Mass transfer coefficient of water vapour in m/s\n", + "\n", + "# Results\n", + "print 'Mass transfer coefficient of water vapour is %3.3f m/s'%(h)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mass transfer coefficient of water vapour is 0.118 m/s\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.4 Page No : 583" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "T = 27;\t\t\t#Temperature of swimming pool in degree C\n", + "h = 0.4;\t\t\t#Relative humidity\n", + "v = 2;\t\t\t#Speed of wind in m/s\n", + "v1 = (15.89*10**-6);\t\t\t#Kinematic viscosity in m**2/s\n", + "p = 0.0436;\t\t\t#Density in kg/m**3\n", + "DAB = (0.26*10**-4)\t\t\t#DAB value in m**2/s\n", + "L = 15;\t\t\t#Length in m\n", + "\n", + "\n", + "# Calculations\n", + "Sc = (v1/DAB);\t\t\t#Schmidt number\n", + "Re = (v*L)/v1;\t\t\t#Reynolds number\n", + "ShL = (((0.037*Re**(4./5))-870)*Sc**(1./3));\t\t\t#Equivalent Schmidt number\n", + "h1 = (ShL*(DAB/L))/10**-3;\t\t\t#Mass transfer coefficient for evaporation in mm/s\n", + "\n", + "# Results\n", + "print 'Mass transfer coefficient for evaporation is %3.1f*10**-3 m/s'%(h1)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mass transfer coefficient for evaporation is 4.4*10**-3 m/s\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.5 Page No : 585" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "T = 25;\t\t\t#Temperature of air in degree C\n", + "v = 3;\t\t\t#Velocity im m/s\n", + "D = 0.01;\t\t\t#Diameter of tube in m\n", + "L = 1;\t\t\t#Length of tube in m\n", + "\n", + "# Calculations\n", + "v1 = (15.7*10**-6);\t\t\t#Kinematic viscosity in m**2/s\n", + "DAB = (0.62*10**-5)\t\t\t#DAB value in m**2/s\n", + "Re = (v*D)/v1;\t\t\t#Reynolds number\n", + "Sh = 3.66;\t\t\t#Schmidt number\n", + "h = (Sh*DAB)/D;\t\t\t#Average mass transfer coefficient in m/s\n", + "\n", + "# Results\n", + "print 'Average mass transfer coefficient is %3.5f m/s'%(h)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Average mass transfer coefficient is 0.00227 m/s\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.6 Page No : 586" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "T = 25;\t\t\t#Temperature of air in degree C\n", + "v = 5;\t\t\t#Velocity in m/s\n", + "D = 0.03;\t\t\t#Diameter of tube in m\n", + "DAB = (0.82*10**-5)\t\t\t#DAB value in m**2/s\n", + "\n", + "# Calculations\n", + "v1 = (15.7*10**-6);\t\t\t#Kinematic viscosity in m**2/s\n", + "Sc = (v1/DAB);\t\t\t#Schnidt number\n", + "Re = (v*D)/v1;\t\t\t#Reynolds number\n", + "h = (0.023*Re**(4./5)*Sc**(1./3)*DAB)/D;\t\t\t#Mass transfer coefficient in m/s\n", + "\n", + "# Results\n", + "print 'Mass transfer coefficient is %3.4f m/s'%(h)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mass transfer coefficient is 0.0119 m/s\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.7 Page No : 589" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# Variables\n", + "Ta = 40.+273;\t\t\t#Temperature of air in K\n", + "w = 100.;\t\t\t #Molecular weight in kg/k.mol\n", + "H = 120.;\t\t\t #Latent heat of vapourisation of volatile liquid in kJ/kg\n", + "p = 3530.;\t\t\t #Saturated vapour pressure in N/m**2\n", + "DAB = (0.2*10**-4);\t\t#DAB value in m**2/s\n", + "\n", + "# Calculations\n", + "p1 = 1.16;\t\t\t#Density in kg/m**2\n", + "Cp = 1.007;\t\t\t#Specific heat in J/kg.K\n", + "a = (22.5*10**-6);\t\t\t#Diffusivity in m**2/s\n", + "X = ((H*100*p*10**-3)/(8.315*p1*Cp*(a/DAB)**(2./3)));\t\t\t#X value for temperature\n", + "T = (Ta+math.sqrt((Ta**2-(4*X))))*0.5;\t\t\t#Temperature in K\n", + "\n", + "# Results\n", + "print 'Steady state temperature of cold water inside the pot is %3.1f K'%(T)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Steady state temperature of cold water inside the pot is 299.5 K\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.8 Page No : 590" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "T = 22. + 273;\t\t\t#Thermometer reading in K\n", + "\n", + "# Calculations\n", + "p = 2617;\t\t\t#Pressure in N/m**2\n", + "hfg = 2449;\t\t\t#Enthalpy in kJ/kg\n", + "p1 = (p*18)/(8315*T);\t\t\t#Density in kg/m**3\n", + "p2 = (1.0132*10**5)/(287*T);\t\t\t#Density in kg/m**3\n", + "Cp = 1.008;\t\t\t#Specific heat in kJ/kg.K\n", + "a = (26.2*10**-6);\t\t\t#Diffusivity in m**2/s\n", + "DAB = (0.26*10**-4);\t\t\t#DAB value in m**2/s\n", + "Ts = ((T-273)+((hfg*1000*p1)/(p2*Cp*1000)));\t\t\t#True air temperature in degree C\n", + "\n", + "# Results\n", + "print 'True air temperature is %3.2f degree C'%(Ts)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "True air temperature is 60.99 degree C\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.9 Page No : 591" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "T = 50.;\t\t\t#Temperature of air stream in degree C\n", + "Tb = 22.;\t\t\t#Bulb temperature in degree C\n", + "\n", + "# Calculations\n", + "Tf = (T+Tb)/2;\t\t\t#Film temperature in degree C\n", + "p = 1.14;\t\t\t#Density in kg/m**3\n", + "Cp = 1.006;\t\t\t#Specific heat in J/kg.K\n", + "Pr = 0.7;\t\t\t#Prantl number\n", + "u = (2*10**-5);\t\t\t#Dynamic viscosity in Ns/m**2\n", + "DAB = (0.26*10**-4);\t\t\t#DAB value in m**2/s\n", + "Sc = (u/(p*DAB));\t\t\t#Schmidt nuber\n", + "Le = (Sc/Pr);\t\t\t#Lewis number\n", + "p1 = 0.01920;\t\t\t#Density in kg/m**3\n", + "hfg = 2449;\t\t\t#Enthalpy in kJ/kg\n", + "pA = 0.0064;\t\t\t#Density in kg/m**3\n", + "psat = (1./12.23);\t\t\t#Saturation density in kg/m**3\n", + "RH = (pA/0.0817)*100;\t\t\t#Relative humidity\n", + "\n", + "# Results\n", + "print 'Relative humidity of the airstream is %3.2f percent'%(RH)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Relative humidity of the airstream is 7.83 percent\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.10 Page No : 592" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Td = 27.;\t\t\t#Dry bulb teperature in degree C\n", + "Tw = 17.;\t\t\t#Wet bulb temperature in degree C\n", + "Pr = 0.74;\t\t\t#Prantl number\n", + "Sc = 0.6;\t\t\t#Schmidt number\n", + "Mv = 18.;\t\t\t#Molecular weight of vapour\n", + "Ma = 29.;\t\t\t#Molecular weight of air\n", + "Cp = 1004.;\t\t\t#Specific heat in J/kg.K\n", + "p = (1.0132*10**5);\t\t\t#Pressure in N/m**2\n", + "\n", + "# Calculations\n", + "pv2 = 1917;\t\t\t#Saturation presusre of air at 17 degree C in N/m**2\n", + "hfg = 2461;\t\t\t#Enthalpy in kJ/kg\n", + "w2 = (Mv*pv2)/(Ma*(p-pv2));\t\t\t#Weight in kg/kg of dry air\n", + "w1 = w2-((Cp*(Pr/Sc)**(2./3)*(Td-Tw))/(hfg*1000));\t\t\t#Specific humidity of air in kg/kg of dry air\n", + "\n", + "# Results\n", + "print 'Specific humidity of air is %3.5f kg/kg of dry air'%(w1)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Specific humidity of air is 0.00728 kg/kg of dry air\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.11 Page No : 592" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "T = 27.;\t\t \t#Temperature of swimming pool in degree C\n", + "Ts = 37.;\t\t\t#Surface temperature in degree C\n", + "h = 0.4;\t\t\t#Relative humidity\n", + "D1 = 5.;\t\t \t#Dimension of swimming pool in m\n", + "D2 = 15.;\t\t\t#Dimension of swimming pool in m\n", + "v = 2.;\t\t \t#Speed of wind in m/s\n", + "v1 = (15.89*10**-6);\t\t\t#Kinematic viscosity in m**2/s\n", + "p = 0.0436;\t\t\t #Density in kg/m**3\n", + "DAB = (0.26*10**-4)\t \t\t#DAB value in m**2/s\n", + "Sc = (v1/DAB);\t\t \t#Schmidt number\n", + "Re = (v*D2)/v1;\t\t \t#Reynolds number\n", + "ShL = (((0.037*Re**(4./5))-870)*Sc**(1./3));\t\t\t#Equivalent Schmidt number\n", + "h1 = (ShL*(DAB/D2));\t\t\t#Mass transfer coefficient for evaporation in m/s\n", + "\n", + "# Calculations\n", + "Psat = 3531.;\t\t\t#Partial pressure of water vapour in N/m**2\n", + "pi = (0.4*6221);\t\t\t#Saturation pressure of water vapour in N/m**2\n", + "pt = 101325.;\t\t\t#Total pressure of air in N/m**2\n", + "pAs = (18*Psat)/(8361*(T+273));\t\t\t#Density at the water surface in kg/m\n", + "pAi = (18*pi)/(8316*(T+273));\t\t\t#Density at the water surface in kg/m\n", + "n = round((h1*(pAs-pAi)*3600*24),);\t\t\t#Rate of evaporation of water in kg/m**2 day\n", + "L = (n*D1*D2);\t\t\t#Total water loss from the swimming pool in kg/day\n", + "\n", + "# Results\n", + "print 'Rate of evaporation of water is %3.1f kg/day'%(L)\n", + "\n", + "# there is a rounding off error in textbook." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rate of evaporation of water is 225.0 kg/day\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch3.ipynb b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch3.ipynb new file mode 100755 index 00000000..930ae930 --- /dev/null +++ b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch3.ipynb @@ -0,0 +1,1535 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a4eb855d2f94bc4d3b88cc418db7aa787fc638da4c248a8ac28f23e0760aff9c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 3 : OneDimensional Steady State Heat Conduction" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.1 Page No : 45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "# Variables\n", + "l = 5;\t\t \t#Length of the wall in m\n", + "h = 4;\t\t\t #Height of the wall in m\n", + "L = 0.25;\t\t\t#Thickness of the wall in m\n", + "T = [110,40];\t\t#Temperature on the inner and outer surface in degree C\n", + "k = 0.7;\t\t\t#Thermal conductivity in W/m.K\n", + "x = 0.20;\t\t\t#Distance from the inner wall in m\n", + "\n", + "# Calculations\n", + "A = l*h;\t\t\t #Arear of the wall in m**2\n", + "Q = (k*A*(T[0]-T[1]))/L;\t \t\t#Heat transfer rate in W\n", + "T = (((T[1]-T[0])*x)/L)+T[0];\t\t\t#Temperature at interior point of the wall, 20 cm distant from the inner wall in degree C\n", + "\n", + "# Results\n", + "print 'a)Heat transfer rate is %i W \\n \\\n", + "b)Temperature at interior point of the wall, 20 cm distant from the inner wall is %i degree C'%(Q, T)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a)Heat transfer rate is 3920 W \n", + " b)Temperature at interior point of the wall, 20 cm distant from the inner wall is 54 degree C\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.2 Page No : 48" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables\n", + "Di = 0.05;\t\t \t#Inner diameter of hollow cylinder in m\n", + "Do = 0.1;\t\t\t #Outer diameter of hollow cylinder in m\n", + "T = [200,100];\t\t\t#Inner and outer surface temperature in degree C\n", + "k = 70;\t\t\t #Thermal conductivity in W/m.K\n", + "\n", + "# Calculations\n", + "ro = (Do/2);\t\t\t#Outer radius of hollow cylinder in m\n", + "ri = (Di/2);\t\t\t#Inner radius of hollow cylinder in m\n", + "Q = ((2*3.14*k*(T[0]-T[1]))/(math.log(ro/ri)));\t\t\t#Heat transfer rate in W\n", + "r1 = (ro+ri)/2;\t\t\t#Radius at halfway between ro and ri in m\n", + "T1 = T[0]-((T[0]-T[1])*(math.log(r1/ri)/(math.log(ro/ri))));\t\t\t#Temperature of the point halfway between the inner and outer surface in degree C\n", + "\n", + "# Results\n", + "print 'Heat transfer rate is %3.1f W /m \\n \\\n", + "Temperature of the point halfway between the inner and outer surface is %3.1f degree C'%(Q,T1)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer rate is 63420.9 W /m \n", + " Temperature of the point halfway between the inner and outer surface is 141.5 degree C\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3 Page No : 51" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Di = 0.1;\t\t\t#Inner diameter of hollow sphere in m\n", + "Do = 0.3;\t\t\t#Outer diameter of hollow sphere in m\n", + "k = 50. \t\t\t#Thermal conductivity in W/m.K\n", + "T = [300,100];\t\t\t#Inner and outer surface temperature in degree C\n", + "\n", + "# Calculations\n", + "ro = (Do/2);\t\t\t#Outer radius of hollow sphere in m\n", + "ri = (Di/2);\t\t\t#Inner radius of hollow sphere in m\n", + "Q = ((4*3.14*ro*ri*k*(T[0]-T[1]))/(ro-ri))/1000;\t\t\t#Heat transfer rate in W\n", + "r = ri+(0.25*(ro-ri));\t\t\t#The value at one-fourth way of te inner and outer surfaces in m\n", + "T = ((ro*(r-ri)*(T[1]-T[0]))/(r*(ro-ri)))+T[0];\t\t\t#Temperature at a point a quarter of the way between the inner and outer surfaces in degree C\n", + "\n", + "# Results\n", + "print 'Heat flow rate through the sphere is %3.2f kW \\n \\\n", + " Temperature at a point a quarter of the way between the inner and outer surfaces is %i degree C'%(Q,T)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat flow rate through the sphere is 9.42 kW \n", + " Temperature at a point a quarter of the way between the inner and outer surfaces is 200 degree C\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.4 Page No : 55" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "L = 0.4;\t\t\t#Thickness of the furnace in m\n", + "T = [300,50];\t\t\t#Surface temperatures in degree C\n", + "#k = 0.005T-5*10**-6T**2\n", + "\n", + "# Calculations\n", + "q = ((1./L)*(((0.005/2)*(T[0]**2-T[1]**2))-((5*10**-6*(T[0]**3-T[1]**3))/3)));\t\t\t#Heat loss per square meter surface area in W/m**2\n", + "\n", + "# Results\n", + "print 'Heat loss per square meter surface area is %3.0f W/m**2'%(q)\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat loss per square meter surface area is 435 W/m**2\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.5 Page No : 55" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "L = 0.2;\t\t\t #Thickness of the wall in m\n", + "T = [1000,200];\t\t\t#Surface temperatures in degree C\n", + "ko = 0.813;\t\t\t #Value of thermal conductivity at T = 0 in W/m.K\n", + "b = 0.0007158;\t\t\t#Temperature coefficient of thermal conductivity in 1./K\n", + "\n", + "# Calculations\n", + "km = ko*(1+((b*(T[0]+T[1]))/2));\t\t\t#Constant thermal conductivity in W/m.K\n", + "q = ((km*(T[0]-T[1]))/L);\t\t\t #Rate of heat flow in W/m**2\n", + "\n", + "# Results\n", + "print 'Rate of heat flow is %3.0f W/m**2'%(q)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rate of heat flow is 4649 W/m**2\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.6 Page No : 58" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# Variables\n", + "r = [0.01,0.02];\t\t\t#Inner and outer radius of a copper cylinder in m\n", + "T = [310,290];\t\t\t#Inner and Outer surface temperature in degree C\n", + "ko = 371.9;\t\t\t#Value of thermal conductivity at T = 0 in W/m.K\n", + "b = (9.25*10**-5);\t\t\t#Temperature coefficient of thermal conductivity in 1./K\n", + "\n", + "# Calculations\n", + "Tm = ((T[0]-150)+(T[1]-150))/2;\t\t\t#Mean temperature in degree C\n", + "km = ko*(1-(b*Tm));\t\t\t #Constant thermal conductivity in W/m.K\n", + "q = ((2*3.14*km*(T[0]-T[1]))/math.log(r[1]/r[0]))/1000;\t\t\t#Heat loss per unit length in kW/m\n", + "\n", + "# Results\n", + "print 'Heat loss per unit length is %3.2f kW/m'%(q)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat loss per unit length is 66.45 kW/m\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.8 Page No : 63" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "L1 = 0.5;\t\t \t#Thickness of the wall in m\n", + "k1 = 1.4;\t\t\t #Thermal conductivity in W/m.K\n", + "k2 = 0.35;\t\t\t #Thermal conductivity of insulating material in W/m.K\n", + "q = 1450.;\t\t\t #Heat loss per square metre in W\n", + "T = [1200,15];\t\t\t#Inner and outer surface temperatures in degree C\n", + "\n", + "# Calculations\n", + "L2 = (((T[0]-T[1])/q)-(L1/k1))*k2;\t\t\t#Thickness of the insulation required in m\n", + "\n", + "# Results\n", + "print 'Thickness of the insulation required is %3.3f m'%(L2)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thickness of the insulation required is 0.161 m\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.9 Page No : 64" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "L1 = 0.006; \t\t\t#Thickness of each glass sheet in m\n", + "L2 = 0.002;\t \t\t#Thickness of air gap in m\n", + "Tb = -20;\t\t \t#Temperature of the air inside the room in degree C\n", + "Ta = 30;\t\t\t #Ambient temperature of air in degree C\n", + "ha = 23.26;\t\t\t #Heat transfer coefficient between glass and air in W/m**2.K\n", + "kglass = 0.75;\t\t\t#Thermal conductivity of glass in W/m.K\n", + "kair = 0.02;\t\t\t#Thermal conductivity of air in W/m.K\n", + "\n", + "# Calculations\n", + "q = ((Ta-Tb)/((1./ha)+(L1/kglass)+(L2/kair)+(L1/kglass)+(1./ha)));\t\t\t#Rate of heat leaking into the room per unit area of the door in W/m**2\n", + "\n", + "# Results\n", + "print 'Rate of heat leaking into the room per unit area of the door is %3.1f W/m**2'%(q)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rate of heat leaking into the room per unit area of the door is 247.5 W/m**2\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.10 Page No : 65" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "LA = 0.05;\t\t\t#Length of section A in m\n", + "LB = 0.1;\t\t\t#Length of section A in m\n", + "LC = 0.1;\t\t\t#Length of section A in m\n", + "LD = 0.05;\t\t\t#Length of section A in m\n", + "LE = 0.05;\t\t\t#Length of section A in m\n", + "kA = 50;\t\t\t#Thermal conductivity of section A in W/m.K\n", + "kB = 10;\t\t\t#Thermal conductivity of section B in W/m.K\n", + "kC = 6.67;\t\t\t#Thermal conductivity of section C in W/m.K\n", + "kD = 20;\t\t\t#Thermal conductivity of section D in W/m.K\n", + "kE = 30;\t\t\t#Thermal conductivity of section E in W/m.K\n", + "Aa = 1;\t\t\t#Area of section A in m**2\n", + "Ab = 0.5;\t\t\t#Area of section B in m**2\n", + "Ac = 0.5;\t\t\t#Area of section C in m**2\n", + "Ad = 1;\t\t\t#Area of section D in m**2\n", + "Ae = 1;\t\t\t#Area of section E in m**2\n", + "T = [800,100];\t\t\t#Temperature at inlet and outlet temperatures in degree C\n", + "\n", + "# Calculations\n", + "Ra = (LA/(kA*Aa)); \t\t\t#Thermal Resistance of section A in K/W\n", + "Rb = (LB/(kB*Ab));\t \t\t#Thermal Resistance of section B in K/W\n", + "Rc = (LC/(kC*Ac));\t\t \t#Thermal Resistance of section C in K/W\n", + "Rd = (LD/(kD*Ad));\t\t\t #Thermal Resistance of section D in K/W\n", + "Re = (LE/(kE*Ae));\t\t \t#Thermal Resistance of section E in K/W\n", + "Rf = ((Rb*Rc)/(Rb+Rc));\t\t\t#Equivalent resistance of section B and section C in K/W\n", + "R = Ra+Rf+Rd+Re;\t\t\t #Equivalent resistance of all sections in K/W\n", + "Q = ((T[0]-T[1])/R)/1000;\t\t\t#Heat transfer through the composite wall in kW\n", + "\n", + "# Results\n", + "print 'Heat transfer through the composite wall is %3.1f kW'%(Q)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer through the composite wall is 40.8 kW\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.11 Page No : 66" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "T1 = 2000;\t\t\t#Temperature of hot gas in degree C\n", + "Ta = 45;\t\t\t#Room air temperature in degree C\n", + "Qr1 = 23.260;\t\t\t#Heat flow by radiation from gases to inside surface of the wall in kW/m**2\n", + "h = 11.63;\t\t\t#Convective heat transfer coefficient in W/m**2.\n", + "C = 58;\t\t\t#Thermal conductance of the wall in W/m**2.K\n", + "Q = 9.3;\t\t\t#Heat flow by radiation from external surface to the surrounding in kW.m**2\n", + "T2 = 1000;\t\t\t#Interior wall temperature in degree C\n", + "\n", + "# Calculations\n", + "qr1 = Qr1;\t\t\t#Haet by radiation in kW/m**2\n", + "qc1 = h*((T1-T2)/1000);\t\t\t#Heat by conduction in kW/m**2\n", + "q = qc1+qr1;\t\t\t#Total heat entering the wall in kW/m**2\n", + "R = (1./C);\t\t\t#Thermal resistance in m**2.K/W\n", + "T3 = T2-(q*R*1000);\t\t\t#External wall temperature in degree C\n", + "Ql = q-Q;\t\t\t#Heat loss due to convection kW/m**2\n", + "h4 = (Ql*1000)/(T3-Ta);\t\t\t#Convective conductance in W/m**2.K\n", + "\n", + "# Results\n", + "print 'The surface temperature is %i degree C \\n \\\n", + "The convective conductance is %3.1f W/m**2.K'%(T3,h4)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The surface temperature is 398 degree C \n", + " The convective conductance is 72.4 W/m**2.K\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.12 Page No : 67" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "L1 = 0.125;\t\t\t#Thickness of fireclay layer in m\n", + "L2 = 0.5;\t\t\t#Thickness of red brick layer in m\n", + "T = [1100,50];\t\t\t#Temperatures at inside and outside the furnaces in degree C\n", + "k1 = 0.533;\t\t\t#Thermal conductivity of fireclay in W/m.K\n", + "k2 = 0.7;\t\t\t#Thermal conductivity of red brick in W/m.K\n", + "\n", + "# Calculations\n", + "R1 = (L1/k1);\t\t\t#Resismath.tance of fireclay per unit area in K/W\n", + "R2 = (L2/k2);\t\t\t#Resistance of red brick per unit area in K/W\n", + "R = R1+R2;\t\t\t#Total resistance in K/W\n", + "q = (T[0]-T[1])/R;\t\t\t#Heat transfer in W/m**2\n", + "T2 = T[0]-(q*R1);\t\t\t#Temperature in degree C\n", + "T3 = T[1]+(q*R2*0.5);\t\t\t#Temperature at the interface between the two layers in degree C\n", + "km = 0.113+(0.00023*((T2+T3)/2));\t\t\t#Mean thermal conductivity in W/m.K\n", + "x = ((T2-T3)/q)*km;\t\t\t#Thickness of diatomite in m\n", + "\n", + "# Results\n", + "print 'Amount of heat loss is %3.1f W/m**2 \\n \\\n", + "Thickness of diatomite is %3.4f m'%(q,x )\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Amount of heat loss is 1106.7 W/m**2 \n", + " Thickness of diatomite is 0.0932 m\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.13 Page No : 70" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# Variables\n", + "Di = 0.1;\t\t\t#I.D of the pipe in m\n", + "L = 0.01;\t\t\t#Thickness of the wall in m\n", + "L1 = 0.03;\t\t\t#Thickness of insulation in m\n", + "Ta = 85;\t\t\t#Temperature of hot liquid in degree C\n", + "Tb = 25;\t\t\t#Temperature of surroundings in degree C\n", + "k1 = 58;\t\t\t#Thermal conductivity of steel in W/m.K\n", + "k2 = 0.2;\t\t\t#Thermal conductivity of insulating material in W/m.K\n", + "ha = 720;\t\t\t#Inside heat transfer coefficient in W/m**2.K\n", + "hb = 9;\t\t\t #Outside heat transfer coefficient in W/m**2.K\n", + "D2 = 0.12;\t\t\t#Inner diameter in m\n", + "r3 = 0.09;\t\t\t#Radius in m\n", + "\n", + "# Calculations\n", + "q = ((2*3.14*(Ta-Tb))/((1./(ha*(Di/2)))+(1./(hb*r3))+(math.log(D2/Di)/k1)+(math.log(r3/(D2/2))/k2)));\t\t\t#Heat loss fro an insulated pipe in W/m\n", + "\n", + "# Results\n", + "print 'Heat loss from an insulated pipe is %3.2f W/m'%(q)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat loss from an insulated pipe is 114.43 W/m\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.14 Page No : 71" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "Di = 0.1;\t\t\t#I.D of the pipe in m\n", + "Do = 0.11;\t\t\t#O.D of the pipe in m\n", + "L = 0.005;\t\t\t#Thickness of the wall in m\n", + "k1 = 50;\t\t\t#Thermal conductivity of steel pipe line in W/m.K\n", + "k2 = 0.06;\t\t\t#Thermal conductivity of first insulating material in W/m.K\n", + "k3 = 0.12;\t\t\t#Thermal conductivity of second insulating material in W/m.K\n", + "T = [250,50];\t\t\t#Temperature at inside tube surface and outside surface of insulation in degree C\n", + "r3 = 0.105;\t\t\t#Radius of r3 in m as shown in fig.3.14 on page no.71\n", + "r4 = 0.155;\t\t\t#Radius of r4 in m as shown in fig.3.14 on page no.71\n", + "\n", + "# Calculations\n", + "r1 = (Di/2);\t\t\t#Radius of the pipe in m\n", + "r2 = (Do/2);\t\t\t#Radius of the pipe in m\n", + "q = ((2*3.14*(T[0]-T[1]))/(((math.log(r2/r1))/k1)+((math.log(r3/r2))/k2)+((math.log(r4/r3))/k3)));\t\t\t#Loss of heat per metre length of pipe in W/m\n", + "T3 = ((q*math.log(r4/r3))/(2*3.14*k3))+T[1];\t\t\t#Interface temperature in degree C\n", + "\n", + "# Results\n", + "print 'Loss of heat per metre length of pipe is %3.1f W/m \\n \\\n", + "Interface temperature is %3.1f degree C'%(q,T3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Loss of heat per metre length of pipe is 89.6 W/m \n", + " Interface temperature is 96.3 degree C\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.15 Page No : 72" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "Di = 0.1;\t\t\t#I.D of the pipe in m\n", + "Do = 0.11;\t\t\t#O.D of the pipe in m\n", + "L = 0.005;\t\t\t#Thickness of the wall in m\n", + "k1 = 50;\t\t\t#Thermal conductivity of steel pipe line in W/m.K\n", + "k3 = 0.06;\t\t\t#Thermal conductivity of first insulating material in W/m.K\n", + "k2 = 0.12;\t\t\t#Thermal conductivity of second insulating material in W/m.K\n", + "T = [250,50];\t\t\t#Temperature at inside tube surface and outside surface of insulation in degree C\n", + "r3 = 0.105;\t\t\t#Radius of r3 in m as shown in fig.3.14 on page no.71\n", + "r4 = 0.155;\t\t\t#Radius of r4 in m as shown in fig.3.14 on page no.71\n", + "\n", + "# Calculations\n", + "r1 = (Di/2);\t\t\t#Radius of the pipe in m\n", + "r2 = (Do/2);\t\t\t#Radius of the pipe in m\n", + "q = ((2*3.14*(T[0]-T[1]))/(((math.log(r2/r1))/k1)+((math.log(r3/r2))/k2)+((math.log(r4/r3))/k3)));\t\t\t#Loss of heat per metre length of pipe in W/m\n", + "\n", + "# Results\n", + "print 'Loss of heat per metre length of pipe is %3.2f W/m'%(q)\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Loss of heat per metre length of pipe is 105.71 W/m\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.16 Page No : 73" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "D1 = 0.1;\t\t\t#O.D of the pipe in m\n", + "P = 1373;\t\t\t#Pressure of saturated steam in kPa\n", + "D2 = 0.2;\t\t\t#Diameter of magnesia in m\n", + "k1 = 0.07;\t\t\t#Thermal conductivity of magnesia in W/m.K\n", + "k2 = 0.08;\t\t\t#Thermal conductivity of asbestos in W/m.K\n", + "D3 = 0.25;\t\t\t#Diameter of asbestos in m\n", + "T3 = 20;\t\t\t#Temerature under the canvas in degree C\n", + "t = 12;\t\t\t #Time for condensation in hours\n", + "l = 150;\t\t\t#Lemgth of pipe in m\n", + "T1 = 194.14;\t\t\t#Saturation temperature of steam in degree C from Table A.6 (Appendix A) at 1373 kPa on page no. 643\n", + "hfg = 1963.15;\t\t\t#Latent heat of steam in kJ/kg from Table A.6 (Appendix A) at 1373 kPa on page no. 643\n", + "\n", + "# Calculations\n", + "r1 = (D1/2);\t\t\t#Radius of the pipe in m\n", + "r2 = (D2/2);\t\t\t#Radius of magnesia in m\n", + "r3 = (D3/2);\t\t\t#Radius of asbestos in m\n", + "Q = (((2*3.14*l*(T1-T3))/((math.log(r2/r1)/k1)+(math.log(r3/r2)/k2)))*(3600./1000));\t\t\t#Heat transfer rate in kJ/h\n", + "m = (Q/hfg);\t\t\t#Mass of steam condensed per hour\n", + "m1 = (m*t);\t\t\t#Mass of steam condensed in 12 hours\n", + "\n", + "# Results\n", + "print 'Mass of steam condensed in 12 hours is %3.2f kg'%(m1)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mass of steam condensed in 12 hours is 284.43 kg\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.17 Page No : 74" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "D1 = 0.1;\t\t\t#I.D of the first pipe in m\n", + "D2 = 0.3;\t\t\t#O.D of the first pipe in m\n", + "k1 = 70;\t\t\t#Thermal conductivity of first material in W/m.K\n", + "D3 = 0.4;\t\t\t#O.D of the second pipe in m\n", + "k2 = 15;\t\t\t#Thermal conductivity of second material in W/m.K\n", + "T = [300,30];\t\t\t#Inside and outside temperatures in degree C\n", + "\n", + "# Calculations\n", + "r1 = (D1/2);\t\t\t#Inner Radius of first pipe in m\n", + "r2 = (D2/2);\t\t\t#Outer Radius of first pipe in m\n", + "r3 = (D3/2);\t\t\t#Radius of second pipe in m\n", + "Q = ((4*3.14*(T[0]-T[1]))/(((r2-r1)/(k1*r1*r2))+((r3-r2)/(k2*r2*r3))))/1000;\t\t\t#Rate of heat flow through the sphere in kW\n", + "\n", + "# Results\n", + "print 'Rate of heat flow through the sphere is %3.2f kW'%(Q)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rate of heat flow through the sphere is 11.24 kW\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.18 Page No : 77" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "Di = 0.1;\t\t\t#I.D of a steam pipe in m\n", + "Do = 0.25;\t\t\t#I.D of a steam pipe in m\n", + "k = 1.;\t\t\t#Thermal conductivity of insulating material in W/m.K\n", + "T = [200.,20];\t\t\t#Steam temperature and ambient temperatures in degree C\n", + "h = 8.;\t\t\t#Convective heat transfer coefficient between the insulation surface and air in W/m**2.K\n", + "\n", + "# Calculations\n", + "ri = (Di/2);\t\t\t#Inner Radius of steam pipe in m\n", + "ro = (Do/2);\t\t\t#Outer Radius of steam pipe in m\n", + "rc = (k/h)*100;\t\t\t#Critical radius of insulation in cm\n", + "q = ((T[0]-T[1])/((math.log(ro/ri)/(2*3.14*k)+(1./(2*3.14*ro*h)))));\t\t\t#Heat loss per metre of pipe at critical radius in W/m\n", + "Ro = (q/(2*3.14*ro*h))+T[1];\t\t\t#Outer surface temperature in degree C\n", + "\n", + "# Results\n", + "print 'Heat loss per metre of pipe at critical radius is %i W/m \\n \\\n", + "Outer surface temperature is %3.2f degree C'%(q,Ro)\n", + "\n", + "# Note : Answer in book is wrong. Please check manually." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat loss per metre of pipe at critical radius is 589 W/m \n", + " Outer surface temperature is 113.93 degree C\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.19 Page No : 78" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "Di = 0.001;\t\t \t#Diameter of copper wire in m\n", + "t = 0.001;\t\t \t#Thickness of insulation in m;\n", + "To = 20;\t\t \t#Temperature of surrondings in degree C\n", + "Ti = 80;\t\t \t#Maximum temperature of the plastic in degree C\n", + "kcopper = 400;\t\t\t#Thermal conductivity of copper in W/m.K\n", + "kplastic = 0.5;\t\t\t#Thermal conductivity of plastic in W/m.K\n", + "h = 8;\t \t\t#Heat transfer coefficient in W/m**2.K\n", + "p = (3*10**-8);\t\t\t#Specific electric resistance of copper in Ohm.m\n", + "\n", + "# Calculations\n", + "r = (Di/2);\t\t\t#Radius of copper tube in m\n", + "ro = (r+t);\t\t\t#Radius in m \n", + "R = (p/(3.14*r*r*0.01));\t\t\t#Electrical resistance per meter length in ohm/m\n", + "Rth = ((1./(2*3.14*ro*h))+(math.log(ro/r)/(2*3.14*kplastic)));\t\t\t#Thermal resistance of convection film insulation per metre length \n", + "Q = ((Ti-To)/Rth);\t\t\t#Heat transfer in W\n", + "I = math.sqrt(Q/R);\t\t\t#Maximum safe current limit in A\n", + "rc = ((kplastic*100)/h);\t\t\t#Critical radius in cm\n", + "\n", + "# Results\n", + "print 'The maximum safe current limit is %3.3f A \\n \\\n", + "As the critical radius of insulation is much greater,the current carrying capacity of the conductor can be raised upto %3.1f cm \\n \\\n", + "considerably in increasing the radius of plastic covering'%(I,rc)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum safe current limit is 1.074 A \n", + " As the critical radius of insulation is much greater,the current carrying capacity of the conductor can be raised upto 6.2 cm \n", + " considerably in increasing the radius of plastic covering\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.20 Page No : 83" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "L = 0.1;\t\t\t#Thickness of the wall in m\n", + "Q = (4*10**4);\t\t\t#Heat transfer rate in W/m**3\n", + "h = 50;\t\t\t#Convective heat transfer coefficient in W/m**2.K\n", + "T = 20;\t\t\t#Ambient air temperature in degree C\n", + "k = 15;\t\t\t#Thermal conductivity of the material in W/m.K\n", + "\n", + "# Calculations\n", + "Tw = (T+((Q*L)/(2*h)));\t\t\t#Surface temperature in degree C\n", + "Tmax = (Tw+((Q*L*L)/(8*k)));\t\t\t#Maximum temperature in the wall in degree C\n", + "\n", + "# Results\n", + "print 'Surface temperature is %i degree C \\nMaximum temperature in the wall is %3.3f degree C'%(Tw,Tmax)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Surface temperature is 60 degree C \n", + "Maximum temperature in the wall is 63.333 degree C\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.21 Page No : 85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "Do = 0.006;\t\t\t#Outer diameter of hallow cylinder in m\n", + "Di = 0.004;\t\t\t#Inner diameter of hallow cylinder in m\n", + "I = 1000;\t\t\t#Current in A\n", + "T = 30;\t\t\t#Temperature of water in degree C\n", + "h = 35000;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "k = 18;\t\t\t#Thermal conductivity of the material in W/m.K\n", + "R = 0.1;\t\t\t#Electrical reisitivity of the material in ohm.mm**2/m\n", + "\n", + "# Calculations\n", + "ro = (Do/2);\t\t\t#Outer radius of hallow cylinder in m\n", + "ri = (Di/2);\t\t\t#Inner radius of hallow cylinder in m\n", + "V = ((3.14*(ro**2-ri**2)));\t\t\t#Vol. of wire in m**2\n", + "Rth = (R/(3.14*(ro**2-ri**2)*10**6));\t\t\t#Resistivity in ohm/mm**2\n", + "q = ((I*I*Rth)/V);\t\t\t#Heat transfer rate in W/m**3\n", + "To = T+(((q*ri*ri)/(4*k))*((((2*k)/(h*ri))-1)*((ro/ri)**2-1)+(2*(ro/ri)**2*math.log(ro/ri))));\t\t\t#Temperature at the outer surface in degree C\n", + "\n", + "# Results\n", + "print 'Temperature at the outer surface is %3.2f degree C'%(To)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temperature at the outer surface is 57.44 degree C\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.22 Page No : 88" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "D = 0.025;\t\t\t#Diameter of annealed copper wire in m\n", + "I = 200;\t\t\t#Current in A\n", + "R = (0.4*10**-4);\t#Resistance in ohm/cm\n", + "T = [200,10];\t\t#Surface temperature and ambient temperature in degree C\n", + "k = 160;\t\t\t#Thermal conductivity in W/m.K\n", + "\n", + "# Calculations\n", + "r = (D/2);\t\t\t#Radius of annealed copper wire in m\n", + "Q = (I*I*R*100);\t#Heat transfer rate in W/m\n", + "V = (3.14*r*r);\t #Vol. of wire in m**2\n", + "q = (Q/V);\t\t\t#Heat loss in conductor in W/m**2\n", + "Tc = T[0]+((q*r*r)/(4*k));\t\t\t#Maximum temperature in the wire in degree C\n", + "h = ((r*q)/(2*(T[0]-T[1])));\t\t#Heat transfer coefficient in W/m**2.K\n", + "\n", + "# Results\n", + "print 'Maximum temperature in the wire is %3.2f degree C \\nHeat transfer coefficient is %3.2f W/m**2.K'%(Tc,h)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum temperature in the wire is 200.08 degree C \n", + "Heat transfer coefficient is 10.73 W/m**2.K\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.23 Page No : 89" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "p = 100.;\t\t\t#Resistivity of nichrome in \u00b5 ohm-cm \n", + "Q = 10000.;\t\t\t#Heat input of a heater in W\n", + "T = 1220.;\t\t\t#Surface temperature of nichrome in degree C\n", + "Ta = 20.;\t\t\t#Temperature of surrounding air in degree C\n", + "h = 1150.;\t\t\t#Outside surface coeffient in W/m**2.K\n", + "k = 17.;\t\t\t#Thermal conductivity of nichrome in W/m.K\n", + "L = 1.; \t\t\t#Length of heater in m\n", + "\n", + "# Calculations\n", + "d = (Q/((T-Ta)*3.14*h))*1000;\t#Diameter of nichrome wire in mm\n", + "A = (3.14*d*d)/4;\t\t\t #Area of the wire in m**2\n", + "R = ((p*10**-8*L)/A);\t\t\t#Resistance of the wire in ohm\n", + "I = math.sqrt(Q/R)/1000;\t\t#Rate of current flow in A\n", + "\n", + "# Results\n", + "print 'Diameter of nichrome wire is %3.4f mm \\n \\\n", + "Rate of current flow is %i A'%(d,I)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diameter of nichrome wire is 2.3078 mm \n", + " Rate of current flow is 204 A\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.24 Page No : 93" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Do = 0.025;\t\t\t#O.D of the rod in m\n", + "k = 20;\t\t\t#Thermal conductivity in W/m.K\n", + "Q = (2.5*10**6);\t\t\t#Rate of heat removal in W/m**2\n", + "\n", + "# Calculations\n", + "ro = (Do/2);\t\t\t#Outer radius of the rod in m\n", + "q = ((4*Q)/(ro));\t\t\t#Heat transfer rate in W/m**3\n", + "T = ((-3*q*ro**2)/(16*k));\t\t\t#Temperature drop from the centre line to the surface in degree C\n", + "\n", + "# Results\n", + "print 'Temperature drop from the centre line to the surface is %3.3f degree C'%(T)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temperature drop from the centre line to the surface is -1171.875 degree C\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.25 Page No : 95" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Q = 300;\t\t\t#Heat produced by the oranges in W/m**2\n", + "s = 0.08;\t\t\t#Size of the orange in m\n", + "k = 0.15;\t\t\t#Thermal conductivity of the sphere in W/m.K\n", + "\n", + "# Calculations\n", + "q = (3*Q)/(s/2);\t\t\t#Heat flux in W/m**2\n", + "Tc = 10+((q*(s/2)**2)/(6*k));\t\t\t#Temperature at the centre of the sphere in degree C\n", + "\n", + "# Results\n", + "print 'Temperature at the centre of the orange is %i degree C'%(Tc)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temperature at the centre of the orange is 50 degree C\n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.26 Page No : 102" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "To = 140;\t\t\t#Temperature at the junction in degree C\n", + "Ti = 15;\t\t\t#Temperature of air in the room in degree C\n", + "D = 0.003;\t\t\t#Diameter of the rod in m\n", + "h = 300;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "k = 150;\t\t\t#Thermal conductivity in W/m.K\n", + "\n", + "# Calculations\n", + "P = (3.14*D);\t\t\t#Perimeter of the rod in m\n", + "A = (3.14*D**2)/4;\t\t\t#Area of the rod in m**2\n", + "Q = math.sqrt(h*P*k*A)*(To-Ti);\t\t\t#Total heat dissipated by the rod in W\n", + "\n", + "# Results\n", + "print 'Total heat dissipated by the rod is %3.3f W'%(Q)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Total heat dissipated by the rod is 6.841 W\n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.27 Page No : 103" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "D = 0.025\t\t\t#Diameter of the rod in m\n", + "Ti = 22.;\t\t\t#Temperature of air in the room in degree C\n", + "x = 0.1;\t\t\t#Dismtance between the points in m\n", + "T = [110.,85.];\t\t#Temperature sat two points in degree C\n", + "h = 28.4;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "\n", + "# Calculations\n", + "m = -math.log((T[1]-Ti)/(T[0]-Ti))/x;\t\t\t#Calculation of m for obtaining k\n", + "P = (3.14*D);\t\t\t#Perimeter of the rod in m\n", + "A = (3.14*D**2)/4;\t\t\t#Area of the rod in m**2\n", + "k = ((h*P)/((m)**2*A));\t\t\t#Thermal conductivity of the rod material in W/m.K\n", + "\n", + "# Results\n", + "print 'Thermal conductivity of the rod material is %3.1f W/m.K'%(k)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thermal conductivity of the rod material is 406.8 W/m.K\n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.28 Page No : 103" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "L = 0.06;\t\t\t#Length of the turbine blade in m\n", + "A = (4.65*10**-4);\t#Cross sectional area in m**2\n", + "P = 0.12;\t\t\t#Perimeter in m\n", + "k = 23.3;\t\t\t#Thermal conductivity of stainless steel in W/m.K\n", + "To = 500;\t\t\t#Temperature at the root in degree C\n", + "Ti = 870;\t\t\t#Temperature of the hot gas in degree C\n", + "h = 442;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "\n", + "# Calculations\n", + "m = math.sqrt((h*P)/(k*A));\t\t\t#Calculation of m for calculating heat transfer rate\n", + "X = (To-Ti)/math.cosh(m*L);\t\t\t#X for calculating tempetarure distribution\n", + "Q = math.sqrt(h*P*k*A)*(To-Ti)*math.tanh(m*L);\t\t\t#Heat transfer rate in W\n", + "\n", + "# Results\n", + "print 'Temperature distribution is given by :\\n T-Ti = %i cosh[%3.2f%3.2f-x)] cosh[%3.2f%3.2f)] \\n \\\n", + "Heat transfer rate is %3.1f W'%(To-Ti,m,L,m,L,Q)\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temperature distribution is given by :\n", + " T-Ti = -370 cosh[69.970.06-x)] cosh[69.970.06)] \n", + " Heat transfer rate is -280.4 W\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.29 Page No : 104" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "W = 1;\t\t\t#Length of the cylinder in m\n", + "D = 0.05;\t\t\t#Diameter of the cylinder in m\n", + "Ta = 45;\t\t\t#Ambient temperature in degree C\n", + "n = 10;\t\t\t#Number of fins\n", + "k = 120;\t\t\t#Thermal conductivity of the fin material in W/m.K\n", + "t = 0.00076;\t\t\t#Thickness of fin in m\n", + "L = 0.0127;\t\t\t#Height of fin in m\n", + "h = 17;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "Ts = 150;\t\t\t#Surface temperature of cylinder in m\n", + "\n", + "# Calculations\n", + "P = (2*W);\t\t\t#Perimeter of cylinder in m\n", + "A = (W*t);\t\t\t#Surface area of cyinder in m**2\n", + "m = round(math.sqrt((h*P)/(k*A)),2);\t\t\t#Calculation of m for determining heat transfer rate\n", + "Qfin = (math.sqrt(h*P*k*A)*(Ts-Ta)*((math.tanh(m*L)+(h/(m*k)))/(1+((h/(m*k))*math.tanh(m*L)))));\t\t\t#Heat transfer through the fin in kW\n", + "Qb = h*((3.14*D)-(n*t))*W*(Ts-Ta);\t\t\t#Heat from unfinned (base) surface in W\n", + "Q = ((Qfin*10)+Qb);\t\t\t#Total heat transfer in W\n", + "Ti = ((Ts-Ta)/(math.cosh(m*L)+((h*math.sinh(m*L))/(m*k))));\t\t\t#Ti to calculate temperature at the end of the fin in degree C\n", + "T = (Ti+Ta);\t\t\t#Temperature at the end of the fin in degree C\n", + "\n", + "# Results\n", + "print 'Rate of heat transfer is %3.2f W \\nTemperature at the end of the fin is %3.2f degree C'%(Q,T )\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rate of heat transfer is 723.99 W \n", + "Temperature at the end of the fin is 146.74 degree C\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.31 Page No : 109" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "t = 0.025;\t\t\t#Thickness of fin in m\n", + "L = 0.1;\t\t\t#Length of fin in m\n", + "k = 17.7;\t\t\t#Thermal conductivity of the fin material in W/m.K\n", + "p = 7850;\t\t\t#Density in kg/m**3\n", + "Tw = 600;\t\t\t#Temperature of the wall in degree C\n", + "Ta = 40;\t\t\t#Temperature of the air in degree C\n", + "h = 20;\t\t\t #Heat transfer coefficient in W/m**2.K\n", + "I0 = 2.1782;\t\t\t#Io value taken from table 3.2 on page no.108\n", + "I1 = 1.48871;\t\t\t#I1 value taken from table 3.2 on page no. 108\n", + "\n", + "# Calculations\n", + "B = math.sqrt((2*L*h)/(k*t));\t\t\t#Calculation of B for determining temperature distribution \n", + "\n", + "X = ((Tw-Ta)/2.1782);\t\t\t#Calculation of X for determining temperature distribution \n", + "Y = (2*B);\t\t\t#Calculation of Y for determining temperature distribution \n", + "Q = (math.sqrt(2*h*k*t)*(Tw-Ta)*((1.48871))/(2.1782));\n", + "m = ((p*t*L)/2);\t\t\t#Mass of the fin per meter of width in kg/m\n", + "q = (Q/m);\t\t\t#Rate of heat flow per unit mass in W/kg\n", + "\n", + "# Results\n", + "print 'Temperature distribution is T = %i+%3.1f(%3.4f\u221ax) \\n \\\n", + "Rate of heat flow per unit \\\n", + " mass of the fin is %3.2f W/kg'%(Ta,X,Y,q)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temperature distribution is T = 40+257.1(6.0132\u221ax) \n", + " Rate of heat flow per unit mass of the fin is 164.10 W/kg\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.32 Page No : 116" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "t = 0.002;\t\t\t#Thickness of fin in m\n", + "L = 0.015;\t\t\t#Length of fin in m\n", + "k1 = 210.;\t\t\t#Thermal conductivity of aluminium in W/m.K\n", + "h1 = 285.;\t\t\t#Heat transfer coefficient of aluminium in W/m**2.K\n", + "k2 = 40.;\t\t\t#Thermal conductivity of steel in W/m.K\n", + "h2 = 510.;\t\t\t#Heat transfer coefficient of steel in W/m**2.K\n", + "\n", + "# Calculations\n", + "Lc = (L+(t/2));\t\t\t#Corrected length of fin in m\n", + "mLc1 = Lc*math.sqrt((2*h1)/(k1*t));\t\t\t#Calculation of mLc for efficiency\n", + "n1 = math.tanh(mLc1)/mLc1;\t\t\t#Efficiency of fin when aluminium is used\n", + "mLc2 = Lc*math.sqrt((2*h2)/(k2*t));\t\t\t#Calculation of mLc for efficiency\n", + "n2 = math.tanh(mLc2)/mLc2;\t\t\t#Efficiency of fin when steel is used\n", + "\n", + "\n", + "# Results\n", + "print 'Efficiency of fin when aluminium is used is %3.4f \\n \\\n", + "Efficiency of fin when steel is used is %3.3f'%(n1,n2)\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Efficiency of fin when aluminium is used is 0.8983 \n", + " Efficiency of fin when steel is used is 0.524\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.33 Page No : 117" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "k = 200;\t\t\t#Thermal conductivity of aluminium in W/m.K\n", + "t = 0.001;\t\t\t#Thickness of fin in m\n", + "L = 0.015;\t\t\t#Width of fin in m\n", + "D = 0.025;\t\t\t#Diameter of the tube in m\n", + "Tb = 170;\t\t\t#Fin base temperature in degree C\n", + "Ta = 25;\t\t\t#Ambient fluid temperature in degree C\n", + "h = 130;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "\n", + "# Calculations\n", + "Lc = (L+(t/2));\t\t\t#Corrected length of fin in m\n", + "r1 = (D/2);\t\t\t#Radius of tube in m\n", + "r2c = (r1+Lc);\t\t\t#Corrected radius in m\n", + "Am = t*(r2c-r1);\t\t\t#Corrected area in m**2\n", + "x = Lc**(3/2)*math.sqrt(h/(k*Am));\t\t\t#x for calculating efficiency\n", + "n = 0.82;\t\t\t#From fig. 3.18 on page no. 112 efficiency is 0.82\n", + "qmax = (2*3.14*(r2c**2-r1**2)*h*(Tb-Ta));\t\t\t#Maximum heat transfer in W\n", + "qactual = (n*qmax);\t\t\t#Actual heat transfer in W\n", + "\n", + "# Results\n", + "print 'Heat loss per fin is %3.2f W'%(qactual)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat loss per fin is 60.94 W\n" + ] + } + ], + "prompt_number": 51 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.34 Page No : 117" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "k = 16;\t\t\t#Thermal conductivity of fin in W/m.K\n", + "L = 0.1;\t\t\t#Length of fin in m\n", + "D = 0.01;\t\t\t#Diameter of fin in m\n", + "h = 5000;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "\n", + "# Calculations\n", + "P = (3.14*D);\t\t\t#Perimeter of fin in m\n", + "A = (3.14*D**2)/4;\t\t\t#Area of fin in m**2\n", + "m = math.sqrt((h*P)/(k*A));\t\t\t#Calculation of m for determining heat transfer rate\n", + "n = math.tanh(m*L)/math.sqrt((h*A)/(k*P));\t\t\t#Calculation of n for checking whether installation of fin is desirable or not\n", + "x = (n-1)*100;\t\t\t#Conversion into percentage\n", + "\n", + "# Results\n", + "print 'This large fin only produces an increase of %i percent in heat dissipation, \\\n", + " so naturally this configuration is undesirable'%(x)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "This large fin only produces an increase of 13 percent in heat dissipation, so naturally this configuration is undesirable\n" + ] + } + ], + "prompt_number": 52 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.35 Page No : 119" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "k = 55.8;\t\t\t#Thermal conductivity of steel in W/m.K\n", + "t = 0.0015;\t\t\t#Thickness of steel tube in m\n", + "L = 0.12;\t\t\t#Length of steel tube in m\n", + "h = 23.3;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "Tl = 84;\t\t\t#Temperature recorded by the thermometer in degree C\n", + "Tb = 40;\t\t\t#Temperature at the base of the well in degree C\n", + "\n", + "# Calculations\n", + "m = math.sqrt(h/(k*t));\t\t\t#Calculation of m for determining the temperature distribution\n", + "x = 1./math.cosh(m*L);\t\t\t#Calculation of x for determining the temperature distribution\n", + "Ti = ((Tl-(x*Tb))/(1-x));\t\t\t#Temperature distribution in degree C\n", + "T = (Ti-Tl);\t\t\t#Measurement error in degree C\n", + "\n", + "# Results\n", + "print 'Measurement error is %3.0f degree C'%(T)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Measurement error is 16 degree C\n" + ] + } + ], + "prompt_number": 16 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch5.ipynb b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch5.ipynb new file mode 100755 index 00000000..e575c46c --- /dev/null +++ b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch5.ipynb @@ -0,0 +1,1137 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:61a219887cda63633e57a8123bccc97a816d976e2f7d7f9c9b9bf006f781c953" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 5 : Transient Heat Conduction" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.1 Page No : 159" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "\n", + "# Variables\n", + "t = 0.5;\t\t\t#Thickness of slab in m\n", + "A = 5;\t\t\t#Area of slab in m**2\n", + "k = 1.2;\t\t\t#Thermal conductivity in W/m.K\n", + "a = 0.00177;\t\t\t#Thermal diffusivity in m**2/h\n", + "\n", + "# Calculations\n", + "x = 0;\n", + "y = -50+(24*x)+(60*x**2)-(60*x**3);\t\t\t#Temperature when x = 0\n", + "Qo = (-k*A*y);\t\t\t#Heat entering the slab in W \n", + "x = 0.5;\n", + "y = -50+(24*x)+(60*x**2)-(60*x**3);\t\t\t#Temperature when x = 0.5\n", + "QL = (-k*A*y);\t\t\t#Heat leaving the slab in W\n", + "R = (Qo-QL);\t\t\t#Rate of heat storage in W\n", + "x = 0;\n", + "z1 = 24+(120*x)-(180*x**2);\t\t\t#T' when x = 0\n", + "p1 = (a*z1);\t\t\t#Rate of temperature change at one side of slab in degree C/h\n", + "x = 0.5;\n", + "z2 = 24+(120*x)-(180*x**2);\t\t\t#T' when x = 0.5\n", + "p2 = (a*z2);\t\t\t#Rate of temperature change at one side of slab in degree C/h\n", + "\t\t\t#For the rate of heating or cooling to be maximum, T''' = 0\n", + "x = (120./360);\n", + "\n", + "# Results\n", + "print 'a)\\n \\\n", + "i)Heat entering the slab is %i W \\n \\\n", + "ii)Heat leaving the slab is %i W \\n\\n\\\n", + "b)Rate of heat storage is %i Wc \\n \\\n", + "i)Rate of temperature change at one side of slab is %3.4f degree C/h \\n \\\n", + "ii)Rate of temperature change at other side of slab is %3.4f degree C/h \\\n", + "\\n\\nd)For the rate of heating or cooling to be maximum x = %3.2f'%(Qo,QL,R,p1,p2,x)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a)\n", + " i)Heat entering the slab is 300 W \n", + " ii)Heat leaving the slab is 183 W \n", + "\n", + "b)Rate of heat storage is 117 Wc \n", + " i)Rate of temperature change at one side of slab is 0.0425 degree C/h \n", + " ii)Rate of temperature change at other side of slab is 0.0690 degree C/h \n", + "\n", + "d)For the rate of heating or cooling to be maximum x = 0.33\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.2 Page No : 164" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "A = (0.4*0.4);\t\t\t#Area of copper slab in m**2\n", + "t = 0.005;\t\t\t#Thickness of copper slab in m\n", + "T = 250.;\t\t\t#Uniform teperature in degree c\n", + "Ts = 30.;\t\t\t#Surface temperature in degree C\n", + "Tsl = 90.;\t\t\t#Slab temperature in degree C\n", + "p = 9000.;\t\t\t#Density in kg/m**3\n", + "c = 380.;\t\t\t#Specific heat in J/kg.K\n", + "k = 370.;\t\t\t#Thermal conductivity in W/m.K\n", + "h = 90.;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "\n", + "# Calculations\n", + "A1 = (2*A);\t\t\t#Area of two sides in m**2\n", + "V = (A*t);\t\t\t#Volume of the slab in m**3\n", + "Lc = (V/A1);\t\t\t#Corrected length in m\n", + "Bi = ((h*Lc)/k);\t\t\t#Biot number\n", + "t = -math.log((Tsl-Ts)/(T-Ts))/((h*A1)/(p*c*V));\t\t\t#Time at which slab temperature becomes 90 degree C in s\n", + "y = (h*A1)/(p*c*V);\n", + "\n", + "# Results\n", + "print 'Time at which slab temperature becomes 90 degree C is %3.2f s'%(t)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time at which slab temperature becomes 90 degree C is 123.43 s\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.3 Page No : 164" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "D = 0.01;\t\t\t#Outer diameter of the rod in m\n", + "T = 320.;\t\t\t#Original temperature in degree C\n", + "Tl = 120.;\t\t\t#Temperature of liquid in degree C\n", + "h = 100.;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "Tf = 200.;\t\t\t#Final temperature of rod in degree C\n", + "k = 40.;\t\t\t#Thermal conductivity in W/m.K\n", + "c = 460.;\t\t\t#Specific heat in J/kg.K\n", + "p = 7800.;\t\t\t#Density in kg/m**3\n", + "\n", + "# Calculations\n", + "V = (3.14*D**2*1)/4;\t\t#Volume of rod in m**3 taking 1m length\n", + "A = (3.14*D*1);\t\t \t#Surface area of rod in m**2 taking 1m length\n", + "Lc = (D/4);\t\t\t #Corrected length in m\n", + "Bi = ((h*Lc)/k);\t\t\t#Biot number\n", + "t = -math.log((Tf-Tl)/(T-Tl))/((h*4)/(p*c*D));\t\t\t#Time at which rod temperature becomes 200 degree C in s\n", + "\n", + "# Results\n", + "print 'Time at which rod temperature becomes 200 degree C is %3.2f s'%(t)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time at which rod temperature becomes 200 degree C is 82.19 s\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.4 Page No : 165" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "w = 5.5;\t\t\t#Weight of the sphere in kg\n", + "Ti = 290.;\t\t\t#Initial temperature in degree C\n", + "Tl = 15.;\t\t\t#Temperature of liquid in degree C\n", + "h = 58.;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "Tf = 95.;\t\t\t#Final temperature in degree C\n", + "k = 205.;\t\t\t#Thermal conductivity in W/m.K\n", + "c = 900.;\t\t\t#Specific heat in J/kg.K\n", + "p = 2700.;\t\t\t#Density in kg/m**3\n", + "\n", + "# Calculations\n", + "V = (w/p);\t\t\t#Volume of the sphere in m**3\n", + "R = ((3*V)/(4*3.14))**(1./3);\t\t\t#Radius of sphere in m\n", + "Lc = (R/3.);\t\t\t#Corrected length in m\n", + "t = -math.log((Tf-Tl)/(Ti-Tl))/((h*3)/(p*c*R));\t\t\t#Time at which rod temperature becomes 95 degree C in s\n", + "\n", + "# Results\n", + "print 'Time at which rod temperature becomes 95 degree C is %3.0f s'%(t)\n", + "\n", + "\n", + "# note : answer is slightly different because of rouding off error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time at which rod temperature becomes 95 degree C is 1356 s\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.5 Page No : 166" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "Ti = 100.;\t\t\t#Temperature of air in degree C\n", + "t = 0.03;\t\t\t#Thickness of slab in m\n", + "To = 210.;\t\t\t#Initial temperature of the plate in degree C\n", + "t = 300.;\t\t\t#Time for attaining temperature in s\n", + "T = 170.;\t\t\t#Temperature decreased in degree C\n", + "c = 380.;\t\t\t#Specific heat in J/kg.K\n", + "p = 9000.;\t\t\t#Density in kg/m**3\n", + "\n", + "# Calculations\n", + "Lc = (t/2);\t\t\t#Corrected length in m\n", + "h = -math.log((T-Ti)/(To-Ti))/((t*10**4)/(p*c*Lc));\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "\n", + "# Results\n", + "print 'Heat transfer coefficient is %3.2f W/m**2.K'%(h)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer coefficient is 77.29 W/m**2.K\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.6 Page No : 167" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "D = 0.00071;\t\t\t#Diameter of thermocouple in m\n", + "h = 600.;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "c = 420.;\t\t\t#Specific heat in J/kg.K\n", + "p = 8600.;\t\t\t#Density in kg/m**3\n", + "\n", + "# Calculations\n", + "t = (p*c*D)/(4*h);\t\t\t#Time period in s\n", + "T = math.exp(-1);\t\t\t#Temperture distribution ratio\n", + "t1 = (4*t);\t\t\t#Total time in s\n", + "\n", + "# Results\n", + "print 'At the end of time period t* = %3.3f s the temperature difference \\\n", + " between the body and the source would be %3.3f of the initial temperature differnce.\\n\\\n", + " To get a true reading of gas temperature, it should be recorded after 4t* = %i seconds after\\\n", + " the thermocouple has been \\nintroduced into the stream'%(t,T,t1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "At the end of time period t* = 1.069 s the temperature difference between the body and the source would be 0.368 of the initial temperature differnce.\n", + " To get a true reading of gas temperature, it should be recorded after 4t* = 4 seconds after the thermocouple has been \n", + "introduced into the stream\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.8 Page No : 177" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "x = 0.2;\t\t\t#Distance of plane from the wall in m\n", + "t = 10;\t\t\t#Time for heat flow in h\n", + "T = [25,800];\t\t\t#Initial and final tempertaure in degree C\n", + "k = 0.8;\t\t\t#Thermal conductivity in W/m.K\n", + "a = 0.003;\t\t\t#Thermal diffusivity in m**2/h\n", + "\n", + "# Calculations\n", + "X = (x*(2*math.sqrt(a*t)));\t\t\t#Calculation of X for erf function\n", + "Y = 0.585 #erf(X) from table 5.1\n", + "Ti = T[1]-((T[1]-T[0]))*Y;\t\t\t#Temperarture of the plane in degree C\n", + "Qi = ((-k*(T[0]-T[1])*math.exp(-x**2/(4*a*t)))/(math.sqrt(3.14*a*t)));\t\t\t#Instanteneous heat flow rate per unit area in W/m**2\n", + "Q = ((2*k*(T[1]-T[0])*3600)/(math.sqrt((3.14*a)/t)))/10**8;\t\t\t #Total heat energy taken up by the wall in 10 hours in J/m**2\n", + "print \n", + "# Results\n", + "print 'Temperarture of the plane is %3.2f degree C\\nInstanteneous heat flow rate per \\\n", + " unit area is %i W/m**2 \\nTotal heat energy taken up by the wall in 10 hours is %3.3f*10**8 J/m**2'%(Ti,Qi,Q)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "Temperarture of the plane is 346.62 degree C\n", + "Instanteneous heat flow rate per unit area is 1447 W/m**2 \n", + "Total heat energy taken up by the wall in 10 hours is 1.454*10**8 J/m**2\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.9 Page No : 177" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Tc = 55;\t\t\t#Tempertaure of the concrete in degree C\n", + "Ts = 35;\t\t\t#Temperature lowered in degree C\n", + "Tf = 45;\t\t\t#Final temperature in degree C\n", + "x = 0.05;\t\t\t#Depth of the slab in m\n", + "k = 1.279;\t\t\t#Thermal conductivity in W/m.K\n", + "a = 0.00177;\t\t\t#Thermal diffusivity in m**2/h\n", + "\n", + "# Calculations\n", + "T = (Tf-Ts)/(Tc-Ts);\t\t\t#Temperature distribution\n", + "X = 0.485;\t\t\t#Taking 0.5 = erf(0.482) from table 5.1 on page no. 175\n", + "t = (x**2)/(4*X**2*a);\t\t\t#Time taken to cool the concrete to 45 degree C in h\n", + "\n", + "# Results\n", + "print 'Time taken to cool the concrete to 45 degree C is %3.2f h'%(t)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time taken to cool the concrete to 45 degree C is 1.50 h\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.10 Page No : 178" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "q = (0.3*10**6);\t\t\t#Heat flux in W/m**2\n", + "t = (10./60);\t\t\t#Time taken for heat transfer in s\n", + "Ti = 30.;\t\t\t#Initial temperature of the slab in degree C\n", + "x = 0.2;\t\t\t#Distance of the plane from the surface in m\n", + "k = 386.;\t\t\t#Thermal conductivity in W/m.K\n", + "a = 0.404;\t\t\t#Thermal diffusivity in m**2/h\n", + "\n", + "# Calculations\n", + "Ts = ((q*math.sqrt(3.14*a*t))/k)+Ti;\t\t\t#Surface temperature in degree C\n", + "X = (x/(2*math.sqrt(a*t)));\t\t\t#X for calculating erf function\n", + "Y = 0.4134;\t\t\t#Taking ref(0.385) = 0.4134 from table 5.1 on page no. 175\n", + "T = Ts-(Y*(Ts-Ti));\t\t\t#Tempertaure at a distance of 20 cm from the surface after 10 min in degree C\n", + "\n", + "# Results\n", + "print 'Tempertaure at a distance of 20 cm from the surface after 10 min is %3.2f degree C'%(T)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Tempertaure at a distance of 20 cm from the surface after 10 min is 239.63 degree C\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.11 Page No : 178" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "a = 0.405;\t\t\t#Thermal diffusivity in m**2/h\n", + "Ti = 100;\t\t\t#Initial temperture in degree C\n", + "Tf = 0;\t\t\t#Final tempertaure in degree C\n", + "Tg = (4*100);\t#Temperature gradient in degree C/m\n", + "t1 = 1;\t\t\t#Time taken in m\n", + "\n", + "# Calculations\n", + "t = (Ti-Tf)**2/(Tg**2*3.14*a);\t\t\t#Time required for the temperature gradient at the surface to reach 4 degree/cm in h\n", + "x = math.sqrt(2*a*(t1/60.));\t\t\t#The depth at which the rate of cooling is maximum after 1 minute in m\n", + "\n", + "# Results\n", + "print 'Time required for the temperature gradient at the surface to reach 4 degree/cm is %3.3f h \\\n", + "\\nThe depth at which the rate of cooling is maximum after 1 minute is %3.4f m'%(t,x)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time required for the temperature gradient at the surface to reach 4 degree/cm is 0.049 h \n", + "The depth at which the rate of cooling is maximum after 1 minute is 0.1162 m\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.12 Page No : 185" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "x = 0.1;\t\t\t#Thickness of the slab in m\n", + "Ti = 500;\t\t\t#Initial temperature in degree C\n", + "Tl = 100;\t\t\t#Liquid temperature in degree C\n", + "h = 1200;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "t = (1*60);\t\t\t#Time for immersion in s\n", + "k = 215;\t\t\t#Thermal conductivity in W/m.K\n", + "a = (8.4*10**-5);\t\t\t#Thermal diffusivity in m**2/h\n", + "c = 900;\t\t\t#Specific heat in J/kg/K\n", + "p = 2700;\t\t\t#Density in kg/m**3\n", + "\n", + "# Calculations\n", + "X = (a*t)/(x/2)**2;\t\t\t#Calculation for input in Heisler charts\n", + "B = (k/(h*(x/2)));\t\t\t#Calculation for input in Heisler charts\n", + "T1 = 0.68;\t\t\t#T value taken from Fig. 5.7 on page no. 183\n", + "Tc1 = (T1*(Ti-Tl));\t\t\t#Temperature in degree C\n", + "To = Tc1+Tl;\t\t\t#Temperature in degree C\n", + "T2 = 0.880;\t\t\t#From Fig 5.8 on page no. 184 at x/L = 1.0 and for k/hL = 3.583, tempertaure in degree C\n", + "Tc2 = (T2*(To-Tl))+Tl;\t\t\t#Temperature in degree C \n", + "Y = (h**2*a*t)/(k**2);\t\t\t#Y to calculate the energy losses\n", + "Bi = (h*(x/2))/k;\t\t\t#Biot number\n", + "U = 0.32;\t\t\t#U/Uo value from Fig. 5.9 on page no.185 \n", + "Uo = (p*c*x*(Ti-Tl));\t\t\t#For unit area in J/m**2\n", + "U1 = (U*Uo)/(10**6);\t\t\t#Heat removed per unit surface area in MJ/m**2\n", + "\n", + "# Results\n", + "print 'Temperature at the centreline and the surface 1 minute after the immersion is %3.2f degree C \\n \\\n", + "Heat removed per unit surface area is %3.1f*10**6 J/m**2'%(Tc2,U1)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temperature at the centreline and the surface 1 minute after the immersion is 339.36 degree C \n", + " Heat removed per unit surface area is 31.1*10**6 J/m**2\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.13 Page No : 186" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "D = 0.12;\t\t\t#Diameter of cylinder in m\n", + "Ti = 20;\t\t\t#Initial temperature in degree C\n", + "Tf = 820;\t\t\t#Temperature of furnace in degree C\n", + "h = 140;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "Ta = 800;\t\t\t#Axis temperature in degree C\n", + "r = 0.054;\t\t\t#Radius in m\n", + "k = 21;\t\t\t#Thermal conductivity in W/m.K\n", + "a = (6.11*10**-6);\t\t\t#Thermal diffusivity in m**2/h\n", + "\n", + "# Calculations\n", + "Bi = (h*(D/2))/(2*k);\t\t\t#Biot number\n", + "T = (Ta-Tf)/(Ti-Tf);\t\t\t#Temperature distribution\n", + "Fo = 5.2;\t\t\t#Umath.sing Fig.5.10, on page no.187 for 1./(2Bi) = 2.5 \n", + "t = (Fo*(D/2)**2)/a;\t\t\t#Time required for the axis temperature to reach 800 degree C in s\n", + "r1 = (r/(D/2));\t\t\t#Ratio at a radius of 5.4 cm\n", + "X = 0.85;\t\t\t#From Fig.5.11 on page no. 188 the temperature at r = 5.4 i sgiven by X\n", + "T1 = X*(Ta-Tf)+Tf;\t\t\t#Temperature at a radius of 5.4 cm at that tim ein degree C\n", + "\n", + "# Results\n", + "print 'Time required for the axis temperature to reach 800 degree C is %3.0f s \\n \\\n", + "Temperature at a radius of 5.4 cm at that time is %i degree C'%(t,T1)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time required for the axis temperature to reach 800 degree C is 3064 s \n", + " Temperature at a radius of 5.4 cm at that time is 803 degree C\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.14 Page No : 189" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "import math\n", + "# Variables\n", + "r = 0.01;\t\t\t#Radius of the mettalic sphere in m\n", + "Ti = 400.;\t\t\t#Initial temperature in degree C\n", + "h1 = 10.;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "Ta = 20.;\t\t\t#Temperature of air in degree C\n", + "Tc = 335.;\t\t\t#Central temperature in degree C\n", + "Tw = 20.;\t\t\t#Temperature of water bath in degree C\n", + "h2 = 6000.;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "Tf = 50.;\t\t\t#Final temperature of the sphere in degree C\n", + "k = 20.;\t\t\t#Thermal conductivity in W/m.K\n", + "a = (6.66*10**-6);\t\t\t#Thermal diffusivity in m**2/h\n", + "c = 1000.;\t\t\t#Specific heat in J/kg/K\n", + "p = 3000.;\t\t\t#Density in kg/m**3\n", + "\n", + "# Calculations\n", + "Bi1 = (h1*r)/(3*k);\t\t\t#Biot number\n", + "t = ((p*r*c)/(3*h1)*math.log((Ti-Ta)/(Tc-Ta)))\t\t\t#Time required for cooling in air in s\n", + "Bi2 = (h2*r)/(3*k);\t\t\t#Biot number\n", + "X = 1./(3*Bi2);\t\t\t#X value for lumped capacity method\n", + "T = (Tf-Ta)/(Tc-Ta);\t\t\t#Temperature distribution\n", + "Fo = 0.5;\t\t\t#Umath.sing Fig.5.13, on page no.190\n", + "t1 = (Fo*r**2)/a;\t\t\t#Time required for cooling in water in s\n", + "Z = 0.33;\t\t\t#Umath.sing Fig.5.14, on page no.191\n", + "Tr = Z*(Tf-Ta)+Ta;\t\t\t#Surface temperature at the end of cooling in degree C\n", + "\n", + "\n", + "# Results\n", + "print 'Time required for cooling in air is %3.0f s \\n \\\n", + "Time required for cooling in water is %3.1f s \\n \\\n", + "Surface temperature at the end of cooling is %3.0f degree C'%(t,t1,Tr)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time required for cooling in air is 188 s \n", + " Time required for cooling in water is 7.5 s \n", + " Surface temperature at the end of cooling is 30 degree C\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.15 Page No : 192" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# Variables\n", + "Ti = 250.;\t\t\t#Temperature of aluminium slab in degree C\n", + "Tc = 50.;\t\t\t#Convective environment temperature in degree C\n", + "h = 500.;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "x = 0.05;\t\t\t#Depth of the plane in m\n", + "t = (1.*3600);\t\t\t#Time in s\n", + "k = 215.;\t\t\t#Thermal conductivity in W/m.K\n", + "a = (8.4*10**-5);\t\t\t#Thermal diffusivity in m**2/h\n", + "\n", + "# Calculations\n", + "X = (h*math.sqrt(a*t))/k;\t\t\t#X for calculating Temperature\n", + "Y = (x/(2*math.sqrt(a*t)));\t\t\t#Y for calculating Temperature\n", + "Z = 0.62;\t\t\t#From Fig. 5.16 on page no.193\n", + "T = (Z*(Tc-Ti)+Ti);\t\t\t#Temperature at a depth of 5 cm after 1 hour in degree C\n", + "\n", + "# Results\n", + "print 'Temperature at a depth of 5 cm after 1 hour is %3.0f degree C'%(T)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temperature at a depth of 5 cm after 1 hour is 126 degree C\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.16 Page No : 196" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "D = 0.08;\t\t\t#Diameter of the cylinder in m\n", + "L = 0.16;\t\t\t#Length of the cylinder in m\n", + "Ti = 800;\t\t\t#Initial tempertaure in degree C\n", + "Tm = 30;\t\t\t#Temperature of the medium in degree C\n", + "h = 120;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "t = (30*60);\t\t\t#Time for cooling in s\n", + "k = 23.5;\t\t\t#Thermal conductivity in W/m.K\n", + "a = 0.022;\t\t\t#Thermal diffusivity in m**2/h\n", + "\n", + "# Calculations\n", + "Bi2 = (h*(D/2))/k;\t\t\t#2 times the Biot number\n", + "X = (a*t)/(D/2)**2;\t\t\t#X for calculating C(R)\n", + "CR = 0.068;\t\t\t#From Fig.5.10 on page no.187\n", + "Bi1 = (k/(h*L));\t\t\t#Biot number\n", + "Y = (a*t)/L**2;\t\t\t#Y for calculating P(X)\n", + "PX = 0.54;\t\t\t#From Fig.5.7 on page no.183\n", + "T = CR*PX;\t\t\t#Temperature at the centre of the cylinder in degree C\n", + "T30 = T*(Ti-Tm)+Tm;\t\t\t#Temperature at the centre of cylinder 30 minutes after cooling is initiated in degree C\n", + "\n", + "# Results\n", + "print 'Temperature at the centre of cylinder 30 minutes after cooling is initiated is %3.2f degree C'%(T30)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temperature at the centre of cylinder 30 minutes after cooling is initiated is 58.27 degree C\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.17 Page No : 197" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "from numpy import *\n", + "\n", + "\n", + "# Variables\n", + "L = array([0.5,0.4,0.2]);\t\t\t#Lengths of sides of a recmath.tangular steel billet in m\n", + "Ti = 30;\t\t\t#Initial temperature in degree C\n", + "Tf = 1000;\t\t\t#Final temperature in degree C\n", + "t = (90*60);\t\t\t#Time for heating in s\n", + "h = 185;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "k = 37;\t\t\t#Thermal conductivity in W/m.K\n", + "a = 0.025;\t\t\t#Thermal diffusivity in m**2/h\n", + "\n", + "# Calculations\n", + "L1 = L/2;\t\t\t#L values of the parallelepiped in m\n", + "Bi1 = (h*L[0])/k;\t\t\t#Biot number\n", + "X1 = (a*t)/L[0]**2;\t\t\t#X1 for calculating P(X1)\n", + "PX1 = 0.68;\t\t\t#P(X1) value from From Fig.5.7 on page no.183\n", + "Bi2 = (h*L[1])/k;\t\t\t#Biot number\n", + "X1 = (a*t)/L[1]**2;\t\t\t#X1 for calculating P(X2)\n", + "PX2 = 0.57;\t\t\t#P(X2) value from From Fig.5.7 on page no.183\n", + "Bi3 = (h*L[2])/k;\t\t\t#Biot number\n", + "Y = (1./Bi3);\t\t\t#Inverse of Biot number\n", + "X1 = (a*t)/L[2]**2;\t\t\t#X1 for calculating P(X3)\n", + "PX3 = 0.22;\t\t\t#P(X3) value from From Fig.5.7 on page no.183\n", + "T = PX1*PX2*PX3;\t\t\t#Temperature at the centre of billet in degree C\n", + "T1 = T*(Ti-Tf)+Tf;\t\t\t#Temperature at the centre of cylinder 90 minutes after heating is initiated in degree C\n", + "\n", + "# Results\n", + "print 'Temperature at the centre of cylinder 90 minutes after heating is initiated is %3.2f degree C'%(T1)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temperature at the centre of cylinder 90 minutes after heating is initiated is 917.29 degree C\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.18 Page No : 202" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Ti = 30;\t\t\t#Initial temperature of the slab in degree C\n", + "q = (2*10**5);\t\t\t#Constant heat flux in W/m**2\n", + "k = 400;\t\t\t#Thermal conductivity in W/m.K\n", + "a = (117*10**-6);\t\t\t#Thermal diffusivity in m**2/h\n", + "n = 0.075;\t\t\t#Nodal spacing in m\n", + "x = 0.15;\t\t\t#Depth in m\n", + "t = (4*60);\t\t\t#Time elapsed in s\n", + "\n", + "#CALCULATION\n", + "R = (x**2/(a*t));\t\t\t#R value for t1\n", + "t1 = (n**2/(R*a));\t\t\t#Value of t1 in s\n", + "To = 121.9;\t\t\t#The surface temperature after 4 min in degree C from the table on page no. 203\n", + "T2 = 64;\t\t\t#Temperature at 0.15 m from the surface after 4 minutes in degree C from the table on page no. 203\n", + "\n", + "# Results\n", + "print 'The surface temperature after 4 min is %3.1f degree C \\n \\\n", + "Temperature at 0.15 m from the surface after 4 minutes is %i degree C'%(To,T2)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The surface temperature after 4 min is 121.9 degree C \n", + " Temperature at 0.15 m from the surface after 4 minutes is 64 degree C\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.19 Page No : 205" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "t = 0.6;\t\t\t#Thickness of the wall in m\n", + "x = 0.1;\t\t\t#x value taken from Fig.Ex. 5.19 on page no. 205\n", + "Ti = 20;\t\t\t#Initial temperature in degree C\n", + "T = [150,300];\t\t\t#Temperatures of the sides of the wall in degree C\n", + "Tf = 150;\t\t\t#Final temperature of the wall in degree C\n", + "a = (1.66*10**-3);\t\t\t#Thermal diffusivity in m**2/h\n", + "\n", + "# Calculations\n", + "t = (x**2/(2*a));\t\t\t#Length of one time increment in h\n", + "t1 = (9*t);\t \t\t#Elapsed time in h\n", + "\n", + "# Results\n", + "print 'Elasped time before the centre of the wall attains a temperature of 150 degree C is %3.0f h'%(t1)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Elasped time before the centre of the wall attains a temperature of 150 degree C is 27 h\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.20 Page No : 206" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "k = 0.175;\t\t\t#Thermal conductivity in W/m.K\n", + "a = (0.833*10**-7);\t\t\t#Thermal diffusivity in m**2/h\n", + "Th = 144;\t\t\t#Heated temeparture in degree C\n", + "Tc = 15;\t\t\t#Cooled temperature in degree C\n", + "x = 0.02;\t\t\t#Thickness of the plate in m\n", + "h = 65;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "t = (4*60);\t\t\t#Tiem elapsed in s\n", + "\n", + "# Calculations\n", + "s = 0.002;\t\t\t#Space increment in m from FIg. Ex. 5.20 on page no. 207\n", + "t1 = (s**2/(2*a));\t\t\t#Time increment for the space increment in s\n", + "x1 = (k/h);\t\t\t#Convective film thickness in mm\n", + "Tn = 114;\t\t\t#Temperature at the centre in degree C from Fig. Ex.5.20 on page no. 207\n", + "Ts = 50;\t\t\t#Surface temperature in degree C from Fig. Ex.5.20 on page no. 207\n", + "\n", + "# Results\n", + "print 'Temperature at the centre is %i degree C \\n \\\n", + "Surface temperature is %i degree C'%(Tn,Ts)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temperature at the centre is 114 degree C \n", + " Surface temperature is 50 degree C\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.21 Page No : 213" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# Variables\n", + "t = 24;\t\t\t#Time period in h\n", + "T = [-10,10];\t\t\t#Range of temperatures in degree C\n", + "x = 0.1;\t\t\t#Depth in m\n", + "c = 1970;\t\t\t#Specific heat in J/kg/K\n", + "p = 1000;\t\t\t#Density in kg/m**3\n", + "k = 0.349;\t\t\t#Thermal conductivity in W/m.K\n", + "ta = 5;\t\t\t#Time in h\n", + "\n", + "# Calculations\n", + "w = (2*3.14)/t;\t\t\t#Angular velocity in rad/h\n", + "Tm = (T[0]+T[1])/2;\t\t\t#Mean teperature in degree C\n", + "Tmax = T[1]-Tm;\t\t\t#Maximum temperature in degree C\n", + "a = ((k*3600)/(p*c));\t\t\t#Thermal diffusivity in m**2/h\n", + "Txmax = Tmax*exp(-math.sqrt(w/(2*a))*x);\t\t\t#Amplitude of temperature variation in degree C\n", + "t1 = math.sqrt(1./(2*a*w))*x;\t\t\t#Time lag of temperature wave at a depth of 0.1 m in h\n", + "t2 = (3.14/w);\t\t\t#Time for surface temperature is minimum in h\n", + "t3 = t2+ta;\t\t\t#Time in h\n", + "Tx = Tmax*exp(-math.sqrt(w/(2*a))*x)*math.cos((w*t3)-(x*x*math.sqrt(w/(2*a))));\t\t\t#Temperature at 0.1m 5 hours after the surface temperature reaches the minimum in degree C\n", + "\n", + "# Results\n", + "print 'Amplitude of temperature variation at a depth of 0.1m is %3.2f degree C \\n \\\n", + "Time lag of temperature wave at a depth of 0.1 m is %3.2f h \\n \\\n", + "Temperature at 0.1m 5 hours after the surface temperature reaches the minimum is %3.3f degree C'%(Txmax,t1,Tx)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Amplitude of temperature variation at a depth of 0.1m is 2.39 degree C \n", + " Time lag of temperature wave at a depth of 0.1 m is 5.47 h \n", + " Temperature at 0.1m 5 hours after the surface temperature reaches the minimum is -0.946 degree C\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.22 Page No : 214" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables\n", + "T = [800.,200.];\t\t\t#Limits in which temperature varies in degree C\n", + "t = 12.;\t\t\t#Cycle time in h\n", + "x = 0.1;\t\t\t#Depth of penetration in m\n", + "k = 1.8;\t\t\t#Thermal conductivity in W/m.K\n", + "a = 0.02;\t\t\t#Thermal diffusivity in m**2/h\n", + "\n", + "# Calculations\n", + "w = (2*3.14)/t;\t\t\t#Angular velocity in rad/h\n", + "t1 = math.sqrt(1./(2*a*w))*x;\t\t\t#Time lag in h\n", + "Tmax = (T[0]-T[1])/2;\t\t\t#Range of maximum temperature in degree C\n", + "q = ((2*k*Tmax)/math.sqrt(math.pi/6*a))*(3600./1000);\t\t\t#Heat flow through the surface in kJ/m**2\n", + "\n", + "# Results\n", + "print 'i)Time lag of the temperature wave at a depth of 10 cm from the inner surface is %3.2f h \\n \\\n", + "ii)The flow through a surface located at a distance of 10 cm from the surface during the first\\\n", + " six hours interval while the temperature is above the mean value is %i kJ/m**2'%(t1,q) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)Time lag of the temperature wave at a depth of 10 cm from the inner surface is 0.69 h \n", + " ii)The flow through a surface located at a distance of 10 cm from the surface during the first six hours interval while the temperature is above the mean value is 37993 kJ/m**2\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.23 Page No : 215" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "N = 2000;\t\t\t#Speed of the engine\n", + "a = 0.06;\t\t\t#Thermal diffusivity in m**2/h\n", + "\n", + "# Calculations\n", + "t = 1./(60*N);\t\t\t#Period of on oscillation in h\n", + "x = (1.6*math.sqrt(3.14*a*t))*1000;\t\t\t#Depth of penetration in mm\n", + "\n", + "# Results\n", + "print 'Depth of penetration of the temperature oscillation into the cylinder wall of a \\\n", + "single acting cylinder two stroke IC engine is%3.0f mm'%(x)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Depth of penetration of the temperature oscillation into the cylinder wall of a single acting cylinder two stroke IC engine is 2 mm\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.24 Page No : 218" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# Variables\n", + "Tc = 55;\t\t\t#Tempaerature of concrete hyway in degree C\n", + "Tl = 35;\t\t\t#Temperature lowered in degree C\n", + "Tf = 45;\t\t\t#Final temperature in degree C\n", + "x = 0.05;\t\t\t#Depth in m\n", + "k = 1.279;\t\t\t#Thermal conductivity in W/m.K\n", + "a = (1.77*10**-3);\t\t\t#Thermal diffusivity in m**2/h\n", + "\n", + "# Calculations\n", + "t = 1.4;\t\t\t#Time taken from page no. 219 in h\n", + "q = 2*(k*(Tl-Tf))/(math.sqrt(3*a*t));\t\t\t#Instantaneous heat removal rate in W/m**2\n", + "\n", + "# Results\n", + "print 'Instantaneous heat removal rate is %3.1f W/m**2'%(q)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Instantaneous heat removal rate is -296.7 W/m**2\n" + ] + } + ], + "prompt_number": 30 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch6.ipynb b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch6.ipynb new file mode 100755 index 00000000..e94dc654 --- /dev/null +++ b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch6.ipynb @@ -0,0 +1,244 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:ffaa9591022d13ef65a41826984b020004775f9f4dff2ea59062ac5a8175fc23" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 6 : Fundamentals of convective heat transfer" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.2 Page No : 241" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "\n", + "# Variables\n", + "L = 1;\t\t\t#Length of the palte in m\n", + "W = 1;\t\t\t#Width of the plate in m\n", + "v = 2.5;\t\t\t#Velocity of air in m/s\n", + "Re = (5*10**5);\t\t\t#Reynolds number at the transition from laminar to turbulant\n", + "p = (0.85*10**-5);\t\t\t#Dynamic vismath.cosity in N.s/m**2\n", + "r = 1.12;\t\t\t#Density in kg/m**3\n", + "\n", + "# Calculations\n", + "x = (p*Re)/(r*v);\t\t\t#Calculated length in m\n", + "\n", + "# Results\n", + "print 'The actual length of the plate is %i m, which is less than %3.2f m'%(L,x)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The actual length of the plate is 1 m, which is less than 1.52 m\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6 Page No : 247" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "p = 0.8;\t\t\t#Dynamic viscosity in N.s/m**2\n", + "k = 0.15;\t\t\t#Thermal conductivity in W/m.K\n", + "Tb = 10;\t\t\t#Temperature of bearing in degree C\n", + "Ts = 30;\t\t\t#Temperature of the shaft in degree C\n", + "C = 0.002;\t\t\t#Clearance between bearig and shaft in m\n", + "U = 6;\t\t\t#Velocity in m/s\n", + "\n", + "# Calculations\n", + "qb = (((-p*U**2)/(2*C))-((k/C)*(Ts-Tb)))/1000;\t\t\t#Surface heat flux at the bearing in kW/m**2\n", + "qs = (((p*U**2)/(2*C))-((k/C)*(Ts-Tb)))/1000;\t\t\t#Surface heat flux at the shaft in kW/m**2\n", + "Tmax = Tb+(((p*U**2)/(2*k))*(0.604-0.604**2))+((Ts-Tb)*0.604);\t\t\t#Maximum temperature in degree C occurs when ymax = 0.604L\n", + "\n", + "# Results\n", + "print 'Maximum temperature rise is %3.3f degree C \\n \\\n", + "Heat fux to the bearing is %3.1f kW/m**2 \\n \\\n", + "Heat fux to the shaft is %3.1f kW/m**2'%(Tmax,qb,qs)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum temperature rise is 45.042 degree C \n", + " Heat fux to the bearing is -8.7 kW/m**2 \n", + " Heat fux to the shaft is 5.7 kW/m**2\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.7 Page No : 257" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "D = 0.02;\t\t\t#I.D of the tube in m\n", + "Q = 1.5;\t\t\t#Flow rate in litres per minute\n", + "k = (1*10**-6);\t\t\t#kinematic vismath.cosity in m**2/s\n", + "\n", + "# Calculations\n", + "um = ((Q/60)*10**-3)/(3.14*(D**2/4));\t\t\t#Average velocity in m/s\n", + "Re = (um*D)/k;\t\t\t#Reynolds number\n", + "x = 0.05*D*Re;\t\t\t#Entry length in m\n", + "\n", + "# Results\n", + "print 'Re which is %3.0f less than 2300, the flow is laminar. \\n \\\n", + "Entry length is %3.3f m'%(Re,x)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Re which is 1592 less than 2300, the flow is laminar. \n", + " Entry length is 1.592 m\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.8 Page No : 257" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "L = 3000;\t\t\t#Distance transported in m\n", + "D = 0.02;\t\t\t#I.D of the tube in m\n", + "Q = 1.5;\t\t\t#Flow rate in litres per minute\n", + "k = (1*10**-6);\t\t\t#kinematic vismath.cosity in m**2/s\n", + "pw = 1000;\t\t\t#Density of water in kg/m**3\n", + "\n", + "# Calculations\n", + "um = ((Q/60)*10**-3)/(3.14*(D**2/4));\t\t\t#Average velocity in m/s\n", + "Re = (um*D)/k;\t\t\t#Reynolds number\n", + "x = 0.05*D*Re;\t\t\t#Entry length in m\n", + "hL = ((64./Re)*L*um**2)/(2*D*9.81)\t\t\t#Head loss in m\n", + "P = (pw*9.81*(3.14/4)*D**2*um*hL);\t\t\t#Power required to maintain this flow rate in W\n", + "\n", + "# Results\n", + "print 'Head loss is %3.2f m \\n \\\n", + "Power required to maintain this flow rate is %3.4f W'%(hL,P)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Head loss is 1.95 m \n", + " Power required to maintain this flow rate is 0.4777 W\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.9 Page No : 258" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "L = 100;\t\t\t#Length of rectangular duct in m\n", + "A = [0.02,0.025];\t\t\t#Area of duct in m**2\n", + "Tw = 40;\t\t\t#Temperature of water in degree C\n", + "v = 0.5;\t\t\t#Velocity of flow in m/s\n", + "k = (0.66*10**-6);\t\t\t#kinematic viscosity in m**2/s\n", + "p = 995;\t\t\t#Density of water in kg/m**3\n", + "\n", + "# Calculations\n", + "P = 2*(A[0]+A[1]);\t\t\t#Perimeter of the duct in m\n", + "Dh = (4*(A[0]*A[1]))/P\t\t\t#Hydraulic diameter of the duct in m\n", + "Re = (v*Dh)/k;\t\t\t#Reynolds number\n", + "f = 0.316*Re**(-0.25);\t\t\t#Friction factor \n", + "hL = (f*L*v**2)/(2*Dh*9.81);\t\t\t#Head loss in m\n", + "P = (hL*9.81*p)/10**4;\t\t\t#Pressure drop in smooth rectangular duct in 10**4 N/m**2\n", + "\n", + "# Results\n", + "print 'Pressure drop in smooth rectangular duct is %3.4f*10**4 N/m**2'%(P)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Pressure drop in smooth rectangular duct is 1.5527*10**4 N/m**2\n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch7.ipynb b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch7.ipynb new file mode 100755 index 00000000..2aa36e2e --- /dev/null +++ b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch7.ipynb @@ -0,0 +1,1284 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:9c4fb6b8e822716d4e29e95db1a4da7b71b979535df0e47c4729b7b08b5165ec" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 7 : Forced Convection Systems" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.1 Page No : 275" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables\n", + "Ta = 20;\t\t\t#Temperature of air in degree C\n", + "Tp = 134;\t\t\t#Temperature of heated plate in degree C\n", + "v = 3;\t\t\t#Velocity of flow in m/s\n", + "L = 2;\t\t\t#Length of plate in m\n", + "W = 1.5;\t\t\t#Width of plate in m\n", + "x = 0.4;\t\t\t#Distance of plane from the plate in m\n", + "k = (15.06*10**-6);\t\t\t#Kinematic vismath.cosity in m**2/s\n", + "\n", + "# Calculations\n", + "Re = (v*x)/k;\t\t\t#Reynold number\n", + "q = ((5*x)/math.sqrt(Re))*1000;\t\t\t#Thickness of boundary layer in mm\n", + "Cfx = (0.664/math.sqrt(Re))/10**-3;\t\t\t#Local skin friction coefficient *10**-3\n", + "\n", + "# Results\n", + "print 'Thickness of boundary layer is %3.1f mm Local skin friction coefficient is %3.2f*10**-3'%(q,Cfx)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thickness of boundary layer is 7.1 mm Local skin friction coefficient is 2.35*10**-3\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.2 Page No : 275" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Ta = 20;\t\t\t#Temperature of air in degree C\n", + "Tp = 134;\t\t\t#Temperature of heated plate in degree C\n", + "v = 3;\t\t\t#Velocity of flow in m/s\n", + "L = 2;\t\t\t#Length of plate in m\n", + "W = 1.5;\t\t\t#Width of plate in m\n", + "x = 0.4;\t\t\t#Distance of plane from the plate in m\n", + "k = (15.06*10**-6);\t\t\t#Kinematic vismath.cosity in m**2/s\n", + "\n", + "# Calculations\n", + "Tf = (Ta+Tp)/2;\t\t\t#Film temperature in degree C\n", + "pw = 0.998;\t\t\t#Density of air at 77 degree C\n", + "Cp = 1009;\t\t\t#Specific heat of air at 77 degree C\n", + "kw = (20.76*10**-6);\t\t\t#Kinematic viscosity of air at 77 degree C\n", + "k = 0.03;\t\t\t#Thermal conductivity of air at 77 degree C\n", + "Pr = 0.697;\t\t\t#prantl number of air at 77 degree C\n", + "Re = (v*x)/kw;\t\t\t#Reynolds number\n", + "Nu = (0.332*Re**0.5*Pr**(1./3));\t\t\t#Nusselts number\n", + "h = (Nu*k)/x;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "h1 = (h*2);\t\t\t#Average value of heat transfer coefficient in W/m**2.K\n", + "Q = h1*x*W*(Tp-Ta);\t\t\t#Heat flow in W\n", + "Q1 = (2*Q);\t\t\t#Heat flow from both sides of the plate in W\n", + "\n", + "# Results\n", + "print 'Heat flow from both sides of the plate is %3.0f W'%(round(Q1,-1))\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat flow from both sides of the plate is 1450 W\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.3 Page No : 282" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# Variables\n", + "Ta = 20;\t\t\t#Temperature of air in degree C\n", + "v = 3;\t\t\t#Velocity of flow in m/s\n", + "L = 2;\t\t\t#Length of plate in m\n", + "W = 1;\t\t\t#Width of plate in m\n", + "x1 = 0.3;\t\t\t#Initial point of the boundary layer in m\n", + "x2 = 0.8;\t\t\t#Final point of the boundary layer in m\n", + "p = 1.17;\t\t\t#Density of air at 20 degree C in kg/m**3\n", + "k = (15*10**-6);\t\t\t#Kinematic vismath.cosity in m**2/s\n", + "Re = (5*10**5);\t\t\t#Reynolds number at the transition frm laminar to turbulant\n", + "\n", + "# Calculations\n", + "x = (k*Re)/v;\t\t \t#Critical length in m\n", + "Rel = (v*L)/k; \t\t\t#Reynolds number\n", + "q = (4.64*L)/math.sqrt(Rel)*1000;\t\t\t#Boundary layer thickness at the trailing edge of plate in mm\n", + "ts = 1.292*(0.5*p*v**2)*math.sqrt(1./Rel);\t#Average shear stress in N/m**2\n", + "F = (2*L*ts);\t\t \t#Drag force on the two sides of the plate in N\n", + "q80 = (4.64*x2)/math.sqrt((v*x2)/k);\t\t\t#Boundray layer thickness at x = 0.8 m\n", + "q30 = (4.64*x1)/math.sqrt((v*x1)/k);\t\t\t#Boundray layer thickness at x = 0.3 m\n", + "m = ((5./8)*p*v*(q80-q30))/10**-3;\t\t\t#Mass flow of air in kg/s\n", + "\n", + "# Results\n", + "print 'Boundary layer thickness at the trailing edge of plate is % 3.2f mm \\n \\\n", + "Drag force on the two sides of the plate is %3.4f N \\n \\\n", + "Mass flow of air is %3.1f*10**-3 kg/s'%(q,F,m)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Boundary layer thickness at the trailing edge of plate is 14.67 mm \n", + " Drag force on the two sides of the plate is 0.0430 N \n", + " Mass flow of air is 7.9*10**-3 kg/s\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.4 Page No : 283" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# Variables\n", + "P = 8.;\t\t\t#Pressure of air in kN/m**2\n", + "Ta = 250.;\t\t\t#Temperature of air in degree C\n", + "L = 1.;\t\t\t#Length of the palte in m\n", + "W = 0.3;\t\t\t#Width of the plate in m\n", + "v = 8.;\t\t\t#Velocity of air in m/s\n", + "Tp = 78.;\t\t\t#Temperature of plate in degree C\n", + "\n", + "# Calculations\n", + "Tf = (Ta+Tp)/2;\t\t\t#Film temperature in degree C\n", + "Cp = 1018;\t\t\t#Specific heat of air at 164 degree C and 1 atm pressure\n", + "kw = (30.8*10**-6);\t\t\t#Kinematic viscosity of air at 164 degree C and 1 atm pressure\n", + "k = 0.0364;\t\t\t#Thermal conductivity of air at 164 degree C and 1 atm pressure\n", + "Pr = 0.69;\t\t\t#prant number of air at 164 degree C and 1 atm pressure\n", + "k1 = kw*(101330/(P*1000));\t\t\t#Kinematic viscosity of air at 164 degree C and 8kN/m**2 pressure\n", + "Re = (v*L)/k1;\t\t\t#Reynolds number\n", + "h = 0.662*(k/L)*math.sqrt(Re)*Pr**(1./3);\t\t\t#Heat transfer coefficient in W/m.K\n", + "Q = 2*h*L*W*(Ta-Tp);\t\t\t#Rate of heat removal in W\n", + "\n", + "\n", + "# Results\n", + "print 'Rate of heat removal is %3.1f W'%(Q)\n", + "\n", + "#note : answer is slightly different because of rounding off error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rate of heat removal is 314.7 W\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.5 Page No : 286" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# Variables\n", + "P = 8.;\t\t\t#Pressure of air in kN/m**2\n", + "Ta = 250.;\t\t\t#Temperature of air in degree C\n", + "L = 1.;\t\t\t#Length of the palte in m\n", + "W = 0.3;\t\t\t#Width of the plate in m\n", + "v = 8.;\t\t\t#Velocity of air in m/s\n", + "Tp = 78.;\t\t\t#Temperature of plate in degree C\n", + "R = 287.;\t\t\t#Universal gas constant in J/kg.K\n", + "\n", + "# Calculations\n", + "Tf = (Ta+Tp)/2;\t\t\t#Film temperature in degree C\n", + "Cp = 1018;\t\t\t#Specific heat of air at 164 degree C and 1 atm pressure\n", + "kw = (30.8*10**-6);\t\t\t#Kinematic viscosity of air at 164 degree C and 1 atm pressure\n", + "k = 0.0364;\t\t\t#Thermal conductivity of air at 164 degree C and 1 atm pressure\n", + "Pr = 0.69;\t\t\t#prant number of air at 164 degree C and 1 atm pressure\n", + "k1 = kw*(101330/(P*1000));\t\t\t#Kinematic viscosity of air at 164 degree C and 8kN/m**2 pressure\n", + "Re = (v*L)/k1;\t\t\t#Reynolds number\n", + "h = 0.662*(k/L)*math.sqrt(Re)*Pr**(1./3);\t\t\t#Heat transfer coefficient in W/m.K\n", + "Q = 2*h*L*W*(Ta-Tp);\t\t\t#Rate of heat removal in W\n", + "p = (P*1000)/(R*(Tf+273));\t\t\t#Density in kg/m**3\n", + "St = (h/(p*Cp*v));\t\t\t#Stanton number\n", + "Cfx2 = (St*Pr**(2/3));\t\t\t#Colburn factor\n", + "ts = (Cfx2*p*v**2);\t\t\t#Average shear stress in N/m**2\n", + "D = (0.0186*W*L);\t\t\t#Drag force on one side of plate in N\n", + "D2 = (2*D)/10**-3;\t\t\t#Total drag force on both sides of plate in N\n", + "\n", + "# Results\n", + "print 'The drag force exerted on the plate is %3.2f*10**-3 N'%(D2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drag force exerted on the plate is 11.16*10**-3 N\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.6 Page No : 289" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "L = 1;\t\t\t#Length of the palte in m\n", + "W = 1;\t\t\t#Width of the plate in m\n", + "Ts = 10;\t\t\t#Temperature of free strem air in degree C\n", + "v = 80;\t\t\t#Velocity of free stream air in m/s\n", + "\n", + "# Calculations\n", + "k = 0.025;\t\t\t #Thermal conductivity of air at 10 degree C \n", + "Pr = 0.72;\t\t\t #prant number of air at 10 degree C\n", + "v1 = (14.15*10**-6);\t\t\t #Kinematic viscosity of air at 10 degree C \n", + "Re = (v*L)/v1;\t\t\t #Reynolds number\n", + "q = 0.381*L*Re**(-1./5);\t\t \t#Thickness of the boundary layer at the trailing edge of the plate in m\n", + "Nu = (0.037*Re**(4./5)*Pr**(1./3));\t\t\t#Nusselts number\n", + "h = (Nu*k)/L; \t\t\t#Mean value of the heat transfer coefficient in W/m**2.K\n", + "\n", + "# Results\n", + "print 'Thickness of the boundary layer at the trailing edge of the plate is %3.4f m \\n \\\n", + "Mean value of the heat transfer coefficient is %3.0f W/m**2.K'%(q,h)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Thickness of the boundary layer at the trailing edge of the plate is 0.0170 m \n", + " Mean value of the heat transfer coefficient is 209 W/m**2.K\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.7 Page No : 290" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Ta = 0.;\t\t \t#Temperature of air stream in degree C\n", + "Tp = 90.; \t\t#Temperature of heated plate in degree C\n", + "v = 75.;\t\t \t#Speed of air in m/s\n", + "L = 0.45;\t\t\t#Length of the palte in m\n", + "W = 0.6;\t\t\t#Width of the plate in m\n", + "Re = (5.