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diff --git a/sample_notebooks/MohdAsif/MohdAsif_version_backup/ch2.ipynb b/sample_notebooks/MohdAsif/MohdAsif_version_backup/ch2.ipynb new file mode 100755 index 00000000..66270ccb --- /dev/null +++ b/sample_notebooks/MohdAsif/MohdAsif_version_backup/ch2.ipynb @@ -0,0 +1,387 @@ +{ + "metadata": { + "name": "", + "signature": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 2 Casting Processes" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 2.1 on page no. 46" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import sqrt\n", + "# Given that\n", + "h=15 # Height of spur in cm\n", + "l= 50 # Length of cast in cm\n", + "w= 25 # weidth of cast in cm\n", + "h1= 15 # Height of cast in cm\n", + "g= 981 # Acceleration due to gravity in cm/sec**2\n", + "Ag= 5 # Cross sectional area of the grate in cm**2\n", + "v3= sqrt(2* g * h)\n", + "V = l*w*h1\n", + "tf1= V/(Ag*v3)\n", + "Am = l*w\n", + "tf2 = (Am/Ag)*(1/sqrt(2*g))*2*(sqrt(h) - sqrt(h-h1))\n", + "print \" Filling time for first design = %f sec, \\n Filling time for second design = %f sec\"% (tf1, tf2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Filling time for first design = 21.859294 sec, \n", + " Filling time for second design = 43.718589 sec\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 2 on page no. 53" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi, sqrt\n", + "# Given that\n", + "h=15 # Height of spur in cm\n", + "l= 50 # Length of cast in cm\n", + "w= 25 # weidth of cast in cm\n", + "h1= 15 # Height of cast in cm\n", + "g= 981 # Acceleration due to gravity in cm/sec**2\n", + "Ag= 5 # Cross sectional area of the grate in cm**2\n", + "Dm = 7800 # Density of molten Fe in Kg/m**3\n", + "Neta = 0.00496 # Kinetic viscosity in Kg/m-sec\n", + "theta = 90 # Angle in degree\n", + "Eq = 25 # (L/D) Equivalent \n", + "v3= sqrt(2* g * h)*(10**(-2))\n", + "d= sqrt((Ag*4)/(pi))*(10**(-2))\n", + "Re = Dm*v3*d/Neta\n", + "f = 0.0791*(Re)**(-1/4)\n", + "L=0.12 # in meter\n", + "Cd= (1+0.45+4*f*((L/d)+Eq))**(-1/2)\n", + "v3_ = Cd*v3\n", + "Re_ = (v3_/v3)*(Re)\n", + "f_ = 0.0791 *(Re_)**(-1/4)\n", + "Cd_ = (1+0.46+4*f_*(L/d + Eq))**(-1/2)\n", + "v3__ = Cd_*v3\n", + "V = l*w*h1\n", + "tf= (V/(Ag*v3__))*(10**-2)\n", + "print \" Filling time for first design = %f sec. \"% tf " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Filling time for first design = 31.918954 sec. \n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem on page no. 56" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given that\n", + "Hi=1.2 # Initial height in m\n", + "H= 0.05 # Height in m\n", + "g= 9.81 # Acceleration due to gravity in m/sec**2\n", + "Dm = 2700 # Density of molten metal in Kg/m**3\n", + "Neta = 0.00273 # Kinetic viscosity in Kg/m-sec \n", + "d= 0.075 # Diameter in m\n", + "D = 1 # Internal diameter of ladle in m\n", + "v3= sqrt(2* g * Hi)\n", + "Re = Dm*v3*d/Neta\n", + "ef=0.075\n", + "Cd= (1+ef)**(-1/2)\n", + "ef_=0.82\n", + "Re_ = (2+ef_)**(-1/2)\n", + "v3_ = sqrt(2*g*H)\n", + "Re_ = Dm*v3_*d/Neta\n", + "At = (pi/4)*D**2\n", + "An = (pi/4)*d**2\n", + "Cd= 0.96\n", + "tf= (sqrt(2/g))*(At/An)*(1/Cd)*sqrt(Hi)\n", + "m = Dm*An*Cd*sqrt(2*g*Hi)\n", + "m_ = Dm*An*Cd*sqrt(2*g*Hi*0.25)\n", + "print \"\"\"Time required to empty the ladle = %f sec,\n", + "Discharge rate are -\n", + "Initially = %f Kg/sec \n", + "When the ladle is 75 percent empty = %f