*10**5);\t\t#Reynolds number at the transition from laminar to turbulant\n", + "\n", + "# Calculations\n", + "Tf = (Ta+Tp)/2;\t\t\t#Film temperature in degree C\n", + "k = 0.028;\t\t\t#Thermal conductivity of air at 10 degree C \n", + "Pr = 0.698;\t\t\t#prant number of air at 10 degree C\n", + "v1 = (17.45*10**-6);\t\t\t#Kinematic viscosity of air at 10 degree C \n", + "x = (Re*v1)/v;\t\t\t#Critical length in m\n", + "Rel = (v*L)/v1;\t\t\t#Reynolds number\n", + "Cfl = ((0.074/Rel**(1./5))-(1740/Rel))/10**-3;\t\t\t#Average value of friction coefficient *10**-3\n", + "Nu = ((0.037*Rel**(4./5))-870)*Pr**(1./3);\t\t\t#Nussults number\n", + "h = (Nu*k)/L;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "Q = (2*h*L*W*Tp);\t\t\t#Rate of energy dissipation in W\n", + "\n", + "# Results\n", + "print 'Average value of friction coefficient is %3.2f*10**-3 \\n \\\n", + "Heat transfer coefficient is %3.0f W/m**2.K \\n \\\n", + "Rate of energy dissipation is %i W'%(Cfl,h,Q)\n", + "\n", + "# note : book answer is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Average value of friction coefficient is 3.19*10**-3 \n", + " Heat transfer coefficient is 170 W/m**2.K \n", + " Rate of energy dissipation is 8281 W\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.8 Page No : 296" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "D = 0.3;\t\t\t#Diameter of cylinder in m\n", + "L = 1.7;\t\t\t#Height of cylinder in m\n", + "Ts = 30.;\t\t\t#Surface temperature in degree C\n", + "v = 10.;\t\t\t#Speed of wind in m/s\n", + "Ta = 10.;\t\t\t#Temperature of air in degree C\n", + "\n", + "# Calculations\n", + "Tf = (Ta+Ts)/2;\t\t\t#Film temperature in degree C\n", + "k = 0.0259;\t\t\t#Thermal conductivity of air at 20 degree C \n", + "Pr = 0.707;\t\t\t#prant number of air at 20 degree C\n", + "v1 = (15*10**-6);\t\t\t#Kinematic viscosity of air at 20 degree C\n", + "Re = (v*D)/v1;\t\t\t#Reynolds number\n", + "Nu = 0.027*Re**0.805*Pr**(1./3)\t\t\t#Nusselts number\n", + "h = (Nu*k)/D;\t\t\t#Heat transfer coefficent in W/m**2.K\n", + "Q = (h*3.14*D*L*(Ts-Ta));\t\t\t#Rate of heat loss in W\n", + "\n", + "# Results\n", + "print 'Rate of heat loss is %3.1f W'%(Q)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rate of heat loss is 1230.9 W\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.9 Page No : 297" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Ta = 27;\t\t\t#Temperature of air stream in degree C\n", + "v = 0.3;\t\t\t#Velodity of air in m/s\n", + "Q = 100;\t\t\t#Poer of electric bulb in W\n", + "Te = 127;\t\t\t#Temperature of electric bulb in degree C\n", + "D = 0.06;\t\t\t#Diameter of sphere in m\n", + "\n", + "# Calculations\n", + "Tf = (Ta+Te)/2;\t\t\t#Film temperature in degree C\n", + "k = 0.03;\t\t\t#Thermal conductivity of air at 77 degree C \n", + "Pr = 0.697;\t\t\t#prant number of air at 77 degree C\n", + "v1 = (2.08*10**-5);\t\t\t#Kinematic viscosity of air at 77 degree C\n", + "Re = (v*D)/v1;\t\t\t#Reynolds number\n", + "h = (k*0.37*Re**0.6)/D;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "Q = (h*3.14*D**2*(Te-Ta));\t\t\t#Heat transfer rate in W\n", + "Qp = (Q*100)/100;\t\t\t#Percentage of heat lost by forced convection \n", + "\n", + "# Results\n", + "print \"Heat transfer rate is %3.2f W \\n \\\n", + "Percentage of power lost due to convection is %3.2f percent\"%(Q,Qp)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer rate is 12.10 W \n", + " Percentage of power lost due to convection is 12.10 percent\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.10 Page No : 297" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# Variables\n", + "D = 0.015;\t\t\t#Diamter of copper bus bar in m\n", + "Ta = 20;\t\t\t#Temperature of air stream in degree C\n", + "v = 1;\t\t\t#Velocity of air in m/s\n", + "Ts = 80;\t\t\t#Surface temperature in degree C\n", + "p = 0.0175;\t\t\t#Resistivity of copper in ohm mm**2/m\n", + "\n", + "# Calculations\n", + "Tf = (Ta+Ts)/2;\t\t\t#Film temperature in degree C\n", + "k = 0.02815;\t\t\t#Thermal conductivity of air at 50 degree C \n", + "Pr = 0.703;\t\t\t#prant number of air at 50 degree C\n", + "v1 = (18.9*10**-6);\t\t\t#Kinematic viscosity of air at 50 degree C\n", + "Re = (v*D)/v1;\t\t\t#Reynolds number\n", + "Nu = 0.3+(((0.62*math.sqrt(Re)*Pr**(1./3))/(1+(0.4/Pr)**(2./3))**(1./4))*(1+(Re/28200.)**(5./8))**(4./5));\t\t\t#Nusselts number\n", + "h = (Nu*k)/D;\t\t\t#Heat transfer coefficent in W/m**2.K\n", + "I = 1000*3.14*D*math.sqrt((h*(Ts-Ta)*D)/(4*p));\t\t\t#Current in A\n", + "\n", + "# Results\n", + "print 'Heat transfer coefficient between the bus bar and cooling air is %3.2f W/m**2.K \\n \\\n", + "Maximum admissible current intensity for the bus bar is %3.0f A'%(h,I)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer coefficient between the bus bar and cooling air is 28.31 W/m**2.K \n", + " Maximum admissible current intensity for the bus bar is 899 A\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.11 Page No : 298" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Ta = 30;\t\t\t#Temperature of air stream in degree C\n", + "v = 25;\t\t\t#Velocity of stream in m/s\n", + "x = 0.05;\t\t\t#Side of a square in m\n", + "D = 0.05;\t\t\t#Diameter of circular cylinder in m\n", + "Ts = 124;\t\t\t#Surface temperature in degree C\n", + "\n", + "# Calculations\n", + "Tf = (Ta+Ts)/2;\t\t\t#Film temperature in degree C\n", + "k = 0.03;\t\t\t#Thermal conductivity of air at 77 degree C \n", + "Pr = 0.7;\t\t\t#prantL number of air at 77 degree C\n", + "v1 = (20.92*10**-6);\t\t\t#Kinematic viscosity of air at 77 degree C\n", + "Re = (v*D)/v1;\t\t\t#Reynolds number\n", + "Nu1 = 0.027*Re**0.805*Pr**(1./3);\t\t\t#Nussults number for circulat tube\n", + "h1 = (Nu1*k)/D;\t\t\t#Heat tansfer coefficient for circular tube in W/m**2.K\n", + "Nu2 = 0.102*Re**0.675*Pr**(1./3);\t\t\t#Nussults number for square tube\n", + "h2 = (Nu2*k)/D;\t\t\t#Heat transfer coefficient for square tube in W/m**2.K\n", + "\n", + "# Results\n", + "print 'Heat transfer coefficient for circular tube is %3.1f W/m**2.K \\n \\\n", + "Heat transfer coefficient for square tube is %3.2f W/m**2.K'%(h1,h2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer coefficient for circular tube is 100.7 W/m**2.K \n", + " Heat transfer coefficient for square tube is 91.02 W/m**2.K\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.12 Page No : 302" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "n = 7;\t\t\t#Number of rows of tube\n", + "Ta = 15;\t\t\t#Temperature of air in degree C\n", + "v = 6;\t\t\t#Velocity of air in m/s\n", + "ST = 0.0205;\t\t\t#Transverse pitch in m\n", + "SD = 0.0205;\t\t\t#Longitudinal pitch in m\n", + "D = 0.0164;\t\t\t#Outside diameter of the tube in m\n", + "Ts = 70;\t\t\t#Surface temperature in degree C\n", + "\n", + "# Calculations\n", + "Tf = (Ta+Ts)/2;\t\t\t#Film temperature in degree C\n", + "k = 0.0274;\t\t\t#Thermal conductivity of air at 42.5 degree C \n", + "Pr = 0.705;\t\t\t#prant number of air at 42.5 degree C\n", + "v1 = (17.4*10**-6);\t\t\t#Kinematic viscosity of air at 42.5 degree C\n", + "p = 1.217;\t\t\t#Density in kg/m**3\n", + "vmax = (v*ST)/(ST-D);\t\t\t#Maximum velocity in m/s\n", + "Re = (vmax*D)/v1;\t\t\t#Reynolds number\n", + "Nu = (1.13*0.518*Re**0.556*Pr**(1./3))*0.97;\t\t\t#Nusselts number\n", + "h = (Nu*k)/D;\t\t\t#Heat transfer coefficent in W/m**2.K\n", + "f = 0.4;\t\t\t#From Fig. 7.10 on page no 303 \n", + "g = 1.04;\t\t\t#From Fig. 7.10 on page no 303\n", + "dp = (n*f*p*vmax**2*g)/2;\t\t\t#Pressure drop in N/m**2\n", + "\n", + "# Results\n", + "print 'Heat transfer coefficent is %3.2f W/m**2.K \\n \\\n", + "Pressure drop is %3.0f N/m**2'%(h,dp)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer coefficent is 252.04 W/m**2.K \n", + " Pressure drop is 1595 N/m**2\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.13 Page No : 304" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "n = 7;\t\t\t#Number of rows of tube\n", + "Ta = 15;\t\t\t#Temperature of air in degree C\n", + "v = 6;\t\t\t#Velocity of air in m/s\n", + "ST = 0.0205;\t\t\t#Transverse pitch in m\n", + "SD = 0.0205;\t\t\t#Longitudinal pitch in m\n", + "D = 0.0164;\t\t\t#Outside diameter of the tube in m\n", + "Ts = 70;\t\t\t#Surface temperature in degree C\n", + "\n", + "# Calculations\n", + "Tf = (Ta+Ts)/2;\t\t\t#Film temperature in degree C\n", + "k = 0.0253;\t\t\t#Thermal conductivity of air at 15 degree C \n", + "Pr = 0.710;\t\t\t#prant number of air at 15 degree C\n", + "v1 = (14.82*10**-6);\t\t\t#Kinematic viscosity of air at 15 degree C\n", + "p = 1.217;\t\t\t#Density in kg/m**3\n", + "Pr1 = 0.701;\t\t\t#prant number of air at 70 degree C\n", + "vmax = (v*ST)/(ST-D);\t\t\t#Maximum velocity in m/s\n", + "Re = (vmax*D)/v1;\t\t\t#Reynolds number\n", + "Nu = 0.35*Re**0.6*(Pr/Pr1)**0.25;\t\t\t#\n", + "h = (Nu*k)/D;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "\n", + "# Results\n", + "print ' Heat transfer coefficient is %3.1f W/m**2 K'%(h) \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Heat transfer coefficient is 279.5 W/m**2 K\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.14 Page No : 305" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "m = 0.314;\t\t\t#Mass flow rate of air in m**3/s\n", + "n1 = 7;\t\t\t#Number of tubes in the direction of flow\n", + "n2 = 8;\t\t\t#Number of tubes perpendicular to the direction of flow\n", + "L = 1.25;\t\t\t#Length of each tube in m\n", + "D = 0.019;\t\t\t#Outer diameter in m\n", + "ST = 0.0286;\t\t\t#Transverse pitch in m\n", + "SD = 0.038;\t\t\t#Longitudinal pitch in m\n", + "Ta = 200;\t\t\t#Temperature of air in degree C\n", + "Ts = 96;\t\t\t#Surface temperature in degree C\n", + "\n", + "# Calculations\n", + "Tf = (Ta+Ts)/2;\t\t\t#Film temperature in degree C\n", + "k = 0.039;\t\t\t#Thermal conductivity of air at 15 degree C \n", + "Pr = 0.688;\t\t\t#prantl number of air at 15 degree C\n", + "v1 = (3*10**-5);\t\t\t#Kinematic vismath.cosity of air at 15 degree C\n", + "vmax = (m/((ST*n2*L)-(D*n2*L)));\t\t\t#Maximum velocity in m/s\n", + "Re = (vmax*D)/v1;\t\t\t#Reynolds number\n", + "Nu = (0.299*Re**0.602*Pr**(1./3));\t\t\t#Nusselts number\n", + "X = 0.96;\t\t\t#From Table 7.5 on page no.302\n", + "Nux = (X*Nu);\t\t\t#Average nusselts number\n", + "h = (Nux*k)/D;\t\t\t#Convective heat transfer coefficient in W/m**2.K\n", + "\n", + "# Results\n", + "print 'Convective heat transfer coefficient is %3.2f W/m**2.K'%(h)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Convective heat transfer coefficient is 51.58 W/m**2.K\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.15 Page No : 310" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "D = 0.2;\t\t\t#Diameter of pipeline in m\n", + "\t\t\t#velocity profile is given by u = 96r-190r**2 m/s\n", + "\t\t\t#Temperature profile is given by T = 100(1-2r) degree C\n", + "\n", + "# Calculations\n", + "vmax = (64*(D/2))-(95*(D/2)**2);\t\t\t#Mean velocity in m/s\n", + "T = (2/(vmax*(D/2)**2))*(((9600*(D/2)**3)/3)-((38200*(D/2)**4)/4)+((38000*(D/2)**5)/5));\t\t\t#Average temperature of the fluid in degree C\n", + "\n", + "# Results\n", + "print 'Average temperature of the fluid is %3.2f degree C'%(T)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Average temperature of the fluid is 85.17 degree C\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.16 Page No : 311" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Di = 0.025;\t\t\t#I.D of the tube in m\n", + "Do = 0.04;\t\t\t#O.D of the tube in m\n", + "m = 5.;\t\t\t#Mass flow rate of water in kg/m\n", + "T = [20.,70.];\t\t\t#Temperature at entry and exit of water in degree C\n", + "Q = 10.**7;\t\t\t#Heat in W/m**3\n", + "Ts = 80.;\t\t\t#Surface temperature in degree C\n", + "Cp = 4179.;\t\t\t#Specific heat of water in J/kg.K\n", + "\n", + "# Calculations\n", + "Tb = (T[0]+T[1])/2;\t\t\t#Film temperature in degree C\n", + "L = ((4*(m/60)*Cp*(T[1]-T[0]))/(3.14*(Do**2-Di**2)*Q));\t\t\t#Length of tube in m\n", + "qs = ((Q*(Do**2-Di**2))/(4*Di));\t\t\t#Heat flux at the surface in W/m**2\n", + "h = (qs/(Ts-T[1]));\t\t\t#Heat transfer coefficient at the outlet in W/m**2.K\n", + "\n", + "\n", + "# Results\n", + "print 'Length of tube is %3.3f m \\nHeat transfer coefficient at the outlet is %3.0f W/m**2.K'%(L,h)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Length of tube is 2.275 m \n", + "Heat transfer coefficient at the outlet is 9750 W/m**2.K\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.17 Page No : 312" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "k = 0.175;\t\t\t#Thermal conductivity in W/m.K\n", + "Di = 0.006;\t\t\t#I.D of the tube in m\n", + "L = 8;\t\t\t#Length of the tube in m\n", + "dT = 50;\t\t\t#Mean temperature difference in degree C\n", + "\n", + "# Calculations\n", + "h = (3.66*k)/Di;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "Q = (h*3.14*Di*L*dT);\t\t\t#Heat transfer rate in W\n", + "\n", + "# Results\n", + "print 'Heat transfer coefficient is %3.2f W/m**2.K Heat transfer rate is %3.0f W'%(h,Q)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer coefficient is 106.75 W/m**2.K Heat transfer rate is 804 W\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.18 Page No : 312" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Ti = 25;\t\t\t#Initial temperature of water in degree C\n", + "D = 0.05;\t\t\t#Diamter of the tube in m\n", + "Re = 1600;\t\t\t#Reynolds number\n", + "q = 800;\t\t\t#Heat flux in W/m\n", + "Tf = 50;\t\t\t#Final temperature of water in degree C\n", + "\n", + "# Calculations\n", + "k = 0.61;\t\t\t#Thermal conductivity of water at 25 degree C in W/m.K\n", + "u = (915*10**-6);\t\t\t#Dynamic viscosity in N.s/m**2\n", + "m = (Re*3.14*D*u)/4;\t\t\t#Mass flow rate of water in kg/s\n", + "h = (4.364*k)/D;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "qs = (q/(3.14*D));\t\t\t#Constant heat flux in W/m**2\n", + "Cp = 4178;\t\t\t#Specific heat of water in J/kg.K\n", + "L = ((m*Cp*(Tf-Ti))/q);\t\t\t#Length of the tube in m\n", + "\n", + "# Results\n", + "print 'Average heat transfer coefficient is %3.2f W/m**2.K \\n \\\n", + "Length of the tube is %3.3f m'%(h,L)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Average heat transfer coefficient is 53.24 W/m**2.K \n", + " Length of the tube is 7.502 m\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.19 Page No : 314" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Di = 0.015;\t\t\t#I.D of the tube in m\n", + "Tb = 60;\t\t\t#Temperature of the tube in degree C\n", + "m = 10;\t\t\t#Flow rate of water in ml/s\n", + "Ti = 20;\t\t\t#Temperature of water at entry in degree C\n", + "x = 1;\t\t\t#Dismath.tance form the plane in m\n", + "Tx = 34;\t\t\t#Temperature of water at 1 m dismath.tance in degree C\n", + "\n", + "# Calculations\n", + "Tbm = (Ti+Tx)/2;\t\t\t#Mean value of bulk temperature in degree C\n", + "pw = 997;\t\t\t#Density of air at 27 degree C in kg/m**3\n", + "Cp = 4180;\t\t\t#Specific heat of air at 27 degree C in J/kg.K\n", + "u = (855*10**-6);\t\t\t#Dynamic vismath.cosity of air at 27 degree C in N.s/m**2\n", + "k = 0.613;\t\t\t#Thermal conductivity of air at 27 degree C in W/m.K\n", + "Pr = 5.83;\t\t\t#prantl number of air at 27 degree C\n", + "us = (464*10**-6);\t\t\t#Dynamic vismath.cosity of air at 60 degree C in Ns/m**2\n", + "um = (m*10**-6)/((3.14/4)*Di**2);\t\t\t#Mean speed in m/s\n", + "Re = (pw*um*Di)/u;\t\t\t#Reynolds number\n", + "Nu = 3.66+((0.0668*(Di/x)*Re*Pr)/(1+(0.04*((Di/x)*Re*Pr)**(2./3))));\t\t\t#Nusselts number in Haussen correlation\n", + "Nux = (1.86*((Re*Pr)/(x/Di))**(1./3)*(u/us)**0.14);\t\t\t#Nusselsts number in Sieder - Tate correlation \n", + "\n", + "# Results\n", + "print 'Nusselts number in Haussen correlation is %3.2f \\n \\\n", + "Nusselsts number in Sieder - Tate correlation is %3.3f'%(Nu,Nux)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Nusselts number in Haussen correlation is 6.90 \n", + " Nusselsts number in Sieder - Tate correlation is 8.964\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.20 Page No : 318" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Tw = 50;\t\t\t#Temperature of water in degree C\n", + "Di = 0.005;\t\t\t#Inner diameter of the tube in m\n", + "L = 0.5;\t\t\t#Length of the tube in m\n", + "v = 1;\t\t\t#Mean velocity in m/s\n", + "Ts = 30;\t\t\t#Surface temperature in degree C\n", + "\n", + "# Calculations\n", + "Tf = (Tw+Ts)/2;\t\t\t#Film temperature in degree C\n", + "k = 0.039;\t\t\t#Thermal conductivity of air at 15 degree C \n", + "Pr = 0.688;\t\t\t#prant number of air at 15 degree C\n", + "p = 990;\t\t\t#Density of air at 50 degree C in kg/m**3\n", + "Cp = 4178;\t\t\t#Specific heat of air at 50 degree C in J/kg.K\n", + "v1 = (5.67*10**-7);\t\t\t#Kinematic viscosity of air at 50 degree C\n", + "v2 = (6.57*10**-7);\t\t\t#Kinematic viscosity of air at 40 degree C\n", + "Re = (v*Di)/v1;\t\t\t#Reynolds number\n", + "h = ((0.316/8)*((v*Di*10)/v2)**(-0.25)*p*Cp*v*(4.34)**(-2./3));\t\t\t#Heat transfer coefficient umath.sing the Colburn anamath.logy in W/m**2.K\n", + "\n", + "# Results\n", + "print 'Heat transfer coefficient using the Colburn analogy is %3.0f W/m**2.K'%(h)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer coefficient using the Colburn analogy is 3697 W/m**2.K\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.21 Page No : 319" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Ti = 50;\t\t\t#Temperature of water at inlet in degree C\n", + "D = 0.015;\t\t\t#Diameter of tube in m\n", + "L = 3;\t\t\t#Length of the tube in m\n", + "v = 1;\t\t\t#Velocity of flow in m/s\n", + "Tb = 90;\t\t\t#Temperature of tube wall in degree C\n", + "Tf = 64;\t\t\t#Exit temperature of water in degree C\n", + "\n", + "# Calculations\n", + "Tm = (Ti+Tf)/2;\t\t\t#Bulk mean temperature in degree C\n", + "p = 990;\t\t\t#Density of air at 57 degree C in kg/m**3\n", + "Cp = 4184;\t\t\t#Specific heat of air at 57 degree C in J/kg.K\n", + "u = (0.517*10**-6);\t\t\t#Kinematic viscosity of air at 57 degree C in m**2/s\n", + "k = 0.65;\t\t\t#Thermal conductivity of air at 57 degree C in W/m.K\n", + "Pr = 3.15;\t\t\t#prantl number of air at 57 degree C \n", + "Re = (v*D)/u;\t\t\t#Reynolds number\n", + "Nu = (0.023*Re**(4./5)*Pr**0.4);\t\t\t#Nusselts number\n", + "h = (Nu*k)/D;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "Q = (h*3.14*D*L*(Tb-Tm))/1000;\t\t\t#Rate of heat transfered in kW\n", + "\n", + "# Results\n", + "print 'Heat transfer coefficient is %3.0f W/m**2.K \\nRate of heat transfered is %3.2f kW'%(h,Q)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer coefficient is 5861 W/m**2.K \n", + "Rate of heat transfered is 27.33 kW\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.22 Page No : 320" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "D = 0.022;\t\t\t#Diamter of the tube in m\n", + "v = 2;\t\t\t#Average velocity in m/s\n", + "Tw = 95;\t\t\t#Temperature of tube wall in degree C\n", + "T = [15,60];\t\t\t#Initial and final temperature of water in degree C\n", + "\n", + "# Calculations\n", + "Tm = (T[0]+T[1])/2;\t\t\t#Bulk mean temperature in degree C\n", + "p = 990;\t\t\t#Density of air at 37.5 degree C in kg/m**3\n", + "Cp = 4160;\t\t\t#Specific heat of air at 37.5 degree C in J/kg.K\n", + "u = (0.69*10**-3);\t\t\t#Dynamic viscosity of air at 37.5 degree C in Ns/m**2\n", + "k = 0.63;\t\t\t#Thermal conductivity of air at 37.5 degree C in W/m.K\n", + "us = (0.3*10**-3);\t\t\t#Dynamic viscosity of air at 37.5 degree C in Ns/m**2\n", + "Re = (p*v*D)/u;\t\t\t#Reynolds number\n", + "Pr = (u*Cp)/k;\t\t\t#Prantl number\n", + "Nu = (0.027*Re**(4./5)*Pr**(1./3)*(u/us)**0.14);\t\t\t#Nusselts number\n", + "h = (Nu*k)/D;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "\n", + "# Results\n", + "print 'Heat transfer coefficient is %3.0f W/m**2.K'%(h)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer coefficient is 9969 W/m**2.K\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.23 Page No : 320" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "D = 0.05;\t\t\t#Diamter of the tube in m\n", + "T = 147;\t\t\t#Average temperature in degree C\n", + "v = 0.8;\t\t\t#Flow vwlocity in m/s\n", + "Tw = 200;\t\t\t#Wall temperature in degree C\n", + "L = 2;\t\t\t#Length of the tube in m\n", + "\n", + "# Calculations\n", + "p = 812.1;\t\t\t#Density in kg/m**3 of oil at 147 degree C\n", + "Cp = 2427;\t\t\t#Specific heat of oil at 147 degree C in J/kg.K\n", + "u = (6.94*10**-6);\t\t\t#Kinematic viscosity of oil at 147 degree C in m**2/s\n", + "k = 0.133;\t\t\t#Thermal conductivity of oil at 147 degree C in W/m.K\n", + "Pr = 103;\t\t\t#prantl number of oil at 147 degree C \n", + "Re = (v*D)/u;\t\t\t#Reynolds number\n", + "Nu = (0.036*Re**0.8*Pr**(1./3)*(D/L)**0.055);\t\t\t#Nussults number\n", + "h = (Nu*k)/D;\t\t\t#Average heat transfer coefficient in W/m**2.K\n", + "\n", + "# Results\n", + "print 'Average heat transfer coefficient is %3.1f W/m**2.K'%(h)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Average heat transfer coefficient is 373.7 W/m**2.K\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.24 Page No : 321" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "D = [0.4,0.8];\t\t\t#Dimensions of the trunk duct in m\n", + "Ta = 20;\t\t\t#Temperature of air in degree C\n", + "v = 7;\t\t\t#Velocity of air in m/s\n", + "v1 = (15.06*10**-6);\t\t\t#Kinematic viscosity in m**2/s\n", + "a = (7.71*10**-2);\t\t\t#Thermal diffusivity in m**2/h\n", + "k = 0.0259;\t\t\t#Thermal conductivity in W/m.K\n", + "\n", + "# Calculations\n", + "Dh = (4*(D[0]*D[1]))/(2*(D[0]+D[1]));\t\t\t#Value of Dh in m\n", + "Re = (v*Dh)/v1;\t\t\t#Reynolds number\n", + "Pr = (v1/a)*3600;\t\t\t#Prantl number\n", + "Nu = (0.023*Re**(4./5)*Pr**0.4);\t\t\t#Nussults number\n", + "h = (Nu*k)/Dh;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "Q = (h*(2*(D[0]+D[1])));\t\t\t#Heat leakage per unit length per unit difference in W\n", + "\n", + "# Results\n", + "print 'Heat leakage per unit length per unit difference is %3.2f W'%(Q)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat leakage per unit length per unit difference is 48.14 W\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.25 Page No : 322" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Di = 0.03125;\t\t\t#I.D of the annulus in m\n", + "Do = 0.05;\t\t\t#O.D of the annulus in m\n", + "Ts = 50;\t\t\t#Outer surface temperature in degree C\n", + "Ti = 16;\t\t\t#Temeperature at which air enters in degree C\n", + "Tf = 32;\t\t\t#Temperature at which air exits in degree C\n", + "v = 30;\t\t\t#Flow rate in m/s\n", + "\n", + "# Calculations\n", + "Tb = (Ti+Tf)/2;\t\t\t#Mean bulk temperature of air in degree C\n", + "p = 1.614;\t\t\t#Density in kg/m**3 of air at 24 degree C\n", + "Cp = 1007;\t\t\t#Specific heat of air at 24 degree C in J/kg.K\n", + "u = (15.9*10**-6);\t\t\t#Kinematic viscosity of air at 24 degree C in m**2/s\n", + "k = 0.0263;\t\t\t#Thermal conductivity of air at 24 degree C in W/m.K\n", + "Pr = 0.707;\t\t\t#prantl number of air at 24 degree C\n", + "Dh = (4*(3.14/4)*(Do**2-Di**2))/(3.14*(Do+Di));\t\t\t#Hydraulic diameter in m\n", + "Re = (v*Dh)/u;\t\t\t#Reynolds number\n", + "Nu = (0.023*Re**0.8*Pr**0.4);\t\t\t#Nussults number\n", + "h = (Nu*k)/Dh;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "\n", + "# Results\n", + "print 'Heat transfer coefficient is %3.1f W/m**2.K'%(h)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer coefficient is 122.3 W/m**2.K\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.26 Page No : 324" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "T = [120,149];\t\t\t#Initail and final temperatures in degree C\n", + "m = 2.3;\t\t\t#Mass flow rate in kg/s\n", + "D = 0.025;\t\t\t#Diameter of the tube in m\n", + "Ts = 200;\t\t\t#Surface temperature in degree C\n", + "\n", + "# Calculations\n", + "Tb = (T[0]+T[1])/2;\t\t\t#Bulk mean temperature in degree C\n", + "p = 916;\t\t\t#Density in kg/m**3 of air at 134.5 degree C\n", + "Cp = 1356.6;\t\t\t#Specific heat of air at 134.5 degree C in J/kg.K\n", + "u = (0.594*10**-6);\t\t\t#Kinematic viscosity of air at 134.5 degree C in m**2/s\n", + "k = 84.9;\t\t\t#Thermal conductivity of air at 134.5 degree C in W/m.K\n", + "Pr = 0.0087;\t\t\t#prantl number of air at 134.5 degree C\n", + "Q = (m*Cp*(T[1]-T[0]))/1000;\t\t\t#Total heat transfer in kW\n", + "v = (m/(p*(3.14/4)*D**2));\t\t\t#Velocity of flow in m/s\n", + "Re = (v*D)/u;\t\t\t#Reynolds number\n", + "Pe = (Pr*Re);\t\t\t#Peclet number\n", + "Nu = (4.82+(0.0185*Pe**0.827));\t\t\t#Nussults number\n", + "h = (Nu*k)/D;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "L = ((Q*1000)/(h*3.14*D*(Ts-Tb)));\t\t\t#Minimum length of the tube in m if the wall temperature is not to exceed 200 degree C\n", + "\n", + "# Results\n", + "print 'Minimum length of the tube if the wall temperature is not to exceed 200 degree C is %3.3f m'%(L)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum length of the tube if the wall temperature is not to exceed 200 degree C is 0.361 m\n" + ] + } + ], + "prompt_number": 34 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch8.ipynb b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch8.ipynb new file mode 100755 index 00000000..532f0847 --- /dev/null +++ b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch8.ipynb @@ -0,0 +1,790 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:ca5009f0cb9dd781f90e694bd80e1f1d823182523e44edb65991cb3279963181" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 8 : Natural Convection" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.1 Page No : 340" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables\n", + "L = 0.3;\t\t\t#Length of the glass plate in m\n", + "Ta = 27;\t\t\t#Temperature of air in degree C\n", + "Ts = 77;\t\t\t#Surface temperature in degree C\n", + "v = 4;\t\t\t#Velocity of air in m/s\n", + "\n", + "# Calculations\n", + "Tf = (Ta+Ts)/2;\t\t\t#Film temperature in degree C\n", + "k = 0.02815;\t\t\t#Thermal conductivity in W/m.K\n", + "v1 = (18.41*10**-6);\t\t\t#Kinematic viscosity in m**2/s\n", + "Pr = 0.7;\t\t\t#Prantl number\n", + "b = (3.07*10**-3);\t\t\t#Coefficient of thermal expansion in 1./K\n", + "Gr = (9.81*b*(Ts-Ta)*L**3)/v1**2;\t\t\t#Grashof number\n", + "q = L*((3.93*(1./math.sqrt(Pr))*(0.952+Pr)**0.25*Gr**(-0.25)));\t\t\t#Boundary layer thickness at the trailing edge of the plate in free convection in m\n", + "Re = (v*L)/v1;\t\t\t#Reynolds number\n", + "q1 = (5*L)/math.sqrt(Re);\t\t\t#Boundary layer thickness at the trailing edge of the plate in forced convection in m\n", + "\n", + "# Results\n", + "print 'Boundary layer thickness at the trailing edge of the plate in free convection is % 3.4f m \\\n", + "\\nBoundary layer thickness at the trailing edge of the plate in forced convection is %3.4f m'%(q,q1) \n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Boundary layer thickness at the trailing edge of the plate in free convection is 0.0153 m \n", + "Boundary layer thickness at the trailing edge of the plate in forced convection is 0.0059 m\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.2 Page No : 341" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# Variables\n", + "L = 0.3;\t\t\t#Length of the glass plate in m\n", + "Ta = 27;\t\t\t#Temperature of air in degree C\n", + "Ts = 77;\t\t\t#Surface temperature in degree C\n", + "v = 4;\t\t\t#Velocity of air in m/s\n", + "\n", + "# Calculations\n", + "Tf = (Ta+Ts)/2;\t\t\t#Film temperature in degree C\n", + "k = 0.02815;\t\t\t#Thermal conductivity in W/m.K\n", + "v1 = (18.41*10**-6);\t\t\t#Kinematic viscosity in m**2/s\n", + "Pr = 0.7;\t\t\t#Prantl number\n", + "b = (3.07*10**-3);\t\t\t#Coefficient of thermal expansion in 1./K\n", + "Gr = (9.81*b*(Ts-Ta)*L**3)/v1**2;\t\t\t#Grashof number\n", + "Re = (v*L)/v1;\t\t\t#Reynolds number\n", + "Nu = (0.677*math.sqrt(Pr)*(0.952+Pr)**(-0.25)*Gr**0.25);\t\t\t#Nusselts number\n", + "h = (Nu*k)/L;\t\t\t#Heat transfer coefficient for natural convection in W/m**2.K\n", + "Nux = (0.664*math.sqrt(Re)*Pr**(1./3));\t\t\t#Nusselts number\n", + "hx = (Nux*k)/L;\t\t\t#Heat transfer coefficient for forced convection in W/m**2.K\n", + "\n", + "# Results\n", + "print 'Heat transfer coefficient for natural convection is %3.1f W/m**2.K \\n \\\n", + "Heat transfer coefficient for forced convection is %3.2f W/m**2.K'%(h,hx)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer coefficient for natural convection is 4.9 W/m**2.K \n", + " Heat transfer coefficient for forced convection is 14.12 W/m**2.K\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.3 Page No : 343" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "L = 0.609;\t\t\t#Height of the metal plate in m\n", + "Ts = 161.;\t\t\t#Temperature of the wall in degree C\n", + "Ta = 93.;\t\t\t#Temperature of air in degree C\n", + "\n", + "# Calculations\n", + "Tf = (Ts+Ta)/2;\t\t\t#Film temperature in degree C\n", + "k = 0.0338;\t\t\t#Thermal conductivity in W/m.K\n", + "v1 = (26.4*10**-6);\t\t\t#Kinematic vismath.cosity in m**2/s\n", + "Pr = 0.69;\t\t\t#Prantl number\n", + "b = 0.0025;\t\t\t#Coefficient of thermal expansion in 1./K\n", + "a = (38.3*10**-6);\t\t\t#Thermal diffusivity in m**2/s\n", + "Ra = ((9.81*b*L**3*(Ts-Ta))/(v1*a));\t\t\t#Rayleigh number\n", + "Nu = (0.68+((0.67*Ra**0.25)/(1+(0.492/Pr)**(9./16))**(4./9)));\t\t\t#Nussults number\n", + "h = (Nu*k)/L;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "Q = (h*L*(Ts-Ta));\t\t\t#Rate of heat transfer in W\n", + "Nul = 0.59 * (3.72*10**8)**(1./4)\n", + "\n", + "# Results\n", + "print 'Heat transfer coefficient is %3.3f W/m**2.K Rate of heat transfer is %3.2f W'%(h,Q)\n", + "print \"NuL = %.2f W\"%Nul" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer coefficient is 3.990 W/m**2.K Rate of heat transfer is 165.24 W\n", + "NuL = 81.94 W\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.4 Page No : 344" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "W = 0.5;\t\t\t#Width of the radiator in m\n", + "L = 1.;\t\t\t#Height of the radiator in m\n", + "Ts = 84.;\t\t\t#Surface temperature in degree C\n", + "Ta = 20.;\t\t\t#Room temperature in degree C\n", + "\n", + "# Calculations\n", + "Tf = (Ts+Ta)/2;\t\t\t#Film temperature in degree C\n", + "k = 0.02815;\t\t\t#Thermal conductivity in W/m.K\n", + "v1 = (18.41*10**-6);\t\t\t#Kinematic viscosity in m**2/s\n", + "Pr = 0.7;\t\t\t#Prantl number\n", + "b = 0.003077;\t\t\t#Coefficient of thermal expansion in 1./K\n", + "Ra = ((9.81*b*L**3*(Ts-Ta)*Pr)/(v1**2));\t\t\t#Rayleigh number\n", + "Nu = (0.825+((0.387*Ra**(1./6))/(1+(0.492/Pr)**(9./16))**(8./27)))**2;\t\t\t#Nussults number\n", + "h = (1170.9*k)/L;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "Q = (h*W*L*(Ts-Ta));\t\t\t#Convective heat loss in W\n", + "\n", + "# Results\n", + "print 'Convective heat loss from the radiator is %3.2f W'%(Q)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Convective heat loss from the radiator is 1054.75 W\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.5 Page No : 345" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "L = 0.8;\t\t\t#Height of the plate in m\n", + "W = 0.08;\t\t\t#Width of the plate in m\n", + "Ts = 170;\t\t\t#Surafce temperature in degree C\n", + "Tw = 70;\t\t\t#Temperature of water in degree C\n", + "Tf = 130;\t\t\t#Final temperature in degree C\n", + "\n", + "# Calculations\n", + "Tb = (Ts+Tw)/2;\t\t\t#Film temperature in degree C\n", + "p = 960.63;\t\t\t#Density in kg/m**3\n", + "k = 0.68;\t\t\t#Thermal conductivity in W/m.K\n", + "v1 = (0.294*10**-6);\t\t\t#Kinematic viscosity in m**2/s\n", + "b = 0.00075;\t\t\t#Coefficient of thermal expansion in 1./K\n", + "Cp = 4216;\t\t\t#Specific heat in J/kg.K\n", + "a = (1.68*10**-7);\t\t\t#Thermal diffusivity in m**2/s\n", + "Lc = (W/2);\t\t\t#Characteristic length in m\n", + "Ra = ((9.81*b*Lc**3*(Tf-Tw))/(v1*a));\t\t\t#Rayleigh number\n", + "Nu1 = (0.15*Ra**(1./3));\t\t\t#Nussults number\n", + "h1 = (Nu1*k)/Lc;\t\t\t#Heat transfer coefficient at top surface in W/m**2.K \n", + "Nu2 = 0.27*(Ra)**(0.25);\t\t\t#Nusselts number\n", + "h2 = (Nu2*k)/Lc;\t\t\t#Heat transfer coefficient at bottom surface in W/m**2.K\n", + "Q = ((h1+h2)*W*L*(Tf-Tw))/1000;\t\t\t#Rate of heat input to the plate in kW\n", + "\n", + "# Results\n", + "print 'Rate of heat input to the plate necessary to maintain the temperature at %3.0f degree C is %3.2f kW'%(Tf,Q)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rate of heat input to the plate necessary to maintain the temperature at 130 degree C is 10.85 kW\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.6 Page No : 346" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "L = 0.3;\t\t\t#Height of the duct in m\n", + "W = 0.6;\t\t\t#Width of the duct in m\n", + "Ts = 15;\t\t\t#Surface temperature in degree C\n", + "Ta = 25;\t\t\t#Temeprature of air in degree C\n", + "\n", + "# Calculations\n", + "Tb = (Ts+Ta)/2;\t\t\t#Film temperature in degree C\n", + "p = 1.205;\t\t\t#Density in kg/m**3\n", + "k = 0.02593;\t\t\t#Thermal conductivity in W/m.K\n", + "v1 = (15.06*10**-6);\t\t\t#Kinematic viscosity in m**2/s\n", + "b = 0.00341;\t\t\t#Coefficient of thermal expansion in 1./K\n", + "Cp = 1005;\t\t\t#Specific heat in J/kg.K\n", + "Pr = 0.705;\t\t\t#Prantl number\n", + "Ra = ((9.81*b*L**3*(Ta-Ts)*Pr)/(v1**2));\t\t\t#Rayleigh number\n", + "Nux = (0.59*Ra**(0.25));\t\t\t#Nusselts number\n", + "hx = (Nux*k)/L;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "Lc = (W/2);\t\t\t#Characteristic length in m\n", + "Nu1 = (0.15*Ra**(1./3));\t\t\t#Nussults number\n", + "h1 = (Nu1*k)/Lc;\t\t\t#Heat transfer coefficient at top surface in W/m**2.K \n", + "Nu2 = 0.27*(Ra)**(0.25);\t\t\t#Nusselts number\n", + "h2 = (Nu2*k)/Lc;\t\t\t#Heat transfer coefficient at bottom surface in W/m**2.K\n", + "Q = ((2*hx*L)+(W*(h1+h2)))*(Ta-Ts);\t\t\t#Rate of heat gained per unit length in W/m\n", + "\n", + "# Results\n", + "print 'Rate of heat gained per unit length is %3.2f W/m'%(Q)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rate of heat gained per unit length is 56.11 W/m\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.7 Page No : 348" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "LH = 0.08;\t\t\t#Horizantal length in m\n", + "LV = 0.12;\t\t\t#Vertical length in m\n", + "Ts = 50;\t\t\t#Surface temperature in degree C\n", + "Ta = 0;\t\t\t#Temeprature of air in degree C\n", + "\n", + "# Calculations\n", + "L = (LH*LV)/(LH+LV);\t\t\t#Characteristic length in m\n", + "Tb = (Ts+Ta)/2;\t\t\t#Film temperature in degree C\n", + "p = 0.707;\t\t\t#Density in kg/m**3\n", + "k = 0.0263;\t\t\t#Thermal conductivity in W/m.K\n", + "v1 = (15.89*10**-6);\t\t\t#Kinematic viscosity in m**2/s\n", + "b = (1./300);\t\t\t#Coefficient of thermal expansion in 1./K\n", + "Pr = 0.707;\t\t\t#Prantl number\n", + "Gr = ((9.81*b*L**3*(Ts-Ta))/(v1**2));\t\t\t#Grashof number\n", + "Nu = 0.55*Gr**(0.25);\t\t\t#Nussults number\n", + "h = (Nu*k)/L;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "\n", + "# Results\n", + "print 'Heat transfer coefficient is %3.2f W/m**2.K'%(h)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer coefficient is 8.77 W/m**2.K\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.8 Page No : 349" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "D = 0.2;\t\t\t#Outer diameter of the pipe in m\n", + "Ts = 100;\t\t\t#Surface temperature in degree C\n", + "Ta = 20;\t\t\t#Temperature of air in degree C\n", + "L = 3;\t\t\t#Length of pipe in m\n", + "\n", + "# Calculations\n", + "Tf = (Ts+Ta)/2;\t\t\t#Film temperature in degree C\n", + "k = 0.02896;\t\t\t#Thermal conductivity in W/m.K\n", + "v1 = (18.97*10**-6);\t\t\t#Kinematic viscosity in m**2/s\n", + "b = (1./333);\t\t\t#Coefficient of thermal expansion in 1./K\n", + "Pr = 0.696;\t\t\t#Prantl number\n", + "Gr = ((9.81*b*L**3*(Ts-Ta))/(v1**2));\t\t\t#Grashof number\n", + "Ra = (Gr*Pr);\t\t\t#Rayleigh number\n", + "Nu = (0.1*Ra**(1./3));\t\t\t#Nussults number\n", + "h = (Nu*k)/L;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "Q = (h*3.14*D*(Ts-Ta));\t\t\t#Rate of heat loss per meter length of pipe in W/m\n", + "\n", + "# Results\n", + "print 'Rate of heat loss per meter length of pipe is %3.2f W/m'%(Q)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rate of heat loss per meter length of pipe is 241.24 W/m\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.9 Page No : 350" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "D = 0.1;\t\t\t#Outer diamter of the pipe in m\n", + "Ta = 30.;\t\t\t#Ambient temperature of air degree C\n", + "Ts = 170.;\t\t\t#Surface temperature in degree C\n", + "e = 0.9;\t\t\t#Emissivity \n", + "\n", + "# Calculations\n", + "Tb = (Ts+Ta)/2;\t\t\t#Film temperature in degree C\n", + "k = 0.0321;\t\t\t#Thermal conductivity in W/m.K\n", + "v1 = (23.13*10**-6);\t\t\t#Kinematic viscosity in m**2/s\n", + "b = 0.00268;\t\t\t#Coefficient of thermal expansion in 1./K\n", + "Pr = 0.688;\t\t\t#Prantl number\n", + "Ra = ((9.81*b*D**3*(Ts-Ta)*Pr)/(v1**2));\t\t\t#Rayleigh number\n", + "Nu = (0.6+((0.387*Ra**(1./6))/(1+(0.559/Pr)**(9./16))**(8./27)))**2;\t\t\t#Nussults number\n", + "h = (Nu*k)/D;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "Q = (h*3.1415*D*(Ts-Ta))+(e*3.1415*D*5.67*10**-8*((Ts+273)**4-(Ta+273)**4));\t\t\t#Total heat loss per meter length of pipe in m\n", + "NuD = 0.48*(4.72*10**6)*0.25\n", + "\n", + "# Results\n", + "print 'Total heat loss per meter length of pipe is %3.2f W/m'%(Q)\n", + "print \"NuD = %.2f\"%NuD\n", + "\n", + "# Note : rounding off error.\n", + "# Note : 2nd answer is wrong in book" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Total heat loss per meter length of pipe is 801.20 W/m\n", + "NuD = 566400.00\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.10 Page No : 351" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# Variables\n", + "Ta = 25;\t\t\t#Temperature of air in degree C\n", + "Ts = 95;\t\t\t#Surface temperature of wire in degree C\n", + "D = 0.0025;\t\t\t#Diameter of wire in m\n", + "R = 6;\t\t\t#Resistivity in ohm/m\n", + "\n", + "# Calculations\n", + "Tf = (Ts+Ta)/2;\t\t\t#Film temperature in degree C\n", + "k = 0.02896;\t\t\t#Thermal conductivity in W/m.K\n", + "v1 = (18.97*10**-6);\t\t\t#Kinematic viscosity in m**2/s\n", + "b = (1./333);\t\t\t#Coefficient of thermal expansion in 1./K\n", + "Pr = 0.696;\t\t\t#Prantl number\n", + "Gr = ((9.81*b*D**3*(Ts-Ta))/(v1**2));\t\t\t#Grashof number\n", + "Ra = (Gr*Pr);\t\t\t#Rayleigh number\n", + "Nu = (1.18*Ra**(1./8));\t\t\t#Nussults number\n", + "h = (Nu*k)/D;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "Q = (h*3.14*D*(Ts-Ta));\t\t\t#Rate of heat loss per unit length of wire in W/m\n", + "I = math.sqrt(Q/R);\t\t\t#Maximum current intensity in A\n", + "\n", + "# Results\n", + "print 'Heat transfer coefficient is %3.2f W/m**2.K \\n \\\n", + "Maximum current intensity is %3.2f A'%(h,I)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer coefficient is 22.91 W/m**2.K \n", + " Maximum current intensity is 1.45 A\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.11 Page No : 352" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "D = 0.01;\t\t\t#Diameter of spherical steel ball in m\n", + "Ts = 260;\t\t\t#Surface temperature in degree C\n", + "Ta = 20;\t\t\t#Temperature of air in degree C\n", + "\n", + "# Calculations\n", + "Tf = (Ts+Ta)/2;\t\t\t#Film temperature in degree C\n", + "k = 0.0349;\t\t\t#Thermal conductivity in W/m.K\n", + "v1 = (27.8*10**-6);\t\t\t#Kinematic viscosity in m**2/s\n", + "b = (1./413);\t\t\t#Coefficient of thermal expansion in 1./K\n", + "Pr = 0.684;\t\t\t#Prantl number\n", + "Ra = ((9.81*b*D**3*(Ts-20)*Pr)/(v1**2));\t\t\t#Rayleigh number\n", + "Nu = (2+(0.43*Ra**0.25));\t\t\t#Nusuults number\n", + "h = (k*Nu)/D;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "Q = (h*3.14*D**2*(Ts-Ta));\t\t\t#Rate of heat loss in W\n", + "\n", + "# Results\n", + "print 'Rate of convective heat loss is %3.2f W'%(Q)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rate of convective heat loss is 1.48 W\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.12 Page No : 353" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "D = 0.1;\t\t\t#Outer diamter of the pipe in m\n", + "Ta = 30;\t\t\t#Ambient temperature of air degree C\n", + "Ts = 170;\t\t\t#Surface temperature in degree C\n", + "e = 0.9;\t\t\t#Emissivity \n", + "\n", + "# Calculations\n", + "h = (1.32*((Ts-Ta)/D)**0.25);\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "q = (h*3.1415*D*(Ts-Ta));\t\t\t#Heat transfer in W/m\n", + "\n", + "# Results\n", + "print 'Heat loss due to free convection is %3.2f W/m'%(q)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat loss due to free convection is 355.12 W/m\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.13 Page No : 355" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "L = 0.015;\t\t\t#Thickness of the slot in m\n", + "D = 2;\t\t\t#Dimension of square plate in m\n", + "T1 = 120;\t\t\t#Temperature of plate 1\n", + "T2 = 20;\t\t\t#Temperature of plate 2\n", + "\n", + "# Calculations\n", + "Tf = (T1+T2)/2;\t\t\t#Film temperature in degree C\n", + "k = 0.0295;\t\t\t#Thermal conductivity in W/m.K\n", + "v1 = (2*10**-5);\t\t\t#Kinematic viscosity in m**2/s\n", + "b = (1./343);\t\t\t#Coefficient of thermal expansion in 1./K\n", + "Gr = ((9.81*b*L**3*(T1-T2))/(v1**2));\t\t\t#Grashof number\n", + "ke = (0.064*k*Gr**(1./3)*(D/L)**(-1./9));\t\t\t#Effective thermal conductivity in W/m.K\n", + "Q = (ke*D**2*(T1-T2))/L;\t\t\t#Rate of heat transfer in W\n", + "\n", + "# Results\n", + "print 'Effective thermal conductivity is %3.4f W/m.K Rate of heat transfer is %3.1f W'%(ke,Q)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Effective thermal conductivity is 0.0317 W/m.K Rate of heat transfer is 844.8 W\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.14 Page No : 356" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "d = 0.0254;\t\t\t#Diamath.tance between the plates in m\n", + "Tl = 60;\t\t\t#Temperature of the lower panel n degree C\n", + "Tu = 15.6;\t\t\t#Temperature of the upper panel in degree C\n", + "\n", + "# Calculations\n", + "Tf = (Tl+Tu)/2;\t\t\t#Film temperature in degree C\n", + "p = 1.121;\t\t\t#Density in kg/m**3\n", + "k = 0.0292;\t\t\t#Thermal conductivity in W/m.K\n", + "v1 = (0.171*10**-4);\t\t\t#Kinematic viscosity in m**2/s\n", + "b = (3.22*10**-3);\t\t\t#Coefficient of thermal expansion in 1./K\n", + "Pr = 0.7;\t\t\t#Prantl number\n", + "Gr = ((9.81*b*d**3*(Tl-Tu))/(v1**2));\t\t\t#Grashof number\n", + "Nu = (0.195*Gr**0.25);\t\t\t#Nussults number\n", + "q = (Nu*k*(Tl-Tu))/d;\t\t\t#Heat flux across the gap in W/m**2\n", + "\n", + "# Results\n", + "print 'Free convection heat transfer is %3.1f W/m**2'%(q)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Free convection heat transfer is 166.7 W/m**2\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.15 Page No : 359" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "p = 3;\t\t\t#Pressure of air in atm\n", + "r1 = 0.075;\t\t\t#Radius of first sphere in m\n", + "r2 = 0.1;\t\t\t#Radius of second sphere in m\n", + "L = 0.025;\t\t\t#Distance in m\n", + "T1 = 325;\t\t\t#Temperature of first sphere in K\n", + "T2 = 275;\t\t\t#Temperature of second sphere in K\n", + "R = 287;\t\t\t#Universal gas consmath.tant in J/\n", + "\n", + "# Calculations\n", + "Tf = (T1+T2)/2;\t\t\t#Film temperature in degree C\n", + "d = (p/(R*Tf));\t\t\t#Desnsity in kg/m**3\n", + "k = 0.0263;\t\t\t#Thermal conductivity in W/m.K\n", + "v1 = (5.23*10**-6);\t\t\t#Kinematic viscosity in m**2/s\n", + "b = (1./300);\t\t\t#Coefficient of thermal expansion in 1./K\n", + "Pr = 0.707;\t\t\t#Prantl numbe\n", + "Gr = ((9.81*b*L**3*(T1-T2))/(v1**2));\t\t\t#Grashof number\n", + "Ra = (Gr*Pr);\t\t\t#Rayleigh number\n", + "Ra1 = ((L/((4*r1*r2)**4))*(Ra/((2*r1)**(-7./5)+(2*r2)**(-7./5))**5))**0.25;\t\t\t#Equivalent Rayleigh's number\n", + "ke = (k*0.74*((Pr*Ra1)/(0.861+Pr))**0.25);\t\t \t#Effective thermal conductivity in W/m.K\n", + "Q = (ke*3.14*4*r1*r2*(T1-T2))/L;\t\t \t#Rate of heat loss in W\n", + "\n", + "# Results\n", + "print 'Convection heat transfer rate is %3.2f W'%(Q)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Convection heat transfer rate is 4.92 W\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.16 Page No : 362" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "p = 1;\t\t\t#Pressure of air in atm\n", + "Ta = 27;\t\t\t#Temperature of air in degree C\n", + "D = 0.02;\t\t\t#Diamter of the tube in m\n", + "v = 0.3;\t\t\t#Velocity of air in m/s\n", + "Ts = 127;\t\t\t#Surface temperature in degree C\n", + "L = 1;\t\t\t#Length of the tube in m\n", + "\n", + "# Calculations\n", + "k = 0.0262;\t\t\t#Thermal conductivity in W/m.K\n", + "v1 = (1.568*10**-5);\t\t\t#Kinematic viscosity in m**2/s\n", + "Pr = 0.708;\t\t\t#Prantl number\n", + "b = (1./300);\t\t\t#Coefficient of thermal expansion in 1./K\n", + "ub = (1.847*10**-5);\t\t\t#Dynamic viscosity in Ns/m**2\n", + "us = (2.286*10**-5);\t\t\t#Viscosity of wall in Ns/m**2\n", + "Re = (v*D)/v1;\t\t\t#Reynolds number\n", + "Gr = ((9.81*b*D**3*(Ts-Ta))/(v1**2));\t\t\t#Grashof number\n", + "Gz = (Re*Pr*(D/L));\t\t\t#Graetz number\n", + "Nu = (1.75*(ub/us)**0.14*(Gz+(0.012*(Gz*Gr**(1./3))**(4./3)))**(1./3));\t\t\t#Nussults number\n", + "h = (k*Nu)/D;\t\t\t#Heat transfer coefficient in W/m**2.K\n", + "Q = (h*3.14*D*L*(Ts-Ta));\t\t\t#Heat transfer in W\n", + "\n", + "# Results\n", + "print 'Heat transfer in the tube is %3.2f W'%(Q)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer in the tube is 40.86 W\n" + ] + } + ], + "prompt_number": 18 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch9.ipynb b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch9.ipynb new file mode 100755 index 00000000..bc6026e1 --- /dev/null +++ b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/ch9.ipynb @@ -0,0 +1,351 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:20bf663c37cc88094e4d6e0d7e23e0cfc96b3cd4eb305559eda1ce27188b9a4a" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 9 : Thermal radiation basic\n", + "relations" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.1 Page No : 378" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables\n", + "T = 5527;\t\t\t#Temperature of black body in degree C\n", + "D = (1.39*10**6);\t\t\t#Diameter of the sun in km\n", + "L = (1.5*10**8);\t\t\t#Distance between the earth and sun in km\n", + "\n", + "# Calculations\n", + "q = (5.67*10**-8*(T+273)**4*D**2)/(4*L**2);\t\t\t#Rate of solar radiation in W/m**2\n", + "\n", + "# Results\n", + "print 'Rate of solar radiation on a plane normal to sun rays is %3.0f W/m**2'%(q)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rate of solar radiation on a plane normal to sun rays is 1377 W/m**2\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.2 Page No : 383" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "T = (727+273);\t\t\t#Temperature of black body in K\n", + "l1 = 1;\t\t\t#Wavelength in micro meter\n", + "l2 = 5;\t\t\t#Wavelength in micro meter\n", + "F1 = 0.0003;\t\t\t#From Table 9.2 on page no. 385\n", + "F2 = 0.6337;\t\t\t#From Table 9.2 on page no. 385\n", + "\n", + "# Calculations\n", + "a = (5.67*10**-8*T**4)/1000;\t\t\t#Heat transfer in kW/m**2\n", + "F = (F2-F1)*a;\t\t\t#Fraction of thermal radiation emitted by the surface in kW/m**2\n", + "\n", + "# Results\n", + "print 'Fraction of thermal radiation emitted by the surface is %3.1f kW/m**2'%(F)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Fraction of thermal radiation emitted by the surface is 35.9 kW/m**2\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.3 Page No : 384" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "t = 0.8;\t\t\t#Transmittivity of glass in the region except in the wave length region [0.4,3]\n", + "T = 5555;\t\t\t#Temperature of black body in K\n", + "\n", + "# Calculations\n", + "ao = 0;\t\t\t#a0 in micro K\n", + "a1 = (0.4*T);\t\t\t#a1 for the wavelength 0.4 micro meter in micro K\n", + "a2 = (3*T);\t\t\t#a1 for the wavelength 3 micro meter in micro K\n", + "F0 = 0;\t\t\t#From Table 9.2 on page no.385\n", + "F1 = 0.10503;\t\t\t#From Table 9.2 on page no.385\n", + "F2 = 0.97644;\t\t\t#From Table 9.2 on page no.385\n", + "t1 = t*(F2-F1);\t\t\t#Average hemispherical transmittivity of glass \n", + "\n", + "# Results\n", + "print 'Average hemispherical transmittivity of glass is %3.2f'%(t1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Average hemispherical transmittivity of glass is 0.70\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.4 Page No : 386" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "l = 0.5;\t\t\t#Wavelength at maximum intensity of radiation in micro meter\n", + "C3 = 0.289*10**-2 #mK\n", + "\n", + "# Calculations\n", + "T = C3/(l*10**-6);\t\t\t#Temperature according to Wien's print lacement law in degree C\n", + "E = (5.67*10**-8*T**4)/10**6;\t\t\t#Emissive power umath.sing Stefan-Boltzmann law in MW/m**2\n", + "\n", + "# Results\n", + "print 'Surface temperature is %3.0f K Emissive power is %3.1f MW/m**2'%(T,E)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Surface temperature is 5780 K Emissive power is 63.3 MW/m**2\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.5 Page No : 389" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Ts = (827+273);\t\t\t#Surface temperature in degree C\n", + "E = (1.37*10**10);\t\t\t#Emmisive power in W/m**3\n", + "\n", + "# Calculations\n", + "Eblmax = (1.307*10**-5*Ts**5);\t\t\t#Maximum emissive power in W/m**3\n", + "e = (E/Eblmax);\t\t\t#Emissivity of the body \n", + "lmax = ((0.289*10**-2)/Ts)/10**-6;\t\t\t#Wavelength correspoing to the maximum spectral intensity of radiation in micro meter\n", + "\n", + "# Results\n", + "print 'Wavelength corresponding to the maximum spectral intensity of radiation is %3.2f micro meter'%(lmax)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength corresponding to the maximum spectral intensity of radiation is 2.63 micro meter\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.6 Page No : 389" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# Variables\n", + "T = (1400+273);\t\t\t#Temperature of the body in K\n", + "l = 0.65;\t\t\t#Wavelength in micro meter\n", + "e = 0.6;\t\t\t#Emissivity\n", + "\n", + "# Calculations\n", + "T = (1./((1./T)-((l*10**-6*math.log(1./e))/(1.439*10**-2))));\t\t\t#Temperature of the body in K\n", + "Tb = (T-273);\t\t\t#Temperature of the body in degree C\n", + "\n", + "# Results\n", + "print 'Temperature of the body is %3.0f degree C'%(Tb)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temperature of the body is 1467 degree C\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.7 Page No : 391" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Ts = (37+273);\t\t\t#Temperature of metallic bar in K\n", + "T = 1100;\t\t\t#Interior temperature in K\n", + "a = 0.52;\t\t\t#Absorptivity at 1100 K\n", + "e = 0.8;\t\t\t#Emissivity at 310 K\n", + "\n", + "# Calculations\n", + "Q = (a*5.67*10**-8*T**4)/1000;\t\t\t#Rate of absorption in kW/m**2\n", + "E = (e*5.67*10**-8*Ts**4)/1000;\t\t\t#Rate of emission in kW/m**2\n", + "\n", + "# Results\n", + "print 'Rate of absorption is %3.2f kW/m**2 \\n \\\n", + "Rate of emission is %3.2f kW/m**2'%(Q,E)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rate of absorption is 43.17 kW/m**2 \n", + " Rate of emission is 0.42 kW/m**2\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.8 Page No : 391" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "e1 = 0.3\t\t\t#Emissivity of glass upto 3 micro meter\n", + "e2 = 0.9;\t\t\t#Emissivity of glass above 3 micro meter\n", + "t = 0.8;\t\t\t#Transmittivity of glass in the region except in the wave length region [0.4,3]\n", + "\n", + "# Calculations\n", + "E = (5.67*10**-8*5780**4)/10**6;\t\t\t#Emissive power in MW/m**2\n", + "F1 = 0.10503;\t\t\t#From Table 9.2 on page no.385\n", + "F2 = 0.97644;\t\t\t#From Table 9.2 on page no.385\n", + "I = (E*10**6*(F2-F1))/10**6;\t\t\t#Total incident radiation in MW/m**2\n", + "T = (t*I);\t\t\t#Total radiation transmitted in MW/m**2\n", + "t1 = (e1*I);\t\t\t#Absorbed radiation in MW/m**2 in wavelength [0.4,3] micro meter\n", + "t2 = (e1*E*F1);\t\t\t#Absorbed radiation in MW/m**2 in wavelength not in the range [0.4,3] micro meter\n", + "t3 = (e2*(1-F2)*E);\t\t\t#Absorbed radiation in MW/m**2 in wavelength greater than 3 micro meter\n", + "R = (t1+t2+t3);\t\t\t#Total radiation absorbed in MW/m**2\n", + "\n", + "# Results\n", + "print 'Total radiation transmitted is %3.2f MW/m**2 \\n \\\n", + "Total radiation absorbed is %3.2f MW/m**2'%(T,R)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Total radiation transmitted is 44.12 MW/m**2 \n", + " Total radiation absorbed is 19.88 MW/m**2\n" + ] + } + ], + "prompt_number": 9 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/screenshots/1.png b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/screenshots/1.png Binary files differnew file mode 100755 index 00000000..b37b0877 --- /dev/null +++ b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/screenshots/1.png diff --git a/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/screenshots/2.png b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/screenshots/2.png Binary files differnew file mode 100755 index 00000000..3740e2b0 --- /dev/null +++ b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/screenshots/2.png diff --git a/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/screenshots/3.png b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/screenshots/3.png Binary files differnew file mode 100755 index 00000000..b452eb4f --- /dev/null +++ b/Fundamentals_Of_Engineering_Heat_And_Mass_Transfer/screenshots/3.png diff --git a/Fundamentals_of_Electrical_Machines/CH_1.ipynb b/Fundamentals_of_Electrical_Machines/CH_1.ipynb new file mode 100755 index 00000000..801b1ed1 --- /dev/null +++ b/Fundamentals_of_Electrical_Machines/CH_1.ipynb @@ -0,0 +1,731 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 1: REVIEW OF ELECTRIC CIRCUITS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.1, Page number 2-3"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "Q = 4.0 #Charge(C) \n",
+ "t = 0.54 #Time(sec) \n",
+ "\n",
+ "#Calculation\n",
+ "I = Q/t #Current(A) \n",
+ "\n",
+ "#Result\n",
+ "print('Value of Current is , I = %.2f A' %I) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of Current is , I = 7.41 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.2, Page number 4-5"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "V = -24.0 #Voltage(V)\n",
+ "I = 3.0 #Current(A)\n",
+ "\n",
+ "#Calculation \n",
+ "P = V*I #Power supplied by the element A(W) \n",
+ "\n",
+ "#Result\n",
+ "print('Power supplied by the element A is , P = %.1f W' %P)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Power supplied by the element A is , P = -72.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.3, Page number 7-9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "R1 = 5.0 #Resistance(ohm)\n",
+ "R2 = 4.0 #Resistance(ohm)\n",
+ "R3 = 9.0 #Resistance(ohm)\n",
+ "R4 = 6.0 #Resistance(ohm)\n",
+ "V1 = 10.0 #Resistance(ohm)\n",
+ "V2 = 6.0 #Resistance(ohm)\n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "R_th = (R1*R4/(R1+R4))+R2 #Thevenin resistance(ohm) by removing R3 & short-circuiting voltage sources\n",
+ "I = (V1-V2)/(R1+R4) #Current(A) by applying KVL\n",
+ "V_th = 6*I+V2 #Thevenin voltage(V) by applying KVL\n",
+ "I_9ohm = V_th/(R_th+R3) #Current through 9 ohm resistor(A)\n",
+ "\n",
+ "#Result\n",
+ "print('Current through 9 ohm resistor , I_9\u03a9 = %.2f A' %I_9ohm) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current through 9 ohm resistor , I_9\u03a9 = 0.52 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.4, Page number 10-11"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from scipy.integrate import quad\n",
+ "\n",
+ "#Variable declaration\n",
+ "V_t1 = 30.0 #Magnitudes of voltages(V) 0 < t1 < 2\n",
+ "V_t2 = -10.0 #Magnitudes of voltages(V) 2 < t2 < 4\n",
+ "T = 4.0 #Time period(sec) from figure\n",
+ "\n",
+ "#Calculation\n",
+ "def integrand(V):\n",
+ " return V**0\n",
+ "\n",
+ "a, err = quad(integrand, 0, 2)\n",
+ "\n",
+ "def integrand(V):\n",
+ " return V**0\n",
+ "\n",
+ "b, err = quad(integrand, 2, 4)\n",
+ "\n",
+ "V_rms = ((a*V_t1**2+b*V_t2**2)/4)**0.5 #RMS value of voltage waveform(V)\n",
+ "\n",
+ "#Result\n",
+ "print('RMS value , V_rms = %.2f V' %V_rms)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "RMS value , V_rms = 22.36 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.5, Page number 15-16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import cmath\n",
+ "\n",
+ "#Variable declaration\n",
+ "V_P = 200.0 #Magnitude of each phase(V) \n",
+ "\n",
+ "#Calculation\n",
+ "V_an = V_P*cmath.exp(1j*0*math.pi/180) #Magnitude of 3-phase voltage(V)\n",
+ "V_bn = V_P*cmath.exp(1j*-120*math.pi/180) #Magnitude of 3-phase voltage(V)\n",
+ "V_cn = V_P*cmath.exp(1j*120*math.pi/180) #Magnitude of 3-phase voltage(V)\n",
+ "V_L = 3**0.5*V_P #Magnitude of line voltage(V)\n",
+ "\n",
+ "#Result\n",
+ "print('Expression of phase voltages are,')\n",
+ "print('\\t\\t\\t V_an = %.f\u2220%.f\u00b0 V' %(abs(V_an),cmath.phase(V_an)))\n",
+ "print('\\t\\t\\t V_bn = %.f\u2220%.f\u00b0 V' %(abs(V_bn),cmath.phase(V_bn)*180/math.pi))\n",
+ "print('\\t\\t\\t V_cn = %.f\u2220%.f\u00b0 V' %(abs(V_cn),cmath.phase(V_cn)*180/math.pi))\n",
+ "print('Magnitude of the line voltage , V_L = %.1f V' %V_L) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Expression of phase voltages are,\n",
+ "\t\t\t V_an = 200\u22200\u00b0 V\n",
+ "\t\t\t V_bn = 200\u2220-120\u00b0 V\n",
+ "\t\t\t V_cn = 200\u2220120\u00b0 V\n",
+ "Magnitude of the line voltage , V_L = 346.4 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.6, Page number 16-17"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import cmath\n",
+ "\n",
+ "#Variable declaration\n",
+ "R = 10.0 #Resistance of each coil(ohm)\n",
+ "X = 15.0 #Inductive reactance of each coil(ohm)\n",
+ "V_L = 420.0 #Line voltage(V)\n",
+ "f = 50.0 #Frequency of supply(Hz)\n",
+ "\n",
+ "#Calculation\n",
+ "V_an = (V_L/3**0.5)*cmath.exp(1j*(0-30)*math.pi/180) #Phase voltage(V)\n",
+ "V_bn = (V_L/3**0.5)*cmath.exp(1j*(-120-30)*math.pi/180) #Phase voltage(V)\n",
+ "V_cn = (V_L/3**0.5)*cmath.exp(1j*(120-30)*math.pi/180) #Phase voltage(V)\n",
+ "Z_P = complex(R,X) #Phase impedance(ohm)\n",
+ "#For case(i)\n",
+ "I_L1 = V_an/Z_P #Line current(A)\n",
+ "I_L2 = V_bn/Z_P #Line current(A)\n",
+ "I_L3 = V_cn/Z_P #Line current(A)\n",
+ "#For case(ii)\n",
+ "pf = R/abs(Z_P) #Power factor\n",
+ "\n",
+ "#Result\n",
+ "print('(i) Values of line currents are,')\n",
+ "print('\\t I_L1 = I_an = %.2f\u2220%.2f\u00b0 A' %(abs(I_L1),cmath.phase(I_L1)*180/math.pi))\n",
+ "print('\\t I_L2 = I_bn = %.2f\u2220%.2f\u00b0 A' %(abs(I_L2),cmath.phase(I_L2)*180/math.pi))\n",
+ "print('\\t I_L3 = I_cn = %.2f\u2220%.2f\u00b0 A' %(abs(I_L3),cmath.phase(I_L3)*180/math.pi))\n",
+ "print('(ii) Power factor is , pf = %.1f lag' %pf)\n",
+ "print('\\nNOTE : I_L2 has an angle -206.31\u00b0 in textbook which is same as 153.69\u00b0 i.e (360-206.31)\u00b0 obtained here')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Values of line currents are,\n",
+ "\t I_L1 = I_an = 13.45\u2220-86.31\u00b0 A\n",
+ "\t I_L2 = I_bn = 13.45\u2220153.69\u00b0 A\n",
+ "\t I_L3 = I_cn = 13.45\u222033.69\u00b0 A\n",
+ "(ii) Power factor is , pf = 0.6 lag\n",
+ "\n",
+ "NOTE : I_L2 has an angle -206.31\u00b0 in textbook which is same as 153.69\u00b0 i.e (360-206.31)\u00b0 obtained here\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.7, Page number 19-20"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import cmath\n",
+ "\n",
+ "#Variable declaration\n",
+ "Z_P = complex(10,15) #Per phase impedance(ohm)\n",
+ "V_L = 420.0 #Voltage(V)\n",
+ "\n",
+ "#Calculation\n",
+ "#For case(i)\n",
+ "V_ab = V_L*cmath.exp(1j*0*math.pi/180) #Phase voltage(V)\n",
+ "V_bc = V_L*cmath.exp(1j*-120*math.pi/180) #Phase voltage(V)\n",
+ "V_ca = V_L*cmath.exp(1j*120*math.pi/180) #Phase voltage(V)\n",
+ "I_ab = V_ab/Z_P #Phase current(A)\n",
+ "I_bc = V_bc/Z_P #Phase current(A)\n",
+ "I_ca = V_ca/Z_P #Phase current(A)\n",
+ "#For case(ii)\n",
+ "I_P = abs(I_ab) #Phase current magnitude(A)\n",
+ "I_L = 3**0.5*I_P #Line current magnitude(A)\n",
+ "\n",
+ "#Result\n",
+ "print('(i) Phase currents are,')\n",
+ "print('\\t\\t I_ab = %.2f\u2220%.2f\u00b0 A' %(abs(I_ab),cmath.phase(I_ab)*180/math.pi))\n",
+ "print('\\t\\t I_bc = %.2f\u2220%.2f\u00b0 A' %(abs(I_bc),cmath.phase(I_bc)*180/math.pi))\n",
+ "print('\\t\\t I_ca = %.2f\u2220%.2f\u00b0 A' %(abs(I_ca),cmath.phase(I_ca)*180/math.pi))\n",
+ "print('(ii) Magnitude of line current , I_L = %.2f A' %I_L)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Phase currents are,\n",
+ "\t\t I_ab = 23.30\u2220-56.31\u00b0 A\n",
+ "\t\t I_bc = 23.30\u2220-176.31\u00b0 A\n",
+ "\t\t I_ca = 23.30\u222063.69\u00b0 A\n",
+ "(ii) Magnitude of line current , I_L = 40.35 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.8, Page number 22-23"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import cmath\n",
+ "\n",
+ "#Variable declaration\n",
+ "V_P = 280.0 #Generator Phase voltage(V)\n",
+ "Z_P = complex(2,3) #Line impedance per phase(ohm)\n",
+ "Z_L = complex(4,5) #Load impedance per phase(ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "V_An = V_P*cmath.exp(1j*0*math.pi/180) #Phase voltage(V)\n",
+ "V_Bn = V_P*cmath.exp(1j*-120*math.pi/180) #Phase voltage(V)\n",
+ "V_Cn = V_P*cmath.exp(1j*120*math.pi/180) #Phase voltage(V)\n",
+ "Z_t = Z_P+Z_L #Total impedance(ohm)\n",
+ "I_Aa = V_An/Z_t #Magnitude of line current for phase A(A)\n",
+ "I_Bb = V_Bn/Z_t #Magnitude of line current for phase B(A)\n",
+ "I_Cc = V_Cn/Z_t #Magnitude of line current for phase C(A)\n",
+ "V_an = I_Aa*Z_L #Phase voltage of load(V)\n",
+ "V_bn = I_Bb*Z_L #Phase voltage of load(V)\n",
+ "V_cn = I_Cc*Z_L #Phase voltage of load(V)\n",
+ "\n",
+ "#Result\n",
+ "print('Line currents are,')\n",
+ "print('\\t\\t I_Aa = %.f\u2220%.f\u00b0 A' %(abs(I_Aa),cmath.phase(I_Aa)*180/math.pi))\n",
+ "print('\\t\\t I_Bb = %.f\u2220%.f\u00b0 A' %(abs(I_Bb),cmath.phase(I_Bb)*180/math.pi))\n",
+ "print('\\t\\t I_Cc = %.f\u2220%.f\u00b0 A' %(abs(I_Cc),cmath.phase(I_Cc)*180/math.pi))\n",
+ "print('\\nLoad phase voltages are,')\n",
+ "print('\\t\\t V_an = %.1f\u2220%.1f\u00b0 V' %(abs(V_an),cmath.phase(V_an)*180/math.pi))\n",
+ "print('\\t\\t V_bn = %.1f\u2220%.1f\u00b0 V' %(abs(V_bn),cmath.phase(V_bn)*180/math.pi))\n",
+ "print('\\t\\t V_cn = %.1f\u2220%.1f\u00b0 V' %(abs(V_cn),cmath.phase(V_cn)*180/math.pi))\n",
+ "print('\\nNOTE : ERROR : Z_L = 6.4\u222038.6\u00b0\u03a9 is taken in textbook solution instead of 6.4\u222051.34\u00b0\u03a9 = (4+j5)\u03a9')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Line currents are,\n",
+ "\t\t I_Aa = 28\u2220-53\u00b0 A\n",
+ "\t\t I_Bb = 28\u2220-173\u00b0 A\n",
+ "\t\t I_Cc = 28\u222067\u00b0 A\n",
+ "\n",
+ "Load phase voltages are,\n",
+ "\t\t V_an = 179.3\u2220-1.8\u00b0 V\n",
+ "\t\t V_bn = 179.3\u2220-121.8\u00b0 V\n",
+ "\t\t V_cn = 179.3\u2220118.2\u00b0 V\n",
+ "\n",
+ "NOTE : ERROR : Z_L = 6.4\u222038.6\u00b0\u03a9 is taken in textbook solution instead of 6.4\u222051.34\u00b0\u03a9 = (4+j5)\u03a9\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.9, Page number 23-24"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import cmath\n",
+ "\n",
+ "#Variable declaration\n",
+ "Z = complex(6,8) #Per phase impedance of load(ohm)\n",
+ "V_AN = 340.0*cmath.exp(1j*0*math.pi/180) #Phase voltage(V)\n",
+ "\n",
+ "#Calculation\n",
+ "V_P = abs(V_AN) #Voltage(V)\n",
+ "V_BN = V_P*cmath.exp(1j*-120*math.pi/180) #Phase voltage(V)\n",
+ "V_CN = V_P*cmath.exp(1j*120*math.pi/180) #Phase voltage(V)\n",
+ "I_an = V_AN/Z #Load current(A)\n",
+ "I_bn = V_BN/Z #Load current(A)\n",
+ "I_cn = V_CN/Z #Load current(A)\n",
+ "I_n = I_an+I_bn+I_cn #Neutral current(A)\n",
+ "\n",
+ "#Result\n",
+ "print('Phase current in each load = Line current in each load are,')\n",
+ "print('\\t\\t\\t\\t I_an = I_Aa = %.f\u2220%.f\u00b0 A' %(abs(I_an),cmath.phase(I_an)*180/math.pi))\n",
+ "print('\\t\\t\\t\\t I_bn = I_Bb = %.f\u2220%.f\u00b0 A' %(abs(I_bn),cmath.phase(I_bn)*180/math.pi))\n",
+ "print('\\t\\t\\t\\t I_cn = I_Cc = %.f\u2220%.f\u00b0 A' %(abs(I_cn),cmath.phase(I_cn)*180/math.pi))\n",
+ "print('Neutral current is , I_n = %.f A' %abs(I_n))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Phase current in each load = Line current in each load are,\n",
+ "\t\t\t\t I_an = I_Aa = 34\u2220-53\u00b0 A\n",
+ "\t\t\t\t I_bn = I_Bb = 34\u2220-173\u00b0 A\n",
+ "\t\t\t\t I_cn = I_Cc = 34\u222067\u00b0 A\n",
+ "Neutral current is , I_n = 0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.10, Page number 25-26"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import cmath\n",
+ "\n",
+ "#Variable declaration\n",
+ "Z = complex(3,4) #Per phase impedance of load(ohm)\n",
+ "V_AN = 200.0*cmath.exp(1j*0*math.pi/180) #Phase voltage(V)\n",
+ "\n",
+ "#Calculation\n",
+ "V_P = abs(V_AN) #Voltage(V)\n",
+ "V_AB = 3**0.5*V_P*cmath.exp(1j*30*math.pi/180) #Line voltage(V)\n",
+ "V_BC = 3**0.5*V_P*cmath.exp(1j*-90*math.pi/180) #Line voltage(V)\n",
+ "V_CA = 3**0.5*V_P*cmath.exp(1j*150*math.pi/180) #Line voltage(V)\n",
+ "#For case(i)\n",
+ "I_ab = V_AB/Z #Load current(A)\n",
+ "I_bc = V_BC/Z #Load current(A)\n",
+ "I_ca = V_CA/Z #Load current(A)\n",
+ "#For case(ii)\n",
+ "I_Aa = I_ab-I_ca #Line current(A)\n",
+ "I_Bb = I_bc-I_ab #Line current(A)\n",
+ "I_Cc = I_ca-I_bc #Line current(A)\n",
+ "\n",
+ "#Result\n",
+ "print('(i) Magnitude of load currents are,')\n",
+ "print('\\t\\t I_ab = %.1f\u2220%.2f\u00b0 A' %(abs(I_ab),cmath.phase(I_ab)*180/math.pi))\n",
+ "print('\\t\\t I_bc = %.1f\u2220%.2f\u00b0 A' %(abs(I_bc),cmath.phase(I_bc)*180/math.pi))\n",
+ "print('\\t\\t I_ca = %.1f\u2220%.2f\u00b0 A' %(abs(I_ca),cmath.phase(I_ca)*180/math.pi))\n",
+ "print('\\n(ii) Magnitude of line currents are,')\n",
+ "print('\\t\\t I_Aa = %.2f\u2220%.2f\u00b0 A' %(abs(I_Aa),cmath.phase(I_Aa)*180/math.pi))\n",
+ "print('\\t\\t I_Bb = %.2f\u2220%.2f\u00b0 A' %(abs(I_Bb),cmath.phase(I_Bb)*180/math.pi))\n",
+ "print('\\t\\t I_Cc = %.2f\u2220%.2f\u00b0 A' %(abs(I_Cc),cmath.phase(I_Cc)*180/math.pi))\n",
+ "print('\\nNOTE : ERROR : Calculation mistakes in textbook')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Magnitude of load currents are,\n",
+ "\t\t I_ab = 69.3\u2220-23.13\u00b0 A\n",
+ "\t\t I_bc = 69.3\u2220-143.13\u00b0 A\n",
+ "\t\t I_ca = 69.3\u222096.87\u00b0 A\n",
+ "\n",
+ "(ii) Magnitude of line currents are,\n",
+ "\t\t I_Aa = 120.00\u2220-53.13\u00b0 A\n",
+ "\t\t I_Bb = 120.00\u2220-173.13\u00b0 A\n",
+ "\t\t I_Cc = 120.00\u222066.87\u00b0 A\n",
+ "\n",
+ "NOTE : ERROR : Calculation mistakes in textbook\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.11, Page number 28-29"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import cmath\n",
+ "\n",
+ "#Variable declaration\n",
+ "Z = complex(3,4) #Per phase impedance of load(ohm)\n",
+ "V_AN = 150.0*cmath.exp(1j*0*math.pi/180) #Phase voltage(V)\n",
+ "\n",
+ "#Calculation\n",
+ "V_P = abs(V_AN) #Voltage(V)\n",
+ "V_BN = V_P*cmath.exp(1j*-120*math.pi/180) #Phase voltage(V)\n",
+ "V_CN = V_P*cmath.exp(1j*120*math.pi/180) #Phase voltage(V)\n",
+ "I_Aa = V_AN/Z #Line current(A)\n",
+ "I_Bb = V_BN/Z #Line current(A)\n",
+ "I_Cc = V_CN/Z #Line current(A)\n",
+ "pf = Z.real/abs(Z) #Power factor\n",
+ "I = abs(I_Aa) #Magnitude of line current(A)\n",
+ "P = V_P*I*pf*10**-3 #Power supplied to each phase(kW)\n",
+ "P_t = 3*P #Total power supplied(kW)\n",
+ "\n",
+ "#Result\n",
+ "print('Line currents are,')\n",
+ "print(' I_Aa = %.f\u2220%.2f\u00b0 A' %(abs(I_Aa),cmath.phase(I_Aa)*180/math.pi))\n",
+ "print(' I_Bb = %.f\u2220%.2f\u00b0 A' %(abs(I_Bb),cmath.phase(I_Bb)*180/math.pi))\n",
+ "print(' I_Cc = %.f\u2220%.2f\u00b0 A' %(abs(I_Cc),cmath.phase(I_Cc)*180/math.pi))\n",
+ "print('Power factor , pf = %.1f ' %pf)\n",
+ "print('Power supplied to each phase , P = %.1f kW' %P)\n",
+ "print('Total Power supplied to the load , P_t = %.1f kW' %P_t)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Line currents are,\n",
+ " I_Aa = 30\u2220-53.13\u00b0 A\n",
+ " I_Bb = 30\u2220-173.13\u00b0 A\n",
+ " I_Cc = 30\u222066.87\u00b0 A\n",
+ "Power factor , pf = 0.6 \n",
+ "Power supplied to each phase , P = 2.7 kW\n",
+ "Total Power supplied to the load , P_t = 8.1 kW\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.12, Page number 32"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "P = 120.0 #Total power(kW)\n",
+ "pf = 0.6 #Power factor \n",
+ "\n",
+ "#Calculation\n",
+ "teta = math.acos(pf) #Power factor angle(radians)\n",
+ "teta_deg = teta*180/math.pi #Power factor angle(degree)\n",
+ "P_2 = 1.0/2*((math.tan(teta)*P/3**0.5)+P) #Second wattmeter reading(kW)\n",
+ "\n",
+ "#Result\n",
+ "print('Second wattmeter reading , P_2 = %.1f kW' %P_2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Second wattmeter reading , P_2 = 106.2 kW\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.13, Page number 35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "P = 5000.0 #Power(W)\n",
+ "pf_1 = 0.8 #Initial Power factor\n",
+ "V = 110.0 #rms Voltage(V)\n",
+ "f = 50.0 #Frequency(Hz)\n",
+ "pf_2 = 0.9 #Final Power factor\n",
+ "\n",
+ "#Calculation\n",
+ "phi_1 = math.acos(pf_1) #Initial Power factor angle(radians)\n",
+ "phi_1_deg = phi_1*180/math.pi #Initial Power factor angle(degree)\n",
+ "phi_2 = math.acos(pf_2) #Final Power factor angle(radians)\n",
+ "phi_2_deg = phi_2*180/math.pi #Final Power factor angle(degree)\n",
+ "C = P*(math.tan(phi_1)-math.tan(phi_2))/(2*math.pi*f*V**2)*10**6 #Parallel capacitance(\u00b5F)\n",
+ "\n",
+ "#Result\n",
+ "print('Capacitance , C = %.1f \u00b5F' %C)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitance , C = 349.5 \u00b5F\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.14, Page number 35-36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "pf_1 = 0.85 #Initial Power factor\n",
+ "kVA = 20.0 #Load(kVA)\n",
+ "f = 50.0 #Frequency(Hz)\n",
+ "pf_2 = 0.95 #Final Power factor\n",
+ "V = 200.0 #Voltage(V)\n",
+ "R = 0.05 #Resistance(ohm)\n",
+ "X = 0.2 #Inductive reactance(ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "phi_1 = math.acos(pf_1) #Initial Power factor angle(radians)\n",
+ "phi_1_deg = phi_1*180/math.pi #Initial Power factor angle(degree)\n",
+ "phi_2 = math.acos(pf_2) #Final Power factor angle(radians)\n",
+ "phi_2_deg = phi_2*180/math.pi #Final Power factor angle(degree)\n",
+ "P = kVA*pf_1 #Load power(kW)\n",
+ "C = P*1000*(math.tan(phi_1)-math.tan(phi_2))/(2*math.pi*f*V**2)*10**6 #Parallel capacitance(\u00b5F)\n",
+ "#Before adding capacitor\n",
+ "I_1 = P*1000/(pf_1*V) #Line current(A)\n",
+ "P_1 = I_1**2*R #Power loss in line(W)\n",
+ "#After adding capacitor\\n\",\n",
+ "S = P*1000/pf_2 #Apparent power(VA)\n",
+ "I_2 = S/V #Line current(A)\n",
+ "P_2 = I_2**2*R #Power loss in line(W)\n",
+ "\n",
+ "#Result\n",
+ "print('Capacitance , C = %.1f \u00b5F' %C)\n",
+ "print('Power loss in the line before adding capacitor , P_1 = %.1f W' %P_1)\n",
+ "print('Power loss in the line after adding capacitor , P_2 = %.1f W' %P_2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitance , C = 393.8 \u00b5F\n",
+ "Power loss in the line before adding capacitor , P_1 = 500.0 W\n",
+ "Power loss in the line after adding capacitor , P_2 = 400.3 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 10: SYNCHRONOUS MOTOR"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.1, Page number 335"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "V = 2.5*10**3 #Supply voltage(V)\n",
+ "R_r = 0.12 #Per phase resistance(ohm)\n",
+ "X_r = 3.2 #Syncronous reactance(ohm)\n",
+ "I_a = 185.0 #Line current(A)\n",
+ "pf = 0.8 #Leading power factor\n",
+ "\n",
+ "#Calculation\n",
+ "phi = math.acos(pf) #Angle(radians)\n",
+ "phi_deg = phi*180/math.pi #Angle(degree)\n",
+ "V_t = V/3**0.5 #Terminal voltage per phase(V)\n",
+ "Z_s = complex(R_r,X_r) #Impedance per phase(ohm)\n",
+ "beta = math.atan(X_r/R_r) #Angle(radians)\n",
+ "beta_deg = beta*180/math.pi #Angle(degree)\n",
+ "E_r = I_a*abs(Z_s) #Resultant voltage due to impedance(V)\n",
+ "E_f = (V_t**2+E_r**2-2*V_t*E_r*math.cos(beta+phi))**0.5 #Excitation voltage per phase(V)\n",
+ "\n",
+ "#Result\n",
+ "print('Excitation voltage per phase , E = %.2f V' %E_f)\n",
+ "print('\\nNOTE : Changes in answer is due to precision i.e more number of decimal places')\n",
+ "print(' ERROR : Line current I_a = 185 A not 180 A as given in textbook question')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Excitation voltage per phase , E = 1846.18 V\n",
+ "\n",
+ "NOTE : Changes in answer is due to precision i.e more number of decimal places\n",
+ " ERROR : Line current I_a = 185 A not 180 A as given in textbook question\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.2, Page number 335-337"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "kVA = 1200.0 #kVA ratings\n",
+ "V = 14.0*10**3 #Supply voltage(V)\n",
+ "R_r = 4.8 #Per phase resistance(ohm)\n",
+ "X_r = 35.0 #Syncronous reactance(ohm)\n",
+ "pf = 0.95 #Leading power factor\n",
+ "\n",
+ "#Calculation\n",
+ "phi = math.acos(pf) #Angle(radians)\n",
+ "phi_deg = phi*180/math.pi #Angle(degree)\n",
+ "Z_s = complex(R_r,X_r) #Impedance per phase(ohm)\n",
+ "I_a = kVA*10**3/(3**0.5*V) #Armature current(A)\n",
+ "E_r = I_a*abs(Z_s) #Resultant voltage due to impedance(V)\n",
+ "V_t = V/3**0.5 #Terminal voltage per phase(V)\n",
+ "b = math.atan(X_r/R_r) #Beta value(radians)\n",
+ "b_deg = b*180/math.pi #Beta value(degree)\n",
+ "E_f = (V_t**2+E_r**2-2*V_t*E_r*math.cos(b-phi))**0.5 #Excitation voltage per phase(V)\n",
+ "sin_delta = (E_r/E_f)*math.sin(b-phi)\n",
+ "delta = math.asin(sin_delta)*180/math.pi #Torque angle(degree)\n",
+ "\n",
+ "#Result\n",
+ "print('Excitation voltage per phase , E_f = %.2f V' %E_f)\n",
+ "print('Torque angle , \u03b4 = %.2f\u00b0' %delta)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Excitation voltage per phase , E_f = 7483.23 V\n",
+ "Torque angle , \u03b4 = 12.12\u00b0\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.3, Page number 343-344"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "V = 440.0 #Supply voltage(V)\n",
+ "R_a = 1.5 #Per phase armature resistance(ohm)\n",
+ "X_a = 8.0 #Synchronous reactance(ohm)\n",
+ "P = 4.0 #Number of poles\n",
+ "f = 50.0 #Supply frequency(Hz)\n",
+ "pf = 0.9 #Leading power factor\n",
+ "I_a = 50.0 #Armature current(A)\n",
+ "\n",
+ "#Calculation\n",
+ "V_t = V/3**0.5 #Terminal voltage per phase(V)\n",
+ "phi = math.acos(pf) #Angle(radians)\n",
+ "phi_deg = phi*180/math.pi #Angle(degree)\n",
+ "Z_s = complex(R_a,X_a) #Impedance per phase(ohm)\n",
+ "E_r = I_a*abs(Z_s) #Resultant voltage due to impedance(V)\n",
+ "beta = math.atan(X_a/R_a) #Beta value(radians)\n",
+ "beta_deg = beta*180/math.pi #Beta value(degree)\n",
+ "E_f = (V_t**2+E_r**2-2*V_t*E_r*math.cos(beta+phi))**0.5 #Excitation voltage per phase(V)\n",
+ "P_dm = (((E_f*V_t)/abs(Z_s))-((E_f**2*R_a)/abs(Z_s)**2)) #Maximum power per phase(W)\n",
+ "\n",
+ "#Result \n",
+ "print('Maximum power per phase , P_dm = %.1f W' %P_dm)\n",
+ "print('\\nNOTE : ERROR : In textbook solution E_f = 513.5 V is taken instead of 533.337089826 V')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum power per phase , P_dm = 10205.3 W\n",
+ "\n",
+ "NOTE : ERROR : In textbook solution E_f = 513.5 V is taken instead of 533.337089826 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.4, Page number 344"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "P = 4.0 #Number of poles\n",
+ "f = 50.0 #Supply frequency(Hz)\n",
+ "V_t = 1500.0 #Terminal voltage per phase(V)\n",
+ "E_f = 1000.0 #Excitation voltage per phase(V)\n",
+ "Z_s = 12.0 #Synchronous impedance per phase(ohm)\n",
+ "R_a = 1.5 #Armature resistance(ohm)\n",
+ "\n",
+ "#Caclulation\n",
+ "P_dm = ((E_f*V_t/Z_s)-(E_f**2*R_a/Z_s**2)) #Maximum power(W)\n",
+ "N_s = 120*f/P #Synchronous speed(rpm)\n",
+ "T_dm = 9.55*P_dm/N_s #Maximum torque(N-m)\n",
+ "\n",
+ "#Result \n",
+ "print('Maximum power developed , P_dm = %.f W' %P_dm)\n",
+ "print('Maximum toruqe , T_dm = %.1f N-m' %T_dm)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum power developed , P_dm = 114583 W\n",
+ "Maximum toruqe , T_dm = 729.5 N-m\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Fundamentals_of_Electrical_Machines/CH_11.ipynb b/Fundamentals_of_Electrical_Machines/CH_11.ipynb new file mode 100755 index 00000000..3c86183e --- /dev/null +++ b/Fundamentals_of_Electrical_Machines/CH_11.ipynb @@ -0,0 +1,354 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 11: SINGLE-PHASE MOTORS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.1, Page number 354-355"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "V = 220.0 #Supply voltage(V)\n",
+ "P = 4.0 #Number of poles\n",
+ "f = 50.0 #Frequency(Hz)\n",
+ "N_l = 1450.0 #Speed(rpm)\n",
+ "R_2 = 10.0 #Rotor resistance at standstill(ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "N_s = 120*f/P #Synchronous speed(rpm)\n",
+ "#For case(i)\n",
+ "s_f = (N_s-N_l)/N_s #Slip due to forward field\n",
+ "#For case(ii)\n",
+ "s_b = 2-s_f #Slip due to backward field\n",
+ "#For case(iii)\n",
+ "R_f = R_2/s_f #Effective rotor resistance due to forward slip(ohm)\n",
+ "R_b = R_2/(2-s_f) #Effective rotor resistance due to backward slip(ohm)\n",
+ "\n",
+ "#Result\n",
+ "print('(i) Slip due to forward field , s_f = %.2f ' %s_f)\n",
+ "print('(ii) Slip due to backward field , s_b = %.2f ' %s_b)\n",
+ "print('(iii) Effective rotor resistance due to forward slip , R_f = %.2f ohm' %R_f)\n",
+ "print(' Effective rotor resistance due to backward slip , R_b = %.2f ohm' %R_b)\n",
+ "print('\\nNOTE : Changes in answer from that of textbook is due to precision')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Slip due to forward field , s_f = 0.03 \n",
+ "(ii) Slip due to backward field , s_b = 1.97 \n",
+ "(iii) Effective rotor resistance due to forward slip , R_f = 300.00 ohm\n",
+ " Effective rotor resistance due to backward slip , R_b = 5.08 ohm\n",
+ "\n",
+ "NOTE : Changes in answer from that of textbook is due to precision\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.2, Page number 357-358"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import cmath\n",
+ "\n",
+ "#Variable declaration\n",
+ "V_t = 220.0 #Supply voltage(V)\n",
+ "R_1 = 6.0 #Resistance(ohm)\n",
+ "R_2 = 6.0 #Resistance(ohm)\n",
+ "X_1 = 10.0 #Inductive reactance(ohm)\n",
+ "X_2 = 10.0 #Inductive reactance(ohm)\n",
+ "N = 1500.0 #Speed(rpm)\n",
+ "s = 0.03 #Slip\n",
+ "X_m = 150.0 #Inductive reactance(ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "Z_f = 0.5*complex(0,X_m)*complex(R_2/s,X_2)/complex(R_2/s,X_2+X_m) #Impedance due to forward field(ohm)\n",
+ "Z_b = 0.5*complex(0,X_m)*complex(R_2/(2-s),X_2)/complex(R_2/(2-s),X_2+X_m) #Impedance due to backward field(ohm)\n",
+ "Z_t = complex(R_1+Z_f+Z_b,X_1) #Total impedance(ohm)\n",
+ "#For case(i)\n",
+ "I_1 = V_t/Z_t #Input current(A)\n",
+ "#For case(ii)\n",
+ "P_i = V_t*abs(I_1) #Input power(W)\n",
+ "#For case(iii)\n",
+ "R_f = Z_f.real\n",
+ "R_b = Z_b.real\n",
+ "P_d = abs(I_1)**2*(R_f-R_b)*(1-s) #Power developed(W)\n",
+ "#For case(iv)\n",
+ "T_d = 9.55*P_d/N #Torque(N-m)\n",
+ "\n",
+ "#Result\n",
+ "print('(i) Input current , I_1 = %.2f\u2220%.1f\u00b0 A' %(abs(I_1),cmath.phase(I_1)*180/math.pi))\n",
+ "print('(ii) Input power , P_i = %.2f W' %P_i)\n",
+ "print('(iii) Power developed , P_d = %.1f W' %P_d)\n",
+ "print('(iv) Torque developed , T_d = %.2f N-m' %T_d)\n",
+ "print('\\nNOTE : Case(ii) is not solved in textbook but solved here')\n",
+ "print(' ERROR : Calculation mistake in Z_b in textbook solution')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Input current , I_1 = 2.94\u2220-56.2\u00b0 A\n",
+ "(ii) Input power , P_i = 646.10 W\n",
+ "(iii) Power developed , P_d = 275.8 W\n",
+ "(iv) Torque developed , T_d = 1.76 N-m\n",
+ "\n",
+ "NOTE : Case(ii) is not solved in textbook but solved here\n",
+ " ERROR : Calculation mistake in Z_b in textbook solution\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.3, Page number 363-364"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import cmath\n",
+ "\n",
+ "#Variable declaration\n",
+ "V_t = 220.0 #Supply voltage(V)\n",
+ "f = 50.0 #Frequency(Hz)\n",
+ "Z_m = complex(3,5) #Main winding impedance of motor(ohm)\n",
+ "Z_s = complex(5,3) #Starting impedance of motor(ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "#For case(i)\n",
+ "alpha_s = cmath.phase(Z_s) #Starting winding impedance angle(radians)\n",
+ "I_s = V_t/Z_s #Starting current(A)\n",
+ "#For case(ii)\n",
+ "alpha_m = cmath.phase(Z_m) #Main winding impedance angle(radians)\n",
+ "I_m = V_t/Z_m #Main winding current(A)\n",
+ "#For case(iii)\n",
+ "a = alpha_m-alpha_s #Angle of line current(radians)\n",
+ "I = (abs(I_s)**2+abs(I_m)**2+2*abs(I_s)*abs(I_m)*math.cos(a))**0.5 #Line current(A)\n",
+ "\n",
+ "#Result\n",
+ "print('(i) Starting current , I_s = %.1f\u2220%.2f\u00b0 A' %(abs(I_s),cmath.phase(I_s)*180/math.pi))\n",
+ "print('(ii) Main winding current , I_m = %.1f\u2220%.f\u00b0 A' %(abs(I_m),cmath.phase(I_m)*180/math.pi))\n",
+ "print('(iii) Line current , I = %.1f A' %I)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Starting current , I_s = 37.7\u2220-30.96\u00b0 A\n",
+ "(ii) Main winding current , I_m = 37.7\u2220-59\u00b0 A\n",
+ "(iii) Line current , I = 73.2 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.4, Page number 364"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import cmath\n",
+ "\n",
+ "#Variable declaration\n",
+ "V_t = 220.0 #Supply voltage(V)\n",
+ "f = 50.0 #Frequency(Hz)\n",
+ "Z_m = complex(4,3.5) #Main winding impedance of motor(ohm)\n",
+ "Z_s = complex(5,3) #Starting impedance of motor(ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "alpha_m = cmath.phase(Z_m) #Main winding impedance angle(radians)\n",
+ "alpha_s = (alpha_m)-(90*math.pi/180) #Angle of starting winding current(radians)\n",
+ "X_c = Z_s.imag-Z_s.real*math.tan(alpha_s) #Reactance connected in series with starting winding(ohm)\n",
+ "C = 1/(2*math.pi*f*X_c)*10**6 #Starting capacitance for getting maximum torque(\u00b5F)\n",
+ "\n",
+ "#Result\n",
+ "print('Starting capacitance for getting maximum torque , C = %.f \u00b5F' %C)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Starting capacitance for getting maximum torque , C = 365 \u00b5F\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.5, Page number 370-371"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "f = 50.0 #Supply frequency(Hz)\n",
+ "V_nl = 100.0 #No-load voltage(V)\n",
+ "I_nl = 2.5 #No-load current(A)\n",
+ "P_nl = 60.0 #No-load power(W)\n",
+ "V_br = 60.0 #Block rotor voltage(V)\n",
+ "I_br = 3.0 #Block rotor current(A)\n",
+ "P_br = 130.0 #Block rotor power(W)\n",
+ "R_1 = 2.0 #Main winding resistance(ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "Z_br = V_br/I_br #Impedance due to blocked rotor test(ohm)\n",
+ "R_br = P_br/I_br**2 #Resistance due to blocked rotor test(ohm)\n",
+ "X_br = (Z_br**2-R_br**2)**0.5 #Reactance under blocked rotor condition(ohm)\n",
+ "X_1 = 0.5*X_br #Leakage reactance(ohm)\n",
+ "X_2 = X_1 #Leakage reactance(ohm)\n",
+ "R_2 = R_br-R_1 #Rotor circuit resistance(ohm)\n",
+ "Z_nl = V_nl/I_nl #Impedance due to no-load(ohm)\n",
+ "R_nl = P_nl/I_nl**2 #Resistance due to no-load(ohm)\n",
+ "X_nl = (Z_nl**2-R_nl**2)**0.5 #Reactance due to no-load(ohm)\n",
+ "X_m = 2*(X_nl-X_1-0.5*X_2) #Magnetizing reactance(ohm)\n",
+ "P_rot = P_nl-I_nl**2*(R_1+(R_2/4)) #Rotational loss(W)\n",
+ "\n",
+ "#Result\n",
+ "print('Equivalent circuit parameters of the motor')\n",
+ "print('Under Blocked rotor test :')\n",
+ "print('Input impedance , Z_br = %.f ohm' %Z_br)\n",
+ "print('Total resistance , R_br = %.1f ohm' %R_br)\n",
+ "print('Total reactance , X_br = %.1f ohm' %X_br)\n",
+ "print('Rotor circuit resistance , R_2 = %.1f ohm' %R_2)\n",
+ "print('Leakage reactances , X_1 = X_2 = %.1f ohm' %X_1)\n",
+ "print('\\nUnder No load test :')\n",
+ "print('Input impedance , Z_nl = %.f ohm' %Z_nl)\n",
+ "print('No-load resistance , R_nl = %.1f ohm' %R_nl)\n",
+ "print('No-load reactance , X_nl = %.1f ohm' %X_nl)\n",
+ "print('Magnetizing reactance , X_m = %.1f ohm' %X_m)\n",
+ "print('Rotational loss , P_rot = %.f W' %P_rot)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equivalent circuit parameters of the motor\n",
+ "Under Blocked rotor test :\n",
+ "Input impedance , Z_br = 20 ohm\n",
+ "Total resistance , R_br = 14.4 ohm\n",
+ "Total reactance , X_br = 13.8 ohm\n",
+ "Rotor circuit resistance , R_2 = 12.4 ohm\n",
+ "Leakage reactances , X_1 = X_2 = 6.9 ohm\n",
+ "\n",
+ "Under No load test :\n",
+ "Input impedance , Z_nl = 40 ohm\n",
+ "No-load resistance , R_nl = 9.6 ohm\n",
+ "No-load reactance , X_nl = 38.8 ohm\n",
+ "Magnetizing reactance , X_m = 56.9 ohm\n",
+ "Rotational loss , P_rot = 28 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 11.6, Page number 372"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "r_t = 36.0 #Number of rotor teeth\n",
+ "N = 4.0 #Number of stator phases\n",
+ "\n",
+ "#Calculation\n",
+ "#For case(i)\n",
+ "T_p = 360/r_t #Tooth pitch(degree)\n",
+ "#For case(ii)\n",
+ "teta = 360/(N*r_t) #Step angle(degree)\n",
+ "\n",
+ "#Result\n",
+ "print('(i) Tooth pitch , T_p = %.f\u00b0 ' %T_p)\n",
+ "print('(ii) Step angle , \u03b8 = %.1f\u00b0 ' %teta)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Tooth pitch , T_p = 10\u00b0 \n",
+ "(ii) Step angle , \u03b8 = 2.5\u00b0 \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Fundamentals_of_Electrical_Machines/CH_2.ipynb b/Fundamentals_of_Electrical_Machines/CH_2.ipynb new file mode 100755 index 00000000..4a1c83c7 --- /dev/null +++ b/Fundamentals_of_Electrical_Machines/CH_2.ipynb @@ -0,0 +1,623 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 2: BASICS OF MAGNETIC CIRCUITS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.1, Page number 53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "l = 4.0 #Length(m)\n",
+ "w = 2.0 #Width(m)\n",
+ "B = 0.12 #Magnetic flux density(Tesla)\n",
+ "\n",
+ "#Calculation \n",
+ "A = l*w #Area(m^2)\n",
+ "flux = B*A #Magnetic flux(Wb)\n",
+ "\n",
+ "#Result\n",
+ "print('Magnetic flux , \u03a6 = %.2f Wb' %flux)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic flux , \u03a6 = 0.96 Wb\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2, Page number 54"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "d_in = 3.0 #Inside diameter(cm)\n",
+ "d_out = 6.0 #Outside diameter(cm)\n",
+ "N = 200.0 #Number of turns\n",
+ "I = 3.0 #Current(A)\n",
+ "flux = 0.015 #Flux(Wb)\n",
+ "\n",
+ "#Calculation\n",
+ "d = d_in+(d_out-d_in)/2 #Distance(cm)\n",
+ "l = math.pi*d #Mean length of core(cm)\n",
+ "A = math.pi*d**2/4*10**-4 #Area(m^2)\n",
+ "B = flux/A #Flux density(Wb/m^2)\n",
+ "MMF = N*I #Magnetomotive force(At)\n",
+ "H = N*I/(l*10**-2) #Magnetic field intensity(At/m)\n",
+ "\n",
+ "#Result\n",
+ "print('Flux density , B = %.2f Wb/m^2' %B)\n",
+ "print('Magnetomotive force , MMF = %.1f At' %MMF)\n",
+ "print('Magnetic field intensity , H = %.2f At/m' %H)\n",
+ "print('\\nNOTE : ERROR : Calculation & unit mistakes in textbook')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Flux density , B = 9.43 Wb/m^2\n",
+ "Magnetomotive force , MMF = 600.0 At\n",
+ "Magnetic field intensity , H = 4244.13 At/m\n",
+ "\n",
+ "NOTE : ERROR : Calculation & unit mistakes in textbook\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.3, Page number 55"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "u_r = 625.0 #Relative permeability of rectangular core\n",
+ "N = 25.0 #Number of turns\n",
+ "I = 2.0 #Current(A)\n",
+ "a = 5.5 #Lenght of rectangular core(cm)\n",
+ "b = 1.5 #Width of rectangular core(cm)\n",
+ "\n",
+ "#Calculation\n",
+ "l = 2*(a+b) #Mean length of core(cm)\n",
+ "H = N*I/(l*10**-2) #Magnetic field intensity(At/m)\n",
+ "u_0 = 4*math.pi*10**-7 #Permeability of free space(H/m)\n",
+ "u = u_0*u_r #Permeabilty\n",
+ "B = u*H #Magnetic flux density(Wb/m^2)\n",
+ "\n",
+ "#Result\n",
+ "print('Magnetic field intensity , H = %.f At/m ' %H)\n",
+ "print('Permeabilty , \u00b5 = %.2e ' %u)\n",
+ "print('Magnetic flux density , B = %.2f Wb/m^2 ' %B)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic field intensity , H = 357 At/m \n",
+ "Permeabilty , \u00b5 = 7.85e-04 \n",
+ "Magnetic flux density , B = 0.28 Wb/m^2 \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.4, Page number 57"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "N = 6.0 #Number of turns\n",
+ "I = 3.0 #Current(A)\n",
+ "flux = 0.056 #Flux(Wb)\n",
+ "\n",
+ "#Calculation\n",
+ "MMF = N*I #Magnetomotive force(At)\n",
+ "R_m = MMF/flux #Reluctance(At/Wb)\n",
+ "\n",
+ "#Result\n",
+ "print('Magnetomotive force , MMF = %.f At' %MMF)\n",
+ "print('Reluctance , R_m = %.1f At/Wb' %R_m)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetomotive force , MMF = 18 At\n",
+ "Reluctance , R_m = 321.4 At/Wb\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.5, Page number 59"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "I = 15.0 #Current through conductor(A)\n",
+ "N = 10.0 #Number of turns\n",
+ "u_0 = 4.0*math.pi*10**-7 #Permeability of free space(H/m)\n",
+ "u_r = 1.0 #Relative permeability of air medium\n",
+ "r = 0.015 #Distance(m)\n",
+ "\n",
+ "#Calculation\n",
+ "B = u_0*u_r*N*I/(2*math.pi*r) #Magnetic flux density(T)\n",
+ "\n",
+ "#Result\n",
+ "print('Magnetic flux density , B = %.1e T' %B)\n",
+ "print('\\nNOTE : ERROR : Distance is 1.5 cm & not 2.5 cm as given in textbook')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic flux density , B = 2.0e-03 T\n",
+ "\n",
+ "NOTE : ERROR : Distance is 1.5 cm & not 2.5 cm as given in textbook\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.6, Page number 60-61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "N = 200.0 #Number of turns \n",
+ "d_in = 7.0 #Inner diameter(cm)\n",
+ "d_out = 10.0 #Outer diameter(cm)\n",
+ "A = 0.005 #Cross sectional area(m^2)\n",
+ "I = 5.0 #Current through coil(A)\n",
+ "\n",
+ "#Calculation\n",
+ "u_0 = 4.0*math.pi*10**-7 #Permeability of free space(H/m)\n",
+ "R = d_out-d_in\n",
+ "l = round(2*math.pi*R/100,2) #Mean circumference length(m)\n",
+ "#For case(i)\n",
+ "H = N*I/l #Magnetic field intensity(At/m)\n",
+ "#For case(ii)\n",
+ "B = u_0*H*1000 #Flux density(mWb/m^2)\n",
+ "#For case(iii)\n",
+ "flux = B*A*1000 #Flux(\u00b5Wb)\n",
+ "\n",
+ "#Result\n",
+ "print('Magnetic field intensity , H = %.1f At/m' %H)\n",
+ "print('Flux density , B = %.1f mWb/m^2' %B)\n",
+ "print('Flux , \u03a6 = %.f \u00b5Wb' %flux)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnetic field intensity , H = 5263.2 At/m\n",
+ "Flux density , B = 6.6 mWb/m^2\n",
+ "Flux , \u03a6 = 33 \u00b5Wb\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.7, Page number 62-63"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "l = 0.1 #Length(m)\n",
+ "w = 0.01 #Width(m)\n",
+ "h = 0.1 #Height(m)\n",
+ "N = 450.0 #Number of turns\n",
+ "I = 0.2 #Current(A)\n",
+ "u_r = 850.0 #Relative permeability\n",
+ "\n",
+ "#Calculation\n",
+ "MMF = N*I #Magnetomotive force(At)\n",
+ "l_c = (h-w)*4 #Mean length of the path(m)\n",
+ "A = w*w #Cross sectional area(m^2)\n",
+ "u_0 = 4.0*math.pi*10**-7 #Permeability of free space(H/m)\n",
+ "R_m = l_c/(u_0*u_r*A) #Reluctance(At/Wb)\n",
+ "flux = MMF/R_m #Flux(Wb)\n",
+ "B = flux/A #Magnetic flux density(Wb/m^2)\n",
+ "H = B/(u_0*u_r) #Magnetic field intensity(At/m)\n",
+ "\n",
+ "#Result\n",
+ "print('Flux , \u03a6 = %.2e Wb' %flux)\n",
+ "print('Flux density , B = %.2f Wb/m^2' %B)\n",
+ "print('Field intensity , H = %.1f At/m' %H)\n",
+ "print('\\nNOTE : Changes in obtained answer from that of textbook is due to more precision')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Flux , \u03a6 = 2.67e-05 Wb\n",
+ "Flux density , B = 0.27 Wb/m^2\n",
+ "Field intensity , H = 250.0 At/m\n",
+ "\n",
+ "NOTE : Changes in obtained answer from that of textbook is due to more precision\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.8, Page number 64-65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "N = 450.0 #Number of turns wound on left side of limb\n",
+ "A = 4.0 #Cross sectional area(m^2)\n",
+ "phi_2 = 3.0 #Flux(Wb) in the right limb\n",
+ "u_r = 500.0 #Relative permeability\n",
+ "l_1 = 0.12 #Length of middle limb(m)\n",
+ "l_2 = 0.24 #Length of right limb(m)\n",
+ "\n",
+ "#Calculation\n",
+ "u_0 = 4.0*math.pi*10**-7 #Permeability of free space(H/m)\n",
+ "phi_1 = phi_2*l_2/l_1 #Flux in middle limb(Wb)\n",
+ "phi = phi_1+phi_2 #Total flux(Wb)\n",
+ "B = phi/A #Flux density in the left limb(Wb/m^2)\n",
+ "H = B/(u_0*u_r) #Magnetic field intensity(At/m)\n",
+ "MMF = H*l_2 #Magnetomotive force(At)\n",
+ "B_2 = phi_2/A #Flux density in the right limb(Wb/m^2)\n",
+ "H_2 = B_2/(u_0*u_r) #Magnetic field(At/m)\n",
+ "MMF_2 = H_2*l_2 #Magnetomotive force(At)\n",
+ "MMF_t = MMF+MMF_2 #Total magnetomotive force(At)\n",
+ "I = MMF_t/N #Current(A)\n",
+ "\n",
+ "#Result\n",
+ "print('Current , I = %.2f A' %I)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current , I = 2.55 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.9, Page number 67-68"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "l = 0.45 #Mean length(m)\n",
+ "A = 25.0*10**-4 #Cross sectional area(m^2)\n",
+ "l_g = 0.8*10**-3 #Length of air gap(m)\n",
+ "N = 500.0 #Number of turns \n",
+ "I = 1.25 #Current(A) \n",
+ "flux = 1.5*10**-3 #Flux(Wb)\n",
+ "\n",
+ "#Calculation\n",
+ "u_0 = 4.0*math.pi*10**-7 #Permeability of free space(H/m)\n",
+ "B = flux/A #Magnetic flux density(Wb/m^2)\n",
+ "MMF = N*I #Magnetomotive force(At)\n",
+ "R_m = MMF/flux #Reluctance(At/Wb)\n",
+ "H = B/u_0 #Magnetizing force(At/m)\n",
+ "MMF_ag = H*l_g #Magnetomotive force(At)\n",
+ "MMF_i = MMF-MMF_ag #Magnetomotive force for iron ring(At)\n",
+ "H_i = MMF_i/l #Magnetic field intensity for iron part(At/m)\n",
+ "u_r = B/(u_0*H_i) #Relative permeability for iron\n",
+ "\n",
+ "#Result\n",
+ "print('Reluctance , R_m = %.2e At/Wb' %R_m)\n",
+ "print('Relative permeability of the iron ring iron , \u00b5_r = %.f ' %u_r)\n",
+ "print('\\nNOTE : Reluctance part is not solved in textbook')\n",
+ "print('ERROR : Current is 1.25A not 2.25A & flux is 1.5 mWb not 2.5 mWb as given in textbook')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Reluctance , R_m = 4.17e+05 At/Wb\n",
+ "Relative permeability of the iron ring iron , \u00b5_r = 884 \n",
+ "\n",
+ "NOTE : Reluctance part is not solved in textbook\n",
+ "ERROR : Current is 1.25A not 2.25A & flux is 1.5 mWb not 2.5 mWb as given in textbook\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.10, Page number 68"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "A = 2.0*10**-4 #Cross sectional area(m^2)\n",
+ "N = 200.0 #Number of turns \n",
+ "flux = 1.5*10**-3 #Flux(Wb)\n",
+ "u_r = 4000.0 #Relative permeability of core\n",
+ "l_g = 0.01 #Length of air gap(m)\n",
+ "a = 9.0 #Length(cm)\n",
+ "w = 3.0 #Width(cm)\n",
+ "\n",
+ "#Calculation\n",
+ "u_0 = 4.0*math.pi*10**-7 #Permeability of free space(H/m)\n",
+ "R_mg = l_g/(u_0*A) #Reluctance of air gap(At/Wb)\n",
+ "l = 4*(a-w-w+(1.5+1.5))-1 #Mean length of iron(cm)\n",
+ "u = u_0*u_r #Permeability\n",
+ "R_mi = l*10**-2/(u*A) #Reluctance of iron(At/Wb)\n",
+ "R_mt = R_mg+R_mi #Total reluctance(At/Wb)\n",
+ "I = R_mt*flux/N #Current(A)\n",
+ "\n",
+ "#Result\n",
+ "print('Total reluctance , R_mt = %.3e AT/Wb' %R_mt)\n",
+ "print('Current flowing through the coil , I = %.1f A' %I)\n",
+ "print('\\nNOTE : ERROR : Total flux is 1.5 mWB & not 2.5 mWB as given in textbook question')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total reluctance , R_mt = 4.002e+07 AT/Wb\n",
+ "Current flowing through the coil , I = 300.1 A\n",
+ "\n",
+ "NOTE : ERROR : Total flux is 1.5 mWB & not 2.5 mWB as given in textbook question\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.11, Page number 70"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "I = 150.0 #Current through conductor(A)\n",
+ "l = 2.0 #Conductor length(m)\n",
+ "B = 0.35 #Magnetic flux density(T)\n",
+ "\n",
+ "#Calculation\n",
+ "F = B*l*I #Force(N)\n",
+ "\n",
+ "#Result\n",
+ "print('Force , F = %.f N' %F)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force , F = 105 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.12, Page number 76"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "l = 25.0*10**-2 #Length of air-core coil(m)\n",
+ "A = 4.0*10**-4 #Cross sectional area(m^2)\n",
+ "N = 65.0 #Number of turns\n",
+ "\n",
+ "#Calculation\n",
+ "u_0 = 4.0*math.pi*10**-7 #Permeability of free space(H/m)\n",
+ "u_r = 1.0\n",
+ "u = u_0*u_r #Permeability\n",
+ "L = N**2*u*A/l*10**6 #Inductance(\u00b5H)\n",
+ "\n",
+ "#Result\n",
+ "print('Inductance of the coil , L = %.1f \u00b5H' %L)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Inductance of the coil , L = 8.5 \u00b5H\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.13, Page number 80"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "k_h = 110.0 #Hysteresis co-efficient(J/m^3)\n",
+ "V_cvol = 0.005 #Volume of core(m^3)\n",
+ "B_m = 1.12 #Maximum flux density(T)\n",
+ "f = 60.0 #Frequency(Hz)\n",
+ "n = 1.6\n",
+ "\n",
+ "#Calculation\n",
+ "P_h = k_h*V_cvol*B_m**n*f #Hysteresis loss(W)\n",
+ "\n",
+ "#Result\n",
+ "print('Hysteresis loss , P_h = %.2f W' %P_h)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Hysteresis loss , P_h = 39.56 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Fundamentals_of_Electrical_Machines/CH_3.ipynb b/Fundamentals_of_Electrical_Machines/CH_3.ipynb new file mode 100755 index 00000000..68c7ac6d --- /dev/null +++ b/Fundamentals_of_Electrical_Machines/CH_3.ipynb @@ -0,0 +1,883 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 3: TRANSFORMER AND PER UNIT SYSTEM"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.1, Page number 90-91"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "V_1 = 2200.0 #Primary voltage of transformer(V)\n",
+ "V_2 = 220.0 #Secondary voltage of transformer(V)\n",
+ "N_2 = 56.0 #Number of turns in the secondary coil of transformer\n",
+ "kVA = 25.0 #Rating of transformer(kVA)\n",
+ "\n",
+ "#Calculation\n",
+ "a = V_1/V_2 #Turns ratio\n",
+ "#For case(i)\n",
+ "N_1 = a*N_2 #Number of primary turns\n",
+ "#For case(ii)\n",
+ "I_1 = kVA*10**3/V_1 #Primary full load current(A)\n",
+ "#For case(iii)\n",
+ "I_2 = kVA*10**3/V_2 #Secondary full load current(A)\n",
+ "\n",
+ "#Result\n",
+ "print('(i) Number of primary turns , N_1 = %.f ' %N_1)\n",
+ "print('(ii) Primary full load current , I_2 = %.2f A' %I_1)\n",
+ "print('(iii) Secondary full load current , I_2 = %.1f A' %I_2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Number of primary turns , N_1 = 560 \n",
+ "(ii) Primary full load current , I_2 = 11.36 A\n",
+ "(iii) Secondary full load current , I_2 = 113.6 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.2, Page number 91"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "V_1 = 220.0 #Voltage(V)\n",
+ "N_1 = 150.0 #Number of turns in primary side\n",
+ "N_2 = 300.0 #Number of turns in secondary side\n",
+ "f = 50.0 #Frequency(Hz)\n",
+ "\n",
+ "#Calculation\n",
+ "#For case(i)\n",
+ "a = N_1/N_2 #Turns ratio\n",
+ "#For case(ii)\n",
+ "phi_m = V_1/(4.44*f*N_1)*10**3 #Mutual flux(mWb)\n",
+ "\n",
+ "#Result\n",
+ "print('(i) Turns ratio , a = %.1f ' %a)\n",
+ "print('(ii) Mutual flux in the core , \u03a6_m = %.2f mWb' %phi_m)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Turns ratio , a = 0.5 \n",
+ "(ii) Mutual flux in the core , \u03a6_m = 6.61 mWb\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3, Page number 92"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "V_1 = 2200.0 #Primary voltage(V)\n",
+ "V_2 = 220.0 #Secondary voltage(V)\n",
+ "I_0 = 0.5 #No-load current(A)\n",
+ "P_0 = 350.0 #Power absorbed(W)\n",
+ "\n",
+ "#Calculation\n",
+ "cos_phi_0 = P_0/(V_1*I_0) #No-load power factor\n",
+ "I_w = I_0*cos_phi_0 #Iron loss component of current(A)\n",
+ "phi_0 = math.acos(cos_phi_0) #Power factor angle\n",
+ "I_m = I_0*math.sin(phi_0) #Magnetizing component of current(A)\n",
+ "\n",
+ "#Result\n",
+ "print('Iron loss component of current , I_w = %.2f A' %I_w)\n",
+ "print('Magnetizing component of current , I_m = %.2f A' %I_m)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Iron loss component of current , I_w = 0.16 A\n",
+ "Magnetizing component of current , I_m = 0.47 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.4, Page number 94"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "N_1 = 450.0 #Number of turns in the primary side\n",
+ "N_2 = 45.0 #Number of turns in the secondary side\n",
+ "Z_L = 3.0 #Load impedance(ohm)\n",
+ "V_1 = 15.0 #Primary coil voltage of transformer(V)\n",
+ "\n",
+ "#Calculation\n",
+ "#For case(i)\n",
+ "a = N_1/N_2 #Turns ratio\n",
+ "#For case(ii)\n",
+ "Z_1 = a**2*Z_L #Load impedance referred to primary(ohm)\n",
+ "#For case(iii)\n",
+ "I_1 = V_1/Z_1 #Primary current(A)\n",
+ "\n",
+ "#Result\n",
+ "print('(i) Turns ratio , a = %.f ' %a)\n",
+ "print('(ii) Load impedance referred to primary , Z_1 = %.f ohm' %Z_1)\n",
+ "print('(iii) Primary current , I_1 = %.2f A' %I_1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Turns ratio , a = 10 \n",
+ "(ii) Load impedance referred to primary , Z_1 = 300 ohm\n",
+ "(iii) Primary current , I_1 = 0.05 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.5, Page number 96-97"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "V_1 = 400.0 #Primary voltage of transformer(V)\n",
+ "V_2 = 100.0 #Secondary voltage of transformer(V)\n",
+ "I_0 = 0.4 #No-load current(A)\n",
+ "I_2 = 100.0 #Current drawn by load(A)\n",
+ "cos_phi_0 = 0.3 #Power factor lagging from the supply\n",
+ "cos_phi_2 = 0.6 #Power factor lagging from the secondary\n",
+ "\n",
+ "#Calculation\n",
+ "phi_0 = math.acos(cos_phi_0) #Power factor angle(radians)\n",
+ "phi_0_deg = phi_0*180/math.pi #Power factor angle(degree)\n",
+ "phi_2 = math.acos(cos_phi_2) #Power factor angle(radians)\n",
+ "phi_2_deg = phi_2*180/math.pi #Power factor angle(degree)\n",
+ "phi_1 = phi_0-phi_2 #Angle(radians)\n",
+ "phi_1_deg = phi_1*180/math.pi #Angle(degree)\n",
+ "a = V_1/V_2 #Turns ratio\n",
+ "I_2_ = I_2/a #Secondary current equivalent to the primary(A) \n",
+ "I_1 = ((I_2_**2)+(I_0**2)+(2*I_2_*I_0*math.cos(phi_1)))**0.5 #Primary current(A)\n",
+ "\n",
+ "#Result\n",
+ "print('Primary current , I_1 = %.1f A' %I_1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Primary current , I_1 = 25.4 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.6, Page number 101"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "V_1 = 2000.0 #Primary voltage of transformer(V)\n",
+ "V_2 = 400.0 #Secondary voltage of transformer(V)\n",
+ "kVA = 200.0 #Rating of transformer(kVA)\n",
+ "R_1 = 3.0 #Primary resistance(ohm)\n",
+ "X_1 = 12.0 #Primary reactance(ohm)\n",
+ "R_2 = 0.3 #Secondary resistance(ohm)\n",
+ "X_2 = 0.1 #Secondary reactance(ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "a = V_1/V_2 #Turns ratio\n",
+ "R_01 = R_1+(a**2*R_2) #Total resistance referred to primary(ohm)\n",
+ "X_01 = X_1+(a**2*X_2) #Total reactance referred to primary(ohm)\n",
+ "Z_01 = ((R_01**2)+(X_01**2))**0.5 #Equivalent impedance referred to primary(ohm)\n",
+ "R_02 = R_2+(R_1/a**2) #Total resistance referred to secondary side(ohm)\n",
+ "X_02 = X_2+(X_1/a**2) #Total reactance referred to secondary side(ohm)\n",
+ "Z_02 = ((R_02**2)+(X_02**2))**0.5 #Equivalent impedance referred to secondary side(ohm)\n",
+ "\n",
+ "#Result\n",
+ "print('Equivalent impedance referred to primary , Z_01 = %.1f ohm' %Z_01)\n",
+ "print('Equivalent impedance referred to secondary , Z_02 = %.2f ohm' %Z_02)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equivalent impedance referred to primary , Z_01 = 17.9 ohm\n",
+ "Equivalent impedance referred to secondary , Z_02 = 0.72 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.7, Page number 103-104"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "V_1 = 200.0 #Primary voltage(V)\n",
+ "V_2 = 400.0 #Secondary voltage(V)\n",
+ "R_1 = 0.3 #Primary resistance(ohm)\n",
+ "X_1 = 0.6 #Primary reactance(ohm)\n",
+ "R_2 = 0.8 #Secondary resistance(ohm)\n",
+ "X_2 = 1.6 #Secondary reactance(ohm)\n",
+ "I_2 = 10.0 #Secondary supply current(A)\n",
+ "cos_phi_2 = 0.8 #Power factor lagging\n",
+ "\n",
+ "#Calculation\n",
+ "a = V_1/V_2 #Turns ratio\n",
+ "R_02 = R_2+(R_1/a**2) #Total resistance referred to secondary(ohm)\n",
+ "X_02 = X_2+(X_1/a**2) #Total reactance referred to primary(ohm)\n",
+ "sin_phi_2 = math.sin(math.acos(cos_phi_2))\n",
+ "E_2 = complex((V_2*cos_phi_2+I_2*R_02),(V_2*sin_phi_2+I_2*X_02)) #No-load voltage(V)\n",
+ "V_reg = (abs(E_2)-V_2)/V_2*100 #Voltage regulation(percent)\n",
+ "\n",
+ "#Result\n",
+ "print('Voltage regulation = %.f percent' %V_reg)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage regulation = 10 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.8, Page number 105-106"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "P_i = 1.0 #Iron loss of transformer(kW)\n",
+ "P_cu = 2.0 #Full load copper loss of transformer(kW)\n",
+ "kVA = 200.0 #Rating of transformer(kVA)\n",
+ "pf_1 = 1.0 #Power factor\n",
+ "pf_2 = 0.95 #Power factor\n",
+ "\n",
+ "#Calculation\n",
+ "P_cu1 = (3.0/4)**2*P_cu #Copper loss at 3/4 full load(kW)\n",
+ "P_cu2 = (1.0/2)**2*P_cu #Copper loss at 1/2 full load(kW)\n",
+ "P_01 = (3.0/4)*kVA*pf_1 #Output power at 3/4 full load and unity pf(kW)\n",
+ "P_in1 = P_01+P_i+P_cu1 #Input power at 3/4 full load and unity pf(kW)\n",
+ "n_1 = (P_01/P_in1)*100 #Efficiency at 3/4 full load and unity pf(percent)\n",
+ "P_02 = (1.0/2)*kVA*pf_2 #Output power at 1/2 full load and 0.95 pf(kW)\n",
+ "P_in2 = P_02+P_i+P_cu2 #Input power at 1/2 full load and 0.95 pf(kW)\n",
+ "n_2 = (P_02/P_in2)*100 #Efficiency at 1/2 full load and 0.95 pf(percent)\n",
+ "\n",
+ "#Result\n",
+ "print('Efficiency at 3/4 full load and unity power factor , \u03b7_1 = %.2f percent' %n_1)\n",
+ "print('Efficiency at 1/2 full load and 0.95 power factor , \u03b7_2 = %.2f percent' %n_2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Efficiency at 3/4 full load and unity power factor , \u03b7_1 = 98.60 percent\n",
+ "Efficiency at 1/2 full load and 0.95 power factor , \u03b7_2 = 98.45 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.9, Page number 108"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "P_i = 350.0 #Iron loss of transformer(W)\n",
+ "P_cu = 650.0 #Full load copper loss of transformer(W)\n",
+ "kVA = 30.0 #Rating of transformer(kVA)\n",
+ "pf = 0.6 #Power factor\n",
+ "\n",
+ "#Calculation\n",
+ "#For case(i)\n",
+ "P_tloss = (P_i+P_cu)*10**-3 #Total full load loss(kW)\n",
+ "P_out = kVA*pf #Output power at full load(kW)\n",
+ "P_in = P_out+P_tloss #Input power at full load(kW)\n",
+ "n_1 = (P_out/P_in)*100 #Efficiency at full load(percent)\n",
+ "#For case(ii)\n",
+ "kVA_out = kVA*(P_i/P_cu)**0.5 #Output kVA corresponding to maximum efficiency(kVA) \n",
+ "P_01 = kVA_out*pf #Output power(kW)\n",
+ "#For case(iii)\n",
+ "P_tloss1 = 2*P_i*10**-3 #Total loss(kW)\n",
+ "P_in1 = P_01+P_tloss1 #Input power(kW)\n",
+ "n_2 = (P_01/P_in1)*100 #Efficiency(percent)\n",
+ "\n",
+ "#Result\n",
+ "print('(i) Full load efficiency , \u03b7 = %.2f percent' %n_1)\n",
+ "print('(ii) Output kVA corresponding to maximum efficiency = %.1f kVA' %kVA_out)\n",
+ "print('(iii) Maximum efficiency , \u03b7 = %.2f percent' %n_2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Full load efficiency , \u03b7 = 94.74 percent\n",
+ "(ii) Output kVA corresponding to maximum efficiency = 22.0 kVA\n",
+ "(iii) Maximum efficiency , \u03b7 = 94.97 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.10, Page number 109-110"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "kVA = 12.0 #Rating of transformer(kVA)\n",
+ "n = 0.97 #Maximum efficiency at unity pf\n",
+ "pf = 1.0 #Power factor\n",
+ "t_1 = 8.0 #Time(hours)\n",
+ "P_1 = 10.0 #Load(kW)\n",
+ "pf_1 = 0.8 #Lagging power factor\n",
+ "t_2 = 10.0 #Time(hours)\n",
+ "P_2 = 15.0 #Load(kW)\n",
+ "pf_2 = 0.90 #Leading power factor\n",
+ "t_3 = 6.0 #Time at no-load(hours)\n",
+ "P_3 = 0 #Load(kW)\n",
+ "\n",
+ "#Calculation\n",
+ "P_01 = kVA*pf #Output power at full load and unity pf(kW)\n",
+ "P_in1 = P_01/n #Input power at full load(kW)\n",
+ "P_tloss = P_in1-P_01 #Total loss(kW)\n",
+ "P_cu = P_tloss/2 #Copper loss at 12 kVA(kW)\n",
+ "P_024 = P_1*t_1+P_2*t_2+P_3*t_3 #All-day output power(kWh)\n",
+ "P_i24 = 24*P_cu #Iron loss for 24 hours(kWh)\n",
+ "P_cu24 = P_cu*t_1*((P_1/pf_1)/kVA)**2+P_cu*t_2*((P_2/pf_2)/kVA)**2 #Copper loss for 24 hours(kW)\n",
+ "P_in24 = P_024+P_i24+P_cu24 #All day input power(kWh)\n",
+ "n_allday = (P_024/P_in24)*100 #All day efficiency(percent)\n",
+ "\n",
+ "#Result\n",
+ "print('All-day efficiency , \u03b7_allday = %.1f percent' %n_allday)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "All-day efficiency , \u03b7_allday = 96.0 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.11, Page number 111"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "V_1 = 200.0 #Voltage(V)\n",
+ "f = 50.0 #Frequency(Hz)\n",
+ "I_0 = 0.6 #Current(A)\n",
+ "P_0 = 80.0 #Power(W)\n",
+ "\n",
+ "#Calculation\n",
+ "cos_phi_0 = P_0/(V_1*I_0) #Power factor\n",
+ "sin_phi_0 = math.sin(math.acos(cos_phi_0))\n",
+ "I_w = I_0*cos_phi_0 #Working component of no-load current(A)\n",
+ "I_m = I_0*sin_phi_0 #Magnetizing component of no-load current(A)\n",
+ "R_0 = V_1/I_w #No-load circuit resistance(ohm)\n",
+ "X_0 = V_1/I_m #No-load circuit reactance(ohm)\n",
+ "\n",
+ "#Result\n",
+ "print('No-load circuit resistance , R_0 = %.1f ohm' %R_0)\n",
+ "print('No-load circuit reactance , X_0 = %.1f ohm' %X_0)\n",
+ "print('\\nNOTE : Changes in obtained answer from that of textbook is due to more precision')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "No-load circuit resistance , R_0 = 500.0 ohm\n",
+ "No-load circuit reactance , X_0 = 447.2 ohm\n",
+ "\n",
+ "NOTE : Changes in obtained answer from that of textbook is due to more precision\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.12, Page number 112-113"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "kVA = 25.0 #Rating of transformer(kVA)\n",
+ "V1 = 2200.0 #Voltage at low-voltage primary side(V)\n",
+ "V2 = 220.0 #Voltage at low-voltage secondary side(V)\n",
+ "V_1 = 40.0 #Voltage at h.v side(V)\n",
+ "I_1 = 5.0 #Current at high voltage side(A)\n",
+ "P = 150.0 #Power at high voltage side(W)\n",
+ "\n",
+ "#Calculation\n",
+ "#For case(i)\n",
+ "Z_01 = V_1/I_1 #Equivalent impedance referred to primary(ohm)\n",
+ "R_01 = P/I_1**2 #Equivalent resistance referred to primary(ohm)\n",
+ "phi = math.acos(R_01/Z_01) #Power factor angle(radians)\n",
+ "phi_deg = phi*180/math.pi #Power factor angle(degree)\n",
+ "X_01 = Z_01*math.sin(phi) #Equivalent reactance referred to primary(ohm)\n",
+ "#For case(ii)\n",
+ "a = V1/V2 #Turns ratio\n",
+ "Z_02 = Z_01/a**2 #Equivalent impedance referred to secondary(ohm)\n",
+ "R_02 = R_01/a**2 #Equivalent resistance referred to secondary(ohm)\n",
+ "X_02 = X_01/a**2 #Equivalent reactance referred to secondary(ohm)\n",
+ "#For case(iii)\n",
+ "I_2 = kVA*10**3/V2 #Secondary side current(A)\n",
+ "E_2 = V2+I_2*Z_02 #Secondary induced voltage(V)\n",
+ "VR = (E_2-V2)/V2*100 #Voltage regulation(percent)\n",
+ "\n",
+ "#Result\n",
+ "print('Case(i)')\n",
+ "print(' Equivalent resistance referred to primary , R_01 = %.1f ohm' %R_01)\n",
+ "print(' Equivalent reactance referred to primary , X_01 = %.1f ohm' %X_01)\n",
+ "print(' Equivalent impedance referred to primary , Z_01 = %.1f ohm' %Z_01)\n",
+ "print('\\nCase(ii)')\n",
+ "print(' Equivalent resistance referred to secondary , R_02 = %.2f ohm' %R_02)\n",
+ "print(' Equivalent reactance referred to secondary , X_02 = %.3f ohm' %X_02)\n",
+ "print(' Equivalent impedance referred to secondary , Z_02 = %.2f ohm' %Z_02)\n",
+ "print('\\nCase(iii)')\n",
+ "print(' Voltage regulation , VR = %.1f percent' %VR)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Case(i)\n",
+ " Equivalent resistance referred to primary , R_01 = 6.0 ohm\n",
+ " Equivalent reactance referred to primary , X_01 = 5.3 ohm\n",
+ " Equivalent impedance referred to primary , Z_01 = 8.0 ohm\n",
+ "\n",
+ "Case(ii)\n",
+ " Equivalent resistance referred to secondary , R_02 = 0.06 ohm\n",
+ " Equivalent reactance referred to secondary , X_02 = 0.053 ohm\n",
+ " Equivalent impedance referred to secondary , Z_02 = 0.08 ohm\n",
+ "\n",
+ "Case(iii)\n",
+ " Voltage regulation , VR = 4.1 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.13, Page number 114-115"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "kVA = 120.0 #Rating of autotransformer(kVA)\n",
+ "V1 = 220.0 #Volteage at upper part of coil(V)\n",
+ "V2 = 2200.0 #Volteage at lower part of coil(V)\n",
+ "\n",
+ "#Calculation\n",
+ "I_pq = kVA*10**3/V1 #Current of upper winding(A)\n",
+ "I_qr = kVA*10**3/V2 #Current of lower winding(A)\n",
+ "I_1 = I_pq+I_qr #Current in primary side(A)\n",
+ "V_2 = V1+V2 #Voltage across the secondary side(V)\n",
+ "kVA_1 = I_1*V2/1000 #Rating of autotransformer(kVA)\n",
+ "kVA_2 = I_pq*V_2/1000 #Rating of autotransformer(kVA)\n",
+ " \n",
+ "\n",
+ "#Result\n",
+ "print('kVA ratings of the autotransformer')\n",
+ "print('\\t kVA_1 = %.f kVA' %kVA_1)\n",
+ "print('\\t kVA_2 = %.f kVA' %kVA_2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "kVA ratings of the autotransformer\n",
+ "\t kVA_1 = 1320 kVA\n",
+ "\t kVA_2 = 1320 kVA\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.14, Page number 119"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import cmath\n",
+ "\n",
+ "#Variable declaration\n",
+ "kVA_1 = 100.0 #Rating of transformer(kVA)\n",
+ "kVA_2 = 200.0 #Rating of transformer(kVA)\n",
+ "E_1 = 500.0 #Secondary induced voltage in 100 kVA transformer(V)\n",
+ "E_2 = 450.0 #Secondary induced voltage in 200 kVA transformer(V)\n",
+ "Z_1 = 0.05 #Impedance of 100 kVA transformer\n",
+ "Z_2 = 0.08 #Impedance of 200 kVA transformer\n",
+ "\n",
+ "#Calculation\n",
+ "Z1 = Z_1*E_1/(kVA_1*10**3/E_1) #Actual impedance of first transformer(ohm)\n",
+ "Z2 = Z_2*E_2/(kVA_1*10**3/E_2) #Actual impedance of second transformer(ohm)\n",
+ "I_c = (E_1-E_2)/complex(0,Z1+Z2) #Circulating current(A)\n",
+ "\n",
+ "#Result\n",
+ "print('Circulating current , I_c = %.2f\u2220%.f\u00b0 A' %(abs(I_c),cmath.phase(I_c)*180/math.pi))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Circulating current , I_c = 174.22\u2220-90\u00b0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.15, Page number 125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "V_L1 = 11.0 #Supply voltage(kV)\n",
+ "I_P1 = 6.0 #Current drawn by transformer(A)\n",
+ "a = 11.0 #Turns ratio\n",
+ "\n",
+ "#Calculation\n",
+ "#For delta-wye connections\n",
+ "V_dP1 = V_L1 #Phase voltage at primary side(kV)\n",
+ "V_dP2 = V_dP1*10**3/a #Phase voltage at secondary side(V)\n",
+ "V_dL2 = 3**0.5*V_dP2 #Line voltage at secondary side(V)\n",
+ "I_dP1 = a/3**0.5 #Phase current in the primary(A)\n",
+ "I_dL2 = a*I_dP1 #Line current in secondary(A)\n",
+ "#For Wye-delta connection \n",
+ "V_wP1 = V_L1*10**3/3**0.5 #Phase voltage at primary side(V)\n",
+ "V_wP2 = V_wP1/a #Phase voltage at secondary(V)\n",
+ "V_wL2 = V_wP2 #Line voltage at secondary(V)\n",
+ "I_wP2 = a*I_P1 #Phase current in secondary(A)\n",
+ "I_wL2 = 3**0.5*I_wP2 #Line current in secondary(A)\n",
+ "\n",
+ "#Result\n",
+ "print('For delta-wye connection')\n",
+ "print(' (i) Line voltage at secondary side , V_L2 = %.f V' %V_dL2)\n",
+ "print(' (ii) Line current in the secondary , I_L2 = %.2f A' %I_dL2)\n",
+ "print('\\nFor wye-delta connection')\n",
+ "print(' (i) Line voltage at secondary side , V_L2 = %.2f V' %V_wL2)\n",
+ "print(' (ii) Line current in the secondary , I_L2 = %.2f A' %I_wL2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "For delta-wye connection\n",
+ " (i) Line voltage at secondary side , V_L2 = 1732 V\n",
+ " (ii) Line current in the secondary , I_L2 = 69.86 A\n",
+ "\n",
+ "For wye-delta connection\n",
+ " (i) Line voltage at secondary side , V_L2 = 577.35 V\n",
+ " (ii) Line current in the secondary , I_L2 = 114.32 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.16, Page number 132"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import cmath\n",
+ "\n",
+ "#Variable declaration\n",
+ "V_b = 220.0 #Voltage(V)\n",
+ "f = 50.0 #Frequency(Hz)\n",
+ "S_b = 600.0 #Base value of rating(VA)\n",
+ "R = 3.0 #Resistance(ohm)\n",
+ "X_L = 5.0 #Inductance(ohm)\n",
+ "Z = complex(R,X_L) #Impedance(ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "I_b = S_b/V_b #Base value of current(A)\n",
+ "Z_b = V_b**2/S_b #Base impedance(ohm)\n",
+ "R_pu = R/Z_b #Per unit value of resistance\n",
+ "X_Lpu = X_L/Z_b #Per unit value of inductance\n",
+ "Z_pu = complex(R_pu,X_Lpu) #Per unit of value of impedance\n",
+ "Z_pu_alt = abs(Z)/Z_b #Per unit of value of impedance-alternative method\n",
+ "\n",
+ "#Result\n",
+ "print('Per unit value of resistance , R_pu = %.3f ' %R_pu)\n",
+ "print('Per unit value of inductance , X_Lpu = %.3f ' %X_Lpu)\n",
+ "print('Per unit of value of impedance , Z_pu = %.3f\u2220%.2f\u00b0 ' %(abs(Z_pu),cmath.phase(Z_pu)*180/math.pi))\n",
+ "print('Per unit of value of impedance by alternative method , Z_pu = %.3f ' %Z_pu_alt)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Per unit value of resistance , R_pu = 0.037 \n",
+ "Per unit value of inductance , X_Lpu = 0.062 \n",
+ "Per unit of value of impedance , Z_pu = 0.072\u222059.04\u00b0 \n",
+ "Per unit of value of impedance by alternative method , Z_pu = 0.072 \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.17, Page number 132-134"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import cmath\n",
+ "\n",
+ "#Variable declaration\n",
+ "MVA_T1 = 100.0 #Rating of transformer T1(MVA)\n",
+ "V_T1_hv = 220.0 #Voltage of h.v side of T1(kV)\n",
+ "V_T1_lv = 132.0 #Voltage of l.v side of T1(kV)\n",
+ "X_T1 = 0.2 #Impedance of T1(pu)\n",
+ "MVA_T2 = 50.0 #Rating of transformer T2(MVA)\n",
+ "V_T2_hv = 132.0 #Voltage of h.v side of T2(kV)\n",
+ "V_T2_lv = 66.0 #Voltage of l.v side of T2(kV)\n",
+ "X_T2 = 0.05 #Impedance of T2(pu)\n",
+ "X_L = 4.0 #Line impedance(ohm)\n",
+ "P = 50.0 #Power absorbed(MW)\n",
+ "pf = 0.6 #Lagging power factor\n",
+ "\n",
+ "#Calculation\n",
+ "S_b = MVA_T1 #Base apparent power(MW)\n",
+ "V_b = V_T1_hv #Base voltage(kV)\n",
+ "a = V_T1_hv/V_T1_lv #Turns ratio for first transformer\n",
+ "V_bline = V_T1_hv/a #Base voltage of line(kV)\n",
+ "Z_bline = V_bline**2/S_b #Base impedance of line(ohm)\n",
+ "X_puline = X_L/Z_bline #Per unit reactance of line\n",
+ "X_pu_T1 = X_T1*(V_T1_hv/V_b)**2*(S_b/MVA_T1) #Per unit reactance of first transformer\n",
+ "V_bload = V_T2_hv/(V_T2_hv/V_T2_lv) #Load side base voltage(kV)\n",
+ "X_pu_load = X_T2*(V_T2_lv/V_bload)**2*(S_b/MVA_T2) #Per unit reactance of second transformer\n",
+ "I_b = S_b*1000/(3**0.5*V_bload) #Base current at the load(A)\n",
+ "I_L = MVA_T2*1000/(3**0.5*V_T2_lv*pf) #Actual current in load(A)\n",
+ "I_Lpu = I_L/I_b #Per unit value of load current(pu)\n",
+ "V_L = V_T2_lv/V_bload #Per unit value of voltage at the load terminal(pu)\n",
+ "V_gb = I_Lpu*cmath.exp(1j*math.acos(pf))*complex(0,X_T1+X_puline+X_pu_load)+V_L #Per unit value of grid to bus voltage(pu)\n",
+ "V_gba = V_gb*V_T1_hv #Actual value of grid to bus voltage(kV)\n",
+ "\n",
+ "#Result\n",
+ "print('Grid to bus voltage , V_gb = %.f\u2220%.2f\u00b0 kV' %(abs(V_gba),cmath.phase(V_gba)*180/math.pi))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Grid to bus voltage , V_gb = 176\u222011.63\u00b0 kV\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Fundamentals_of_Electrical_Machines/CH_4.ipynb b/Fundamentals_of_Electrical_Machines/CH_4.ipynb new file mode 100755 index 00000000..76dec5aa --- /dev/null +++ b/Fundamentals_of_Electrical_Machines/CH_4.ipynb @@ -0,0 +1,499 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 4: DIRECT CURRENT GENERATORS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.1, Page number 143"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from sympy import *\n",
+ "\n",
+ "#Variable declaration\n",
+ "N = 100.0 #Number of turns\n",
+ "\n",
+ "#Calculation\n",
+ "t = Symbol('t')\n",
+ "e = N*diff(0.05*sin(314*t), t, 1)\n",
+ "\n",
+ "#Result\n",
+ "print('Induced voltage at the coil terminals , e = ' + repr(e) + ' V')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Induced voltage at the coil terminals , e = 1570.0*cos(314*t) V\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.2, Page number 145"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "l = 0.65 #Length of conductor(m)\n",
+ "v = 35.0 #Speed of conductor(m/s)\n",
+ "B = 0.8 #Magnetic flux density(Tesla)\n",
+ "\n",
+ "#Calculation\n",
+ "e = B*l*v #Induced voltage at the conductor(V)\n",
+ "\n",
+ "#Result\n",
+ "print('Induced voltage at the conductor , e = %.1f V' %e)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Induced voltage at the conductor , e = 18.2 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.3, Page number 145"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "l = 1.5 #Length of conductor(m)\n",
+ "v = 20.0 #Velocity of conductor(m/s)\n",
+ "theta = 35.0*math.pi/180 #Angle(radians)\n",
+ "B = 0.9 #Magnetic flux density(Wb/m^2)\n",
+ "\n",
+ "#Calculation\n",
+ "e = B*l*v*math.sin(theta) #Induced voltage at the conductor(V)\n",
+ "\n",
+ "#Result\n",
+ "print('Induced voltage at the conductor , e = %.1f V' %e)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Induced voltage at the conductor , e = 15.5 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4, Page number 152-153"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "P = 4.0 #Number of poles\n",
+ "S = 40.0 #Number of slots\n",
+ "C = 10.0 #Number of conductors per slot\n",
+ "phi = 0.02 #Flux per pole(Wb)\n",
+ "N = 1200.0 #Speed(rpm)\n",
+ "\n",
+ "#Calculation\n",
+ "Z = S*C #Total number of conductors\n",
+ "A = 2.0 #Number of parallel paths for Wave winding\n",
+ "E_g = P*phi*Z*N/(60*A) #Generated emf(V)\n",
+ "\n",
+ "#Result\n",
+ "print('Generated emf , E_g = %.f V' %E_g)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Generated emf , E_g = 320 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.5, Page number 153"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "P = 6.0 #Number of poles\n",
+ "Z = 600.0 #Number of conductors\n",
+ "phi = 0.05 #Flux per pole(Wb)\n",
+ "N = 1000.0 #Speed of generator(rpm)\n",
+ "I_a = 120.0 #Current supplied by generator(A)\n",
+ "\n",
+ "#Calculation\n",
+ "A = P #Number of parallel paths for lap winding\n",
+ "E_g = P*phi*Z*N/(60*A) #Generated voltage(V)\n",
+ "T_em = (P*Z*phi)/(2*math.pi*A)*I_a #Electromagnetic torque(N-m)\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print('Generated voltage , E_g = %.f V' %E_g)\n",
+ "print('Electromagnetic torque , T_em = %.2f N-m' %T_em)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Generated voltage , E_g = 500 V\n",
+ "Electromagnetic torque , T_em = 572.96 N-m\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.6, Page number 156"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "V_t = 220.0 #Shunt generator voltage(V)\n",
+ "I_L = 250.0 #Load current(A)\n",
+ "R_sh = 50.0 #Shunt field resistance(ohm)\n",
+ "R_a = 0.02 #Armature resistance(ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "I_sh = V_t/R_sh #Shunt field current(A)\n",
+ "I_a = I_L+I_sh #Armature current(A)\n",
+ "E_g = V_t+I_a*R_a #Generated voltage(V)\n",
+ "\n",
+ "#Result\n",
+ "print('Generated voltage , E_g = %.2f V' %E_g)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Generated voltage , E_g = 225.09 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.7, Page number 158-160"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "E = 25.0 #Power of compound generator(kW)\n",
+ "V_t = 220.0 #Terminal voltage(V)\n",
+ "R_a = 0.07 #Armature resistance(ohm)\n",
+ "R_se = 0.05 #Series resistance(ohm)\n",
+ "R_sh = 55.0 #Shunt field resistance(ohm)\n",
+ "V_brush = 1.0 #Voltage drop per brush(V)\n",
+ "\n",
+ "#Calculation\n",
+ "I_L = E*10**3/V_t #Load current in A\n",
+ "I_sh = V_t/R_sh #Shunt field current(A)\n",
+ "I_a = I_sh+I_L #Armature current(A)\n",
+ "#For case(i)\n",
+ "E_g1 = V_t+I_a*(R_a+R_se)+2*V_brush #Generated emf(V)\n",
+ "#For case(ii)\n",
+ "V_ab = V_t+I_L*R_se #Voltage across the shunt field(V)\n",
+ "I_sh2 = V_ab/R_sh #Current in the shunt field(A)\n",
+ "I_a2 = I_sh2+I_L #Armature current(A)\n",
+ "E_g2 = V_ab+I_a2*R_a+2*V_brush #Generated emf(V)\n",
+ "\n",
+ "#Result\n",
+ "print('(i) Generated emf when generator is connected in long shunt , E_g = %.f V' %E_g1)\n",
+ "print('(ii) Generated emf when generator is connected in short shunt , E_g = %.1f V' %E_g2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Generated emf when generator is connected in long shunt , E_g = 236 V\n",
+ "(ii) Generated emf when generator is connected in short shunt , E_g = 235.9 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.8, Page number 160-161"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "V_t = 220.0 #Shunt generator voltage(V)\n",
+ "I_L = 146.0 #Current delivered by generator(A)\n",
+ "R_sh = 55.0 #Shunt field resistance(ohm)\n",
+ "R_a = 0.012 #Armature resistance(ohm)\n",
+ "R_se = 0.02 #Series field resistance(ohm)\n",
+ "R_di = 0.03 #Diverter field resistance(ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "I_sh = V_t/R_sh #Shunt field current(A)\n",
+ "I_a = I_L+I_sh #Armature current(A)\n",
+ "R_com = R_se*R_di/(R_se+R_di) #Combined resistance(ohm)\n",
+ "E_g = V_t+I_a*(R_a+R_com) #Generated voltage(V)\n",
+ "P_lsd = I_a**2*R_com #Power loss in series field and diverter(W)\n",
+ "P_la = I_a**2*R_a #Power loss in the armature circuit resistance(W)\n",
+ "P_lsh = V_t*I_sh #Power loss in shunt field resistance(W)\n",
+ "P_dl = I_L*V_t #Power delivered(W)\n",
+ "\n",
+ "#Result\n",
+ "print('Generated voltage , E_g = %.1f V' %E_g)\n",
+ "print('Power loss in the series field and diverter , P_lsd = %.1f W' %P_lsd)\n",
+ "print('Power loss in the armature circuit resistance , P_la = %.1f W' %P_la)\n",
+ "print('Power loss in the shunt field resistance , P_lsh = %.f W' %P_lsh)\n",
+ "print('Power delivered to the load , P_dl = %.f W' %P_dl)\n",
+ "print('\\nNOTE : ERROR : Shunt field resistance is taken as 50 ohm while solving I_sh in textbook but it is 55 ohm as per textbook question')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Generated voltage , E_g = 223.6 V\n",
+ "Power loss in the series field and diverter , P_lsd = 270.0 W\n",
+ "Power loss in the armature circuit resistance , P_la = 270.0 W\n",
+ "Power loss in the shunt field resistance , P_lsh = 880 W\n",
+ "Power delivered to the load , P_dl = 32120 W\n",
+ "\n",
+ "NOTE : ERROR : Shunt field resistance is taken as 50 ohm while solving I_sh in textbook but it is 55 ohm as per textbook question\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.9, Page number 169"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "P = 4.0 #Number of poles\n",
+ "Z = 500.0 #Number of conductors\n",
+ "I_a = 30.0 #Current delivered by generator(A)\n",
+ "alpha = 6.0 #Angle at which brushes are displaced angle(degree)\n",
+ "\n",
+ "#Calculation\n",
+ "A = 2.0 #Number of parallel paths for Wave winding\n",
+ "I_c = I_a/A #Current per conductor(A)\n",
+ "#For case(i)\n",
+ "AT_d = Z*I_c*alpha/360 #Demagnetizing ampere-turns per pole(At)\n",
+ "#For case(ii)\n",
+ "AT_c = Z*I_c*((1/(2*P))-(alpha/360)) #Cross magnetizing ampere-turns per pole(At)\n",
+ "\n",
+ "#Result\n",
+ "print('(i) Demagnetizing ampere-turns , AT_d = %.f At' %AT_d)\n",
+ "print('(ii) Cross-magnetizing ampere-turns , AT_c = %.1f At' %AT_c)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Demagnetizing ampere-turns , AT_d = 125 At\n",
+ "(ii) Cross-magnetizing ampere-turns , AT_c = 812.5 At\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.10, Page number 176"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "Power = 12.0 #Power(kW)\n",
+ "P = 4.0 #Number of poles\n",
+ "Z = 500.0 #Number of conductors\n",
+ "V_t = 250.0 #Generator voltage(V)\n",
+ "N = 1000.0 #Speed(rpm)\n",
+ "P_cu = 600 #Full load copper loss(W)\n",
+ "brush_drop = 2.0 #Total brush drop(V)\n",
+ "\n",
+ "#Calculation\n",
+ "A = P #Number of parallel paths for lap winding\n",
+ "I_a = Power*10**3/V_t #Armature current(A)\n",
+ "R_a = P_cu/I_a**2 #Armature resistance(ohm)\n",
+ "E_g = V_t+I_a*R_a+brush_drop #Generated voltage(V)\n",
+ "phi = E_g*60*A/(P*Z*N) #Flux per pole(Wb)\n",
+ "\n",
+ "#Result\n",
+ "print('Flux per pole , \u03a6 = %.3f Wb' %phi)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Flux per pole , \u03a6 = 0.032 Wb\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.11, Page number 176-177"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "P = 4.0 #Number of poles\n",
+ "I_L = 25.0 #Current delivered by generator(A)\n",
+ "V_t = 230.0 #Generator terminal voltage(V)\n",
+ "R_a = 0.2 #Armature resistance(ohm)\n",
+ "R_sh = 55.0 #Shunt field resistance(ohm)\n",
+ "V_brush = 1.0 #Voltage drop per brush(V)\n",
+ "\n",
+ "#Calculation\n",
+ "I_sh = V_t/R_sh #Shunt field current(A)\n",
+ "I_a = I_L+I_sh #Armature current(A)\n",
+ "E_g = V_t+I_a*R_a+2*V_brush #Induced voltage(V)\n",
+ "P_arm = E_g*I_a #Power generated in armature(W)\n",
+ "P_L = V_t*I_L #Power absorbed by load(W)\n",
+ "n = (P_L/P_arm)*100 #Efficiency(percent)\n",
+ "\n",
+ "#Result\n",
+ "print('Induced voltage , E_g = %.1f V' %E_g)\n",
+ "print('Efficiency of generator , \u03b7 = %.1f percent' %n)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Induced voltage , E_g = 237.8 V\n",
+ "Efficiency of generator , \u03b7 = 82.8 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Fundamentals_of_Electrical_Machines/CH_5.ipynb b/Fundamentals_of_Electrical_Machines/CH_5.ipynb new file mode 100755 index 00000000..27fbc256 --- /dev/null +++ b/Fundamentals_of_Electrical_Machines/CH_5.ipynb @@ -0,0 +1,435 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 5: DIRECT CURRENT MOTORS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.1, Page number 182"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "l = 10.0 #Conductor length(m)\n",
+ "B = 0.56 #Magnetic flux density(T)\n",
+ "I = 2.0 #Current through conductor(A)\n",
+ "\n",
+ "#Calculation\n",
+ "F = B*I*l #Magnitude of force(N)\n",
+ "\n",
+ "#Result\n",
+ "print('Magnitude of force , F = %.1f N' %F)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of force , F = 11.2 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.2, Page number 189"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "I = 20.0 #Total current(A)\n",
+ "V_t = 250.0 #Supply voltage(V) \n",
+ "R_sh = 200.0 #Shunt field resistance(ohm)\n",
+ "R_a = 0.3 #Armature resistance(ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "I_sh = V_t/R_sh #Shunt field current(A)\n",
+ "I_a = I-I_sh #Armature current(A)\n",
+ "#For case(i)\n",
+ "E_b = V_t-R_a*I_a #Back emf(V)\n",
+ "#For case(ii)\n",
+ "P_md = E_b*I_a #Mechanical power developed(W) \n",
+ "\n",
+ "#Result\n",
+ "print('(i) Value of back emf , E_b = %.1f V' %E_b) \n",
+ "print('(ii) Mechanical power developed in the armature motor , P_md = %.1f W' %P_md)\n",
+ "print('\\nNOTE : ERROR : Armature current I_a = 18.13 A is taken in textbook solution instead of I_a = 18.75 A')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Value of back emf , E_b = 244.4 V\n",
+ "(ii) Mechanical power developed in the armature motor , P_md = 4582.0 W\n",
+ "\n",
+ "NOTE : ERROR : Armature current I_a = 18.13 A is taken in textbook solution instead of I_a = 18.75 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.3, Page number 189"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "R_a = 0.7 #Armature circuit resistance(ohm)\n",
+ "V_t = 5.0 #Applied voltage(V)\n",
+ "I_anl = 5.0 #No-load armature current(A)\n",
+ "I_afl = 35.0 #Full-load armature current(A)\n",
+ "\n",
+ "#Calculation\n",
+ "E_bnl = V_t-R_a*I_anl #Back emf under no-load(V)\n",
+ "E_bfl = V_t-R_a*I_afl #Back emf under full-load(V)\n",
+ "E_bc = E_bnl-E_bfl #Change in back emf from no-load to full load(V)\n",
+ " \n",
+ "#Result\n",
+ "print('Change in back emf from no-load to full load , E_bc = %.f V' %E_bc)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Change in back emf from no-load to full load , E_bc = 21 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.4, Page number 191"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "I = 40.0 #Current(A)\n",
+ "V_t = 230.0 #Supply voltage(V)\n",
+ "N = 1100.0 #Speed(rpm)\n",
+ "R_a = 0.25 #Armature resistance(ohm)\n",
+ "R_sh = 230.0 #Shunt field resistance(ohm) \n",
+ "\n",
+ "#Calculation\n",
+ "I_sh = V_t/R_sh #Shunt field current(A)\n",
+ "I_a = I-I_sh #Armature current(A)\n",
+ "E_b = V_t-I_a*R_a #Back emf(V) \n",
+ "T_a = 9.55*E_b*I_a/N #Armature torque(N-m)\n",
+ " \n",
+ "#Result\n",
+ "print('Torque developed by the armature , T_a = %.2f N-m' %T_a)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Torque developed by the armature , T_a = 74.57 N-m\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5, Page number 191-192"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "P = 6.0 #Number of poles\n",
+ "V_t = 230.0 #Supply voltage to shunt motor(V)\n",
+ "Z = 450.0 #Number of conductors\n",
+ "R_a = 0.8 #Armature resistance(ohm)\n",
+ "I = 30.0 #Current drawn from supply(A)\n",
+ "P_0 = 5560.0 #Output power(W)\n",
+ "I_f = 3.0 #Current through field winding(A)\n",
+ "phi = 25*10**-3 #Flux per pole(Wb)\n",
+ "\n",
+ "#Calculation\n",
+ "A = P #Number of parallel paths in lap winding\n",
+ "I_a = I-I_f #Armature current(A)\n",
+ "E_b = V_t-I_a*R_a #Back emf(V)\n",
+ "N = 60*A*E_b/(P*Z*phi) #Speed(rpm)\n",
+ "T_sh = 9.55*P_0/N #Shaft torque(N-m)\n",
+ "\n",
+ "#Result\n",
+ "print('Speed , N = %.1f rpm' %N)\n",
+ "print('Shaft torque , T_sh = %.1f N-m' %T_sh)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Speed , N = 1111.5 rpm\n",
+ "Shaft torque , T_sh = 47.8 N-m\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.6, Page number 193-194"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "I_Lnl = 5.0 #Current drawn at no-load(A)\n",
+ "V_t = 230.0 #Terminal voltage at no-load(V)\n",
+ "N_nl = 1000.0 #Speed at no-load(rpm)\n",
+ "R_a = 0.2 #Armature resistance(ohm)\n",
+ "R_f = 230.0 #Field resistance(ohm)\n",
+ "I_Lfl = 30.0 #Current drawn at full-load(A)\n",
+ "\n",
+ "#Calculation\n",
+ "#Under No-load condition\n",
+ "I_sh = V_t/R_f #Shunt field current(A)\n",
+ "I_a1 = I_Lnl-I_sh #Armature current(A)\n",
+ "E_b1 = V_t-I_a1*R_a #Back emf(V)\n",
+ "#Under Full-load condition\n",
+ "I_a2 = I_Lfl-I_sh #Armature current(A)\n",
+ "E_b2 = V_t-I_a2*R_a #Back emf(V)\n",
+ "N_2 = (E_b2/E_b1)*N_nl #Motor speed under load condition(rpm)\n",
+ "\n",
+ "#Result\n",
+ "print('Motor speed under load condition , N_2 = %.1f rpm' %N_2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Motor speed under load condition , N_2 = 978.2 rpm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.7, Page number 194-195"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "I_a1 = 65.0 #Current drawn(A)\n",
+ "V_t = 230.0 #Supply voltage(V)\n",
+ "N_1 = 900.0 #Speed(rpm)\n",
+ "R_a = 0.2 #Armature resistance(ohm)\n",
+ "R_sh = 0.25 #Field resistance(ohm)\n",
+ "I_a2 = 15.0 #Line current(A)\n",
+ "phi_1 = 1.0 #Assumtion of flux(Wb)\n",
+ "phi_2 = 0.4*phi_1 #Flux(Wb)\n",
+ "\n",
+ "#Calculation\n",
+ "E_b1 = V_t-I_a1*(R_a+R_sh) #Initial back emf(V)\n",
+ "E_b2 = V_t-I_a2*(R_a+R_sh) #Final back emf(V)\n",
+ "N_2 = N_1*E_b2*phi_1/(E_b1*phi_2) #Speed of motor(rpm)\n",
+ "\n",
+ "#Result\n",
+ "print('Speed of motor when line current is 15 A , N_2 = %.f rpm' %N_2)\n",
+ "print('\\nNOTE : ERROR : In textbook question current I_a1 = 56 A is given but it should be I_a1 = 65 A')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Speed of motor when line current is 15 A , N_2 = 2502 rpm\n",
+ "\n",
+ "NOTE : ERROR : In textbook question current I_a1 = 56 A is given but it should be I_a1 = 65 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.8, Page number 197-198"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "I_Lnl = 5.0 #Current drawn at no-load(A)\n",
+ "V_t = 230.0 #Supply voltage at no-load(V)\n",
+ "R_a = 0.25 #Armature circuit resistance(ohm)\n",
+ "R_sh = 115 #Field circuit resistance(ohm)\n",
+ "I_L = 40.0 #Current under load condition(A)\n",
+ "\n",
+ "#Calculation\n",
+ "#Under No-load condition\n",
+ "P_in1 = V_t*I_Lnl #Input power(W)\n",
+ "I_sh = V_t/R_sh #Shunt field current(A)\n",
+ "I_a1 = I_Lnl-I_sh #Armature current(A)\n",
+ "P_acu1 = I_a1**2*R_a #Armature copper loss(W)\n",
+ "P_shcu = I_sh**2*R_sh #Shunt field copper loss(W)\n",
+ "P_iron_friction = P_in1-(P_acu1+P_shcu) #Iron and friction losses(W)\n",
+ "#Under load condition\n",
+ "I_a2 = I_L-I_sh #Armature current(A)\n",
+ "P_acu2 = I_a2**2*R_a #Armature copper loss(W)\n",
+ "P_loss = P_iron_friction+P_shcu+P_acu2 #Total losses(W)\n",
+ "P_in2 = V_t*I_L #Input power(W)\n",
+ "P_0 = P_in2-P_loss #Output power(W)\n",
+ "n = (P_0/P_in2)*100 #Efficiency(percent)\n",
+ "\n",
+ "#Result\n",
+ "print('Iron and friction losses , P_iron_friction = %.2f W' %P_iron_friction)\n",
+ "print('Efficiency , \u03b7 = %.f percent' %n)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Iron and friction losses , P_iron_friction = 687.75 W\n",
+ "Efficiency , \u03b7 = 84 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.9, Page number 198-199"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "I_L = 80.0 #Current drawn(A)\n",
+ "V_t = 220.0 #Supply voltage(V)\n",
+ "N = 800.0 #Speed(rpm)\n",
+ "R_a = 0.1 #Armature resistance(ohm)\n",
+ "R_sh = 50.0 #Shunt field resistance(ohm)\n",
+ "P_if = 1600.0 #Iron and friction losses(W)\n",
+ "\n",
+ "#Calculation\n",
+ "I_sh = V_t/R_sh #Shunt field current(A)\n",
+ "I_a = I_L-I_sh #Armature current(A)\n",
+ "E_b = V_t-I_a*R_a #Back emf(V)\n",
+ "#For case(i)\n",
+ "P_in = V_t*I_L #Input power(W)\n",
+ "P_md = E_b*I_a #Mechanical power developed in the armature(W)\n",
+ "P_cu = P_in-P_md #Copper loss(W)\n",
+ "#For case(ii)\n",
+ "T_a = 9.55*E_b*I_a/N #Armature torque(N-m)\n",
+ "#For case(iii)\n",
+ "P_0 = P_md-P_if #Output power(W)\n",
+ "T_sh = 9.55*P_0/N #Shaft torque(N-m)\n",
+ "#For case(iv)\n",
+ "n = (P_0/P_in)*100 #Efficiency(percent)\n",
+ "\n",
+ "#Result\n",
+ "print('(i) Copper losses , P_cu = %.2f W' %P_cu)\n",
+ "print('(ii) Armature torque , T_a = %.2f N-m' %T_a)\n",
+ "print('(iii) Shaft torque , T_sh = %.2f N-m' %T_sh)\n",
+ "print('(iv) Efficiency , \u03b7 = %.f percent' %n)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Copper losses , P_cu = 1539.54 W\n",
+ "(ii) Armature torque , T_a = 191.72 N-m\n",
+ "(iii) Shaft torque , T_sh = 172.62 N-m\n",
+ "(iv) Efficiency , \u03b7 = 82 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Fundamentals_of_Electrical_Machines/CH_6.ipynb b/Fundamentals_of_Electrical_Machines/CH_6.ipynb new file mode 100755 index 00000000..84ed6f6b --- /dev/null +++ b/Fundamentals_of_Electrical_Machines/CH_6.ipynb @@ -0,0 +1,311 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 6: CONTROL AND STARTING OF A DC MOTOR"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.1, Page number 212"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "V_t = 220.0 #Supply voltage(V)\n",
+ "I_a1 = 10.0 #Armature current(A)\n",
+ "N_1 = 900.0 #Speed(rpm)\n",
+ "R_a = 1.0 #Armature resistance(ohm)\n",
+ "N_2 = 500.0 #Reduced speed(rpm)\n",
+ "\n",
+ "#Calculation\n",
+ "E_b1 = V_t-I_a1*R_a #Initial back emf(V)\n",
+ "R = E_b1/I_a1*(1-(N_2/N_1)) #Value of additional resistance(ohm)\n",
+ "\n",
+ "#Result\n",
+ "print('Value of additional resistance , R = %.1f ohm' %R)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of additional resistance , R = 9.3 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.2, Page number 212-213"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "V_t = 230.0 #Supply voltage(V)\n",
+ "I_a1 = 15.0 #Armature current(A)\n",
+ "N_1 = 650.0 #Speed(rpm)\n",
+ "R_a = 0.4 #Armature resistance(ohm)\n",
+ "R = 1.0 #Variable resistance in series with the armature(ohm)\n",
+ "\n",
+ "#Calculation \n",
+ "E_b1 = V_t-I_a1*R_a #Initial back emf(V)\n",
+ "#At full load torque\n",
+ "E_b2 = V_t-I_a1*(R+R_a) #Final back emf(V)\n",
+ "N_2 = N_1*(E_b2/E_b1) #Speed at full load torque(rpm)\n",
+ "\n",
+ "#At half load torque\n",
+ "I_a2hl = I_a1/2 #Armature current(A)\n",
+ "E_b2hl = V_t-I_a2hl*(R+R_a) #Back emf(V)\n",
+ "N_2hl = N_1*(E_b2hl/E_b1) #Speed at half load torque(rpm)\n",
+ "\n",
+ "#Result\n",
+ "print('Speed at full load torque , N_2 = %.1f rpm' %N_2)\n",
+ "print('Speed at half load torque , N_2 = %.1f rpm' %N_2hl)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Speed at full load torque , N_2 = 606.5 rpm\n",
+ "Speed at half load torque , N_2 = 636.9 rpm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.3, Page number 215-216"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "V_t = 230.0 #Supply voltage(V)\n",
+ "R_af = 0.2 #Total resistance of armature & field(ohm)\n",
+ "I_a1 = 10.0 #Armature current for certain load(A)\n",
+ "N = 1000.0 #Motor speed for certain load(rpm)\n",
+ "R_1 = 0 #Variable resistance for certain load(ohm)\n",
+ "I_a2 = 8.0 #Armature current for other load(A)\n",
+ "R_2 = 2.0 #Variable resistance for other load(ohm)\n",
+ "phi_1 = 1.0 #Assuming flux in certain load(Wb)\n",
+ "phi_2 = 0.8*phi_1 #Flux in other load(Wb)\n",
+ "\n",
+ "#Calculation\n",
+ "#For certain load\n",
+ "R_a1 = R_af+R_1 #New armature resistance(ohm)\n",
+ "E_b1 = V_t-I_a1*R_a1 #Back emf(V)\n",
+ "#For other load\n",
+ "R_a2 = R_af+R_2 #New armature resistance(ohm)\n",
+ "E_b2 = V_t-I_a2*R_a2 #Back emf(V)\n",
+ "N_2 = (E_b2/E_b1)*(phi_1/phi_2)*N #New speed(rpm) \n",
+ "\n",
+ "#Result\n",
+ "print('New speed , N_2 = %.1f rpm' %N_2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "New speed , N_2 = 1164.5 rpm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.4, Page number 216-218"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "N = 1100.0 #Speed of dc series motor(rpm)\n",
+ "P = 4.0 #Number of poles\n",
+ "path = 4.0 #Number of parallel paths\n",
+ "I_a1 = 15.0 #Supply current to dc series motor(A)\n",
+ "V_t = 220.0 #Supply voltage(V)\n",
+ "R_a = 0.9 #Series armature resistance(ohm)\n",
+ "R_se = 0.6 #Series field resistance(ohm)\n",
+ "I_a2 = 25 #Supply current to dc series motor(A)\n",
+ "phi_1 = 1.0 #Assuming flux for 15 A case(Wb)\n",
+ "phi_2 = 0.8*phi_1 #Flux in 25 A case(Wb)\n",
+ "\n",
+ "#Calculation\n",
+ "#First case\n",
+ "R_se1 = R_se #Total series field resistance(ohm)\n",
+ "E_b1 = V_t-I_a1*(R_a+R_se1) #Back emf(V)\n",
+ "#Second case\n",
+ "R_se2 = R_se1/path #Total series field resistance(ohm)\n",
+ "E_b2 = V_t-I_a2*(R_a+R_se2) #Back emf(V)\n",
+ "N_2 = (E_b2/E_b1)*(phi_1/phi_2)*N #New speed(rpm) \n",
+ "\n",
+ "#Result\n",
+ "print('Speed for second case , N_2 = %.1f rpm = %.f rpm' %(N_2,N_2))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Speed for second case , N_2 = 1348.9 rpm = 1349 rpm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.5, Page number 233-234"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "V_t = 230.0 #Shunt motor supply voltage(V)\n",
+ "R_a = 0.4 #Armature resistance(ohm)\n",
+ "I_a = 30.0 #Armature current(A)\n",
+ "n = 3.0 #Number of steps\n",
+ "\n",
+ "#Calculation\n",
+ "R_1 = V_t/I_a #Maximum resistance(ohm)\n",
+ "k = (R_1/R_a)**(1.0/3) #Constant\n",
+ "R_2 = R_1/k #Resistance(ohm)\n",
+ "R_3 = R_2/k #Resistance(ohm)\n",
+ "R_4 = R_3/k #Resistance(ohm)\n",
+ "R_1step = R_1-R_2 #Resistance of the first step(ohm)\n",
+ "R_2step = R_2-R_3 #Resistance of the second step(ohm)\n",
+ "R_3step = R_3-R_4 #Resistance of the third step(ohm)\n",
+ "\n",
+ "#Result\n",
+ "print('Resistance of the first step , R_1step = %.1f ohm' %R_1step)\n",
+ "print('Resistance of the second step , R_2step = %.1f ohm' %R_2step)\n",
+ "print('Resistance of the third step , R_3step = %.2f ohm' %R_3step)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance of the first step , R_1step = 4.8 ohm\n",
+ "Resistance of the second step , R_2step = 1.8 ohm\n",
+ "Resistance of the third step , R_3step = 0.67 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6, Page number 234-236"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "V_t = 220.0 #Shunt motor supply voltage(V)\n",
+ "P_0 = 3550.0 #Output power(W) \n",
+ "n = 0.85 #Efficiency\n",
+ "\n",
+ "#Calculation\n",
+ "P_in = P_0/n #Input power(W)\n",
+ "P_tloss = P_in-P_0 #Total loss(W)\n",
+ "I_a = P_in/V_t #Armature current(A)\n",
+ "P_cu = P_tloss/2 #Copper loss(W)\n",
+ "R_a = P_cu/I_a**2 #Armature resistance(ohm)\n",
+ "I_1 = 2*I_a #Maximum starting current(A)\n",
+ "R_1 = V_t/I_1 #Maximum resistance(ohm)\n",
+ "k = (R_1/R_a)**(1.0/4) #Constant\n",
+ "R_2 = R_1/k #Resistance(ohm)\n",
+ "R_3 = R_2/k #Resistance(ohm)\n",
+ "R_4 = R_3/k #Resistance(ohm)\n",
+ "R_5 = R_4/k #Resistance(ohm)\n",
+ "R_1step = R_1-R_2 #Resistance of the first step(ohm)\n",
+ "R_2step = R_2-R_3 #Resistance of the second step(ohm)\n",
+ "R_3step = R_3-R_4 #Resistance of the third step(ohm)\n",
+ "R_4step = R_4-R_5 #Resistance of the fourth step(ohm)\n",
+ "\n",
+ "#Result\n",
+ "print('Resistance of the first step , R_1step = %.1f ohm' %R_1step)\n",
+ "print('Resistance of the second step , R_2step = %.2f ohm' %R_2step)\n",
+ "print('Resistance of the third step , R_3step = %.2f ohm' %R_3step)\n",
+ "print('Resistance of the fourth step , R_4step = %.2f ohm' %R_4step)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance of the first step , R_1step = 2.2 ohm\n",
+ "Resistance of the second step , R_2step = 1.36 ohm\n",
+ "Resistance of the third step , R_3step = 0.85 ohm\n",
+ "Resistance of the fourth step , R_4step = 0.53 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Fundamentals_of_Electrical_Machines/CH_7.ipynb b/Fundamentals_of_Electrical_Machines/CH_7.ipynb new file mode 100755 index 00000000..167a7464 --- /dev/null +++ b/Fundamentals_of_Electrical_Machines/CH_7.ipynb @@ -0,0 +1,400 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 7: THREE-PHASE INDUCTION MOTOR"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.1, Page number 246-247"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "V = 230.0 #Supply voltage(V)\n",
+ "P = 4.0 #Number of poles\n",
+ "f = 50.0 #Frequency(Hz)\n",
+ "N_l = 1445.0 #Full load speed(rpm)\n",
+ "\n",
+ "#Calculation\n",
+ "#For case(i)\n",
+ "N_s = 120*f/P #Synchronous speed(rpm)\n",
+ "#For case(ii)\n",
+ "s = (N_s-N_l)/N_s #Slip\n",
+ "#For case(iii)\n",
+ "f_r = s*f #Rotor frequency(Hz)\n",
+ "\n",
+ "#Result\n",
+ "print('(i) Synchronous speed , N_s = %.f rpm' %N_s)\n",
+ "print('(ii) Slip , s = %.4f ' %s)\n",
+ "print('(iii) Rotor frequency , f_r = %.1f Hz' %f_r)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Synchronous speed , N_s = 1500 rpm\n",
+ "(ii) Slip , s = 0.0367 \n",
+ "(iii) Rotor frequency , f_r = 1.8 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.2, Page number 247-248"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "E_BR = 120.0 #Voltage under blocked condition(V)\n",
+ "P = 4.0 #Number of poles\n",
+ "f = 50.0 #Frequency(Hz)\n",
+ "N_l = 1450.0 #Speed(rpm)\n",
+ "\n",
+ "#Calculation\n",
+ "N_s = 120*f/P #Synchronous speed(rpm)\n",
+ "s = (N_s-N_l)/N_s #Slip\n",
+ "f_r = s*f #Rotor frequency(Hz)\n",
+ "E_r = s*E_BR #Rotor voltage(V)\n",
+ "\n",
+ "#Result\n",
+ "print('Synchronous speed , N_s = %.f rpm' %N_s)\n",
+ "print('Rotor frequency , f_r = %.2f Hz' %f_r)\n",
+ "print('Rotor voltage , E_r = %.2f V' %E_r)\n",
+ "print('\\nNOTE : Changes in answer is due to precision i.e more number of decimal places')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Synchronous speed , N_s = 1500 rpm\n",
+ "Rotor frequency , f_r = 1.67 Hz\n",
+ "Rotor voltage , E_r = 4.00 V\n",
+ "\n",
+ "NOTE : Changes in answer is due to precision i.e more number of decimal places\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.3, Page number 250"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "V_0 = 230.0 #Supply voltage(V)\n",
+ "P = 4.0 #Number of poles\n",
+ "T_0 = 230.0 #Original torque(N-m)\n",
+ "V_s = 150.0 #Stator voltage(V)\n",
+ "I_0 = 560.0 #Starting current(A)\n",
+ "\n",
+ "#Calculation\n",
+ "T_st = (V_s/V_0)**2*T_0 #Starting torque(N-m)\n",
+ "I_st = I_0*(V_s/V_0) #Starting current(A)\n",
+ "\n",
+ "#Result\n",
+ "print('Starting torque , T_st = %.1f N-m' %T_st)\n",
+ "print('Starting current , I_st = %.1f A' %I_st)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Starting torque , T_st = 97.8 N-m\n",
+ "Starting current , I_st = 365.2 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.4, Page number 254"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "f = 50.0 #Frequency(Hz)\n",
+ "P = 8.0 #Number of poles\n",
+ "a = 0.03 #Full load slip\n",
+ "R_2 = 0.01 #Rotor resistance per phase(ohm)\n",
+ "X_2 = 0.1 #Standstill reactance per phase(ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "#For case(i)\n",
+ "N_s = 120*f/P #Synchronous speed(rpm)\n",
+ "s = R_2/X_2 #Slip at maximum torque\n",
+ "N_l = (1-s)*N_s #Rotor speed at maximum torque(rpm)\n",
+ "#For case(ii)\n",
+ "T = (s**2+a**2)/(2*a*s) #Ratio of maximum torque to full load torque\n",
+ "\n",
+ "#Result\n",
+ "print('(i) Speed at which maximum torque occurs , N_l = %.f rpm' %N_l)\n",
+ "print('(ii) Ratio of the maximum torque to full load torque , T_max/T_f = %.2f ' %T)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Speed at which maximum torque occurs , N_l = 675 rpm\n",
+ "(ii) Ratio of the maximum torque to full load torque , T_max/T_f = 1.82 \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.5, Page number 260"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "V = 440.0 #Supply voltage(V)\n",
+ "P = 4.0 #Number of poles\n",
+ "P_ag = 1500.0 #Rotor input(W)\n",
+ "P_rcu = 250.0 #Copper loss(W)\n",
+ "f = 50.0 #Frequency(Hz)\n",
+ "\n",
+ "#Calculation\n",
+ "#For case(i)\n",
+ "s = P_rcu/P_ag #Slip\n",
+ "#For case(ii)\n",
+ "N_s = 120*f/P #Synchronous speed(rpm)\n",
+ "#For case(iii)\n",
+ "N_l = (1-s)*N_s #Shaft speed(rpm)\n",
+ "#For case(iv)\n",
+ "P_mech = (1-s)*P_ag #Mechanical power developed(W)\n",
+ "\n",
+ "#Result\n",
+ "print('(i) Slip , s = %.2f' %s)\n",
+ "print('(ii) Synchronous speed , N_s = %.f rpm' %N_s)\n",
+ "print('(iii) Shaft speed , N_l = %.f rpm' %N_l)\n",
+ "print('(iv) Mechanical power developed , P_mech = %.f W' %P_mech)\n",
+ "print('\\nNOTE : Changes in answer is due to precision i.e more number of decimal places')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Slip , s = 0.17\n",
+ "(ii) Synchronous speed , N_s = 1500 rpm\n",
+ "(iii) Shaft speed , N_l = 1250 rpm\n",
+ "(iv) Mechanical power developed , P_mech = 1250 W\n",
+ "\n",
+ "NOTE : Changes in answer is due to precision i.e more number of decimal places\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.6, Page number 264"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "V_1 = 150.0 #Supply voltage(V)\n",
+ "P = 4.0 #Number of poles\n",
+ "f = 50.0 #Frequency(Hz)\n",
+ "Z_1 = complex(0.12,0.16) #Per phase standstill stator impedance(ohm)\n",
+ "Z_2 = complex(0.22,0.28) #Per phase standstill rotor impedance(ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "Z_eq = Z_1+Z_2 #Equivalent impedance(ohm)\n",
+ "R_eq = Z_eq.real\n",
+ "P_mech = 3*V_1**2/(2*(R_eq+abs(Z_eq)))*10**-3 #Maximum mechanical power developed(kW)\n",
+ "R_2 = Z_2.real\n",
+ "s_mp = R_2/(abs(Z_eq)+R_2) #Slip\n",
+ "W_s = 2*math.pi*2*f/P #Synchronous speed(rad/s)\n",
+ "W = (1-s_mp)*W_s #Speed of rotor(rad/s)\n",
+ "T_mxm = P_mech*1000/W #Maximum torque(N-m) \n",
+ "\n",
+ "#Result\n",
+ "print('Maximum mechanical power , P_mech = %.2f kW' %P_mech)\n",
+ "print('Maximum torque , T_mxm = %.2f N-m' %T_mxm)\n",
+ "print('Slip , s_mp = %.2f ' %s_mp)\n",
+ "print('\\nNOTE : Changes in answer is due to precision i.e more number of decimal places')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum mechanical power , P_mech = 37.66 kW\n",
+ "Maximum torque , T_mxm = 334.65 N-m\n",
+ "Slip , s_mp = 0.28 \n",
+ "\n",
+ "NOTE : Changes in answer is due to precision i.e more number of decimal places\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.7, Page number 265-266"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "V = 440.0 #Supply voltage(V)\n",
+ "P = 6.0 #Number of poles\n",
+ "f = 50.0 #Frequency(Hz)\n",
+ "P_a = 45000.0 #Input power(W)\n",
+ "N_l = 900.0 #Speed(rpm)\n",
+ "P_tloss = 2000.0 #Total stator losses(W)\n",
+ "P_fw = 1000.0 #Friction and windage losses(W)\n",
+ "\n",
+ "#Calculation\n",
+ "N_s = 120*f/P #Synchronous speed(rpm)\n",
+ "s = (N_s-N_l)/N_s #Slip\n",
+ "P_ag = (P_a-P_tloss) #Air gap power(W)\n",
+ "P_rcu = s*P_ag #Rotor copper loss(W)\n",
+ "P_mech = P_ag-P_rcu #Mechanical power(W)\n",
+ "P_0 = P_mech-(P_tloss+P_fw) #Output power(W)\n",
+ "n = (P_0/P_ag)*100 #Efficiency(percent)\n",
+ "\n",
+ "#Result\n",
+ "print('(i) Slip , s = %.1f ' %s)\n",
+ "print('(ii) Rotor copper loss , P_rcu = %.f W' %P_rcu)\n",
+ "print('(iii) Shaft or Output power , P_0 = %.f W' %P_0)\n",
+ "print('(iv) Efficiency , \u03b7 = %.f percent' %n)\n",
+ "print('\\nNOTE : ERROR : Friction & windage losses are 1 kW not 1.5 kW as given in textbook question')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Slip , s = 0.1 \n",
+ "(ii) Rotor copper loss , P_rcu = 4300 W\n",
+ "(iii) Shaft or Output power , P_0 = 35700 W\n",
+ "(iv) Efficiency , \u03b7 = 83 percent\n",
+ "\n",
+ "NOTE : ERROR : Friction & windage losses are 1 kW not 1.5 kW as given in textbook question\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.8, Page number 268-269"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "v_s = 120.0 #Train speed(km/h)\n",
+ "f = 50.0 #Stator frequency(Hz)\n",
+ "\n",
+ "#Calculation\n",
+ "v_s1 = v_s*1000/(60*60) #Train speed(m/s)\n",
+ "w = v_s1/(2*f) #Length of the pole-pitch(m)\n",
+ "\n",
+ "#Result\n",
+ "print('Length of the pole-pitch of linear induction motor , w = %.2f m' %w)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Length of the pole-pitch of linear induction motor , w = 0.33 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Fundamentals_of_Electrical_Machines/CH_8.ipynb b/Fundamentals_of_Electrical_Machines/CH_8.ipynb new file mode 100755 index 00000000..ae1358cd --- /dev/null +++ b/Fundamentals_of_Electrical_Machines/CH_8.ipynb @@ -0,0 +1,337 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 8: STARTING, CONTROL AND TESTING OF AN INDUCTION MOTOR"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.1, Page number 273"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "T_st = '1.5*T_f' #Starting torque\n",
+ "s = 0.03 #Slip\n",
+ "\n",
+ "#Calculation\n",
+ "I_sc_I_f = (1.5/s)**0.5 #I_sc/I_f\n",
+ "\n",
+ "#Result\n",
+ "print('Short circuit current , I_sc = %.2f*I_f' %I_sc_I_f)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Short circuit current , I_sc = 7.07*I_f\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.2, Page number 274-275"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "T_ratio = 50.0/100 #Ratio of starting torque to full load torque T_st/T_f\n",
+ "s_f = 0.03 #Full load slip\n",
+ "I_ratio = 5.0 #Ratio of short circuit current to full load current I_sc/I_f\n",
+ "\n",
+ "#Calculation\n",
+ "x = (1/I_ratio)*(T_ratio/s_f)**0.5 #Percentage of taping\n",
+ "\n",
+ "#Result\n",
+ "print('Percentage tapings required on the autotransformer , x = %.3f ' %x)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Percentage tapings required on the autotransformer , x = 0.816 \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.3, Page number 277"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "T_ratio = 25.0/100 #Ratio of starting torque to full load torque T_st/T_f\n",
+ "I_ratio = 3.0*120/100 #Ratio of short circuit current to full load current I_sc/I_f\n",
+ "\n",
+ "#Calculation\n",
+ "s_f = T_ratio*3/I_ratio**2 #Full load slip\n",
+ "\n",
+ "#Result\n",