Kg/sec. \"\"\"%(tf,m,m_)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time required to empty the ladle = 91.596179 sec,\n", + "Discharge rate are -\n", + "Initially = 55.563236 Kg/sec \n", + "When the ladle is 75 percent empty = 27.781618 Kg/sec. \n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 5 on page no. 66" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import symbols, solve\n", + "# Given that\n", + "thetaF= 1540 # Temperature of mould face in degree centigrade\n", + "Theta0 = 28 # Initial temperature of mould in Degree centigrade\n", + "L= 272e3 # Latent heat of liquid metal in J/Kg\n", + "Dm = 7850 # Density of liquid metal in Kg/m**3\n", + "c = 1.17e+3 #Specific heat of sand in J/Kg-K\n", + "k = 0.8655 # Conductivity of sand in W/m-K\n", + "D= 1600 # Density of sand in Kg/m**3\n", + "h = 0.1 # Height in m\n", + "b = 10 # Thickness of slab in cm\n", + "r =h/2# V/A in meter\n", + "lamda = (thetaF - Theta0)*(D*c)/(Dm*L)\n", + "Beta1 = 2*lamda/sqrt(pi)\n", + "Alpha = k /(D*c)\n", + "ts1 = r**2 /((Beta1**2)*Alpha)#In sec\n", + "ts1_=ts1/3600 # In hour\n", + "Beta= symbols('Beta') \n", + "p=Beta**2 - lamda*(2/sqrt(pi))*Beta -lamda/3\n", + "Beta2 = solve(p)[0]\n", + "print \"The value of Beta2 is %f \"%Beta2\n", + "print \"We only take the positive value of Beta2 ,\\nHence Beta2=1.75\" \n", + "r1 = r/3\n", + "ts2 = (r1**2)/((1.75**2)*Alpha) # in sec\n", + "ts2_=ts2/3600#in Hour\n", + "print \"Solidification time for slab-shaped casting = %f hr,\\nSolidification time for sphere = %f hr\"%(ts1_,ts2_)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of Beta2 is -0.252713 \n", + "We only take the positive value of Beta2 ,\n", + "Hence Beta2=1.75\n", + "Solidification time for slab-shaped casting = 0.671318 hr,\n", + "Solidification time for sphere = 0.054495 hr\n" + ] + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 7 on page no. 75" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given that\n", + "thetaF= 1540 # Temperature of mould face in degree centigrade\n", + "thetaO = 28 # Initial temperature of mould in Degree centigrade\n", + "L= 272e3 # Latent heat of iron in J/Kg\n", + "Dm = 7850 # Density of iron in Kg/m**3\n", + "Cs = 0.67e+3 #Specific heat of iron in J/Kg-K\n", + "C = 0.376e3 #Specific heat of copper in J/Kg-K\n", + "Ks = 83 # Conductivity of iron in W/m-K\n", + "K = 398 # Conductivity of copper in W/m-K\n", + "D= 8960 # Density of copper in Kg/m**3\n", + "h = .1 # Height in m\n", + "hF = 1420 # Total heat transfer coefficient across the casting-mould interface in W/m**2-\u00b0C\n", + "AlphaS = K /(D*C)\n", + "thetaS = 982 #In \u00b0C as in example 2.6\n", + "h1= (1+(sqrt((Ks*Dm*Cs)/(K*D*C))))*hF\n", + "a = 1/2 + (sqrt((1/4)+Cs*(thetaF-thetaS)/(3*L)))\n", + "delta=h/2\n", + "ts = (delta+((h1*delta**2)/(2*Ks)))/((h1*(thetaF-thetaS))/(Dm*L*a)) # in sec\n", + "ts_ = ts/3600 # in hours\n", + "h2= (1+(sqrt((K*D*C)/(Ks*Dm*Cs))))*hF\n", + "gama= ((h2**2)/(K**2))*AlphaS*ts\n", + "thetaS_ = thetaO + (thetaS-thetaO)*(1-((exp(gama))))\n", + "print \" Solidification time = %f hr,\\n The surface temperature of the mould = %f \u00b0 C\"%(ts_,thetaS_)\n", + "# The value of the surface temperature of the mould in the book is given as 658.1\u00b0 C, Which is wrong." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Solidification time = 0.026965 hr,\n", + " The surface temperature of the mould = -1901.439242 \u00b0 C\n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 8 on page no. 77" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given that\n", + "A= 60*7.5 # Cross sectional area in cm**2\n", + "v=0.05 # Withdrawal rate in m/sec\n", + "t = 0.0125 # Thickness in m\n", + "thetaF= 1500 # Temperature of mould face in degree centigrate\n", + "thetaP = 1550 # \n", + "thetaO = 20 # Initial temperature of mould in Degree centigrate\n", + "L= 268e3 # Latent heat of molten metal in J/Kg\n", + "Dm = 7680 # Density of molten metal in Kg/m**3\n", + "Cs = 0.67e+3 #Specific heat of molten metal in J/Kg-K\n", + "Cm = 0.755e3 #Specific heat of mould in J/Kg-K\n", + "Ks = 76 # Conductivity of molten metal in W/m-K\n", + "hF = 1420 # Heat transfer coefficient at the casting-mould interface in W/m**2-\u00b0C\n", + "Dtheta = 10 # Maximum temperature of cooling water in \u00b0 C\n", + "L_ = L+Cm*(thetaP-thetaF)\n", + "x=L_ / (Cs*(thetaF-thetaO))\n", + "y= hF*t/Ks\n", + "print \"L_/(Cs(thetaF-thetaO)) = %f,\\nhF*t/Ks = %f\"%(x,y)\n", + "z=0.11 # Where z=hF**2 * lm / (v*Ks*Dm*Cs)\n", + "lm= (z*v*Ks*Dm*Cs)/(hF**2)\n", + "Z=0.28 # Where Z=Q/(lm*(thetaF-thetaO)*sqrt(lm*v*Dm*Cs*Ks))\n", + "Q = Z*lm*(thetaF-thetaO)*sqrt(lm*v*Dm*Cs*Ks)\n", + "m = Q / (4.2e3*Dtheta)\n", + "print \"The mould length = %f meter,\\nThe cooling water requirement = %f Kg/sec\"%(lm,m)\n", + "# Answer for The cooling water requirement in the book is given as 5.05 Kg/sec, Which is wrong." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "L_/(Cs(thetaF-thetaO)) = 0.308340,\n", + "hF*t/Ks = 0.233553\n", + "The mould length = 1.066684 meter,\n", + "The cooling water requirement = 48.065525 Kg/sec\n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 9 on page no. 81" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import floor\n", + "# Given that\n", + "a = 15 # Side of the aluminium cube in cm\n", + "Sh = 0.065 # Volume shrinkage of aluminium during solidification\n", + "Vc = a**3\n", + "Vr = 3*Sh*Vc\n", + "h = ((4*Vr)/pi)**(1/3)\n", + "Rr = 6.0/h # Where Rr= (A/V)r\n", + "Rc = 6.0/a # Where Rc = (A/V)c\n", + "print \"(A/V)r=%f, (A/V)c=%f\\n Hence Rr is greater than Rc\"%(Rr,Rc)\n", + "dmin = 6.0/Rc\n", + "Vr_ = (pi/4)*dmin**3\n", + "print \"\"\" With minimum value of d Vr=%d cm**3 .\n", + "This valume is much more than the minimum Vr necessary. \n", + "Let us now consider the top riser when the optimum cylindrical shape is obtained with h=d/2 \n", + "and again (A/V)r = 6/d. However, with a large top riser,\\n the cube loses its top surface for the purpose of heat dissipation.\"\"\"%Vr_\n", + "Rc_ = 5.0/a\n", + "dmin_=6.0/Rc_\n", + "print \" d should be greater than or equal to %d cm\"%dmin_\n", + "Vr__ = (pi/4)*dmin_**2 *floor(h)\n", + "print \" The riser volume with minimum diameter is %d cm**3\"%Vr__" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(A/V)r=6.000000, (A/V)c=0.400000\n", + " Hence Rr is greater than Rc\n", + " With minimum value of d Vr=2650 cm**3 .\n", + "This valume is much more than the minimum Vr necessary. \n", + "Let us now consider the top riser when the optimum cylindrical shape is obtained with h=d/2 \n", + "and again (A/V)r = 6/d. However, with a large top riser,\n", + " the cube loses its top surface for the purpose of heat dissipation.\n", + " d should be greater than or equal to 18 cm\n", + " The riser volume with minimum diameter is 254 cm**3\n" + ] + } + ], + "prompt_number": 40 + } + ], + "metadata": {} + } + ] +} |