+ "print('Full load slip , s_f = %.2f ' %s_f)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Full load slip , s_f = 0.06 \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.4, Page number 281-282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "Z_icr = complex(0.04,0.5) #Inner cage impedance per phase at standstill(ohm)\n",
+ "Z_ocr = complex(0.4,0.2) #Outer cage impedance per phase at standstill(ohm)\n",
+ "V = 120.0 #Per phase rotor induced voltage at standstill(V)\n",
+ "\n",
+ "#Calculation\n",
+ "#For case(i)\n",
+ "Z_com = (Z_icr*Z_ocr)/(Z_icr+Z_ocr) #Combined impedance(ohm)\n",
+ "I_2 = V/abs(Z_com) #Rotor current per phase(A)\n",
+ "R_2 = Z_com.real #Combined rotor resistance(ohm)\n",
+ "T = I_2**2*R_2 #Torque at stand still condition(synchronous watts)\n",
+ "#For case(ii)\n",
+ "s = 0.06 #Slip\n",
+ "R_ocr = Z_ocr.real\n",
+ "X_ocr = Z_ocr.imag\n",
+ "R_icr = Z_icr.real\n",
+ "X_icr = Z_icr.imag\n",
+ "Z_com6 = complex(R_ocr/s,X_ocr)*complex(R_icr/s,X_icr)/complex(R_ocr/s+R_icr/s,X_ocr+X_icr) #Combined impedance(ohm)\n",
+ "I2_6 = V/abs(Z_com6) #Rotor current per phase(A)\n",
+ "R2_6 = Z_com6.real #Combined rotor resistance(ohm)\n",
+ "T_6 = I2_6**2*R2_6 #Torque at 6% slip(synhronous watts)\n",
+ "\n",
+ "#Result\n",
+ "print('(i) Torque at standstill condition , T = %.2f syn.watt' %T)\n",
+ "print('(ii) Torque at 6 percent slip , T_6 = %.2f syn.watt' %T_6)\n",
+ "print('\\nNOTE : Changes in answer is due to precision i.e more number of decimal places')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Torque at standstill condition , T = 31089.35 syn.watt\n",
+ "(ii) Torque at 6 percent slip , T_6 = 15982.06 syn.watt\n",
+ "\n",
+ "NOTE : Changes in answer is due to precision i.e more number of decimal places\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.5, Page number 285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "V = 210.0 #Supply voltage(V)\n",
+ "f = 50.0 #Supply frequency(Hz)\n",
+ "P = 50.0 #Input power(W)\n",
+ "I_br = 2.5 #Line current(A)\n",
+ "V_L = 25.0 #Line voltage(V)\n",
+ "R_1 = 2.4 #DC resistance between any two terminal(ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "V_br = V_L/3**0.5 #Phase voltage(V)\n",
+ "P_br = P/3 #Power per phase(W)\n",
+ "R_eq = P_br/I_br**2 #Equivalent resistance(ohm)\n",
+ "R_2 = R_eq-(R_1/2) #Per phase rotor resistance(ohm)\n",
+ "Z_eq = V_br/I_br #Equivalent impedance(ohm)\n",
+ "X_eq = (Z_eq**2-R_2**2)**0.5 #Equivalent reactance(ohm)\n",
+ "X_1 = 0.5*X_eq #For practical cases reactances(ohm)\n",
+ "\n",
+ "#Result\n",
+ "print('Equivalent resistance , R_eq = %.1f ohm' %R_eq)\n",
+ "print('Equivalent impedance , Z_eq = %.1f ohm' %Z_eq)\n",
+ "print('Equivalent reactance , X_eq = %.1f ohm' %X_eq)\n",
+ "print('Per phase rotor resistance , R_2 = %.1f ohm' %R_2)\n",
+ "print('Reactances for practical cases , X_1 = X_2 = %.1f ohm' %X_1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equivalent resistance , R_eq = 2.7 ohm\n",
+ "Equivalent impedance , Z_eq = 5.8 ohm\n",
+ "Equivalent reactance , X_eq = 5.6 ohm\n",
+ "Per phase rotor resistance , R_2 = 1.5 ohm\n",
+ "Reactances for practical cases , X_1 = X_2 = 2.8 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.6, Page number 287"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "V = 210.0 #Supply voltage(V)\n",
+ "f = 50.0 #Supply frequency(Hz)\n",
+ "P = 4.0 #Number of poles\n",
+ "P_0 = 400.0 #Input power(W)\n",
+ "I_0 = 1.2 #Line current(A)\n",
+ "V_0 = 210.0 #Line voltage(V)\n",
+ "P_fw = 150.0 #Total friction and windage losses(W)\n",
+ "R = 2.2 #Stator resistance between any two terminals(ohm)\n",
+ " \n",
+ "#Calculation\n",
+ "R_1 = R/2 #Per phase stator resistance(ohm)\n",
+ "P_scu = 3*I_0**2*R_1 #Stator copper loss(W)\n",
+ "P_core = P_0-P_fw-P_scu #Stator core loss(W)\n",
+ "R_0 = (V_0/3**0.5)**2/(P_core/3) #No-load resistance(ohm)\n",
+ "#Alternate approach\\n\",\n",
+ "phi_0 = math.acos(P_core/(3**0.5*V_0*I_0)) #Power factor angle(radians)\n",
+ "phi_0_deg = phi_0*180/math.pi #Power factor angle(degree)\n",
+ "R_01 = (V_0/3**0.5)/(I_0*math.cos(phi_0)) #No-load circuit resistance per phase(ohm)\n",
+ "X_0 = (V_0/3**0.5)/(I_0*math.sin(phi_0)) #Magnetizing reactance per phase(ohm)\n",
+ " \n",
+ "#Result\n",
+ "print('Stator core loss , P_core = %.1f W' %P_core)\n",
+ "print('No-load circuit resistance per phase , R_0 = %.1f ohm' %R_01)\n",
+ "print('Magnetizing reactance per phase , X_0 = %.f ohm' %X_0)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stator core loss , P_core = 245.2 W\n",
+ "No-load circuit resistance per phase , R_0 = 179.8 ohm\n",
+ "Magnetizing reactance per phase , X_0 = 122 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.7, Page number 290"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "P_1 = 6.0 #Number of pole\n",
+ "P_2 = 4.0 #Number of pole\n",
+ "f = 50.0 #Supply frequency(Hz)\n",
+ "P = 60.0 #Power(kW)\n",
+ "\n",
+ "#Calculation\n",
+ "#For case(i)\n",
+ "s = P_2/(P_1+P_2) #Combined slip\n",
+ "#For case(ii)\n",
+ "N_cs = 120*f/(P_1+P_2) #Combined synchronous speed(rpm)\n",
+ "#For case(iii)\n",
+ "P_0 = P*P_2/(P_1+P_2) #Output of 4-pole motor(kW)\n",
+ "\n",
+ "#Result\n",
+ "print('(i) Combined slip , s = %.1f ' %s)\n",
+ "print('(ii) Combined synchronous speed , N_cs = %.f rpm' %N_cs)\n",
+ "print('(iii) Output of the 4-pole motor , P_0 = %.f kW' %P_0)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Combined slip , s = 0.4 \n",
+ "(ii) Combined synchronous speed , N_cs = 600 rpm\n",
+ "(iii) Output of the 4-pole motor , P_0 = 24 kW\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Fundamentals_of_Electrical_Machines/CH_9.ipynb b/Fundamentals_of_Electrical_Machines/CH_9.ipynb new file mode 100755 index 00000000..35fe3167 --- /dev/null +++ b/Fundamentals_of_Electrical_Machines/CH_9.ipynb @@ -0,0 +1,625 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 9: SYNCHRONOUS GENERATOR"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.1, Page number 295"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "N = 300.0 #Speed of water turbine(rpm)\n",
+ "f = 50.0 #Frequency of induced voltage(Hz)\n",
+ "\n",
+ "#Calculation\n",
+ "P = 120*f/N #Number of poles\n",
+ "\n",
+ "#Result\n",
+ "print('Number of poles of the generator , P = %.f poles' %P)\n",
+ "print('Alternatively , %.f pairs of north(N) and south(S) poles' %(P/2))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Number of poles of the generator , P = 20 poles\n",
+ "Alternatively , 10 pairs of north(N) and south(S) poles\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.2, Page number 299-300"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "P = 8.0 #Number of poles\n",
+ "m = 3.0 #Number of phase \n",
+ "S = 144.0 #Number of slots\n",
+ "\n",
+ "#Calculation\n",
+ "T_p = S/P #Pole pitch(slots)\n",
+ "slots_1 = 180/T_p #Pole pitch per slot(degree)\n",
+ "gamma = 2*slots_1 #Short pitch angle(degree)\n",
+ "y = gamma*math.pi/180 #Short pitch angle(radian)\n",
+ "k_p = math.cos(y/2) #Pitch factor\n",
+ "\n",
+ "#Result\n",
+ "print('Pitch factor , k_p = %.2f ' %k_p)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Pitch factor , k_p = 0.98 \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.3, Page number 300-301"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "P = 4.0 #Number of poles\n",
+ "m = 3.0 #Number of phase \n",
+ "S = 40.0 #Number of slots\n",
+ "s_1 = 1.0 #Coil span\n",
+ "s_2 = 9.0 #Coil span\n",
+ "\n",
+ "#Calculation\n",
+ "T_p = S/P #Pole pitch(slots)\n",
+ "T_c = s_2-s_1 #Coil pitch for coil spans 1 to 9(slots)\n",
+ "slots = 180/T_p #Pole pitch per slot(degree)\n",
+ "y = T_p-T_c #Short pitch angle(slots)\n",
+ "gamma = y*slots #Short pitch angle(degree)\n",
+ "y = gamma*math.pi/180 #Short pitch angle(radian)\n",
+ "k_p = math.cos(y/2) #Pitch factor\n",
+ "\n",
+ "#Result\n",
+ "print('Pitch factor , k_p = %.2f ' %k_p)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Pitch factor , k_p = 0.95 \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.4, Page number 302"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "P = 4.0 #Number of poles\n",
+ "S = 48.0 #Number of slots\n",
+ "m = 3.0 #Number of phase \n",
+ "\n",
+ "#Calculation\n",
+ "T_p = S/P #Pole pitch(slots)\n",
+ "slot = 180/T_p #Pole pitch per slot(degree)\n",
+ "a = slot*math.pi/180 #Pole pitch per slot(radian)\n",
+ "n = S/(P*m) #Number of slots or coils per pole per phase\n",
+ "k_d = math.sin(n*a/2)/(n*math.sin(a/2)) #Distribution factor\n",
+ "\n",
+ "#Result\n",
+ "print('Distribution factor , k_d = %.2f ' %k_d)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distribution factor , k_d = 0.96 \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.5, Page number 304-305"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "P = 12.0 #Number of poles\n",
+ "S = 180.0 #Number of slots\n",
+ "phi_m = 0.05 #Flux per pole(Wb)\n",
+ "N = 600.0 #Speed of machine(rpm)\n",
+ "m = 3.0 #Number of phase\n",
+ "\n",
+ "#Calculation\n",
+ "T_p = S/P #Pole pitch(slots)\n",
+ "slot = 180/T_p #Slots per pole(degree)\n",
+ "n = S/(P*m) #Number of slots or coils per pole per phase\n",
+ "a = slot*math.pi/180 #Pole pitch per slot(radian)\n",
+ "k_d = math.sin(n*a/2)/(n*math.sin(a/2)) #Distribution factor\n",
+ "k_p = 1.0 #Pitch factor\n",
+ "Z = (180/m)*slot #Number of conductors per phase\n",
+ "T = Z/2 #Number of turns per phase\n",
+ "f = P*N/120 #Frequency(Hz)\n",
+ "E = 4.44*k_p*k_d*f*phi_m*T #Induced voltage(V)\n",
+ "E_L = 3**0.5*E #Line voltage(V)\n",
+ "\n",
+ "#Result\n",
+ "print('Line voltage , E_L = %.1f V' %E_L)\n",
+ "print('\\nNOTE : Changes in obtained answer from that of textbook answer is due to precision i.e more number of decimal places')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Line voltage , E_L = 7945.7 V\n",
+ "\n",
+ "NOTE : Changes in obtained answer from that of textbook answer is due to precision i.e more number of decimal places\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.6, Page number 305-307"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "P = 4.0 #Number of poles\n",
+ "m = 3.0 #Number of phase \n",
+ "f = 50.0 #Frequency(Hz)\n",
+ "phi_m = 0.05 #Flux per pole(Wb)\n",
+ "n = 6.0 #Number of slots per pole per phase\n",
+ "cond = 5.0 #Conductors per layer\n",
+ "no_layer = 2.0 #Number of layer winding\n",
+ "\n",
+ "#Calculation\n",
+ "T_p = n*m #Slots per pole\n",
+ "slot = 180/T_p #Slots per pole(degree)\n",
+ "a = slot*math.pi/180 #Pole pitch per slot(radian)\n",
+ "T_c = (5.0/6)*T_p #Coil pitch is 5/6 of full pitch\n",
+ "gamma = T_p-T_c #Short pitch angle(slots)\n",
+ "y_angle = gamma*slot #Short pitch(angle)\n",
+ "y = y_angle*math.pi/180 #Short pitch(radians)\n",
+ "k_p = math.cos(y/2) #Pitch factor\n",
+ "k_d = math.sin(n*a/2)/(n*math.sin(a/2)) #Distribution factor\n",
+ "T = 1.0/2*n*P*cond*no_layer #Number of turns in any phase\n",
+ "E = 4.44*k_p*k_d*f*phi_m*T #Voltage per phase(V)\n",
+ "\n",
+ "#Result\n",
+ "print('Voltage per phase , E = %.2f V' %E)\n",
+ "print('\\nNOTE : Changes in obtained answer from that of textbook answer is due to precision i.e more number of decimal places')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage per phase , E = 1230.19 V\n",
+ "\n",
+ "NOTE : Changes in obtained answer from that of textbook answer is due to precision i.e more number of decimal places\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.7, Page number 307-308"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "P = 10.0 #Number of poles\n",
+ "m = 3.0 #Number of phase \n",
+ "f = 50.0 #Frequency(Hz)\n",
+ "n = 3.0 #Number of slots per pole per phase\n",
+ "phi_m1 = 0.05 #Fundamental component of flux(Wb)\n",
+ "phi_m3 = 0.006 #Third harmonic component of flux(Wb)\n",
+ "T_c = 150.0 #Coil span(degree)\n",
+ "cond = 5.0 #Conductors per layer\n",
+ "no_layer = 2.0 #Number of layer winding\n",
+ "\n",
+ "#Calculation\n",
+ "T_p = n*m #Slots per pole\n",
+ "slot = 180/T_p #Slots per pole(degree)\n",
+ "a = slot*math.pi/180 #Pole pitch per slot(radian)\n",
+ "gamma = 180-T_c #Short pitch angle(degree)\n",
+ "y = gamma*math.pi/180 #Short pitch angle(radian)\n",
+ "T = 1.0/2*P*n*cond*no_layer #Number of turns\n",
+ "k_p1 = math.cos(y/2) #Fundamental pitch factor\n",
+ "k_d1 = math.sin(n*a/2)/(n*math.sin(a/2)) #Fundamental distribution factor\n",
+ "E_1 = 4.44*k_p1*k_d1*f*phi_m1*T #Fundamental emf per phase(V)\n",
+ "k_p3 = math.cos(3*y/2) #Third harmonic pitch factor\n",
+ "k_d3 = math.sin(3*n*a/2)/(n*math.sin(3*a/2)) #Third harmonic distribution factor\n",
+ "E_3 = 4.44*k_p3*k_d3*f*phi_m3*T #Voltage(V)\n",
+ "E = (E_1**2+E_3**2)**0.5 #Induced voltage per phase(V)\n",
+ "\n",
+ "#Result\n",
+ "print('rms value of induced voltage per phase , E = %.1f V' %E)\n",
+ "print('\\nNOTE : Changes in obtained answer from that of textbook answer is due to precision i.e more number of decimal places')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "rms value of induced voltage per phase , E = 1546.5 V\n",
+ "\n",
+ "NOTE : Changes in obtained answer from that of textbook answer is due to precision i.e more number of decimal places\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.8, Page number 312"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "kVA = 50.0 #Ratings(kVA)\n",
+ "V_t = 220.0 #Voltage(V)\n",
+ "R_a = 0.011 #Effective resistance(ohm)\n",
+ "X_s = 0.09 #Synchronous reactance(ohm)\n",
+ "pf = 0.85 #Lagging power factor\n",
+ "\n",
+ "#Calculation\n",
+ "phi = math.acos(pf) #Power factor angle(radians)\n",
+ "I_a = kVA*10**3/V_t #Armature current(A)\n",
+ "E_f = ((V_t*pf+I_a*R_a)**2+(V_t*math.sin(phi)+I_a*X_s)**2)**0.5 #Induced voltage per phase(V)\n",
+ "VR = ((E_f-V_t)/V_t)*100 #Voltage regulation(percent)\n",
+ "\n",
+ "#Result\n",
+ "print('No-load induced voltage per phase , E_f = %.1f V' %E_f)\n",
+ "print('Voltage regulation , VR = %.1f percent' %VR)\n",
+ "print('\\nNOTE : Changes in obtained answer from that of textbook answer is due to precision i.e more number of decimal places')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "No-load induced voltage per phase , E_f = 233.5 V\n",
+ "Voltage regulation , VR = 6.1 percent\n",
+ "\n",
+ "NOTE : Changes in obtained answer from that of textbook answer is due to precision i.e more number of decimal places\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9, Page number 314"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "kVA = 200.0 #Rating(kVA)\n",
+ "V_t = 33.0*10**3 #Voltage(V)\n",
+ "R_a = 0.54 #Armature resistance(ohm)\n",
+ "V_L = 415.0 #Voltage between lines for SC test(V)\n",
+ "I_sh = 25.0 #Short circuit current(A)\n",
+ "pf = 0.9 #Lagging power factor\n",
+ "\n",
+ "#Calculation\n",
+ "#For case(i)\n",
+ "V_P = V_L/3**0.5 #Phase voltage during short circuit test(V)\n",
+ "Z_s = V_P/I_sh #Synchronous impedance(ohm)\n",
+ "#For case(ii)\n",
+ "X_s = (Z_s**2-R_a**2)**0.5 #Synchronous reactance(ohm)\n",
+ "#For case(iii)\n",
+ "I_a = kVA*1000/(3**0.5*V_t) #Full load current(A)\n",
+ "V_ta = V_t/3**0.5 #Voltage per phase of alternator(V)\n",
+ "phi = math.acos(pf) #Power factor angle(radians)\n",
+ "E_f = ((V_ta*pf+I_a*R_a)**2+(V_ta*math.sin(phi)+I_a*X_s)**2)**0.5 #No-load voltage per phase(V) \n",
+ "VR = ((E_f-V_ta)/V_ta)*100 #Voltage regulation\n",
+ "\n",
+ "#Result\n",
+ "print('(i) Synchronous impedance , Z_s = %.1f ohm' %Z_s)\n",
+ "print('(ii) Synchronous reactance , X_s = %.2f ohm' %X_s)\n",
+ "print('(iii) Voltage regulation , VR = %.2f percent' %VR)\n",
+ "print('\\nNOTE : ERROR : In textbook calculation , R_a is taken instead of X_s in calculation of E_f')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Synchronous impedance , Z_s = 9.6 ohm\n",
+ "(ii) Synchronous reactance , X_s = 9.57 ohm\n",
+ "(iii) Voltage regulation , VR = 0.09 percent\n",
+ "\n",
+ "NOTE : ERROR : In textbook calculation , R_a is taken instead of X_s in calculation of E_f\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.10, Page number 317"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "MVA = 30.0 #Rating(MVA)\n",
+ "V = 20.0 #Supply voltage(kV)\n",
+ "N = 1800.0 #Speed(rpm)\n",
+ "V_t = 15.0 #Voltage per phase(kV)\n",
+ "E_f = 10.0 #Per phase terminal voltage(kV)\n",
+ "delta = 40.0 #Power angle(degree)\n",
+ "X_s = 6.0 #Per phase synchronous reactance(ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "#For case(i)\n",
+ "d = delta*math.pi/180 #Power angle(radians)\n",
+ "P = 3*V_t*E_f*math.sin(d)/X_s #3-phase power delivered to the load(MW)\n",
+ "#For case(ii)\n",
+ "P_max = 3*V_t*E_f/X_s #Three phase maximum power(MW)\n",
+ "\n",
+ "#Result\n",
+ "print('(i) Three-phase real power delivered to the load , P = %.2f MW' %P)\n",
+ "print('(ii) Three-phase maximum power , P_max = %.f MW' %P_max)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Three-phase real power delivered to the load , P = 48.21 MW\n",
+ "(ii) Three-phase maximum power , P_max = 75 MW\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.11, Page number 321-322"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "kVA = 25.0 #Rating(kVA)\n",
+ "V = 440.0 #Suppy voltage(V)\n",
+ "f = 50.0 #Supply frequency(Hz)\n",
+ "pf = 0.8 #Lagging power factor\n",
+ "R_a = 0.3 #Resistance of machine per phase(ohm)\n",
+ "X_d = 5.0 #Reactance of machine per phase(ohm)\n",
+ "X_q = 3.0 #Reactance of machine per phase(ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "#For case(i)\n",
+ "phi = math.acos(pf) #Power factor angle(radians)\n",
+ "phi_deg = phi*180/math.pi #Power factor angle(degree)\n",
+ "V_t = V/3**0.5 #Terminal voltage per phase(V)\n",
+ "I_a = kVA*10**3/(3**0.5*V) #Armature current(A)\n",
+ "tan_d = (I_a*X_q*pf/(V_t+I_a*X_q*math.sin(phi)))\n",
+ "d = math.atan(tan_d) #Torque angle(radians)\n",
+ "d_angle = d*180/math.pi #Torque angle(degree)\n",
+ "#For case(ii)\n",
+ "I_d = I_a*math.sin(d+phi) #Direct axis component of the current(A)\n",
+ "E_f = V_t*math.cos(d)+I_d*X_d #Induced voltage per phase(V)\n",
+ "#For case(iii)\n",
+ "VR = ((E_f-V_t)/V_t)*100 #Voltage regulation(percent)\n",
+ "\n",
+ "#Result\n",
+ "print('(i) Torque angle , \u03b4 = %.2f\u00b0 ' %d_angle)\n",
+ "print('(ii) Induced voltage per phase , E_f = %.2f V' %E_f)\n",
+ "print('(iii) Voltage regulation , VR = %.2f percent' %VR)\n",
+ "print('\\nNOTE : Changes in obtained answer from that of textbook answer is due to precision i.e more number of decimal places')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Torque angle , \u03b4 = 14.12\u00b0 \n",
+ "(ii) Induced voltage per phase , E_f = 373.80 V\n",
+ "(iii) Voltage regulation , VR = 47.15 percent\n",
+ "\n",
+ "NOTE : Changes in obtained answer from that of textbook answer is due to precision i.e more number of decimal places\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.12, Page number 324-325"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import cmath\n",
+ "\n",
+ "#Variable declaration\n",
+ "E_1 = 220.0 #Induced voltage per phase by alternator 1(V)\n",
+ "E_2 = 220*cmath.exp(1j*5*math.pi/180) #Induced voltage per phase by alternator 2(V)\n",
+ "Z_1 = complex(0,3) #Impedance of alternator 1(ohm)\n",
+ "Z_2 = complex(0,4) #Impedance of alternator 2(ohm)\n",
+ "Z = 5.0 #Load(ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "#For case(i)\n",
+ "I = (E_1*Z_2+E_2*Z_1)/(Z_1*Z_2+Z*(Z_1+Z_2)) #Load current(A)\n",
+ "#For case(ii)\n",
+ "V_t = I*Z #Terminal voltage(V)\n",
+ "#For case(iii)\n",
+ "I_a1 = ((E_1-E_2)*Z+E_1*Z_2)/(Z_1*Z_2+Z*(Z_1+Z_2)) #Armature current(A)\n",
+ "P_1 = abs(V_t*I_a1)*math.cos(cmath.phase(V_t)-cmath.phase(I_a1))*10**-3 #Power per phase delivered by the first alternator(W)\n",
+ "\n",
+ "#Result\n",
+ "print('(i) Load current , I = %.1f\u2220%.f\u00b0 A' %(abs(I),cmath.phase(I)*180/math.pi))\n",
+ "print('(ii) Terminal voltage , V_t = %.f\u2220%.f\u00b0 V' %(abs(V_t),cmath.phase(V_t)*180/math.pi))\n",
+ "print('(iii) Power per phase delivered by the first alternator , P_1 = %.1f kW' %P_1)\n",
+ "print('\\nNOTE : ERROR : In textbook case(iii) power calculation current I is taken instead of I_a1')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Load current , I = 41.6\u2220-17\u00b0 A\n",
+ "(ii) Terminal voltage , V_t = 208\u2220-17\u00b0 V\n",
+ "(iii) Power per phase delivered by the first alternator , P_1 = 4.4 kW\n",
+ "\n",
+ "NOTE : ERROR : In textbook case(iii) power calculation current I is taken instead of I_a1\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Fundamentals_of_Electrical_Machines/README.txt b/Fundamentals_of_Electrical_Machines/README.txt new file mode 100755 index 00000000..eab345e3 --- /dev/null +++ b/Fundamentals_of_Electrical_Machines/README.txt @@ -0,0 +1,10 @@ +Contributed By: KAVAN A B +Course: be +College/Institute/Organization: SRI JAYACHAMARAJENDRA COLLEGE OF ENGINEERING +Department/Designation: ELECTRICAL & ELECTRONICS +Book Title: Fundamentals of Electrical Machines +Author: M A Salam +Publisher: Narosa Publishing House, New Delhi - 110002 +Year of publication: 2009 +Isbn: 978-81-8487-163-0 +Edition: 2
\ No newline at end of file diff --git a/Fundamentals_of_Electrical_Machines/screenshots/1.png b/Fundamentals_of_Electrical_Machines/screenshots/1.png Binary files differnew file mode 100755 index 00000000..338e1e9b --- /dev/null +++ b/Fundamentals_of_Electrical_Machines/screenshots/1.png diff --git a/Fundamentals_of_Electrical_Machines/screenshots/2.png b/Fundamentals_of_Electrical_Machines/screenshots/2.png Binary files differnew file mode 100755 index 00000000..0f649c1a --- /dev/null +++ b/Fundamentals_of_Electrical_Machines/screenshots/2.png diff --git a/Fundamentals_of_Electrical_Machines/screenshots/3.png b/Fundamentals_of_Electrical_Machines/screenshots/3.png Binary files differnew file mode 100755 index 00000000..7ec6174d --- /dev/null +++ b/Fundamentals_of_Electrical_Machines/screenshots/3.png diff --git a/Theory_Of_Machines/README.txt b/Theory_Of_Machines/README.txt new file mode 100755 index 00000000..0967ddbe --- /dev/null +++ b/Theory_Of_Machines/README.txt @@ -0,0 +1,10 @@ +Contributed By: Maulik Rathod +Course: be +College/Institute/Organization: Alpha College of Engineering & Tech., Kalol +Department/Designation: Electronics & Communication +Book Title: Theory Of Machines +Author: R. S. Khurmi And J. K. Gupta +Publisher: Eurasia Publishing House (Pvt.) Ltd., New Delhi +Year of publication: 2009 +Isbn: 81-219-2524-X +Edition: 14
\ No newline at end of file diff --git a/Theory_Of_Machines/ch10.ipynb b/Theory_Of_Machines/ch10.ipynb new file mode 100755 index 00000000..dc3b7316 --- /dev/null +++ b/Theory_Of_Machines/ch10.ipynb @@ -0,0 +1,2084 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:4504e5e6afcec1f6614278d6f9600661d0ec5a3c47dfce6a30afddafccb88875" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 10 : Friction" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.1 Page No : 263" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from numpy import linalg\n", + "\n", + "# Variables:\n", + "theta = 30. \t\t\t#degrees\n", + "P1 = 180. \t\t\t#Pulling force N\n", + "P2 = 220. \t\t\t#Pushing force N\n", + "\n", + "#Solution:\n", + "#Resolving the forces horizontally for the pull of 180N\n", + "F1 = P1*math.cos(math.radians(theta)) \t\t\t#N\n", + "#Resolving the forces for the push of 220 N\n", + "F2 = P2*math.cos(math.radians(theta)) \t\t\t#N\n", + "#Calculating the coefficient of friction\n", + "#For the pull of 180N F1 = mu*W-90*mu or F1/mu-W = -90 .....(i)\n", + "#For the push of 220N F2 = W*mu+110*mu or F2/mu-W = 110 .....(ii)\n", + "A = [[F1, -1],[ F2, -1]]\n", + "B = [-90., 110.]\n", + "V = linalg.solve(A, B)\n", + "mu = 1/V[0]\n", + "W = V[1]\n", + "\n", + "#Results:\n", + "print \" The weight of the body, W = %d N.\"%(round(W,-2))\n", + "print \" The coefficient of friction, mu = %.4f.\"%(mu)\n", + "\n", + "# note : rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The weight of the body, W = 1000 N.\n", + " The coefficient of friction, mu = 0.1732.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.2 Page No : 268" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "from numpy import linalg\n", + "\n", + "# Variables:\n", + "P1 = 1500.\n", + "P2 = 1720. \t\t\t#N\n", + "alpha1 = 12.\n", + "alpha2 = 15. \t\t\t#degrees\n", + "\n", + "#Solution:\n", + "#Refer Fig. 10.10\n", + "#Effort applied parallel to the plane P1 = W*(math.sin(alpha1)+mu*math.cos(alpha1)) or P1/W-mu*math.cos(alpha1) = math.sin(alpha1) .....(i)\n", + "#Effort applied parallel to the plane P2 = W*(math.sin(alpha2)+mu*math.cos(alpha2)) or P2/W-mu*math.cos(alpha2) = math.sin(alpha2) .....(ii)\n", + "A = [[P1, -math.cos(math.radians(alpha1))],[ P2, -math.cos(math.radians(alpha2))]]\n", + "B = [math.sin(math.radians(alpha1)), math.sin(math.radians(alpha2))]\n", + "V = linalg.solve(A, B)\n", + "W = 1.0/V[0]\n", + "mu = V[1]\n", + "\n", + "#Results:\n", + "print \" Coefficient of friction, mu = %.3f.\"%(mu)\n", + "print \" Weight of the body, W = %d N.\"%(W)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Coefficient of friction, mu = 0.131.\n", + " Weight of the body, W = 4462 N.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.3 Page No : 272" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "W = 75.*1000 \t\t\t#W\n", + "v = 300. \t\t\t#mm/min\n", + "p = 6.\n", + "d0 = 40. \t \t\t#mm\n", + "mu = 0.1\n", + "\n", + "#Solution:\n", + "#Calculating the mean diameter of the screw\n", + "d = (d0-p/2)/1000 \t\t\t#m\n", + "#Calculating the helix angle\n", + "alpha = math.tan(p/(math.pi*d*1000)) \t\t\t#radians\n", + "#Calculating the force required at the circumference of the screw\n", + "phi = math.tan(mu) \t\t\t#Limiting angle of friction radians\n", + "P = W*math.tan(alpha+phi) \t\t\t#N\n", + "#Calculating the torque required to overcome the friction\n", + "T = P*d/2 \t\t\t#N-m\n", + "#Calculating the speed of the screw\n", + "N = v/p \t\t\t#rpm\n", + "#Calculating the angular speed\n", + "omega = round(2*math.pi*N/60,2) \t\t\t#rad/s\n", + "#Calculating the power of the motor\n", + "Power = T*omega/1000 \t\t\t#Power of the motor kW\n", + "\n", + "#Results:\n", + "print \" Power of the motor required = %.3f kW.\"%(Power)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Power of the motor required = 1.114 kW.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.4 Page No : 273" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "p = 12.\n", + "d = 40. \t\t\t#mm\n", + "mu = 0.16\n", + "W = 2500. \t\t\t#N\n", + "\n", + "#Solutiom:\n", + "#Work done in drawing the wagons together agianst a steady load of 2500 N:\n", + "#Calculating the helix angle\n", + "alpha = math.tan(p/(math.pi*d)) \t\t\t#radians\n", + "#Calculating the effort required at the circumference of the screw\n", + "phi = math.tan(mu) \t\t\t#Limiting angle of friction radians\n", + "P = W*math.tan(alpha+phi) \t\t\t#N\n", + "#Calculating the torque required to overcome friction between the screw and nut\n", + "T = P*d/(2*1000) \t\t\t#N-m\n", + "#Calculating the number of turns required\n", + "N = 240/(2*p)\n", + "#Calculating the work done\n", + "W1 = T*2*math.pi*N \t\t\t#Work done N-m\n", + "#Work done in drawing the wagons together when the load increases from 2500 N to 6000 N:\n", + "W2 = W1*(6000.-2500)/2500.0 \t\t\t#Work done N-m\n", + "\n", + "#Results:\n", + "print \" Work done in drawing the wagons together agianst a steady load of 2500 N = %.1f N-m.\"%(W1)\n", + "print \" Work done in drawing the wagons together when the load increases from 2500 N to 6000 N = %.1f N-m.\"%(W2)\n", + "\n", + "# note : answer in book is wrong. " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Work done in drawing the wagons together agianst a steady load of 2500 N = 826.2 N-m.\n", + " Work done in drawing the wagons together when the load increases from 2500 N to 6000 N = 1156.7 N-m.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.5 Page No : 274" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "D = 150./1000 \t\t\t#m\n", + "ps = 2.*10**6 \t\t\t#N/m**2\n", + "d0 = 50.\n", + "p = 6. \t\t\t#mm\n", + "mu = 0.12\n", + "\n", + "#Solution:\n", + "#Calculating the load on the valve\n", + "W = ps*math.pi/4*D**2 \t\t\t#N\n", + "#Calculating the mean diameter of the screw\n", + "d = (d0-p/2)/1000 \t\t\t#m\n", + "#Calculating the helix angle\n", + "alpha = math.tan(p/(math.pi*d*1000))\n", + "#Calculating the force required to turn the handle\n", + "phi = math.tan(mu) \t\t\t#Limiting angle of friction radians\n", + "P = W*math.tan(alpha+phi) \t\t\t#N\n", + "#Calculating the torque required to turn the handle\n", + "T = P*d/2 \t\t\t#N-m\n", + "\n", + "#Results:\n", + "print \" The torque required to turn the handle, T = %.1f N-m.\"%(T)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The torque required to turn the handle, T = 135.1 N-m.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.6 Page No : 274" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "dc = 22.5 #mm\n", + "p = 5. #mm\n", + "D = 50. #mm\n", + "R = D/2 #mm\n", + "l = 500. \t #mm\n", + "mu = 0.1\n", + "mu1 = 0.16\n", + "W = 10.*1000 \t\t\t#N\n", + "\n", + "#Solution:\n", + "#Calculating the mean diameter of the screw\n", + "d = dc+p/2 \t\t\t#mm\n", + "#Calculating the helix angle\n", + "alpha = p/(math.pi*d) \t\t\t#radians\n", + "#Calculating the force required at the circumference of the screw\n", + "phi = math.tan(mu) \t\t\t#Limiting angle of friction radians\n", + "P = W*math.tan(alpha+phi) \t\t\t#N\n", + "#Calculating the total torque required\n", + "T = P*d/2+mu1*W*R \t\t\t#N-mm\n", + "#Calculating the force required at the end of a spanner\n", + "P1 = T/l \t\t\t#N\n", + "\n", + "#Results:\n", + "print \" Force required at the end of a spanner, P1 = %.2f N.\"%(P1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Force required at the end of a spanner, P1 = 121.37 N.\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.7 Page No : 275" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "d = 50. #mm\n", + "p = 12.5 #mm\n", + "D = 60. #mm\n", + "R = D/2 \t\t\t#mm\n", + "W = 10.*1000 #N\n", + "P1 = 100. \t\t\t#N\n", + "mu = 0.15\n", + "mu1 = 0.18\n", + "\n", + "#Solution:\n", + "#Calculating the helix angle\n", + "alpha = math.tan(p/(math.pi*d)) \t\t\t#radians\n", + "#Calculating the math.tangential force required at the circumference of the screw\n", + "phi = math.tan(mu) \t\t\t#Limiting angle of friction radians\n", + "P = W*math.tan(alpha+phi) \t\t\t#N\n", + "#Calculating the total torque required to turn the hand wheel\n", + "T = P*d/2+mu1*W*R \t\t\t#N-mm\n", + "#Calculating the diameter of the hand wheel\n", + "D1 = T/(2*P1*1000)*2 \t\t\t#m\n", + "\n", + "#Results:\n", + "print \" Diameter of the hand wheel, D1 = %.3f m.\"%(D1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Diameter of the hand wheel, D1 = 1.128 m.\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.8 Page No : 276" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "d0 = 55. #mm\n", + "D2 = 60. #mm\n", + "R2 = D2/2 #mm\n", + "D1 = 90. #mm\n", + "R1 = D1/2 \t\t #mm\n", + "p = 10./1000 \t #m\n", + "W = 400. \t\t #N\n", + "mu = 0.15\n", + "v = 6. \t\t\t#Cutting speed m/min\n", + "\n", + "#Solution:\n", + "#Calculating the mean diameter of the screw\n", + "d = d0-p/2 \t\t\t#mm\n", + "#Calculating the helix angle\n", + "alpha = p/(math.pi*d) \t\t\t#radians\n", + "#Calculating the force required at the circumference of the screw\n", + "phi = math.tan(mu) \t\t\t#Limiting angle of friction radians\n", + "P = W*math.tan(alpha+phi) \t\t\t#N\n", + "#Calculating the mean radius of the flat surface\n", + "R = (R1+R2)/2 \t\t\t#mm\n", + "#Calculating the torque required\n", + "T = (P*d/2+mu1*W*R)/1000 \t\t\t#N-m\n", + "#Calculating the speed of the screw\n", + "N = v/p \t\t\t#rpm\n", + "#Calculating the angular speed\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the power required to operate the nut\n", + "Power = T*omega/1000 \t\t\t#Power required to operate the nut kW\n", + "\n", + "#Results:\n", + "print \" Power required to operate the nut = %.3f kW.\"%(Power)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Power required to operate the nut = 0.275 kW.\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.9 Page No : 277" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "d = 50./1000\n", + "l = 0.7 \t\t\t#m\n", + "p = 10. \t\t\t#mm\n", + "mu = 0.15\n", + "W = 20.*1000 \t\t\t#N\n", + "\n", + "#Solution:\n", + "#Calculating the helix angle\n", + "alpha = math.tan(p/(math.pi*d*1000)) \t\t\t#radians\n", + "#Force required to raise the load:\n", + "#Calculating the force required at the circumference of the screw\n", + "phi = math.tan(mu) \t\t\t#Limiting angle of friction radians\n", + "P1 = W*math.tan(alpha+phi) \t\t\t#N\n", + "#Calculating the force required at the end of the lever\n", + "P11 = P1*d/(2*l) \t\t\t#N\n", + "#Calculating the force required at the circumference of the screw\n", + "P2 = W*(phi-alpha) \t\t\t#N\n", + "#Foce rewuired to lower the load:\n", + "#Calculating the force required at the end of the lever\n", + "P21 = P2*d/(2*l) \t\t\t#N\n", + "\n", + "#Results:\n", + "print \" Force required at the end of the lever to raise the load, P1 = %d N.\"%(P11)\n", + "print \" Force required at the end of the lever to lower the load, P1 = %d N.\"%(P21)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Force required at the end of the lever to raise the load, P1 = 155 N.\n", + " Force required at the end of the lever to lower the load, P1 = 62 N.\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.10 Page No : 279" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "d = 50.\n", + "p = 12.5 \t\t\t#mm\n", + "mu = 0.13\n", + "W = 25.*1000 \t\t\t#N\n", + "\n", + "#Solution:\n", + "#Calculating the helix angle\n", + "alpha = round(math.tan(p/(math.pi*d)),2) \t\t\t#radians\n", + "#Calculating the force required on the screw to raise the load\n", + "phi = round(math.tan(mu),2) \t\t\t#Limiting angle of friction radians\n", + "P1 = W*(alpha+phi)/(1-(alpha*phi)) \t\t\t#N\n", + "#Calculating the torque required on the screw to raise the load\n", + "T1 = P1*d/2 \t\t\t#N-mm\n", + "#Calculating the force required on the screw to lower the load\n", + "P2 = W*math.tan(phi-alpha) \t\t\t#N\n", + "#Calculating the torque required to lower the load\n", + "T2 = P2*d/2 \t\t\t#N\n", + "#Calculating the ratio of the torques required\n", + "r = T1/T2 \t\t\t#Ratio of the torques required N-mm\n", + "#Calculating the efficiency of the machine\n", + "eta = math.tan(alpha)/math.tan(alpha+phi)*100 \t\t\t#%\n", + "\n", + "#Results:\n", + "print \" Torque required on the screw to raise the load, T1 = %d N-mm.\"%(T1)\n", + "print \" Ratio of the torque required to raise the load to the torque required to lower the load = %.1f.\"%(r)\n", + "print \" Efficiency of the machine, eta = %.1f %%.\"%(eta)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Torque required on the screw to raise the load, T1 = 132629 N-mm.\n", + " Ratio of the torque required to raise the load to the torque required to lower the load = 4.2.\n", + " Efficiency of the machine, eta = 37.6 %.\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.11 Page No : 280" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "p = 10. #mm\n", + "d = 50. #mm\n", + "D2 = 60. #mm\n", + "R2 = D2/2 #mm\n", + "D1 = 10. #mm\n", + "R1 = D1/2 \t\t#mm\n", + "W = 20.*1000 \t#N\n", + "mu = 0.08\n", + "mu1 = mu\n", + "\n", + "#Solution:\n", + "#Calculating the helix angle\n", + "alpha = round(math.tan(p/(math.pi*d)),4) \t\t\t#radians\n", + "#Calculating the force required at the circumference of the screw to lift the load\n", + "phi = round(math.tan(mu),2) \t\t\t#Limiting angle of friction radians\n", + "P = round(W*(alpha+phi)/(1 - (alpha*phi)),-1) #math.tan(alpha+phi) \t\t\t#N\n", + "#Calculating the torque required to overcome friction at the screw\n", + "T = P*d/(2*1000) \t\t\t#N-m\n", + "#Calculating the number of rotations made by the screw\n", + "N = 170/p\n", + "#When the load rotates with the screw:\n", + "#Calculating the work done in lifting the load\n", + "W1 = T*2*math.pi*N \t\t\t#Work done in lifting the load N-m\n", + "#Calculating the efficiency of the screw jack\n", + "eta1 = math.tan(alpha)/math.tan(alpha+phi)*100 \t\t\t#%\n", + "#When the load does not rotate with the screw:\n", + "#Calculating the mean radius of the bearing surface\n", + "R = (R1+R2)/2 \t\t\t#mm\n", + "#Calculating the torque required to overcome friction at the screw and the collar\n", + "T = (P*d/2+mu1*W*R)/1000 \t\t\t#N-m\n", + "#Calculating the work done by the torque in lifting the load\n", + "W2 = T*2*math.pi*N \t\t\t#Work done by the torque in lifting the load N-m\n", + "#Calculating the torque required to lift the load neglecting frition\n", + "T0 = (W*math.tan(alpha)*d/2)/1000 \t\t\t#N-m\n", + "#Calculating the efficiency of the screw jack\n", + "eta2 = T0/T*100 \t\t\t#%\n", + "\n", + "#Results:\n", + "print \" When the load rotates with the screw, work done in lifting the load = %.f N-m.\"%(W1)\n", + "print \" Efficiency of the screw jack, eta = %.1f %%.\"%(eta1)\n", + "print \" When the load does not rotate with the screw, work done in lifting the load = %d N-m.\"%(round(W2,-1))\n", + "print \" Efficiency of the screw jack, eta = %.1f %%.\"%(eta2)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " When the load rotates with the screw, work done in lifting the load = 7717 N-m.\n", + " Efficiency of the screw jack, eta = 44.1 %.\n", + " When the load does not rotate with the screw, work done in lifting the load = 10710 N-m.\n", + " Efficiency of the screw jack, eta = 31.8 %.\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.12 Page No : 282" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "# Variables:\n", + "W = 10.*1000 #N\n", + "P1 = 100. \t\t\t#N\n", + "p = 12. #mm\n", + "d = 50. \t\t\t#mm\n", + "mu = 0.15\n", + "\n", + "#Solution:\n", + "#Calculating the helix angle\n", + "alpha = math.tan(p/(math.pi*d)) \t\t\t#radians\n", + "#Calculating the effort required at the circumference of the screw to raise the load\n", + "phi = math.tan(mu) \t\t\t#Limiting angle of friction radians\n", + "P = W*math.tan(alpha+phi) \t\t\t#N\n", + "#Calculating the torque required to overcome friction\n", + "T = P*d/2 \t\t\t#N-mm\n", + "#Calculating the length of the lever\n", + "l = T/P1 \t\t\t#mm\n", + "#Calculating the mechanical advantage\n", + "MA = W/P1\n", + "#Calculating the efficiency of the screw jack\n", + "eta = math.tan(alpha)/math.tan(alpha+phi)*100 \t\t\t#%\n", + "\n", + "#Results:\n", + "print \" The length of the lever to be used, l = %.1f mm.\"%(l)\n", + "print \" Mechanical advantage obtained, M.A. = %d.\"%(MA)\n", + "if eta<50:\n", + " print \" The screw is a self locking screw.\";\n", + "else:\n", + " print \" The screw is not a self locking screw.\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The length of the lever to be used, l = 579.2 mm.\n", + " Mechanical advantage obtained, M.A. = 100.\n", + " The screw is a self locking screw.\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.13 Page No : 284" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "d = 22. #mm\n", + "p = 3. \t\t\t#mm\n", + "beta = 60./2 \t\t\t#degrees\n", + "W = 40.*1000 \t\t\t#N\n", + "mu = 0.15\n", + "\n", + "#Solution:\n", + "#Calculating the helix angle\n", + "alpha = round(math.tan(p/(math.pi*d)),4) \t\t\t#radians\n", + "#Calculating the virtual coefficient of friction\n", + "mu1 = round(mu/math.cos(math.radians(beta)),3)\n", + "#Calculating the force required at the circumference of the screw\n", + "P = W*((alpha+mu1)/(1 - alpha * mu1))\n", + "#Calculating the torque on one rod\n", + "T = P*d/(2.*1000) \t\t\t#N-m\n", + "#Calculating the torque required on the nut\n", + "T1 = 2*T \t\t\t#N-m\n", + "\n", + "\n", + "#Results:\n", + "print \" The torque required on the nut, T1 = %.2f N-m.\"%(T1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The torque required on the nut, T1 = 191.87 N-m.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.14 Page No : 284 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "d = 25. #mm\n", + "p = 5. #mm\n", + "R = 25. \t\t\t#mm\n", + "beta = 27.5 \t\t#degrees\n", + "mu = 0.1\n", + "mu2 = 0.16\n", + "l = 0.5 \t\t\t#m\n", + "W = 10.*1000 \t\t#N\n", + "\n", + "#Solution:\n", + "#Calculating the virtual coefficient of friction\n", + "mu1 = mu/math.cos(math.radians(beta))\n", + "#Calculating the helix angle\n", + "alpha = math.tan(p/(math.pi*d)) \t\t\t#radians\n", + "#Calculating the force on the screw\n", + "phi1 = math.tan(mu1) \t\t\t#Virtual limiting angle of frcition radians\n", + "P = W*math.tan(alpha+phi1) \t\t\t#N\n", + "#Calculating the total torque transmitted\n", + "T = (P*d/2+mu2*W*R)/1000 \t\t\t#N-m\n", + "#Calculating the force required at the end of a spanner\n", + "P1 = T/l \t\t\t#N\n", + "\n", + "#Results:\n", + "print \" Force required at the end of a spanner, P1 = %.1f N.\"%(P1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Force required at the end of a spanner, P1 = 124.7 N.\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.15 Page No : 286" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "d = 60.\n", + "r = d/2. \t\t\t#mm\n", + "W = 2000. \t\t\t#N\n", + "mu = 0.03\n", + "N = 1440. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the shaft\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the torque transmitted\n", + "T = mu*W*(r/1000) \t\t\t#N-m\n", + "#Calculating the power transmitted\n", + "P = T*omega \t\t\t#W\n", + "\n", + "#Results:\n", + "print \" The power transmitted, P = %.1f W.\"%(P)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The power transmitted, P = 271.4 W.\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.16 Page No : 288" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "D = 150./1000 #m\n", + "R = D/2 \t\t\t#m\n", + "N = 100. \t\t\t#rpm\n", + "W = 20.*1000 \t\t\t#N\n", + "mu = 0.05\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the shaft\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the total frictional torque for uniform pressure distribution\n", + "T = 2./3*mu*W*R \t\t\t#N-m\n", + "#Calculating the power lost in friction\n", + "P = T*omega \t\t\t#W\n", + "\n", + "#Results:\n", + "print \" Power lost in friction, P = %.1f W.\"%(P)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Power lost in friction, P = 523.6 W.\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.17 Page No : 292" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "W = 20.*1000 \t\t\t#N\n", + "alpha = 120./2 \t\t\t#degrees\n", + "Pn = 0.3 \t\t\t#N/mm**2\n", + "N = 200. \t \t\t#rpm\n", + "mu = 0.1\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the shaft\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the inner radius of the bearing surface\n", + "r2 = math.sqrt(W/(3*math.pi*Pn)) \t\t\t#mm\n", + "#Calculating the outer radius of the bearing surface\n", + "r1 = 2*r2 \t\t\t#mm\n", + "#Calculating the total frictional torque assuming uniform pressure\n", + "T = 2./3*mu*W*(1/math.sin(math.radians(alpha)))*(r1**3-r2**3)/(r1**2-r2**2)/1000.0 \t\t\t#N-m\n", + "#Calculating the power absorbed in friction\n", + "P = T*omega/1000.0 \t\t\t#kW\n", + "\n", + "#Results:\n", + "print \"External Diameters r1= %.f mm\"%r1\n", + "print \"Internal Diameters r2= %.f mm\"%r2\n", + "print \" Power absorbed in friction, P = %.3f kW.\"%(P)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "External Diameters r1= 168 mm\n", + "Internal Diameters r2= 84 mm\n", + " Power absorbed in friction, P = 6.328 kW.\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.18 Page No : 292" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "D = 200./1000\n", + "R = D/2 \t\t\t#m\n", + "W = 30.*1000 \t\t\t#N\n", + "alpha = 120./2 \t\t\t#degrees\n", + "mu = 0.025\n", + "N = 140. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the shaft\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Power lost in friction assuming uniform pressure:\n", + "#Calculating the total frictional torque\n", + "T = 2./3*mu*W*R*(1/math.sin(math.radians(alpha))) \t\t\t#N-m\n", + "#Calculating the power lost in friction\n", + "P1 = T*omega \t\t\t#Power lost in friction W\n", + "#Power lost in friction assuming uniform wear:\n", + "#Calculating the total frictional torque\n", + "T = 1./2*mu*W*R*(1./math.sin(math.radians(alpha))) \t\t\t#N-m\n", + "#Calculating the power lost in friction\n", + "P2 = T*omega \t\t\t#Power lost in friction W\n", + "\n", + "#Resluts:\n", + "print \" Power lost in friction assuming uniform pressure, P = %d W.\"%(P1)\n", + "print \" Power lost in friction assuming uniform wear, P = %.1f W.\"%(P2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Power lost in friction assuming uniform pressure, P = 846 W.\n", + " Power lost in friction assuming uniform wear, P = 634.8 W.\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.19 Page No : 295" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "n = 6.\n", + "d1 = 600. #mm\n", + "r1 = d1/2 #mm\n", + "d2 = 300. #mm\n", + "r2 = d2/2 \t\t #mm\n", + "W = 100.*1000 \t #N\n", + "mu = 0.12\n", + "N = 90. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the engine\n", + "omega = 2*math.pi*N/60.0 \t\t\t#rad/s\n", + "#Power absorbed in friction assuming uniform pressure:\n", + "#Calculating the total frictional torque transmitted\n", + "T = 2./3*mu*W*(r1**3-r2**3)/(r1**2-r2**2)/1000.0 \t\t\t#N-m\n", + "#Calculating the power absorbed in friction\n", + "P1 = T*omega/1000.0 \t\t\t#Power absorbed in friction assuming uniform pressure kW\n", + "#Power absorbed in friction assuming uniform wear:\n", + "#Calculating the total frictional torque transmitted\n", + "T = 1./2*mu*W*(r1+r2)/1000.0 \t\t\t#N-m\n", + "#Calculating the power absorbed in friction\n", + "P2 = T*omega/1000.0 \t\t\t#Power absorbed in friction assuming uniform wear kW\n", + "\n", + "#Results:\n", + "print \" Power absorbed in friction assuming uniform pressure, P = %.1f kW.\"%(P1)\n", + "print \" Power absorbed in friction assuming uniform wear, P = %.2f kW.\"%(P2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Power absorbed in friction assuming uniform pressure, P = 26.4 kW.\n", + " Power absorbed in friction assuming uniform wear, P = 25.45 kW.\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.20 Page No : 296" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "d1 = 400. #mm\n", + "r1 = d1/2 #mm\n", + "d2 = 250. #mm\n", + "r2 = d2/2 \t\t\t#mm\n", + "p = 0.35 \t\t\t#N/mm**2\n", + "mu = 0.05\n", + "N = 105. \t\t\t#rpm\n", + "W = 150.*1000 \t\t\t#N\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the shaft\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the total frictional torque transmitted for uniform pressure\n", + "T = 2./3*mu*W*(r1**3-r2**3)/(r1**2-r2**2)/1000 \t\t\t#N-m\n", + "#Calculating the power absorbed\n", + "P = T*omega/1000 \t\t\t#kW\n", + "#Calculating the number of collars required\n", + "n = W/(p*math.pi*(r1**2-r2**2))\n", + "\n", + "#Results:\n", + "print \" Power absorbed, P = %.2f kW.\"%(P)\n", + "print \" Number of collars required, n = %d.\"%(n+1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Power absorbed, P = 13.64 kW.\n", + " Number of collars required, n = 6.\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.21 Page No : 296" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "d2 = 300./1000 #mm\n", + "r2 = d2/2 \t\t\t#m\n", + "W = 200.*1000 \t\t#N\n", + "N = 75. \t\t\t#rpm\n", + "mu = 0.05\n", + "p = 0.3 \t\t\t#N/mm**2\n", + "P = 16.*1000 \t\t#W\n", + "\n", + "#Solution:\n", + "#Calculating the angular velocity of the shaft\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the total frictional torque transmitted\n", + "T = P/omega \t\t\t#N-m\n", + "#Calculating the external diameter of the collar\n", + "#We have T = 2/3*mu*W*(r1**3-r2**3)/(r1**2-r2**2) or (2*mu*W)*r1**2-(3*T-2*mu*W*r2)*r1+(2*mu*W*r2**2-3*T*r2) = 0\n", + "A = 2*mu*W\n", + "B = -(3*T-2*mu*W*r2)\n", + "C = 2*mu*W*r2**2-3*T*r2\n", + "r1 = (-B+math.sqrt(B**2-4*A*C))/(2*A)*1000 \t\t\t#mm\n", + "d1 = 2*r1 \t\t\t#mm\n", + "#Calculating the number of collars\n", + "n = W/(p*math.pi*(r1**2-(r2*1000)**2))\n", + "\n", + "#Results:\n", + "print \" External diameter of the collar, d1 = %d mm.\"%(d1)\n", + "print \" Number of collars, n = %d.\"%(n+1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " External diameter of the collar, d1 = 498 mm.\n", + " Number of collars, n = 6.\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.22 Page No : 302" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "W = 4.*1000 \t\t#N\n", + "r2 = 50.\n", + "r1 = 100. \t\t\t#mm\n", + "\n", + "#Solution:\n", + "#Calculating the maximum pressure\n", + "pmax = W/(2*math.pi*r2*(r1-r2)) \t\t\t#N/mm**2\n", + "#Calculating the minimum pressure\n", + "pmin = W/(2*math.pi*r1*(r1-r2)) \t\t\t#N/mm**2\n", + "#Calculating the average pressure\n", + "pav = W/(math.pi*(r1**2-r2**2)) \t\t\t#N/mm**2\n", + "\n", + "#Results:\n", + "print \" Maximum pressure, pmax = %.4f N/mm**2.\"%(pmax)\n", + "print \" Minimum pressure, pmin = %.4f N/mm**2.\"%(pmin)\n", + "print \" Average pressure, pav = %.2f N/mm**2.\"%(pav)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Maximum pressure, pmax = 0.2546 N/mm**2.\n", + " Minimum pressure, pmin = 0.1273 N/mm**2.\n", + " Average pressure, pav = 0.17 N/mm**2.\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.23 Page No : 303" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "d1 = 300. #mm\n", + "r1 = d1/2 #mm\n", + "d2 = 200. #mm\n", + "r2 = d2/2 \t\t\t#mm\n", + "p = 0.1 \t\t\t#N/mm**2\n", + "mu = 0.3\n", + "N = 2500. \t\t\t#rpm\n", + "n = 2.\n", + "\n", + "#Solution:\n", + "#Calculating the radial speed of the clutch\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the intensity of pressure\n", + "C = p*r2 \t\t\t#N/mm\n", + "#Calculating the axial thrust\n", + "W = 2*math.pi*C*(r1-r2) \t\t\t#N\n", + "#Calculating the mean radius of the friction surfaces for uniform wear\n", + "R = (r1+r2)/(2*1000) \t\t\t#m\n", + "#Calculating the torque transmitted\n", + "T = n*mu*W*R \t\t\t#N-m\n", + "#Calculating the power transmitted by a clutch\n", + "P = T*omega/1000 \t\t\t#kW\n", + "\n", + "#Results:\n", + "print \" Power transmitted by a clutch, P = %.3f kW.\"%(P)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Power transmitted by a clutch, P = 61.685 kW.\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.24 Page No : 303" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "n = 2.\n", + "mu = 0.255\n", + "P = 25.*1000 \t\t#W\n", + "N = 3000. \t\t\t#rpm\n", + "r = 1.25 \t\t\t#Ratio of radii r1/r2\n", + "p = 0.1 \t\t\t#N/mm**2\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the clutch\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the torque transmitted\n", + "T = P/omega*1000 \t\t\t#N-mm\n", + "#Calculating the inner radius\n", + "r2 = (T/(n*mu*2*math.pi*0.1*(1.25-1)*(1.25+1)/2))**(1./3) \t\t\t#mm\n", + "#Calculating the outer radius\n", + "r1 = r*r2 \t\t\t#mm\n", + "#Calculating the axial thrust to be provided by springs\n", + "C = 0.1*r2 \t\t\t#Intensity of pressure N/mm\n", + "W = 2*math.pi*C*(r1-r2) \t\t\t#N\n", + "\n", + "#Results:\n", + "print \" Outer radius of the frictional surface, r1 = %.f mm.\"%(r1)\n", + "print \" Inner radius of the frictional surface, r2 = %.f mm.\"%(r2)\n", + "print \" Axial thrust to be provided by springs, W = %.f N.\"%(W)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Outer radius of the frictional surface, r1 = 120 mm.\n", + " Inner radius of the frictional surface, r2 = 96 mm.\n", + " Axial thrust to be provided by springs, W = 1446 N.\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.25 Page No : 304" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "import math \n", + "\n", + "# Variables:\n", + "P = 7.5*1000 \t\t#W\n", + "N = 900. \t\t\t#rpm\n", + "p = 0.07 \t\t\t#N/mm**2\n", + "mu = 0.25\n", + "n = 2.\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the clutch\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the torque transmitted\n", + "T = P/omega*1000 \t\t\t#N-mm\n", + "#Calculating the mean radius of the friction lining\n", + "R = (T/(math.pi/2*n*mu*p))**(1./3) \t\t\t#mm\n", + "#Calculating the face width of the friction lining\n", + "w = R/4 \t\t\t#mm\n", + "#Calculating the outer and inner radii of the clutch plate\n", + "#We have w = r1-r2 or r1-r2 = w .....(i) \n", + "#Also R = (r1+r2)/2 or r1+r2 = 2*R .....(ii) \n", + "A = [[1, -1],[ 1, 1]]\n", + "B = [w,2*R]\n", + "V = linalg.solve(A,B)\n", + "r1 = V[0]\n", + "r2 = V[1]\n", + "\n", + "#Results:\n", + "print \" Mean radius of the friction lining, R = %d mm.\"%(R)\n", + "print \" Face width of the friction lining, w = %.2f mm.\"%(w)\n", + "print \" Outer radius of the clutch plate, r1 = %.3f mm.\"%(r1)\n", + "print \" Inner radius of the clutch plate, r2 = %.3f mm.\"%(r2)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Mean radius of the friction lining, R = 113 mm.\n", + " Face width of the friction lining, w = 28.28 mm.\n", + " Outer radius of the clutch plate, r1 = 127.258 mm.\n", + " Inner radius of the clutch plate, r2 = 98.979 mm.\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.26 Page No : 305" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "P = 100. \t\t\t#kW\n", + "N = 2400. \t\t\t#rpm\n", + "T = 500.*1000 \t\t#N-mm\n", + "p = 0.07 \t\t\t#N/mm**2\n", + "mu = 0.3\n", + "Ns = 8. \t\t\t#Number of springs\n", + "k = 40. \t\t\t#Stiffness N/mm\n", + "n = 2.\n", + "\n", + "#Solution:\n", + "#Calculating the inner radius of the friction plate\n", + "r2 = round((T/(n*mu*2*math.pi*p*(1.25-1)*(1.25+1)/2))**(1./3),-1) \t\t\t#mm\n", + "#Calculating the outer radius of the friction plate\n", + "r1 = 1.25*r2 \t\t\t#mm\n", + "#Calculating the total stiffness of the springs\n", + "s = k*Ns \t\t\t#N/mm\n", + "#Calculating the intensity of pressure\n", + "C = p*r2 \t\t\t#N/mm\n", + "#Calculating the axial force required to engage the clutch\n", + "W = 2*math.pi*C*(r1-r2) \t\t\t#N\n", + "#Calculating the initial compression in the springs\n", + "IC = W/s \t\t\t#Initial compression in the springs mm\n", + "\n", + "#Results:\n", + "print \" Outer radius of the friction plate, r1 = %.1f mm.\"%(r1)\n", + "print \" Inner radius of the friction plate, r2 = %.f mm.\"%(r2)\n", + "print \" Initial compression in the springs = %.1f mm.\"%(IC)\n", + "\n", + "# book answer is wrong." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Outer radius of the friction plate, r1 = 237.5 mm.\n", + " Inner radius of the friction plate, r2 = 190 mm.\n", + " Initial compression in the springs = 12.4 mm.\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.27 Page No : 306" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "from numpy import linalg\n", + "\n", + "# Variables:\n", + "d1 = 220. #mm\n", + "r1 = d1/2 #mm\n", + "d2 = 160. #mm\n", + "r2 = d2/2 \t\t\t#mm\n", + "W = 570. \t\t\t#N\n", + "m1 = 800. #kg\n", + "m2 = 1300. \t\t\t#kg\n", + "k1 = 200./1000 #m\n", + "k2 = 180./1000 \t\t#m\n", + "mu = 0.35\n", + "N1 = 1250. \t\t\t #rpm\n", + "n = 2.\n", + "\n", + "#Solution:\n", + "#Calculating the initial angular speed of the motor shaft\n", + "omega1 = 2*math.pi*N1/60 \t\t\t#rad/s\n", + "#Calculating the moment of inertia for the motor armature and shaft\n", + "I1 = m1*k1**2 \t\t\t#kg-m**2\n", + "#Calculating the moment of inertia for the rotor\n", + "I2 = m2*k2**2 \t\t\t#kg-m**2\n", + "#Calculating the final speed of the motor and rotor\n", + "omega2 = 0\n", + "omega3 = (I1*omega1+I2*omega2)/(I1+I2) \t\t\t#rad/s\n", + "#Calculating the mean radius of the friction plate\n", + "R = (r1+r2)/(2*1000) \t\t\t#m\n", + "#Calculating the frictional torque\n", + "T = n*mu*W*R \t\t\t#N-m\n", + "#Calculating the angular acceleration of the rotor\n", + "alpha2 = T/I2 \t\t\t#rad/s**2\n", + "#Calculating the time to reach the speed of omega3\n", + "omegaF = omega3\n", + "omegaI = omega2\n", + "t = (omegaF-omegaI)/alpha2 \t\t\t#seconds\n", + "#Calculating the angular kinetic energy before impact\n", + "E1 = 1./2*I1*omega1**2+1./2*I2*omega2**2 \t\t\t#N-m\n", + "#Calculating the angular kinetic energy after impact\n", + "E2 = 1./2*(I1+I2)*omega3**2 \t\t\t#N-m\n", + "#Calculating the kinetic energy lost during the period of slipping\n", + "E = round(E1-E2,-3) \t\t\t#N-m\n", + "#Calculating the torque on armature shaft\n", + "T1 = -60-T \t\t\t#N-m\n", + "#Calculating the torque on rotor shaft\n", + "T2 = T \t\t\t#N-m\n", + "#Calculating the time of slipping assuming constant resisting torque:\n", + "#Considering armature shaft omega3 = omega1+alpha1*t1 or omega3-(T1/I1)*t1 = omega1 .....(i)\n", + "#Considering rotor shaft omega3 = alpha2*t1 or omega3-(T2/I2)*t1 = 0 .....(ii)\n", + "A = [[1, -T1/I1],[ 1, -T2/I2]]\n", + "B = [omega1, 0]\n", + "V = linalg.solve(A,B)\n", + "t11 = V[1] \t\t\t#Time of slipping assuming constant resisting torque seconds\n", + "#Calculating the time of slipping assuming constant driving torque:\n", + "#Calculating the torque on armature shaft\n", + "T1 = 60-T \t\t\t#N-m\n", + "t12 = (omega2-omega1)/(T1/I1-T2/I2) \t\t\t#Time of slipping assuming constant driving torque seconds\n", + "\n", + "#Results:\n", + "print \" Final speed of the motor and rotor, omega3 = %.2f rad/s.\"%(omega3)\n", + "print \" Time to reach the speed of %.2f rad/s, t = %.1f s.\"%(omega3,t)\n", + "print \" Kinetic energy lost during the period of slipping = %d N-m.\"%(E)\n", + "print \" Time of slipping assuming constant resisting torque, t1 = %.1f s.\"%(t11)\n", + "print \" Time of slipping assuming constant driving torque, t1 = %d s.\"%(t12)\n", + "\n", + "# rounding off error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Final speed of the motor and rotor, omega3 = 56.51 rad/s.\n", + " Time to reach the speed of 56.51 rad/s, t = 62.8 s.\n", + " Kinetic energy lost during the period of slipping = 156000 N-m.\n", + " Time of slipping assuming constant resisting torque, t1 = 33.1 s.\n", + " Time of slipping assuming constant driving torque, t1 = 624 s.\n" + ] + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.28 Page No : 308" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "n = 4.\n", + "mu = 0.3\n", + "p = 0.127 \t\t\t#N/mm**2\n", + "N = 500. \t\t\t#rpm\n", + "r1 = 125. #mm\n", + "r2 = 75. \t\t\t#mm\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the clutch\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the maximum intensity of pressure\n", + "C = p*r2 \t\t\t#N/mm\n", + "#Calculating the axial force required to engage the clutch\n", + "W = 2*math.pi*C*(r1-r2) \t\t\t#N\n", + "#Calculating the mean radius of the friction surfaces\n", + "R = (r1+r2)/(2*1000) \t\t\t#m\n", + "#Calculating the torque transmitted\n", + "T = n*mu*W*R \t\t\t#N-m\n", + "#Calculating the power transmitted\n", + "P = T*omega/1000 \t\t\t#kW\n", + "\n", + "#Results:\n", + "print \" Power transmitted, P = %.1f kW.\"%(P)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Power transmitted, P = 18.8 kW.\n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.29 Page No : 308" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "n1 = 3.\n", + "n2 = 2.\n", + "mu = 0.3\n", + "d1 = 240. #mm\n", + "r1 = d1/2 #mm\n", + "d2 = 120. #mm\n", + "r2 = d2/2 \t\t\t#mm\n", + "P = 25.*1000 \t\t#W\n", + "N = 1575. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the shaft\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the torque transmitted\n", + "T = P/omega \t\t\t#N-m\n", + "#Calculating the number of pairs of friction surfaces\n", + "n = n1+n2-1\n", + "#Calculating the mean radius of friction surfaces for uniform wear\n", + "R = (r1+r2)/(2*1000) \t\t\t#m\n", + "#Calculating the axial force on each friction surface\n", + "W = T/(n*mu*R) \t\t\t#N\n", + "#Calculating the maximum axial intensity of pressure\n", + "p = W/(2*math.pi*r2*(r1-r2)) \t\t\t#N/mm**2\n", + "\n", + "#Results:\n", + "print \" Maximum axial intensity of pressure, p = %.3f N/mm**2.\"%(p)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Maximum axial intensity of pressure, p = 0.062 N/mm**2.\n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.30 Page No : 309" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "n1 = 3.\n", + "n2 = 2.\n", + "n = 4.\n", + "mu = 0.3\n", + "d1 = 240. #mm\n", + "r1 = d1/2 #mm\n", + "d2 = 120. #mm\n", + "r2 = d2/2 \t\t\t#mm\n", + "P = 25.*1000 \t\t#W\n", + "N = 1575. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the shaft\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the torque transmitted\n", + "T = P/omega \t\t\t#N-m\n", + "#Calculating the mean radius of the contact surface for uniform pressure\n", + "R = 2./3*(r1**3-r2**3)/(r1**2-r2**2)/1000 \t\t\t#m\n", + "#Calculating the total spring load\n", + "W1 = T/(n*mu*R) \t\t\t#N\n", + "#Calculating the maximum power transmitted:\n", + "\n", + "# Variables:\n", + "ns = 6. \t\t\t#Number of springs\n", + "c = 8. \t\t\t#Contact surfaces of the spring\n", + "w = 1.25 \t\t\t#Wear on each contact surface mm\n", + "k = 13.*1000 \t\t\t#Stiffness of each spring N/m\n", + "#Calculating the total wear\n", + "Tw = c*w/1000 \t\t\t#Total wear m\n", + "#Calculating the reduction in spring force\n", + "Rs = Tw*k*ns \t\t\t#N\n", + "#Calculating the new axial load\n", + "W2 = W1-Rs \t\t\t#N\n", + "#Calculating the mean radius of the contact surfaces for uniform wear\n", + "R = (r1+r2)/(2*1000) \t\t\t#m\n", + "#Calculating the torque transmitted\n", + "T = n*mu*W2*R \t\t\t#N-m\n", + "#Calculating the maximum power transmitted\n", + "P = T*omega/1000 \t\t\t#kw\n", + "\n", + "#Results:\n", + "print \" Total spring load, W = %d N.\"%(W1)\n", + "print \" Maximum power that can be transmitted, P = %.2f kW.\"%(P)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Total spring load, W = 1353 N.\n", + " Maximum power that can be transmitted, P = 10.21 kW.\n" + ] + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.31 Page No : 314" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "import math \n", + "\n", + "# Variables:\n", + "P = 90.*1000 \t\t#W\n", + "N = 1500. \t\t\t#rpm\n", + "alpha = 20. \t\t#degrees\n", + "mu = 0.2\n", + "D = 375.\n", + "R = D/2. \t\t\t#mm\n", + "pn = 0.25 \t\t\t#N/mm**2\n", + "\n", + "#SOlution:\n", + "#Calculating the angular speed of the clutch\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the torque transmitted\n", + "T = P*1000/156 \t\t\t#N-mm\n", + "#Calculating the width of the bearing surface\n", + "b = T/(2*math.pi*mu*pn*R**2) \t\t\t#mm\n", + "#Calculating the external and internal radii of the bearing surface\n", + "#We know that r1+r2 = 2*R and r1-r2 = b*math.sin(math.radians(alpha)\n", + "A = [[1, 1],[1, -1]]\n", + "B = [2*R, b*math.sin(math.radians(alpha))]\n", + "V = linalg.solve(A,B)\n", + "r1 = V[0] \t\t\t#mm\n", + "r2 = V[1] \t\t\t#mm\n", + "#Calculating the intensity of pressure\n", + "C = round(pn*r2,1) \t\t\t#N/mm\n", + "#Calculating the axial load required\n", + "W = 2*math.pi*C*(r1-r2) \t\t\t#N\n", + "\n", + "#Results:\n", + "print \" Width of the bearing surface, b = %.1f mm.\"%(b)\n", + "print \" External radius of the bearing surface, r1 = %.1f mm.\"%(r1)\n", + "print \" Internal radius of the bearing surface, r2 = %.1f mm.\"%(r2)\n", + "print \" Axial load required, W = %.f N.\"%(W)\n", + "\n", + "# rounding off error. Please check." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Width of the bearing surface, b = 52.2 mm.\n", + " External radius of the bearing surface, r1 = 196.4 mm.\n", + " Internal radius of the bearing surface, r2 = 178.6 mm.\n", + " Axial load required, W = 5006 N.\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.32 Page No : 315" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "P = 45.*1000 \t\t#W\n", + "N = 1000. \t\t\t#rpm\n", + "alpha = 12.5 \t\t#degrees\n", + "D = 500./1000\n", + "R = D/2 \t\t\t#m\n", + "mu = 0.2\n", + "pn = 0.1 \t\t\t#N/mm**2\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the shaft\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the torque developed by the clutch\n", + "T = P/omega \t\t\t#N-m\n", + "#Calculating the normal load acting on the friction surface\n", + "Wn = T/(mu*R) \t\t\t#N\n", + "#Calculating the axial spring force necessary to engage the clutch\n", + "We = Wn*(math.sin(math.radians(alpha))+mu*math.cos(math.radians(alpha))) \t\t\t#N\n", + "#Calculating the face width required\n", + "b = Wn/(pn*2*math.pi*R*1000) \t\t\t#mm\n", + "\n", + "#Results:\n", + "print \" Axial force necessary to engage the clutch, We = %d N.\"%(round(We,-1))\n", + "print \" Face width required, b = %.1f mm.\"%(b)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Axial force necessary to engage the clutch, We = 3540 N.\n", + " Face width required, b = 54.7 mm.\n" + ] + } + ], + "prompt_number": 45 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.33 Page No : 315" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "import math \n", + "\n", + "# Variables:\n", + "alpha = 30./2 \t\t#degrees\n", + "pn = 0.35 \t\t\t#N/mm**2\n", + "P = 22.5*1000 \t\t#W\n", + "N = 2000. \t\t\t#rpm\n", + "mu = 0.15\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the clutch\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the torque transmitted by the clutch\n", + "T = P/omega*1000 \t\t\t#N-mm\n", + "#Calculating the mean radius of the contact surface\n", + "R = (T/(2*math.pi*mu*pn/3))**(1./3) \t\t\t#mm\n", + "#Calculating the face width of the contact surface\n", + "b = R/3\n", + "#Calculating the outer and inner radii of the contact surface\n", + "#Refer Fig. 10.27\n", + "#We have r1-r2 = b*math.sin(math.radians(alpha) and r1+r2 = 2*R\n", + "A = [[1, -1],[ 1, 1]]\n", + "B = [b*math.sin(math.radians(alpha)), 2*R]\n", + "V = linalg.solve(A, B)\n", + "r1 = V[0] \t\t\t#mm\n", + "r2 = V[1] \t\t\t#mm\n", + "\n", + "#Results:\n", + "print \" Mean radius of the contact surface, R = %d mm.\"%(R)\n", + "print \" Outer radius of the contact surface, r1 = %.2f mm.\"%(r1)\n", + "print \" Inner radius of the contact surface, r2 = %.2f mm.\"%(r2)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Mean radius of the contact surface, R = 99 mm.\n", + " Outer radius of the contact surface, r1 = 103.51 mm.\n", + " Inner radius of the contact surface, r2 = 94.95 mm.\n" + ] + } + ], + "prompt_number": 48 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.34 Page No : 316" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "D = 75./1000 #mm\n", + "R = D/2 \t\t\t#m\n", + "alpha = 15 \t\t\t#degrees\n", + "mu = 0.3\n", + "W = 180. \t\t\t#N\n", + "NF = 1000. \t\t\t#rpm\n", + "m = 13.5 \t\t\t#kg\n", + "k = 150./1000 \t\t#m\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the flywheel\n", + "omegaF = 2*math.pi*NF/60 \t\t\t#rad/s\n", + "#Calculating the torque required to produce slipping\n", + "T = round(mu*W*R*(1/math.sin(math.radians(alpha))),1) \t\t\t#N-m\n", + "#Calculating the mass moment of inertia of the flywheel\n", + "IF = m*k**2 \t\t\t#kg-m**2\n", + "#Calculating the angular acceleration of the flywheel\n", + "alphaF = T/IF \t\t\t#rad/s**2\n", + "#Calculating the time required for the flywheel to attain full speed\n", + "tF = round(omegaF/alphaF,1) \t\t\t#seconds\n", + "#Calculating the angle turned through by the motor and flywheel in time tF\n", + "theta = 1./2*omegaF*tF \t\t\t#rad\n", + "#Calculating the energy lost in slipping of the clutch\n", + "E = T*theta \t\t\t#Energy lost in slipping of the clutch N-m\n", + "\n", + "#Results:\n", + "print \" Torque required to produce slipping, T = %.1f N-m.\"%(T)\n", + "print \" Time required for the flywheel to attain full speed, tF = %.1f s.\"%(tF)\n", + "print \" Energy lost in slipping of the clutch = %d N-m.\"%(E)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Torque required to produce slipping, T = 7.8 N-m.\n", + " Time required for the flywheel to attain full speed, tF = 4.1 s.\n", + " Energy lost in slipping of the clutch = 1674 N-m.\n" + ] + } + ], + "prompt_number": 55 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.35 Page No : 319" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "P = 15.*1000 \t\t\t#W\n", + "N = 900. \t\t\t #rpm\n", + "n = 4.\n", + "mu = 0.25\n", + "R = 150./1000\n", + "r = 120./1000 \t\t\t#m\n", + "theta = 60. \t\t\t#degrees\n", + "p = 0.1 \t\t\t #N/mm**2\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the clutch\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the speed at which the engagement begins\n", + "omega1 = 3./4*omega \t\t\t#rad/s\n", + "#Calculating the torque transmitted at the running speed\n", + "T = P/omega \t\t\t#N-m\n", + "#Calculating the mass of the shoes\n", + "m = T/(n*mu*(omega**2*r-omega1**2*r)*R) \t\t\t#kg\n", + "#Calculating the contact length of shoes\n", + "l = (theta*math.pi/180)*R*1000 \t\t\t#mm\n", + "#Calculating the centrifugal force acting on each shoe\n", + "Pc = m*omega**2*r \t\t\t#N\n", + "#Calculating the inward force on each shoe exerted by the spring\n", + "Ps = m*omega1**2*r \t\t\t#N\n", + "#Calculating the width of the shoes\n", + "b = (Pc-Ps)/(l*p) \t\t\t#mm\n", + "\n", + "#Results:\n", + "print \" Mass of the shoes, m = %.2f kg.\"%(m)\n", + "print \" Width of the shoes, b = %.1f mm.\"%(b)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Mass of the shoes, m = 2.28 kg.\n", + " Width of the shoes, b = 67.5 mm.\n" + ] + } + ], + "prompt_number": 56 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.36 Page No : 320" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "n = 4.\n", + "mu = 0.3\n", + "c = 5.\n", + "r = 160. \t\t\t#mm\n", + "S = 500. \t\t\t#N\n", + "D = 400./1000\n", + "R = D/2 \t\t\t#m\n", + "m = 8. \t\t\t #kg\n", + "s = 50. \t\t\t#N/mm\n", + "N = 500. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the clutch\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the operating radius\n", + "r1 = (r+c)/1000 \t\t\t#m\n", + "#Calculating the centrifugal force on each shoe\n", + "Pc = m*omega**2*r1 \t\t\t#N\n", + "#Calculating the inward force exerted by the spring\n", + "Ps = S+c*s \t\t\t#N\n", + "#Calculating the frictional force acting math.tangentially on each shoe\n", + "F = mu*(Pc-Ps) \t\t\t#N\n", + "#Calculating the total frictional torque transmitted by the clutch\n", + "T = n*F*R \t\t\t#N-m\n", + "#Calculating the power transmitted\n", + "P = T*omega/1000 \t\t\t#kW\n", + "\n", + "#Results:\n", + "print \" Power transmitted, P = %.1f kW.\"%(P)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Power transmitted, P = 36.1 kW.\n" + ] + } + ], + "prompt_number": 45 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Theory_Of_Machines/ch11.ipynb b/Theory_Of_Machines/ch11.ipynb new file mode 100755 index 00000000..dab30d35 --- /dev/null +++ b/Theory_Of_Machines/ch11.ipynb @@ -0,0 +1,1499 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:090249f83a260ba2b6a1983f6426576be55ba0e881a9fae7d9c5cbc674e22e36" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 11 : Belt, Rope and Chain Drives" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.1 Page No : 333" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "N1 = 150. \t\t\t#rpm\n", + "d1 = 750. #mm\n", + "d2 = 450. #mm\n", + "d3 = 900. #mm\n", + "d4 = 150. \t\t\t#mm\n", + "\n", + "#Solution:\n", + "#Calculating the speed of the dynamo shaft when there is no slip\n", + "N41 = N1*(d1*d3)/(d2*d4) \t\t\t#rpm\n", + "#Calculating the speed of the dynamo shaft whne there is a slip of 2% at each drive\n", + "s1 = 2.\n", + "s2 = 2. \t\t\t#%\n", + "N42 = N1*(d1*d3)/(d2*d4)*(1-s1/100)*(1-s2/100) \t\t\t#rpm\n", + "\n", + "#Results:\n", + "print \" Speed of the dynamo shaft when there is no slip, N4 = %d rpm.\"%(N41)\n", + "print \" Speed of the dynamo shaft when there is a slip of 2 %% at each drive, N4 = %d rpm.\"%(N42)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Speed of the dynamo shaft when there is no slip, N4 = 1500 rpm.\n", + " Speed of the dynamo shaft when there is a slip of 2 % at each drive, N4 = 1440 rpm.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.2 Page No : 334" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "d1 = 1.\n", + "d2 = 2.25 \t\t\t#m\n", + "N1 = 200. \t\t\t#rpm\n", + "sigma1 = 1.4*10**6\n", + "sigma2 = 0.5*10**6\n", + "E = 100.*10**6 \t\t\t#N/m**2\n", + "\n", + "#Solution:\n", + "#Calculating the speed of the driven pulley\n", + "N21 = round(N1*(d1/d2),1) \t\t\t#rpm\n", + "\n", + "#Calculating the speed of the shaft considering creep\n", + "N22 = N1*(d1/d2)*(E+math.sqrt(sigma2))/(E+math.sqrt(sigma1))\t\t\t#rpm\n", + "#Calculating the speed lost by the driven pulley due to creep\n", + "Nl = N21-N22 \t\t\t#Speed lost by the driven pulley due to creep rpm\n", + "\n", + "#Results:\n", + "print \" Speed lost by the driven pulley due to creep = %.3f rpm.\"%(Nl)\n", + "\n", + "# note : answer is accurate. please check using calculator." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Speed lost by the driven pulley due to creep = 0.012 rpm.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.3 Page No : 337" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "from scipy.optimize import fsolve \n", + "import math \n", + "\n", + "# Variables:\n", + "N1 = 160. #rpm\n", + "N3 = N1 #rpm\n", + "N5 = N3 #rpm\n", + "N2 = 60. #rpm\n", + "N4 = 80. #rpm\n", + "N6 = 100. \t\t\t#rpm\n", + "x = 720.\n", + "r1 = 40. \t\t\t#mm\n", + "\n", + "#Solution:\n", + "#For a crossed belt:\n", + "#Calcluating the radii of pulleys 2 3 4 5 and 6 \n", + "r2 = r1*(N1/N2) \t\t\t#mm\n", + "#For pulleys 3 and 4 r4 = r3*(N3/N4) or r3*(N3/N4)-r4 = 0\n", + "#For a crossed belt drive r3+r4 = r1+r2\n", + "A = [[N3/N4, -1],[ 1, 1]]\n", + "B = [0, r1+r2]\n", + "V = linalg.solve(A,B)\n", + "r3 = V[0] \t\t\t#mm\n", + "r4 = V[1] \t\t\t#mm\n", + "#For pulleys 5 and 6 r6 = r5*(N5/N6) or r5*(N5/N6)-r6 = 0\n", + "#For a crossed belt drive r5+r6 = r1+r2\n", + "A = [[N5/N6, -1],[ 1, 1]]\n", + "B = [0, r1+r2]\n", + "V = linalg.solve(A,B)\n", + "r5 = V[0] \t\t\t#mm\n", + "r6 = V[1] \t\t\t#mm\n", + "\n", + "#Results:\n", + "print \" For a crossed belt, r2 = %.1fmm;\"%(r2)\n", + "print \" r3 = %.1f mm;\"%(r3)\n", + "print \" r4 = %.1f mm;\"%(r4)\n", + "print \" r5 = %.1f mm;\"%(r5)\n", + "print \" r6 = %.1f mm.\"%(r6)\n", + "#For an open belt:\n", + "#Calcluating the radii of pulleys 2 3 4 5 and 6\n", + "r2 = r1*(N1/N2) \t\t\t#mm\n", + "#Calculating the length of belt for an open belt drive\n", + "L = math.pi*(r1+r2)+(r2-r1)**2/x+2*x \t\t\t#mm\n", + "#For pulleys 3 and 4 r4 = r3*(N3/N4) or r3*(N3/N4)-r4 = 0\n", + "#Since L is constant for pulleys 3 and 4 pi*(r3+r4)+(r4-r3)**2/x+2*x-L = 0\n", + "def f(a):\n", + " r3 = a[0]\n", + " r4 = a[1]\n", + " y = [0,0]\n", + " y[0] = r3*(N3/N4)-r4\n", + " y[1] = math.pi*(r3+r4)+(r4-r3)**2/x+2*x-L\n", + " return y\n", + "\n", + "z = fsolve(f,[1,1])\n", + "r3 = z[0] \t\t\t#mm\n", + "r4 = z[1] \t\t\t#mm\n", + "#For pulleys 5 and 6 r6 = r5*(N5/N6) or r5*(N5/N6)-r6 = 0\n", + "#Since L is constant for pulleys 5 and 6 pi*(r5+r6)+(r6-r5)**2/x+2*x-L = 0\n", + "def f1(a):\n", + " r5 = a[0]\n", + " r6 = a[1]\n", + " y = [0,0]\n", + " y[0] = r5*(N5/N6)-r6\n", + " y[1] = math.pi*(r5+r6)+(r6-r5)**2/x+2*x-L\n", + " return y\n", + "\n", + "z = fsolve(f1,[1,1])\n", + "r5 = z[0] \t\t\t#mm\n", + "r6 = z[1] \t\t\t#mm\n", + "\n", + "#Results:\n", + "print \" For an open belt, r2 = %.1fmm\"%(r2)\n", + "print \" r3 = %.1f mm;\"%(r3)\n", + "print \" r4 = %.1f mm;\"%(r4)\n", + "print \" r5 = %d mm;\"%(round(r5,-1))\n", + "print \" r6 = %d mm.\"%(r6)\n", + "\n", + "# note : answers are slightly different because of fsolve function and rounding off error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " For a crossed belt, r2 = 106.7mm;\n", + " r3 = 48.9 mm;\n", + " r4 = 97.8 mm;\n", + " r5 = 56.4 mm;\n", + " r6 = 90.3 mm.\n", + " For an open belt, r2 = 106.7mm\n", + " r3 = 49.2 mm;\n", + " r4 = 98.4 mm;\n", + " r5 = 60 mm;\n", + " r6 = 91 mm.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.4 Page No : 342" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "d = 600./1000 \t\t\t#m\n", + "N = 200. \t\t\t #rpm\n", + "mu = 0.25\n", + "theta = 160*math.pi/180 \t\t\t#radians\n", + "T1 = 2500. \t\t\t#N\n", + "\n", + "#Solution:\n", + "#Calcluating the velocity of the belt\n", + "v = math.pi*d*N/60 \t\t\t#m/s\n", + "#Calcluating the tension in the slack side of the belt\n", + "T2 = T1/math.exp(mu*theta) \t\t\t#N\n", + "#Calcluating the power transmitted by the belt\n", + "P = (T1-T2)*v/1000 \t\t\t#kW\n", + "\n", + "#Results:\n", + "print \" Power transmitted by the belt, P = %.2f kW.\"%(P)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Power transmitted by the belt, P = 7.89 kW.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.5 Page No : 343" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "W = 9.*1000\n", + "T1 = W \t\t\t#N\n", + "d = 300./1000 \t#m\n", + "N = 20. \t\t#rpm\n", + "mu = 0.25\n", + "\n", + "#Solution:\n", + "#Force required by the man:\n", + "#Calculating the angle of contact\n", + "theta = 2.5*2*math.pi \t\t\t#rad\n", + "#Calculating the force required by the man\n", + "T2 = T1/math.exp(mu*theta) \t\t\t#N\n", + "#Power to raise the casting:\n", + "#Calculating the velocity of the rope\n", + "v = math.pi*d*N/60 \t\t\t#m/s\n", + "#Calculating the power to raise the casting\n", + "P = (T1-T2)*v/1000 \t\t\t#kW\n", + "\n", + "#Results:\n", + "print \" Force required by the man, T2 = %.2f N.\"%(T2)\n", + "print \" Power to raise the casting, P = %.3f kW.\"%(P)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Force required by the man, T2 = 177.33 N.\n", + " Power to raise the casting, P = 2.772 kW.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.6 Page No : 344" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "d1 = 450./1000 #mm\n", + "r1 = d1/2 #m\n", + "d2 = 200./1000 #m\n", + "r2 = d2/2\n", + "x = 1.95 \t\t\t#m\n", + "N1 = 200. \t\t\t#rpm\n", + "T1 = 1.*1000 \t\t\t#N\n", + "mu = 0.25\n", + "\n", + "#Solution:\n", + "#Calculating the speed of the belt\n", + "v = math.pi*d1*N1/60 \t\t\t#m/s\n", + "#Length of the belt:\n", + "#Calculating the length of the crossed belt\n", + "L = math.pi*(r1+r2)+2*x+(r1+r2)**2/x \t\t\t#m\n", + "#Angle of contact between the belt and each pulley:\n", + "#Calculating the angle alpha\n", + "alpha = math.sin((r1+r2)/x)*180/math.pi \t\t\t#degrees\n", + "#Calculating the angle of contact between the belt and each pulley\n", + "theta = (180+2*alpha)*math.pi/180 \t\t\t#radians\n", + "#Power transmitted:\n", + "#Calculating the tension in the slack side of the belt\n", + "T2 = T1/math.exp(mu*theta) \t\t\t#N\n", + "#Calculating the power transmitted\n", + "P = (T1-T2)*v/1000 \t\t\t#kW\n", + "\n", + "#Results:\n", + "print \" Length of the belt, L = %.3f m.\"%(L)\n", + "print \" Angle of contact between the belt and each pulley, theta = %.3f rad.\"%(theta)\n", + "print \" Power transmitted, P = %.2f kW.\"%(P)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Length of the belt, L = 4.975 m.\n", + " Angle of contact between the belt and each pulley, theta = 3.473 rad.\n", + " Power transmitted, P = 2.73 kW.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.7 Page No : 347" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "import math \n", + "\n", + "# Variables:\n", + "N1 = 200.\n", + "N2 = 300. \t\t\t#rpm\n", + "P = 6.*1000 \t\t#W\n", + "b = 100.\n", + "t = 10. \t\t\t#mm\n", + "x = 4.\n", + "d2 = 0.5 \t\t\t#m\n", + "mu = 0.3\n", + "\n", + "#Solution:\n", + "#Stress in the belt for an open belt drive:\n", + "#Calculating the diameter of the larger pulley\n", + "d1 = d2*(N2/N1) \t\t\t#m\n", + "#Calculating the velocity of the belt\n", + "v = math.pi*d2*N2/60 \t\t\t#m/s\n", + "#Calculating the angle alpha for an open belt drive\n", + "alphao = math.sin((r1-r2)/x)*180/math.pi \t\t\t#degrees\n", + "#Calculating the angle of contact on the smaller pulley\n", + "thetao = (180-2*alphao)*math.pi/180 \t\t\t#radians\n", + "#Calculating the tensions in the belt\n", + "#Ratio of the tensions in the belt T1/T2 = math.exp(mu*thetao) or T1-T2*math.exp(mu*thetao) = 0\n", + "#Power transmitted P = (T1-T2)*v or T1-T2 = P/v\n", + "A = [[1, -math.exp(mu*thetao)],[ 1, -1]]\n", + "B = [0, P/v]\n", + "V = linalg.solve(A,B)\n", + "T1o = V[0] \t\t\t#N\n", + "T2o = V[1] \t\t\t#N\n", + "#Calculating the stress in the belt\n", + "sigmao = T1o/(b*t) \t\t\t#MPa\n", + "#Stress in the belt for a cross belt drive:\n", + "#Calculating the angle alpha for a cross belt drive\n", + "alphac = math.sin((d1+d2)/(2*x))*180/math.pi \t\t\t#degrees\n", + "#Calculating the angle of contact\n", + "thetac = (180+2*alphac)*math.pi/180 \t\t\t#radians\n", + "#Calculating the tensions in the belt \n", + "#Ratio of the tensions in the belt T1/T2 = math.exp(mu*thetac) or T1-T2*math.exp(mu*thetac) = 0\n", + "#Power transmitted P = (T1-T2)*v or T1-T2 = P/v\n", + "A = [[1, -math.exp(mu*thetac)],[ 1, -1]]\n", + "B = [0, P/v]\n", + "V = linalg.solve(A,B)\n", + "T1c = V[0] \t\t\t#N\n", + "T2c = V[1] \t\t\t#N\n", + "#Calculating the stress in the belt\n", + "sigmac = T1c/(b*t) \t\t\t#MPa\n", + "\n", + "#Results:\n", + "print \" Stress in the belt for an open belt drive, sigma = %.3f MPa.\"%(sigmao)\n", + "print \" Stress in the belt for a cross belt drive, sigma = %.3f MPa.\"%(sigmac)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Stress in the belt for an open belt drive, sigma = 1.267 MPa.\n", + " Stress in the belt for a cross belt drive, sigma = 1.184 MPa.\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.8 Page No : 348" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "P = 7.5*1000 \t\t#W\n", + "d = 1.2 #m\n", + "t = 10./1000 \t\t#m\n", + "N = 250. \t\t\t#rpm\n", + "theta = 165.*math.pi/180 \t\t\t#radians\n", + "mu = 0.3\n", + "sigma = 1.5*10**6 \t\t\t#N/m**2\n", + "rho = 1.*10**3 \t\t\t#kg/m**3\n", + "\n", + "#Solution:\n", + "#Calculating the velocity of the belt\n", + "v = math.pi*d*N/60 \t\t\t#m/s\n", + "#Calculating the tensions in the belt\n", + "#Power transmitted P = (T1-T2)*v or T1-T2 = P/v\n", + "#Ratio of tensions in the belt math.log(T1/T2) = mu*theta or T1-T2*math.exp(mu*theta) = 0\n", + "A = [[1, -1],[1, -math.exp(mu*theta)]]\n", + "B = [P/v, 0]\n", + "V = linalg.solve(A,B)\n", + "T1 = V[0] \t\t\t#N\n", + "T2 = V[1] \t\t\t#N\n", + "#Calculating the width of the belt\n", + "b = T1/(sigma*t-t*1*rho*v**2)*1000 \t\t\t#mm\n", + "\n", + "#Results:\n", + "print \" Width of the belt, b = %.1f mm.\"%(b)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Width of the belt, b = 65.9 mm.\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.9 Page No : 349" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "import math \n", + "\n", + "# Variables:\n", + "t = 9.75/1000\n", + "d1 = 300./1000\n", + "x = 3. \t\t\t #m\n", + "P = 15.*1000 \t\t\t#W\n", + "N1 = 900.\n", + "N2 = 300. \t\t\t#rpm\n", + "rho = 1000. \t\t\t#kg/m**3\n", + "sigma = 2.5*10**6 \t\t\t#N/m**2\n", + "mu = 0.3\n", + "\n", + "#Solution:\n", + "#Calculating the diameter of the driven pulley\n", + "d2 = d1*(N1/N2) \t\t\t#m\n", + "#Calculating the velocity of the belt\n", + "v = math.pi*d1*N1/60 \t\t\t#m/s\n", + "#Calculating the angle alpha for an open belt drive\n", + "alpha = math.sin((d2-d1)/(2*x))*180/math.pi \t\t\t#degrees\n", + "#Calculating the angle of lap\n", + "theta = (180-2*alpha)*math.pi/180 \t\t\t#radians\n", + "#Calculating the tensions in the belt\n", + "#Ratio of tensions math.log(T1/T2) = mu*theta or T1-T2*math.exp(mu*theta) = 0\n", + "#Power transmitted P = (T1-T2)*v or T1-T2 = P/v\n", + "A = [[1, -math.exp(mu*theta)],[ 1, -1]]\n", + "B = [0, P/v]\n", + "V = linalg.solve(A,B)\n", + "T1 = V[0] \t\t\t#N\n", + "T2 = V[1] \t\t\t#N\n", + "#Calculating the width of the belt\n", + "b = T1/(sigma*t-t*1*rho*v**2)*1000 \t\t\t#mm\n", + "\n", + "#Results:\n", + "print \" Width of the belt, b = %d mm.\"%(b)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Width of the belt, b = 80 mm.\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.10 Page No : 350" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "theta = 120*math.pi/180 #degrees\t\t\t#radians\n", + "b = 100./1000\n", + "t = 6./1000 \t\t\t#m\n", + "rho = 1000. \t\t\t#kg/m**3\n", + "mu = 0.3\n", + "sigma = 2*10**6 \t\t\t#N/m**2\n", + "\n", + "#Solution:\n", + "#Speed of the belt for greatest power:\n", + "#Calculating the maximum tension in the belt\n", + "T = sigma*b*t \t\t\t#N\n", + "#Calculating the mass of the belt per metre length\n", + "l = 1. \t\t\t#m\n", + "m = b*t*l*rho \t\t\t#kg/m\n", + "#Calculating the speed of the belt for greatest power\n", + "v = math.sqrt(T/(3*m)) \t\t\t#m/s\n", + "#Greatest power which the belt can transmit\n", + "#Calculating the centrifugal tension for maximum power to be transmitted\n", + "TC = T/3 \t\t\t#N\n", + "#Calculating the tension in the tight side of the belt\n", + "T1 = T-TC \t\t\t#N\n", + "#Calculating the tension in the slack side of the belt\n", + "T2 = T1/math.exp(mu*theta) \t\t\t#N\n", + "#Calculating the greatest power which the belt can transmit\n", + "P = (T1-T2)*v/1000 \t\t\t#kW\n", + "\n", + "#Results:\n", + "print \" Speed of the belt for greatest power, v = %.2f m/s.\"%(v)\n", + "print \" Greatest power which the belt can transmit, P = %.2f kW.\"%(P)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Speed of the belt for greatest power, v = 25.82 m/s.\n", + " Greatest power which the belt can transmit, P = 9.64 kW.\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.11 Page No : 351" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "d1 = 1.2 #m\n", + "r1 = d1/2 #m\n", + "d2 = 0.5 #m\n", + "r2 = d2/2 #m\n", + "x = 4. \t\t\t#m\n", + "m = 0.9 \t\t\t#kg/m\n", + "T = 2000. \t\t\t#N\n", + "mu = 0.3\n", + "N1 = 200. #rpm\n", + "N2 = 450. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Calculating the velocity of the belt\n", + "v = math.pi*d1*N1/60 \t\t\t#m/s\n", + "#Calculating the centrifugal tension\n", + "TC = m*v**2 \t\t\t#N\n", + "#Calculating the tension in the tight side of the belt\n", + "T1 = T-TC \t\t\t#N\n", + "#Calculating the angle alpha for an open belt drive\n", + "alpha = math.sin((r1-r2)/x)*180/math.pi \t\t\t#degrees\n", + "#Calculating the angle of lap on the smaller pulley\n", + "theta = (180-2*alpha)*math.pi/180 \t\t\t#radians\n", + "#Calculating the tension in the slack side of the belt\n", + "T2 = T1/math.exp(mu*theta) \t\t\t#N\n", + "#Calculating the torque on the shaft of larger pulley\n", + "TL = (T1-T2)*r1 \t\t\t#N-m\n", + "#Calculating the torque on the shaft of smaller pulley\n", + "TS = (T1-T2)*r2 \t\t\t#N-m\n", + "#Calculating the power transmitted\n", + "P = (T1-T2)*v/1000 \t\t\t#kW\n", + "#Power lost in friction:\n", + "#Calculating the input power\n", + "P1 = TL*2*math.pi*N1/(60*1000) \t\t\t#kW\n", + "#Calculating the output power\n", + "P2 = TS*2*math.pi*N2/(60*1000) \t\t\t#kW\n", + "#Calculating the power lost in friction\n", + "Pf = P1-P2 \t\t\t#Power lost in friction kW\n", + "#Calculating the efficiency of the drive\n", + "eta = P2/P1*100 \t\t\t#%\n", + "\n", + "#Results:\n", + "print \" Torque on the shaft of larger pulley, TL = %.1f N-m.\"%(TL)\n", + "print \" Torque on the shaft of smaller pulley, TS = %d N-m.\"%(TS)\n", + "print \" Power transmitted, P = %.2f kW.\"%(P)\n", + "print \" Power lost in friction = %.2f kW.\"%(Pf)\n", + "print \" Efficiency of the drive, eta = %.1f %%.\"%(eta)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Torque on the shaft of larger pulley, TL = 657.0 N-m.\n", + " Torque on the shaft of smaller pulley, TS = 273 N-m.\n", + " Power transmitted, P = 13.76 kW.\n", + " Power lost in friction = 0.86 kW.\n", + " Efficiency of the drive, eta = 93.8 %.\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.12 Page No : 353" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "import math \n", + "\n", + "# Variables:\n", + "T0 = 2000. \t\t\t#N\n", + "mu0 = 0.3\n", + "theta = 150.*math.pi/180 \t\t\t#radians\n", + "r2 = 200./1000\n", + "d2 = 2*r2 \t\t\t#m\n", + "N2 = 500.\t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Calculating the velocity of the belt\n", + "v = math.pi*d2*N2/60 \t\t\t#m/s\n", + "#Calculating the tensions in the belt\n", + "#Initial tension T0 = (T1+T2)/2 or T1+T2 = 2*T0\n", + "#Ratio of the tensions in the belt math.log(T1/T2) = mu*theta or T1-T2*math.exp(mu*theta) = 0\n", + "A = [[1, 1],[ 1 ,-math.exp(mu*theta)]]\n", + "B = [2*T0, 0]\n", + "V = linalg.solve(A,B)\n", + "T1 = V[0] \t\t\t#N\n", + "T2 = V[1] \t\t\t#N\n", + "#Calculating the power transmitted\n", + "P = (T1-T2)*v/1000 \t\t\t#kW\n", + "\n", + "#Results:\n", + "print \" Power transmitted, P = %.1f kW.\"%(P)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Power transmitted, P = 15.7 kW.\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.13 Page No : 354" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "import math \n", + "\n", + "# Variables:\n", + "x = 4.8\n", + "d1 = 1.5 #m\n", + "d2 = 1. \t\t\t#m\n", + "T0 = 3.*1000 \t\t#N\n", + "m = 1.5 \t\t\t#kg/m\n", + "mu = 0.3\n", + "N2 = 400. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Calculating the velocity of the belt\n", + "v = math.pi*d2*N2/60 \t\t\t#m/s\n", + "#Calculating the centrifugal tension\n", + "TC = m*v**2 \t\t\t#N\n", + "#Calculating the angle alpha\n", + "alpha = math.sin((d1-d2)/(2*x))*180/math.pi \t\t\t#degrees\n", + "#Calculating the angle of lap for the smaller pulley\n", + "theta = (180-2*alpha)*math.pi/180 \t\t\t#radians\n", + "#Calculating the tensions in the belt\n", + "#Initial tension T0 = (T1+T2+2*TC)/2 or T1+T2 = 2*(T0-TC)\n", + "#Ratio of tensions in the belt math.log(T1/T2) = mu*theta or T1-T2*math.exp(mu*theta) = 0\n", + "A = [[1, 1],[1, -math.exp(mu*theta)]]\n", + "B = [2*(T0-TC), 0]\n", + "V = linalg.solve(A,B)\n", + "T1 = V[0] \t\t\t#N\n", + "T2 = V[1] \t\t\t#N\n", + "#Calculating the power transmitted\n", + "P = (T1-T2)*v/1000 \t\t\t#kW\n", + "\n", + "#Results:\n", + "print \" Power transmitted, P = %.f kW.\"%(P)\n", + "\n", + "# answer are slightly different because of solve function" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Power transmitted, P = 42 kW.\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.14 Page No : 355" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "x = 1.2 #m\n", + "d2 = 400./1000 #m\n", + "t = 5./1000 #m\n", + "b = 80./1000 \t\t#m\n", + "N1 = 350. #rpm\n", + "N2 = 140. \t\t\t#rpm\n", + "mu = 0.3\n", + "sigma = 1.4*10**6 \t\t\t#N/m**2\n", + "\n", + "#Solution:\n", + "#Calculating the diameter of the driving pulley\n", + "d1 = d2*(N2/N1) \t\t\t#m\n", + "#Maximum power transmitted by the belting:\n", + "#Refer Fig. 11.18\n", + "#Calculating the angle alpha\n", + "alpha = math.sin((d2-d1)/(2*x))*180/math.pi \t\t\t#degrees\n", + "#Calculating the angle of contact of the belt on the driving pulley\n", + "theta = (180-2*alpha)*math.pi/180 \t\t\t#radians\n", + "#Calculating the maximum tension to which the belt can be subjected\n", + "T1 = sigma*b*t \t\t\t#N\n", + "#Calculating the tension in the slack side of the belt\n", + "T2 = T1/math.exp(mu*theta) \t\t\t#N\n", + "#Calculating the velocity of the belt\n", + "v = math.pi*d1*N1/60 \t\t\t#m/s\n", + "#Calculating the power transmitted\n", + "P = (T1-T2)*v/1000 \t\t\t#kW\n", + "#Calculating the required initial belt tension\n", + "T0 = (T1+T2)/2 \t\t\t#N\n", + "\n", + "#Results:\n", + "print \" Diameter of the driving pulley, d1 = %.2f m.\"%(d1)\n", + "print \" Maximum power transmitted by the belting, P = %.3f kW.\"%(P)\n", + "print \" Required initial belt tension, T0 = %.1f N.\"%(T0)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Diameter of the driving pulley, d1 = 0.16 m.\n", + " Maximum power transmitted by the belting, P = 0.963 kW.\n", + " Required initial belt tension, T0 = 395.8 N.\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.15 Page No : 356" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "import math \n", + "\n", + "# Variables:\n", + "d2 = 240./1000 #m\n", + "d1 = 600./1000 #m\n", + "x = 3. \t\t\t #m\n", + "P = 4.*1000 \t\t#W\n", + "N2 = 300. \t\t\t#rpm\n", + "mu = 0.3\n", + "T1s = 10. \t\t\t#Safe working tension N/mm width\n", + "\n", + "#Solution:\n", + "#Minimum width of the belt:\n", + "#Calculating the velocity of the belt\n", + "v = math.pi*d2*N2/60 \t\t\t#m/s\n", + "#Calculating the angle alpha for an open belt drive\n", + "alpha = math.sin((d1-d2)/(2*x))*180/math.pi \t\t\t#degrees\n", + "#Calculating the angle of lap on the smaller pulley\n", + "theta = (180-2*alpha)*math.pi/180 \t\t\t#radians\n", + "#Calculating the tensions in the belt\n", + "#Power transmitted P = (T1-T2)*v or T1-T2 = P/v\n", + "#Ratio of tensions math.log(T1/T2) = mu*theta or T1-T2*math.exp(mu*theta) = 0\n", + "A = [[1, -1],[ 1, -math.exp(mu*theta)]]\n", + "B = [P/v, 0]\n", + "V = linalg.solve(A, B)\n", + "T1 = V[0] \t\t\t#N\n", + "T2 = V[1] \t\t\t#N\n", + "#Calculating the minimum width of the belt\n", + "b = T1/T1s \t\t\t#mm\n", + "#Calculating the initial belt tension\n", + "T0 = (T1+T2)/2 \t\t\t#N\n", + "#Calculating the length of the belt required\n", + "L = math.pi/2*(d1+d2)+2*x+(d1-d2)**2/(4*x) \t\t\t#m\n", + "\n", + "#Results:\n", + "print \" Minimum width of the belt, b = %.1f mm.\"%(b)\n", + "print \" Initial belt tension, T0 = %.1f N.\"%(T0)\n", + "print \" Length of the belt required, L = %.2f m.\"%(L)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Minimum width of the belt, b = 178.0 mm.\n", + " Initial belt tension, T0 = 1249.5 N.\n", + " Length of the belt required, L = 7.33 m.\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.16 Page No : 357" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "import math\n", + "\n", + "# Variables:\n", + "d1 = 400./1000 #m\n", + "d2 = 250./1000 #m\n", + "x = 2.\n", + "mu = 0.4 \t\t\t#m\n", + "T = 1200. \t\t\t#N\n", + "v = 10. \t\t\t#m/s\n", + "\n", + "#Solution:\n", + "#Power transmitted:\n", + "#Calculating the angle alpha for an open belt drive\n", + "alpha = math.sin((d1-d2)/(2*x))*180/math.pi \t\t\t#degrees\n", + "#Calculating the angle of contact\n", + "theta = (180-2*alpha)*math.pi/180 \t\t\t#radians\n", + "#Calculating the tension in the tight side of the belt\n", + "T1 = T \t\t\t#Neglecting centrifugal tension N\n", + "#Calculating the tension in the slack side of the belt\n", + "T2 = T1/math.exp(mu*theta) \t\t\t#N\n", + "#Calculating the power transmitted\n", + "P = (T1-T2)*v/1000 \t\t\t#kW\n", + "\n", + "#Results:\n", + "print \" Power transmitted, P = %.2f kW.\"%(P)\n", + "#Power transmitted when initial tension is increased by 10%:\n", + "#Calculating the initial tension\n", + "T0 = (T1+T2)/2 \t\t\t#N\n", + "#Calculating the increased initial tension\n", + "T0dash = T0+10./100*T0 \t\t\t#N\n", + "#Calculating the corresponding tensions in the belt\n", + "#We have T0dash = (T1+T2)/2 or T1+T2 = 2*T0dash\n", + "#Ratio of the tensions math.log(T1/T2) = mu*theta or T1-T2*math.exp(mu*theta) = 0\n", + "A = [[1, 1],[ 1, -math.exp(mu*theta)]]\n", + "B = [2*T0dash, 0]\n", + "V = linalg.solve(A,B)\n", + "T1 = V[0] \t\t\t#N\n", + "T2 = V[1] \t\t\t#N\n", + "#Calculating the power transmitted\n", + "P1 = (T1-T2)*v/1000 \t\t\t#kW\n", + "#Power transmitted when coefficient of friction is increased by 10%:\n", + "#Calculating the increased coefficient of friction\n", + "mudash = mu+10./100*mu\n", + "#Calculating the corresponding tensions in the belt\n", + "#Ratio of the tensions math.log(T1/T2) = mudash*theta or T1-T2*math.exp(mudash*theta) = 0\n", + "#Initial tension T0 = (T1+T2)/2 or T1+T2 = 2*T0\n", + "A = [[1, -math.exp(mudash*theta)],[ 1, 1]]\n", + "B = [0, 2*T0]\n", + "V = linalg.solve(A,B)\n", + "T1 = V[0] \t\t\t#N\n", + "T2 = V[1] \t\t\t#N\n", + "#Calculating the power transmitted\n", + "P2 = (T1-T2)*v/1000 \t\t\t#kW\n", + "\n", + "#Results:\n", + "if P1>P2:\n", + " print \" Since the power transmitted by increasing the initial tension is more\\\n", + " therefore in order to increase the power transmitted we shall adopt\\\n", + " the method of increasing the initial tension.\"\n", + "else:\n", + " print \" Since the power transmitted by increasing the coefficient of friction is more,\\\n", + " therefore in order to increase the power transmitted we shall adopt the method of increasing\\\n", + " the coefficient of friction.\"\n", + "#Percentage increase in power:\\\n", + "#Calculating the percentage increase in power when the initial tension is increased\n", + "I1 = (P1-P)/P*100. \t\t\t#Percentage increase in power when the initial tension is increased %\n", + "#Calculating the percentage increase in power when coefficient of friction is increased\n", + "I2 = (P2-P)/P*100. \t\t\t#Percentage increase in power when coefficient of friction is increased %\n", + "\n", + "\n", + "#Results:\n", + "print \" Percentage increase in power when the initial tension is increased = %.2f %%.\"%(I1)\n", + "print \" Percentage increase in power when coefficient of friction is increased = %.1f %%.\"%(I2)\n", + "\n", + "#rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Power transmitted, P = 8.48 kW.\n", + " Since the power transmitted by increasing the initial tension is more therefore in order to increase the power transmitted we shall adopt the method of increasing the initial tension.\n", + " Percentage increase in power when the initial tension is increased = 10.00 %.\n", + " Percentage increase in power when coefficient of friction is increased = 7.6 %.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.17 Page No : 362" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "beta = 30./2 \t\t\t#degrees\n", + "alpha = 750.*10**-6 \t#mm**2\n", + "mu = 0.12\n", + "rho = 1.2*1000 \t\t\t#kg/m**3\n", + "sigma = 7.*10**6 \t\t#N/m**2\n", + "d = 300./1000 \t\t\t#m\n", + "N = 1500. \t\t\t #rpm\n", + "\n", + "#Solution:\n", + "#Power transmitted:\n", + "#Calculating the velocity of the belt\n", + "v = math.pi*d*N/60 \t\t\t#m/s\n", + "#Calculating the mass of the belt per metre length\n", + "l = 1 \t\t\t#m\n", + "m = alpha*l*rho \t\t\t#kg/m\n", + "#Calculating the centrifugal tension\n", + "TC = m*v**2 \t\t\t#N\n", + "#Calculating the maximum tension in the belt\n", + "T = sigma*alpha \t\t\t#N\n", + "#Calculating the tension in the tight side of the belt\n", + "T1 = T-TC \t\t\t#N\n", + "#Calculating the tension in the slack side of the belt\n", + "theta = math.pi \t\t\t#Angle of contact radians\n", + "T2 = T1/math.exp(mu*theta*(1/math.sin(math.radians(beta)))) \t\t\t#N\n", + "#Calculating the power transmitted\n", + "P = (T1-T2)*v*2/1000 \t\t\t#kW\n", + "#Shaft speed:\n", + "#Calculating the belt speed for maximum power transmitted\n", + "v1 = math.sqrt(T/(3*m)) \t\t\t#m/s\n", + "#Calculating the shaft speed for maximum power transmitted\n", + "N1 = v1*60/(math.pi*d) \t\t\t#rpm\n", + "\n", + "#Results:\n", + "print \" Power transmitted, P = %.3f kW.\"%(P)\n", + "print \" Shaft speed at which the power transmitted would be maximum, N1 = %d rpm.\"%(N1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Power transmitted, P = 171.690 kW.\n", + " Shaft speed at which the power transmitted would be maximum, N1 = 2807 rpm.\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.18 Page No : 363" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "beta = 30./2 \t\t\t#degrees\n", + "t = 20./1000\n", + "b = 20./1000 \t\t\t#m\n", + "m = 0.35 \t\t\t #kg/m\n", + "sigma = 1.4*10**6 \t\t\t#N/m**2\n", + "theta = 140.*math.pi/180 \t\t\t#radians\n", + "mu = 0.15\n", + "\n", + "#Solution:\n", + "#Calculating the maximum tension in the belt\n", + "T = sigma*b*t \t\t\t#N\n", + "#Calculating the velocity of the belt for maximum power to be transmitted\n", + "v = math.sqrt(T/(3*m)) \t\t\t#m/s\n", + "#Calculating the centrifugal tension\n", + "TC = T/3 \t\t\t#N\n", + "#Calculating the tension in the tight side of the belt\n", + "T1 = T-TC \t\t\t#N\n", + "#Calculating the tension in the slack side of the belt\n", + "T2 = T1/math.exp(mu*theta*(1/math.sin(math.radians(beta)))) \t\t\t#N\n", + "#Calculating the maximum power transmitted\n", + "P = (T1-T2)*v/1000 \t\t\t#kW\n", + "\n", + "#Results:\n", + "print \" Maximum power transmitted, P = %.2f kW.\"%(P)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Maximum power transmitted, P = 6.53 kW.\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.19 Page No : 363" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "P = 90. \t\t\t#kW\n", + "N2 = 250.\n", + "N1 = 750. \t\t\t#rpm\n", + "d2 = 1.\n", + "x = 1.75 \t\t\t#m\n", + "v = 1600./60 \t\t\t#m/s\n", + "a = 375.*10**-6 \t\t\t#m**2\n", + "rho = 1000. \t\t\t#kg/m**3\n", + "sigma = 2.5*10**6 \t\t\t#N/m**2\n", + "beta = 35./2 \t\t\t#degrees\n", + "mu = 0.25\n", + "\n", + "#Solution:\n", + "#Calculating the diameter of the pulley on the motor shaft\n", + "d1 = d2*(N2/N1) \t\t\t#m\n", + "#Calculating the mass of the belt per metre length\n", + "l = 1 \t\t\t #m\n", + "m = a*l*rho \t\t\t#kg/m\n", + "#Calculating the centrifugal tension\n", + "TC = m*v**2 \t\t\t#N\n", + "#Calculating the maximum tension in the belt\n", + "T = sigma*a \t\t\t#N\n", + "#Calculating the tension in the tight side of the belt\n", + "T1 = T-TC \t\t\t #N\n", + "#Refer Fig. 11.21\n", + "#Calculating the angle alpha\n", + "alpha = math.sin(math.radians((d2-d1)/(2*x)))*180/math.pi \t\t\t#degrees\n", + "#Calculating the angle of lap on smaller pulley\n", + "theta = (180-2*alpha)*math.pi/180 \t\t\t#radians\n", + "#Calculating the tension in the slack side of the belt\n", + "T2 = T1/math.exp(mu*theta*(1/math.sin(math.radians(beta)))) \t\t\t#N\n", + "#Number of V-belts:\n", + "#Calculating the power transmitted per belt\n", + "P1 = (T1-T2)*v/1000 \t\t\t#Power transmitted per belt kW\n", + "#Calculating the number of V-belts\n", + "n = P/16.086 \t\t\t#Number of V-belts\n", + "#Calculating the length each of belt for an open belt drive\n", + "L = math.pi/2*(d2+d1)+2*x+(d2-d1)**2/(4*x) \t\t\t#m\n", + "\n", + "#Results:\n", + "print \" Number of V-belts = %.f.\"%(n)\n", + "print \" Length of each belt, L = %.2f m.\"%(L)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Number of V-belts = 6.\n", + " Length of each belt, L = 5.66 m.\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.20 Page No : 367" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "P = 600. \t\t\t#kW\n", + "d = 4. \t\t\t #m\n", + "N = 90. \t\t\t#rpm\n", + "theta = 160.*math.pi/180 \t\t\t#radians\n", + "beta = 45./2 \t\t\t#degrees\n", + "mu = 0.28\n", + "m = 1.5 \t\t\t#kg/m\n", + "T = 2400. \t\t\t#N\n", + "\n", + "#Solution:\n", + "#Calculating the velocity of the rope\n", + "v = math.pi*d*N/60 \t\t\t#m/s\n", + "#Calculating the centrifugal tension\n", + "TC = m*v**2 \t\t\t#N\n", + "#Calculating the tension in the tight side of the rope\n", + "T1 = T-TC \t\t\t#N\n", + "#Calculating the tension in the slack side of the belt\n", + "T2 = T1/math.exp(mu*theta*(1/math.sin(math.radians(beta)))) \t\t\t#N\n", + "#Calculating the power transmitted per rope\n", + "P1 = (T1-T2)*v/1000 \t\t\t#Power transmitted per rope kW\n", + "#Calculating the number of ropes\n", + "n = P/P1 \t\t\t#Number of ropes\n", + "\n", + "#Results:\n", + "print \" Number of ropes required = %d.\"%(n+1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Number of ropes required = 20.\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.21 Page No : 367" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "d = 3.6 \t\t\t#m\n", + "n = 15. \t\t\t#Number of grooves\n", + "beta = 45./2 \t\t#degrees\n", + "theta = 170.*math.pi/180 \t\t\t#radians\n", + "mu = 0.28\n", + "T = 960. \t\t\t#N\n", + "m = 1.5 \t\t\t#kg/m\n", + "\n", + "#Solution:\n", + "#Speed of the pulley:\n", + "#Calculating the velocity of the rope\n", + "v = math.sqrt(T/(3*m)) \t\t\t#m/s\n", + "#Calculating the speed of the pulley\n", + "N = v*60/(math.pi*d) \t\t\t#rpm\n", + "#Power transmitted\n", + "#Calculating the centrifugal tension for maximum power\n", + "TC = T/3 \t\t\t#N\n", + "#Calculating the tension in the tight side of the rope\n", + "T1 = T-TC \t\t\t#N\n", + "#Calculating the tension in the slack side of the rope\n", + "T2 = T1/math.exp(mu*theta*(1/math.sin(math.radians(beta)))) \t\t\t#N\n", + "#Calculating the power transmitted per rope\n", + "P1 = (T1-T2)*v/1000 \t\t\t#Power transmitted per rope kW\n", + "#Calculating the total power transmitted\n", + "P = P1*n \t\t\t#Total power transmitted kW\n", + "\n", + "#Results:\n", + "print \" Speed of the pulley for maximum power, N = %.1f rpm.\"%(N)\n", + "print \" Power transmitted = %.2f kW.\"%(P)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Speed of the pulley for maximum power, N = 77.5 rpm.\n", + " Power transmitted = 124.22 kW.\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.22 Page No : 368" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "import math \n", + "\n", + "# Variables:\n", + "PT = 24. \t\t\t#kW\n", + "d = 400./1000 \t\t#m\n", + "N = 110. \t\t\t#rpm\n", + "beta = 45./2 \t\t#degrees\n", + "theta = 160.*math.pi/180 \t\t\t#radians\n", + "mu = 0.28\n", + "n = 10.\n", + "\n", + "#Solution:\n", + "#Initial tension:\n", + "#Calculating the power transmitted per rope\n", + "P = PT/n*1000 \t\t\t#W\n", + "#Calculating the velocity of the rope\n", + "v = math.pi*d*N/60 \t\t\t#m/s\n", + "#Calculating the tensions in the rope\n", + "#Power transmitted P = (T1-T2)*v or T1-T2 = P/v\n", + "#Ratio of tensions math.log(T1/T2) = mu*theta*(1/math.sin(math.radians(beta)) or T1-T2*math.exp(mu*theta*(1/math.math.sin(math.radians(beta))) = 0\n", + "A = [[1, -1],[ 1, -math.exp(mu*theta*(1/math.sin(math.radians(beta))))]]\n", + "B = [P/v, 0]\n", + "V = linalg.solve(A,B)\n", + "T1 = V[0] \t\t\t#N\n", + "T2 = V[1] \t\t\t#N\n", + "#Calculating the initial tension in each rope\n", + "T0 = (T1+T2)/2 \t\t\t#N\n", + "#Diameter of each rope:\n", + "#Calculating the girth of rope\n", + "C = math.sqrt(T1/(122.*10**3-53.*v**2))*1000 \t\t\t#mm\n", + "#Calculating the diameter of each rope\n", + "d1 = C/math.pi \t\t\t#mm\n", + "\n", + "#Results:\n", + "print \" Initial tension, T0 = %.2f N.\"%(T0)\n", + "print \" Diameter of each rope, d1 = %.2f mm.\"%(d1)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Initial tension, T0 = 676.00 N.\n", + " Diameter of each rope, d1 = 31.56 mm.\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.23 Page No : 376" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "N1 = 240. #rpm\n", + "N2 = 120. \t\t\t#rpm\n", + "T1 = 20.\n", + "d2 = 600./1000 #m\n", + "r2 = d2/2 #m\n", + "x = 800./1000 \t\t#m\n", + "\n", + "#SOlution:\n", + "#Calculating the number of teeth on the drive sprocket\n", + "T2 = T1*(N1/N2)\n", + "#Calculating the pitch of the chain\n", + "p = r2*2*math.sin(math.radians(180/T2))*1000 \t\t\t#mm\n", + "#Length of the chain:\n", + "m = x*1000/p\n", + "#Calculating the multiplying factor\n", + "K = (T1+T2)/2+2*m+(1/math.sin(math.radians(180/T1))-1/math.sin(math.radians(180/T2)))**2/(4*m)\n", + "#Calculating the length of the chain\n", + "L = p*K/1000 \t\t\t#m\n", + "\n", + "#Results:\n", + "print \" Number of teeth on the driven sprocket, T2 = %d.\"%(T2)\n", + "print \" Pitch of the chain, p = %.1f mm.\"%(p)\n", + "print \" Length of the chain, L = %.4f m.\"%(L)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Number of teeth on the driven sprocket, T2 = 40.\n", + " Pitch of the chain, p = 47.1 mm.\n", + " Length of the chain, L = 3.0402 m.\n" + ] + } + ], + "prompt_number": 23 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Theory_Of_Machines/ch12.ipynb b/Theory_Of_Machines/ch12.ipynb new file mode 100755 index 00000000..22e6fba9 --- /dev/null +++ b/Theory_Of_Machines/ch12.ipynb @@ -0,0 +1,1145 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:844073ad5c928bfdaf4248256fc5562a2dc71b1c2156a1230e38f2c366a18bf7" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 12 : Toothed Gearing" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.1 Page No : 393" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "P = 120.*1000 \t\t\t#W\n", + "d = 250./1000\n", + "r = d/2 \t\t\t#m\n", + "N = 650. \t\t\t#rpm\n", + "phi = 20. \t\t\t#degrees\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the gear\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the torque transmitted\n", + "T = P/omega \t\t\t#N-m\n", + "#Calculating the math.tangential load on the pinion\n", + "FT = T/r \t\t\t#N\n", + "#Calculating the total load due to power transmitted\n", + "F = FT/(math.cos(math.radians(phi))*1000) \t\t\t#kN\n", + "\n", + "#Results:\n", + "print \" Total load due to power transmitted, F = %.2f kN.\"%(F)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Total load due to power transmitted, F = 15.01 kN.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.2 Page No : 397" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "T = 40.\n", + "t = T\n", + "phi = 20. \t\t\t#degrees\n", + "m = 6. \t\t\t#mm\n", + "\n", + "#Solution:\n", + "#Calculating the circular pitch\n", + "pc = math.pi*m \t\t\t#mm\n", + "#Calculating the length of arc of contact\n", + "Lac = 1.75*pc \t\t\t#Length of arc of contact mm\n", + "#Calculating the length of path of contact\n", + "Lpc = Lac*math.cos(phi) \t\t\t#Length of path of contact mm\n", + "#Calculating the pitch circle radii of each wheel\n", + "R = m*T/2 \t\t\t#mm\n", + "r = R \t\t\t#mm\n", + "#Calculating the radius of the addendum circle of each wheel\n", + "RA = math.sqrt(R**2*(math.cos(phi))**2+(Lpc/2+R*math.sin(phi))**2) \t\t\t#mm\n", + "#Calculating the addendum of the wheel\n", + "Ad = RA-R \t\t\t#Addendum of the wheel mm\n", + "\n", + "#Results:\n", + "print \" Addendum of the wheel = %.2f mm.\"%(Ad)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Addendum of the wheel = 6.17 mm.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.3 Page No : 398" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "t = 30.\n", + "T = 80.\n", + "phi = 20. \t\t\t#degrees\n", + "m = 12. \t\t\t#mm\n", + "Addendum = 10. \t\t#mm\n", + "\n", + "#Solution:\n", + "#Length of path of contact:\n", + "#Calculating the pitch circle radius of pinion\n", + "r = m*t/2 \t\t\t#mm\n", + "#Calculating the pitch circle radius of gear\n", + "R = m*T/2 \t\t\t#mm\n", + "#Calculating the radius of addendum circle of pinion\n", + "rA = r+Addendum \t\t\t#mm\n", + "#Calculating the radius of addendum circle of gear\n", + "RA = R+Addendum \t\t\t#mm\n", + "#Calculating the length of path of approach\n", + "#Refer Fig. 12.11\n", + "KP = math.sqrt(RA**2-R**2*(math.cos(math.radians(phi)))**2)-R*math.sin(math.radians(phi)) \t\t\t#mm\n", + "#Calculating the length of path of recess\n", + "PL = math.sqrt(rA**2-r**2*(math.cos(math.radians(phi)))**2)-r*math.sin(math.radians(phi)) \t\t\t#mm\n", + "#Calculating the length of path of contact\n", + "KL = KP+PL \t\t\t#mm\n", + "#Calculating the length of arc of contact\n", + "Lac = KL/math.cos(math.radians(phi)) \t\t\t#Length of arc of contact mm\n", + "#Contact ratio:\n", + "#Calculating the circular pitch\n", + "Pc = math.pi*m \t\t\t#mm\n", + "#Calculating the contact ratio\n", + "CR = Lac/Pc \t\t\t#Contact ratio\n", + "\n", + "#Results:\n", + "print \" Length of path of contact, KL = %.1f mm.\"%(KL)\n", + "print \" Length of arc of contact = %.2f mm.\"%(Lac)\n", + "print \" Contact ratio = %.1f.\"%(CR)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Length of path of contact, KL = 52.3 mm.\n", + " Length of arc of contact = 55.61 mm.\n", + " Contact ratio = 1.5.\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.4 Page No : 399" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "phi = 20. \t\t\t#degrees\n", + "t = 20.\n", + "G = 2.\n", + "m = 5. \t\t\t#mm\n", + "v = 1.2 \t\t\t#m/s\n", + "addendum = 1*m \t\t\t#mm\n", + "\n", + "#Solution:\n", + "#Angle turned through by pinion when one pair of teeth is in mesh:\n", + "#Calculating the pitch circle radius of pinion\n", + "r = m*t/2 \t\t\t#mm\n", + "#Calculating the pitch circle radius of wheel\n", + "R = m*G*t/2 \t\t\t#mm\n", + "#Calculating the radius of addendum circle of pinion\n", + "rA = r+addendum \t\t\t#mm\n", + "#Calculating the radius of addendum circle of wheel\n", + "RA = R+addendum \t\t\t#mm\n", + "#Calculating the length of path of approach\n", + "KP = math.sqrt(RA**2-R**2*(math.cos(math.radians(phi)))**2)-R*math.sin(math.radians(phi)) \t\t\t#mm\n", + "#Calculating the length of path of recess\n", + "PL = math.sqrt(rA**2-r**2*(math.cos(math.radians(phi)))**2)-r*math.sin(math.radians(phi)) \t\t\t#mm\n", + "#Calculating the length of path of contact\n", + "KL = KP+PL \t\t\t#mm\n", + "#Calculating the length of arc of contact\n", + "Lac = KL/math.cos(math.radians(phi)) \t\t\t#mm\n", + "#Calculating the angle turned by the pinion\n", + "angle = Lac*360/(2*math.pi*r) \t\t\t#Angle turned by the pinion degrees\n", + "#Maximum velocity of sliding:\n", + "#Calculating the angular speed of pinion\n", + "omega1 = v*1000/r \t\t\t#rad/s\n", + "#Calculating the angular speed of wheel\n", + "omega2 = v*1000/R \t\t\t#rad/s\n", + "#Calculating the maximum velocity of sliding\n", + "vS = (omega1+omega2)*KP \t\t\t#mm/s\n", + "\n", + "#Results:\n", + "print \" Angle turned through by pinion when one pair of teeth is in mesh = %.2f degrees.\"%(angle)\n", + "print \" Maximum velocity of sliding, vS = %.1f mm/s.\"%(vS)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Angle turned through by pinion when one pair of teeth is in mesh = 29.43 degrees.\n", + " Maximum velocity of sliding, vS = 455.3 mm/s.\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.5 Page No : 400" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "T = 40.\n", + "t = 20.\n", + "N1 = 2000. \t\t\t#rpm\n", + "phi = 20. \t\t\t#degrees\n", + "addendum = 5.\n", + "m = 5. \t\t\t#mm\n", + "\n", + "#Solution:\n", + "#Calculating the angular velocity of the smaller gear\n", + "omega1 = 2*math.pi*N1/60 \t\t\t#rad/s\n", + "#Calculating the angular velocity of the larger gear\n", + "omega2 = omega1*t/T \t\t\t#rad/s\n", + "#Calculating the pitch circle radius of the smaller gear\n", + "r = m*t/2 \t\t\t#mm\n", + "#Calculating the pitch circle radius of the larger gear\n", + "R = m*T/2 \t\t\t#mm\n", + "#Calculating the radius of aaddendum circle of smaller gear\n", + "rA = r+addendum \t\t\t#mm\n", + "#Calculating the radius of addendum circle of larger gear\n", + "RA = R+addendum \t\t\t#mm\n", + "#Calculating the length of path of approach\n", + "KP = math.sqrt(RA**2-R**2*(math.cos(math.radians(phi)))**2)-R*math.sin(math.radians(phi)) \t\t\t#mm\n", + "#Calculating the length of path of recess\n", + "PL = math.sqrt(rA**2-r**2*(math.cos(math.radians(phi)))**2)-r*math.sin(math.radians(phi)) \t\t\t#mm\n", + "#Calculating the velocity of sliding at the point of engagement\n", + "vSK = (omega1+omega2)*KP \t\t\t#mm/s\n", + "#Calculating the velocity of sliding at the point of disengagement\n", + "vSL = (omega1+omega2)*PL \t\t\t#mm/s\n", + "#Angle through which the pinion turns:\n", + "#Calculating the length of path of contact\n", + "KL = KP+PL \t\t\t#mm\n", + "#Calculating the length of arc of contact\n", + "Lac = KL/math.cos(math.radians(phi)) \t\t\t#Length of arc of contact mm\n", + "#Calculating the circumference of pinion\n", + "C = 2*math.pi*r \t\t\t#Circumference of pinion mm\n", + "#Calculating the angle through which the pinion turns\n", + "angle = Lac*360/C \t\t\t#Angle through which the pinion turns degrees\n", + "\n", + "#Results:\n", + "print \" Velocity of sliding at the point of engagement, vSK = %.f mm/s.\"%(vSK)\n", + "print \" Velocity of sliding at the point of disengagement, vsL = %.f mm/s.\"%(vSL)\n", + "print \" Angle through which the pinion turns = %.2f degrees.\"%(angle)\n", + "\n", + "# answers differ due to rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Velocity of sliding at the point of engagement, vSK = 3973 mm/s.\n", + " Velocity of sliding at the point of disengagement, vsL = 3610 mm/s.\n", + " Angle through which the pinion turns = 29.43 degrees.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.6 Page No : 401" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# Variables:\n", + "phi = 20. \t\t\t#degrees\n", + "m = 6.\n", + "addendum = 1*m \t\t\t#mm\n", + "t = 17.\n", + "T = 49.\n", + "\n", + "#Solution:\n", + "#Number of pairs of teeth in contact:\n", + "#Calculating the pitch circle radius of pinion\n", + "r = m*t/2 \t\t\t#mm\n", + "#Calculating the pitch circle radius of gear\n", + "R = m*T/2 \t\t\t#mm\n", + "#Calculating the radius of addendum circle of pinion\n", + "rA = r+addendum \t\t\t#mm\n", + "#Calculating the radius of addendum circle of gear\n", + "RA = R+addendum \t\t\t#mm\n", + "#Calculating the length of path of approach\n", + "#Refer Fig. 12.11\n", + "KP = math.sqrt(RA**2-R**2*(math.cos(math.radians(phi)))**2)-R*math.sin(math.radians(phi)) \t\t\t#mm\n", + "#Calculating the length of path of recess\n", + "PL = math.sqrt(rA**2-r**2*(math.cos(math.radians(phi)))**2)-r*math.sin(math.radians(phi)) \t\t\t#mm\n", + "#Calculating the length of path of contact\n", + "KL = KP+PL \t\t\t#mm\n", + "#Calculating the length of arc of contact\n", + "Lac = KL/math.cos(math.radians(phi)) \t\t\t#Length of arc of contact mm\n", + "#Calculating the circular pitch\n", + "pc = math.pi*m \t\t\t#mm\n", + "#Calculating the number of pairs of teeth in contact\n", + "n = Lac/pc \t\t\t#Number of pairs of teeth in contact\n", + "#Angle turned by the pinion and gear wheel when one pair of teeth is in contact:\n", + "#Calculating the angle turned through by the pinion\n", + "anglep = Lac*360/(2*math.pi*r) \t\t\t#Angle turned through by the pinion degrees\n", + "#Calculating the angle turned through by the wheel\n", + "angleg = Lac*360/(2*math.pi*R) \t\t\t#Angle turned through by the gear wheel degrees\n", + "#Ratio of sliding to rolling motion:\n", + "#At the instant when the tip of a tooth on the larger wheel is just making contact with its mating teeth\n", + "r1 = ((1+t/T)*KP)/r \t\t\t#Ratio of sliding velocity to rolling velocity\n", + "#At the instant when the tip of a tooth on a larger wheel is just leaving contact with its mating teeth\n", + "r2 = ((1+t/T)*PL)/r \t\t\t#Ratio of sliding velocity to rolling velocity\n", + "\n", + "\n", + "#Results:\n", + "print \" Number of pairs of teeth in contact = %.f.\"%(n)\n", + "print \" Angle turned through by the pinion = %.1f degrees.\"%(anglep)\n", + "print \" Angle turned through by the gear wheel = %.f degrees.\"%(angleg)\n", + "print \" At the instant when the tip of a tooth on the larger wheel is just\\\n", + " making contact with its mating teeth, ratio of sliding \\nvelocity to rolling velocity = %.2f.\"%(r1)\n", + "print \" At the instant when the tip of a tooth on a larger wheel is just leaving contact\\\n", + " with its mating teeth, ratio of sliding velocity to rolling velocity = %.3f.\"%(r2)\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Number of pairs of teeth in contact = 2.\n", + " Angle turned through by the pinion = 34.6 degrees.\n", + " Angle turned through by the gear wheel = 12 degrees.\n", + " At the instant when the tip of a tooth on the larger wheel is just making contact with its mating teeth, ratio of sliding \n", + "velocity to rolling velocity = 0.41.\n", + " At the instant when the tip of a tooth on a larger wheel is just leaving contact with its mating teeth, ratio of sliding velocity to rolling velocity = 0.354.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.7 Page No : 403" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "t = 18.\n", + "T = 72.\n", + "phi = 20. \t\t\t#degrees\n", + "m = 4. \t\t\t #mm\n", + "addendump = 8.5 \t\t\t#Addendum on pinion mm\n", + "addendumg = 3.5 \t\t\t#Addendum on gear mm\n", + "\n", + "#SOlution:\n", + "#Refer Fig. 12.12\n", + "#Calculating the pitch circle radius of the pinion\n", + "r = m*t/2 \t\t\t#mm\n", + "#Calculating the pitch circle radius of the gear\n", + "R = m*T/2 \t\t\t#mm\n", + "#Calculating the radius of addendum circle of the pinion\n", + "rA = r+addendump \t\t\t#mm\n", + "#Calculating the radius of addendum circle of the gear\n", + "RA = R-addendumg \t\t\t#mm\n", + "#Calculating the radius of the base circle of the pinion\n", + "O1M = r*math.cos(math.radians(phi)) \t\t\t#mm\n", + "#Calculating the radius of the base circle of the gear\n", + "O2N = R*math.cos(math.radians(phi)) \t\t\t#mm\n", + "#Calculating the length of path of approach\n", + "KP = R*math.sin(math.radians(phi))-math.sqrt(RA**2-O2N**2) \t\t\t#mm\n", + "#Calculating the length of path of recess\n", + "PL = math.sqrt(rA**2-O1M**2)-r*math.sin(math.radians(phi)) \t\t\t#mm\n", + "#Calculating the length of the path of contact\n", + "KL = KP+PL \t\t\t#mm\n", + "\n", + "#Results:\n", + "print \" Length of the path of contact, KL = %.2f mm.\"%(KL)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Length of the path of contact, KL = 28.04 mm.\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.8 Page No : 406" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "t = 20.\n", + "T = 40.\n", + "m = 10. \t\t\t#mm\n", + "phi = 20. \t\t\t#degrees\n", + "\n", + "#Solution:\n", + "#Addendum height for each gear wheel:\n", + "#Calculating the pitch circle radius of the smaller gear wheel\n", + "r = m*t/2 \t\t\t#mm\n", + "#Calculating the pitch circle radius of the larger wheel\n", + "R = m*T/2 \t\t\t#mm\n", + "#Calculating the radius of addendum circle for the larger gear wheel\n", + "RA = math.sqrt((r*math.sin(math.radians(phi))/2+R*math.sin(math.radians(phi)))**2+R**2*(math.cos(math.radians(phi)))**2) \t\t\t#mm\n", + "#Calculating the addendum height for larger gear wheel\n", + "addendumg = RA-R \t\t\t#mm\n", + "#Calculating the radius of addendum circle for the smaller gear wheel\n", + "rA = math.sqrt((R*math.sin(math.radians(phi))/2+r*math.sin(math.radians(phi)))**2+r**2*(math.cos(math.radians(phi)))**2) \t\t\t#mm\n", + "#Calculating the addendum height for smaller gear wheel\n", + "addendump = rA-r \t\t\t#mm\n", + "#Calculating the length of the path of contact\n", + "Lpc = (r+R)*math.sin(math.radians(phi))/2 \t\t\t#Length of the path of contact mm\n", + "#Calculating the length of the arc of contact\n", + "Lac = Lpc/math.cos(math.radians(phi)) \t\t\t#Length of the arc of contact mm\n", + "#Contact ratio:\n", + "#Calculating the circular pitch\n", + "pc = math.pi*m \t\t\t#mm\n", + "#Calculating the contact ratio\n", + "CR = Lpc/pc \t\t\t#Contact ratio\n", + "\n", + "#Results:\n", + "print \" Addendum height for larger gear wheel = %.1f mm.\"%(addendumg)\n", + "print \" Addendum height for smaller gear wheel = %.1f mm.\"%(addendump)\n", + "print \" Length of the path of contact = %.1f mm.\"%(Lpc)\n", + "print \" Length of the arc of contact = %.1f mm.\"%(Lac)\n", + "print \" Contact ratio = %d.\"%(CR+1)\n", + "\n", + "# book answer is wrong for 2nd " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Addendum height for larger gear wheel = 6.5 mm.\n", + " Addendum height for smaller gear wheel = 16.2 mm.\n", + " Length of the path of contact = 51.3 mm.\n", + " Length of the arc of contact = 54.6 mm.\n", + " Contact ratio = 2.\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.9 Page No : 410" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "G = 3.\n", + "phi = 20. \t\t\t#degrees\n", + "Aw = 1. \t\t\t#module\n", + "\n", + "#Solution:\n", + "#Calculating the minimum number of teeth for a gear ratio of 3:1\n", + "t1 = (2*Aw)/(G*(math.sqrt(1+1/G*(1/G+2)*(math.sin(math.radians(phi)))**2)-1))\n", + "#Calculating the minimum number of teeth for equal wheel\n", + "t2 = (2*Aw)/(math.sqrt(1+3*(math.sin(math.radians(phi)))**2)-1)\n", + "\n", + "#Results:\n", + "print \" Minimum number of teeth for a gear ratio of 3:1, t = %.f.\"%(t1+1)\n", + "print \" Minimum number of teeth for equal wheel, t = %d.\"%(t2+1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Minimum number of teeth for a gear ratio of 3:1, t = 16.\n", + " Minimum number of teeth for equal wheel, t = 13.\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.10 Page No : 410" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "import numpy\n", + "\n", + "# Variables:\n", + "G = 4.\n", + "phi = 14.5 \t\t\t#degrees\n", + "\n", + "#Solution:\n", + "#Least number of teeth on each wheel:\n", + "#Calculating the least number of teeth on the pinion\n", + "t = 2*math.pi/(math.tan(math.radians(phi)))\n", + "#Calculating the least number of teeth on the gear\n", + "T = G*t\n", + "\n", + "#Results:\n", + "print \" Least number of teeth on the pinion, t = %.1f.\"%(t)\n", + "print \" Least number of teeth on the gear, T = %.f.\"%(round(T,-1))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Least number of teeth on the pinion, t = 24.3.\n", + " Least number of teeth on the gear, T = 100.\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.11 Page No : 411" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "phi = 16. \t\t\t#degrees\n", + "m = 6. \t\t\t #mm\n", + "t = 16.\n", + "G = 1.75\n", + "T = G*t\n", + "N1 = 240. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the pinion\n", + "omega1 = 2*math.pi*N1/60 \t\t\t#rad/s\n", + "#Addenda on pinion and gear wheel:\n", + "#Calculating the addendum on pinion\n", + "addendump = m*t/2*(math.sqrt(1+T/t*(T/t+2)*(math.sin(math.radians(phi)))**2)-1) \t\t\t#Addendum on pinion mm\n", + "#Calculating the addendum on wheel\n", + "addendumg = m*T/2*(math.sqrt(1+t/T*(t/T+2)*(math.sin(math.radians(phi)))**2)-1) \t\t\t#Addendum on wheel mm\n", + "#Length of path of contact:\n", + "#Calculating the pitch circle radius of wheel\n", + "R = m*T/2 \t\t\t#mm\n", + "#Calculating the pitch circle radius of pinion\n", + "r = m*t/2 \t\t\t#mm\n", + "#Calculating the addendum circle radius of wheel\n", + "RA = R+addendump \t\t\t#mm\n", + "#Calculating the addendum circle radius of pinion\n", + "rA = r+addendumg \t\t\t#mm\n", + "#Calculating the length of path of approach\n", + "KP = math.sqrt(RA**2-R**2*(math.cos(math.radians(phi)))**2)-R*math.sin(math.radians(phi)) \t\t\t#mm\n", + "#Calculating the length of path of recess\n", + "PL = math.sqrt(rA**2-r**2*(math.cos(math.radians(phi)))**2)-r*math.sin(math.radians(phi)) \t\t\t#mm\n", + "#Calculating the length of path of contact\n", + "KL = KP+PL \t\t\t#mm\n", + "#Maximum velocity of sliding of teeth on either side of pitch point:\n", + "#Calculating the angular speed of gear wheel\n", + "omega2 = omega1/G \t\t\t#rad/s\n", + "#Calculating the maximum velocity of sliding of teeth on the left side of pitch point\n", + "vmaxl = (omega1+omega2)*KP \t\t\t#Maximum velocity of sliding of teeth on the left side of pitch point mm/s\n", + "#Calculating the maximum velocity of sliding of teeth on the right side of pitch point\n", + "vmaxr = (omega1+omega2)*PL \t\t\t#Maximum velocity of sliding of teeth on the right side of pitch point mm/s\n", + "\n", + "#Results:\n", + "print \" Addendum on pinion = %.2f mm.\"%(addendump)\n", + "print \" Addendum on wheel = %.2f mm.\"%(addendumg)\n", + "print \" Length of path of contact, KL = %.2f mm.\"%(KL)\n", + "print \" Maximum velocity of sliding of teeth on the left side of pitch point = %d mm/s.\"%(vmaxl)\n", + "print \" Maximum velocity of sliding of teeth on the right side of pitch point = %d mm/s.\"%(vmaxr)\n", + "\n", + "# rounding error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Addendum on pinion = 10.76 mm.\n", + " Addendum on wheel = 4.56 mm.\n", + " Length of path of contact, KL = 38.39 mm.\n", + " Maximum velocity of sliding of teeth on the left side of pitch point = 1044 mm/s.\n", + " Maximum velocity of sliding of teeth on the right side of pitch point = 471 mm/s.\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.12 Page No : 412" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "phi = 20. \t\t\t#degrees\n", + "t = 30.\n", + "T = 50.\n", + "m = 4.\n", + "N1 = 1000. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of thr pinion\n", + "omega1 = 2*math.pi*N1/60 \t\t\t#rad/s\n", + "#Sliding velocities at engagement and at disengagement of a pair of teeth:\n", + "#Calculating the addendum of the smaller gear\n", + "addendump = m*t/2*(math.sqrt(1+T/t*(T/t+2)*(math.sin(math.radians(phi)))**2)-1) \t\t\t#Addendum of the smaller gear mm\n", + "#Calculating the addendum of the larger gear\n", + "addendumg = m*T/2*(math.sqrt(1+t/T*(t/T+2)*(math.sin(math.radians(phi)))**2)-1) \t\t\t#Addendum of the larger gear mm\n", + "#Calculating the pitch circle radius of the smaller gear\n", + "r = m*t/2 \t\t\t#mm\n", + "#Calculating the radius of addendum circle of the smaller gear\n", + "rA = r+addendump \t\t\t#mm\n", + "#Calculating the pitch circle radius of the larer gear\n", + "R = m*T/2 \t\t\t#mm\n", + "#Calculating the radius of addendum circle of the larger gear\n", + "RA = R+addendumg \t\t\t#mm\n", + "#Calculating the path of approach\n", + "KP = math.sqrt(RA**2-R**2*(math.cos(math.radians(phi)))**2)-R*math.sin(math.radians(phi)) \t\t\t#mm\n", + "#Calculating the path of recess\n", + "PL = math.sqrt(rA**2-r**2*(math.cos(math.radians(phi)))**2)-r*math.sin(math.radians(phi)) \t\t\t#mm\n", + "#Calculating the angular speed of the larger gear\n", + "omega2 = omega1*t/T \t\t\t#rad/s\n", + "#Calculating the sliding velocity at engagement of a pair of teeth\n", + "v1 = (omega1+omega2)*KP \t\t\t#Sliding velocity at engagement of a pair of teeth mm/s\n", + "#Calculating the sliding velocity at disengagement of a pair of teeth\n", + "v2 = (omega1+omega2)*PL \t\t\t#Sliding velocity at disengagement of a pair of teeth mm/s\n", + "#Contact ratio:\n", + "#Calculating the length of the arc of contact\n", + "Lac = (KP+PL)/math.cos(math.radians(phi)) \t\t\t#mm\n", + "#Calculating the circular pitch\n", + "pc = math.pi*m \t\t\t#Circular pitch mm\n", + "#Calculating the contact ratio\n", + "CR = Lac/pc \t\t\t#Contact ratio\n", + "\n", + "#Results:\n", + "print \" Sliding velocity at engagement of a pair of teeth = %.3f m/s.\"%(v1/1000)\n", + "print \" Sliding velocity at disengagement of a pair of teeth = %.3f m/s.\"%(v2/1000)\n", + "print \" Contact ratio = %d.\"%(CR+1)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Sliding velocity at engagement of a pair of teeth = 3.438 m/s.\n", + " Sliding velocity at disengagement of a pair of teeth = 5.731 m/s.\n", + " Contact ratio = 5.\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.13 Page No : 414" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "G = 3.\n", + "m = 6.\n", + "AP = 1*m\n", + "AW = AP \t\t\t#mm\n", + "phi = 20. \t\t\t#degrees\n", + "N1 = 90. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the pinion\n", + "omega1 = 2*math.pi*N1/60 \t\t\t#rad/s\n", + "#Calculating the number of teeth on the pinion to avoid interference on it\n", + "t = 2*AP/(math.sqrt(1+G*(G+2)*(math.sin(math.radians(phi)))**2)-1)\n", + "#Calculating the corresponding number of teeth on the wheel\n", + "T = G*t\n", + "#Length of path and arc of contact:\n", + "#Calculating the pitch circle radius of pinion\n", + "r = m*t/2 \t\t\t#mm\n", + "#Calculating the radius of addendum circle of pinion\n", + "rA = r+AP \t\t\t#mm\n", + "#Calculating the pitch circle radius of wheel\n", + "R = m*T/2 \t\t\t#mm\n", + "#Calculating the radius of addendum circle of wheel\n", + "RA = R+AW \t\t\t#mm\n", + "#Calculating the path of approach\n", + "KP = math.sqrt(RA**2-R**2*(math.cos(math.radians(phi)))**2)-R*math.sin(math.radians(phi)) \t\t\t#mm\n", + "#Calculating the path of recess\n", + "PL = math.sqrt(rA**2-r**2*(math.cos(math.radians(phi)))**2)-r*math.sin(math.radians(phi)) \t\t\t#mm\n", + "#Calculating the length of path of contact\n", + "KL = KP+PL \t\t\t#mm\n", + "#Calculating the length of arc of contact\n", + "Lac = KL/math.cos(math.radians(phi)) \t\t\t#Length of arc of contact mm\n", + "#Number of pairs of teeth in contact:\n", + "#Calculating the circular pitch\n", + "pc = math.pi*m \t\t\t#mm\n", + "#Calculating the number of pairs of teeth in contact\n", + "n = Lac/pc \t\t\t#Number of pairs of teeth in contact\n", + "#Maximum velocity of sliding:\n", + "#Calculating the angular speed of wheel\n", + "omega2 = omega1*t/T \t\t\t#rad/s\n", + "#Calculating the maximum velocity of sliding\n", + "vs = (omega1+omega2)*KP \t\t\t#mm/s\n", + "\n", + "#Results:\n", + "print \" Number of teeth on the pinion to avoid interference, t = %d.\"%(t+1)\n", + "print \" Corresponding number of teeth on the wheel, T = %.F.\"%(T+1)\n", + "print \" Length of path of contact, KL = %.2f mm.\"%(KL)\n", + "print \" Length of arc of contact = %.2f mm.\"%(Lac)\n", + "print \" Number of pairs of teeth in contact = %d.\"%(n+1)\n", + "print \" Maximum velocity of sliding, vs = %.f mm/s.\"%(vs)\n", + "\n", + "# ROUNDING ERROR" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Number of teeth on the pinion to avoid interference, t = 19.\n", + " Corresponding number of teeth on the wheel, T = 56.\n", + " Length of path of contact, KL = 29.24 mm.\n", + " Length of arc of contact = 31.12 mm.\n", + " Number of pairs of teeth in contact = 2.\n", + " Maximum velocity of sliding, vs = 197 mm/s.\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.14 Page No : 416" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "T = 20.\n", + "d = 125. #mm\n", + "r = d/2\n", + "OP = r\n", + "LH = 6.25 \t\t\t#mm\n", + "#Calculating the least pressure angle to avoid interference\n", + "phi = math.sin(math.sqrt(LH/r))*180/math.pi \t\t\t#degrees\n", + "#Length of arc of contact:\n", + "#Calculating the length of path of contact\n", + "KL = math.sqrt((OP+LH)**2-(OP*math.cos(math.radians(phi)))**2) \t\t\t#mm\n", + "#Calculating the length of arc of contact\n", + "Lac = KL/math.cos(math.radians(phi)) \t\t\t#Length of arc of contact mm\n", + "#Minimum number of teeth:\n", + "#Calculating the circular pitch\n", + "pc = math.pi*d/T \t\t\t#mm\n", + "#Calculating the number of pairs of teeth in contact\n", + "n = Lac/pc \t\t\t#Number of pairs of teeth in contact\n", + "#Calculating the minimum number of teeth in contact\n", + "nmin = n \t\t\t#Mimimum number of teeth in contact\n", + "\n", + "#Results:\n", + "print \" Least pressure angle to avoid interference, phi = %.3f degrees.\"%(phi)\n", + "print \" Length of arc of contact = %.2f mm.\"%(Lac)\n", + "print \" Minimum number of teeth in contact = %d or %d pair.\"%(nmin+1,(nmin+1)/2)\n", + "\n", + "# rounding error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Least pressure angle to avoid interference, phi = 17.818 degrees.\n", + " Length of arc of contact = 36.17 mm.\n", + " Minimum number of teeth in contact = 2 or 1 pair.\n" + ] + } + ], + "prompt_number": 41 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.15 Page No : 421" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "from scipy.optimize import fsolve \n", + "import math \n", + "\n", + "# Variables:\n", + "L = 175./1000\n", + "d2 = 100./1000 #m\n", + "r2 = d2/2 \t\t\t #m\n", + "theta = 70. \t\t\t#degrees\n", + "G = 1.5\n", + "T2 = 80.\n", + "Tf = 75. \t\t\t#Torque on faster wheel N-m\n", + "\n", + "#Solution:\n", + "#Spiral angles for each wheel:\n", + "#Calculating the number of teeth on slower wheel\n", + "T1 = T2*G\n", + "#Calculating the pitch circle diameter of the slower wheel\n", + "d1 = (L*2)-d2 \t\t\t#m\n", + "#Calculating the spiral angles\n", + "#We have d2/d1 = (T2*math.cos(alpha1))/(T1*math.cos(alpha2)) or T2*d1*math.cos(alpha1)-T1*d2*math.cos(alpha2) = 0 .....(i)\n", + "#Also alpha1+alpha2 = theta or alpha1+alpha2-theta = 0 .....(ii)\n", + "def f(x):\n", + " alpha1 = x[0]\n", + " alpha2 = x[1]\n", + " y = [0,0]\n", + " y[0] = T2*d1*math.cos(alpha1)-T1*d2*math.cos(alpha2)\n", + " y[1] = alpha1+alpha2-theta*math.pi/180\n", + " return y\n", + " \n", + "z = fsolve(f,[1,1])\n", + "alpha1 = z[0]*180/math.pi \t\t\t#Spiral angle for slower wheel degrees \n", + "alpha2 = z[1]*180/math.pi \t\t\t#Spiral angle for faster wheel degrees\n", + "#Axial thrust on each shaft:\n", + "#Calculating the math.tangential force at faster wheel\n", + "F2 = Tf/r2 \t\t\t#N\n", + "#Calculating the normal reaction at the point of contact\n", + "RN = F2/math.cos(math.radians(alpha2)) \t\t\t#N\n", + "#Calculating the axial thrust on the shaft of slower wheel\n", + "Fa1 = RN*math.sin(math.radians(alpha1)) \t\t\t#N\n", + "#Calculating the axial thrust on the shaft of faster wheel\n", + "Fa2 = RN*math.sin(math.radians(alpha2)) \t\t\t#N\n", + "\n", + "#Results:\n", + "print \" Spiral angle for slower wheel, alpha1 = %.2f degrees.\"%(alpha1)\n", + "print \" Spiral angle for faster wheel, alpha2 = %.2f degrees.\"%(alpha2)\n", + "print \" Axial thrust on the shaft of slower wheel, Fa1 = %d N.\"%(Fa1+1)\n", + "print \" Axial thrust on the shaft of faster wheel, Fa2 = %d N.\"%(Fa2+1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Spiral angle for slower wheel, alpha1 = 54.65 degrees.\n", + " Spiral angle for faster wheel, alpha2 = 15.35 degrees.\n", + " Axial thrust on the shaft of slower wheel, Fa1 = 1269 N.\n", + " Axial thrust on the shaft of faster wheel, Fa2 = 412 N.\n" + ] + } + ], + "prompt_number": 45 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.16 Page No : 422" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "L = 400./1000 \t\t\t#m\n", + "G = 3.\n", + "theta = 50.\n", + "phi = 6. \t\t\t#degrees\n", + "pN = 18. \t\t\t#mm\n", + "\n", + "#Solution:\n", + "#Number of teeth on each wheel:\n", + "#Calculating the spiral angles of the driving and driven wheels\n", + "alpha1 = theta/2 \t\t\t#degrees\n", + "alpha2 = alpha1 \t\t\t#degrees\n", + "#Calculating the number of teeth on driver wheel\n", + "T1 = L*1000*2*math.pi/(pN*(1/math.cos(math.radians(alpha1))+G/math.cos(math.radians(alpha2))))\n", + "#Calculating the number of teeth on driven wheel\n", + "T2 = G*T1\n", + "#Calculating the exact centre distance\n", + "#L1 = pN*T1/(2*math.pi)*(1/math.cos(math.radians(alpha1))+G/math.cos(math.radians(alpha2))) \t\t\t#mm\n", + "L1 = pN*T1/(2*math.pi)*((1+G)/math.cos(math.radians(alpha1))) \t\t\t#mm\n", + "#Calculating the efficiency of the drive\n", + "eta = (math.cos(math.radians(alpha2+phi))*math.cos(math.radians(alpha1)))/(math.cos(math.radians(alpha1-phi))*math.cos(math.radians(alpha2)))*100 \t\t\t#%\n", + "\n", + "#Results:\n", + "print \" Number of teeth on driver wheel, T1 = %d.\"%(T1+1)\n", + "print \" Number of teeth on driven wheel, T2 = %.f.\"%(T2+1)\n", + "print \" Exact centre distance, L1 = %.1f mm.\"%(L1)\n", + "print \" Efficiency of the drive, eta = %.1f %%.\"%(eta)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Number of teeth on driver wheel, T1 = 32.\n", + " Number of teeth on driven wheel, T2 = 96.\n", + " Exact centre distance, L1 = 400.0 mm.\n", + " Efficiency of the drive, eta = 90.7 %.\n" + ] + } + ], + "prompt_number": 51 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.17 Page No : 423" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "from scipy.optimize import fsolve \n", + "import math \n", + "\n", + "# Variables:\n", + "pN = 12.5\n", + "L = 134. \t\t\t#mm\n", + "theta = 80.\n", + "phi = 6. \t\t\t#degrees\n", + "G = 1.25\n", + "\n", + "#Solution:\n", + "#Spiral angle of each wheel:\n", + "#Calculating the spiral angles of wheels 1 and 2\n", + "#We have d2/d1 = (T2*math.cos(alpha1))/(T1*math.cos(alpha2)) or math.cos(alpha1)-G*math.cos(alpha2) = 0 .....(i)\n", + "#Also alpha1+alpha2 = theta or alpha1+alpha2-theta = 0 .....(ii)\n", + "def f(x):\n", + " alpha1 = x[0]\n", + " alpha2 = x[1]\n", + " y = [0,0]\n", + " y[0] = math.cos(alpha1)-G*math.cos(alpha2)\n", + " y[1] = alpha1+alpha2-theta*math.pi/180\n", + " return y\n", + "\n", + "z = fsolve(f,[1,1])\n", + "alpha1 = z[0]*180/math.pi \t\t\t#Spiral angle for slower wheel degrees\n", + "alpha2 = z[1]*180/math.pi \t\t\t#Spiral angle for faster wheel degrees\n", + "#Number of teeth on each wheel:\n", + "#Calculating the diameters of the wheels\n", + "d1 = L\n", + "d2 = d1 \t\t\t#mm\n", + "#Calculating the number of teeth on wheel 1\n", + "T1 = d1*math.pi*math.cos(math.radians(alpha1))/pN\n", + "#Calculating the number of teeth on wheel 2\n", + "T2 = T1/G\n", + "#Calculating the efficiency of the drive\n", + "eta = (math.cos(math.radians(alpha2+phi))*math.cos(math.radians(alpha1)))/(math.cos(math.radians(alpha1-phi))*math.cos(math.radians(alpha2)))*100 \t\t\t#%\n", + "#Calculating the maximum efficiency\n", + "etamax = (math.cos(math.radians(theta+phi))+1)/(math.cos(math.radians(theta-phi))+1)*100 \t\t\t#%\n", + "\n", + "#Results:\n", + "print \" Spiral angle for slower wheel, alpha1 = %.2f degrees.\"%(alpha1)\n", + "print \" Spiral angle for faster wheel, alpha2 = %.2f degrees.\"%(alpha2)\n", + "print \" Number of teeth on wheel 1, T1 = %.1f.\"%(T1)\n", + "print \" Number of teeth on wheel 2, T2 = %.f.\"%(T2+1)\n", + "print \" Efficiency of the drive, eta = %d %%.\"%(eta+1)\n", + "print \" Maximum efficiency, etamax = %.1f %%.\"%(etamax)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Spiral angle for slower wheel, alpha1 = 32.46 degrees.\n", + " Spiral angle for faster wheel, alpha2 = 47.54 degrees.\n", + " Number of teeth on wheel 1, T1 = 28.4.\n", + " Number of teeth on wheel 2, T2 = 24.\n", + " Efficiency of the drive, eta = 83 %.\n", + " Maximum efficiency, etamax = 83.9 %.\n" + ] + } + ], + "prompt_number": 57 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Theory_Of_Machines/ch13.ipynb b/Theory_Of_Machines/ch13.ipynb new file mode 100755 index 00000000..d6a66198 --- /dev/null +++ b/Theory_Of_Machines/ch13.ipynb @@ -0,0 +1,1411 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:e7234bd223a2d57f8ee3c0d73106a260e10c7534a13d5c3bf25633720527eb77" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 13 : Gear Trains" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.1 Page No : 432" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables:\n", + "NA = 975 \t\t\t#rpm\n", + "TA = 20.\n", + "TB = 50.\n", + "TC = 25\n", + "TD = 75\n", + "TE = 26\n", + "TF = 65\n", + "\n", + "#Solution:\n", + "#Calculating the speed of gear F\n", + "NF = NA*(TA*TC*TE)/(TB*TD*TF) \t\t\t#rpm\n", + "\n", + "#Results:\n", + "print \" Speed of gear F, NF = %d rpm.\"%(NF)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Speed of gear F, NF = 52 rpm.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.2 Page No : 433" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "from numpy import linalg\n", + "\n", + "# Variables:\n", + "x = 600.\n", + "pc = 25. \t\t\t#mm\n", + "N1 = 360.\n", + "N2 = 120. \t\t\t#rpm\n", + "#Solution:\n", + "#Calculating the pitch circle diameters of each gear\n", + "#Speed ratio N1/N2 = d2/d1 or N1*d1-N2*d2 = 0 .....(i)\n", + "#Centre distance between the shafts x = 1/2*(d1+d2) or d1+d2 = 600*2 .....(ii)\n", + "A = [[N1, -N2],[ 1, 1]]\n", + "B = [0, 600*2]\n", + "V = linalg.solve(A,B)\n", + "d1 = V[0] \t\t\t#mm\n", + "d2 = V[1] \t\t\t#mm\n", + "#Calculating the number of teeth on the first gear\n", + "T1 = round(math.pi*d1/pc)\n", + "#Calculating the number of teeth on the second gear\n", + "T2 = int(math.pi*d2/pc+1)\n", + "#Calculating the pitch circle diameter of the first gear\n", + "d1dash = T1*pc/math.pi \t\t\t#mm\n", + "#Calculating the pitch circle diameter of the second gear\n", + "d2dash = T2*pc/math.pi \t\t\t#mm\n", + "#Calculating the exact distance between the two shafts\n", + "xdash = (d1dash+d2dash)/2 \t\t\t#mm\n", + "\n", + "#Results:\n", + "print \" The number of teeth on the first and second gear must be %d and %d and their pitch\\\n", + " circle diameters must be %.2f mm and %.1f mm respectively.\"%(T1,T2,d1dash,d2dash)\n", + "print \" The exact distance between the two shafts must be %.2f mm.\"%(xdash)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The number of teeth on the first and second gear must be 38 and 114 and their pitch circle diameters must be 302.39 mm and 907.2 mm respectively.\n", + " The exact distance between the two shafts must be 604.79 mm.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.3 Page No : 435" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "import math \n", + "\n", + "# Variables:\n", + "rAD = 12. \t\t\t#Speed ratio NA/ND\n", + "mA = 3.125 #mm\n", + "mB = mA #mm\n", + "mC = 2.5 #mm\n", + "mD = mC #mm\n", + "x = 200. \t\t\t#mm\n", + "\n", + "#Solution:\n", + "#Calculating the speed ratio between the gears A and B and C and D \n", + "rAB = math.sqrt(rAD) \t\t\t#Speed ratio between the gears A and B\n", + "rCD = math.sqrt(rAB) \t\t\t#Speed ratio between the gears C and D\n", + "#Calculating the ratio of teeth on gear B to gear A\n", + "rtBA = rAB \t\t\t#Ratio of teeth on gear B to gear A\n", + "#Calculating the ratio of teeth on gear D to gear C\n", + "rtDC = rCD \t\t\t#Ratio of teeth on gear D to gear C\n", + "#Calculating the number of teeth on the gears A and B\n", + "#Distance between the shafts x = mA*TA/2+mB*TB/2 or (mA/2)*TA+(mB/2)*TB = x .....(i)\n", + "#Ratio of teeth on gear B to gear A TB/TA = math.sqrt(12) or math.sqrt(12)*TA-TB = 0 .....(ii)\n", + "A = [[mA/2, mB/2],[math.sqrt(12) ,-1]]\n", + "B = [x, 0]\n", + "V = linalg.solve(A,B)\n", + "TA = int(V[0])\n", + "TB = round(V[1])\n", + "#Calculating the number of teeth on the gears C and D\n", + "#Dismath.tance between the shafts x = mC*TC/2+mD*TD/2 or (mC/2)*TC+(mD/2)*TD = x .....(iii)\n", + "#Ratio of teeth on gear D to gear C TD/TC = math.sqrt(12) or math.sqrt(12)*TC-TD = 0 .....(iv)\n", + "A = [[mC/2, mD/2],[ math.sqrt(12) ,-1]]\n", + "B = [x, 0]\n", + "V = linalg.solve(A,B)\n", + "TC = round(V[0])\n", + "TD = int(V[1])\n", + "\n", + "#Results:\n", + "print \" Number of teeth on gear A, TA = %d.\"%(TA)\n", + "print \" Number of teeth on gear B, TB = %d.\"%(TB)\n", + "print \" Number of teeth on gear C, TC = %d.\"%(TC)\n", + "print \" Number of teeth on gear D, TD = %d.\"%(TD)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Number of teeth on gear A, TA = 28.\n", + " Number of teeth on gear B, TB = 99.\n", + " Number of teeth on gear C, TC = 36.\n", + " Number of teeth on gear D, TD = 124.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.4 Page No : 438" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "TA = 36.\n", + "TB = 45.\n", + "NC = 150. \t\t\t#rpm anticlockwise\n", + "\n", + "#Solution:\n", + "#Refer Fig. 13.7\n", + "#Algebraic method:\n", + "#Calculating the speed of gear B when gear A is fixed\n", + "NA = 0.\n", + "NC = 150. \t\t\t#rpm\n", + "NB1 = (-TA/TB)*(NA-NC)+NC \t\t\t#rpm\n", + "#Calculating the speed of gear B when gear A makes 300 rpm clockwise\n", + "NA = -300. \t\t\t#rpm\n", + "NB2 = (-TA/TB)*(NA-NC)+NC \t\t\t#rpm\n", + "\n", + "#Results:\n", + "print \" Speed of gear B when gear A is fixed, NB = %d rpm.\"%(NB1)\n", + "print \" Speed of gear B when gear A makes 300 rpm clockwise, NB = %d rpm.\"%(NB2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Speed of gear B when gear A is fixed, NB = 270 rpm.\n", + " Speed of gear B when gear A makes 300 rpm clockwise, NB = 510 rpm.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.5 Page No : 440" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "TB = 75.\n", + "TC = 30.\n", + "TD = 90.\n", + "NA = 100. \t\t\t#rpm clockwise\n", + "\n", + "#Solution:\n", + "#Refer Table 13.3\n", + "#Calculating the number of teeth on gear E\n", + "TE = TC+TD-TB\n", + "#Calculating the speed of gear C\n", + "y = -100.\n", + "x = y*(TB/TE)\n", + "NC = y-x*(TD/TC) \t\t\t#rpm\n", + "\n", + "#Results:\n", + "print \" Speed of gear C, NC = %d rpm, anticlockwise.\"%(NC)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Speed of gear C, NC = 400 rpm, anticlockwise.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.6 Page No : 443" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "TA = 72.\n", + "TC = 32.\n", + "NEF = 18. \t\t\t#Speed of arm EF rpm\n", + "\n", + "#Solution:\n", + "#Refer Table 13.5\n", + "#Speed of gear C:\n", + "y = 18. \t\t\t#rpm\n", + "x = y*(TA/TC)\n", + "NC = x+y \t\t\t#Speed of gear C rpm\n", + "#Speed of gear B:\n", + "#Calculating the number of teeth on gear B\n", + "TB = (TA-TC)/2\n", + "#Calculating the speed of gear B\n", + "NB = y-x*(TC/TB) \t\t\t#Speed of gear B rpm\n", + "\n", + "#Solution:\n", + "print \" Speed of gear C = %.1f rpm.\"%(NC)\n", + "print \" Speed of gear B = %.1f rpm in the opposite direction of arm.\"%(-NB)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Speed of gear C = 58.5 rpm.\n", + " Speed of gear B = 46.8 rpm in the opposite direction of arm.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.7 Page No : 444" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "import math \n", + "\n", + "# Variables:\n", + "TA = 40.\n", + "TD = 90.\n", + "\n", + "#Solution:\n", + "#Calculating the number of teeth on gears B and C\n", + "#From geometry of the Fig. 13.11 dA+2*dB = dD\n", + "#Since the number of teeth are proportional to their pitch circle diameters\n", + "TB = (TD-TA)/2\n", + "TC = TB\n", + "#Refer Table 13.6\n", + "#Speed of arm when A makes 1 revolution clockwise and D makes half revolution anticlockwise:\n", + "#Calculating the values of x and y\n", + "#From the fourth row of the table -x-y = -1 or x+y = 1 .....(i)\n", + "#The gear D makes half revolution anticlockwise i.e. x*(TA/TD)-y = 1/2 .....(ii)\n", + "A = [[1, 1],[TA/TD, -1]]\n", + "B = [1, 1./2]\n", + "V = linalg.solve(A,B)\n", + "x = V[0]\n", + "y = V[1]\n", + "#Calculating the speed of arm\n", + "varm = -y \t\t\t#Speed of arm revolutions\n", + "\n", + "#Results:\n", + "print \" Speed of arm when A makes 1 revolution clockwise and D makes half revolution\\\n", + " anticlockwise = %.2f revolution anticlockwise.\"%(varm)\n", + "#Speed of arm when A makes 1 revolution clockwise and D is stationary:\n", + "#Calculating the values of x and y\n", + "#From the fourth row of the table -x-y = -1 or x+y = 1 .....(iii)\n", + "#The gear D is stationary i.e. x*(TA/TD)-y = 0 .....(iv)\n", + "A = [[1, 1],[ TA/TD, -1]]\n", + "B = [1, 0]\n", + "V = linalg.solve(A,B)\n", + "x = V[0]\n", + "y = V[1]\n", + "#Calculating the speed of arm\n", + "varm = -y \t\t\t#Speed of arm revolutions\n", + "\n", + "#Results:\n", + "print \" Speed of arm when A makes 1 revolution clockwise and D is stationary = %.3f revolution\\\n", + " clockwise.\"%(-varm)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Speed of arm when A makes 1 revolution clockwise and D makes half revolution anticlockwise = 0.04 revolution anticlockwise.\n", + " Speed of arm when A makes 1 revolution clockwise and D is stationary = 0.308 revolution clockwise.\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.8 Page No : 446" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "TC = 28.\n", + "TD = 26.\n", + "TE = 18.\n", + "TF = TE\n", + "\n", + "#Solution:\n", + "#The sketch is as in Fig. 13.12\n", + "#Number of teeth on wheels A and B:\n", + "#From geometry dA = dC+2*dE and dB = dD+2*dF\n", + "#Since the number of teeth are proportional to their pitch circle diameters\n", + "TA = TC+2*TE\n", + "TB = TD+2*TF\n", + "#Speed of wheel B when arm G makes 100 rpm clockwise and wheel A is fixed:\n", + "#Since the arm G makes 100 rpm clockwise therefore from the fourth row of Table 13.7\n", + "y = -100\n", + "x = -y\n", + "#Calculating the speed of wheel B\n", + "NB1 = y+x*(TA/TC)*(TD/TB) \t\t\t#Speed of wheel B when arm G makes 100 rpm clockwise and wheel A is fixed rpm\n", + "#Speed of wheel B when arm G makes 100 rpm clockwise and wheel A makes 10 rpm counter clockwise:\n", + "#Since the arm G makes 100 rpm clockwise therefore from the fourth row of Table 13.7\n", + "y = -100\n", + "x = 10-y\n", + "#Calculating the speed of wheel B\n", + "NB2 = y+x*(TA/TC)*(TD/TB) \t\t\t#Speed of wheel B when arm G makes 100 rpm clockwise and wheel A makes 10 rpm counter clockwise rpm\n", + "\n", + "#Solution:\n", + "print \" Number of teeth on wheel A, TA = %d.\"%(TA)\n", + "print \" Number of teeth on wheel B, TB = %d.\"%(TB)\n", + "print \" Speed of wheel B when arm G makes 100 rpm clockwise and wheel A is fixed = %.1f rpm, clockwise.\"%(-NB1)\n", + "print \" Speed of wheel B when arm G makes 100 rpm clockwise and wheel A makes 10 rpm counter\\\n", + " clockwise = %.1f rpm, counter clockwise.\"%(NB2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Number of teeth on wheel A, TA = 64.\n", + " Number of teeth on wheel B, TB = 62.\n", + " Speed of wheel B when arm G makes 100 rpm clockwise and wheel A is fixed = 4.1 rpm, clockwise.\n", + " Speed of wheel B when arm G makes 100 rpm clockwise and wheel A makes 10 rpm counter clockwise = 5.4 rpm, counter clockwise.\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.9 Page No : 447" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "dD = 224.\n", + "m = 4. \t\t\t#mm\n", + "\n", + "#Solution:\n", + "#Refer Table 13.8\n", + "#Calculating the values of x and y\n", + "y = 1.\n", + "x = 5-y\n", + "#Calculating the number of teeth on gear D\n", + "TD = dD/m\n", + "#Calculating the number of teeth on gear B\n", + "TB = y/x*TD\n", + "#Calculating the number of teeth on gear C\n", + "TC = (TD-TB)/2\n", + "\n", + "#Results:\n", + "print \" Number of teeth on gear D, TD = %d.\"%(TD)\n", + "print \" Number of teeth on gear B, TB = %d.\"%(TB)\n", + "print \" Number of teeth on gear C, TC = %d.\"%(TC)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Number of teeth on gear D, TD = 56.\n", + " Number of teeth on gear B, TB = 14.\n", + " Number of teeth on gear C, TC = 21.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.10 Page No : 448" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "import math \n", + "\n", + "# Variables:\n", + "TC = 50.\n", + "TD = 20.\n", + "TE = 35.\n", + "NA = 110. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Calculating the number of teeth on internal gear G\n", + "TG = TC+TD+TE\n", + "#Speed of shaft B:\n", + "#Calculating the values of x and y\n", + "#From the fourth row of Table 13.9\n", + "#y-x*(TC/TD)*(TE/TG) = 0 .....(i)\n", + "#Also x+y = 110 or y+x = 110 .....(ii)\n", + "A = [[1, -(TC/TD)*(TE/TG)],[ 1, 1]]\n", + "B = [0, 110]\n", + "V = linalg.solve(A,B)\n", + "x = V[1]\n", + "y = V[0]\n", + "#Calculating the speed of shaft B\n", + "NB = round(+y) \t\t\t#Speed of shaft B rpm\n", + "\n", + "#Results:\n", + "print \" Number of teeth on internal gear G, TG = %d.\"%(TG)\n", + "print \" Speed of shaft B = %d rpm, anticlockwise.\"%(NB)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Number of teeth on internal gear G, TG = 105.\n", + " Speed of shaft B = 50 rpm, anticlockwise.\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.11 Page No : 450" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "TA = 12.\n", + "TB = 30.\n", + "TC = 14.\n", + "NA = 1.\n", + "ND = 5. \t\t\t#rps\n", + "\n", + "#Solution:\n", + "#Number of teeth on wheels D and E:\n", + "#Calculating the number of teeth on wheel E\n", + "TE = TA+2*TB\n", + "#Calculating the number of teeth on wheel E\n", + "TD = TE-(TB-TC)\n", + "#Magnitude and direction of angular velocities of arm OP and wheel E:\n", + "#Calculating the values of x and y\n", + "#From the fourth row of Table 13.10 -x-y = -1 or x+y = 1 .....(i)\n", + "#Also x*(TA/TB)*(TC/TD)-y = 5 .....(ii)\n", + "A = [[1, 1],[(TA/TB)*(TC/TD) ,-1]]\n", + "B = [1, 5]\n", + "V = linalg.solve(A,B)\n", + "x = V[0]\n", + "y = V[1]\n", + "#Calculating the angular velocity of arm OP\n", + "omegaOP = -y*2*math.pi \t\t\t#Angular velocity of arm OP rad/s\n", + "#Calculating the angular velocity of wheel E\n", + "omegaE = (x*TA/TE-y)*2*math.pi \t\t\t#Angular velocity of wheel E rad/s\n", + "\n", + "#Results:\n", + "print \" Number of teeth on wheel E, TE = %d.\"%(TE)\n", + "print \" Number of teeth on wheel D, TD = %d.\"%(TD)\n", + "print \" Angular velocity of arm OP = %.3f rad/s, counter clockwise.\"%(omegaOP)\n", + "print \" Angular velocity of wheel E = %.2f rad/s, counter clockwise.\"%(omegaE)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Number of teeth on wheel E, TE = 72.\n", + " Number of teeth on wheel D, TD = 56.\n", + " Angular velocity of arm OP = 27.989 rad/s, counter clockwise.\n", + " Angular velocity of wheel E = 33.70 rad/s, counter clockwise.\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.12 Page No : 451" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "TB = 80.\n", + "TC = 82.\n", + "TD = 28.\n", + "NA = 500. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Calculating the number of teeth on wheel E\n", + "TE = TB+TD-TC\n", + "#Calculating the values of x and y\n", + "y = 800.\n", + "x = -y*(TE/TB)*(TC/TD)\n", + "#Calculating the speed of shaft F\n", + "NF = x+y \t\t\t#Speed of shaft F rpm\n", + "\n", + "#Results:\n", + "print \" Speed of shaft F = %d rpm, anticlockwise.\"%(NF)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Speed of shaft F = 38 rpm, anticlockwise.\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.13 Page No : 452" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# variables\n", + "\n", + "TA = 100. ; # Gear A teeth\n", + "TC = 101. ; # Gear C teeth\n", + "TD = 99. ; # Gear D teeth\n", + "TP = 20. # Gear planet teeth\n", + "y = 1\n", + "x = 0 - y\n", + "\n", + "# calculations\n", + "NC = y + x * TA/TC\n", + "ND = y + x * TA/TD\n", + "\n", + "# results\n", + "print \"revolutions of gear C : %.4f\"%NC\n", + "print \"revolutions of gear D : %.4f \"%ND\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "revolutions of gear C : 0.0099\n", + "revolutions of gear D : -0.0101 \n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.14 Page No : 453" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "import math \n", + "\n", + "# Variables:\n", + "NA = 300. \t\t\t#rpm\n", + "TD = 40.\n", + "TE = 30.\n", + "TF = 50.\n", + "TG = 80.\n", + "TH = 40.\n", + "TK = 20.\n", + "TL = 30.\n", + "\n", + "#Solution:\n", + "#Refer Fig. 13.18 and Table 13.13\n", + "#Calculating the speed of wheel E\n", + "NE = NA*(TD/TE) \t\t\t#rpm\n", + "#Calculating the number of teeth on wheel C\n", + "TC = TH+TK+TL\n", + "#Speed and direction of rotation of shaft B:\n", + "#Calculating the values of x and y\n", + "#We have -x-y = -400 or x+y = 400 .....(i)\n", + "#Also x*(TH/TK)*(TL/TC)-y = 0 .....(ii) \n", + "A = [[1, 1],[ (TH/TK)*(TL/TC), -1]]\n", + "B = [400, 0]\n", + "V = linalg.solve(A,B)\n", + "x = V[0]\n", + "y = V[1]\n", + "#Calculating the speed of wheel F\n", + "NF = -y \t\t\t#rpm\n", + "#Calculating the speed of shaft B\n", + "NB = -NF*(TF/TG) \t\t\t#Speed of shaft B rpm\n", + "\n", + "#Results:\n", + "print \" Number of teeth on wheel C, TC = %.1f.\"%(TC)\n", + "print \" Speed of shaft B = %.1f rpm, anticlockwise.\"%(NB)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Number of teeth on wheel C, TC = 90.0.\n", + " Speed of shaft B = 100.0 rpm, anticlockwise.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.15 Page No : 455" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "import math \n", + "\n", + "# Variables:\n", + "T1 = 80.\n", + "T8 = 160.\n", + "T4 = 100.\n", + "T3 = 120.\n", + "T6 = 20.\n", + "T7 = 66.\n", + "\n", + "#Solution:\n", + "#Refer Fig. 13.19 and Table 13.14\n", + "#Calculating the number of teeth on wheel 2\n", + "T2 = (T3-T1)/2\n", + "#Calculating the values of x and y\n", + "#Assuming that wheel 1 makes 1 rps anticlockwise x+y = 1 .....(i)\n", + "#Also y-x*(T1/T3) = 0 or x*(T1/T3)-y = 0 .....(ii)\n", + "A = [[1, 1],[ 1, T1/T3]]\n", + "B = [1, 0]\n", + "V = linalg.solve(A,B)\n", + "x = V[0]\n", + "y = V[1]\n", + "#Calculating the speed of casing C\n", + "NC = y \t\t\t#Speed of casing C rps\n", + "#Calculating the speed of wheel 2\n", + "N2 = y-x*(T1/T2) \t\t\t#Speed of wheel 2 rps\n", + "#Calculating the number of teeth on wheel 5\n", + "T5 = (T4-T6)/2\n", + "#Calculating the values of x1 and y1\n", + "y1 = -2\n", + "x1 = (y1-0.4)*(T4/T6)\n", + "#Calculating the speed of wheel 6\n", + "N6 = x1+y1 \t\t\t#Speed of wheel 6 rps\n", + "#Calculating the values of x2 and y2\n", + "y2 = 0.4\n", + "x2 = -(14+y2)*(T7/T8)\n", + "#Calculating the speed of wheel 8\n", + "N8 = x2+y2 \t\t\t#Speed of wheel 8 rps\n", + "#Calculating the velocity ratio of the output shaft B to the input shaft A\n", + "vr = N8/1 \t\t\t#Velocity ratio\n", + "\n", + "#Results:\n", + "print \" Velocity ratio of the output shaft B to the input shaft A = %.2f.\"%(vr)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Velocity ratio of the output shaft B to the input shaft A = -5.54.\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.16 Page No : 459" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "TA = 40.\n", + "TB = 30.\n", + "TC = 50.\n", + "NX = 100.\n", + "NA = NX \t\t\t #rpm\n", + "Narm = 100. \t\t\t#Speed of armrpm\n", + "\n", + "#Solution:\n", + "#Refer Fig. 13.22 and Table 13.18\n", + "#Calculating the values of x and y\n", + "y = +100\n", + "x = -100-y\n", + "#Calculating the speed of the driven shaft\n", + "NY = y-x*(TA/TB) \t\t\t#rpm\n", + "\n", + "#Results:\n", + "print \" Speed of the driven shaft, NY = %.1f rpm, anticlockwise.\"%(NY)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Speed of the driven shaft, NY = 366.7 rpm, anticlockwise.\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.17 Page No : 460" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "import math \n", + "\n", + "# Variables:\n", + "TB = 20.\n", + "TC = 80.\n", + "TD = 80.\n", + "TE = 30.\n", + "TF = 32.\n", + "NB = 1000. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Refer Fig. 13.23 and Table 13.19\n", + "#Speed of the output shaft when gear C is fixed:\n", + "#Calculating the values of x and y\n", + "#From the fourth row of the table y-x*(TB/TC) = 0 .....(i) \n", + "#Also x+y = +1000 or y+x = 1000 .....(ii)\n", + "A = [[1, -TB/TC],[ 1, 1]]\n", + "B = [0, 1000]\n", + "V = linalg.solve(A,B)\n", + "x = V[1]\n", + "y = V[0]\n", + "#Calculating the speed of output shaft\n", + "NF1 = y-x*(TB/TD)*(TE/TF) \t\t\t#Speed of the output shaft when gear C is fixed rpm\n", + "#Speed of the output shaft when gear C is rotated at 10 rpm counter clockwise:\n", + "#Calculating the values of x and y\n", + "#From the fourth row of te table y-x*(TB/TC) = +10 .....(iii)\n", + "#Also x+y = +1000 or y+x = 1000 .....(iv)\n", + "A = [[1, -TB/TC],[1, 1]]\n", + "B = [10, 1000]\n", + "V = linalg.solve(A,B)\n", + "x = V[1]\n", + "y = V[0]\n", + "#Calculating the speed of output shaft\n", + "NF2 = y-x*(TB/TD)*(TE/TF) \t\t\t#Speed of the output shaft when gear C is rotated at 10 rpm counter clockwise rpm\n", + "\n", + "#Results:\n", + "print \" Speed of the output shaft when gear C is fixed = %.1f rpm, counter clockwise.\"%(NF1)\n", + "print \" Speed of the output shaft when gear C is rotated at 10 rpm counter clockwise = %.1f rpm, \\\n", + " counter clockwise.\"%(NF2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Speed of the output shaft when gear C is fixed = 12.5 rpm, counter clockwise.\n", + " Speed of the output shaft when gear C is rotated at 10 rpm counter clockwise = 22.4 rpm, counter clockwise.\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.18 Page No : 461" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "TA = 10.\n", + "TB = 60.\n", + "NA = 1000.\n", + "NQ = 210.\n", + "ND = NQ \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Refer Fig. 13.24 and Table 13.20\n", + "#Calculating the speed of crown gear B\n", + "NB = NA*(TA/TB) \t\t\t#rpm\n", + "#Calculating the values of x and y\n", + "y = 200.\n", + "x = y-210.\n", + "#Calculating the speed of road wheel attached to axle P\n", + "NC = x+y \t\t\t#Speed of road wheel attached to axle P rpm\n", + "\n", + "#Results:\n", + "print \" Speed of road wheel attached to axle P = %d rpm.\"%(NC)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Speed of road wheel attached to axle P = 190 rpm.\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.19 Page No : 463" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "import math \n", + "\n", + "# Variables:\n", + "TA = 15.\n", + "TB = 20.\n", + "TC = 15.\n", + "NA = 1000. \t\t\t#rpm\n", + "Tm = 100. \t\t\t#Torque developed by motor N-m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 13.26 and Table 13.21\n", + "#Calculating the number of teeth on gears E and D\n", + "TE = TA+2*TB\n", + "TD = TE-(TB-TC)\n", + "#Speed of the machine shaft:\n", + "#From the fourth row of the table x+y = 1000 or y+x = 1000 .....(i)\n", + "#Also y-x*(TA/TE) = 0 .....(ii) \n", + "A = [[1, 1],[1, -TA/TE]]\n", + "B = [1000, 0]\n", + "V = linalg.solve(A,B)\n", + "y = round(V[0])\n", + "x = round(V[1])\n", + "#Calculating the speed of machine shaft\n", + "ND = y-x*(TA/TB)*(TC/TD) \t\t\t#rpm\n", + "#Calculating the torque exerted on the machine shaft\n", + "Ts = Tm*NA/ND \t\t\t#Torque exerted on the machine shaft N-m\n", + "\n", + "#Results:\n", + "print \" Speed of machine shaft, ND = %.2f rpm, anticlockwise.\"%(ND)\n", + "print \" Torque exerted on the machine shaft = %.f N-m.\"%(Ts)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Speed of machine shaft, ND = 37.15 rpm, anticlockwise.\n", + " Torque exerted on the machine shaft = 2692 N-m.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.20 Page No : 465" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "Ts = 100 \t\t\t#Torque on the sun wheel N-m\n", + "r = 5 \t\t\t#Ratio of speeds of gear S to C NS/NC\n", + "#Refer Fig. 13.27 and Table 13.22\n", + "#Number of teeth on different wheels:\n", + "#Calculating the values of x and y\n", + "y = 1.\n", + "x = 5-y\n", + "\n", + "#Calculating the number of teeth on wheel E\n", + "TS = 16.\n", + "TE = 4*TS\n", + "#Calculating the number of teeth on wheel P\n", + "TP = (TE-TS)/2\n", + "#Torque necessary to keep the internal gear stationary:\n", + "Tc = Ts*r \t\t\t#Torque on CN-m\n", + "#Caluclating the torque necessary to keep the internal gear stationary\n", + "Ti = Tc-Ts \t\t\t#Torque necessary to keep the internal gear stationary N-m\n", + "\n", + "#Results:\n", + "print \" Number of teeth on different wheels, TE = %d.\"%(TE)\n", + "print \" Torque necessary to keep the internal gear stationary = %d N-m.\"%(Ti)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Number of teeth on different wheels, TE = 64.\n", + " Torque necessary to keep the internal gear stationary = 400 N-m.\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.21 Page No : 466" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "from numpy import linalg\n", + "\n", + "# Variables:\n", + "TA = 14.\n", + "TC = 100.\n", + "r = 98./41 \t\t\t#TE/TD\n", + "PA = 1.85*1000 \t\t\t#W\n", + "NA = 1200. \t\t\t#rpm\n", + "TB = 43\n", + "\n", + "#Solution:\n", + "#Refer Fig. 13.28 and Table 13.23\n", + "#Calculating the number of teeth on wheel B TB = (TC-TA)/2\n", + "#Calculating the values of x and y\n", + "#From the fourth row of the table -y+x*(TA/TC) = 0 or x*(TA/TC)-y = 0 .....(i)\n", + "#Also x-y = 1200 or x+y = -1200 .....(ii)\n", + "A = [[TA/TC, -1],[ 1, 1]]\n", + "B = [0, -1200]\n", + "V = linalg.solve(A,B)\n", + "x = V[0]\n", + "y = V[1]\n", + "#Calculating the speed of gear E\n", + "NE = round(-y+x*(TA/TB)*(1./r)) \t\t\t#rpm\n", + "#Fixing torque required at C:\n", + "#Calculating the torque on A\n", + "Ta = PA*60./(2*math.pi*NA) \t\t\t#Torque on A N-m\n", + "#Calculating the torque on E\n", + "Te = PA*60./(2*math.pi*NE) \t\t\t#Torque on E\n", + "#Calculating the fixing torque required at C\n", + "Tc = Te-Ta \t\t\t#Fixing torque at C N-m\n", + "\n", + "#Results:\n", + "print \" Speed and direction of rotation of gear E, NE = %d rpm, anticlockwise.\"%(NE)\n", + "print \" Fixing torque required at C = %.1f N-m.\"%(Tc)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Speed and direction of rotation of gear E, NE = 4 rpm, anticlockwise.\n", + " Fixing torque required at C = 4401.8 N-m.\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.22 Page No : 468" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "import math \n", + "\n", + "# Variables:\n", + "TB = 15.\n", + "TA = 60.\n", + "TC = 20.\n", + "omegaY = 740.\n", + "omegaA = omegaY \t\t\t#rad/s\n", + "P = 130*1000. \t\t\t#W\n", + "\n", + "#Solution:\n", + "#Refer Fig. 13.29 and Table 13.24\n", + "#Calculating the number of teeth on wheel D\n", + "TD = TA-(TC+TB)\n", + "#Calculating the values of x and y\n", + "#From the fourth row of the table y-x*(TD/TC)*(TB/TA) = 740 .....(i)\n", + "#Also x+y = 0 or y+x = 0 .....(ii)\n", + "A = [[1, -(TD/TC)*(TB/TA)],[ 1, 1]]\n", + "B = [740, 0]\n", + "V = linalg.solve(A,B)\n", + "x = V[1]\n", + "y = V[0]\n", + "#Calculating the speed of shaft X\n", + "omegaX = y \t\t\t#rad/s\n", + "#Holding torque on wheel D:\n", + "#Calculating the torque on A\n", + "Ta = P/omegaA \t\t\t#Torque on A N-m\n", + "#Calculating the torque on X\n", + "Tx = P/omegaX \t\t\t#Torque on X N-m\n", + "#Calculating the holding torque on wheel D\n", + "Td = Tx-Ta \t\t\t#Holding torque on wheel D N-m\n", + "\n", + "#Results:\n", + "print \" Speed of shaft X, omegaX = %.1f rad/s.\"%(omegaX)\n", + "print \" Holding torque on wheel D = %.1f N-m.\"%(Td)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Speed of shaft X, omegaX = 563.8 rad/s.\n", + " Holding torque on wheel D = 54.9 N-m.\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.23 Page No : 469" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "import math \n", + "\n", + "# Variables:\n", + "TP = 144.\n", + "TQ = 120.\n", + "TR = 120.\n", + "TX = 36.\n", + "TY = 24.\n", + "TZ = 30.\n", + "NI = 1500. \t\t\t#rpm\n", + "P = 7.5*1000 \t\t\t#W\n", + "eta = 0.8\n", + "\n", + "#Solution:\n", + "#Refer Fig. 13.30 and Table 13.25\n", + "#Calculating the values of x and y\n", + "#From the fourth row of the table x+y = -1500 .....(i) \n", + "#Also y-x*(TZ/TR) = 0 or -x*(TZ/TR)+y = 0 .....(ii)\n", + "A = [[1, 1],[-TZ/TR, 1]]\n", + "B = [-1500, 0]\n", + "V = linalg.solve(A,B)\n", + "x = V[0]\n", + "y = V[1]\n", + "\n", + "#Calculating the values of x1 and y1\n", + "#We have y1-x1*(TY/TQ) = y .....(iii)\n", + "#Also x1+y1 = x+y or y1+x1 = x+y .....(iv)\n", + "A = [[1, -TY/TQ],[ 1, 1]]\n", + "B = [y, x+y]\n", + "V = linalg.solve(A , B)\n", + "x1 = V[1]\n", + "y1 = V[0]\n", + "#Speed and direction of the driven shaft O and the wheel P:\n", + "#Calculating the speed of shaft O\n", + "NO = y1 \t\t\t#rpm\n", + "#Calculating the speed of wheel P\n", + "NP = y1+x1*(TY/TQ)*(TX/TP) \t\t\t#rpm\n", + "#Torque tending to rotate the fixed wheel R:\n", + "#Calculating the torque on shaft I\n", + "T1 = P*60/(2*math.pi*NI) \t\t\t#N-m\n", + "#Calculating the torque on shaft O\n", + "T2 = eta*P*60/(2*math.pi*(-NO)) \t\t\t#N-m\n", + "#Calculating the torque tending to rotate the fixed wheel R\n", + "T = T2-T1 \t\t\t#Torque tending to rotate the fixed wheel R N-m\n", + "\n", + "#Results:\n", + "print \" Speed of the driven shaft O, NO = %d rpm, clockwise.\"%(-NO)\n", + "print \" Speed of the wheel P, NP = %d rpm, clockwise.\"%(-NP)\n", + "print \" Torque tending to rotate the fixed wheel R = %.2f N-m.\"%(T)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Speed of the driven shaft O, NO = 500 rpm, clockwise.\n", + " Speed of the wheel P, NP = 550 rpm, clockwise.\n", + " Torque tending to rotate the fixed wheel R = 66.85 N-m.\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.24 Page No : 471" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "import math \n", + "\n", + "# Variables:\n", + "TA = 34.\n", + "TB = 120.\n", + "TC = 150.\n", + "TD = 38.\n", + "TE = 50.\n", + "PX = 7.5*1000 \t\t\t#W\n", + "NX = 500. \t\t\t#rpm\n", + "m = 3.5 \t\t\t#mm\n", + "\n", + "#Solution:\n", + "#Refer Fig. 13.31 and Table 13.27\n", + "#Output torque of shaft Y:\n", + "#Calculating the values of x and y\n", + "#From the fourth row of the table x+y = 500 or y+x = 500 .....(i)\n", + "#Alsoy-x*(TA/TC) = 0 .....(ii)\n", + "A = [[1, 1],[ 1, -TA/TC]]\n", + "B = [500, 0]\n", + "V = linalg.solve(A, B)\n", + "y = round(V[0],1) \t\t\t#rpm\n", + "x = round(V[1],1) \t\t\t#rpm\n", + "#Calculating the speed of output shaft Y\n", + "NY = y-x*(TA/TB)*(TD/TE) \t\t\t#rpm\n", + "#Calculating the speed of wheel E\n", + "NE = NY \t\t\t#rpm\n", + "#Calculating the input power assuming 100 per cent efficiency\n", + "PY = PX \t\t\t#W\n", + "#Calculating the output torque of shaft Y\n", + "Ty = PY*60/(2*math.pi*NY*1000) \t\t\t#Output torque on shaft Y kN-m\n", + "#Tangential force between wheels D and E:\n", + "#Calculating the pitch circle radius of wheel E\n", + "rE = m*TE/(2*1000) \t\t\t#m\n", + "#Calculating the tangential force between wheels D and E\n", + "FtDE = Ty/rE \t\t\t#Tangential force between wheels D and E kN\n", + "#Tangential force between wheels B and C:\n", + "#Calculating the input torque on shaft X\n", + "Tx = PX*60/(2*math.pi*NX) \t\t\t#Input torque on shaft X N-m\n", + "#Calculating the fixing torque on the fixed wheel C\n", + "Tf = Ty-Tx/1000 \t\t\t#Fixing torque on the fixed wheelC kN-m\n", + "#Calculating the pitch circle radius of wheel C\n", + "rC = m*TC/(2*1000) \t\t\t#m\n", + "#Calculating the tangential forces between wheels B and C\n", + "FtBC = Tf/rC \t\t\t#kN\n", + "\n", + "#Results:\n", + "print \" Output torque of shaft Y = %.3f kN-m.\"%(Ty)\n", + "print \" Tangential force between wheels D and E = %.1f kN.\"%(FtDE)\n", + "print \" Tangential force between wheels B and C = %.f kN.\"%(FtBC)\n", + "\n", + "# note : answers are slightly different because of solve function and rounding off errors." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Output torque of shaft Y = 15.468 kN-m.\n", + " Tangential force between wheels D and E = 176.8 kN.\n", + " Tangential force between wheels B and C = 58 kN.\n" + ] + } + ], + "prompt_number": 6 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Theory_Of_Machines/ch14.ipynb b/Theory_Of_Machines/ch14.ipynb new file mode 100755 index 00000000..90379bfd --- /dev/null +++ b/Theory_Of_Machines/ch14.ipynb @@ -0,0 +1,1036 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:6f6f9834675f8099d2eeff645403e77a31c86a5aa089d3f5df9c6f15fe0a122e" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 14 : Gyroscopic Couple and Precessional Motion" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.1 Page No : 484" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "d = 300./1000\n", + "r = d/2\n", + "l = 600./1000 \t\t\t#m\n", + "m = 5. \t\t\t#kg\n", + "N = 300. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the disc\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the mass moment of inertia of the disc\n", + "#about an axis through its centre of gravity and perpendicular to the plane of the disc\n", + "I = m*r**2/2 \t\t\t#kg-m**2\n", + "#Calculating the couple due to mass of disc\n", + "C = m*9.81*l \t\t\t#N-m\n", + "#Calculating the speed of precession\n", + "omegaP = C/(I*omega) \t\t\t#rad/s\n", + "\n", + "#Results:\n", + "print \" Speed of precession, omegaP = %.1f rad/s.\"%(omegaP)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Speed of precession, omegaP = 16.7 rad/s.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.2 Page No : 485" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "d = 150./1000\n", + "r = d/2\n", + "x = 100./1000 \t\t\t#m\n", + "m = 5. \t\t\t #kg\n", + "N = 1000.\n", + "NP = 60. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the disc\n", + "omega = round(2*math.pi*N/60,1) \t\t\t#rad/s\n", + "#Calculating the speed of precession of the axle\n", + "omegaP = round(2*math.pi*NP/60,3) \t\t\t#rad/s\n", + "#Calculating the mass moment of inertia of the disc\n", + "# about an axis through its centre of gravity and perpendicular to the plane of disc\n", + "I = round(m*r**2/2,3) \t\t\t#kg-m**2\n", + "#Calculating the gyroscopic couple acting on the disc\n", + "C = round(I*omega*omegaP,1) \t\t\t#N-m\n", + "#Calculating the force at each bearing due to the gyroscopic couple\n", + "F = C/x \t\t\t#N\n", + "#Calculating the reactions at the bearings A and B\n", + "RA = m/2*9.81 \t\t\t#N\n", + "RB = round(RA,1) \t\t\t#N\n", + "#Resulmath.tant reaction at each bearing:\n", + "#Calculating the resultant reaction at the bearing A\n", + "RA1 = F+RA \t\t\t#N\n", + "#Calculating the resultant reaction at the bearing B\n", + "RB1 = F-RB \t\t\t#N\n", + "\n", + "#Results:\n", + "print \" Resultant reaction at the bearing A RA1 = %.1f N upwards.\"%(RA1)\n", + "print \" Resultant reaction at the bearing B RB1 = %.1f N downwards.\"%( RB1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Resultant reaction at the bearing A RA1 = 116.5 N upwards.\n", + " Resultant reaction at the bearing B RB1 = 67.5 N downwards.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.3 Page No : 487" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "R = 50.\n", + "k = 0.3 \t\t\t#m\n", + "v = 200.*1000/3600 \t\t\t#m/s\n", + "m = 400. \t\t\t#kg\n", + "N = 2400. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the engine\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the mass moment of inertia of the engine and the propeller\n", + "I = m*k**2 \t\t\t#kg-m**2\n", + "#Calculating the angular velocity of precession\n", + "omegaP = v/R \t\t\t#rad/s\n", + "#Calculating the gyroscopic couple acting on the aircraft\n", + "C = I*omega*omegaP/1000 \t\t\t#kN-m\n", + "\n", + "#Results:\n", + "print \" Gyroscopic couple acting on the aircraft, C = %.3f kN-m.\"%(C)\n", + "print \" The effect of the gyroscopic couple is to lift the nose upwards and tail downwards.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Gyroscopic couple acting on the aircraft, C = 10.053 kN-m.\n", + " The effect of the gyroscopic couple is to lift the nose upwards and tail downwards.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.4 Page No : 491" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m = 8.*1000 \t\t\t#kg\n", + "k = 0.6\n", + "R = 75. \t\t\t#m\n", + "N = 1800. \t\t\t#rpm\n", + "v = 100.*1000/3600 \t\t\t#m/s\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the rotor\n", + "omega = round(2*math.pi*N/60,1) \t\t\t#rad/s\n", + "#Calculating the mass moment of inertia of the rotor\n", + "I = m*k**2 \t\t\t#kg-m**2\n", + "#Calculating the angular velocity of precession\n", + "omegaP = round(v/R,2) \t\t\t#rad/s\n", + "#Calculating the gyroscopic couple\n", + "C = I*omega*omegaP/1000 \t\t\t#kN-m\n", + "\n", + "#Results:\n", + "print \" Gyroscopic couple, C = %.3f kN-m.\"%(C)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Gyroscopic couple, C = 200.866 kN-m.\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.5 Page No : 491" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "N = 1500. \t\t\t#rpm\n", + "m = 750. \t\t\t#kg\n", + "omegaP = 1. \t\t\t#rad/s\n", + "k = 250./1000 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the rotor\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the mass moment of inertia of the rotor\n", + "I = m*k**2 \t\t\t#kg-m**2\n", + "#Calculating the gyroscopic couple transmitted to the hull\n", + "C = I*omega*omegaP/1000 \t\t\t#kN-m\n", + "\n", + "#Results:\n", + "print \" Gyroscopic couple transmitted to the hull, C = %.3f kN-m.\"%(C)\n", + "print \" When the pitching is upward, the relative gyroscopic couple acts in the clockwise direction.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Gyroscopic couple transmitted to the hull, C = 7.363 kN-m.\n", + " When the pitching is upward, the relative gyroscopic couple acts in the clockwise direction.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.6 Page No : 492" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m = 3500. \t\t\t#kg\n", + "k = 0.45 \t\t\t#m\n", + "N = 3000. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the rotor\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#When the ship is steering to the left:\n", + "R = 100. \t\t\t#m\n", + "v = 36.*1000/3600 \t\t\t#m/s\n", + "\n", + "#Calculating the mass moment of inertia of the rotor\n", + "I = m*k**2 \t\t\t#kg-m**2\n", + "#Calculating the angular velocity of precession\n", + "omegaP = v/R \t\t\t#rad/s\n", + "#Calculating the gyroscopic couple\n", + "C = I*omega*omegaP/1000 \t\t\t#kN-m\n", + "\n", + "#Results:\n", + "print \" Gyroscopic couple when the ship is steering to the left. C = %.2f kN-m.\"%(C)\n", + "print \" When the rotor rotates clockwise and the ship takes a left turn.\\\n", + " the effect of the reactive gyroscopic couple is to raise the bow and lower the stern.\"\n", + "#When the ship is pitching with the bow falling:\n", + "tp = 40. \t\t\t#s\n", + "#Calculating the amplitude of swing\n", + "phi = 12./2*math.pi/180 \t\t\t#rad\n", + "#Calculating the angular velocity of the simple harmonic motion\n", + "omega1 = 2*math.pi/tp \t\t\t#rad/s\n", + "#Calculating the maximum angular velocity of precession\n", + "omegaP = phi*omega1 \t\t\t#rad/s\n", + "#Calculating the gyroscopic couple\n", + "C = I*omega*omegaP/1000 \t\t\t#kN-m\n", + "\n", + "#Results:\n", + "print \" Gyroscopic couple when the ship is pitching with the bow falling, C = %.3f kN-m.\"%(C)\n", + "print \" When the bow is falling, the effect of the reactive gyroscopic couple is to move\\\n", + " the ship towards port side.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Gyroscopic couple when the ship is steering to the left. C = 22.27 kN-m.\n", + " When the rotor rotates clockwise and the ship takes a left turn. the effect of the reactive gyroscopic couple is to raise the bow and lower the stern.\n", + " Gyroscopic couple when the ship is pitching with the bow falling, C = 3.663 kN-m.\n", + " When the bow is falling, the effect of the reactive gyroscopic couple is to move the ship towards port side.\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.7 Page No : 492" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m = 20.*1000 \t\t\t#kg\n", + "k = 0.6 \t\t\t#m\n", + "N = 2000. \t\t\t#rpm\n", + "phi = 6.*math.pi/180 \t\t\t#rad\n", + "tp = 30. \t\t\t#s\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the rotor\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Maximum gyroscopic couple:\n", + "#Calculating the mass moment of inertia of the rotor\n", + "I = m*k**2 \t\t\t#kg-m**2\n", + "#Calculating the angular velocity of the simple harmonic motion\n", + "omega1 = 2*math.pi/tp \t\t\t#rad/s\n", + "#Calculating the maximum angular velocity of precession\n", + "omegaPmax = phi*omega1 \t\t\t#rad/s\n", + "#Calculating the maximum gyroscopic couple\n", + "Cmax = I*omega*omegaPmax/1000 \t\t\t#kN-m\n", + "#Calculating the maximum angular acceleration during pitching\n", + "alphamax = phi*omega1**2 \t\t\t#Maximum angular acceleration during pitching rad/s**2\n", + "\n", + "#Results:\n", + "print \" Maximum gyroscopic couple, Cmax = %.3f kN-m.\"%(Cmax)\n", + "print \" Maximum angular acceleration during pitching = %.4f rad/s**2.\"%(alphamax)\n", + "print \" When the rotation of the rotor is clockwise when looking from the left and when\\\n", + " the bow is rising, then the reactive gyroscopic couple acts in the direction which\\\n", + " tends to turn the bow towrds right.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Maximum gyroscopic couple, Cmax = 33.073 kN-m.\n", + " Maximum angular acceleration during pitching = 0.0046 rad/s**2.\n", + " When the rotation of the rotor is clockwise when looking from the left and when the bow is rising, then the reactive gyroscopic couple acts in the direction which tends to turn the bow towrds right.\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.8 Page No : 493" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m = 5.*1000 \t\t\t#kg\n", + "N = 1000. \t\t\t#rpm\n", + "k = 0.5 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the rotor\n", + "omega = 2.*math.pi*2100/60 \t\t\t#rad/s\n", + "#When the ship steers to the left:\n", + "v = 30.*1000/3600 \t\t\t#m/s\n", + "R = 60. \t\t\t#m\n", + "#Calculating the angular velocity of precession\n", + "omegaP = round(v/R,2) \t\t\t#rad/s\n", + "#Calculating the mass moment of inertia of the rotor\n", + "I = m*k**2 \t\t\t#kg-m**2\n", + "#Calculating the gyroscopic couple\n", + "C = I*omega*omegaP/1000 \t\t\t#kN-m\n", + "\n", + "#Results:\n", + "print \" C = %.1f k N-m\"%C\n", + "\n", + "#When the ship pitches with the bow descending:\n", + "phi = round(6*math.pi/180,3) \t\t\t#rad\n", + "tp = 20. \t\t\t#s\n", + "#Calculating the angular velocity of simple harmonic motion\n", + "omega1 = round(2*math.pi/tp,4) \t\t\t#rad/s\n", + "#Calculating the maximum velocity of precession\n", + "omegaPmax = round(phi*omega1,3) \t\t\t#rad/s\n", + "#Calculating the maximum gyroscopic couple\n", + "Cmax = I*omega*omegaPmax \t\t\t#N-m\n", + "\n", + "#Results:\n", + "print \" Cmax = %.f N-m\"%Cmax\n", + "#When the ship rolls:\n", + "omegaP = 0.03 \t\t\t#rad/s\n", + "#Calculating the gyroscopic couple\n", + "C = I*omega*omegaP \t\t\t#N-m\n", + "\n", + "#Results:\n", + "print \" C = %.2f N-m\"%(round(C,-1))\n", + "#Calculating the maximum angular acceleration during pitching\n", + "alphamax = phi*omega1**2 \t\t\t#rad/s**2\n", + "\n", + "#Results:\n", + "print \" Maximum angular acceleration during pitching, alphamax = %.2f rad/s**2.\"%(alphamax)\n", + "\n", + "# rounding off error. please check." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " C = 38.5 k N-m\n", + " Cmax = 9071 N-m\n", + " C = 8250.00 N-m\n", + " Maximum angular acceleration during pitching, alphamax = 0.01 rad/s**2.\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.9 Page No : 494" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m = 2000. \t\t\t#kg\n", + "N = 3000. \t\t\t#rpm\n", + "k = 0.5\n", + "R = 100. \t\t\t#m\n", + "v = 16.1*1855/3600 \t\t\t#m/s\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the rotor\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Gyroscopic couple:\n", + "#Calculating the mass moment of inertia of the rotor\n", + "I = m*k**2 \t\t\t#kg-m**2\n", + "#Calculating the angular velocity of precession\n", + "omegaP = v/R \t\t\t#rad/s\n", + "#Calculating the gyroscopic couple\n", + "C = I*omega*omegaP/1000 \t\t\t#kN-m\n", + "#Torque during pitching:\n", + "tp = 50. \t\t\t#s\n", + "phi = 12./2*math.pi/180 \t\t\t#rad\n", + "#Calculating the angular velocity of simple harmonic motion\n", + "omega1 = 2*math.pi/tp \t\t\t#rad/s\n", + "#Calculating the maximum angular velocity of precession\n", + "omegaPmax = phi*omega1 \t\t\t#rad/s\n", + "#Calculating the maximum gyroscopic couple during pitching\n", + "Cmax = I*omega*omegaPmax \t\t\t#N-m\n", + "#Calculating the maximum acceleration during pitching\n", + "alphamax = phi*omega1**2 \t\t\t#rad/s**2\n", + "\n", + "#Results:\n", + "print \" Torque during pitching, Cmax = %d N-m.\"%(Cmax)\n", + "print \" Maximum acceleration during pitching, alphamax = %.5f rad/s**2.\"%(alphamax)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Torque during pitching, Cmax = 2067 N-m.\n", + " Maximum acceleration during pitching, alphamax = 0.00165 rad/s**2.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.10 Page No : 497" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m = 2500. \t\t\t#kg\n", + "x = 1.5\n", + "R = 30.\n", + "dW = 0.75\n", + "rW = dW/2\n", + "h = 0.9 \t\t\t#m\n", + "v = 24.*1000/3600 \t\t\t#m/s\n", + "G = 5.\n", + "IW = 18.\n", + "IE = 12. \t\t\t#kg-m**2\n", + "\n", + "#Solution:\n", + "#Calculating the road reaction on each wheel\n", + "r = m*9.81/4 \t\t\t#Road reaction on each wheel N\n", + "#Calculating the angular velocity o the wheels\n", + "omegaW = round(v/rW,1) \t\t\t#rad/s\n", + "#Calculating the angular velocity of precession\n", + "omegaP = round(v/R,2) \t\t\t#rad/s\n", + "#Calculating the gyroscopic couple due to one pair of wheels and axle\n", + "CW = round(2*IW*omegaW*omegaP) \t\t\t#N-m\n", + "#Calculating the gyroscopic couple due to the rotating parts of the motor and gears\n", + "CE = round(2*IE*G*omegaW*omegaP) \t\t\t#N-m\n", + "#Calculating the net gyroscopic couple\n", + "C = CW-CE \t\t\t#N-m\n", + "#Calculating the reaction due to gyroscopic couple at each of the outer or inner wheels\n", + "P = round((-C)/(2*x),1) \t\t\t#N\n", + "#Calculating the centrifugal force\n", + "FC = round(m*v**2/R,1) \t\t\t#N\n", + "#Calculating the overturning couple\n", + "CO = FC*h \t\t\t#N-m\n", + "#Calculating the reaction due to overturning couple at each of the outer and inner wheels\n", + "Q = 2*CO/(2*x) \t\t\t#N\n", + "#Calculating the vertical force exerted on each outer wheel\n", + "PO = m*9.81/4-P/2+Q/2 \t\t\t#N\n", + "#Calculating the vertical force exerted on each inner wheel\n", + "PI = m*9.81/4+P/2-Q/2 \t\t\t#N\n", + "\n", + "#Results:\n", + "print \" Vertical force exerted on each outer wheel, PO = %.2f N.\"%(PO)\n", + "print \" Vertical force exerted on each inner wheel, PI = %.2f N.\"%(PI)\n", + "\n", + "# note : value of Fc is calculated wrongly. please check using calculator. so answers are different." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Vertical force exerted on each outer wheel, PO = 7187.51 N.\n", + " Vertical force exerted on each inner wheel, PI = 5074.99 N.\n" + ] + } + ], + "prompt_number": 41 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.11 pageno : 498" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# variables\n", + "R = 100 #m \n", + "IW = 2.5 # kg-m 2 \n", + "dW = 0.6 #m \n", + "rW = 0.3 #m \n", + "IE = 1.2 #kg-m 2 ;\n", + "G = 3. \n", + "m = 1600. # kg \n", + "h = 0.5 #m ; \n", + "x = 1.5 #m\n", + "v = 1.\n", + "\n", + "# calculations\n", + "road_reaction = m*9.81/4 #N\n", + "wW = v / rW\n", + "wP = v/R\n", + "cW = 4 * IW*wW*wP\n", + "cE = IE*G*wW*wP\n", + "C = cW + cE\n", + "Pby2 = C/(2*x)\n", + "Fc = m*v**2/R\n", + "Co = Fc * h\n", + "Qby2 = Co/(2*x)\n", + "v_2 = road_reaction/(Pby2 + Qby2)\n", + "v = math.sqrt(v_2)\n", + "ans = v*3600./1000\n", + "\n", + "# result\n", + "print \"V <= %.1f m/s = %.1f * 3600 /1000 = %.2f km/h\"%(v,v,ans)\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "V <= 37.3 m/s = 37.3 * 3600 /1000 = 134.34 km/h\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.12 Page No : 500" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m = 2000. #kg\n", + "mE = 75. \t\t\t#kg\n", + "b = 2.5 #m\n", + "x = 1.5 #m\n", + "h = 500./1000 #m\n", + "L = 1. #m\n", + "dW = 0.8 #m\n", + "rW = round(dW/2,1) #m\n", + "kE = 100./1000 #m\n", + "R = 60. \t\t\t#m\n", + "IW = 0.8 \t\t\t#kg-m**2\n", + "G = 4.\n", + "v = round(60.*1000/3600,2) \t#m/s\n", + "\n", + "#Solution:\n", + "#Refer Fig. 14.12\n", + "#Calculating the weight on the rear wheels\n", + "W2 = (m*9.81*1)/b \t\t\t#N\n", + "#Calculating the weight on the front wheels\n", + "W1 = m*9.81-W2 \t\t\t#N\n", + "#Calculating the weight on each of the front wheels\n", + "Wf = W1/2 \t\t\t#Weight on each of the front wheels N\n", + "#Calculating the weight on each of the rear wheels\n", + "Wr = W2/2 \t\t\t#Weight on each of the rear wheels N\n", + "#Calculating the angular velocity of wheels\n", + "omegaW = v/rW \t\t\t#rad/s\n", + "#Calculating the angular velocity of precession\n", + "omegaP = round(v/R,3) \t\t\t#rad/s\n", + "#Calculating the gyroscopic couple due to four wheels\n", + "CW = round(4*IW*omegaW*omegaP,1) \t\t\t#N-m\n", + "#Calculating the magnitude of reaction due to gyroscopic couple due to four wheels at each of the inner or outer wheel\n", + "P = round((CW/(2*x)),2) \t\t\t#N\n", + "#Calculating the mass moment of inertia of rotating parts of the engine\n", + "IE = mE*(kE)**2 \t\t\t#kg-m**2\n", + "#Calculating the gyroscopic couple due to rotating parts of the engine\n", + "CE = round(IE*(kE)**2*G*omegaW*omegaP*100,1)\t\t\t#N-m\n", + "#Calculating the magnitude of reaction due to gyroscopic couple due to rotating parts of the engine at each of the inner or outer wheel\n", + "F = (CE/(2*b)) \t\t\t#N\n", + "#Calculating the centrifugal force\n", + "FC = round(m*v**2/R) \t\t\t#N\n", + "#Calculating the centrifugal couple tending to overturn the car\n", + "CO = FC*h \t\t\t#N-m\n", + "#Calculating the magnitude of reaction due to overturning couple at each of the inner or outer wheel\n", + "Q = round((CO/(2*x)),2) \t\t\t#N\n", + "#Calculating the load on front wheel 1\n", + "Fw1 = (W1/2)-(P/2)-(F/2)-(Q/2) \t\t\t#Load on front wheel 1 N\n", + "#Calculating the load on front wheel 2\n", + "Fw2 = W1/2+P/2-F/2+Q/2 \t\t\t#Load on front wheel 2 N\n", + "#Calculating the load on rear wheel 3\n", + "Rw3 = W2/2-P/2+F/2-Q/2 \t\t\t#Load on rear wheel 3 N\n", + "#Calculating the load on rear wheel 4\n", + "Rw4 = W2/2+P/2+F/2+Q/2 \t\t\t#Load on rear wheel 4 N\n", + "\n", + "#Results:\n", + "print \" Load on front wheel 1 = %.2f N.\"%(Fw1)\n", + "print \" Load on front wheel 2 = %.2f N.\"%(Fw2)\n", + "print \" Load on rear wheel 3 = %.2f N.\"%(Rw3)\n", + "print \" Load on rear wheel 4 = %.2f N.\"%(Rw4)\n", + "\n", + "# note : incorrect answers in the textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Load on front wheel 1 = 5104.42 N.\n", + " Load on front wheel 2 = 6660.62 N.\n", + " Load on rear wheel 3 = 3149.38 N.\n", + " Load on rear wheel 4 = 4705.58 N.\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.13 Page No : 502" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m = 2000. #kg\n", + "mI = 200. \t\t#kg\n", + "x = 1.6 #m\n", + "R = 30. #m\n", + "dW = 0.7 #m\n", + "rW = dW/2 #m\n", + "k = 0.3 #m\n", + "h = 1. \t\t\t#m\n", + "v = 54.*1000/3600 \t\t\t#m/s\n", + "theta = 8. \t\t\t#degrees\n", + "\n", + "#Solution:\n", + "#Refer Fig. 14.13\n", + "#Calculating the reactions at the wheels:\n", + "#Taking moments about B\n", + "RA = (m*9.81*math.cos(math.radians(theta))+m*v**2/R*math.sin(math.radians(theta)))*1/2+(m*9.81*math.sin(math.radians(theta)) \\\n", + " -m*v**2/R*math.cos(math.radians(theta)))*h/x \t\t\t#N\n", + "#Resolving the forces perpendicular to the track\n", + "RB = (m*9.81*math.cos(math.radians(theta))+m*v**2/R*math.sin(math.radians(theta)))-RA \t\t\t#N\n", + "#Calculating the angular velocity of wheels\n", + "omegaW = v/rW \t\t\t#rad/s\n", + "#Calculating the angular velocity of precession\n", + "omegaP = v/R \t\t\t#rad/s\n", + "#Calculating the gyroscopic couple\n", + "C = mI*k**2*omegaW*math.cos(math.radians(theta))*omegaP \t\t\t#N-m\n", + "#Calculating the force at each pair of wheels due to the gyroscopic couple\n", + "P = C/x \t\t\t#N\n", + "#Calculating the pressure on the inner rail\n", + "PI = RA-P \t\t\t#N\n", + "#Calculating the pressure o the outer rail\n", + "PO = RB+P \t\t\t#N\n", + "\n", + "#Results:\n", + "print \" Pressure on the inner rail, PI = %.2f N.\"%(PI)\n", + "print \" Pressure on the outer rail, PO = %.2f N.\"%(PO)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Pressure on the inner rail, PI = 2942.45 N.\n", + " Pressure on the outer rail, PO = 18574.21 N.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.14 Page No : 503" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "I = 180. \t\t\t#kg-m**2\n", + "D = 1.8 #m\n", + "R = D/2 #m\n", + "x = 1.5 \t\t\t#m\n", + "v = 95.*1000/3600 \t#m/s\n", + "t = 0.1 \t\t\t#s\n", + "\n", + "#Solution:\n", + "#Gyroscopic couple set up:\n", + "#Calculating the angular velocity of the locomotive\n", + "omega = v/R \t\t\t#rad/s\n", + "#Calculating the amplitude\n", + "A = 1./2*6 \t\t\t#mm\n", + "#Calculating the maximum velocity while falling\n", + "vmax = 2*math.pi/t*A/1000 \t\t\t#m/s\n", + "#Calculating the maximum angular velocity of tilt of the axle or angular velocity of precession\n", + "omegaPmax = vmax/x \t\t\t#rad/s\n", + "#Calculating the gyroscopic couple set up\n", + "C = I*omega*omegaPmax \t\t\t#N-m\n", + "#Calculating the reaction between the wheel and rail due to the gyroscopic couple\n", + "P = C/x \t\t\t#N\n", + "\n", + "#Results:\n", + "print \" Gyroscopic couple set up, C = %.1f N-m.\"%(C)\n", + "print \" Reaction between the wheel and rail due to the gyroscopic couple, P = %d N.\"%(P)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Gyroscopic couple set up, C = 663.2 N-m.\n", + " Reaction between the wheel and rail due to the gyroscopic couple, P = 442 N.\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.15 Page No : 506" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m = 250. \t\t\t#kg\n", + "IE = 0.3 #kg-m**2\n", + "IW = 1. \t\t\t#kg-m**2\n", + "G = 5.\n", + "h = 0.6 #m\n", + "rW = 300./1000 #m\n", + "R = 50. \t\t\t#m\n", + "v = 90.*1000/3600 \t#m/s\n", + "\n", + "#Solution:\n", + "#Calculating the angle of inclination with respect to the vertical of a two wheeler\n", + "#Equating total overturning couple to balancing couple\n", + "tantheta = math.atan(1/(m*9.81*h))*((v**2/(R*rW)*(2*IW+G*IE))+(m*v**2/R*h)) \t\t\t#degrees\n", + "theta = math.degrees(math.atan(tantheta))\n", + "\n", + "#Results:\n", + "print \" Angle of inclination with respect to the vertical of a two wheeler, tan theta = %.2f .\"%(theta)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Angle of inclination with respect to the vertical of a two wheeler, tan theta = 53.94 .\n" + ] + } + ], + "prompt_number": 62 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.16 Page No : 507" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m1 = 0.5 #kg\n", + "m2 = 0.3 \t\t\t#kg\n", + "k = 20./1000 #m\n", + "OG = 10./1000 #m\n", + "h = OG #m\n", + "R = 50. \t\t\t#m\n", + "N = 3000. \t\t\t#rpm\n", + "v = 15. \t\t\t#m/s\n", + "\n", + "#Solution:\n", + "#Refer Fig. 14.15 and Fig. 14.16\n", + "#Calculating the angular speed of the wheel\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the mass moment of inertia of the gyrowheel\n", + "I = m1*k**2 \t\t\t#kg-m**2\n", + "#Calculating the angular velocity of precession\n", + "omegaP = v/R \t\t\t#rad/s\n", + "#When the vehicle moves in the direction of arrow X taking a left turn along the curve:\n", + "#Calculating the angle of inclination of the gyrowheel from the vertical\n", + "#Equating the overturning couple to the balancing couple for equilibrium condition\n", + "tantheta1 = (1/(m2*9.81*h))*(I*omega*omegaP-m2*v**2/R*h) \t\t\t#degrees\n", + "theta1 = math.degrees(math.atan(tantheta1))\n", + "#When the vehicle reverses at the same speed in the direction of arrow Y along the same path:\n", + "#Calculating the angle of inclination of the gyrowheel from the vertical\n", + "#Equating the overturning couple to the balancing couple for equilibrium condition\n", + "tantheta2 = round((1/(m2*9.81*h))*(I*omega*omegaP+m2*v**2/R*h),1) \t\t\t#degrees\n", + "theta2 = math.degrees(math.atan(tantheta2))\n", + "\n", + "#Results:\n", + "print \" Angle of inclination of the gyrowheel from the vertical when the vehicle moves\\\n", + " in the direction of arrow X taking a left turn along the curve, theta = %.2f degrees.\"%(theta1)\n", + "print \" Angle of inclination of the gyrowheel from the vertical when the vehicle reverses \\\n", + "at the same speed in the direction of arrow Y along the same path, theta = %.f degrees.\"%(theta2)\n", + "\n", + "# rounding error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Angle of inclination of the gyrowheel from the vertical when the vehicle moves in the direction of arrow X taking a left turn along the curve, theta = 10.30 degrees.\n", + " Angle of inclination of the gyrowheel from the vertical when the vehicle reverses at the same speed in the direction of arrow Y along the same path, theta = 48 degrees.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.17 Page No : 510" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables: \n", + "d = 0.6 #m\n", + "r = d/2 \t\t\t#m\n", + "m = 30. \t\t\t#kg\n", + "theta = 1. \t\t\t#degree\n", + "N = 1200. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the shaft\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the gyroscopic couple acting on the bearings\n", + "C = round(m/8*omega**2*r**2*math.sin(math.radians(2*theta))) \t\t\t#N-m\n", + "\n", + "#Results:\n", + "print \" Gyroscopic couple acting on the bearings, C = %d N-m.\"%(C)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Gyroscopic couple acting on the bearings, C = 186 N-m.\n" + ] + } + ], + "prompt_number": 25 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Theory_Of_Machines/ch15.ipynb b/Theory_Of_Machines/ch15.ipynb new file mode 100755 index 00000000..05026c42 --- /dev/null +++ b/Theory_Of_Machines/ch15.ipynb @@ -0,0 +1,1526 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:13844e2491f60bc4c521c5dd5ea680eaf26c07059f7a5c77baf4e368f712f152" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 15 : Inertia Forces in Reciprocating Parts" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.1 Page No : 521" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "OC = 200./1000 #m\n", + "PC = 700./1000 \t\t\t#m\n", + "omega = 120. \t\t\t#rad/s\n", + "\n", + "#Solution:\n", + "#Refer Fig. 15.5\n", + "OM = 127./1000\n", + "CM = 173./1000\n", + "QN = 93./1000\n", + "NO = 200./1000 \t\t\t#m\n", + "\n", + "#Velocity and acceleration of the piston:\n", + "#Calculating the velocity of the piston P\n", + "vP = omega*OM \t\t\t#m/s\n", + "#Calculating the acceleration of the piston P\n", + "aP = omega**2*NO \t\t\t#m/s**2\n", + "#Velocity and acceleration of the mid-point of the connecting rod:\n", + "#By measurement\n", + "OD1 = 140./1000\n", + "OD2 = 193./1000 \t\t\t#m\n", + "#Calculating the velocity of D\n", + "vD = omega*OD1 \t\t\t#m/s\n", + "#Calculating the acceleration of D\n", + "aD = omega**2*OD2 \t\t\t#m/s**2\n", + "#Angular velocity and angular acceleration of the connecting rod:\n", + "#Calculating the velocity of the connecting rod PC\n", + "vPC = omega*CM \t\t\t#m/s\n", + "#Calculating the angular velocity of the connecting rod PC\n", + "omegaPC = vPC/PC \t\t\t#rad/s\n", + "#Calculating the math.tangential component of the acceleration of P with respect to C\n", + "atPC = omega**2*QN \t\t\t#m/s**2\n", + "#Calculating the angular acceleration of the connecting rod PC\n", + "alphaPC = atPC/PC \t\t\t#ra/s**2\n", + "\n", + "#Results:\n", + "print \" Velocity of the piston P, vP = %.2f m/s.\"%(vP)\n", + "print \" Acceleration of the piston P, aP = %d m/s**2.\"%( aP)\n", + "print \" Velocity of D, vD = %.1f m/s.\"%(vD)\n", + "print \" Acceleration of D, aD = %.1f m/s**2.\"%(aD)\n", + "print \" Angular velocity of the connecting rod PC, omegaPC = %.2f rad/s.\"%(omegaPC)\n", + "print \" Angular acceleration of the connecting rod PC alphaPC = %.2f rad/s**2.\"%(alphaPC)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Velocity of the piston P, vP = 15.24 m/s.\n", + " Acceleration of the piston P, aP = 2880 m/s**2.\n", + " Velocity of D, vD = 16.8 m/s.\n", + " Acceleration of D, aD = 2779.2 m/s**2.\n", + " Angular velocity of the connecting rod PC, omegaPC = 29.66 rad/s.\n", + " Angular acceleration of the connecting rod PC alphaPC = 1913.14 rad/s**2.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.2 Page No : 522" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "OC = 150./1000 #m\n", + "PC = 600./1000 #m\n", + "CD = 150./1000 \t\t#m\n", + "N = 450. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Refer Fig. 15.6\n", + "#Calculating the angular speed of the crank\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#By measurement\n", + "OM = 145./1000\n", + "CM = 78./1000\n", + "QN = 130./1000\n", + "NO = 56./1000 \t\t\t#m\n", + "\n", + "#Velocity and acceleration of alider:\n", + "#Calculating the velocity of the slider P\n", + "vP = omega*OM \t\t\t#m/s\n", + "#Calculating the acceleration of the slider P\n", + "aP = omega**2*NO \t\t\t#m/s**2\n", + "#Velocity and acceleration of point D on the connecting rod:\n", + "#Calculating the length od CD1\n", + "CD1 = CD/PC*CM \t\t\t#m\n", + "#By measurement\n", + "OD1 = 145./1000\n", + "OD2 = 120./1000 \t\t\t#m\n", + "\n", + "#Calculating the velocity of point D\n", + "vD = omega*OD1 \t\t\t#m/s\n", + "#Calculating the acceleration of point D\n", + "aD = omega**2*OD2 \t\t\t#m/s**2\n", + "#Angular velocity and angular acceleration of the connecting rod:\n", + "#Calculating the velocity of the connecting rod PC\n", + "vPC = omega*CM \t\t\t#m/s\n", + "#Calculating the angular velocity of the connecting rod\n", + "omegaPC = vPC/PC \t\t\t#rad/s\n", + "#Calculating the tangential component of the acceleration of P with respect to C\n", + "atPC = omega**2*QN \t\t\t#m/s**2\n", + "#Calculating the angular acceleration of the connecting rod PC\n", + "alphaPC = atPC/PC \t\t\t#rad/s**2\n", + "\n", + "#Results:\n", + "print \" Velocity of the slider P vP = %.3f m/s.\"%(vP)\n", + "print \" Acceleration of the slider P aP = %.1f m/s**2.\"%(aP)\n", + "print \" Velocity of point D vD = %.3f m/s.\"%(vD)\n", + "print \" Acceleration of point D aD = %.2f m/s**2.\"%(aD)\n", + "print \" Angular velocity of the connecting rod omegaPC = %.3f rad/s.\"%(omegaPC)\n", + "print \" Angular acceleration of the connecting rod PC alphaPC = %.2f rad/s**2.\"%(alphaPC)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Velocity of the slider P vP = 6.833 m/s.\n", + " Acceleration of the slider P aP = 124.4 m/s**2.\n", + " Velocity of point D vD = 6.833 m/s.\n", + " Acceleration of point D aD = 266.48 m/s**2.\n", + " Angular velocity of the connecting rod omegaPC = 6.126 rad/s.\n", + " Angular acceleration of the connecting rod PC alphaPC = 481.14 rad/s**2.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.3 Page No : 527" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "r = 300./1000\n", + "l = 1. \t\t\t#m\n", + "N = 200. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the crank\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Crank angle at which the maximum velocity occurs:\n", + "#Calculating the ratio of length of connecting rod to crank radius\n", + "n = l/r\n", + "#Velocity of the piston vP = omega*r*(math.sin(math.radians(theta)+math.sin(math.radians(2*theta)/(2*n)) .....(i)\n", + "#For maximum velocity d(vP)/d(theta) = 0 .....(ii)\n", + "#Substituting (i) in (ii) we get 2(math.cos(theta))**2+n*math.cos(theta)-1 = 0\n", + "a = 2.\n", + "b = n\n", + "c = -1.\n", + "costheta = (-b+math.sqrt(b**2-4*a*c))/(2*a)\n", + "#Calculating the crank angle from the inner dead centre at which the maximum velocity occurs\n", + "theta = round(math.degrees(math.acos(costheta))) \t\t\t#degrees\n", + "#Calculating the maximum velocity of the piston:\n", + "vPmax = omega*r*(math.sin(math.radians(theta))+math.sin(math.radians(2*theta))/(2*n)) \t\t\t#m/s\n", + "#Results:\n", + "print \" Crank angle from the inner dead centre at which the maximum velocity occurs theta = %.2f degrees.\"%(theta)\n", + "print \" Maximum velocity of the piston( vPmax) = %.2f m/s.\"%(vPmax)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Crank angle from the inner dead centre at which the maximum velocity occurs theta = 75.00 degrees.\n", + " Maximum velocity of the piston( vPmax) = 6.54 m/s.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.4 Page No : 528" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "r = 0.3 #m\n", + "l = 1.5 \t\t\t#m\n", + "N = 180. \t\t\t#rpm\n", + "theta = 40. \t\t#degrees\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the piston\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Velocity of the piston:\n", + "#Calculating the ratio of lengths of the connecting rod and crank\n", + "n = l/r\n", + "#Calculating the velocity of the piston\n", + "vP = omega*r*(math.sin(math.radians(theta))+math.sin(math.radians(2*theta))/(2*n)) \t\t\t#m/s\n", + "#Calculating the acceleration of the piston\n", + "aP = omega**2*r*(math.cos(math.radians(theta))+math.cos(math.radians(2*theta))/n) \t\t\t#m/s**2\n", + "#Position of the crank for zero acceleration of the piston:\n", + "ap1 = 0\n", + "#Calculating the position of the crank from the inner dead centre for zero acceleration of the piston\n", + "#We have ap1 = omega**2*r*(math.cos(theta1)+math.cos(2*theta1)/n) or 2*(math.cos(theta1))**2+n*math.cos(theta1)-1 = 0\n", + "a = 2.\n", + "b = n\n", + "c = -1.\n", + "costheta1 = (-b+math.sqrt(b**2-4*a*c))/(2*a)\n", + "#Calculating the crank angle from the inner dead centre for zero acceleration of the piston\n", + "theta1 = math.degrees(math.acos(costheta1)) \t\t\t#degrees\n", + "\n", + "#Results:\n", + "print \" Velocity of the piston vP = %.2f m/s.\"%( vP)\n", + "print \" Acceleration of the piston aP = %.2f m/s**2.\"%(aP)\n", + "print \" Position of the crank for zero acceleration of the piston theta1 = %.2f degrees or\\\n", + " %.2f degrees.\"%(theta1,360-theta1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Velocity of the piston vP = 4.19 m/s.\n", + " Acceleration of the piston aP = 85.36 m/s**2.\n", + " Position of the crank for zero acceleration of the piston theta1 = 79.27 degrees or 280.73 degrees.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.5 Page No : 528" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "r = 150./1000\n", + "l = 600./1000 \t\t\t#m\n", + "theta = 60. \t\t\t#degrees\n", + "N = 450. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the crank\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Velocity and acceleration of the slider:\n", + "#Calculating the ratio of length of connecting rod and crank\n", + "n = l/r\n", + "#Calculating the velocity of the slider\n", + "vP = omega*r*(math.sin(math.radians(theta))+math.sin(math.radians(2*theta))/(2*n)) \t\t\t#m/s\n", + "#Calculating the acceleration of the slider\n", + "aP = omega**2*r*(math.cos(math.radians(theta))+math.cos(math.radians(2*theta))/n) \t\t\t#m/s**2\n", + "#Angular velocity and angular acceleration of the connecting rod:\n", + "#Calculating the angular velocity of the connecting rod\n", + "omegaPC = omega*math.cos(math.radians(theta))/n \t\t\t#rad/s\n", + "#Calculating the angular acceleration of the connecting rod\n", + "alphaPC = round(omega**2*math.sin(math.radians(theta))/n) \t\t\t#rad/s**2\n", + "\n", + "#Results:\n", + "print \" Velocity of the slider vP = %.1f m/s.\"%(vP)\n", + "print \" Acceleration of the slider aP = %.2f m/s**2.\"%(aP)\n", + "print \" Angular velocity of the connecting rod omegaPC = %.1f rad/s.\"%(omegaPC)\n", + "print \" Angular acceleration of the connecting rod alphaPC = %d rad/s**2.\"%(alphaPC)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Velocity of the slider vP = 6.9 m/s.\n", + " Acceleration of the slider aP = 124.91 m/s**2.\n", + " Angular velocity of the connecting rod omegaPC = 5.9 rad/s.\n", + " Angular acceleration of the connecting rod alphaPC = 481 rad/s**2.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.6 Page No : 532" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "D = 175./1000 #m\n", + "L = 200./1000 #m\n", + "r = L/2 #m\n", + "l = 400./1000 \t #m\n", + "N = 500. \t\t #rpm\n", + "mR = 180. \t\t #kg\n", + "theta = 60 #degrees\n", + "#Solution:\n", + "omega = round(2*math.pi*N/60,1) \t\t\t#rad/s\n", + "\n", + "# Graphical method\n", + "ON = 0.038 # m\n", + "aR = omega**2 * ON\n", + "FI = mR * aR/1000\n", + "print \" Inertia force FI = %.2f kN.\"%(FI)\n", + "\n", + "#Calculating the angular speed of the crank\n", + "\n", + "#Analytical method:\n", + "#Calculating the ratio of lengths of connecting rod and crank\n", + "n = l/r\n", + "#Calculating the inertia force\n", + "FI = mR*omega**2*r*(math.cos(math.radians(theta))+math.cos(math.radians(2*theta))/n)/1000 \t\t\t#kN\n", + "\n", + "#Results:\n", + "print \" Inertia force FI = %.2f kN.\"%(FI)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Inertia force FI = 18.78 kN.\n", + " Inertia force FI = 18.53 kN.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.7 Page No : 533" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import numpy\n", + "\n", + "# Variables:\n", + "r = 300./1000 #m\n", + "l = 1.2 #m\n", + "D = 0.5 \t\t\t#m\n", + "mR = 250. \t\t\t#kg\n", + "theta = 60. \t\t#degrees\n", + "dp = 0.35 \t\t\t#p1-p2 N/mm**2\n", + "N = 250. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the crank\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the net load on the piston\n", + "FL = (dp)*math.pi/4*(D*1000)**2 \t\t\t#N\n", + "#Calculating the ratio of length of connecting rod and crank\n", + "n = l/r\n", + "#Calculating the accelerating or inertia force on reciprocating parts\n", + "FI = mR*omega**2*r*(math.cos(math.radians(theta))+math.cos(math.radians(2*theta))/n) \t\t\t#N\n", + "#Calculating the piston effort\n", + "FP = (FL-FI)/1000 \t\t\t#kN\n", + "#Pressure on slide bars:\n", + "#Calculating the angle of inclination of the connecting rod to the line of stroke\n", + "sinphi = math.sin(math.radians(theta))/n \t\t\t#degrees\n", + "phi = math.degrees(math.asin(sinphi))\n", + "#Calculating the pressure on the slide bars\n", + "FN = FP*math.tan(math.radians(phi)) \t\t\t#kN\n", + "#Calculating the thrust in the connecting rod\n", + "FQ = FP/math.cos(math.radians(phi)) \t\t\t#kN\n", + "#Calculating the tangential force on the crank pin\n", + "FT = FQ*math.sin(math.radians(theta+phi)) \t\t\t#kN\n", + "#Calculating the turning moment on the crank shaft\n", + "T = FT*r \t\t\t#kN-m\n", + "\n", + "#Results:\n", + "print \" Pressure on the slide bars FN = %.2f kN.\"%(FN)\n", + "print \" Thrust in the connecting rod FQ = %.2f kN.\"%(FQ)\n", + "print \" Tangential force on the crank-pin FT = %.2f kN.\"%(FT)\n", + "print \" Turning moment on the crank shaft T = %.3f kN-m.\"%(T)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Pressure on the slide bars FN = 10.97 kN.\n", + " Thrust in the connecting rod FQ = 50.65 kN.\n", + " Tangential force on the crank-pin FT = 48.30 kN.\n", + " Turning moment on the crank shaft T = 14.491 kN-m.\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.8 Page No : 534" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "D = 300./1000 #m\n", + "L = 450./1000 #m\n", + "r = L/2 #m\n", + "d = 50./1000 #m\n", + "l = 1.2 \t\t\t#m\n", + "N = 200. \t\t\t#rpm\n", + "mR = 225. \t\t\t#kg\n", + "theta = 125. \t\t\t#degrees\n", + "p1 = 30*1000. #N/m**2\n", + "p2 = 1.5*1000. \t\t\t#N/m**2\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the crank\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the area of the piston\n", + "A1 = math.pi/4*D**2 \t\t\t#m**2\n", + "#Calculating the area of the piston rod\n", + "a = math.pi/4*d**2 \t\t\t#m**2\n", + "#Calculating the force on the piston due to steam pressure\n", + "FL = round(p1*A1-p2*(A1-a)) \t\t\t#N\n", + "#Calculating the ratio of lengths of connecting rod and crank\n", + "n = l/r\n", + "#Calculating the inertia force on the reciprocating parts\n", + "FI = mR*omega**2*r*(math.cos(math.radians(theta))+math.cos(math.radians(2*theta))/n) \t\t\t#N\n", + "#Calculating the net force on the piston or piston effort\n", + "FP = FL-FI+mR*9.81 \t\t\t#N\n", + "#Calculating the angle of inclination of the connecting rod to the line of stroke\n", + "sinphi = math.sin(math.radians(theta))/n \t\t\t#degrees\n", + "phi = math.degrees(math.asin(sinphi))\n", + "#Calculating the effective turning moment on the crank shaft\n", + "T = FP*math.sin(math.radians(theta+phi))/math.cos(math.radians(phi))*r \t\t\t#N-m\n", + "\n", + "#Results:\n", + "print \" Effective turning moment of the crank shaft T = %.f N-m.\"%(T)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Effective turning moment of the crank shaft T = 3020 N-m.\n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.9 Page No : 534" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "N = 1800. \t\t\t#rpm\n", + "r = 50./1000 #m\n", + "l = 200./1000 #m\n", + "D = 80./1000 #m\n", + "x = 10./1000 \t\t#m\n", + "mR = 1. \t\t\t#kg\n", + "p = 0.7 \t\t\t#N/mm**2\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the crank\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Net load on the gudgeon pin:\n", + "#Calculating the load on the piston\n", + "FL = round(math.pi/4*(D*1000)**2*p) \t\t\t#N\n", + "#Refer Fig. 15.10\n", + "#By measurement\n", + "theta = 33. \t\t\t#degrees\n", + "#Calculating the ratio of lengths of connecting rod and crank\n", + "n = l/r\n", + "#Calculating the inertia force on the reciprocating parts\n", + "FI = mR*omega**2*r*(math.cos(math.radians(theta))+math.cos(math.radians(2*theta))/n) \t\t\t#N\n", + "#Calculating the net load on the gudgeon pin\n", + "FP = FL-FI \t\t\t#N\n", + "#Thrust in the connecting rod:\n", + "#Calculating the angle of inclination of the connecting rod to the line of stroke\n", + "sinphi = math.sin(math.radians(theta))/n \t\t\t#degrees\n", + "phi = math.degrees(math.asin(sinphi))\n", + "#Calculating the thrust in the connecting rod\n", + "FQ = FP/math.cos(math.radians(phi)) \t\t\t#N\n", + "#Calculating the reaction between the piston and cylinder\n", + "FN = FP*math.tan(math.radians(phi)) \t\t\t#N\n", + "#Engine speed at which the abov values will become zero:\n", + "#Calculating the speed at which FI = FL\n", + "omega1 = math.sqrt((math.pi/4*(D*1000)**2*p)/(mR*r*(math.cos(math.radians(theta))+math.cos(math.radians(2*theta))/n))) \t\t\t#rad/s\n", + "#Calculating the corresponding speed in rpm\n", + "N1 = omega1*60/(2*math.pi) \t\t\t#rpm\n", + "\n", + "print phi\n", + "#Results:\n", + "print \" Net load on the gudgeon pin FP = %.f N.\"%(FP)\n", + "print \" Thrust in the connecting rod FQ = %.1f N.\"%(FQ)\n", + "print \" Reaction between the piston and cylinder FN = %d N.\"%(FN)\n", + "print \" Engine speed at which the above values will become zero N1 = %d rpm.\"%(N1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "7.82568845753\n", + " Net load on the gudgeon pin FP = 1848 N.\n", + " Thrust in the connecting rod FQ = 1865.8 N.\n", + " Reaction between the piston and cylinder FN = 254 N.\n", + " Engine speed at which the above values will become zero N1 = 2612 rpm.\n" + ] + } + ], + "prompt_number": 50 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.10 Page No : 536" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "aP = 36. \t\t\t#m/s**2\n", + "theta = 30. \t\t#degrees\n", + "p = 0.5 \t\t\t#N/mm**2\n", + "RF = 600. \t\t\t#N\n", + "D = 300./1000 #m\n", + "r = 300./1000 \t\t#m\n", + "mR = 180. \t\t\t#kg\n", + "n = 4.5\n", + "\n", + "#Solution:\n", + "#Reaction on the guide bars:\n", + "#Calculating the load on the piston\n", + "FL = round(p*math.pi/4*(D*1000)**2) \t\t\t#N\n", + "#Calculating the inertia force due to reciprocating parts\n", + "FI = mR*aP \t\t\t#N\n", + "#Calculating the piston effort\n", + "FP = (FL-FI-RF)/1000 \t\t\t#kN\n", + "#Calculating the angle of inclination of the connecting rod to the line of stroke\n", + "sinphi = math.sin(math.radians(theta))/n \t\t\t#degrees\n", + "phi = math.degrees(math.asin(sinphi))\n", + "#Calculating the reaction on the guide bars\n", + "FN = FP*math.tan(phi) \t\t\t#kN\n", + "#Calculating the thrust on the crank shaft bearing\n", + "FB = (FP*math.cos(math.radians(phi+theta)))/math.cos(math.radians(phi)) \t\t\t#kN\n", + "#Calculating the turning moment on the crank shaft\n", + "T = (FP*math.sin(math.radians(theta+phi)))/math.cos(math.radians(phi))*r \t\t\t#kN-m\n", + "\n", + "\n", + "#Results:\n", + "print \" Reaction on the guide bars FN = %.f kN.\"%(FN)\n", + "print \" Thrust on the crank shaft bearing FB = %.1f kN.\"%(FB)\n", + "print \" Turning moment on the crank shaft T = %.2f kN-m.\"%(T)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Reaction on the guide bars FN = 3 kN.\n", + " Thrust on the crank shaft bearing FB = 22.9 kN.\n", + " Turning moment on the crank shaft T = 5.06 kN-m.\n" + ] + } + ], + "prompt_number": 60 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.11 Page No : 537" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "D = 100./1000 #m\n", + "L = 120./1000 #m\n", + "r = L/2 #m\n", + "l = 250./1000 \t\t\t#m\n", + "mR = 1.1 \t\t\t#kg\n", + "N = 2000. \t \t\t#rpm\n", + "theta = 20. \t\t\t#degrees\n", + "p = 700. \t\t \t#kN/m**2\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the crank\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Net force on the piston:\n", + "#Calculating the force due to gas pressure\n", + "FL = p*math.pi/4*D**2 \t\t\t#kN\n", + "#Calculating the ratio of lengths of the connecting rod and crank\n", + "n = l/r\n", + "#Calculating the inertia force on the piston\n", + "FI = round(mR*omega**2*r*(math.cos(math.radians(theta))+math.cos(math.radians(2*theta))/n)) \t\t\t#N\n", + "#Calculating the net force on the piston\n", + "FP = (FL*1000)-FI+mR*9.81 \t\t\t#N\n", + "#Resulmath.tant force on the gudgeon pin:\n", + "#Calculating the angle of inclination of the connecting rod to the line of stroke\n", + "sinphi = math.sin(theta)/n \t\t\t#degrees\n", + "phi = math.degrees(math.asin(sinphi))\n", + "#Calculating the resultant load on the gudgeon pin\n", + "FQ = round(FP/math.cos(phi)) \t\t\t#N\n", + "#Calculating the thrust on the cylinder walls\n", + "FN = FP*math.tan(math.radians(4.7)) \t\t\t#N \n", + "#Speed above which the gudgeon pin load would be reversed in direction:\n", + "#Calculating the minimum speed for FP to be negative\n", + "omega1 = math.sqrt((FL*1000+mR*9.81)/(mR*r*(math.cos(theta)+math.cos(2*theta)/n))) \t\t\t#rad/s\n", + "#Calculating the corresponding speed in rpm\n", + "N1 = 273*60/(2*math.pi) \t\t\t#rpm\n", + "\n", + "#Results:\n", + "print \" Net force on the piston FP = %.1f N.\"%(FP)\n", + "print \" Resultant load on the gudgeon pin FQ = %d N.\"%(FQ)\n", + "print \" Thrust on the cylinder walls FN = %.1f N.\"%(FN)\n", + "print \" Speed above which the gudgeon pin load would be reversed in direction N1 > %d rpm.\"%(N1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Net force on the piston FP = 2255.6 N.\n", + " Resultant load on the gudgeon pin FQ = 2265 N.\n", + " Thrust on the cylinder walls FN = 185.4 N.\n", + " Speed above which the gudgeon pin load would be reversed in direction N1 > 2606 rpm.\n" + ] + } + ], + "prompt_number": 71 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.12 Page No : 538" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "N = 120. \t\t\t#rpm\n", + "D = 250./1000 #m\n", + "L = 400./1000 #m\n", + "r = L/2 #m\n", + "l = 0.6 #m\n", + "d = 50./1000 \t\t#m\n", + "mR = 60. \t\t\t#kg\n", + "theta = 45. \t\t#degrees\n", + "p1 = 550.*1000 #N/m**2\n", + "p2 = 70.*1000 \t\t\t#N/m**2\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the crank\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Turning moment on the crankshaft:\n", + "#Calculating the area of the piston on the cover end side\n", + "A1 = math.pi/4*D**2 \t\t\t#m**2\n", + "#Calculating the area of the piston rod\n", + "a = math.pi/4*d**2 \t\t\t#m**2\n", + "#Calculating the net load on the piston\n", + "FL = p1*A1-p2*(A1-a) \t\t\t#N\n", + "#Calculating the ratio of lengths of the connecting rod and crank\n", + "n = l/r\n", + "#Calculating the inertia force on the reciprocating parts\n", + "FI = mR*omega**2*r*(math.cos(math.radians(theta))+math.cos(math.radians(2*theta))/n) \t\t\t#N\n", + "#Calculating the net force on the piston or piston effort\n", + "FP = (FL-FI)/1000 \t\t\t#kN\n", + "#Calculating the angle of inclination of the connecting rod to the line of stroke\n", + "sinphi = math.sin(math.radians(theta))/n \t\t\t#degrees\n", + "phi = math.degrees(math.asin(sinphi))\n", + "#Calculating the turning moment on the crank shaft\n", + "T = (FP*math.sin(math.radians(theta+phi)))/math.cos(math.radians(phi))*r*1000\t\t\t#N-m\n", + "#Calculating the thrust on the bearings\n", + "FB = (FP*math.cos(math.radians(theta+phi)))/math.cos(math.radians(phi)) \t\t\t#kN\n", + "#Acceleration of the flywheel:\n", + "P = 20.*1000 \t\t\t#W\n", + "m = 60. \t\t\t#kg\n", + "k = 0.6 \t\t\t#m\n", + "#Calculating the mass moment of inertia of the flywheel\n", + "I = m*k**2 \t\t\t#kg-m**2\n", + "#Calculating the resisting torque\n", + "TR = P*60/(2*math.pi*N) \t\t\t#N-m\n", + "#Calculating the acceleration of the flywheel\n", + "alpha = (T-TR)/I \t\t\t#rad/s**2\n", + "\n", + "\n", + "#Results:\n", + "print \" Turning moment on the crank shaft T = %f N-m.\"%(T) # rounding off error \n", + "print \" Thrust on the bearings FB = %.2f kN.\"%(FB)\n", + "print \" Acceleration of the flywheel alpha = %.1f rad/s**2.\"%(alpha)\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Turning moment on the crank shaft T = 3929.026139 N-m.\n", + " Thrust on the bearings FB = 11.98 kN.\n", + " Acceleration of the flywheel alpha = 108.2 rad/s**2.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.13 Page No : 540" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "D = 300./1000 #m\n", + "L = 500./1000 #m\n", + "r = L/2 \t\t\t#m\n", + "n = 4.5\n", + "N = 180. \t\t\t#rpm\n", + "mR = 280. \t\t\t#kg\n", + "theta = 45. \t\t#degrees\n", + "p1 = 0.1 \t\t\t#N/mm**2\n", + "CR = 14. \t\t\t#Compression ration V1/V2\n", + "p4 = 2.2\n", + "\n", + "#Solution:\n", + "#Refer Fig. 15.12\n", + "#Calculating the angular speed of the crank\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the pressure corresponding to point 2\n", + "p2 = p1*(CR)**1.35 \t\t\t#N/mm**2\n", + "#Calculating the swept volume\n", + "VS = math.pi/4*D**2*L \t\t\t#m**3\n", + "VC = VS/(CR-1) \t\t\t#m**3\n", + "#Calculating the volume corresponding to point 3\n", + "V3 = VC+(1/10*VS) \t\t\t#m**3\n", + "#Calculating the print lacement of the piston corresponding to crank print lacement of 45 degrees\n", + "x = r*((1-math.cos(math.radians(theta)))+(math.sin(math.radians(theta)))**2/(2*n)) \t\t\t#m\n", + "#Calculating the volume corresponding to point 4'\n", + "V4dash = VC+(math.pi/4*D**2*x) \t\t\t#m**2\n", + "#Calculating the pressure corresponding to point 4'\n", + "p3 = p2\n", + "p4dash = p3*(V3/V4dash)**1.35 \t\t\t#N/mm**2\n", + "\n", + "#Calculating the difference of pressures on two sides of the piston\n", + "p = (p4-p1)*10**6 \t\t\t#N/m**2\n", + "#Calculating the net load on the piston\n", + "FL = p*math.pi/4*D**2 \t\t\t#N\n", + "#Calculating the inertia force on the reciprocating parts\n", + "FI = mR*omega**2*r*(math.cos(math.radians(theta))+math.cos(math.radians(2*theta))/n) \t\t\t#N\n", + "#Calculating the net force on the piston or piston effort\n", + "FP = FL-FI+mR*9.81 \t\t\t#N\n", + "#Crank-pin effort:\n", + "#Calculating the angle of inclination of the connecting rod to the line of stroke\n", + "sinphi = math.sin(math.radians(theta))/n \t\t\t#degrees\n", + "phi = math.degrees(math.asin(sinphi))\n", + "#Calculating the crank-pin effort\n", + "FT = (FP*math.sin(math.radians(theta+phi)))/(math.cos(math.radians(phi))*1000) \t\t\t#kN\n", + "#Calculating the thrust on the bearings\n", + "FB = (FP*math.cos(math.radians(theta+phi)))/(math.cos(math.radians(phi))*1000) \t\t\t#kN\n", + "#Calculating the turning moment on the crankshaft\n", + "T = FT*r \t\t\t#kN-m\n", + "\n", + "#Results:\n", + "print \" Crank-pin effort FT = %.3f kN.\"%(FT)\n", + "print \" Thrust on the bearings FB = %.3f kN.\"%(FB)\n", + "print \" Turning moment on the crankshaft T = %.2f kN-m.\"%(T)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Crank-pin effort FT = 109.501 kN.\n", + " Thrust on the bearings FB = 79.438 kN.\n", + " Turning moment on the crankshaft T = 27.38 kN-m.\n" + ] + } + ], + "prompt_number": 94 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.14 Page No : 542" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "D = 240./1000 #m\n", + "L = 360./1000 #m\n", + "r = L/2 #m\n", + "l = 0.6 \t\t\t#m\n", + "N = 300. \t\t\t#rpm\n", + "mR = 160. \t\t\t#kg\n", + "pA = (8+1.03)*10**5\n", + "pE = (-0.75+1.03)*10**5 \t\t\t#N/m**2\n", + "FR = 500. \t \t\t#N\n", + "theta = 75. \t\t\t#degrees\n", + "\n", + "#Solution:\n", + "#Refer Fig. 15.13\n", + "#Calculating the angular speed of the crank\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the stroke volume\n", + "VS = math.pi/4*D**2*L \t\t\t#m**3\n", + "#Calculating the volume of steam at cut-off\n", + "VB = VS/3 \t\t\t#m**3\n", + "#Calculating the ratio of lengths of the connecting rod and crank\n", + "n = l/r\n", + "#Calculating the print lacement of the piston when the crank position is 75 degrees from the top dead centre\n", + "x = r*((1-math.cos(math.radians(theta)))+(math.sin(math.radians(theta)))**2/(2*n)) \t\t\t#m**3\n", + "#Calculating the volume corresponding to point C'\n", + "VCdash = VS*x/L \t\t\t#m**3\n", + "#Calculating the pressure corresponding to point C'\n", + "pB = pA\n", + "pCdash = round((pB*VB)/VCdash) \t\t\t#N/m**2\n", + "#Calculating the difference of pressures on the two sides of the piston\n", + "p = round(pCdash-pE) \t\t\t#N/m**2\n", + "#Calculating the net load on the piston\n", + "FL = round(math.pi/4*D**2*p) \t\t\t#N\n", + "#Calculating the inertia force on the reciprocating parts\n", + "FI = round(mR*omega**2*r*(math.cos(math.radians(theta))+(math.cos(math.radians(2*theta))/n))) \t\t\t#N\n", + "#Calculating the piston effort\n", + "FP = FL-FI+mR*9.81-FR \t\t\t#N\n", + "#Turning moment on the crankshaft:\n", + "#Calculating the angle of inclination of the connecting rod to the line of stroke\n", + "sinphi = math.sin(math.radians(theta))/n \t\t\t#degrees\n", + "phi = round(math.degrees(math.asin(sinphi)),3)\n", + "\n", + "#Calculating the turning moment on the crankshaft\n", + "T = (FP*math.sin(math.radians(theta+phi)))/math.cos(math.radians(phi))*r \t\t\t#N-m\n", + "\n", + "#Results:\n", + "print \" Turning moment on the crankshaft T = %d N-m.\"%(T)\n", + "\n", + "# note : rounding error. please check using calculator" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Turning moment on the crankshaft T = 5778 N-m.\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.15 Page No : 545" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "l = 300. #mm\n", + "l1 = 200. \t\t\t#mm\n", + "m = 15. \t\t\t#kg\n", + "I = 7000. \t\t\t#kg-mm**2\n", + "\n", + "#Solution:\n", + "#Refer Fig. 15.16 and Fig. 15.17\n", + "#Calculating the radius of gyration of the connecting rod about an axis pasmath.sing through its centre of gravity\n", + "kG = math.sqrt(I/m) \t\t\t#mm\n", + "#Calculating the distance of other mass from the centre of gravity\n", + "l2 = (kG)**2/l1 \t\t\t#mm\n", + "#Calculating the magnitude of mass placed at the small end centre\n", + "m1 = (l2*m)/(l1+l2) \t\t\t#kg\n", + "#Calculating the magnitude of the mass placed at a dismath.tance l2 from the centre of gravity\n", + "m2 = (l1*m)/(l1+l2) \t\t\t#kg\n", + "\n", + "#Results:\n", + "print \" Mass placed at the small end centre m1 = %.2f kg.\"%(m1)\n", + "print \" Mass placed at a distance %.2f mm from the centre of gravity m2 = %.2f kg.\"%(l2,m2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Mass placed at the small end centre m1 = 0.17 kg.\n", + " Mass placed at a distance 2.33 mm from the centre of gravity m2 = 14.83 kg.\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.16 Page No : 546" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables: \n", + "h = 650./1000 #m\n", + "l1 = (650.-25)/1000 #m\n", + "m = 37.5 \t\t\t#kg\n", + "tp = 1.87 \t\t\t#seconds\n", + "\n", + "#Solution:\n", + "#Refer Fig. 15.18 and Fig. 15.19\n", + "#Calculating the radius of gyration of the connecting rod about an axis pasmath.sing through its centre of gravity\n", + "kG = math.sqrt((tp/(2*math.pi))**2*(9.81*h)-h**2) \t\t\t#m\n", + "#Calculating the dismath.tance of mass m2 from the centre of gravity\n", + "l2 = (kG)**2/l1 \t\t\t#m\n", + "#Calculating the magnitude of mass placed at the small end centre\n", + "m1 = (l2*m)/(l1+l2) \t\t\t#kg\n", + "#Calculating the magnitude of mass placed at a dismath.tance l2 from centre of gravity\n", + "m2 = (l1*m)/(l1+l2) \t\t\t#kg\n", + "\n", + "#Results:\n", + "print \" Mass placed at the small end centre A m1 = %d kg.\"%(m1)\n", + "print \" Mass placed at a distance %.3f m from G m2 = %.1f kg.\"%(l2,m2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Mass placed at the small end centre A m1 = 10 kg.\n", + " Mass placed at a distance 0.228 m from G m2 = 27.5 kg.\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.17 Page No : 547" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from scipy.optimize import fsolve \n", + "import math \n", + "\n", + "# Variables:\n", + "m = 55. \t\t\t#kg\n", + "l = 850./1000 #m\n", + "d1 = 75./1000 #m\n", + "d2 = 100./1000 \t\t#m\n", + "tp1 = 1.83 #sec\n", + "tp2 = 1.68 \t\t\t#seconds\n", + "\n", + "#Solution:\n", + "#Refer Fig. 15.20\n", + "#Calculating the length of equivalent simple pendulum when suspended from the top of small end bearing\n", + "L1 = 9.81*(tp1/(2*math.pi))**2 \t\t\t#m\n", + "#Calculating the length of equivalent simple pendulum when suspended from the top of big end bearing\n", + "L2 = 9.81*(tp2/(2*math.pi))**2 \t\t\t#m\n", + "#Radius of gyration of the rod about an axis pasmath.sing through the centre of gravity and perpendicular to the plane of oscillation:\n", + "#Calculating the distances of centre of gravity from the top of small end and big end bearings\n", + "#We have h1*(L1-h1) = h2*(L2-h2) or h1**2-h2**2+h2*L2-h1*L1 = 0 .....(i)\n", + "#Also h1+h2 = d1/2+l+d2/2 or h1+h2-d1/2-l-d2/2 = 0 .....(ii)\n", + "def f(x):\n", + " y = [0,0]\n", + " h1 = x[0]\n", + " h2 = x[1]\n", + " y[0] = h1**2-h2**2+h2*L2-h1*L1\n", + " y[1] = h1+h2-d1/2-l-d2/2\n", + " return y\n", + "\n", + "z = fsolve(f,[1,1])\n", + "h1 = z[0]\n", + "h2 = z[1] \t\t\t#m\n", + "\n", + "#Calculating the required radius of gyration of the rod\n", + "kG = math.sqrt(h1*(L1-h1)) \t\t\t#m\n", + "#Calculating the moment of inertia of the rod\n", + "I = m*(kG)**2 \t\t\t#kg-m**2\n", + "#Dynamically equivalent system for the rod:\n", + "#Calculating the distance of the mass situated at the centre of small end bearing from the centre of gravity\n", + "l1 = h1-d1/2 \t\t\t#m\n", + "#Calculating the distance of the second mass from the centre of gravity towards big end bearing\n", + "l2 = (kG)**2/l1 \t\t\t#m\n", + "#Calculating the magnitude of the mass situated at the centre of small end bearing\n", + "m1 = (l2*m)/(l1+l2) \t\t\t#kg\n", + "#Calculating the magnitude of the second mass\n", + "m2 = (l1*m)/(l1+l2) \t\t\t#kg\n", + "\n", + "#Results:\n", + "print \" Radius of gyration of the rod about an axis passing through the centre of\\\n", + " gravity and perpendicular to the plane of oscillation, kG = %.3f m.\"%(kG)\n", + "print \" Moment of inertia of the rod, I = %.2f kg-m**2.\"%(I)\n", + "print \" Magnitude of the mass situated at the centre of small end bearing, m1 = %.2f kg.\"%(m1)\n", + "print \" Magnitude of the second mass, m2 = %.2f kg.\"%(m2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Radius of gyration of the rod about an axis passing through the centre of gravity and perpendicular to the plane of oscillation, kG = 0.345 m.\n", + " Moment of inertia of the rod, I = 6.56 kg-m**2.\n", + " Magnitude of the mass situated at the centre of small end bearing, m1 = 13.32 kg.\n", + " Magnitude of the second mass, m2 = 41.68 kg.\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.18 Page No : 550" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m = 2. \t\t\t#kg\n", + "l = 250./1000 #m\n", + "l1 = 100./1000 #m\n", + "kG = 110./1000 \t\t\t#m\n", + "alpha = 23000. \t\t\t#rad/s**2\n", + "\n", + "#Solution:\n", + "#Equivalent dynamical system:\n", + "#Calculating the distance of the second mass from the centre of gravity\n", + "l2 = (kG)**2/l1 \t\t\t#m\n", + "#Calculating the magnitude of the mass placed at the gudgeon pin\n", + "m1 = (l2*m)/(l1+l2) \t\t\t#kg\n", + "#Calculating the magnitude of the mass placed at a distance l2 from centre of gravity\n", + "m2 = (l1*m)/(l1+l2) \t\t\t#kg\n", + "#Correction couple:\n", + "#Calculating the magnitude of l3\n", + "l3 = l-l1 \t\t\t#m\n", + "#Calculating the new radius of gyration\n", + "k1 = math.sqrt(l1*l3) \t\t\t#m**2\n", + "#Calculating the correction couple\n", + "Tdash = m*(k1**2-kG**2)*alpha \t\t\t#N-m\n", + "\n", + "#Results:\n", + "print \" Mass placed at the gudgeon pin, m1 = %.1f kg.\"%(m1)\n", + "print \" Mass placed at a distance %.3f m from the centre of gravity m2 = %.1f kg.\"%(l2,m2)\n", + "print \" Correction couple Tdash = %.1f N-m.\"%(Tdash)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Mass placed at the gudgeon pin, m1 = 1.1 kg.\n", + " Mass placed at a distance 0.121 m from the centre of gravity m2 = 0.9 kg.\n", + " Correction couple Tdash = 133.4 N-m.\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.19 Page No : 554" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "r = 125. #mm\n", + "OC = r #mm\n", + "l = 500. #mm\n", + "PC = l #mm\n", + "PG = 275. #mm\n", + "kG = 150. \t\t\t#mm\n", + "mC = 60. \t\t\t#kg\n", + "N = 600. \t\t\t#rpm\n", + "theta = 45. \t\t\t#degrees\n", + "\n", + "#Solution:\n", + "#Refer Fig. 15.24\n", + "#Calculating the angular speed of the crank\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Acceleration of the piston:\n", + "#By measurement\n", + "NO = 90./1000 \t\t\t#m\n", + "#Calculating the acceleration of the piston\n", + "aP = omega**2*NO \t\t\t#m/s**2\n", + "#The magnitude position and direction of inertia force due to the mass of the connecting rod:\n", + "#By measurement\n", + "gO = 103./1000 \t\t\t#m\n", + "#Calculating the magnitude of the inertia force of the connecting rod\n", + "FC = mC*omega**2*gO/1000 \t\t\t#kN\n", + "\n", + "#Results:\n", + "print \" Acceleration of the piston aP = %.1f m/s**2.\"%(aP)\n", + "print \" The magnitude of inertia force due to the mass of the connecting rod FC = %.1f kN.\"%(FC)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Acceleration of the piston aP = 355.3 m/s**2.\n", + " The magnitude of inertia force due to the mass of the connecting rod FC = 24.4 kN.\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.20 Page No : 555" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "D = 240./1000 #m\n", + "L = 600./1000 #m\n", + "r = L/2 #m\n", + "l = 1.5 #m\n", + "GC = 500./1000 #m\n", + "kG = 650./1000 \t #m\n", + "mR = 300.\n", + "mC = 250. \t\t\t#kg\n", + "N = 125. \t\t\t#rpm\n", + "theta = 30. \t\t\t#degrees\n", + "\n", + "#Solution:\n", + "#Refer Fig. 15.25\n", + "#Calculating the angular speed of the crank\n", + "omega = round(2*math.pi*N/60,1) \t\t\t#rad/s\n", + "#Analytical method:\n", + "#Calculating the distance of centre of gravity of the connecting rod from P\n", + "l1 = l-GC \t\t\t#m\n", + "#Calculating the ratio of lengths of the connecting rod and crank\n", + "n = l/r\n", + "#Calculating the inertia force due to total mass of the reciprocating parts at P\n", + "FI = int((mR+(l-l1)/l*mC)*omega**2*r*(math.cos(math.radians(theta))+math.cos(math.radians(2*theta))/n)) \t\t\t#N\n", + "#Calculating the corresponding torque due to FI\n", + "TI = int(FI*r*(math.sin(math.radians(theta))+math.sin(math.radians(2*theta))/ \\\n", + "(2*math.sqrt(n**2-(math.sin(math.radians(theta)))**2)))) \t\t\t#N-m\n", + "#Calculating the equivalent length of a simple pendulum when swung about an axis through P\n", + "L = ((kG)**2+(l1)**2)/l1 \t\t\t#m\n", + "#Calculating the correcting torque\n", + "TC = mC*l1*(l-L) *(omega**2 * math.sin(math.radians(60)))/(2*n**2)\t\t\t#N-m\n", + "#Calculating the torque due to the weight of the connecting rod at C\n", + "TW = mC*9.81*(l1/n)*math.cos(math.radians(theta)) \t\t\t#N-m\n", + "#Calculating the total torque exerted on the crankshaft\n", + "Tt = TI+TC+TW \t\t\t#Total torque exerted on the crankshaft N-m\n", + "\n", + "\n", + "#Results:\n", + "print \" Total torque exerted on the crankshaft = %.1f N-m.\"%(round(Tt,-1))\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Total torque exerted on the crankshaft = 3840.0 N-m.\n" + ] + } + ], + "prompt_number": 110 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.21 Page No : 558" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "N = 1200. \t\t\t#rpm\n", + "L = 110./1000\n", + "r = L/2\n", + "l = 250./1000\n", + "PC = l\n", + "CG = 75./1000 \t\t\t#m\n", + "mC = 1.25 \t\t\t#kg\n", + "theta = 40. \t\t\t#degrees\n", + "\n", + "#Solution:\n", + "#Refer Fig. 15.26\n", + "#Calculating the angular speed of the crank\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Radius of gyration of the connecting rod about an axis through its mass centre:\n", + "#Calculating the distance of the centre of gravity from the point of suspension\n", + "l1 = l-CG \t\t\t#m\n", + "PG = l1\n", + "#Calculating the frequency of oscillation\n", + "n = 21./20 \t\t\t#Hz\n", + "#Calculating the radius of gyration of the connecting rod about an axis through its mass centre\n", + "kG = round(math.sqrt((9.81*l1/(2*math.pi*n)**2)-l1**2)*1000) \t\t\t#mm\n", + "#Acceleration of the piston:\n", + "#Calculating the ratio of lengths of the connecting rod and crank\n", + "n = l/r\n", + "#Calculating the acceleration of the piston\n", + "aP = omega**2*r*(math.cos(math.radians(theta))+math.cos(math.radians(2*theta))/n) \t\t\t#m/s**2\n", + "#Calculating the angular acceleration of the connecting rod\n", + "alphaPC = (-omega**2*math.sin(math.radians(theta)))/n \t\t\t#rad/s**2\n", + "#Inertia torque exerted on the crankshaft:\n", + "#Calculating the mass of the connecting rod at P\n", + "m1 = (l-l1)/l*mC \t\t\t#kg\n", + "#Calculating the vertical inertia force\n", + "FI = round(m1*aP) \t\t\t#N\n", + "#By measurement\n", + "OM = 0.0425\n", + "NC = 0.035 \t\t\t#m\n", + "#Calculating the corresponding torque due to FI\n", + "TI = FI*OM \t\t\t#N-m\n", + "#Calculating the equivalent length of a simple pendulum when swung about an axis pasmath.sing through P\n", + "L = ((kG/1000)**2+(l1)**2)/l1 \t\t\t#m\n", + "#Calculating the correction couple\n", + "Tdash = mC*l1*(l-L)*alphaPC \t\t\t#N-m\n", + "#Calculating the corresponding torque on the crankshaft\n", + "TC = -Tdash*math.cos(math.radians(theta))/n \t\t\t#N-m\n", + "#Calculating the torque due to mass at P\n", + "TP = m1*9.81*OM \t\t\t#N-m\n", + "#Calculating the equivalent mass of the connecting rod at C\n", + "m2 = mC*(l1/l) \t\t\t#kg\n", + "#Calculating the torque due to mass at C\n", + "TW = m2*9.81*NC \t\t\t#N-m\n", + "#Calculating the inertia force exerted on the crankshaft\n", + "Ti = TI+TC-TP-TW \t\t\t#Inertia torque exerted on the crankshaft N-m\n", + "\n", + "#Results:\n", + "print \" Radius of gyration of the connecting rod about an axis through its mass centre kG = %d mm.\"%(kG)\n", + "print \" Acceleration of the piston aP = %.1f m/s**2.\"%(aP)\n", + "print \" Angular acceleration of the connecting rod alphaPC = %.1f rad/s**2.\"%(alphaPC)\n", + "print \" Inertia torque exerted on the crankshaft = %.3f N-m.\"%(Ti)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Radius of gyration of the connecting rod about an axis through its mass centre kG = 94 mm.\n", + " Acceleration of the piston aP = 698.5 m/s**2.\n", + " Angular acceleration of the connecting rod alphaPC = -2233.1 rad/s**2.\n", + " Inertia torque exerted on the crankshaft = 12.696 N-m.\n" + ] + } + ], + "prompt_number": 112 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.22 Page No : 559" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "l = 225./1000 #m\n", + "PC = l #m\n", + "L = 150./1000 #m\n", + "r = L/2 #m\n", + "D = 112.5/1000 #m\n", + "PG = 150./1000 #m\n", + "kG = 87.5/1000 \t #m\n", + "mC = 1.6\n", + "mR = 2.4 \t\t\t#kg\n", + "theta = 40 \t\t\t#degrees\n", + "p = 1.8*10**6 \t\t#N/m**2\n", + "N = 2000. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Refer Fig. 15.27\n", + "#Calculating the angular speed of the crank\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#By measurement\n", + "NO = 0.0625\n", + "gO = 0.0685\n", + "IC = 0.29\n", + "IP = 0.24\n", + "IY = 0.148\n", + "IX = 0.08 \t\t\t#m\n", + "#Calculating the force due to gas pressure\n", + "FL = math.pi/4*D**2*p \t\t\t#N\n", + "#Calculating the inertia force due to mass of the reciprocating parts\n", + "FI = mR*omega**2*NO \t\t\t#N\n", + "#Calculating the net force on the piston\n", + "FP = FL-FI \t\t\t#N\n", + "#Calculating the inertia force due to mass of the connecting rod\n", + "FC = mC*omega**2*gO \t\t\t#N\n", + "#Calculating the force acting perpendicular to the crank OC\n", + "FT = ((FP*IP)-((mC*9.81*IY)+(FC*IX)))/IC \t\t\t#N\n", + "#By measurement\n", + "FN = 3550.\n", + "FR = 7550.\n", + "FQ = 13750. \t\t\t#N\n", + "\n", + "#Results:\n", + "print \" Resultant force on the crank pin FQ = %d N.\"%(FQ)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Resultant force on the crank pin FQ = 13750 N.\n" + ] + } + ], + "prompt_number": 36 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Theory_Of_Machines/ch16.ipynb b/Theory_Of_Machines/ch16.ipynb new file mode 100755 index 00000000..1bb89129 --- /dev/null +++ b/Theory_Of_Machines/ch16.ipynb @@ -0,0 +1,1466 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:18b62f9d05dddb3fa3e34cb2cfa86e210f5eaa5cdedab202d52d0e65a17b4192" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 16 : Turning Moment Diagrams and Flywheel" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.1 Page No : 573" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "from numpy import linalg\n", + "\n", + "# Variables:\n", + "m = 6.5*1000 \t\t#kg\n", + "k = 1.8 \t\t\t#m\n", + "deltaE = 56.*1000 \t#N-m\n", + "N = 120. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Calculating the maximum and minimum speeds\n", + "#We know that fluctuation of energy deltaE = math.pi**2/900*m*k**2*N*(N1-N2) or N1-N2 = (deltaE/(math.pi**2/900*m*k**2*N)) .....(i)\n", + "#Also mean speed N = (N1+N2)/2 or N1+N2 = 2*N .....(ii)\n", + "A = [[1, -1],[ 1, 1]]\n", + "B = [deltaE/(math.pi**2/900*m*k**2*N), 2*N]\n", + "V = linalg.solve(A,B)\n", + "N1 = round(V[0]) \t\t\t#rpm\n", + "N2 = round(V[1]) \t\t\t#rpm\n", + "\n", + "#Results:\n", + "print \" Maximum speed N1 = %d rpm.\"%(N1)\n", + "print \" Minimum speed N2 = %d rpm.\"%(N2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Maximum speed N1 = 121 rpm.\n", + " Minimum speed N2 = 119 rpm.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.2 Page No : 573" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "k = 1. \t\t\t#m\n", + "m = 2500. \t\t\t#kg\n", + "T = 1500. \t\t\t#N-m\n", + "\n", + "#Solution:\n", + "#Angular acceleration of the flywheel:\n", + "#Calculating the mass moment of inertia of the flywheel\n", + "I = m*k**2 \t\t\t#kg-m**2\n", + "#Calculating the angular acceleration of the flywheel\n", + "alpha = T/I \t\t\t#rad/s**2\n", + "#Kinetic energy of the flywheel:\n", + "omega1 = 0 \t\t\t#Angular speed at rest\n", + "#Calculating the angular speed after 10 seconds\n", + "omega2 = omega1+alpha*10 \t\t\t#rad/s\n", + "#Calculating the kinetic energy of the flywheel\n", + "KE = 1./2*I*(omega2)**2/1000 \t\t\t#Kinetic energy of the flywheel kN-m\n", + "\n", + "#Results:\n", + "print \" Angular acceleration of the flywheel alpha = %.1f rad/s**2.\"%(alpha)\n", + "print \" Kinetic energy of the flywheel = %.1f kN-m.\"%(KE)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Angular acceleration of the flywheel alpha = 0.6 rad/s**2.\n", + " Kinetic energy of the flywheel = 45.0 kN-m.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.3 Page No : 574" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "P = 300.*1000 \t#W\n", + "N = 90. \t\t#rpm\n", + "CE = 0.1\n", + "k = 2. \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Calculating the mean angular speed\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the coefficient of fluctuation of speed\n", + "CS = 1./100\n", + "#Calculating the work done per cycle\n", + "WD = P*60/N \t\t\t#Work done per cycle N-m\n", + "#Calculating the maximum fluctuation of energy\n", + "deltaE = WD*CE \t\t\t#N-m\n", + "#Calculating the mass of the flywheel\n", + "m = deltaE/(k**2*omega**2*CS) \t\t\t#kg\n", + "#Results:\n", + "print \" Mass of the flywheel, m = %d kg.\"%(m)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Mass of the flywheel, m = 5628 kg.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.4 Page No : 574" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m = 36. \t\t\t#kg\n", + "k = 150./1000 \t\t#m\n", + "N = 1800. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Refer Fig. 16.6\n", + "#Calculating the angular speed of the crank\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the value of 1 mm**2 on the turning moment diagram\n", + "c = 5*math.pi/180 \t\t\t#Value of 1 mm**2 on turning miment diagram N-m\n", + "#Calculating the maximum fluctuation of energy\n", + "#From the turning moment diagram maximum energy = E+295 and minimum energy = E-690\n", + "deltaE = (285-(-690))*c \t\t\t#N-m\n", + "#Calculating the coefficient of fluctuation of energy\n", + "CS = deltaE/(m*k**2*omega**2)*100 \t\t\t#%\n", + "\n", + "#Results:\n", + "print \" Coefficient of fluctuation of speed CS = %.1f %%.\"%(CS)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Coefficient of fluctuation of speed CS = 0.3 %.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.5 Page No : 575" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "N = 600. \t\t\t#rpm\n", + "R = 0.5 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 16.7\n", + "#Calculating the angular speed of the crank\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the coefficient of fluctuation of speed\n", + "CS = 3./100\n", + "#Calculating the value of 1 mm**2 on turning moment diagram\n", + "c = 600*math.pi/60 \t\t\t#Value of 1 mm**2 on turning moment diagram N-m\n", + "#Calculating the maximum fluctuation of energy\n", + "#From the turning moment diagram maximum fluctuation = E+52 and minimum fluctuation = E-120\n", + "deltaE = (52.-(-120))*c \t\t\t#N-m\n", + "#Calculating the mass of the flywheel\n", + "m = deltaE/(R**2*omega**2*CS) \t\t\t#kg\n", + "\n", + "#Results:\n", + "print \" Mass of the flywheel m = %d kg.\"%(m)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Mass of the flywheel m = 182 kg.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.6 Page No : 584\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "N = 250. \t\t\t#rpm\n", + "m = 500. \t\t\t#kg\n", + "k = 600./1000 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 16.8\n", + "#Calculating the angular speed of the crank\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the torque required for one complete cycle\n", + "T = (6*math.pi*750)+(1./2*math.pi*(3000-750))+(2*math.pi*(3000-750))+(1./2*math.pi*(3000-750)) \t\t\t#N-m\n", + "#Calculating the mean torque\n", + "Tmean = T/(6*math.pi) \t\t\t#N-m\n", + "#Calculating the power required to drive the machine\n", + "P = Tmean*omega/1000 \t\t\t#kW\n", + "#Coefficient of fluctuation of speed:\n", + "#Calculating the value of LM\n", + "LM = math.pi*((3000.-1875)/(3000-750.))\n", + "#Calculating the value of NP\n", + "NP = math.pi*((3000.-1875)/(3000-750))\n", + "#Calculating the value of BM\n", + "BM = 3000-1875. \t\t\t#N-m CN = BM\n", + "#Calculating the value of MN\n", + "MN = 2*math.pi\n", + "#Calculating the maximum fluctuation of energy\n", + "deltaE = (1./2*LM*BM)+(MN*BM)+(1./2*NP*BM) \t\t\t#N-m\n", + "#Calculating the coefficient of fluctuation of speed\n", + "CS = deltaE/(m*k**2*omega**2)\n", + "\n", + "#Results:\n", + "print \" Power required to drive the machine P = %.3f kW.\"%(P)\n", + "print \" Coefficient of speed CS = %.3f.\"%(CS)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Power required to drive the machine P = 49.087 kW.\n", + " Coefficient of speed CS = 0.072.\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.7 Page No : 578" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "N = 100. \t\t\t#rpm\n", + "k = 1.75 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 16.9\n", + "#Calculating the angular speed of the crank\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the coefficient of fluctuation of speed\n", + "CS = 1.5/100\n", + "#Coefficient of fluctuation of energy:\n", + "AB = 2000.\n", + "LM = 1500. \t\t\t#N-m\n", + "#Calculating the work done per cycle\n", + "WD = (1./2*math.pi*AB)+(1./2*math.pi*LM) \t\t\t#Work done per cycle N-m\n", + "#Calculating the mean resisting torque\n", + "Tmean = WD/(2*math.pi) \t\t\t#N-m\n", + "#Calculating the value of CD\n", + "CD = math.pi/2000*(2000-875) \t\t\t#rad\n", + "#Calculating the maximum fluctuation of energy\n", + "deltaE = 1./2*CD*(2000-875) \t\t\t#N-m\n", + "#Calculating the coefficient of fluctuation of energy\n", + "Ce = deltaE/WD*100 \t\t\t#%\n", + "#Calculating the mass of the flywheel\n", + "m = deltaE/(k**2*omega**2*CS) \t\t\t#kg\n", + "#Crank angles for minimum and maximum speeds:\n", + "#Calculating the value of CE\n", + "CE = (2000.-875)/2000*(4*math.pi/9) \t\t\t#rad\n", + "#Calculating the crank angle for minimum speed\n", + "thetaC = ((4.*math.pi/9)-CE)*180/math.pi \t\t\t#degrees\n", + "#Calculating the value of ED\n", + "ED = (2000.-875)/2000*(math.pi-(4*math.pi/9)) \t\t\t#rad\n", + "#Calculating the crank angle for maximum speed\n", + "thetaD = ((4.*math.pi/9)+ED)*180/math.pi \t\t\t#degrees\n", + "\n", + "#Results:\n", + "print \" Coefficient of fluctuation of energy CE = %d %%.\"%(Ce)\n", + "print \" Mass of the flywheel, m = %.1f kg.\"%(m)\n", + "print \" Crank angle from IDC for the minimum speed, thetaC = %d degrees.\"%(thetaC)\n", + "print \" Crank angle from IDC for the maximum speed, thetaD = %d degrees.\"%(thetaD)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Coefficient of fluctuation of energy CE = 18 %.\n", + " Mass of the flywheel, m = 197.3 kg.\n", + " Crank angle from IDC for the minimum speed, thetaC = 35 degrees.\n", + " Crank angle from IDC for the maximum speed, thetaD = 136 degrees.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.8 Page No : 580" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "N = 600. \t\t\t#rpm\n", + "Tmax = 90. \t\t\t#N-m\n", + "m = 12. \t\t\t#kg\n", + "k = 80./1000 \t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 16.10\n", + "#Calculating the angular speed of the crank\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Power developed:\n", + "#Calculating the work done per cycle\n", + "WD = 3*1./2*math.pi*90 \t\t\t#Work done per cycle N-m\n", + "#Calculating the mean torque\n", + "Tmean = WD/(2*math.pi) \t\t\t#N-m\\\n", + "#Calculating the power developed\n", + "P = Tmean*omega/1000 \t\t\t#Power developed kW\n", + "#Coefficient of fluctuation of speed:\n", + "#Calculating the maximum fluctuation of energy\n", + "#From the torque-crank angle diagram maximum energy = E+5.89 and minimum energy = E-5.89\n", + "deltaE = 5.89-(-5.89) \t\t\t#N-m\n", + "#Calculating the coefficient of fluctuation of speed\n", + "CS = round(deltaE/(m*k**2*omega**2)*100) \t\t\t#%\n", + "#Calculating the coefficient of fluctuation of energy\n", + "CE = deltaE/WD*100 \t\t\t#%\n", + "#Calculating the maximum angular acceleration of the flywheel\n", + "alpha = (Tmax-Tmean)/(m*k**2) \t\t\t#rad/s**2\n", + "\n", + "#Results:\n", + "print \" Power developed = %.2f kW.\"%(P)\n", + "print \" Coefficient of fluctuation of speed CS = %d %%.\"%(CS)\n", + "print \" Coefficient of fluctuation of energy CE = %.2f %%.\"%(CE)\n", + "print \" Maximum angular acceleration of the flywheel alpha = %d rad/s**2.\"%(alpha)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Power developed = 4.24 kW.\n", + " Coefficient of fluctuation of speed CS = 4 %.\n", + " Coefficient of fluctuation of energy CE = 2.78 %.\n", + " Maximum angular acceleration of the flywheel alpha = 292 rad/s**2.\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.9 Page No : 582" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "P = 20.*1000 \t\t\t#W\n", + "N = 300. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Refer Fig. 16.11\n", + "#Calculating the angular speed of the crank\n", + "omega = 2*math.pi*N/60 \t\t\t#ra/s\n", + "#Calculating the coefficient of fluctuation of speed\n", + "CS = 4./100\n", + "#Calculating the number of working strokes per cycle for a four stroke engine\n", + "n = N/2\n", + "#Calculating the work done per cycle\n", + "WD = P*60/n \t\t\t#Work done per cycle N-m\n", + "#Calculating the work done during expansion cycle\n", + "WE = WD*3./2 \t\t\t#N-m\n", + "#Calculating the maximum turning moment\n", + "Tmax = WE*2/math.pi \t\t\t#N-m\n", + "#Calculating the mean turning moment\n", + "Tmean = WD/(4*math.pi) \t\t\t#N-m\n", + "#Calculating the excess turning moment\n", + "Texcess = Tmax-Tmean \t\t\t#N-m\n", + "#Calculating the value of DE\n", + "DE = Texcess/Tmax*math.pi \t\t\t#rad\n", + "#Calculating the maximum fluctuation of energy\n", + "deltaE = (1./2*DE*Texcess) \t\t\t#N-m\n", + "#Calculating the moment of inertia of the flywheel\n", + "I = deltaE/(omega**2*CS) \t\t\t#kg-m**2\n", + "\n", + "#Results:\n", + "print \" Moment of inertia of the flywheel I = %.1f kg-m**2.\"%(I)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Moment of inertia of the flywheel I = 255.4 kg-m**2.\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.10 Page No : 584" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "a1 = 0.45*10**-3\n", + "a2 = 1.7*10**-3\n", + "a3 = 6.8*10**-3\n", + "a4 = 0.65*10**-3 \t\t\t#m**2\n", + "N1 = 202.\n", + "N2 = 198. \t\t\t#rpm\n", + "R = 1.2 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 16.12\n", + "#Calculating the net area\n", + "a = a3-(a1+a2+a4) \t\t\t#Net area m**2\n", + "#Calculating the energy scale constant\n", + "c = 3*10**6 \t\t\t#Energy scale constant N-m\n", + "#Calculating the net work done per cycle\n", + "WD = a*c \t\t\t#Net work done per cycle N-m\n", + "#Calculating the mean torque\n", + "Tmean = round(WD/(4*math.pi)) \t\t\t#N-m\n", + "#Calculating the value of FG\n", + "FG = Tmean \t\t\t#N-m\n", + "#Calculating the work done during expansion stroke\n", + "WDe = a3*c \t\t\t#Work done during expansion stroke N-m\n", + "#Calculating the value of AG\n", + "AG = WDe/(1./2*math.pi) \t\t\t#N-m\n", + "#Calculating the excess torque\n", + "Texcess = round(AG-FG,-1) \t\t\t#N-m\n", + "#Calculating the value of AF\n", + "AF = Texcess \t\t\t#N-m\n", + "#Calculating the value of DE\n", + "DE = round(AF/AG*math.pi,1) \t\t\t#rad\n", + "#Calculating the maximum fluctuation of energy\n", + "deltaE = 1./2*DE*AF \t\t\t#N-m\n", + "#Mass of the rim of a flywheel:\n", + "#Calculating the mean speed of the flywheel\n", + "N = (N1+N2)/2 \t\t\t#rpm\n", + "#Calculating the mass of the rim of a flywheel\n", + "m = deltaE/(math.pi**2/900*R**2*N*(N1-N2)) \t\t\t#kg\n", + "\n", + "#Results:\n", + "print \" Mass of the rim of the flywheel m = %.f kg.\"%(m)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Mass of the rim of the flywheel m = 1381 kg.\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.11 page no : 585" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from scipy.integrate import quad\n", + "\n", + "# variables\n", + "w = math.pi*2*180./60 # rad/s\n", + "T = 180 # rpm\n", + "Cs = 0.01 # speed\n", + "\n", + "# Calculations\n", + "work_done_r = 20000 * 2 # pi N-m\n", + "Tmean = work_done_r/2 # N-m\n", + "power = round(Tmean * w,-3)/1000\n", + "deltaE = 11078 \n", + "energy = deltaE/round((w**2*Cs),2)\n", + "excess = 9500*math.sin(math.radians(90)) - 5700*math.cos(math.radians(90))\n", + "alpha = excess/energy\n", + "\n", + "# results\n", + "print \"power developed by the engine : %.f kW\"%power\n", + "print \"maximum fluctuation of energy : %.f kg-m**2\"%energy\n", + "print \"Alpha a = %.3f rad/s**2\"%alpha" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "power developed by the engine : 377 kW\n", + "maximum fluctuation of energy : 3121 kg-m**2\n", + "Alpha a = 3.044 rad/s**2\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.12 Page No : 587" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "from scipy.integrate import quad \n", + "\n", + "# Variables:\n", + "m = 500. \t\t\t#kg\n", + "k = 0.4 \t\t\t#m\n", + "N = 150. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Refer Fig. 16.14\n", + "#Calculating the angular speed of the crank\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Fluctuation of energy:\n", + "#Equating the change in torque to zero and calculating the value of theta\n", + "thetaA = math.sin(math.radians(0))\n", + "thetaC = math.sin(math.radians(0))+180\n", + "thetaE = math.sin(math.radians(0))+360 \t\t\t#degrees\n", + "thetaB = 65.4\n", + "thetaD = 294.6\n", + "\n", + "#Calculating the maximum fluctuation of energy\n", + "def f4(theta): \n", + " return (5000+600*math.sin(2*theta))-(5000+500*math.sin(theta))\n", + "\n", + "deltaE = round( quad(f4 ,thetaC*math.pi/180,thetaD*math.pi/180)[0])\n", + "\n", + "#Calculating the total percentage fluctuation of speed\n", + "CS = deltaE/(m*k**2*omega**2)*100 \t\t\t#%\n", + "#Maximum and minimum angular acceleration of the flywheel and the corresponding shaft positions:\n", + "#Calculating the maximum or minimum values of theta\n", + "#Differentiating (600*math.sin(2*theta))-500*math.sin(theta) = 0 with respect to theta and equating to zero\n", + "#we get 12*2*(math.cos(theta))**2-5*math.cos(theta)-12 = 0\n", + "a = 12.*2\n", + "b = -5.\n", + "c = -12.\n", + "costheta1 = (-b+math.sqrt(b**2-4*a*c))/(2*a)\n", + "costheta2 = (-b-math.sqrt(b**2-4*a*c))/(2*a)\n", + "theta1 = math.degrees(math.acos(costheta1))\n", + "theta2 = math.degrees(math.acos(costheta2)) \t\t\t#degrees\n", + "#Calculating the maximum torque\n", + "Tmax = 600*math.sin(math.radians(2*theta1))-500*math.sin(math.radians(theta1)) \t\t\t#N-m\n", + "#Calculating the minimum torque\n", + "Tmin = 600*math.sin(math.radians(2*theta2))-500*math.sin(math.radians(theta2)) \t\t\t#N-m\n", + "#Calculating the maximum acceleration\n", + "alphamax = Tmax/(m*k**2) \t\t\t#rad/s**2\n", + "#Calculating the minimum acceleration\n", + "alphamin = abs(Tmin)/(m*k**2) \t\t\t#rad/s**2\n", + "\n", + "\n", + "#Results:\n", + "print \" Fluctuation of energy deltaE = %d N-m.\"%(deltaE)\n", + "print \" Total percentage fluctuation of speed CS = %.1f %%.\"%(CS)\n", + "print \" Shaft position corresponding to maximum and minimum accelerations\\\n", + " theta = %d degrees and %.1f degrees.\"%(theta1,theta2)\n", + "print \" Maximum acceleration, alphamax = %.2f rad/s**2.\"%(alphamax)\n", + "print \" Minimum acceleration alphamin = %.1f rad/s**2.\"%(alphamin)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Fluctuation of energy deltaE = 1204 N-m.\n", + " Total percentage fluctuation of speed CS = 6.1 %.\n", + " Shaft position corresponding to maximum and minimum accelerations theta = 35 degrees and 127.6 degrees.\n", + " Maximum acceleration, alphamax = 3.46 rad/s**2.\n", + " Minimum acceleration alphamin = 12.2 rad/s**2.\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.13 Page No : 589" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "from scipy.integrate import quad \n", + "\n", + "# Variables:\n", + "I = 1000. \t\t\t#kg-m**2\n", + "N = 300. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Refer Fig. 16.15 and Fig. 16.16\n", + "#Calculating the angular speed of the crank\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Power of the engine:\n", + "#Calculating the work done per revolution\n", + "def f0(theta): \n", + " return 5000+1500*math.sin(3*theta)\n", + "\n", + "WD = quad(f0,0,2*math.pi)[0]\n", + "\n", + "#Calculating the mean resisting torque\n", + "Tmean = WD/(2*math.pi) \t\t\t#N-m\n", + "#Calculating the power of the engine\n", + "P = Tmean*omega/1000 \t\t\t#kW\n", + "#Maximum fluctuation of the speed of the flywheel when resisting torque is consmath.tant:\n", + "#Calculating the value of theta \n", + "theta = (5000-5000)/1500\n", + "theta = 1./3*(math.sin(math.radians((theta)))+180) \t\t\t#degrees\n", + "#Calculating the maximum fluctuation of energy\n", + "def f1(theta): \n", + " return 5000+1500*math.sin(3*theta)-5000\n", + "\n", + "deltaE = quad(f1,0,60*math.pi/180)[0]\n", + "\n", + "#Calculating the maximum fluctuation of speed of the flywheel\n", + "CS1 = deltaE/(I*omega**2)*100 \t\t\t#%\n", + "#Maximum fluctuation of speed of the flywheel when resisting torque (5000+600*math.sin(theta)) N-m:\n", + "#Calculating the values of theta thetaB and thetaC\n", + "thetaB = math.sin(math.radians(math.sqrt((1./4*(3-600./1500))))) \t\t\t#degrees\n", + "thetaC = 180-thetaB \t\t\t#degrees\n", + "#Calculating the maximum fluctuation of energy\n", + "\n", + "def f2(theta): \n", + " return (5000+1500*math.sin(3*theta))-(5000+600*math.sin(theta))\n", + "\n", + "deltaE = round( quad(f2,thetaB*math.pi/180,thetaC*math.pi/180)[0])\n", + "\n", + "#Calculating the maximum fluctuation of speed of the flywheel\n", + "CS2 = abs(deltaE)/(I*omega**2)*100 \t\t\t#%\n", + "\n", + "#Results:\n", + "print \" Power of the engine P = %.1f kW.\"%(P)\n", + "print \" Maximum fluctuation of the speed of the flywheel when resisting torque\\\n", + " is constant, CS = %.1f %%.\"%(CS1)\n", + "print \" Maximum fluctuation of speed of the flywheel when resisting torque \\\n", + " 5000+600*sintheta N-m CS = %.3f %%.\"%(CS2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Power of the engine P = 157.1 kW.\n", + " Maximum fluctuation of the speed of the flywheel when resisting torque is constant, CS = 0.1 %.\n", + " Maximum fluctuation of speed of the flywheel when resisting torque 5000+600*sintheta N-m CS = 0.020 %.\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.14 Page No : 592" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "N = 800. \t\t\t#rpm\n", + "stroke = 300. \t\t\t#mm\n", + "sigma = 7.*10**6 \t\t\t#N/m**2\n", + "rho = 7200. \t\t\t#kg/m**3\n", + "\n", + "#Solution:\n", + "#Refer Fig. 16.18\n", + "#Calculating the angular speed of the engine\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the coefficient of fluctuation of speed\n", + "CS = 4./100\n", + "#Diameter of the flywheel rim:\n", + "#Calculating the peripheral velocity of the flywheel rim\n", + "v = math.sqrt(sigma/rho) \t\t\t#m/s\n", + "#Calculating the diameter of the flywheel rim\n", + "D = v*60/(math.pi*N) \t\t\t#m\n", + "#Cross-section of the flywheel rim:\n", + "#Calculating the value of 1 mm**2 on the turning moment diagram\n", + "c = 500.*math.pi/30 \t\t\t#Value of 1 mm**2 on the turning moment diagram N-m\n", + "#Calculating the maximum fluctuation of energy\n", + "deltaE = round((420.-(-30))*c) \t\t\t#N-m\n", + "#Calculating the mass of the flywheel rim\n", + "m = deltaE/(v**2*CS) \t\t\t#kg\n", + "#Calculating the thickness of the flywheel rim\n", + "t = math.sqrt(m/(math.pi*D*5*rho))*1000 \t\t\t#mm\n", + "#Calculating the width of the flywheel rim\n", + "b = 5*t \t\t\t #mm\n", + "\n", + "#Results:\n", + "print \" Diameter of the flywheel rim D = %.3f m.\"%(D)\n", + "print \" Thickness of the flywheel rim t = %d mm.\"%(t)\n", + "print \" Width of the flywheel rim b = %d mm.\"%(b)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Diameter of the flywheel rim D = 0.744 m.\n", + " Thickness of the flywheel rim t = 84 mm.\n", + " Width of the flywheel rim b = 424 mm.\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.15 Page No : 594" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "P = 150.*1000 \t\t\t#W\n", + "N = 80. \t\t\t#rpm\n", + "CE = 0.1\n", + "D = 2.\n", + "R = D/2. \t\t\t#m\n", + "rho = 7200. \t\t\t#kg/m**3\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the engine\n", + "omega = round(2*math.pi*N/60,1) \t\t\t#rad/s\n", + "#Calculating the coefficient of fluctuation of speed\n", + "CS = 4./100\n", + "#Mass of the flywheel rim:\n", + "#Calculating the work done per cycle\n", + "WD = P*60/N \t\t\t#Work done per cycle N-m\n", + "#Calculating the maximum fluctuation of energy\n", + "deltaE = WD*CE \t\t\t#N-m\n", + "#Calculating the mass moment of inertia of the flywheel\n", + "I = deltaE/(omega**2*CS) \t\t\t#kg-m**2\n", + "#Calculating the mass moment of inertia of the flywheel rim\n", + "Irim = 0.95*I \t\t\t#kg-m**2\n", + "#Calculating the mass of the flywheel rim\n", + "k = R \t\t\t#Radius of gyration m\n", + "m = Irim/k**2 \t\t\t#kg\n", + "#Calculating the cross-sectional area of the flywheel rim\n", + "A = m/(2*math.pi*R*rho) \t\t\t#m**2\n", + "\n", + "#Resilts:\n", + "print \" Mass of the flywheel rim m = %.f kg.\"%(m)\n", + "print \" Cross-sectional area of the flywheel rim A = %.3f m**2.\"%(A)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Mass of the flywheel rim m = 3787 kg.\n", + " Cross-sectional area of the flywheel rim A = 0.084 m**2.\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.16 Page No : 595" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "N = 600. \t\t\t#rpm\n", + "rho = 7250. \t\t\t#kg/m**3\n", + "sigma = 6.*10**6 \t\t\t#N/m**2\n", + "\n", + "#Solution:\n", + "#Refer Fig. 16.19\n", + "#Calculating the angular speed of the engine\n", + "omega = 2.*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the total fluctuation of speed\n", + "CS = 2./100\n", + "#Moment of inertia of the flywheel:\n", + "#Calculating the value of 1 mm**2 of turning moment diagram\n", + "c = 250.*math.pi/60 \t\t\t#Value of 1 mm**2 of turning moment diagram N-m\n", + "#Calculating the maximum fluctuation of energy\n", + "deltaE = round((162.-(-35))*c) \t\t\t#N-m\n", + "#Calculating the moment of inertia of the flywheel\n", + "I = deltaE/(omega**2*CS) \t\t\t#kg-m**2\n", + "#Dimensions of the flywheel rim:\n", + "#Calculating the peripheral velocity of the flywheel\n", + "v = math.sqrt(sigma/rho) \t\t\t#m/s\n", + "#Calculating the mean diameter of the flywheel\n", + "D = v*60/(math.pi*N) \t\t\t#m\n", + "#Calculating the maximum fluctuation of energy of the flywheel rim\n", + "deltaErim = 0.92*deltaE \t\t\t#N-m\n", + "#Calculating the mass of the flywheel rim\n", + "m = deltaErim/(v**2*CS) \t\t\t#kg\n", + "#Calculating the thickness of the flywheel rim\n", + "t = math.sqrt(m/(math.pi*D*2*rho))*1000 \t\t\t#mm\n", + "#Calculating the breadth of the flywheel rim\n", + "b = 2*t \t\t\t#mm\n", + "\n", + "#Results:\n", + "print \" Moment of inertia of the flywheel I = %.1f kg-m**2.\"%(I)\n", + "print \" Thickness of the flywheel rim t = %.1f mm.\"%(t)\n", + "print \" Breadth of the flywheel rim b = %.1f mm.\"%(b)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Moment of inertia of the flywheel I = 32.7 kg-m**2.\n", + " Thickness of the flywheel rim t = 58.6 mm.\n", + " Breadth of the flywheel rim b = 117.2 mm.\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.17 Page No : 596" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "a1 = 5.*10**-5 #m**2\n", + "a2 = 21.*10**-5 #m**2\n", + "a3 = 85.*10**-5 #m**2\n", + "a4 = 8.*10**-5 \t\t\t#m**2\n", + "N2 = 98.\n", + "N1 = 102. \t\t\t#rpm\n", + "rho = 8150. \t\t\t#kg/m**3\n", + "sigma = 7.5*10**6 \t\t\t#N/m**2\n", + "\n", + "#Solution:\n", + "#Refer Fig. 16.20\n", + "#Calculating the net area\n", + "a = a3-(a1+a2+a4) \t\t\t#Net area m**2\n", + "#Calculating the value of 1 m**2 on the turning moment diagram in terms of work\n", + "c = 14*10**6 \t\t\t#Value of 1 m**2 on the turning moment diagram N-m\n", + "#Calculating the net work done per cycle\n", + "WD = a*c \t\t\t#Net work done per cycle N-m\n", + "#Calculating the mean torque on the flywheel\n", + "Tmean = round(WD/(4*math.pi)) \t\t\t#N-m\n", + "FG = Tmean \t\t\t#N-m\n", + "#Calculating the work done during expansion stroke\n", + "WDe = int(a3*c) \t\t\t#Work done during expansion stroke N-m\n", + "#Calculating the value of AG\n", + "AG = int(WDe/(1./2*math.pi)) \t\t\t#N-m\n", + "#Calculating the excess torque\n", + "Texcess = AG-FG \t\t\t#Excess torque N-m\n", + "AF = Texcess \t\t\t#N-m\n", + "#Calculating the value of DE\n", + "DE = round(AF/AG*math.pi,1) \t\t\t#rad\n", + "#Calculating the maximum fluctuation of energy\n", + "deltaE = 1./2*DE*AF \t\t\t#N-m\n", + "#Moment of inertia of the flywheel:\n", + "#Calculating the mean speed during the cycle\n", + "N = (N1+N2)/2 \t\t\t#rpm\n", + "#Calculating the corresponding angular mean speed\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the coefficient of fluctuation of speed\n", + "CS = (N1-N2)/N\n", + "#Calculating the moment of inertia of the flywheel\n", + "I = deltaE/(omega**2*CS) \t\t\t#kg-m**2\n", + "#Size of flywheel:\n", + "#Calculating the peripheral velocity of the flywheel\n", + "v = math.sqrt(sigma/rho) \t\t\t#m/s\n", + "#Calculating the mean diameter of the flywheel\n", + "D = v*60/(math.pi*N) \t\t\t#m\n", + "#Calculating the mass of the flywheel rim\n", + "m = deltaE/(v**2*CS) \t\t\t#kg\n", + "#Calculating the thickness of the flywheel rim\n", + "t = math.sqrt(m/(math.pi*D*4*rho))*1000 \t\t\t#mm\n", + "#Calculating the width of the flywheel rim\n", + "b = 4*t \t\t\t#mm\n", + "\n", + "#Results:\n", + "print \" Moment of inertia of the flywheel I = %.f kg-m**2.\"%(I)\n", + "print \" Thickness of the flywheel rim t = %.1f mm.\"%(t)\n", + "print \" Width of the flywheel rim b = %.1f mm.\"%(b)\n", + "\n", + "# rounding off error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Moment of inertia of the flywheel I = 2316 kg-m**2.\n", + " Thickness of the flywheel rim t = 21.6 mm.\n", + " Width of the flywheel rim b = 86.3 mm.\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.18 Page No : 599" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "P = 50.*1000 \t\t\t#W\n", + "N = 150. \t\t\t#rpm\n", + "n = 75.\n", + "sigma = 4.*10**6 \t\t\t#N/m**2\n", + "rho = 7200. \t\t\t#kg/m**3\n", + "\n", + "#Solution:\n", + "#Refer Fig. 16.21\n", + "#Calculating the angular speed of the engine\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the mean torque transmitted by the flywheel\n", + "Tmean = P/omega \t\t\t#N-m\n", + "FG = Tmean \t\t\t#N-m\n", + "#Calculating the work done per cycle\n", + "WD = Tmean*4*math.pi \t\t\t#Work done per cycle N-m\n", + "#Calculating the work done during power stroke\n", + "WDp = 1.4*WD \t\t\t#Work done during power stroke N-m\n", + "#Calculating the maximum torque transmitted by the flywheel\n", + "Tmax = WDp/(1./2*math.pi) \t\t\t#N-m\n", + "BF = Tmax \t\t\t#N-m\n", + "#Calculating the excess torque\n", + "Texcess = Tmax-Tmean \t\t\t#N-m\n", + "BG = Texcess \t\t\t#N-m\n", + "#Calculating the value of DE\n", + "DE = BG/BF*math.pi \t\t\t#N-m\n", + "#Calculating the maximum fluctuation of energy\n", + "deltaE = 1./2*DE*BG \t\t\t#N-m\n", + "#Mean diameter of the flywheel:\n", + "#Calculating the peripheral velocity of the flywheel\n", + "v = math.sqrt(sigma/rho) \t\t\t#m/s\n", + "#Calculating the mean diameter of the flywheel\n", + "D = v*60./(math.pi*N) \t\t\t#m\n", + "#Cross-sectional dimensions of the rim:\n", + "#Calculating the coefficient of fluctuation of speed\n", + "CS = 1./100\n", + "#Calculating the total energy of the flywheel\n", + "E = deltaE/(2*CS) \t\t\t#N-m\n", + "#Calculating the energy of the rim\n", + "Erim = 15./16*E \t\t\t#N-m\n", + "#Calculating the mass of the flywheel rim\n", + "m = Erim/(1./2*v**2) \t\t\t#kg\n", + "#Calculating the thickness of the rim\n", + "t = round(math.sqrt(m/(math.pi*D*4*rho))*1000) \t\t\t#mm\n", + "#Calculating the width of the rim\n", + "b = 4*t \t\t\t#mm\n", + "\n", + "#Results:\n", + "print \" Mean diameter of the flywheel D = %d m.\"%(D)\n", + "print \" Thickness of the flywheel rim t = %d mm.\"%(t)\n", + "print \" Width of the flywheel rim b = %d mm.\"%(b)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Mean diameter of the flywheel D = 3 m.\n", + " Thickness of the flywheel rim t = 170 mm.\n", + " Width of the flywheel rim b = 680 mm.\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.19 Page No : 603" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "N1 = 225.\n", + "N2 = 200. \t\t\t#rpm\n", + "k = 0.5 \t\t\t#m\n", + "E1 = 15.*1000 \t\t\t#N-m\n", + "HolePunched = 720. \t\t\t#per hour\n", + "\n", + "#Solution:\n", + "#Power of the motor:\n", + "#Calculating the total energy required per second\n", + "E = E1*HolePunched/3600 \t\t\t#N-m/s\n", + "#Calculating the power of the motor\n", + "P = E/1000 \t\t\t#kW\n", + "#Minimum mass of the flywheel:\n", + "#Calculating the energy supplied by the motor in 2 seconds\n", + "E2 = E*2 \t\t\t#N-m\n", + "#Calculating the energy supplied by the flywheel during punching\n", + "deltaE = E1-E2 \t\t\t#N-m\n", + "#Calculating the mean speed of the flywheel\n", + "N = (N1+N2)/2 \t\t\t#rpm\n", + "#Calculating the minimum mass of the flywheel\n", + "m = round(deltaE*900/(math.pi**2*k**2*N*(N1-N2))) \t\t\t#kg\n", + "\n", + "#Results:\n", + "print \" Power of the motor P = %d kW.\"%(P)\n", + "print \" Minimum mass of the flywheel m = %d kg.\"%(m)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Power of the motor P = 3 kW.\n", + " Minimum mass of the flywheel m = 618 kg.\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.20 Page No : 603" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "d = 38. #mm\n", + "t = 32. #mm\n", + "s = 100. \t\t\t#mm\n", + "E1 = 7. \t\t\t#N-m/mm**2 of sheared area\n", + "v = 25. \t\t\t#m/s\n", + "\n", + "#Solution:\n", + "#Power of the motor required:\n", + "#Calculating the sheared area\n", + "A = round(math.pi*d*t) \t\t\t#mm**2\n", + "#Calculating the total energy required per hole\n", + "E1 = E1*A \t\t\t#N-m\n", + "#Calculating the energy required for punching work per second\n", + "E = E1/10 \t\t\t#Energy required for punching work per second N-m/s\n", + "#Calculating the power of the motor required\n", + "P = E/1000 \t\t\t#Power of the motor required kW\n", + "#Mass of the flywheel required:\n", + "#Calculating the time required to punch a hole in a 32 mm thick plate\n", + "t32 = 10/(2*s)*t \t\t\t#Time required to punch a hole in 32 mm thick plate seconds\n", + "#Calculating the energy supplied by the motor in t32 seconds\n", + "E2 = E*t32 \t\t\t#N-m\n", + "#Calculating the energy to be supplied by the flywheel during punching\n", + "deltaE = E1-E2 \t\t\t#N-m\n", + "#Calculating the coefficient of fluctuation of speed\n", + "CS = 3/100.\n", + "#Calculating the mass of the flywheel required\n", + "m = round(deltaE/(v**2*CS)) \t\t\t#kg\n", + "\n", + "#Results:\n", + "print \" Power of the motor required P = %.3f kW.\"%(P)\n", + "print \" Mass of the flywheel required m = %d kg.\"%(m)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Power of the motor required P = 2.674 kW.\n", + " Mass of the flywheel required m = 1198 kg.\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.21 Page No : 604" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "P = 3. \t\t\t#kW\n", + "m = 150. \t\t\t#kg\n", + "k = 0.6 \t\t\t#m\n", + "N1 = 300. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Calculating the angular speed of the flywheel before riveting\n", + "omega1 = 2*math.pi*N1/60 \t\t\t#rad/s\n", + "#Speed of the flywheel immediately after riveting:\n", + "#Calculating the energy supplied by the motor\n", + "E2 = P*1000 \t\t\t#N-m/s\n", + "#Calculating the energy absorbed during one riveting operation which takes 1 second\n", + "E1 = 10000 \t\t\t#N-m\n", + "#Calculating the energy to be supplied by the flywheel for each riveting operation per second\n", + "deltaE = E1-E2 \t\t\t#N-m\n", + "#Calculating the angular speed of the flywheel immediately after riveting\n", + "omega2 = math.sqrt(omega1**2-(2*deltaE/(m*k**2))) \t\t\t#rad/s\n", + "#Calculating the corresponding speed in rpm\n", + "N2 = omega2*60/(2*math.pi) \t\t\t#rpm\n", + "#Calculating the number of rivets that can be closed per minute\n", + "n = E2/E1*60 \t\t\t#Number of rivets that can be closed per minute\n", + "\n", + "#Results:\n", + "print \" Speed of the flywheel immediately after riveting N2 = %.1f rpm.\"%(N2)\n", + "print \" Number of rivets that can be closed per minute = %d rivets.\"%(n)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Speed of the flywheel immediately after riveting N2 = 257.6 rpm.\n", + " Number of rivets that can be closed per minute = 18 rivets.\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.22 Page No : 605" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "d = 40. #mm\n", + "t = 15. \t\t\t#mm\n", + "NoofHoles = 30. \t\t\t#per minute\n", + "EnergyRequired = 6. \t\t\t#N-m/mm**2\n", + "Time = 1./10 \t\t\t#seconds\n", + "N1 = 160.\n", + "N2 = 140. \t\t\t#rpm\n", + "k = 1. \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Calculating the sheared area per hole\n", + "A = round(math.pi*d*t) \t\t\t#Sheared area per hole mm**2\n", + "#Calculating the energy required to punch a hole\n", + "E1 = EnergyRequired*A \t\t\t#N-m\n", + "#Calculating the energy required for punching work per second\n", + "E = E1*NoofHoles/60 \t\t\t#Energy required for punching work per second N-m/s\n", + "#Calculating the energy supplied by the motor during the time of punching\n", + "E2 = E*Time \t\t\t#N-m\n", + "#Calculating the energy to be supplied by the flywheel during punching a hole\n", + "deltaE = E1-E2 \t\t\t#N-m\n", + "#Calculating the mean speed of the flywheel\n", + "N = (N1+N2)/2 \t\t\t#rpm\n", + "#Calculating the mass of the flywheel required\n", + "m = round(deltaE*900/(math.pi**2*k**2*N*(N1-N2))) \t\t\t#kg\n", + "\n", + "#Results:\n", + "print \" Mass of the flywheel required m = %d kg.\"%(m)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Mass of the flywheel required m = 327 kg.\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.23 Page No : 606" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "n = 25.\n", + "d1 = 25./1000 #m\n", + "t1 = 18./1000 #m\n", + "D = 1.4\n", + "R = D/2 \t\t\t#m\n", + "touu = 300.*10**6 \t\t\t#N/m**2\n", + "etam = 95./100\n", + "CS = 0.1\n", + "sigma = 6.*10**6 \t\t\t#N/m**2\n", + "rho = 7250. \t\t\t#kg/m**3\n", + "\n", + "#Solution:\n", + "#Power needed for the driving motor:\n", + "#Calculating the area of the plate sheared\n", + "AS = math.pi*d1*t1 \t\t\t#m**2\n", + "#Calculating the maximum shearing force required for punching\n", + "FS = AS*touu \t\t\t#N\n", + "#Calculating the energy required per stroke\n", + "E = 1./2*FS*t1 \t\t\t#Energy required per stroke N-m\n", + "#Calculating the energy required per minute\n", + "E1 = E*n \t\t\t#Energy required per minute N-m\n", + "#Calculating the power required for the driving motor\n", + "P = E1/(60*etam)/1000 \t\t\t#Energy required for the driving motor kW\n", + "#Dimensions for the rim cross-section:\n", + "#Calculating the maximum fluctuation of energy\n", + "deltaE = 9./10*E \t\t\t#N-m\n", + "#Calculating the maximum fluctuation of energy provided by the rim\n", + "deltaErim = 0.95*deltaE \t\t\t#N-m\n", + "#Calculating the mean speed of the flywheel\n", + "N = 9.*25 \t\t\t#rpm\n", + "#Calculating the mean angular speed\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the mass of the flywheel\n", + "m = round(deltaErim/(R**2*omega**2*CS)) \t\t\t#kg\n", + "#Calculating the thickness of rim\n", + "t = math.sqrt(m/(math.pi*D*2*rho))*1000 \t\t\t#mm\n", + "#Calculating the width of rim\n", + "b = 2*t \t\t\t#mm\n", + "\n", + "#Results:\n", + "print \" Power needed for the driving motor = %.3f kW.\"%(P)\n", + "print \" Thickness of the flywheel rim t = %d mm.\"%(t)\n", + "print \" Width of the flywheel rim b = %d mm.\"%(b)\n", + "#Answers vary due to rounding-off errors" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Power needed for the driving motor = 1.674 kW.\n", + " Thickness of the flywheel rim t = 43 mm.\n", + " Width of the flywheel rim b = 86 mm.\n" + ] + } + ], + "prompt_number": 32 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Theory_Of_Machines/ch2.ipynb b/Theory_Of_Machines/ch2.ipynb new file mode 100755 index 00000000..cfa99e45 --- /dev/null +++ b/Theory_Of_Machines/ch2.ipynb @@ -0,0 +1,361 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:1f1031fc7a31a26f2df358a4d88897e7d62132cd9ac575d363637d46b39b8c4e" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 2 : Kinematics of Motion" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.1 Page No: 13 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "u1 = 0\n", + "v1 = 72.*1000./3600 \t\t\t#m/s\n", + "s1 = 500. \t\t\t #m\n", + "\n", + "# Solution:\n", + "# Calculating the initial acceleration of the car\n", + "a1 = (v1**2-u1**2)/(2*s1) \t\t\t#m/s**2\n", + "#Calculating time taken by the car to attain the speed\n", + "t1 = (v1-u1)/a1 \t\t\t#seconds\n", + "#Parameters for the second case\n", + "u2 = v1\n", + "v2 = 90.*1000/3600 \t\t\t#m/s\n", + "t2 = 10. \t\t\t#seconds\n", + "\n", + "#Calculating the acceleration for the second case\n", + "a2 = (v2-u2)/t2 \t\t\t#m/s**2\n", + "#Calculating the distance moved by the car in the second case\n", + "s2 = (u2*t2)+(a2/2*t2**2)\n", + "#Parameters for the third case\n", + "u3 = v2\n", + "v3 = 0 \t\t\t#m/s\n", + "t3 = 5 \t\t\t#seconds\n", + "#Calculating the distance moved by the car\n", + "s3 = (u3+v3)*t3/2 \t\t\t#m\n", + "\n", + "#Results:\n", + "print \" The acceleration of the car, a = %.1f m/s**2. \"%(a1)\n", + "print \" The car takes t = %d s to attain the speed.\"%(t1)\n", + "print \" The acceleration of the car in the second case, a = %.1f m/s**2.\"%(a2)\n", + "print \" The distance moved by the cars = %d m.\"%(s2)\n", + "print \" The distance travelled by the car during braking, s = %.1f m.\"%(s3)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The acceleration of the car, a = 0.4 m/s**2. \n", + " The car takes t = 50 s to attain the speed.\n", + " The acceleration of the car in the second case, a = 0.5 m/s**2.\n", + " The distance moved by the cars = 225 m.\n", + " The distance travelled by the car during braking, s = 62.5 m.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.2 Page no : 14" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# variables\n", + "t = 1. # second\n", + "v = 6.25 # m/s\n", + "\n", + "# calculations and results\n", + "C1 = v - 0.25 +t -5\n", + "\n", + "# when t = 2\n", + "t = 2.\n", + "v = t**4/4 - t**3 + 5*t + 2\n", + "print \"Velocity at t=2 seconds, V = %.f m/s\"%v\n", + "\n", + "# when t = 1 seconds and s = 8.30 m.\n", + "t = 1.\n", + "s = 8.30\n", + "C2 = s - 1./20 + 1./4 - 5./2 - 2\n", + "t = 2. # seconds\n", + "s = t**5/20 - t**4/4 + 5*t**2/2 + 2*t + 4\n", + "print \"Displacement = %.1f m\"%s\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Velocity at t=2 seconds, V = 8 m/s\n", + "Displacement = 15.6 m\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.3 Page No: 15" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "from scipy.integrate import quad \n", + "\n", + "# Variables:\n", + "#Initial parameters\n", + "v0 = 100. \t\t\t#kmph\n", + "t0 = 0\n", + "#Parameters at the end of 40 seconds\n", + "v1 = 90./100*v0 \t\t\t#kmph\n", + "t1 = 40. \t\t\t#seconds\n", + "\n", + "#Solution:\n", + "#The acceleration is given by\n", + "#a = (-dv/dt) = k*v\n", + "#Integrating\n", + "#we get ln(v) = -k*t+C\n", + "#Calculating the constant of integration\n", + "def f3(v): \n", + " return 1./v\n", + "\n", + "C = quad(f3,1,100)[0]\n", + "\n", + "#Calculating the constant of proportionality\n", + "k = (C-2.3*math.log10(90))/40\n", + "#Time after 120 seconds\n", + "t2 = 120. \t\t\t#seconds\n", + "#Calculating the velocity after 120 seconds\n", + "v120 = 10**((-k*t2+C)/2.29)\n", + "\n", + "#Results:\n", + "print \" The velocity at the end of 120 seconds = %.1f kmph.\"%(v120)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The velocity at the end of 120 seconds = 73.5 kmph.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.5 Page No: 17" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%pylab inline\n", + "\n", + "import math \n", + "from matplotlib.pyplot import *\n", + "\n", + "# Variables:\n", + "s = 500. #mm\n", + "s1 = 125. #mm\n", + "s2 = 250. #mm\n", + "s3 = 125. \t\t#mm\n", + "t = 1. \t\t\t#second\n", + "\n", + "#Solution:\n", + "#Matrices for the velocity vs. time graph\n", + "V = [0 ,750.,750.,0] \t\t\t#The velocity matrix\n", + "T = [0,1./3,2./3,1] \t\t\t#The time matrix\n", + "plot(T,V)\n", + "xlabel(\"Time\")\n", + "ylabel(\"Velocity\")\n", + "#Calculating the time of uniform acceleration\n", + "\n", + "#Equating the time taken to complete the stroke to 1 second\n", + "v = (125/(1./2)+250/1+125/(1./2))/1 \t\t\t#mm/s\n", + "\n", + "#Results:\n", + "show()\n", + "print \" The maximum cutting speed v = %d mm/s.\"%(v)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Populating the interactive namespace from numpy and matplotlib\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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+ "text": [ + "<matplotlib.figure.Figure at 0x7f5ef004da50>" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The maximum cutting speed v = 750 mm/s.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.6 Page No: 19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables:\n", + "N0 = 0\n", + "N = 2000. \t\t\t#rpm\n", + "t = 20. \t\t\t#seconds\n", + "\n", + "#Solution:\n", + "#Calculating the angular velocities\n", + "omega0 = 0\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the angular acceleration\n", + "alpha = (omega-omega0)/t \t\t\t#rad/s**2\n", + "#Calculating the angular distance moved by the wheel during 2000 rpm\n", + "theta = (omega0+omega)*t/2 \t\t\t#rad\n", + "#Calculating the number of revolutions made by the wheel\n", + "n = theta/(2*math.pi)\n", + "\n", + "#Results:\n", + "print \" The angular acceleration of the wheel, alpha = %.3f rad/s**2.\"%(alpha)\n", + "print \" The wheel makes n = %.1f revolutions.\"%(n)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The angular acceleration of the wheel, alpha = 10.472 rad/s**2.\n", + " The wheel makes n = 333.3 revolutions.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.7 Page No: 21" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "r = 1.5 \t\t\t#m\n", + "N0 = 1200.\n", + "N = 1500. \t\t\t#rpm\n", + "t = 5. \t\t\t#seconds\n", + "\n", + "#Solution:\n", + "#Calculating the angular velocities\n", + "omega0 = 2*math.pi*N0/60\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating the linear velocity at the beginning\n", + "v0 = r*omega0 \t\t\t#m/s\n", + "#Calculating the linear velocity at the end of 5 seconds\n", + "v5 = r*omega \t\t\t#m/s\n", + "#Calculating the angular acceleration\n", + "alpha = (omega-omega0)/t \t\t\t#ad/s**2\n", + "#Calculating the math.tangential acceleration after 5 seconds\n", + "TangentialAcceleration = alpha*(r/2) \t\t\t#m/s**2\n", + "#Calculating the radial acceleration after 5 seconds\n", + "RadialAcceleration = (round(omega)**2)*(r/2) \t\t\t#m/s**2\n", + "\n", + "#Results:\n", + "print \" The linear velocity at the beginning, v0 = %.1f m/s.\"%(v0)\n", + "print \" The linear velocity after 5 seconds, v5 = %.1f m/s.\"%(v5)\n", + "print \" The tangential acceleration after 5 seconds is %.1f m/s**2.\"%(TangentialAcceleration)\n", + "print \" The radial acceleration after 5 seconds is %.f m/s**2.\"%(RadialAcceleration)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The linear velocity at the beginning, v0 = 188.5 m/s.\n", + " The linear velocity after 5 seconds, v5 = 235.6 m/s.\n", + " The tangential acceleration after 5 seconds is 4.7 m/s**2.\n", + " The radial acceleration after 5 seconds is 18487 m/s**2.\n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Theory_Of_Machines/ch3.ipynb b/Theory_Of_Machines/ch3.ipynb new file mode 100755 index 00000000..281d238d --- /dev/null +++ b/Theory_Of_Machines/ch3.ipynb @@ -0,0 +1,1380 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:102997571ab69dde264cca49f55aff87dce1c51132ce271fb2095c0f1179d655" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 3 : Kinetics of Motion" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.1 Page No : 32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "k = 1. \t\t\t#m\n", + "m = 2500. \t\t\t#kg\n", + "T = 1500. \t\t\t#N-m\n", + "\n", + "#Solution:\n", + "#Calculating the mass moment of inertia of the flywheel\n", + "I = m*k**2 \t\t\t#kg-m**2\n", + "#Calculating the angular acceleration of the flywheel\n", + "alpha = T/I \t\t\t#rad/s**2\n", + "#The angular speed at start\n", + "omega1 = 0\n", + "t = 10. \t\t\t#seconds\n", + "#Calculating the angular speed of the flywheel after t = 10 seconds from start\n", + "omega2 = omega1+alpha*t \t\t\t#rad/s\n", + "#Calculating the kinetic energy of the flywheel after 10 seconds from start\n", + "E = 1./2*I*omega2**2/1000 \t\t\t#kJ\n", + "\n", + "#Results:\n", + "print \" The angular acceleration of the flywheel, alpha = %.1f rad/s**2.\"%(alpha)\n", + "print \" The kinetic energy of the flywheel, E = %d kJ.\"%(E)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The angular acceleration of the flywheel, alpha = 0.6 rad/s**2.\n", + " The kinetic energy of the flywheel, E = 45 kJ.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.2 Page No : 32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "mC = 500. #kg\n", + "mD = 250. \t\t\t#kg\n", + "s = 100. #m\n", + "r = 0.5 #m\n", + "k = 0.35 \t\t\t#m\n", + "m = 3. \t\t\t#kg/m\n", + "\n", + "#Solution:\n", + "#Velocities of the cage\n", + "u1 = 0.\n", + "v1 = 10.\n", + "v2 = 10.\n", + "u3 = 10.\n", + "v3 = 0. \t\t\t#m/s\n", + "#Accelerations of the cage\n", + "a1 = 1.5\n", + "a3 = -6. \t\t\t#m/s**2\n", + "s = 100. \t\t\t#m\n", + "\n", + "#Calculating the time taken by the cage to reach the top\n", + "t1 = (v1-u1)/a1 \t\t\t#seconds\n", + "#Calculating the dismath.tance moved by the cage during time t1\n", + "s1 = (v1+u1)/2*t1 \t\t\t#m\n", + "#Calculating the time taken for the cage from initial velocity u3 = 10 m/s to final velocity of v3 = 0\n", + "t3 = (v3-u3)/a3 \t\t\t#seconds\n", + "#Calculating the dismath.tance moved by the cage during time t3\n", + "s3 = (v3+u3)/2*t3 \t\t\t#m\n", + "#Calculating the dismath.tance travelled during consmath.tant velocity of v2 = 10 m/s\n", + "s2 = s-s1-s3 \t\t\t#m\n", + "#Calculating the time taken for the cage during consmath.tant velocity\n", + "t2 = s2/v2 \t\t\t#seconds\n", + "#Calculating the time taken for the cage to reach the top\n", + "t = t1+t2+t3 \t\t\t#seconds\n", + "#Calculating the total mass of the rope for 100 metres\n", + "mR = m*s \t\t\t#kg\n", + "#Calculating the force to raise the cage and rope at uniform speed\n", + "F1 = (mC+mR)*9.81 \t\t\t#N\n", + "#Calculating the torque to raise the cage and rope at uniform speed\n", + "T1 = F1*r \t\t\t#N-m\n", + "#Calculating the force to accelerate the cage and rope\n", + "F2 = (mC+mR)*a1 \t\t\t#N\n", + "#Calculating the torque to accelerate the cage and rope\n", + "T2 = F2*r \t\t\t#N-m\n", + "#Calculating the mass moment of inertia of the drum\n", + "I = mD*k**2 \t\t\t#kg-m**2\n", + "#Calculating the angular acceleration of the drum\n", + "alpha = a1/r \t\t\t#rad/s**2\n", + "#Calculating the torque to accelerate the drum\n", + "T3 = I*alpha \t\t\t#N-m\n", + "#Calculating the total torque which must be applied to the drum at starting\n", + "T = T1+T2+T3 \t\t\t#N-m\n", + "#Calculating the mass of 33.35 m rope\n", + "m1 = m*33.35 \t\t\t#kg\n", + "#Calculating the reduction of torque\n", + "T4 = (m1*9.81+m1*a1)*r \t\t\t#N-m\n", + "#Calculating the angular velocity of drum\n", + "omega = v2/(2*math.pi*r) \t\t\t#rad/s\n", + "#Calculating the power\n", + "P = T4*omega/1000 \t\t\t#Power kW\n", + "\n", + "#Results:\n", + "print \" The time taken for the cage to reach the top t = %.2f s.\"%(t)\n", + "print \" The total torque which must be applied to the drum during starting T = %.1f N-m.\"%(T)\n", + "print \" The power required is %.3f kW.\"%(P)\n", + "#Answers differ due to rounding-off values in textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The time taken for the cage to reach the top t = 14.17 s.\n", + " The total torque which must be applied to the drum during starting T = 4615.9 N-m.\n", + " The power required is 1.801 kW.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3 Page No : 34" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "P = 4.*1000 \t\t\t#W\n", + "I = 140. \t\t\t#kg-m**2\n", + "N1 = 240. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Calculating the angular acceleration at the commencement of operation\n", + "omega1 = 2*math.pi*N1/60 \t\t\t#rad/s\n", + "#Calculating the energy supplied by the motor (E1) and the energy consumed in closing a revet in 1 second\n", + "E1 = 4000.\n", + "E2 = 10000. \t\t\t#N-m\n", + "#Calculating the loss of kinetic energy of the flywheel during the operation\n", + "E = E2-E1 \t\t\t#N-m\n", + "#Calculating the kinetic energy of the flywheel at the commencement of operation\n", + "KEc = 1./2*I*omega1**2 \t\t\t#Kinetic energy at the commencement N-m\n", + "#Calculating the kinetic energy of the flywheel at the end of operation\n", + "KEe = KEc-E \t\t\t#Kinetic energy at the end N-m\n", + "#Calculating the angular speed of the flywheel immediately after closing a revet\n", + "omega2 = math.sqrt(KEe*2/I) \t\t\t#rad/s\n", + "#Calculating the reduction of speed\n", + "ReductionofSpeed = (omega1-omega2)*60/(2*math.pi) \t\t\t#rpm\n", + "#Calculating the maximum rate at which the revets can be closed per minute\n", + "Rate = P*60/E2 \t\t\t#Maximum rate at which the revets can be closed per minute\n", + "\n", + "#Results:\n", + "print \" The reduction of speed is %.1f rpm.\"%(ReductionofSpeed)\n", + "print \" The maximum rate at which rivets can be closed per minute is %d.\"%(Rate)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The reduction of speed is 16.9 rpm.\n", + " The maximum rate at which rivets can be closed per minute is 24.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.4 Page No : 35" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m = 14.*1000 #kg\n", + "m1 = 1.25*1000 #kg\n", + "m2 = 110. \t\t\t#kg\n", + "d = 1. #m\n", + "r = d/2 #m\n", + "k1 = 450./1000 #m\n", + "k2 = 125./1000 \t\t#m\n", + "F = 1.2*1000 \t\t#N\n", + "eta = 0.85\n", + "v = 1.8 \t\t\t#m/s\n", + "a = 0.1 \t\t\t#m/s**2\n", + "\n", + "#Solution:\n", + "#Calculating the forces oppomath.sing the motion\n", + "P1 = m*9.81*1/20+m*a+F \t\t\t#N\n", + "#Calculating the torque on the drum shaft to accelerate the load\n", + "T1 = P1*r \t\t\t#N-m\n", + "#Calculating the mass moment of inertia of the drum\n", + "I1 = m1*k1**2 \t\t\t#kg-m**2\n", + "#Calculating the angular acceleration of the drum\n", + "alpha1 = a/r \t\t\t#rad/s\n", + "#Calculating the torque on the drum to accelerate the drum shaft\n", + "T2 = I1*alpha1 \t\t\t#N-m\n", + "#Calculating the torque on the armature to accelerate drum and load\n", + "T3 = (T1+T2)/(40*eta) \t\t\t#N-m\n", + "#Calculating the mass moment of inertia of the armature\n", + "I2 = m2*k2**2 \t\t\t#kg-m**2\n", + "#Calculating the angular acceleration of the armature\n", + "alpha2 = a/r*40 \t\t\t#rad/s**2\n", + "#Calculating the torque on the armature to accelerate armature shaft\n", + "T4 = I2*alpha2 \t\t\t#N-m\n", + "#Calculating the torque on the motor shaft\n", + "T = T3+T4 \t\t\t#N-m\n", + "#Calculating the angular speed of the motor\n", + "omega = v/r*40 \t\t\t#rad/s\n", + "#Calculating the power developed by the motor\n", + "P = T*omega/1000 \t\t\t#Power developed by the motor kW\n", + "\n", + "#Results:\n", + "print \" The torque on the motor shaft T = %.2f N-m.\"%(T)\n", + "print \" The power developed by the motor is %.2f kW.\"%(P)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The torque on the motor shaft T = 154.46 N-m.\n", + " The power developed by the motor is 22.24 kW.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.5 Page No : 37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m = 12.0*1000\n", + "m1 = 2.0*1000\n", + "m2 = 2.5*1000 \t\t\t#kg\n", + "k1 = 0.4\n", + "d1 = 1.2\n", + "r1 = d1/2.0\n", + "k2 = 0.6\n", + "d2 = 1.5\n", + "r2 = d2/2.0\n", + "s = 6.0 \t\t\t#m\n", + "v = 9.0*1000/3600 \t\t\t#m/s\n", + "\n", + "#Solution:\n", + "#Calculating the mass moment of inertia of the front roller\n", + "I1 = m1*k1**2 \t\t\t#kg-m**2\n", + "#Calculating the mass moment of inertia of the rear axle together with its wheels\n", + "I2 = m2*k2**2 \t\t\t#kg-m**2\n", + "#Calculating the angular speed of the front roller\n", + "omega1 = round(v/r1,2) \t\t\t#rad/s\n", + "#Calculating the angular speed of rear wheels\n", + "omega2 = round(v/r2,1) \t\t\t#rad/s\n", + "#Calculating the kinetic energy of rotation of the front roller\n", + "E1 = 1.0/2*I1*4.16**2 \t\t\t#N-m\n", + "#Calculating the kinetic energy of rotation of the rear axle with its wheels\n", + "E2 = 1.0/2*I2*omega2**2 \t\t\t#N-m\n", + "#Calculating the total kinetic energy of rotation of the wheels\n", + "E = round(E1+E2,-1) \t\t\t#N-m\n", + "#Calculating the kinetic energy of translation of the road roller\n", + "E3 = 1.0/2*m*v**2 \t\t\t#N-m\n", + "#Calculating the total kinetic energy of the road roller\n", + "E4 = E3+E \t\t\t#N-m\n", + "#Calculating the braking force to bring the roller to rest\n", + "F = E4/s \t\t\t#N\n", + "\n", + "#Results:\n", + "print \" The total kinetic energy of rotation of the wheels E = %.f N-m.\"%(E)\n", + "print \" The total kinetic energy of the road roller E4 = %d N-m.\"%(E4)\n", + "print \" The braking force required to bring the roller to rest F = %.1f N.\"%(F)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The total kinetic energy of rotation of the wheels E = 7670 N-m.\n", + " The total kinetic energy of the road roller E4 = 45170 N-m.\n", + " The braking force required to bring the roller to rest F = 7528.3 N.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.6 Page no : 38" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import Symbol,solve\n", + "import math\n", + "\n", + "# variables\n", + "s = 4. # N/m\n", + "m = 4000. # N/m ; \n", + "x1 = 0.075 # m \n", + "x2 = 0.03 # m;\n", + "m = 5. # kg ; \n", + "R = 70. # N\n", + "g = 9.81\n", + "\n", + "# calculations\n", + "Q = m*(x1 - x)\n", + "P = Q + m*g - R\n", + "x = 0.045\n", + "t = Symbol(\"t\")\n", + "ans = -solve(0.07*(1 - math.cos(math.sqrt(800)) * t) - 0.045)[0]\n", + "a = math.acos(ans)\n", + "t = a/math.sqrt(800)\n", + "\n", + "# result\n", + "print \"t = %.4f s\"%t\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "t = 0.0426 s\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.7 Page No : 41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "from numpy import linalg\n", + "\n", + "# Variables:\n", + "r = 500./1000\n", + "k = 450./1000 \t\t\t#m\n", + "m1 = 500.\n", + "m2 = 1250. \t\t\t#kg\n", + "u = 0.75 \t\t\t#m/s\n", + "\n", + "#Solution:\n", + "#Calculating the mass moment of inertia of drum\n", + "I2 = m2*k**2 \t\t\t#kg-m**2\n", + "#Calculating the speed of truck\n", + "#Impulse\n", + "#F = m1*v or\n", + "#F-m1*v = 0 .....(i)\n", + "#Moment of impulse\n", + "#F*r = I2*(omega2-omega2) or\n", + "#F*r = I2*(u-v)/r or\n", + "#F*r+I2*v/r = I2*u/r .....(ii)\n", + "#Solving (i) and (ii)\n", + "A = [[1, -m1],[ r, I2/r]]\n", + "B = [0, I2*u/r]\n", + "V = linalg.solve(A,B)\n", + "v = V[1]\n", + "#Calculating the energy lost to the system\n", + "E = 1./2*I2*(u**2-v**2)/r**2-1./2*m1*v**2 \t\t\t#Energy lost to the system, N-m\n", + "\n", + "#Results:\n", + "print \" The speed of the truck when the motion becomes steady, v = %.3f m/s.\"%(v)\n", + "print \" The energy lost to the system is %d N-m.\"%(E)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The speed of the truck when the motion becomes steady, v = 0.502 m/s.\n", + " The energy lost to the system is 94 N-m.\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.8 Page No : 42" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from scipy.optimize import fsolve \n", + "import math \n", + "\n", + "# Variables:\n", + "s = 0.7*10**6 \t\t\t#N/m\n", + "m1 = 10.*10**3\n", + "m2 = 15.*10**3 \t\t\t#kg\n", + "v1 = 1.8\n", + "v2 = 0.6 \t\t\t#m/s\n", + "\n", + "#Solution:\n", + "#Calculating the common velocity when moving together during impact\n", + "v = (m1*v1+m2*v2)/(m1+m2)\n", + "#Calculating the kinetic energy lost to the system\n", + "E = (1./2*m1*v1**2+1./2*m2*v2**2)-1./2*(m1+m2)*v**2\n", + "#Calculating the compression of each buffer spring\n", + "x = math.sqrt(E/(2*s))\n", + "#Calculating the velocity of each truck on separation\n", + "#Final KE after separation = KE at common velocity+Half of energy stored in springs.\n", + "#And initial and final momentum must be equal.\n", + "#Simplifying the two equations\n", + "# we get\n", + "\n", + "#1/2*m1*v3**2+1/2*m2*v4**2 = 1/2*(m1+m2)*v**2+1/2*E .....(i)\n", + "#m1*v3+m2*v4 = (m1+m2)*v\n", + "def f(x):\n", + " v3 = x[0]\n", + " v4 = x[1]\n", + " y = [0,0]\n", + " y[0] = 1./2*m1*v3**2+1./2*m2*v4**2-1./2*(m1+m2)*v**2-1./2*E\n", + " y[1] = m1*v3+m2*v4-(m1+m2)*v\n", + " return y\n", + " \n", + "z = fsolve(f,[1,1])\n", + "v3 = z[1]\n", + "v4 = z[0]\n", + "\n", + "#Results:\n", + "print \" The common velocity when moving together during impact, v = %.2f m/s.\"%(v)\n", + "print \" The kinetic energy lost to the system is %.2f kN-m.\"%(E/1000.)\n", + "print \" The compression of each buffer spring, x = %.f mm.\"%(x*1000.)\n", + "print \" The velocity of separation for 10 tonnes truck, v3 = %.1f m/s.\"%(v3)\n", + "print \" The velocity of separation for 15 tonnes truck, v4 = %.1f m/s.\"%(v4)\n", + "\n", + "# note : rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The common velocity when moving together during impact, v = 1.08 m/s.\n", + " The kinetic energy lost to the system is 4.32 kN-m.\n", + " The compression of each buffer spring, x = 56 mm.\n", + " The velocity of separation for 10 tonnes truck, v3 = 0.7 m/s.\n", + " The velocity of separation for 15 tonnes truck, v4 = 1.6 m/s.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.9 Page No : 43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m1 = 300. #kg\n", + "m2 = 500. \t\t\t#kg\n", + "s = 1. #m\n", + "x = 150./1000 \t\t#m\n", + "\n", + "#Solution:\n", + "#Calculating the velocity with which mass m1 hits the pile\n", + "u = 0\n", + "v1 = math.sqrt(2*9.81*s+u**2) \t\t\t#m/s\n", + "#Calculating the common velocity after impact\n", + "v2 = 0\n", + "v = (m1*v1+m2*v2)/(m1+m2) \t\t\t#m/s\n", + "#Calculating the kinetic energy before impact\n", + "KEb = m1*9.81*s \t\t\t#Kinetic energy before impact N-m\n", + "#Calculating the kinetic energy after impact\n", + "KEa = 1./2*(m1+m2)*v**2 \t\t\t#Kinetic energy after impact N-m\n", + "#Calculating the energy lost in the blow\n", + "E = KEb-KEa \t\t\t#Energy lost in the blow N-m\n", + "#Calculating the average resistance against the pile\n", + "R = KEa/x+m1*9.81+m2*9.81\n", + "\n", + "#Results:\n", + "print \" The energy lost in the blow is %d N-m.\"%(E)\n", + "print \" The average resistance against the pile R = %.3f kN.\"%(R/1000)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The energy lost in the blow is 1839 N-m.\n", + " The average resistance against the pile R = 15.206 kN.\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.10 Page No : 44" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "from numpy import linalg\n", + "\n", + "# Variables:\n", + "m1 = 0.7\n", + "m2 = 2.4 \t\t\t#kg\n", + "k1 = 270./1000\n", + "k2 = 185./1000\n", + "h1 = 0.25\n", + "DL = 0.2\n", + "CM = 0.275 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Calculating the angular velocity of hammer just before impact\n", + "h = h1*(1-math.cos(20*math.pi/180))\n", + "omega = math.sqrt(m1*9.81*h*2/(m1*k1**2)) \t\t\t#rad/s\n", + "#Calculating the relative linear velocity\n", + "RLV = 0.8*omega*CM\n", + "#Calculating the values of angular velocities\n", + "#The two equations we get in terms of omegaA and omegaB are\n", + "#DL*omegaA-CM*omegaB = RLV .....(i)\n", + "#m1*k1**2*(omega-omegaB) = .275/.2*m2*k2**2*omegaA or\n", + "#2.21*omegaA+omegaB = 2.01 .....(ii)\n", + "A = [[DL, -CM],[ 2.21, 1]]\n", + "B = [RLV,2.01]\n", + "V = linalg.solve(A,B)\n", + "\n", + "#Results:\n", + "print \" The angular velocity of the anvil A, omegaA = %.2f rad/s.\"%(V[0])\n", + "print \" The angular velocity of the hammer B, omegaB = %.2f rad/s, i.e. %.2f rad/s, in the reverse direction.\"%(V[1],\n", + "V[1]*-1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The angular velocity of the anvil A, omegaA = 1.23 rad/s.\n", + " The angular velocity of the hammer B, omegaB = -0.71 rad/s, i.e. 0.71 rad/s, in the reverse direction.\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.11 Page No : 46" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "import math \n", + "\n", + "# Variables:\n", + "m = 30. \t\t\t#kg\n", + "AG = 1\n", + "GB = 150./1000\n", + "k1 = 1.1\n", + "k2 = 350./1000 \t\t\t#m\n", + "theta = 60.*math.pi/180 \t\t\t#rad\n", + "t = 0.005 \t\t\t#s\n", + "a = AG\n", + "b = GB\n", + "\n", + "#Solution:\n", + "#Calculating the mass moment of inertia of the pendulum about the point of suspension A\n", + "IA = m*k1**2 \t\t\t#kg-m**2\n", + "#Calculating the mass moment of inertia ofthe pendulum about centre of gravity G\n", + "IG = m*k2**2 \t\t\t#kg-m**2\n", + "#Calculating the angular velocity of the pendulum\n", + "h1 = a-a*math.cos(theta)\n", + "omega = math.sqrt(2*m*9.81*h1/IA) \t\t\t#rad/s\n", + "#Calculating the striking velocity of the pendulum\n", + "v = omega*(a+b) \t\t\t#m/s\n", + "#Calculating the angular velocity of the pendulum just after the breakage of the specimen\n", + "omega1 = math.sqrt(omega**2-2*54/IA)\n", + "#Calculating the linear velocity of G just before the breakage of specimen\n", + "vG = omega*AG \t\t\t#m/s\n", + "#Calculating the linear velocity of G just after the breakage of specimen\n", + "vGdash = omega1*AG \t\t\t#m/s\n", + "#Calculating the impulses at pivot A and knife edge B\n", + "#F1+F2 = m*(vG-vGdash) .....(i)\n", + "#b*F2-a*F1 = IG*(omega-omega1) .....(ii)\n", + "A = [[1, 1],[-a, b]]\n", + "B = [[m*(vG-vGdash)], [IG*(omega-omega1)]]\n", + "V = linalg.solve(A,B)\n", + "F1 = V[0]\n", + "F2 = V[1]\n", + "\n", + "#Calculating the angle of swing of the pendulum after impact\n", + "theta1 = math.cos(a-1./2*IA*omega1**2/(m*9.81))/a \t\t\t#radians\n", + "#Calculating the average force exerted at the pivot\n", + "Fp = F1/t \t\t\t#N\n", + "#Calculating the average force exerted at the knife edge\n", + "Fk = F2/t \t\t\t#N\n", + "\n", + "#Results:\n", + "print \" The striking velocity of the pendulum, v = %.2f m/s.\"%(v)\n", + "print \" Impulse at the pivot A, F1 = %.1f N.\"%(F1)\n", + "print \" Impulse at the knife edge B, F2 = %.1f N.\"%(F2)\n", + "print \" Angle of swing of the pendulum after impact, theta = %.2f degree.\"%(theta1*180/math.pi)\n", + "print \" Average force exerted at the pivot is %d N.\"%(round(Fp,-1))\n", + "print \" Average force exerted at the knife edge is %d N.\"%(Fk)\n", + "\n", + "# note : python linalg solve gives slightly different answer but accurate. " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The striking velocity of the pendulum, v = 3.27 m/s.\n", + " Impulse at the pivot A, F1 = 0.4 N.\n", + " Impulse at the knife edge B, F2 = 17.0 N.\n", + " Angle of swing of the pendulum after impact, theta = 44.43 degree.\n", + " Average force exerted at the pivot is 80 N.\n", + " Average force exerted at the knife edge is 3407 N.\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.12 Page No : 47" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "import math \n", + "\n", + "# Variables:\n", + "T = 150. \t\t\t#N-m\n", + "m1 = 60. #kg\n", + "m2 = 20. \t\t\t#kg\n", + "k1 = 140./1000 #m\n", + "k2 = 80./1000 \t\t#m\n", + "N1 = 750. #rpm\n", + "N2 = 0. \t\t\t#rpm\n", + "\n", + "#Sloution:\n", + "#Calculating the angular speeds\n", + "omega1 = 2*math.pi*N1/60\n", + "omega2 = 0 \t\t\t#rad/s\n", + "#Calculating the mass moment of inertia of the rotor on motor\n", + "I1 = m1*k1**2 \t\t\t#kg-m**2\n", + "#Calculating the mass moment of inertia of the parts attached to machine\n", + "I2 = m2*k2**2 \t\t\t#kg-m**2\n", + "#Calculating the speed after engagement of the clutch and the time taken\n", + "#We know that impulsive torque = change in angular momentum\n", + "#T*t = I1*(omega1-omega) or I1*omega+T*t = I1*omega1 .....(i)\n", + "#T*t = I2*(omega-omega2) or I2*omega-T*t = I2*omega2 .....(ii)\n", + "A = [[I1, T],[ I2, -T]]\n", + "B = [I1*omega1,I2*omega2]\n", + "V = linalg.solve(A,B)\n", + "omega = V[0] \t\t\t#rad/s\n", + "t = V[1] \t\t\t#s\n", + "#Calculating the kinetic energy lost during the operation\n", + "E = I1*I2*(omega1-omega2)**2/(2*(I1+I2)) \t\t\t#N-m\n", + "\n", + "#Results:\n", + "print \" The speed after engagement, omega = %.1f rad/s.\"%(omega)\n", + "print \" The time taken, t = %.2f s.\"%(t)\n", + "print \" The kinetic energy lost during the operation, E = %d N-m.\"%(E)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The speed after engagement, omega = 70.8 rad/s.\n", + " The time taken, t = 0.06 s.\n", + " The kinetic energy lost during the operation, E = 356 N-m.\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.13 Page No : 50" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "M = 75. \t\t\t#kg\n", + "r = 0.3 \t\t\t#m\n", + "G = 6.\n", + "IA = 100. #kg-m**2\n", + "IB = 5. \t\t\t#kg-m**2\n", + "eta = 90./100 \t\t\n", + "\n", + "#Solution:\n", + "#Calculating the equivalent mass of the geared system\n", + "me = 1/r**2*(IA+G**2*IB) \t\t\t#kg\n", + "#Calculating the total equivalent mass to be accelerated\n", + "Me = me+M \t\t\t#kg\n", + "#Calculating the acceleration when it is allowed to fall freely\n", + "F = M*9.81 \t\t\t#Accelerating force provided by the pull of gravity N\n", + "a = F/Me \t\t\t#m/s**2\n", + "#Calculating the equivalent mass of the geared system when the efficiency is 90%\n", + "me1 = 1/r**2*(IA+G**2*IB/eta) \t\t\t#kg\n", + "#Calculating the total equivalent mass to be accelerated\n", + "Me1 = me1+M \t\t\t#kg\n", + "#Calculating the acceleration when the efficiency is 90%\n", + "F1 = M*9.81 \t\t\t#Accelerating force by the pull of gravity N\n", + "a1 = F1/Me1 \t\t\t#m/s**2\n", + "\n", + "#Results:\n", + "print \" The acceleration of the mass M if it is allowed to fall freely, a = %.3f m/s**2.\"%(a)\n", + "print \" The acceleration of the mass M when the efficiency of the gearing system is 0.9, a = %.3f m/s**2.\"%(a1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The acceleration of the mass M if it is allowed to fall freely, a = 0.231 m/s**2.\n", + " The acceleration of the mass M when the efficiency of the gearing system is 0.9, a = 0.216 m/s**2.\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.14 pageno : 51" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# variables\n", + "T = 100. #N-m ; \n", + "IA = 2. # kg-m**2 ;\n", + "IB = 32. # kg-m**2\n", + "\n", + "# calculations\n", + "G = math.sqrt(IB/IA)\n", + "alphaB = G*100/(IA*G**2 + 32)\n", + "alphaA = G*alphaB\n", + "\n", + "# results\n", + "print \"G = %.f\"%G\n", + "print \"alpha B = %.2f rad/s**2\"%alphaB\n", + "print \"alpha A = %.f rad/s**2\"%alphaA" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "G = 4\n", + "alpha B = 6.25 rad/s**2\n", + "alpha A = 25 rad/s**2\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.15 page no : 52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# variables\n", + "m = 1500. #kg ; \n", + "d = 600. \n", + "mm = 0.6 #m \n", + "r = 0.3 #m ; \n", + "IA = 8. #kg-m 2 ;\n", + "IB = 1. # kg-m 2 ; \n", + "n = 0.85 \n", + "v = 24. #km/h ; \n", + "F = 300. #N ; \n", + "TB = 200. #N-m ; \n", + "sintheta = 0.25\n", + "\n", + "# calculatins\n", + "tW = n * TB\n", + "P = 170./r\n", + "G = round((14 + math.sqrt(14**2 + 4 * 168.4))/2)\n", + "Amax = (P*G - 3980)/ (1590+9.44*G**2)\n", + "v = 6.67 #m/s\n", + "speed = v/r\n", + "W = G*speed\n", + "power = TB*W/1000\n", + "\n", + "#Result\n", + "print \"G = %.f \"%G\n", + "print \"maximum acceleration : %.2f m/s\"%Amax\n", + "print \"speed of the road wheels : %.f rad/s\"%W\n", + "print \"power of the engine : %.1f kW\"%power" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "G = 22 \n", + "maximum acceleration : 1.38 m/s\n", + "speed of the road wheels : 489 rad/s\n", + "power of the engine : 97.8 kW\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.16 page no : 54" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# variables\n", + "TB = 1000. # N-m ; \n", + "v1 = 27.8 #m/s ; \n", + "v2 = 76.4 #m /s ; \n", + "d = 0.9 # m \n", + "r = 0.45 # m ; \n", + "G = 3.3 ; \n", + "v = 47.2 # m/s ; \n", + "P = 50 * 10**3 # W ;\n", + "M = 1000 #kg ; \n", + "m = 40. # kg ; \n", + "k = 0.25 #m ; \n", + "IB = 1. # kg-m**2\n", + "\n", + "# calculation\n", + "IA = 4 * m* k**2\n", + "F = round(P/v,-1)\n", + "Tw = G*1000\n", + "FT = Tw/r\n", + "C1 = round(-(2325./(2*121.5)*math.log((121.5+v1)/(121.5-v1))),1)\n", + "t = round((2325./(2*121.5)*math.log((121.5+v2)/(121.5-v2)))-4.5,1)\n", + "\n", + "# result\n", + "print \"the time taken for the speed to rise : %.1f s\"%t\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the time taken for the speed to rise : 9.6 s\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.17 pageno : 55" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# variables\n", + "G = 9\n", + "IA = 0.6 #kg-m2 \n", + "IB = 45. # kg-m 2 ; \n", + "TB = 100. # N-m;\n", + "n = 0.95; \n", + "N = 160. # r.p.m. ; \n", + "N1 = 0 \n", + "N2 = 60. # r.p.m. ; \n", + "TA = 30. # N-m\n", + "W1 = 0\n", + "# calculations\n", + "P = 2*math.pi*N*TB/(60*n)\n", + "\n", + "t = 60. # time\n", + "TA = 30.\n", + "TB1 = G*TA*n\n", + "B = TB1 - TB\n", + "alphaB = B/91.2\n", + "W2 = 2*math.pi*N2/60\n", + "t = (W2 - W1)/alphaB\n", + "G1 = (7.27 + math.sqrt(7.27**2+4*78.95))/2\n", + "\n", + "# result\n", + "print \"the power which the motor must develop : %.f W\"% P\n", + "print \"final angular speed : %.1f s\"%t\n", + "print \"maximum angular acceleration : %.3f \"%G1" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the power which the motor must develop : 1764 W\n", + "final angular speed : 3.7 s\n", + "maximum angular acceleration : 13.235 \n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.18 Page No : 57" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "d = 1.5 # m\n", + "r = d/2 # m\n", + "d1 = 1. # m\n", + "kM = 90./1000\n", + "kI = 225./1000\n", + "kD = 600./1000\n", + "kP = 450./1000 \t\t\t#m\n", + "NM = 900. # rpm\n", + "N1 = 275. # rpm\n", + "ND = 50. \t\t\t #rpm\n", + "mM = 200. # kg\n", + "mI = 375. # kg\n", + "mD = 2250. # kg\n", + "mP = 200. # kg\n", + "m1 = 1150. # kg\n", + "m2 = 650. \t\t\t#kg\n", + "FI = 150. # N-m\n", + "FD = 1125. # N-m\n", + "FP = 150. \t\t\t#N-m\n", + "F1 = 500. # N\n", + "F2 = 350. \t\t\t#N\n", + "a = 0.9 \t\t\t#m/s**2\n", + "\n", + "#Solution:\n", + "#Calculating the speed of guide pulley\n", + "NP = ND*d/d1 \t\t\t#rpm\n", + "#Calculating the gear ratio for intermediate gear and motor\n", + "G1 = N1/NM\n", + "#Calculating the gear ratio for drum and motor\n", + "G2 = round(ND/NM,3)\n", + "#Calculating the gear ratio for the guide pulley and motor\n", + "G3 = NP/NM\n", + "#Calculating the mass moment of inertia of the motor\n", + "IM = mM*kM**2 \t\t\t#kg-m**2\n", + "#Calculating the mass moment of inertia of the intermediate gear\n", + "II = mI*kI**2 \t\t\t#kg-m**2\n", + "#Calculating the mass moment of inertia of the drum and shaft\n", + "ID = mD*kD**2 \t\t\t#kg-m**2\n", + "#Calculating the mass moment of inertia of the guide pulley\n", + "IP = mP*kP**2 \t\t\t#kg-m**2\n", + "#Calculating the angular acceleration of the drum\n", + "alphaD = a/r \t\t\t#rad/s**2\n", + "#Calculating the angular acceleration of the motor\n", + "alphaM = alphaD*NM/ND \t\t\t#rad/s**2\n", + "#Calculating the equivalent mass moment of inertia of the system\n", + "I = IM+G1**2*II+G2**2*ID+2*G3**2*IP \t\t\t#kg-m**2\n", + "#Calculating the torque at motor to accelerate the system\n", + "T1 = round(I*alphaM,1) \t\t\t#N-m\n", + "#Calculating the torque at motor to overcome friction at intermediate gear\n", + "#drum and two guide pulleys\n", + "T2 = round(G1*FI+G2*FD+2*G3*FP,1) \t\t\t#N-m\n", + "#Calculating the tension in the rimath.sing rope between the pulley and drum\n", + "Q1 = m1*9.81+m1*a+F1 \t\t\t#N\n", + "#Calculating the tension in the falling rope between the pulley and drum\n", + "Q2 = m2*9.81-m2*a-F2 \t\t\t#N\n", + "#Calculating the torque at drum\n", + "TD = round((Q1-Q2)*r) \t\t\t#N-m\n", + "#Calculating the torque at motor to raise and lower cages and ropes and to overcome frictional resistance\n", + "T3 = G2*TD \t\t\t#N-m\n", + "#Calculating the total motor torque required\n", + "T = T1+T2+T3 \t\t\t#N-m\n", + "\n", + "#Results:\n", + "print \" The total motor torque required, T = %.1f N-m.\"%(T)\n", + "\n", + "# rounding off error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The total motor torque required, T = 583.8 N-m.\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.19 Page No : 65" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m1 = 50. #kg\n", + "m2 = 25. \t\t\t#kg\n", + "u1 = 3. #m/s\n", + "u2 = 1.5 \t\t\t#m/s\n", + "\n", + "#Solution:\n", + "#When the impact is inelastic\n", + "#Calculating the common velocity after impact\n", + "v = (m1*u1+m2*u2)/(m1+m2) \t\t\t#m/s\n", + "#Calculating the loss of kinetic energy during impact\n", + "EL = m1*m2/(2*(m1+m2))*(u1-u2)**2 \t\t\t#N-m\n", + "#When the impact is elastic\n", + "#Calculating the velocity of the first sphere immediately after impact\n", + "v1 = 2*v-u1 \t\t\t#m/s\n", + "#Calculating the velocity of the second sphere immediately after impact\n", + "v2 = 2*v-u2 \t\t\t#m/s\n", + "#Calculating the loss of kinetic energy\n", + "EL1 = 0\n", + "#When the coefficient of restitution e = 0.6\n", + "e = 0.6\n", + "#Calculating the velocity of the first sphere immediately after impact\n", + "v12 = (1+e)*v-e*u1 \t\t\t#m/s\n", + "#Calculating the velocity of the second sphere immediately after impact\n", + "v22 = (1+e)*v-e*u2 \t\t\t#m/s\n", + "#Calculating the loss of kinetic energy\n", + "EL2 = m1*m2/(2*(m1+m2))*(u1-u2)**2*(1-e**2) \t\t\t#N-m\n", + "\n", + "#Results:\n", + "print \" The common velocity after impact when the impact is inelastic, v = %.1f m/s.\"%(v)\n", + "print \" The loss of kinetic energy during impact, EL = %.2f N-m.\"%(EL)\n", + "print \" The velocity of the first sphere immediately after impact when the impact is elastic, v1 = %d m/s.\"%(v1)\n", + "print \" The velocity of the second sphere immediately after impact, v2 = %.1f m/s.\"%(v2)\n", + "print \" The loss of kinetic energy, EL = %d.\"%(EL1)\n", + "print \" The velocity of the first sphere immediately after impact When the coefficient of \\\n", + "restitution is 0.6, v1 = %.1f m/s.\"%(v12)\n", + "print \" The velocity of the second sphere immediately after impact, v2 = %.1f m/s.\"%(v22)\n", + "print \" The loss of kinetic energy during impactm EL = %d N-m.\"%(EL2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The common velocity after impact when the impact is inelastic, v = 2.5 m/s.\n", + " The loss of kinetic energy during impact, EL = 18.75 N-m.\n", + " The velocity of the first sphere immediately after impact when the impact is elastic, v1 = 2 m/s.\n", + " The velocity of the second sphere immediately after impact, v2 = 3.5 m/s.\n", + " The loss of kinetic energy, EL = 0.\n", + " The velocity of the first sphere immediately after impact When the coefficient of restitution is 0.6, v1 = 2.2 m/s.\n", + " The velocity of the second sphere immediately after impact, v2 = 3.1 m/s.\n", + " The loss of kinetic energy during impactm EL = 12 N-m.\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.20 Page No : 66" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m1 = 15.*1000 #kg\n", + "m2 = 5.*1000 \t\t\t#kg\n", + "u1 = 20.*1000/3600 #m/s\n", + "u2 = 12.*1000/3600 \t\t#m/s\n", + "s = 1000.*10**3 \t\t#N/m\n", + "e = 0.5\n", + "\n", + "#Solution:\n", + "#Calculating the common speed\n", + "v = (m1*u1+m2*u2)/(m1+m2) \t\t\t#m/s\n", + "#Calculating the difference in kinetic energies before impact and during impact\n", + "d = m1*m2/(2*(m1+m2))*(u1-u2)**2 \t\t\t#Difference in kinetic energies N-m\n", + "#Equating the difference between kinetic energies to the strain energy stored in the springs\n", + "x = math.sqrt(d*2/(4*s))*1000 \t\t\t#mm\n", + "#Calculating the speed of the loaded wagon immediately after impact ends\n", + "v11 = 2*v-u1 \t\t\t#m/s\n", + "#Calculating the speed of the empty wagon immediately after impact ends\n", + "v21 = 2*v-u2 \t\t\t#m/s\n", + "#Calculating the speeds of the wagons taking into account the coefficient of restitution e = 0.5\n", + "v12 = (1+e)*v-e*u1 \t\t\t#m/s\n", + "v22 = (1+e)*v-e*u2 \t\t\t#m/s\n", + "#Calculating the amount of energy dissipated during impact\n", + "EL = m1*m2/(2*(m1+m2))*(u1-u2)**2*(1-e**2) \t\t\t#N-m\n", + "\n", + "#Results:\n", + "print \" The magnitude of common speed v = %d m/s.\"%(v)\n", + "print \" The maximum deflection of each buffer spring during impact x = %d mm.\"%(x)\n", + "print \" The speed of the loaded wagon immediately after the impact ends v1 = %.2f m/s.\"%(v11)\n", + "print \" The speed of the empty wagon immediately after the impact ends v2 = %.2f m/s.\"%(v21)\n", + "print \" When coefficient of restitution is taken into account v1 = %.3f m/s.\"%(v12)\n", + "print \" When coefficient of restitution is taken into account v2 = %.3f m/s.\"%(v22)\n", + "print \" The amount of energy dissipated during impact EL = %d N-m.\"%(EL)\n", + "\n", + "# rounding off error. please check." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The magnitude of common speed v = 5 m/s.\n", + " The maximum deflection of each buffer spring during impact x = 68 mm.\n", + " The speed of the loaded wagon immediately after the impact ends v1 = 4.44 m/s.\n", + " The speed of the empty wagon immediately after the impact ends v2 = 6.67 m/s.\n", + " When coefficient of restitution is taken into account v1 = 4.722 m/s.\n", + " When coefficient of restitution is taken into account v2 = 5.833 m/s.\n", + " The amount of energy dissipated during impact EL = 6944 N-m.\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.21 Page No : 67" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "IA = 22.5 #kg-m**2\n", + "IB = 67.5 \t\t\t#kg-m**2\n", + "q = 225. \t\t\t#N-m/rad\n", + "NA = 150. #rpm\n", + "NB = 0. \t\t\t#rpm\n", + "\n", + "#Calculating the angular speed of the flywheel\n", + "omegaA = 2*math.pi*NA/60 \t\t\t#rad/s\n", + "#Calculating the angular speed of both the flywheels at the instant their speeds are equal\n", + "omega = IA*omegaA/(IA+IB) \t\t\t#rad/s\n", + "#Calculating the kinetic energy of the system at that instant\n", + "E2 = 1./2*(IA+IB)*omega**2 \t\t\t#N-m\n", + "#Calculating the kinetic energy of the flywheel A\n", + "E1 = 1./2*IA*omegaA**2 \t\t\t#N-m\n", + "#Calculating the strain energy stored in the spring\n", + "E = E1-E2 \t\t\t#Strain energy stored in the spring N-m\n", + "#Calculating the maximum twist of the spring\n", + "theta = math.sqrt(E*2/q) \t\t\t#radians\n", + "thetad = theta*180/math.pi \t\t\t#Maximum twist degrees\n", + "#Calculating the speed of each flywheel when the spring regains its initial unstrained condition\n", + "N = 60*omega/(2*math.pi)\n", + "NA1 = 2*N-NA \t\t\t#rpm\n", + "NB1 = 2*N-NB \t\t\t#rpm\n", + "\n", + "#Results:\n", + "print \" The strain energy stored in the spring is %d N-m.\"%(E)\n", + "print \" The maximum twist of the spring theta = %.1f degrees.\"%(thetad)\n", + "print \" The speed of flywheel A when the spring regains its initial unstrained condition NA1 = %d rpm \\\n", + " \\ni.e. %d rpm in the opposite direction.\"%(NA1,-NA1)\n", + "print \" The speed of flywheel B when the spring regains its initial unstrained condition NB1 = %d rpm.\"%(NB1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The strain energy stored in the spring is 2081 N-m.\n", + " The maximum twist of the spring theta = 246.5 degrees.\n", + " The speed of flywheel A when the spring regains its initial unstrained condition NA1 = -75 rpm \n", + "i.e. 75 rpm in the opposite direction.\n", + " The speed of flywheel B when the spring regains its initial unstrained condition NB1 = 75 rpm.\n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Theory_Of_Machines/ch4.ipynb b/Theory_Of_Machines/ch4.ipynb new file mode 100755 index 00000000..845b97ee --- /dev/null +++ b/Theory_Of_Machines/ch4.ipynb @@ -0,0 +1,601 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:cff1e23e14e854a300136760b464ae325e3afc2c7094a19b255704b9aeba45ea" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 4 : Simple Harmonic Motion" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.1 Page No : 75" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "N = 120. \t\t\t#rpm\n", + "r = 1. #m\n", + "x = 0.75 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Calculating Angular Velocity\n", + "omega = 2*math.pi*N/60 \t\t\t#rad/s\n", + "#Calculating Velocity of the Piston\n", + "v = omega*math.sqrt(r**2-x**2) \t\t\t#m/s\n", + "#Calculating Acceleration of the Piston\n", + "a = omega**2*x\n", + "\n", + "#Results:\n", + "print \" The Velocity of the Piston, v = %.2f m/s.\"%(v)\n", + "print \" The Acceleration of the Piston, a = %.2f m/s**2.\"%(a)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The Velocity of the Piston, v = 8.31 m/s.\n", + " The Acceleration of the Piston, a = 118.44 m/s**2.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.2 Page No : 75" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "import math \n", + "\n", + "# Variables:\n", + "x1 = .75\n", + "x2 = 2. \t\t\t#m\n", + "v1 = 11.\n", + "v2 = 3.\t\t\t #m/s\n", + "\n", + "#Solution:\n", + "#We have11 = omega*math.sqrt(r**2-.75**2) and 3 = omega*math.sqrt(r**2-2**2).\n", + "#These upon solving yield r**2-(121/omega**2)-0.5625 = 0 and r**2-(9/omega**2)-4 = 0.\n", + "#Take r**2 = x and (1/omega**2) = y and the equation become x-121y = 0.5625 and x-9y = 4.\n", + "#Variables Matrix\n", + "A = [[1, -121],[ 1, -9]]\n", + "#Constants Matrix\n", + "B = [.5625, 4]\n", + "V = linalg.solve(A,B)\n", + "#Calculating Amplitude of the Particle\n", + "r = math.sqrt(V[0]) \t \t\t#m\n", + "#Calculating Angular Velocity of the Particle\n", + "omega = math.sqrt(1./V[1]) \t\t\t#rad/s\n", + "#Calculating Periodic Time\n", + "tp = 2*math.pi/omega \t\t \t#seconds\n", + "#Calculating Maximum Acceleration\n", + "amax = omega**2*r \t\t\t#m/s**2\n", + "\n", + "#Results:\n", + "print \" The Angular Velocity omega = %.1f rad/s.\"%(omega)\n", + "print \" The Periodic Time tp = %.1f s.\"%(tp)\n", + "print \" The Maximum Acceleration amax = %.2f m/s**2.\"%(amax)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The Angular Velocity omega = 5.7 rad/s.\n", + " The Periodic Time tp = 1.1 s.\n", + " The Maximum Acceleration amax = 67.38 m/s**2.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.3 Page No : 79" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m = 60. \t\t\t#kg\n", + "r = 0.0125 #m\n", + "x = 0.005 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Calculating the Extension of the Spring\n", + "delta = (.25/1.5)*60*10**-3 \t\t\t#m\n", + "#Calculating the Frequency of the System\n", + "n = 1./(2*math.pi)*math.sqrt(9.81/delta) \t\t\t#Hz\n", + "#Calculating the Angular Velocity of the Mass\n", + "omega = math.sqrt(9.81/delta) \t\t\t#rad/s\n", + "#Calculating the Linear Velocity of the Mass\n", + "v = omega*math.sqrt(r**2-x**2)\n", + "\n", + "#Results:\n", + "print \" The Frequency of Natural Vibration n = %.2f Hz.\"%(n)\n", + "print \" The Velocity of the Mass v = %.2f m/s.\"%(v)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The Frequency of Natural Vibration n = 4.98 Hz.\n", + " The Velocity of the Mass v = 0.36 m/s.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.4 Page No : 82" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m = 1. #kg\n", + "m1 = 2.5 \t\t\t#kg\n", + "s = 1.8*10**3 \t\t#N/m\n", + "l = (300.+300)*10**-3\t\t\t#m\n", + "\n", + "#Solution:\n", + "#Calculating the Mass Moment of Inertia of the System\n", + "IA = (m*l**2/3)+(m1*l**2) \t\t\t#kg-m**2\n", + "#Calculating the Ratio of Alpha to Theta\n", + "#delta = 0.3*theta\n", + "#Restoring Force = s*delta = 540*theta\n", + "#Restoring torque about A = 540*theta*0.3 = 162*theta N-m ...(i)\n", + "#Torque about A = IA*alpha = 1.02*alpha N-m ...(ii)\n", + "#Equating (i) and (ii) 1.02*alpha = 162*theta \n", + "alphabytheta = 162/1.02\n", + "#Calculating the Frequency of Oscillation\n", + "n = 1/(2*math.pi)*math.sqrt(alphabytheta)\n", + "\n", + "#Results:\n", + "print \" The Frequency of Oscillation n = %.2f Hz.\"%(n)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The Frequency of Oscillation n = 2.01 Hz.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.5 Page No : 83" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m = 85. \t\t\t#kg\n", + "h = 0.1 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Calculating the Frequency of Oscillation\n", + "n = 100./145 \t\t\t#Hz\n", + "#Calculating the Equivalent Length of Simple Pendulum\n", + "L = (1/(2*math.pi)/.69*math.sqrt(9.81))**2\n", + "#Calculating the Radius of Gyration\n", + "kG = math.sqrt((L-h)*h)\n", + "#Calculating the Moment of Inertia of the Flywheel through the Centre of Gravity\n", + "I = m*kG**2 \t\t\t#kg-m**2\n", + "\n", + "#Results:\n", + "print \" The Moment of Inertia of the Flywheel Through its c.g. I = %.1f kg-m**2.\"%(I)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The Moment of Inertia of the Flywheel Through its c.g. I = 3.6 kg-m**2.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.6 Page No : 84" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from scipy.optimize import fsolve \n", + "import math \n", + "\n", + "# Variables:\n", + "m = 60. \t\t\t#kg\n", + "d1 = 75.\n", + "d2 = 102. \t\t\t#mm\n", + "\n", + "#Solution:\n", + "#Calculating the Frequencies of Oscillation\n", + "n1 = 100./190\n", + "n2 = 100./165 \t\t\t#Hz\n", + "\n", + "#Calculating the Equivalent Lengths of Simple Pendulum\n", + "L1 = 9.81/(2*math.pi*n1)**2 \t\t\t#m\n", + "L2 = 9.81/(2*math.pi*n2)**2 \t\t\t#m\n", + "\n", + "#Calculating Dismath.tance of c.g. from the Small and Big End Centres (h1 and h2) and the Radius of Gyration\n", + "def f(x):\n", + " h1 = x[0]\n", + " h2 = x[1]\n", + " kG = x[2]\n", + " y = [0,0,0]\n", + " y[0] = L1*h1-h1**2-kG**2\n", + " y[1] = L2*h2-h2**2-kG**2\n", + " y[2] = h1+h2-1\n", + " return y\n", + "\n", + "z = fsolve(f,[1,1,1])\n", + "\n", + "h1 = z[0]\n", + "h2 = z[1]\n", + "kG = z[2]\n", + "\n", + "#Calculating the Mass Moment of Inertia of the Rod\n", + "I = m*kG**2 \t\t\t#kg-m**2\n", + "\n", + "#Results:\n", + "print \" The Moment of Inertia of the Rod I = %d kg-m**2.\"%(I)\n", + "print \" The C.G is at a Distance of h1 = %.3f m from the Small End Centre.\"%(h1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The Moment of Inertia of the Rod I = 6 kg-m**2.\n", + " The C.G is at a Distance of h1 = 0.759 m from the Small End Centre.\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.7 Page No : 85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "l = 1.2 \t\t\t#m\n", + "theta = 3*math.pi/180 \t\t\t#rad\n", + "\n", + "#Solution:\n", + "#Calculating the Distance Between the Knife Edge and C.G. of the Rod\n", + "h = 1.2/2-.05 \t\t\t#m\n", + "#Calculating the Radius of Gyration of the Rod About C.G.\n", + "kG = l/math.sqrt(12) \t\t\t#m\n", + "#Calculating the Time of Swing of the Rod\n", + "tp = 2*math.pi*math.sqrt((kG**2+h**2)/(9.81*h)) \t\t\t#seconds\n", + "#Calculating the Minimum Time of Swing\n", + "tpmin = 2*math.pi*math.sqrt((2*kG)/9.81) \t\t\t#seconds\n", + "#Calculating Angular Velocity\n", + "omega = 2*math.pi/tp \t\t\t#rad/s\n", + "#Calculating Maximum Angular Velocity\n", + "omegamax = omega*theta \t\t\t#rad/s\n", + "#Calculating Maximum Angular Acceleration\n", + "alphamax = omega**2*theta \t\t\t#rad/s**2\n", + "\n", + "#Results:\n", + "print \" The Time of Swing of the Rod tp = %.2f seconds.\"%(tp)\n", + "print \" The Minimum Time of Swing tpmin = %.2f seconds.\"%(tpmin)\n", + "print \" The Maximum Angular Velocity omegamax = %.4f rad/s.\"%(omegamax)\n", + "print \" The Maximum Angular Acceleration (alphamax) = %.3f rad/s**2.\"%( alphamax)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The Time of Swing of the Rod tp = 1.76 seconds.\n", + " The Minimum Time of Swing tpmin = 1.67 seconds.\n", + " The Maximum Angular Velocity omegamax = 0.1871 rad/s.\n", + " The Maximum Angular Acceleration (alphamax) = 0.669 rad/s**2.\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.8 Page No : 86" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import linalg\n", + "import math \n", + "\n", + "# Variables:\n", + "m = 30. \t\t\t#kg\n", + "OG = 1.05 #m\n", + "h = OG #m\n", + "AG = 0.15 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Calculating the Frequency of Oscillation\n", + "n = 20/43.5 \t\t\t#Hz\n", + "#Calculating the Equivalent Length of Simple Pendulum\n", + "L = 9.81/(2*math.pi*n)**2 \t\t\t#m\n", + "#Calculating the Distance of Centre of Percussion (C) from the Centre of Gravity (G)\n", + "CG = L-OG \t\t\t#m\n", + "#Calculating the Distance of Centre of Percussion (C) from the Knife Edge A\n", + "AC = AG-CG \t\t\t#m\n", + "#Calculating the Radius of Gyration of the Pendulum About O\n", + "kO = math.sqrt(L*h) \t\t\t#m\n", + "h1 = h*(1-math.cos(60*math.pi/180)) \t\t\t#m\n", + "#Calculating the Angular Velocity of the Pendulum\n", + "omega = math.sqrt(2*m*9.81*h1/(m*kO**2)) \t\t\t#rad/s\n", + "OA = OG+AG\n", + "#Calculating the Velocity of Striking\n", + "v = omega*(OA) \t\t\t#Velocity of Striking\n", + "#Calculating the Angular Velocity of the Pendulum Immediately After Impact\n", + "I = m*kO**2\n", + "LKE = 55. \t\t\t#Loss of Kinetic Energy N-m\n", + "omega1 = math.sqrt(omega**2-LKE*2/I)\n", + "#Calculating the Impulses at Knife Edge A and at Pivot O (P and Q)\n", + "CLM = m*h*(omega-omega1) \t\t\t#Change of Linear Momentum\n", + "CAM = m*(kO**2-h**2)*(omega-omega1) \t\t\t#Change of Angular Momentum\n", + "#P+Q = Change of Linear Momentum and 0.15P-1.05Q = Change of Angular Momentum.\n", + "#i.e. P+Q = CLM and 0.15P-1.05Q = CAM\n", + "#Variables Matrix\n", + "A = [[1, 1],[0.15,-1.05]]\n", + "B = [CLM, CAM]\n", + "V = linalg.solve(A,B)\n", + "P = V[0]\n", + "Q = V[1]\n", + "#Calculating the Change in Axis Reaction When the Pendulum is Vertical\n", + "CAR = m*(omega**2-omega1**2)*h \t\t\t#Change in Axis Reaction, N\n", + "\n", + "#Results:\n", + "print \" The Distance of Centre of Percussion, AC = %.3f m.\"%(AC)\n", + "print \" The Velocity of Striking = %.2f m/s.\"%(v)\n", + "print \" The Impulse at the Knife Edge P = %.1f N-s.\"%(P)\n", + "print \" The Impulse at the Pivot Q = %.2f N-s.\"%(Q)\n", + "print \" The Change in Axis Reaction When the Pendulum is Vertical = %d N.\"%(CAR)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The Distance of Centre of Percussion, AC = 0.024 m.\n", + " The Velocity of Striking = 3.47 m/s.\n", + " The Impulse at the Knife Edge P = 17.6 N-s.\n", + " The Impulse at the Pivot Q = 0.37 N-s.\n", + " The Change in Axis Reaction When the Pendulum is Vertical = 93 N.\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.9 Page No : 89" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables\n", + "m = 1.5 \t\t\t#kg\n", + "l = 1.25 #m\n", + "x = 120.*10**-3 #m\n", + "y = x \t\t\t #m\n", + "\n", + "#Solution:\n", + "#Calculating the Frequency of Oscillation\n", + "n = 20./40 \t\t\t#Hz\n", + "#Calculating the Radius of Gyration of the Connecting Rod\n", + "kG = 1/(2*math.pi*n)*math.sqrt(9.81*x*y/l) \t\t\t#m\n", + "#Calculating the Moment of Inertia of the Connecting Rod\n", + "I = m*kG**2 \t\t\t#kg-m**2\n", + "\n", + "#Results:\n", + "print \" The Radius of Gyration kG = %d mm.\"%(kG*1000)\n", + "print \" The Mass Moment of Inertia I = %.3f kg-m**2.\"%(I)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The Radius of Gyration kG = 107 mm.\n", + " The Mass Moment of Inertia I = 0.017 kg-m**2.\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.10 Page No : 90" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "l = 2.5\n", + "r = 250.*10**-3 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Calculating the Frequency of Oscillation\n", + "n = 50./170 \t\t\t#Hz\n", + "#Calculating the Radius of Gyration of the Wheel\n", + "kG = r/(2*math.pi*n)*math.sqrt(9.81/l) \t\t\t#m\n", + "\n", + "#Results:\n", + "print \" The Radius of Gyration kG = %d mm.\"%(kG*10**3)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The Radius of Gyration kG = 267 mm.\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.11 Page No : 91" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "m1 = 5.5 #kg\n", + "m2 = 1.5 \t\t\t#kg\n", + "l = 1.25 #m\n", + "r = 125.*10**-3 \t#m\n", + "\n", + "#Solution:\n", + "#Calculating the Frequency of Oscillation\n", + "n = 10./30 \t\t\t#Hz\n", + "#Calculating the Radius of Gyration About an Axis Through the c.g.\n", + "kG = r/(2*math.pi*n)*math.sqrt(9.81/l) \t\t\t#m\n", + "#Calculating the Mass Moment of Inertia About an Axis Through its c.g.\n", + "m = m1+m2 \t\t\t#Total Mass kg\n", + "I = m*kG**2 \t\t\t#kg-m**2\n", + "\n", + "#Results:\n", + "print \"The Mass Moment of Inertia About an Axis Through its c.g. I = %.3f kg-m**2.\"%(I)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Mass Moment of Inertia About an Axis Through its c.g. I = 0.196 kg-m**2.\n" + ] + } + ], + "prompt_number": 15 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Theory_Of_Machines/ch5.ipynb b/Theory_Of_Machines/ch5.ipynb new file mode 100755 index 00000000..2ebeaf21 --- /dev/null +++ b/Theory_Of_Machines/ch5.ipynb @@ -0,0 +1,218 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a49d3045051d18d945c16a76ba66d562e97f6c140e0b1e26e143c1e463c579f5" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 5 : Simple Mechanisms" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.1 Page No : 110" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "AC = 300.\n", + "CB1 = 120. \t\t\t#mm\n", + "\n", + "#Solution:\n", + "#Refer Fig. 5.28\n", + "#Calculating the sine of inclination of the slotted bar with the vertical\n", + "sineCAB1 = CB1/AC\n", + "#Calculating the inclination of the slotted bar with the vertical\n", + "angleCAB1 = math.sin(sineCAB1)*180/math.pi \t\t\t#degrees\n", + "#Calculating the angle alpha\n", + "alpha = 2*(90-angleCAB1) \t\t\t#degrees\n", + "#Calculating the ratio of time of cutting stroke to time of return stroke\n", + "r = (360-alpha)/alpha \t\t\t#Ratio of time of cutting stroke to time of return stroke\n", + "\n", + "#Results:\n", + "print \" The ratio of the time of cutting stroke to the time of return stroke is %.1f\"%(r)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The ratio of the time of cutting stroke to the time of return stroke is 1.7\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.2 Page No : 110" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "AC = 240.\n", + "CB1 = 120.\n", + "AP1 = 450. \t\t\t#mm\n", + "\n", + "#Solution:\n", + "#Refer Fig. 5.29\n", + "#Calculating the math.sine of inclination of the slotted bar with the vertical\n", + "sineCAB1 = CB1/AC\n", + "#Calculating the inclination of the slotted bar with the vertical\n", + "angleCAB1 = math.sin(sineCAB1)*180/math.pi \t\t\t#degrees\n", + "#Calculating the angle alpha\n", + "alpha = 2*(90-angleCAB1) \t\t\t#degrees\n", + "#Calculating the time ratio of cutting stroke to the return stroke\n", + "r = (360-alpha)/alpha \t\t\t#Time ratio of cutting stroke to the return stroke\n", + "#Calculating the length of the stroke\n", + "R1R2 = 2*AP1*round(math.sin(math.pi/2-alpha/2*math.pi/180),1) \t\t\t#mm\n", + "\n", + "#Results:\n", + "print \" The time ratio of cutting stroke to the return stroke is %.f.\"%(r)\n", + "print \" The length of the stroke R1R2 = P1P2 = %d mm.\"%(R1R2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The time ratio of cutting stroke to the return stroke is 2.\n", + " The length of the stroke R1R2 = P1P2 = 450 mm.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.3 Page No : 112" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "#Refer Fig. 5.30 and Fig. 5.31\n", + "BC = 30.\n", + "R1R2 = 120. \t\t\t#mm\n", + "r = 1.7 \t\t\t#Time ratio of working stroke to the return stroke\n", + "\n", + "#Solution:\n", + "#Calculating the angle alpha\n", + "alpha = 360/(1.7+1) \t\t\t#degrees\n", + "#Calculating the length of the link AC\n", + "B1C = BC\n", + "AC = B1C/math.cos(math.radians(alpha/2)) \t\t\t#mm\n", + "#Calculating the length of the link AP\n", + "AP1 = R1R2/(2*math.cos(math.radians(alpha/2))) \t\t\t#mm\n", + "AP = AP1\n", + "\n", + "#Results:\n", + "print \" The length of AC = %.1f mm.\"%(AC)\n", + "print \" The length of AP = %.2f mm.\"%(AP)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The length of AC = 75.7 mm.\n", + " The length of AP = 151.48 mm.\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.4 Page No : 112" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "CD = 50. #mm\n", + "CA = 75. #mm\n", + "PA = 150. #mm\n", + "PR = 135. #mm\n", + "\n", + "#Solution:\n", + "#Refer Fig. 5.32 and Fig. 5.33\n", + "#Calculating the cosine of angle beta\n", + "CA2 = CA\n", + "cosbeta = CD/CA2\n", + "#Calculating the angle beta\n", + "beta = 2*math.degrees(math.acos(cosbeta)) \t\t\t#degrees\n", + "#Calculating the ratio of time of cutting stroke to time of return stroke\n", + "r = (360-beta)/beta \t\t\t#Ratio of time of cutting stroke to time of return stroke\n", + "#Calculating the length of effective stroke\n", + "R1R2 = 87.5 \t\t\t#mm\n", + "\n", + "#Results:\n", + "print \" The ratio of time of cutting stroke to time of return stroke is %.3f.\"%(r)\n", + "print \" The length of effective stroke R1R2 = %.1f mm.\"%(R1R2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The ratio of time of cutting stroke to time of return stroke is 2.735.\n", + " The length of effective stroke R1R2 = 87.5 mm.\n" + ] + } + ], + "prompt_number": 5 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Theory_Of_Machines/ch6.ipynb b/Theory_Of_Machines/ch6.ipynb new file mode 100755 index 00000000..dfc6f728 --- /dev/null +++ b/Theory_Of_Machines/ch6.ipynb @@ -0,0 +1,347 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:1b2bc75a9da3f88ed85d2991e9ff321b52f8f56e21ca05cba6bbfb84f1e84dc2" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 6 : Velocity in Mechanisms(Instantaneous Centre Method)" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.1 Page No : 126" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "NAB = 100. \t\t\t#rpm\n", + "AB = 300./1000\n", + "BC = 360./1000\n", + "CD = BC \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 6.9\n", + "#Calculating the angular speed of link AB\n", + "omegaAB = 2*math.pi*NAB/60 \t\t\t#rad/s\n", + "#Calculating the velocity of point B on link AB\n", + "vB = omegaAB*AB \t\t\t#m/s\n", + "#Calculating the angular velocity of link BC\n", + "#By measurement from instantaneous centre diagram Fig. 6.10\n", + "\n", + "I13B = 500./1000 \t\t\t#m\n", + "omegaBC = vB/I13B \t\t\t#rad/s\n", + "\n", + "#Results:\n", + "print \" The angular velocity of the link BC omegaBC = %.3f rad/s.\"%(omegaBC)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The angular velocity of the link BC omegaBC = 6.283 rad/s.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.2 Page No : 128" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "omegaOB = 10. \t\t\t#rad/s\n", + "OB = 100/1000. \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 6.12\n", + "#Calculating the velocity of the crank OB\n", + "vOB = omegaOB*OB \t\t\t#m/s\n", + "vB = vOB\n", + "#By measurement from the instantaneous cemtre diagram Fig. 6.13\n", + "\n", + "I13A = 460./1000\n", + "I13B = 560./1000 \t\t\t#m\n", + "#Calculating the velocity of slider A\n", + "vA = vB/I13B*I13A\n", + "#Calculating the angular velocity of the connecting rod AB\n", + "omegaAB = vB/I13B \t\t\t#rad/s\n", + "\n", + "#Results:\n", + "print \"The velocity of slider A vA = %.2f m/s.\"%(vA)\n", + "print \"The angular velocity of connecting rod AB omegaAB = %.2f rad/s.\"%(omegaAB)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The velocity of slider A vA = 0.82 m/s.\n", + "The angular velocity of connecting rod AB omegaAB = 1.79 rad/s.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.3 Page No : 130" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "NOA = 120. \t\t\t#rpm\n", + "OA = 200./1000 \t\t\t#m/s\n", + "\n", + "#Results:\n", + "#Refer Fig. 6.15\n", + "#Calculating the angular velocity of the crank OA\n", + "omegaOA = 2*math.pi*NOA/60 \t\t\t#rad/s\n", + "#Calculating the velocity of crank OA\n", + "vOA = omegaOA*OA \t\t\t#m/s\n", + "vA = vOA\n", + "#By measurement from the instantaneous cemtre diagram Fig. 6.16\n", + "\n", + "I13A = 840./1000\n", + "I13B = 1070./1000\n", + "I14B = 400./1000\n", + "I14C = 200./1000\n", + "I15C = 740./1000\n", + "I15D = 500./1000 \t\t\t#m\n", + "\n", + "#Calculating the velocity of point B\n", + "vB = vA/I13A*I13B \t\t\t#m/s\n", + "#Calculating the velocity of point C\n", + "vC = vB/I14B*I14C \t\t\t#m/s\n", + "#Calculating the velocity of point B\n", + "vD = vC/I15C*I15D \t\t\t#m/s\n", + "#Calculating the angular velocity of the link AB\n", + "omegaAB = vA/I13A \t\t\t#rad/s\n", + "#Calculating the angular velocity of the link BC\n", + "omegaBC = vB/I14B \t\t\t#rad/s\n", + "#Calculating the angular velocity of the link CD\n", + "omegaCD = vC/I15C \t\t\t#rad/s\n", + "\n", + "#Results:\n", + "print \" The velocity of point B vB = %.1f m/s.\"%(vB)\n", + "print \" The velocity of point C vC = %.1f m/s.\"%(vC)\n", + "print \" The velocity of point D vD = %.2f m/s.\"%(vD)\n", + "print \" The angular velocity of the link AB omegaAB = %.2f rad/s.\"%(omegaAB)\n", + "print \" The angular velocity of the link BC omegaBC = %d rad/s.\"%(omegaBC)\n", + "print \" The angular velocity of the link CD omegaCD = %.2f rad/s.\"%(omegaCD)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The velocity of point B vB = 3.2 m/s.\n", + " The velocity of point C vC = 1.6 m/s.\n", + " The velocity of point D vD = 1.08 m/s.\n", + " The angular velocity of the link AB omegaAB = 2.99 rad/s.\n", + " The angular velocity of the link BC omegaBC = 8 rad/s.\n", + " The angular velocity of the link CD omegaCD = 2.16 rad/s.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.4 Page No : 133" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "omegaO1A = 100. \t\t\t#rad/s\n", + "O1A = 100/1000. \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 6.18\n", + "#Calculating the linear velocity of crank O1A\n", + "vO1A = omegaO1A*O1A \t\t\t#m/s\n", + "vA = vO1A\n", + "#By measurement from the instantaneous cemtre diagram Fig. 6.19\n", + "I13A = 910./1000\n", + "I13B = 820./1000\n", + "I15B = 130./1000\n", + "I15D = 50./1000\n", + "I16D = 200./1000\n", + "I16E = 400./1000 \t\t\t#m\n", + "#Calculating the velocity of point B\n", + "vB = vA/I13A*I13B \t\t\t#m/s\n", + "#Calculating the velocity of point D\n", + "vD = vB/I15B*I15D \t\t\t#m/s\n", + "#Calculating the velocity of point E\n", + "vE = vD/I16D*I16E \t\t\t#m/s\n", + "\n", + "#Results:\n", + "print \" The velocity of point B vB = %.2f m/s.\"%(vB)\n", + "print \" The velocity of point D vD = %.2f m/s.\"%(vD)\n", + "print \" The velocity of point E vE = %.2f m/s.\"%(vE)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The velocity of point B vB = 9.01 m/s.\n", + " The velocity of point D vD = 3.47 m/s.\n", + " The velocity of point E vE = 6.93 m/s.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.5 Page No : 135" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "NO1A = 400. \t\t\t#rpm\n", + "O1A = 16./1000 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 6.21\n", + "#Calculating the angular velocity of the crank O1A\n", + "omegaO1A = 2*math.pi*NO1A/60 \t\t\t#rad/s\n", + "#Calculating the linear velocity of the crank O1A\n", + "vO1A = omegaO1A*O1A \t\t\t#m/s\n", + "vA = vO1A\n", + "#By measurement from the instantaneous cemtre diagram Fig. 6.22\n", + "I13A = 41./1000\n", + "I13B = 50./1000\n", + "I14B = 23./1000\n", + "I14C = 28./1000\n", + "I15C = 65./1000\n", + "I15D = 62./1000 \t\t\t#m\n", + "#Calculating the velocity of point B\n", + "vB = vA/I13A*I13B \t\t\t#m/s\n", + "#Calculating the velocity of point C\n", + "vC = vB/I14B*I14C \t\t\t#m/s\n", + "#Calculating the velocity of of the needle at D\n", + "vD = vC/I15C*I15D \t\t\t#m/s\n", + "\n", + "#Results:\n", + "print \"The velocity of the needle at D vD = %.2f m/s.\"%(vD)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The velocity of the needle at D vD = 0.95 m/s.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6 Page No : 137" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "NOA = 120. \t\t\t#rpm\n", + "#Solution:\n", + "#Refer Fig. 6.24\n", + "#Calculating the angular speed of crank OA\n", + "omegaOA = 2*math.pi*NOA/60 \t\t\t#rad/s\n", + "#By measurement from the instantaneous cemtre diagram Fig. 6.25\n", + "I12I26 = 65./1000 \t\t\t#m\n", + "#Calculating the velocity of the ram\n", + "vD = omegaOA*I12I26 \t\t\t#m/s\n", + "\n", + "#Results:\n", + "print \" The velocity of ram, vD = %.3f m/s.\"%(vD)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The velocity of ram, vD = 0.817 m/s.\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Theory_Of_Machines/ch7.ipynb b/Theory_Of_Machines/ch7.ipynb new file mode 100755 index 00000000..83081a67 --- /dev/null +++ b/Theory_Of_Machines/ch7.ipynb @@ -0,0 +1,788 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:c5834020c85eb4ee9dd0f62630cfe11c8ed3ce693f7ae9cc407ce5e3a8a74085" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 7 : Velocity in Mechanisms (Relative Velocity Method)" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.1 Page No : 148" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "NBA = 120. \t\t\t#rpm\n", + "AB = 40./1000\n", + "CD = 80./1000 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 7.7\n", + "#Calculating the angular velocity of the crank AB\n", + "omegaBA = 2*math.pi*NBA/60 \t\t\t#rad/s\n", + "#Calculating the velocity of B with respect to A\n", + "vBA = omegaBA*AB \t\t\t#m/s\n", + "vB = vBA\n", + "#By measurement from the velocity diagram Fig. 7.7(b)\n", + "vCD = 0.385 \t\t\t#m/s\n", + "vC = vCD\n", + "#Calculating the angular velocity of link CD\n", + "omegaCD = vCD/CD \t\t\t#rad/s\n", + "\n", + "#Results:\n", + "print \" The angular velocity of link CD omegaCD = %.1f rad/s clockwise about D.\"%(omegaCD)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The angular velocity of link CD omegaCD = 4.8 rad/s clockwise about D.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.2 Page No : 148" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "NBO = 180. \t\t\t#rpm\n", + "OB = 0.5\n", + "PB = 2.\n", + "dO = 50./1000\n", + "dB = 60./1000\n", + "dC = 30./1000 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 7.8\n", + "#Calculating the angular velocity of the crank BO\n", + "omegaBO = 2*math.pi*NBO/60 \t\t\t#rad/s\n", + "#Calculating the linear velocity of B with respect to O\n", + "vBO = omegaBO*OB \t\t\t#m/s\n", + "vB = vBO\n", + "#By measurement from the velocity diagram Fig. 7.8(b)\n", + "vP = 8.15\n", + "vPB = 6.8\n", + "vE = 8.5\n", + "bg = 5.\n", + "bp = vPB\n", + "vG = 8. \t\t\t#m/s\n", + "#Calculating the angular velocity of the connecting rod PB\n", + "omegaPB = vPB/PB \t\t\t#rad/s\n", + "#Calculating the velocity of rubbing at the pin of crank-shaft\n", + "vCS = dO/2*omegaBO \t\t\t#Velocity of rubbing at the pin of crank-shaft m/s\n", + "#Calculating the velocity of rubbing at the pin of crank\n", + "vC = dB/2*(omegaBO+omegaPB) \t\t\t#Velocity of rubbing at the pin of crank m/s\n", + "#Calculating the velocity of rubbing at the pin of cross-head\n", + "vPCH = dC/2*omegaPB \t\t\t#Velocity of rubbing at the pin of cross-head m/s\n", + "#Calculating the position of point G on the connecting rod\n", + "BG = bg/bp*PB \t\t\t#m\n", + "\n", + "#Results:\n", + "print \" The velocity of piston P vP = %.2f m/s.\"%(vP)\n", + "print \" The angular velocity of connecting rod omegaPB = %.1f rad/s anticlockwise.\"%(omegaPB)\n", + "print \" The velocity of point E on the connecting rod vE = %.1f m/s.\"%(vE)\n", + "print \" The velocity of rubbing at the pin of crank-shaft is %.2f m/s.\"%(vCS)\n", + "print \" The velocity of rubbing at the pin of crank is %.4f m/s.\"%(vC)\n", + "print \" The velocity of rubbing at the pin of cross-head is %.3f m/s.\"%(vPCH)\n", + "print \" The position of point G on the connecting rod BG = %.2f m.\"%(BG)\n", + "print \" The linear velocity of point G vG = %d m/s.\"%(vG)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The velocity of piston P vP = 8.15 m/s.\n", + " The angular velocity of connecting rod omegaPB = 3.4 rad/s anticlockwise.\n", + " The velocity of point E on the connecting rod vE = 8.5 m/s.\n", + " The velocity of rubbing at the pin of crank-shaft is 0.47 m/s.\n", + " The velocity of rubbing at the pin of crank is 0.6675 m/s.\n", + " The velocity of rubbing at the pin of cross-head is 0.051 m/s.\n", + " The position of point G on the connecting rod BG = 1.47 m.\n", + " The linear velocity of point G vG = 8 m/s.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.3 Page No : 150" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "NAO = 600. \t\t\t#rpm\n", + "OA = 28./1000 #m\n", + "BD = 46./1000 \t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 7.10\n", + "#Calculating the angular velocity of crank AO\n", + "omegaAO = 2*math.pi*NAO/60 \t\t\t#rad/s\n", + "#Calculating the velocity of A with respect to O\n", + "vAO = omegaAO*OA \t\t\t#m/s\n", + "vA = vAO\n", + "#By measurement from the velocity diagram Fig. 7.10(b)\n", + "vD = 1.6\n", + "vDB = 1.7 \t\t\t#m/s\n", + "#Calculating the angular velocity of D with respect to B\n", + "omegaBD = vDB/BD \t\t\t#rad/s\n", + "\n", + "#Results:\n", + "print \" The velocity of the slider D vD = %.1f m/s.\"%(vD)\n", + "print \" The angular velocity of the link BD omegaBD = %.2f rad/s clockwise sbout B.\"%(omegaBD)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The velocity of the slider D vD = 1.6 m/s.\n", + " The angular velocity of the link BD omegaBD = 36.96 rad/s clockwise sbout B.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.4 Page No : 151" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "NBA = 120. \t\t\t#rpm\n", + "AB = 150./1000 #mm\n", + "DC = 450./1000 #mm\n", + "BC = 450./1000 #mm\n", + "dC = 50./1000\n", + "rC = dC/2 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 7.12\n", + "#Calculating the angular velocity of the crank AB\n", + "omegaBA = 2*math.pi*NBA/60 \t\t\t#rad/s\n", + "#Calculating the linear velocity of B with respect to A\n", + "vBA = omegaBA*AB \t\t\t#m/s\n", + "vB = vBA\n", + "#By measurement from the velocity diagram Fig. 7.12(b)\n", + "vF = 0.7\n", + "vCD = 2.25\n", + "vCB = 2.25 \t\t\t#m/s\n", + "#Calculating the angular velocity of DC\n", + "omegaDC = vCD/DC \t\t\t#rad/s\n", + "#Calculating the angular velocity of BC\n", + "omegaCB = vCB/BC \t\t\t#rad/s\n", + "#Calculating the rubbing speed at the pin C\n", + "vr = (omegaCB-omegaDC)*rC \t\t\t#The rubbing speed at the pin C m/s\n", + "\n", + "#Results:\n", + "print \" The velocity of block F vF = %.1f m/s.\"%(vF)\n", + "print \" The angular velocity of DC omegaDC = %d rad/s anticlockwise about D.\"%(omegaDC)\n", + "print \" The rubbing speed at the pin C is %d m/s.\"%(vr)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The velocity of block F vF = 0.7 m/s.\n", + " The angular velocity of DC omegaDC = 5 rad/s anticlockwise about D.\n", + " The rubbing speed at the pin C is 0 m/s.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.5 Page No : 153" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "NAO = 120. \t\t\t#rpm\n", + "OA = 100./1000\n", + "CE = 350./1000 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 7.13\n", + "#Calculating the angular speed of the crank OA\n", + "omegaAO = 2*math.pi*NAO/60 \t\t\t#rad/s\n", + "#Calculating the velocity of A with respect to O\n", + "vAO = omegaAO*OA \t\t\t#m/s\n", + "vA = vAO\n", + "#By measurement from the velocity diagram Fig. 7.14(b)\n", + "vF = 0.53\n", + "od = 1.08\n", + "vCE = 0.44 \t\t\t#m/s\n", + "#Calculating the angular velocity of CE\n", + "omegaCE = vCE/CE \t\t\t#rad/s\n", + "\n", + "#Results:\n", + "print \" The velocity of F vF = %.2f m/s.\"%(vF)\n", + "print \" The velocity of sliding of CE in the trunnion is %.2f m/s.\"%(od)\n", + "print \" The angular velocity of CE omegaCE = %.2f rad/s clockwise about E.\"%(omegaCE)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The velocity of F vF = 0.53 m/s.\n", + " The velocity of sliding of CE in the trunnion is 1.08 m/s.\n", + " The angular velocity of CE omegaCE = 1.26 rad/s clockwise about E.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.6 Page No : 155" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "NCO = 120. \t\t\t#rpm\n", + "OC = 125./1000 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 7.15\n", + "#Calculating the angular velocity of the crank CO\n", + "omegaCO = 2*math.pi*NCO/60 \t\t\t#rad/s\n", + "#Calculating the linear velocity of C with respect to O\n", + "vCO = omegaCO*OC \t\t\t#m/s\n", + "vC = vCO\n", + "#By measurement from the velocity diagram Fig. 7.16(b)\n", + "\n", + "vCO = 1.57\n", + "vE = 0.7 \t\t\t#m/s\n", + "\n", + "#Results:\n", + "print \" The absolute velocity of point E of the lever vE = %.1f m/s.\"%(vE)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The absolute velocity of point E of the lever vE = 0.7 m/s.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.7 Page No : 156" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "NBO1 = 40. \t\t\t#rpm\n", + "O1O2 = 800./1000\n", + "O1B = 300./1000\n", + "O2D = 1300./1000\n", + "DR = 400./1000 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 7.18\n", + "#Calculating the angular speed of the crank BO\n", + "omegaBO1 = 2*math.pi*NBO1/60 \t\t\t#rad/s\n", + "#Calculating the velocity of B with respect to O1\n", + "vBO1 = omegaBO1*O1B \t\t\t#m/s\n", + "vB = vBO1\n", + "#By measurement from the velocity diagram Fig. 7.18(b)\n", + "vR = 1.44\n", + "vDO2 = 1.32 \t\t\t#m/s\n", + "vD = vDO2\n", + "#Calculating the angular velocity of the link O2D\n", + "omegaDO2 = vDO2/O2D \t\t\t#rad/s\n", + "\n", + "#Results:\n", + "print \" The velocity of the ram R vR = %.2f m/s.\"%(vR)\n", + "print \" The angular velocity of the link O2D omegaDO2 = %.3f rad/s anticlockwise about O2.\"%(omegaDO2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The velocity of the ram R vR = 1.44 m/s.\n", + " The angular velocity of the link O2D omegaDO2 = 1.015 rad/s anticlockwise about O2.\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.8 Page No : 158" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "NAO1 = 60. \t\t\t#rpm\n", + "O1A = 85.\n", + "rQ = 50. \t\t\t#mm\n", + "\n", + "#Solution:\n", + "#Refer Fig. 7.20 and Fig. 7.21\n", + "#Calculating the angular velocity of AO1\n", + "omegaAO1 = 2*math.pi*NAO1/60 \t\t\t#rad/s\n", + "#Calculating the velocity of A with respect to O1\n", + "vAO1 = omegaAO1*O1A \t\t\t#mm/s\n", + "vA = vAO1\n", + "#By measurement from the velocity diagram Fig. 7.20(b)\n", + "vDO2 = 410. \t\t\t#mm/s\n", + "O2D = 264. \t\t\t#mm\n", + "angleB1O2B2 = 60*math.pi/180 \t\t\t#rad\n", + "#To vary the Scilab function 'beta'\n", + "alpha = 120\n", + "beta = 240 \t\t\t#degrees\n", + "#Calculating the angular velocity of the quadant Q\n", + "omegaQ = vDO2/O2D \t\t\t#rad/s\n", + "#Calculating the linear speed of the rack\n", + "vR = omegaQ*rQ \t\t\t#mm/s\n", + "#Calculating the ratio of times of lowering and raimath.sing the rack\n", + "r = beta/alpha\n", + "#Calculating the length of stroke of the rack\n", + "L = rQ*angleB1O2B2 \t\t\t#mm\n", + "\n", + "#Results:\n", + "print \" The linear speed of the rack vR = %.1f mm/s.\"%(vR)\n", + "print \" The ratio of times of lowering and raising the rack is %d.\"%(r)\n", + "print \" The length of the stroke of the rack is %.2f mm.\"%(L)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The linear speed of the rack vR = 77.7 mm/s.\n", + " The ratio of times of lowering and raising the rack is 2.\n", + " The length of the stroke of the rack is 52.36 mm.\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.9 Page No : 160" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "NPO = 120. \t\t\t#rpm\n", + "OQ = 100./1000 #mm\n", + "OP = 200./1000 #mm\n", + "RQ = 150./1000 #mm\n", + "RS = 500./1000 \t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 7.22\n", + "#Calculating the angular speed of the crank PO\n", + "omegaPO = 2*math.pi*NPO/60 \t\t\t#rad/s\n", + "#Calculating the velocity of P with respect to O\n", + "vPO = omegaPO*OP \t\t\t#m/s\n", + "vP = vPO\n", + "#By measurement from the velocity diagram Fig. 7.23(b)\n", + "vS = 0.8\n", + "vSR = 0.96\n", + "vTP = 0.85 \t\t\t#m/s\n", + "#Calculating the angular velocity of link RS\n", + "omegaRS = vSR/RS \t\t\t#rad/s\n", + "\n", + "#Results:\n", + "print \" The velocity of the slider S cutting tool vS = %.1f m/s.\"%(vS)\n", + "print \" The angular velocity of the link RS omegaRS = %.2f rad/s clockwise about R.\"%(omegaRS)\n", + "print \" The velocity of the sliding block T on the slotted lever QT vTP = %.2f m/s.\"%(vTP)\n", + "\n", + "# note : answer in book is wrong\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The velocity of the slider S cutting tool vS = 0.8 m/s.\n", + " The angular velocity of the link RS omegaRS = 1.92 rad/s clockwise about R.\n", + " The velocity of the sliding block T on the slotted lever QT vTP = 0.85 m/s.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.10 Page No : 162" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "NAD = 100. \t\t\t#rpm\n", + "TA = 50. \t\t\t#N-m\n", + "DA = 300./1000\n", + "CB = 360./1000\n", + "AB = CB\n", + "DC = 600./1000 \t\t\t#m\n", + "eta = 70./100 \t\t\t#%\n", + "\n", + "#Solution:\n", + "#Refer Fig. 7.25\n", + "#Calculating the angular velocity of the crank AD\n", + "omegaAD = 2*math.pi*NAD/60 \t\t\t#rad/s\n", + "#Calculating the velocity of A with respect to D\n", + "vAD = omegaAD*DA \t\t\t#m/s\n", + "vA = vAD\n", + "#By measurement from the velocity diagram Fig. 7.25(b)\n", + "vBC = 2.25 \t\t\t#m/s\n", + "vB = vBC\n", + "#Calculating the angular velocity of the driven link CB\n", + "omegaBC = vBC/CB \t\t\t#rad/s\n", + "#Calculating the actual mechanical advantage\n", + "omegaA = omegaAD\n", + "omegaB = omegaBC\n", + "MAactual = eta*omegaA/omegaB\n", + "#Calculating the resisting torque\n", + "TB = eta*TA*omegaA/omegaB \t\t\t#N-m\n", + "\n", + "#Results:\n", + "print \" The velocity of the point B. vB = %.2f m/s.\"%(vB)\n", + "print \" The angular velocity of the driven link CB. omegaBC = %.2f rad/s.\"%(omegaBC)\n", + "print \" The actual mechanical advantage. M.A.actual) = %.2f.\"%(MAactual)\n", + "print \" The resisting torque. TB = %.1f N-m.\"%(TB)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The velocity of the point B. vB = 2.25 m/s.\n", + " The angular velocity of the driven link CB. omegaBC = 6.25 rad/s.\n", + " The actual mechanical advantage. M.A.actual) = 1.17.\n", + " The resisting torque. TB = 58.6 N-m.\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.11 Page No : 163" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "WC = 2.5*1000\n", + "WD = 4.*1000 \t\t\t#N\n", + "OA = 175./1000 #mm\n", + "AB = 180./1000 #mm\n", + "AD = 500./1000 #mm\n", + "BC = 325./1000 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 7.26\n", + "#Assuming the speed of crank OA to be 'N'\n", + "#Calculating the angular velocity of crank OA\n", + "#Assume the vector oa (i.e. velocity of A) as 20 m/s\n", + "N = 20/(2*math.pi/60*OA) \t\t\t#mm\n", + "#By measurement from the velocity diagram Fig. 7.27(b)\n", + "vC = 35.\n", + "vD = 21. \t\t\t#mm\n", + "#Calculating the velocity ratio between C and the ram D\n", + "r = vC/vD \t\t\t#The velocity ratio between C and the ram D\n", + "#Calculating the efficiency of the machine\n", + "eta = (WD*vD)/(WC*vC)*100 \t\t\t#%\n", + "\n", + "#Results:\n", + "print \" The velocity ratio between C and the ram D is %.2f.\"%(r)\n", + "print \" The efficiency of the machine eta = n %d %%.\"%(eta)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The velocity ratio between C and the ram D is 1.67.\n", + " The efficiency of the machine eta = n 96 %.\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.12 Page No : 165" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "NAO = 180. \t\t\t#rpm\n", + "OA = 180./1000 #mm\n", + "CB = 240./1000 #mm\n", + "AB = 360./1000 #mm\n", + "BD = 540./1000 \t\t#mm\n", + "FD = 2.*1000 \t\t\t#N\n", + "DA = 30./1000\n", + "DD = DA\n", + "rA = DA/2\n", + "rD = DD/2 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 7.28\n", + "#Calculating the angular velocity of the crank OA\n", + "omegaAO = 2*math.pi*NAO/60 \t\t\t#rad/s\n", + "#Calculating the velocity of A with respect to O\n", + "vAO = omegaAO*OA\n", + "vA = vAO\n", + "#By measurement fro the velocity diagram Fig. 7.29(b)\n", + "vD = 2.05\n", + "vBA = 0.9\n", + "vBC = 2.8\n", + "vDB = 2.4 \t\t\t#m/s\n", + "#Calculating the angular velocity of the link AB\n", + "omegaAB = vBA/AB \t\t\t#rad/s\n", + "#Calculating the angular velocity of the link CB\n", + "omegaCB = vBC/CB \t\t\t#rad/s\n", + "#Calculating the angular velocity of the link BD\n", + "omegaBD = vDB/BD \t\t\t#rad/s\n", + "#Calculating the relative angular velocity at A\n", + "rvA = omegaCB-omegaAB+omegaBD \t\t\t#The relative angular velocity at A rad/s\n", + "#Calculating the relative angular velocity at D\n", + "rvD = omegaBD \t\t\t#The relative angular velocity at D rad/s\n", + "#Calculating the velocity of rubbing on the pin A\n", + "vrA = rvA*rA*1000 \t\t\t#The velocity of rubbing on the pin A mm/s\n", + "#Calculating the velocity of rubbing on the pin D\n", + "vrD = rvD*rD*1000 \t\t\t#The velocity of rubbing on the pin D mm/s\n", + "#Calculating the torque applied to crank OA\n", + "TA = FD*vD/omegaAO \t\t\t#N-m\n", + "\n", + "#Results:\n", + "print \" The velocity of slider D vD = %.2f m/s.\"%(vD)\n", + "print \" The angular velocity of the link AB, omegaAB = %.1f rad/s, anticlockwise about A.\"%(omegaAB)\n", + "print \" The angular velocity of the link CB, omegaCB = %.2f rad/s, anticlockwise about C.\"%(omegaCB)\n", + "print \" The angular velocity of the link BD, omegaBD = %.2f rad/s, clockwise about B.\"%(omegaBD)\n", + "print \" The velocity of rubbing on the pin A is %d mm/s.\"%(vrA)\n", + "print \" The velocity of rubbing on the pin D is %d mm/s.\"%(vrD)\n", + "print \" The torque applied to the crank OA, TA = %.1f N-m.\"%(TA)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The velocity of slider D vD = 2.05 m/s.\n", + " The angular velocity of the link AB, omegaAB = 2.5 rad/s, anticlockwise about A.\n", + " The angular velocity of the link CB, omegaCB = 11.67 rad/s, anticlockwise about C.\n", + " The angular velocity of the link BD, omegaBD = 4.44 rad/s, clockwise about B.\n", + " The velocity of rubbing on the pin A is 204 mm/s.\n", + " The velocity of rubbing on the pin D is 66 mm/s.\n", + " The torque applied to the crank OA, TA = 217.5 N-m.\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.13 Page No : 167" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "NBA = 180 \t\t\t#rpm\n", + "AB = 0.45 #m\n", + "BD = 1.5 #m\n", + "BC = 0.9 #m\n", + "CE = BC \t\t\t#m\n", + "FD = 500.\n", + "FE = 750. \t\t\t#N\n", + "\n", + "#Solution:\n", + "#Refer Fig. 7.31\n", + "#Calculating the angular velocity of the crank AB\n", + "omegaBA = 2*math.pi*NBA/60 \t\t\t#rad/s\n", + "#Calculating the velocity of B with respect to A\n", + "vBA = omegaBA*AB \t\t\t#m/s\n", + "vB = vBA\n", + "#By measurement from the velocity diagram Fig. 7.31(b)\n", + "vD = 9.5\n", + "vE = 1.7 \t\t\t#m/s\n", + "#Calculating the power input\n", + "Pi = FD*vD-FE*vE \t\t\t#N-m/s\n", + "#Calculating the turning moment at A\n", + "TA = Pi/omegaBA \t\t\t#N-m\n", + "\n", + "#Results:\n", + "print \" The velocity of slider D, vD = %.1f m/s.\"%(vD)\n", + "print \" The velocity of slider E, vE = %.1f m/s.\"%(vE)\n", + "print \" The turning moment at A, TA = %.1f N-m.\"%(TA)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The velocity of slider D, vD = 9.5 m/s.\n", + " The velocity of slider E, vE = 1.7 m/s.\n", + " The turning moment at A, TA = 184.4 N-m.\n" + ] + } + ], + "prompt_number": 16 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Theory_Of_Machines/ch8.ipynb b/Theory_Of_Machines/ch8.ipynb new file mode 100755 index 00000000..30669b70 --- /dev/null +++ b/Theory_Of_Machines/ch8.ipynb @@ -0,0 +1,1243 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:8b757cfb7a3d46c26051163fc5ae963591320e2fd8a4c339acfb031be7305afd" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 8 : Acceleration in Mechanisms" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.1 Page No : 177" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "NBO = 300. \t\t\t#rpm\n", + "OB = 150./1000\n", + "BA = 600./1000 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 8.4\n", + "#Calculating the angular velocity of BO\n", + "omegaBO = 2*math.pi*NBO/60 \t\t\t#rad/s\n", + "#Calculating the linear velocity of B with respect to O\n", + "vBO = omegaBO*OB \t\t\t#m/s\n", + "vB = vBO\n", + "#By measurement from the velocity diagram Fig. 8.4(b)\n", + "vAB = 3.4\n", + "vD = 4.1 \t\t\t#m/s\n", + "#Calculating the radial component of the acceleration of B with respect of O\n", + "arBO = vBO**2/OB \t\t\t#m/s**2\n", + "aB = arBO\n", + "#Calculating the radisla component of the accaleration of A with respect to B\n", + "arAB = vAB**2/BA \t\t\t#m/s**2\n", + "#By measurement from the acceleration diagram Fig. 8.4(c)\n", + "aD = 117.\n", + "adashAB = 103. \t\t\t#m/s**2\n", + "#Calculating the angular velocity of the connecting rod\n", + "omegaAB = vAB/BA \t\t\t#rad/s**2\n", + "#Calculating the angular acceleration of the connecting rod\n", + "alphaAB = adashAB/BA \t\t\t#rad/s**2\n", + "\n", + "#Results:\n", + "print \" The linear velocity of the midpoint of the connecting rod, vD = %.1f m/s.\"%(vD)\n", + "print \" The linear acceleration of the midpoint of the connecting rod, aD = %d m/s**2.\"%(aD)\n", + "print \" The angular velocity of the connecting rod, omegaAB = %.2f rad/s, anticlockwise about B.\"%(omegaAB)\n", + "print \" The angular acceleration of the connecting rod, alphaAB = %.2f rad/s**2, clockwise about B.\"%(alphaAB)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The linear velocity of the midpoint of the connecting rod, vD = 4.1 m/s.\n", + " The linear acceleration of the midpoint of the connecting rod, aD = 117 m/s**2.\n", + " The angular velocity of the connecting rod, omegaAB = 5.67 rad/s, anticlockwise about B.\n", + " The angular acceleration of the connecting rod, alphaAB = 171.67 rad/s**2, clockwise about B.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.2 Page No : 180" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "omegaBC = 75. \t \t\t#rad/s\n", + "alphaBC = 1200. \t\t\t#rad/s**2\n", + "CB = 100/1000. #m\n", + "BA = 300/1000. \t\t\t#m/\n", + "\n", + "#Solution:\n", + "#Refer Fig. 8.5\n", + "#Calculating the linear velocity of B with respect to C\n", + "vBC = omegaBC*CB \t\t\t#m/s\n", + "#Calculating the math.tangential component of the acceleration of B with respect to C\n", + "alphatBC = alphaBC*CB \t\t\t#m/s**2\n", + "#By measurement from the velocity diagram Fig. 8.6(b)\n", + "vG = 6.8\n", + "vAB = 4 \t\t\t#m/s\n", + "#Calculating the angular velocity of AB\n", + "omegaAB = vAB/BA \t\t\t#rad/s\n", + "#Calculating the radial component of the acceleration of B with respect to C\n", + "arBC = vBC**2/CB \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of A with respect to B\n", + "arAB = vAB**2/BA \t\t\t#m/s**2\n", + "#By measurement from the acceleration diagram Fig. 8.6(c)\n", + "arBC = 120.\n", + "arAB = 53.3\n", + "aG = 414.\n", + "atAB = 546. \t\t\t#m/s**2\n", + "#Calculating the angular acceleration of AB\n", + "alphaAB = atAB/BA \t\t\t#rad/s**2\n", + "\n", + "#Results:\n", + "print \" The velocity of G, vG = %.1f m/s.\"%(vG)\n", + "print \" The angular velocity of AB, omegaAB = %.1f rad/s, clockwise.\"%(omegaAB)\n", + "print \" The acceleration of G, aG = %d m/s**2.\"%(aG)\n", + "print \" The angular accaleration of AB, alphaAB = %d rad/s**2.\"%(alphaAB)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The velocity of G, vG = 6.8 m/s.\n", + " The angular velocity of AB, omegaAB = 13.3 rad/s, clockwise.\n", + " The acceleration of G, aG = 414 m/s**2.\n", + " The angular accaleration of AB, alphaAB = 1820 rad/s**2.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.3 Page No : 182" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "vC = 1.\n", + "vCD = vC \t\t\t#m/s\n", + "aC = 2.5 \t\t\t#m/s**2\n", + "AB = 3. #m\n", + "BC = 1.5 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 8.8\n", + "#By measurement from the velocity diagram Fig. 8.8(b)\n", + "vBA = 0.72\n", + "vBC = 0.72 \t\t\t#m/s\n", + "#Calculating the radial component of acceleration of B with respect to C\n", + "arBC = vBC**2/BC \t\t\t#m/s**2\n", + "#Calculating the radial component of acceleration of B with respect to A\n", + "arBA = vBA**2/AB \t\t\t#m/s**2\n", + "#By measurement from the acceleration diagram Fig. 8.8(c)\n", + "aCD = 2.5\n", + "aC = aCD\n", + "arBC = 0.346\n", + "arBA = 0.173\n", + "atBA = 1.41\n", + "atBC = 1.94\n", + "vectorbb = 1.13\n", + "vectorab = 0.9 \t\t\t#m/s**2\n", + "#Calculating the angular accaleration of AB\n", + "alphaAB = atBA/AB \t\t\t#rad/s**2\n", + "#Calculating the angular acceleration of BC\n", + "alphaBC = atBC/BC \t\t\t#rad/s**2\n", + "\n", + "#Results:\n", + "print \" The magnitude of vertical component of the acceleration of the point B is %.2f m/s**2.\"%(vectorbb)\n", + "print \" The magnitude of horizontal component of the acceleration of the point B is %.1f m/s**2.\"%(vectorab)\n", + "print \" The angular acceleration of the link AB, alphaAB = %.2f rad/s**2.\"%(alphaAB)\n", + "print \" The angular acceleration of the link BC, alphaBC = %.1f rad/s**2.\"%(alphaBC)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The magnitude of vertical component of the acceleration of the point B is 1.13 m/s**2.\n", + " The magnitude of horizontal component of the acceleration of the point B is 0.9 m/s**2.\n", + " The angular acceleration of the link AB, alphaAB = 0.47 rad/s**2.\n", + " The angular acceleration of the link BC, alphaBC = 1.3 rad/s**2.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.4 Page No : 184" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "omegaQP = 10. \t\t\t#rad/s\n", + "PQ = 62.5/1000 #m\n", + "QR = 175./1000 #m\n", + "RS = 112.5/1000 #m\n", + "PS = 200./1000 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 8.9\n", + "#Calculating the velocity of Q with respect to P\n", + "vQP = omegaQP*PQ \t\t\t#m/s\n", + "vQ = vQP\n", + "#By measurement from the velocity diagram Fig. 8.9(b)\n", + "vRQ = 0.333\n", + "vRS = 0.426\n", + "vR = vRS \t\t\t#m/s\n", + "#Calculating the angular velocity of link QR\n", + "omegaQR = vRQ/QR \t\t\t#rad/s\n", + "#Calculating the angular velocity of link RS\n", + "omegaRS = vRS/RS \t\t\t#rad/s\n", + "#Calculating the radial component of the acceleration of Q with respect to P\n", + "arQP = vQP**2/PQ \t\t\t#m/s**2\n", + "aQP = arQP\n", + "aQ = aQP\n", + "#Calculating the radial component of the acceleration of R with respect to Q\n", + "arRQ = vRQ**2/QR \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of R with respect to S\n", + "arRS = vRS**2/RS \t\t\t#m/s**2\n", + "aRS = arRS\n", + "aR = aRS\n", + "#By measurement from the acceleration diagram Fig. 8.9(c)\n", + "atRQ = 4.1\n", + "atRS = 5.3 \t\t\t#m/s**2\n", + "#Calculating the angular acceleration of link QR\n", + "alphaQR = atRQ/QR \t\t\t#rad/s**2\n", + "#Calculating the angular acceleration of link RS\n", + "alphaRS = atRS/RS \t\t\t#rad/s**2\n", + "\n", + "#Results:\n", + "print \" The angular velocity of link QR, omegaQR = %.1f rad/s anticlockwise.\"%(omegaQR)\n", + "print \" The angular velocity of link RS, omegaRS = %.2f rad/s clockwise.\"%(omegaRS)\n", + "print \" The angular acceleration of link QR, alphaQR = %.2f rad/s**2 anticlockwise.\"%(alphaQR)\n", + "print \" The angular acceleration of link RS, alphaRS = %.1f rad/s**2 anticlockwise.\"%(alphaRS)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The angular velocity of link QR, omegaQR = 1.9 rad/s anticlockwise.\n", + " The angular velocity of link RS, omegaRS = 3.79 rad/s clockwise.\n", + " The angular acceleration of link QR, alphaQR = 23.43 rad/s**2 anticlockwise.\n", + " The angular acceleration of link RS, alphaRS = 47.1 rad/s**2 anticlockwise.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.5 Page No : 186" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "omegaAP1 = 10. \t\t\t#rad/s\n", + "alphaAP1 = 30. \t\t\t#rad/s**2\n", + "P1A = 300./1000 #m\n", + "P2B = 360./1000 #m\n", + "AB = P2B \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 8.10\n", + "#Calculating the velocity of A with respect to P1\n", + "vAP1 = omegaAP1/P1A \t\t\t#m/s\n", + "vA = vAP1\n", + "#By measurement from the velocity diagram Fig. 8.11(b)\n", + "vBP2 = 2.2\n", + "vBA = 2.05 \t\t\t#m/s\n", + "#Calculating the angular velocity of P2B\n", + "omegaP2B = vBP2/P2B \t\t\t#rad/s\n", + "#Calculating the angular velocity of AB\n", + "omegaAB = vBA/AB \t\t\t#rad/s\n", + "#Calculating the math.tangential component of the acceleration of A with respect to P1\n", + "atAP1 = alphaAP1*P1A \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of A with respect to P1\n", + "arAP1 = vAP1**2/P1A \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of B with respect to A\n", + "arBA = vBA**2/AB \t\t\t#m/s**2\n", + "#Calculating the radial component of B with respect to P2\n", + "arBP2 = vBP2**2/P2B \t\t\t#m/s**2\n", + "#By measurement from the acceleration diagram Fig. 8.11(c)\n", + "aBP2 = 29.6\n", + "aB = aBP2\n", + "atBA = 13.6\n", + "atBP2 = 26.6 \t\t\t#m/s**2\n", + "#Calculating the angular acceleration of P2B\n", + "alphaP2B = atBP2/P2B \t\t\t#rad/s**2\n", + "#Calculating the angular acceleration of AB\n", + "alphaAB = atBA/AB \t\t\t#rad/s**2\n", + "\n", + "#Results:\n", + "print \" The velocity of P2B, vBP2 = %.1f m/s.\"%(vBP2)\n", + "print \" The angular velocity of P2B, omegaP2B = %.1f rad/s, clockwise.\"%(omegaP2B)\n", + "print \" The angular velocity of AB, omegaAB = %.1f rad/s, anticlockwise.\"%(omegaAB)\n", + "print \" The acceleration of the joint B, aB = %.1f m/s**2.\"%(aB)\n", + "print \" The angular acceleration of P2B, alphaP2B = %.1f rad/s**2, anticlockwise.\"%(alphaP2B)\n", + "print \" The angular acceleration of AB, alphaAB = %.1f rad/s**2, anticlockwise.\"%(alphaAB)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The velocity of P2B, vBP2 = 2.2 m/s.\n", + " The angular velocity of P2B, omegaP2B = 6.1 rad/s, clockwise.\n", + " The angular velocity of AB, omegaAB = 5.7 rad/s, anticlockwise.\n", + " The acceleration of the joint B, aB = 29.6 m/s**2.\n", + " The angular acceleration of P2B, alphaP2B = 73.9 rad/s**2, anticlockwise.\n", + " The angular acceleration of AB, alphaAB = 37.8 rad/s**2, anticlockwise.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.6 Page No : 188" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "NAO = 20. \t\t\t#rpm\n", + "OA = 300./1000 #m\n", + "AB = 1200./1000 #m\n", + "BC = 450./1000 #m\n", + "CD = BC \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 8.13\n", + "#Calculating the angular velocity of crank AO\n", + "omegaAO = 2*math.pi*NAO/60 \t\t\t#rad/s\n", + "#Calculating the linear velocity of A with respect to O\n", + "vAO = omegaAO*OA \t\t\t#m/s\n", + "vA = vAO\n", + "#By measurement from the velocity diagram Fig. 8.13(b)\n", + "vB = 0.4\n", + "vD = 0.24\n", + "vDC = 0.37\n", + "vBA = 0.54 \t\t\t#m/s\n", + "#Calculating the angular velocity of CD\n", + "omegaCD = vDC/CD \t\t\t#rad/s\n", + "#Calculating the radial component of the acceleration of A with respect to O\n", + "arAO = vAO**2/OA \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of B with respect to A\n", + "arBA = vBA**2/AB \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of D with respect to C\n", + "arDC = vDC**2/CD \t\t\t#m/s**2\n", + "#By measurement from the acceleration diagram Fig. 8.13(c)\n", + "aD = 0.16\n", + "atDC = 1.28 \t\t\t#m/s**2\n", + "#Calculating the angular acceleration of CD\n", + "alphaCD = atDC/CD \t\t\t#rad/s**2\n", + "\n", + "#Results:\n", + "print \" Velocity of sliding at B, vB = %.1f m/s.\"%(vB)\n", + "print \" Velocity of sliding at D, vD = %.2f m/s.\"%(vD)\n", + "print \" Angular velocity of CD, omegaCD = %.2f rad/s.\"%(omegaCD)\n", + "print \" Linear acceleration of D, aD = %.2f m/s**2.\"%(aD)\n", + "print \" Angular acceleration of CD, alphaCD = %.2f rad/s**2, clockwise.\"%(alphaCD)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Velocity of sliding at B, vB = 0.4 m/s.\n", + " Velocity of sliding at D, vD = 0.24 m/s.\n", + " Angular velocity of CD, omegaCD = 0.82 rad/s.\n", + " Linear acceleration of D, aD = 0.16 m/s**2.\n", + " Angular acceleration of CD, alphaCD = 2.84 rad/s**2, clockwise.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.7 Page No : 191" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "NAO = 180. \t\t\t#rpm\n", + "OA = 150./1000 #m\n", + "AB = 450./1000 #m\n", + "PB = 240./1000 #m\n", + "CD = 660./1000 \t\t#m\n", + "\n", + "#solution:\n", + "#Refer Fig. 8.15\n", + "#Calculating the angular speed of crank AO\n", + "omegaAO = 2*math.pi*NAO/60 \t\t\t#rad/s\n", + "#Calculating the velocity of A with respect to O\n", + "vAO = omegaAO*OA \t\t\t#m/s\n", + "vA = vAO\n", + "#By measurement from the velocity diagram Fig. 8.15(b)\n", + "vD = 2.36\n", + "vDC = 1.2\n", + "vBA = 1.8\n", + "vBP = 1.5 \t\t\t#m/s\n", + "#Calculating the radial component of the acceleration of B with respect to A\n", + "arAO = vBA**2/AB \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of B with respect to A\n", + "arBA = vBA**2/AB \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of B with respect to P\n", + "arBP = vBP**2/PB \t\t\t#m/s**2\n", + "#Calculating the radial component of D with respect to C\n", + "arDC = vDC**2/CD \t\t\t#m/s**2\n", + "#By measurement from the acceleration diagram Fig. 8.15(c)\n", + "aD = 69.6\n", + "atDC = 17.4 \t\t\t#m/s**2\n", + "#Calculating the angular acceleration of CD\n", + "alphaCD = atDC/CD \t\t\t#rad/s**2\n", + "\n", + "#Results:\n", + "print \" Acceleration of slider D, aD = %.1f m/s**2.\"%(aD)\n", + "print \" Angular acceleration of link CD, alphaCD = %.1f rad/s**2.\"%(alphaCD)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Acceleration of slider D, aD = 69.6 m/s**2.\n", + " Angular acceleration of link CD, alphaCD = 26.4 rad/s**2.\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.8 Page No : 193" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables:\n", + "NAO = 180. \t\t\t#rpm\n", + "OA = 180./1000\n", + "CB = 240./1000\n", + "AB = 360./1000\n", + "BD = 540./1000 \t\t\t#m\n", + "alphaAO = 50. \t\t\t#rad/s**2\n", + "\n", + "#Solution:\n", + "#Refer Fig. 8.17\n", + "#Calculating the angular speed of crank AO\n", + "omegaAO = 2*math.pi*NAO/60 \t\t\t#rad/s\n", + "#Calculating the velcoity of A with respect to O\n", + "vAO = omegaAO*OA \t\t\t#m/s\n", + "vA = vAO\n", + "#By measurement from the velocity diagram Fig. 8.17(b)\n", + "vBA = 0.9\n", + "vBC = 2.4\n", + "vDB = 2.4\n", + "vD = 2.05 \t\t\t#m/s\n", + "\n", + "#Calculating the angular velocity of BD\n", + "omegaBD = vDB/BD \t\t\t#rad/s\n", + "#Calculating the tangential component of the acceleration of A with respect to O\n", + "atAO = alphaAO*OA \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of A with respect to O\n", + "arAO = vAO**2/OA \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of B with respect to A\n", + "arBA = vBA**2/AB \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of B with respect to C\n", + "arBC = vBC**2/AB \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of D with respect to B\n", + "arDB = vDB**2/BD \t\t\t#m/s**2\n", + "#By measurement from the acceleration diagram Fig. 8.17(c)\n", + "aD = 13.3\n", + "atDB = 38.5 \t\t\t#m/s**2\n", + "\n", + "#Calculating the angular acceleration of BD\n", + "alphaBD = atDB/BD \t\t\t#rad/s**2\n", + "\n", + "#Results:\n", + "print \" Velocity of slider D, vD = %.2f m/s.\"%(vD)\n", + "print \" Angular velocity of BD, omegaBD = %.1f rad/s.\"%(omegaBD)\n", + "print \" Acceleration of slider D, aD = %.1f m/s**2.\"%(aD)\n", + "print \" Angular acceleration of BD, alphaBD = %.1f rad/s**2, clockwise.\"%(alphaBD)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Velocity of slider D, vD = 2.05 m/s.\n", + " Angular velocity of BD, omegaBD = 4.4 rad/s.\n", + " Acceleration of slider D, aD = 13.3 m/s**2.\n", + " Angular acceleration of BD, alphaBD = 71.3 rad/s**2, clockwise.\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.9 Page No : 196" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "omegaAO1 = 100. \t\t\t#rad/s\n", + "O1A = 100./1000 #m\n", + "AC = 700./1000 #m\n", + "BC = 200./1000 #m\n", + "BD = 150./1000 #m\n", + "O2D = 200./1000 #m\n", + "O2E = 400./1000 #m\n", + "O3C = 200./1000 \t\t\t#m\n", + "m=0.0;\n", + "#Solution:\n", + "#Refer Fig. 8.19\n", + "#Calculating the linear velocity of A with respect to O1\n", + "vAO1 = omegaAO1/O1A \t\t\t#m/s\n", + "vA = vAO1\n", + "#By measurement from the velocity diagram Fig. 8.19(b)\n", + "vCA = 7.\n", + "vCO3 = 10.\n", + "vC = vCO3\n", + "vDB = 10.2\n", + "vDO2 = 2.8\n", + "vD = vDO2\n", + "vE = 5.8\n", + "vEO2 = vE \t\t\t#m/s\n", + "#Calculating the radial component of the acceleration of A with respect to O1\n", + "arAO1 = vAO1**2/O1A \t\t\t#m/s**2\n", + "aAO1 = arAO1\n", + "aA = aAO1\n", + "#Calculating the radial component of the acceleration of C with respect to A\n", + "arCA = vCA**2/AC \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of C with respect to O3\n", + "arCO3 = vCO3**2/O3C \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of D with respect to B\n", + "arDB = vDB**2/BD \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of D with respect to O2\n", + "arDO2 = vDO2**2/O2D \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of E with respect to O2\n", + "arEO2 = vEO2**2/O2E \t\t\t#m/s**2\n", + "#By measurement from the acceleration diagram Fig. 8.19(c)\n", + "aE = 1200.\n", + "atDO2 = 610. \t\t\t#m/s**2\n", + "aEO2 = aE\n", + "aB = 440. \t\t\t#Acceleration of point B\n", + "#m/s**2\n", + "#Calculating the angular acceleration of the bell crank lever\n", + "alpha = atDO2/O2D \t\t\t#The angular acceleration of the bell crank lever rad/s**2\n", + "\n", + "#Results:\n", + "print \" Velocity of the point E on the bell crank lever, vE = %.1f m/s.\"%(vE)\n", + "print \" Acceleration of point B = %d m/s**2.\"%(aB)\n", + "print \" Acceleration of point E, aE = %d m/s**2.\"%(aE)\n", + "print \" Angular acceleration of the bell crank lever = %d rad/s**2, anticlockwise.\"%(alpha)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Velocity of the point E on the bell crank lever, vE = 5.8 m/s.\n", + " Acceleration of point B = 440 m/s**2.\n", + " Acceleration of point E, aE = 1200 m/s**2.\n", + " Angular acceleration of the bell crank lever = 3050 rad/s**2, anticlockwise.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.10 Page No : 199" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "NAO = 100. \t\t\t#rpm\n", + "OA = 150./1000 #m\n", + "AB = 600./1000 #m\n", + "BC = 350./1000 #m\n", + "CD = 150./1000 #m\n", + "DE = 500./1000 \t\t#m\n", + "dA = 50./1000\n", + "dB = dA\n", + "rA = dA/2\n", + "rB = dB/2 \t\t\t#m\n", + "pF = 0.35 \t\t\t#N/mm**2\n", + "DF = 250. \t\t\t#mm\n", + "\n", + "#Solution:\n", + "#Refer Fig. 8.21\n", + "#Calculating the angular speed of the crank AO\n", + "omegaAO = 2*math.pi*NAO/60 \t\t\t#rad/s\n", + "#Calculating the velocity of A with respect to O\n", + "vAO = omegaAO*OA \t\t\t#m/s\n", + "vA = vAO\n", + "#By measurement from the velocity diagram Fig. 8.21(b)\n", + "vBA = 1.65\n", + "vBC = 0.93\n", + "vB = vBC\n", + "vED = 0.18\n", + "vEO = 0.36\n", + "vE = vEO\n", + "vF = vE \t\t\t#m/s\n", + "\n", + "#Calculating the velocity of D with respect to C\n", + "vDC = vBC*CD/BC \t\t\t#m/s\n", + "#Calculating the angular velocity of B with respect to A\n", + "omegaBA = vBA/AB \t\t\t#rad/s\n", + "#Calculating the angular velocity of B with respect to C\n", + "omegaBC = vBC/BC \t\t\t#rad/s\n", + "#Calculating the rubbing velocity of pin at A\n", + "vrA = (omegaAO-omegaBA)*rA \t\t\t#The rubbing velocity of pin at A m/s\n", + "#Calculating the rubbing velocity of pin at B\n", + "vrB = (omegaBA+omegaBC)*rB \t\t\t#The rubbing velocity of pin at B m/s\n", + "#Calculating the force at the pump piston at F\n", + "FF = pF*math.pi/4*DF**2 \t\t\t#N\n", + "#Calculating the force required at the crankshaft A\n", + "FA = FF*vF/vA \t\t\t#N\n", + "#Calculating the torque required at the crankshaft\n", + "TA = FA*OA \t\t\t#N-m\n", + "#Calculating the radial component of the acceleration of A with respect to O\n", + "arAO = vAO**2/OA \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of B with respect to A\n", + "arBA = vBA**2/AB \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of B with respect to C\n", + "arBC = vBC**2/BC \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of E with respect to D\n", + "arED = vED**2/DE \t\t\t#m/s**2\n", + "#By measurement from the acceleration diagram Fig. 8.21(c)\n", + "aBC = 9.2\n", + "aB = aBC\n", + "aBA = 9\n", + "aE = 3.8 \t\t\t#m/s**2\n", + "#Calculating the acceleration of D\n", + "aD = aBC*CD/BC \t\t\t#m/s**2\n", + "\n", + "#Results:\n", + "print \" The velocity of the cross-head E, vE = %.2f m/s.\"%(vE)\n", + "print \" The rubbing velocity of pin at A = %.3f m/s.\"%(vrA)\n", + "print \" The rubbing velocity of pin at B = %.3f m/s.\"%(vrB)\n", + "print \" The torque required at the crankshaft, TA = %d N-m.\"%(TA)\n", + "print \" The acceleration of the crosshead E, aE = %.1f m/s**2.\"%(aE)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The velocity of the cross-head E, vE = 0.36 m/s.\n", + " The rubbing velocity of pin at A = 0.193 m/s.\n", + " The rubbing velocity of pin at B = 0.135 m/s.\n", + " The torque required at the crankshaft, TA = 590 N-m.\n", + " The acceleration of the crosshead E, aE = 3.8 m/s**2.\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.11 Page No : 203" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "NAO = 150. \t\t\t#rpm\n", + "OA = 150./1000 #m\n", + "AB = 550./1000 #m\n", + "AC = 450./1000 #m\n", + "DC = 500./1000 #m\n", + "BE = 350./1000 \t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 8.23\n", + "#Calculating the angular speed of the crank AO\n", + "omegaAO = 2*math.pi*NAO/60 \t\t\t#rad/s\n", + "#Calculating the linear velocity of A with respect to O\n", + "vAO = omegaAO*OA \t\t\t#m/s\n", + "vA = vAO\n", + "#By measurement from the velocity diagram Fig. 8.23(b)\n", + "vCA = 0.53\n", + "vCD = 1.7\n", + "vC = vCD\n", + "vEB = 1.93\n", + "vE = 1.05 \t\t\t#m/s\n", + "#Calculating the radial component of the acceleration of A with respect to O\n", + "arAO = vAO**2/OA \t\t\t#m/s**2\n", + "aA = arAO\n", + "#Calculating the radial component of the acceleration of C with respect to A\n", + "arCA = vCA**2/AC \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of C with respect to D\n", + "arCD = vCD**2/DC \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of E with respect to B\n", + "arEB = vEB**2/BE \t\t\t#m/s**2\n", + "#By measurement from the acceleration diagram Fig. 8.23(c)\n", + "aE = 3.1 \t\t\t#m/s**2\n", + "\n", + "#Results:\n", + "print \" Velocity of the ram E, vE = %.2f m/s.\"%(vE)\n", + "print \" Acceleration of the ram E, aE = %.1f m/s**2.\"%(aE)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Velocity of the ram E, vE = 1.05 m/s.\n", + " Acceleration of the ram E, aE = 3.1 m/s**2.\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.12 Page No : 205" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "NDC = 1140. \t\t\t#rpm\n", + "AB = 80./1000 #m\n", + "CD = 40./1000 #m\n", + "BE = 150./1000 #m\n", + "DE = BE #m\n", + "EP = 200./1000 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 8.25\n", + "#Calculating the angular speed of the link CD\n", + "omegaDC = 2*math.pi*NDC/60 \t\t\t#rad/s\n", + "#Calculating the velocity of D with respect to C\n", + "vDC = omegaDC*CD \t\t\t#m/s\n", + "vD = vDC\n", + "#Calculating the angular speed of the larger wheel\n", + "omegaBA = omegaDC*CD/AB \t\t\t#rad/s\n", + "#Calculating the velocity of B with respect to A\n", + "vBA = omegaBA*AB \t\t\t#m/s\n", + "vB = vBA\n", + "#By measurement from the velocity diagram Fig. 8.25(b)\n", + "vEB = 8.1\n", + "vED = 0.15\n", + "vPE = 4.7\n", + "vP = 0.35 \t\t\t#m/s\n", + "#Calculating the radial component of the acceleration of B with respect to A\n", + "arBA = vBA**2/AB \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of D with respect to C\n", + "arDC = vDC**2/CD \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of E with respect to B\n", + "arEB = vEB**2/BE \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of E with respect to D\n", + "arED = vED**2/DE \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of P with respect to E\n", + "arPE = vPE**2/EP \t\t\t#m/s**2\n", + "#By measurement from the acceleration diagram Fig. 8.25(c)\n", + "aP = 655. \t\t\t#m/s**2\n", + "\n", + "#Results:\n", + "print \" Velocity of P, vP = %.2f m/s.\"%(vP)\n", + "print \" Acceleration of the piston P, aP = %d m/s**2.\"%(aP)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Velocity of P, vP = 0.35 m/s.\n", + " Acceleration of the piston P, aP = 655 m/s**2.\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.13 Page No : 211" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "NBA = 120. \t\t\t#rpm\n", + "AB = 150./1000 #m\n", + "OC = 700./1000 #m\n", + "CD = 200./1000 \t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 8.29\n", + "#Calculating the angular speed of the crank AB\n", + "omegaAB = 2*math.pi*NBA/AB \t\t\t#rad/s\n", + "#Calculating the velocity of B with respect to A\n", + "vBA = omegaBA*AB \t\t\t#m/s\n", + "#By measurement from the velocity diagram Fig. 8.29(b)\n", + "vD = 2.15\n", + "vBBdash = 1.05\n", + "vDC = 0.45\n", + "vBdashO = 1.55\n", + "vCO = 2.15 \t\t\t#m/s\n", + "BdashO = 0.52 \t\t\t#m\n", + "#Calculating the angular velocity of the link OC or OB'\n", + "omegaCO = vCO/OC \t\t\t#rad/s\n", + "omegaBdashO = omegaCO \t\t\t#rad/s\n", + "#Calculating the radial component of the acceleration of B with respect to A\n", + "arBA = omegaAB**2/AB \t\t\t#m/s**2\n", + "#Calculating the coriolis component of the acceleration of slider B with respect to the coincident point B'\n", + "acBBdash = 2*omegaCO*vBBdash \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of D with respect to C\n", + "arDC = vDC**2/CD \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of B' with respect to O\n", + "arBdashO = vBdashO**2/BdashO \t\t\t#m/s**2\n", + "#By measurement fro the acceleration diagram Fig. 8.29(c)\n", + "aD = 8.4\n", + "atBdashO = 6.4 \t\t\t#m/s**2\n", + "#Calculating the angular acceleration of the slotted lever\n", + "alpha = atBdashO/BdashO \t\t\t#The angular acceleration of the slotted lever rad/s**2\n", + "\n", + "#Results:\n", + "print \" Velocity of the ram D, vD = %.2f m/s.\"%(vD)\n", + "print \" Acceleration of the ram D, aD = %.1f m/s**2.\"%(aD)\n", + "print \" Angular acceleration of the slotted lever = %.1f rad/s**2, anticlockwise.\"%(alpha)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Velocity of the ram D, vD = 2.15 m/s.\n", + " Acceleration of the ram D, aD = 8.4 m/s**2.\n", + " Angular acceleration of the slotted lever = 12.3 rad/s**2, anticlockwise.\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.14 Page No : 214" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "NBA = 200. \t\t\t#rpm\n", + "AB = 75./1000 #m\n", + "PQ = 375./1000 #m\n", + "QR = 500./1000 \t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 8.31\n", + "#Calculating the angular velocity of the crank AB\n", + "omegaBA = 2*math.pi*NBA/60 \t\t\t#rad/s\n", + "#Calculating the velocity of B with respect to A\n", + "vBA = omegaBA*AB \t\t\t#m/s\n", + "#By measurement from the velocity diagram Fig. 8.31(b)\n", + "vR = 1.6\n", + "vBdashB = 1.06\n", + "vBdashP = 1.13\n", + "vRQ = 0.4\n", + "vQP = 1.7 \t\t\t#m/s\n", + "PBdash = 248./1000 \t\t\t#m\n", + "#Calculating the angular velocity of the link PQ\n", + "omegaPQ = vQP/PQ \t\t\t#rad/s\n", + "#Calculating the radial component of the acceleration of B with respect to A\n", + "arBA = omegaBA**2*AB \t\t\t#m/s**2\n", + "#Calculating the coriolis component of the acceleration of B with respect to coincident point B'\n", + "acBBdash = 2*omegaPQ*vBdashB \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of R with respect to Q\n", + "arRQ = vRQ**2/QR \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of B' with respect to P\n", + "arBdashP = vBdashP**2/PBdash \t\t\t#m/s**2\n", + "#By measurement from the acceleration diagram Fig. 8.31(d)\n", + "aR = 22.\n", + "aBBdash = 18. \t\t\t#m/s**2\n", + "\n", + "#Results:\n", + "print \" Velocity of the tool-box R, vR = %.1f m/s.\"%(vR)\n", + "print \" Acceleration of the tool-box R, aR = %d m/s**2.\"%(aR)\n", + "print \" The acceleration of sliding of the block B along the slotted lever PQ, aBBdash = %d m/s**2.\"%(aBBdash)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Velocity of the tool-box R, vR = 1.6 m/s.\n", + " Acceleration of the tool-box R, aR = 22 m/s**2.\n", + " The acceleration of sliding of the block B along the slotted lever PQ, aBBdash = 18 m/s**2.\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.15 Page No : 218" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "NAO = 30. \t\t\t#rpm\n", + "OA = 150./1000 #m\n", + "OC = 100./1000 #m\n", + "CD = 125./1000 #m\n", + "DR = 500./1000 \t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 8.33\n", + "#Calculating the angular speed of the crank OA\n", + "omegaAO = 2*math.pi*NAO/60 \t\t\t#rad/s\n", + "#Calculating the velocity of A with respect to O\n", + "vAO = omegaAO*OA \t\t\t#m/s\n", + "vA = vAO\n", + "#By measurement from the velocity diagram Fig. 8.33(b)\n", + "vBC = 0.46\n", + "vAB = 0.15\n", + "vRD = 0.12 \t\t\t#m/s\n", + "CB = 240./1000 \t\t\t#m\n", + "#Calculating the angular velocity of the link BC\n", + "omegaBC = vBC/CB \t\t\t#rad/s\n", + "#Calculating the radial component of the acceleration of A with respect to O\n", + "arAO = vAO**2/OA \t\t\t#m/s**2\n", + "#Calculating the coriolis component of the acceleration of A with respect to coincident point B\n", + "acAB = 2*omegaBC*vAB \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of B with respect to C\n", + "arBC = vBC**2/CB \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of R with respect to D\n", + "arRD = vRD**2/DR \t\t\t#m/s**2\n", + "#By measurement from the acceleration diagram Fig. 8.33(c)\n", + "aR = 0.18\n", + "atBC = 0.14 \t\t\t#m/s**2\n", + "#Calculating the angular acceleration of the slotted lever CA\n", + "alphaCA = atBC/CB \t\t\t#rad/s**2\n", + "alphaBC = alphaCA\n", + "\n", + "#Results:\n", + "print \" Acceleration of the sliding block R, aR = %.2f m/s**2.\"%(aR)\n", + "print \" Angular acceleration of the slotted lever CA, alphaCA = %.3f rad/s**2, anticlockwise.\"%(alphaCA)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Acceleration of the sliding block R, aR = 0.18 m/s**2.\n", + " Angular acceleration of the slotted lever CA, alphaCA = 0.583 rad/s**2, anticlockwise.\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.16 Page No : 221" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "AB = 125./1000 \t\t\t#m\n", + "NCO = 300. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Refer Fig. 8.35\n", + "#By measurement from the space diagram Fig. 8.35(a)\n", + "OC = 85./1000 \t\t\t#m\n", + "#Calculating the angular velocity of the link CO\n", + "omegaCO = 2*math.pi*NCO/60 \t\t\t#rad/s\n", + "#Calculating the velocity of C with respect to O\n", + "vCO = omegaCO*OC \t\t\t#m/s\n", + "vC = vCO\n", + "#By measurement from the velocity diagram Fig. 8.35(b)\n", + "vBC = 0.85\n", + "vBA = 2.85\n", + "vB = vBA \t\t\t#m/s\n", + "#Calculating the radial component of of the acceleration of C with respect to O\n", + "arCO = vCO**2/OC \t\t\t#m/s**2\n", + "#Calculating the coriolis component of of acceleration of the piston B with respect to the cylinder or the coincident point C\n", + "acBC = 2*omegaCO*vBC \t\t\t#m/s**2\n", + "#Calculating the radial component of of the acceleration of B with respect to A\n", + "arBA = vBA**2/AB \t\t\t#m/s**2\n", + "#By measurement from the acceleration diagram Fig. 8.35(d)\n", + "aBC = 73.2\n", + "atBA = 37.6 \t\t\t#m/s**2\n", + "#Calculating the angular acceleration of the connecting rod AB\n", + "alphaAB = atBA/AB \t\t\t#rad/s**2\n", + "\n", + "#Results:\n", + "print \" Acceleration of the piston inside the cylinder, aBC = %.1f m/s**2.\"%(aBC)\n", + "print \" Angular acceleration of the connecting rod AB, alphaAB = %d rad/s**2, clockwise.\"%(alphaAB)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Acceleration of the piston inside the cylinder, aBC = 73.2 m/s**2.\n", + " Angular acceleration of the connecting rod AB, alphaAB = 300 rad/s**2, clockwise.\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.17 Page No : 223" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "NAO = 100. \t\t\t#rpm\n", + "OA = 50./1000 #m\n", + "AB = 350./1000 #m\n", + "DE = 250./1000 #m\n", + "EF = DE #m\n", + "CB = 125./1000 \t\t#m\n", + "\n", + "#Solution:\n", + "#Refer Fig. 8.37\n", + "#Calculating the angular velocity of the crank AO\n", + "omegaAO = 2*math.pi*NAO/60 \t\t\t#rad/s\n", + "#Calculating the velocity of A with respect to O\n", + "vAO = omegaAO*OA \t\t\t#m/s\n", + "vA = vAO\n", + "#By measurement from the velocity diagram Fig. 8.37(b)\n", + "vBA = 0.4\n", + "vBC = 0.485\n", + "vB = vBC\n", + "vSD = 0.265\n", + "vQS = 0.4\n", + "vED = 0.73\n", + "vFE = 0.6\n", + "vF = 0.27 \t\t\t#m/s\n", + "DS = 85./1000 \t\t\t#m\n", + "#Calculating the angular velocity of the link DE\n", + "omegaDE = vED/DE \t\t\t#rad/s\n", + "#Calculating the velocity of sliding of the link DE in the swivel block\n", + "vS = vQS \t\t\t#m/s\n", + "#Calculating the radial component of the acceleration of A with respect to O\n", + "arAO = vAO**2/OA \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of B with respect to A\n", + "arBA = vBA**2/AB \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of B with respect to C\n", + "arBC = vBC**2/CB \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of S with respect to D\n", + "arSD = vSD**2/DS \t\t\t#m/s**2\n", + "#Calculating the coriolis component of the acceleration of Q with respect to S\n", + "acQS = 2*omegaDE*vQS \t\t\t#m/s**2\n", + "#Calculating the radial component of the acceleration of F with respect to E\n", + "arFE = vFE**2/EF \t\t\t#m/s**2\n", + "#By measurement from the acceleration diagram Fig. 8.37(d)\n", + "arQS = 1.55 \t\t\t#m/s**2\n", + "\n", + "#Results:\n", + "print \" Velocity of the slider block F, vF = %.2f m/s.\"%(vF)\n", + "print \" Angular velocity of the link DE, omegaDE = %.2f rad/s, anticlockwise.\"%(omegaDE)\n", + "print \" Velocity of sliding of the link DE in the swivel block, vS = %.1f m/s.\"%(vS)\n", + "print \" Acceleration of sliding of the link DE in the trunnion, arQS = %.2f m/s**2.\"%(arQS)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Velocity of the slider block F, vF = 0.27 m/s.\n", + " Angular velocity of the link DE, omegaDE = 2.92 rad/s, anticlockwise.\n", + " Velocity of sliding of the link DE in the swivel block, vS = 0.4 m/s.\n", + " Acceleration of sliding of the link DE in the trunnion, arQS = 1.55 m/s**2.\n" + ] + } + ], + "prompt_number": 18 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Theory_Of_Machines/ch9.ipynb b/Theory_Of_Machines/ch9.ipynb new file mode 100755 index 00000000..bdb83c80 --- /dev/null +++ b/Theory_Of_Machines/ch9.ipynb @@ -0,0 +1,349 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:0872fbc3ae91821d4330e3facbf2559c866bc6b64416332df388f5fee66480e0" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 9 : Mechanisms with Lower Pairs" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.1 Page No : 245" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "c = 1.2\n", + "b = 2.7 \t\t\t#m\n", + "\n", + "#Solution:\n", + "#Calculating the inclination of the track arm to the longitudinal axis\n", + "alpha = math.tan(c/(2*b))*180/math.pi \t\t\t#degrees\n", + "\n", + "#Results:\n", + "print \" Inclination of the track arm to the longitudinal axis, alpha = %.1f degrees.\"%(alpha)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Inclination of the track arm to the longitudinal axis, alpha = 12.9 degrees.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.2 Page No : 251\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# variables\n", + "a = 180. - 160. # degrees\n", + "N = 1500. # r.p.m.; \n", + "m = 12. # kg ; \n", + "k = 0.1 # m\n", + "\n", + "# calculations\n", + "w = round(2*math.pi*N/60)\n", + "I = m*k**2\n", + "cos2theta = 2*math.sin(math.radians(a))**2/(2 - math.sin(math.radians(a))**2)\n", + "theta = math.degrees(math.acos(cos2theta))/2\n", + "dw1bydt = w**2*math.cos(math.radians(a)) * math.sin(math.radians(2*theta)) * math.sin(math.radians(a))**2 / ( 1 - math.cos(math.radians(theta))**2 * math.sin(math.radians(a))**2)**2\n", + "max_t = I * dw1bydt\n", + "\n", + "# results\n", + "print \"Maximum angular acceleration of the driven shaft : %.f rad/s**2\"%dw1bydt\n", + "print \"maximum torque required : %.f N-m\"%max_t\n", + "\n", + "\n", + "# answers are different because of rounding error. please check using calculator." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum angular acceleration of the driven shaft : 3080 rad/s**2\n", + "maximum torque required : 370 N-m\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.3 Page No : 252" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "alpha = 18*math.pi/180 \t\t\t#radians\n", + "\n", + "#Solution:\n", + "#Maximum velocity is possible when\n", + "theta1 = 0.\n", + "theta2 = 180. \t\t\t#degrees\n", + "\n", + "#Calculating the angle turned by the driving shaft when the velocity ratio is unity\n", + "theta3 = math.cos(math.sqrt((1-math.cos(alpha))/(math.sin(alpha)**2)))*180/math.pi \t\t\t#degrees\n", + "theta4 = 180-theta3 \t\t\t#degrees\n", + "\n", + "#Results:\n", + "print \" Angle turned by the driving shaft when the velocity ratio is maximum, theta = %d degrees\\\n", + " or %d degrees.\"%(theta1,theta2)\n", + "print \" Angle turned by the driving shaft when the velocity ratio is unity, theta = %.1f degrees or\\\n", + " %.1f degrees.\"%(theta3,theta4)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Angle turned by the driving shaft when the velocity ratio is maximum, theta = 0 degrees or 180 degrees.\n", + " Angle turned by the driving shaft when the velocity ratio is unity, theta = 43.2 degrees or 136.8 degrees.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.4 Page No : 252" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "N = 500. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Calculating the angular velocity of the driving shaft\n", + "omega = 2*math.pi*N/60.0 \t\t\t#rad/s\n", + "#Calculating the total fluctuation of speed of the driven shaft\n", + "q = 12./100*omega \t\t\t#rad/s\n", + "#Calculating the greatest permissible angle between the centre lines of the shafts\n", + "#alpha = math.cos((-(q/omega)+math.sqrt(0.12**2+4))/2.0)*180/math.pi\t\t\t#degrees\n", + "cosalpha =((-(q/omega)+math.sqrt(0.12**2+4))/2.0)\t\t\t#degrees\n", + "alpha = math.degrees(math.acos(cosalpha))\n", + "\n", + "#Results:\n", + "print \" Greatest permissible angle between the centre lines of the shafts, alpha = %.2f degrees.\"%(alpha)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Greatest permissible angle between the centre lines of the shafts, alpha = 19.64 degrees.\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.5 Page No : 252" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "N = 1200.\n", + "q = 100. \t\t\t#rpm\n", + "#Solution:\n", + "#Calculating the greatest permissible angle between the centre lines of the shafts\n", + "cosalpha = ((-(100./1200)+math.sqrt(0.083**2+4))/2)\n", + "alpha = math.degrees(math.acos(cosalpha)) #degrees\n", + "#Calculating the maximum speed of the driven shaft\n", + "N1max = N/cosalpha \t\t\t#rpm\n", + "#Calculating the minimum speed of the driven shaft\n", + "N1min = N*cosalpha\t\t\t#rpm\n", + "\n", + "#Results:\n", + "print \" Greatest permissible angle between the centre lines of the shafts, alpha = %.1f degrees.\"%(alpha)\n", + "print \" Maximum speed of the driven shaft, N1max = %d rpm.\"%(N1max)\n", + "print \" Minimum speed of the driven shaft, N1min = %d rpm.\"%(N1min)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Greatest permissible angle between the centre lines of the shafts, alpha = 16.4 degrees.\n", + " Maximum speed of the driven shaft, N1max = 1251 rpm.\n", + " Minimum speed of the driven shaft, N1min = 1151 rpm.\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.6 page no : 253" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "# variables\n", + "N = 240. # r.p.m \n", + "w = 2 * math.pi * 240./60 #rad/s \n", + "alpha = 20 \n", + "m = 55. # kg ;\n", + "k = .150 \n", + "mm = 0.15 #m ; \n", + "T1 = 200. #N-m ; \n", + "theta = 45. # \u00b0 ; \n", + "q = 24. # r.p.m.\n", + "\n", + "# calculations\n", + "I = round(m * k**2,2) \n", + "dw1bydt = round(-(w**2)*math.cos(math.radians(alpha))*math.sin(math.radians(2*theta))*math.sin(math.radians(alpha))**2 / (1- math.cos(math.radians(theta))**2 * math.sin(math.radians(alpha))**2)**2,2)\n", + "T2 = I * dw1bydt\n", + "T = T1 + T2\n", + "Tdash = T*math.cos(math.radians(alpha))/(1-math.cos(math.radians(theta))**2 * math.sin(math.radians(alpha))**2)\n", + "cosapha = (-0.1+math.sqrt((0.1**2)+4))/2\n", + "alpha = math.degrees(math.acos(cosapha))\n", + "\n", + "# result\n", + "print \"T' = %.1f N-m\"%Tdash\n", + "print \"Alpha a = %.1f degrees\"%alpha\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "T' = 102.7 N-m\n", + "Alpha a = 18.0 degrees\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.7 Page No : 254" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables:\n", + "alpha = 20. \t\t\t#degrees\n", + "NA = 500. \t\t\t#rpm\n", + "\n", + "#Solution:\n", + "#Calculating the maximum speed of the intermediate shaft\n", + "NBmax = NA/math.cos(math.radians(alpha)) \t\t\t#rpm\n", + "#Calculating the minimum speed of the intermediate shaft\n", + "NBmin = NA*math.cos(math.radians(alpha)) \t\t\t#rpm\n", + "#Calculating the maximum speed of the driven shaft\n", + "NCmax = NBmax/math.cos(math.radians(alpha)) \t\t\t#rpm\n", + "#Calculating the minimum speed of the driven shaft\n", + "NCmin = NBmin*math.cos(math.radians(alpha)) \t\t\t#rpm\n", + "\n", + "#Results:\n", + "print \" Maximum speed of the intermediate shaft( NBmax) = %.1f rad/s.\"%(NBmax)\n", + "print \" Minimum speed of the intermediate shaft( NBmin) = %.2f rad/s.\"%(NBmin)\n", + "print \" Maximum speed of the driven shaft( NCmax) = %.2f rad/s.\"%(NCmax)\n", + "print \" Minimum speed of the driven shaft( NCmin) = %.1f rad/s.\"%(NCmin)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Maximum speed of the intermediate shaft( NBmax) = 532.1 rad/s.\n", + " Minimum speed of the intermediate shaft( NBmin) = 469.85 rad/s.\n", + " Maximum speed of the driven shaft( NCmax) = 566.24 rad/s.\n", + " Minimum speed of the driven shaft( NCmin) = 441.5 rad/s.\n" + ] + } + ], + "prompt_number": 27 + } + ], + "metadata": {} + } + ] +}
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