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| author | Trupti Kini | 2016-06-01 23:30:10 +0600 |
|---|---|---|
| committer | Trupti Kini | 2016-06-01 23:30:10 +0600 |
| commit | 5b6fb21462221711dd740ab4a57b303b3c4660a1 (patch) | |
| tree | cf8b44770b1712c432901e077ee8743ac36c5cb4 | |
| parent | dfe37e794e1b33b0404d411cd96a0a734c25aedd (diff) | |
| download | Python-Textbook-Companions-5b6fb21462221711dd740ab4a57b303b3c4660a1.tar.gz Python-Textbook-Companions-5b6fb21462221711dd740ab4a57b303b3c4660a1.tar.bz2 Python-Textbook-Companions-5b6fb21462221711dd740ab4a57b303b3c4660a1.zip | |
Added(A)/Deleted(D) following books
A Engineering_Mechanics_of_Solids_by_Popov_E_P/Chapter1_5.ipynb
A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter10_5.ipynb
A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter11_5.ipynb
A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter12_5.ipynb
A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter13_4.ipynb
A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter2_5.ipynb
A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter4_5.ipynb
A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter5_5.ipynb
A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter6_5.ipynb
A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter7_5.ipynb
A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter8_5.ipynb
A Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter9_5.ipynb
A Engineering_Mechanics_of_Solids_by_Popov_E_P/charpter_3_6.ipynb
A Engineering_Mechanics_of_Solids_by_Popov_E_P/screenshots/1_3.PNG
A Engineering_Mechanics_of_Solids_by_Popov_E_P/screenshots/2_3.PNG
A Engineering_Mechanics_of_Solids_by_Popov_E_P/screenshots/3_3.PNG
A Thermal_Engineering_by_A._V._Arasu/README.txt
A Thermal_Engineering_by_A._V._Arasu/ch1.ipynb
A Thermal_Engineering_by_A._V._Arasu/ch2.ipynb
A Thermal_Engineering_by_A._V._Arasu/ch3.ipynb
A Thermal_Engineering_by_A._V._Arasu/ch4.ipynb
A Thermal_Engineering_by_A._V._Arasu/ch5.ipynb
A Thermal_Engineering_by_A._V._Arasu/ch6.ipynb
A Thermal_Engineering_by_A._V._Arasu/ch7.ipynb
A Thermal_Engineering_by_A._V._Arasu/screenshots/1.png
A Thermal_Engineering_by_A._V._Arasu/screenshots/2.png
A Thermal_Engineering_by_A._V._Arasu/screenshots/3.png
27 files changed, 11501 insertions, 0 deletions
diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/Chapter1_5.ipynb b/Engineering_Mechanics_of_Solids_by_Popov_E_P/Chapter1_5.ipynb new file mode 100644 index 00000000..f31c3f93 --- /dev/null +++ b/Engineering_Mechanics_of_Solids_by_Popov_E_P/Chapter1_5.ipynb @@ -0,0 +1,393 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 1 : Stress, Axial loads and Safety concepts" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.1 page number 24" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The bearing stress at C is 0.875 MPA\n", + "The maximum normal stress in BD bolt is: 62.0 MPA\n", + "The tensile strss at shank of the bolt is: 40.0 MPA\n" + ] + } + ], + "source": [ + "#Given\n", + "import math\n", + "d_bolt = 20.0 #mm,diameter,This is not the minimum area\n", + "d_bolt_min = 16.0 #mm This is at the roots of the thread \n", + "#This yealds maximum stress \n", + "A_crossection = (math.pi)*(d_bolt**2)/4 #mm*2\n", + "A_crossection_min = (math.pi)*(d_bolt_min**2)/4 #mm*2 ,This is minimum area which yeilds maximum stress\n", + "load = 10.0 #KN\n", + "BC = 1.0 #m\n", + "CF = 2.5 #m\n", + "contact_area = 200*200 # mm*2 , The contact area at c\n", + "\n", + "#caliculations \n", + "#Balancing forces in the x direction:\n", + "# Balncing the moments about C and B:\n", + "Fx = 0 \n", + "R_cy = load*(BC+CF) #KN , Reaction at C in y-direction\n", + "R_by = load*(CF) #KN , Reaction at B in y-direction\n", + "#Because of 2 bolts\n", + "stress_max = (R_by/(2*A_crossection_min))*(10**3) # MPA,maximum stess records at minimum area\n", + "stress_shank = (R_by/(2*A_crossection))*(10**3) # MPA\n", + "Bearing_stress_c = (R_cy/contact_area)*(10**3) #MPA, Bearing stress at C\n", + "\n", + "print\"The bearing stress at C is \",(Bearing_stress_c) ,\"MPA\"\n", + "print\"The maximum normal stress in BD bolt is: \",round(stress_max),\"MPA\"\n", + "print\"The tensile strss at shank of the bolt is: \",round(stress_shank),\"MPA\"\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.2 page number 26" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total weight of pier is 25.0 KN\n", + "The stress at 1 m above is 28.75 MPA\n" + ] + } + ], + "source": [ + "#Given \n", + "load_distributed = 20 #KN/m*2, This is the load distributed over the pier\n", + "H = 2 # m, Total height \n", + "h = 1 #m , point of investigation \n", + "base = 1.5 #m The length of crossection in side veiw \n", + "top = 0.5 #m ,The length where load is distributed on top\n", + "base_inv = 1 #m , the length at the point of investigation \n", + "area = 0.5*1 #m ,The length at a-a crossection \n", + "density_conc = 25 #KN/m*2\n", + "#caliculation of total weight \n", + "\n", + "v_total = ((top+base)/2)*top*H #m*2 ,The total volume \n", + "w_total = v_total* density_conc #KN , The total weight\n", + "R_top = (top**2)*load_distributed #KN , THe reaction force due to load distribution \n", + "reaction_net = w_total + R_top\n", + "\n", + "#caliculation of State of stress at 1m \n", + "v_inv = ((top+base_inv)/2)*top*h #m*2 ,The total volume from 1m to top\n", + "w_inv = v_inv*density_conc #KN , The total weight from 1m to top\n", + "reaction_net = w_inv + R_top #KN\n", + "Stress = reaction_net/area #KN/m*2\n", + "print\"The total weight of pier is\",w_total,\"KN\"\n", + "print\"The stress at 1 m above is\",Stress,\"MPA\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.3 page number 27" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Tensile stress in main bar AB: 17.89 Ksi\n", + "Tensile stress in clevis of main bar AB: 11.18 Ksi\n", + "Comprensive stress in main bar BC: 12.93 Ksi\n", + "Bearing stress in pin at C: 18.86 Ksi\n", + "torsion stress in pin at C: -25.62 Ksi\n" + ] + } + ], + "source": [ + "#Given\n", + "from math import pow\n", + "d_pins = 0.375 #inch\n", + "load = 3 #Kips\n", + "AB_x = 6 #inch,X-component\n", + "AB_y = 3 #inch,Y-component \n", + "BC_y = 6 #inch,Y-component\n", + "BC_x = 6 #inch,X-component\n", + "area_AB = 0.25*0.5 #inch*2 \n", + "area_net = 0.20*2*(0.875-0.375) #inch*2 \n", + "area_BC = 0.875*0.25 #inch*2 \n", + "area_pin = d_pins*2*0.20 #inch*2 \n", + "area_pin_crossection = 3.14*((d_pins/2)**2)\n", + "#caliculations\n", + "\n", + "slope = AB_y/ AB_x #For AB\n", + "slope = BC_y/ BC_x #For BC\n", + "\n", + "#momentum at point C:\n", + "F_A_x = (load*AB_x )/(BC_y + AB_y ) #Kips, F_A_x X-component of F_A\n", + "\n", + "#momentum at point A:\n", + "F_C_x = -(load*BC_x)/(BC_y + AB_y ) #Kips, F_C_x X-component of F_c\n", + "\n", + "#X,Y components of F_A\n", + "F_A= (pow(5,0.5)/2)*F_A_x #Kips\n", + "F_A_y = 0.5*F_A_x #Kips\n", + "\n", + "#X,Y components of F_C \n", + "F_C= pow(2,0.5)*F_C_x #Kips\n", + "F_C_y = F_C_x #Kips\n", + "\n", + "T_stress_AB = F_A/area_AB #Ksi , Tensile stress in main bar AB\n", + "stress_clevis = F_A/area_net #Ksi ,Tensile stress in clevis of main bar AB\n", + "c_strees_BC = F_C/area_BC #Ksi , Comprensive stress in main bar BC\n", + "B_stress_pin = F_C/area_pin #Ksi , Bearing stress in pin at C\n", + "To_stress_pin = F_C/area_pin_crossection #Ksi , torsion stress in pin at C\n", + "\n", + "print\"Tensile stress in main bar AB:\",round(T_stress_AB,2),\"Ksi\"\n", + "print\"Tensile stress in clevis of main bar AB:\",round(stress_clevis,2),\"Ksi\"\n", + "print\"Comprensive stress in main bar BC:\",round(-c_strees_BC,2),\"Ksi\"\n", + "print\"Bearing stress in pin at C:\",round(-B_stress_pin,2),\"Ksi\"\n", + "print\"torsion stress in pin at C:\",round(To_stress_pin,2),\"Ksi\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.4 page number 38" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The factor 2.5 is less than assumed factor 2.7 so this can be considered\n" + ] + } + ], + "source": [ + "#Given\n", + "strength_steel = 120 #Ksi\n", + "factor = 2.5\n", + "F_C = 2.23 #Ksi\n", + "\n", + "#caliculations\n", + "\n", + "stress_allow = strength_steel/factor #Ksi\n", + "A_net = F_C/strength_steel #in*2 , \n", + "#lets adopt 0.20x0.25 in*2 and check wether we are correct or not? \n", + "\n", + "A_net_assumption = 0.25*0.20 #in*2 , this is assumed area which is near to A_net\n", + "stress = 2.23/A_net_assumption #Ksi\n", + "factor_assumed = strength_steel/stress \n", + "\n", + "if factor_assumed > factor :\n", + " print \"The factor\",factor,\"is less than assumed factor\",round(factor_assumed,1),\"so this can be considered\"\n", + "else:\n", + " print \"The assumed factor\",factor, \"is more than assumed factor\",factor_assumed,\"factor_assumed\"\n", + " \n", + " \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.6 page number 35" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The required size of rod is: 49.35 m*2\n" + ] + } + ], + "source": [ + "#Given\n", + "mass = 5 #Kg\n", + "frequency = 10 #Hz\n", + "stress_allow = 200 #MPa\n", + "R = 0.5 #m\n", + "\n", + "#caliculations \n", + "from math import pi\n", + "w = 2*pi*frequency #rad/sec\n", + "a = (w**2)*R #m*2/sec\n", + "F = mass*a #N\n", + "A_req = F/stress_allow #m*2 , The required area for aloowing stress\n", + "print\"The required size of rod is:\",round(A_req,2),\"m*2\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.7 page number 45" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the allowable area for live load 1.0 is 0.273 in*2\n", + "the allowable area for live load 15 is 0.909 in*2\n", + "the crossection area for live load 1.0 is 0.235 in*2\n", + "the crossection area for live load 15 is 0.926 in*2\n" + ] + } + ], + "source": [ + "#Given\n", + "D_n = 5.0 #kips, dead load\n", + "L_n_1 = 1.0 #kips ,live load 1\n", + "L_n_2 = 15 #kips ,live load 2\n", + "stress_allow = 22 #ksi\n", + "phi = 0.9 #probalistic coefficients\n", + "y_stress = 36 #ksi,Yeild strength\n", + "#According to AISR \n", + "\n", + "#a\n", + "p_1 = D_n + L_n_1 #kips since the total load is sum of dead load and live load\n", + "p_2 = D_n + L_n_2 #kips, For second live load\n", + "\n", + "Area_1 = p_1/stress_allow #in*2 ,the allowable area for the allowed stress\n", + "Area_2 = p_2/stress_allow #in*2\n", + "print \"the allowable area for live load\",L_n_1,\"is\",round(Area_1,3),\"in*2\"\n", + "print \"the allowable area for live load\",L_n_2,\"is\",round(Area_2,3),\"in*2\"\n", + "\n", + "#b\n", + "#area_crossection= (1.2*D_n +1.6L_n)/(phi*y_stress)\n", + "\n", + "area_crossection_1= (1.2*D_n +1.6*L_n_1)/(phi*y_stress) #in*2,crossection area for first live load\n", + "area_crossection_2= (1.2*D_n +1.6*L_n_2)/(phi*y_stress) #in*2,crossection area for second live load\n", + "print \"the crossection area for live load\",L_n_1,\"is\",round(area_crossection_1,3),\"in*2\"\n", + "print \"the crossection area for live load\",L_n_2,\"is\",round(area_crossection_2,3),\"in*2\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.8 page number 51" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Length of the Weld 1: 2.54 in\n", + "Length of the Weld 2: 4.65 in\n" + ] + } + ], + "source": [ + "#Given\n", + "A_angle = 2 #in*2 \n", + "stress_allow = 20 #ksi, The maximum alowable stress\n", + "F = stress_allow*A_angle #K, The maximum force\n", + "AD = 3 #in, from the figure\n", + "DC = 1.06 #in, from the figure\n", + "strength_AWS = 5.56 # kips/in,Allowable strength according to AWS\n", + "\n", + "#caliculations \n", + "#momentum at point \"d\" is equal to 0\n", + "R_1 = (F*DC)/AD #k,Resultant force developed by the weld\n", + "R_2 = (F*(AD-DC))/AD #k,Resultant force developed by the weld\n", + "\n", + "l_1 = R_1/strength_AWS #in,Length of the Weld 1\n", + "l_2 = R_2/strength_AWS #in,Length of the Weld 2\n", + " \n", + "print \"Length of the Weld 1:\",round(l_1,2),\"in\"\n", + "print \"Length of the Weld 2:\",round(l_2,2),\"in\" \n", + " \n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter10_5.ipynb b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter10_5.ipynb new file mode 100644 index 00000000..738d912f --- /dev/null +++ b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter10_5.ipynb @@ -0,0 +1,781 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 10:Deflections of beams " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.1 page number 501" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum bending stress developed in the saw 300.0 Mpa\n" + ] + } + ], + "source": [ + "#Given \n", + "dia = 400 #mm - The diameter of a pulley\n", + "E = 200 #Gpa - Youngs modulus\n", + "t = 0.6 #mm - The thickness of band\n", + "c = t/2 #mm - The maximum stress is seen \n", + "#Caliculations\n", + "\n", + "stress_max = E*c*(10**3)/(dia/2) #Mpa - The maximum stress on the crossection occurs at the ends\n", + "print \"The maximum bending stress developed in the saw \",stress_max,\"Mpa\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.3 page number 512" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) The maximum displacement in y direction is -0.0130208333333 W(l**4)/EI \n", + "a) The maximum deflection occured at 0.5 L\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0xc31ce48>" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "b) The above graph is bending moment graph\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0xbfd30f0>" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "b)The above graph is beam displacement graph\n", + "b)The maximum occures in the middle from the above graph \n" + ] + } + ], + "source": [ + "#Given\n", + "import numpy\n", + "l_ab = 1.0 #L in - The length of the beam\n", + "F_D = 1.0 #W lb/in - The force distribution \n", + "F = F_D*l_ab #WL - The force applied\n", + "#Beause of symmetry the moment caliculations can be neglected\n", + "#F_Y = 0\n", + "R_A = F/2 #wl - The reactive force at A\n", + "R_B = F/2 #wl - The reactive force at B\n", + "#EI - The flxure rigidity is constant and 1/EI =1 # k\n", + "\n", + "#part - A\n", + "#section 1--1\n", + "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.1L distance \n", + "M_1 = [0,0,0,0,0,0,0,0,0,0,0]\n", + "v = [0,0,0,0,0,0,0,0,0,0,0]\n", + "for i in range(10):\n", + " v[i] = R_A - F_D*l_1[i] \n", + " M_1[i] = R_A*l_1[i] - F_D*(l_1[i]**2)/2\n", + "# (EI)y'' = M_1[i] we will integrate M_1[i] twice where variable is l_1[i]\n", + "#(EI)y'- \n", + "\n", + "M_1_intg1 = R_A*(l_1[i]**2)/4 - F_D*(l_1[i]**3)/6 - F_D*(l_ab**3)*l_1[i]/24 #integration of x**n = x**n+1/n+1\n", + "#(EI)y- Using end conditions for caliculating constants \n", + "\n", + "M_1_intg2 = R_A*(l_1[i]**3)/12.0 - F_D*(l_1[i]**4)/24.0 + F_D*(l_ab**3)*l_1[i]/24.0 \n", + "#Equations \n", + "\n", + "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.1L distance \n", + "M_1_intg2 = [0,0,0,0,0,0,0,0,0,0,0]\n", + "Y = [0,0,0,0,0,0,0,0,0,0,0]\n", + "for i in range(10):\n", + " M_1_intg2[i] = (l_1[i]**3)/12.0 - (l_1[i]**4)/24.0 - l_1[i]/24.0 # discluding every term for ruling out float values\n", + " Y[i] = M_1_intg2[i] #W(l**4)/EI k = 1/EI\n", + "#The precision is very less while caliculating through this equation because the least count in X direction is 0.1\n", + "print \"a) The maximum displacement in y direction is\",min(Y),\"W(l**4)/EI \"\n", + "print \"a) The maximum deflection occured at\",l_1[Y.index(min(Y))],\"L\"\n", + "\n", + "#Part - B\n", + "#Graphs\n", + "import numpy as np\n", + "values = M_1\n", + "y = np.array(values)\n", + "t = np.linspace(0,1,11)\n", + "poly_coeff = np.polyfit(t, y, 2)\n", + "import matplotlib.pyplot as plt\n", + "%matplotlib inline\n", + "plt.plot(t, y, 'o')\n", + "plt.plot(t, np.poly1d(poly_coeff)(t), '-')\n", + "plt.show()\n", + "print \"b) The above graph is bending moment graph\"\n", + "values = Y \n", + "y = np.array(values)\n", + "t = np.linspace(0,1,11)\n", + "poly_coeff = np.polyfit(t, y, 2)\n", + "import matplotlib.pyplot as plt\n", + "plt.plot(t, y, 'o')\n", + "plt.plot(t, np.poly1d(poly_coeff)(t), '-')\n", + "plt.show()\n", + "print \"b)The above graph is beam displacement graph\"\n", + "print \"b)The maximum occures in the middle from the above graph \"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.5 page number 517" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The reaction at A is 0.375 WL\n", + "The reaction at B is 0.625 WL\n", + "The reaction at C is 0.375 WL\n" + ] + } + ], + "source": [ + "#Given \n", + "#because of symmetry the problem can be solved by considering first half\n", + "#Given\n", + "import numpy\n", + "\n", + "l_ab = 1.0 #L in - The length of the beam\n", + "F_D = 1.0 #W lb/in - The force distribution \n", + "F = F_D*l_ab #WL - The force applied\n", + "#Beause of symmetry the moment caliculations can be neglected\n", + "#EI - The flxure rigidity is constant and 1/EI =1 # k\n", + "\n", + "#part - A\n", + "#section 1--1\n", + "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.1L distance \n", + "M_1 = [0,0,0,0,0,0,0,0,0,0,0]\n", + "v = [0,0,0,0,0,0,0,0,0,0,0]\n", + "\n", + "\n", + "# M_1_intg2[10] = 0, the displacement at the end of rod is 0 since its rigid \n", + "R_A = (F_D*(l_1[10]**4)/24.0 + F_D*(l_ab**3)*l_1[10]/48.0)/((l_1[10]**3)/6.0)\n", + "R_C = R_A #WL - symmetry\n", + "R_B = 1-R_A # WL - F_Y = 0, the equilibrium in Y direction\n", + "print \"The reaction at A is\",R_A ,\"WL\"\n", + "print \"The reaction at B is\",R_B ,\"WL\"\n", + "print \"The reaction at C is\",R_C ,\"WL\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.7 page number 521 " + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false, + "scrolled": true + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0xb2150b8>" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "b) The shape from x belongs to 4<x<5\n" + ] + } + ], + "source": [ + "import numpy as np\n", + "%matplotlib inline \n", + "import matplotlib.pyplot as plt\n", + "\n", + "l_ac = 5 #m - The length of the beam \n", + "l_ab = 4 #m - The length of ac on beam \n", + "l_bc = 1 #m - The length of bc on beam \n", + "F = 20 #N - force applied on beam at 'b'\n", + "R_A = F/2 #wl - The reactive force at A\n", + "R_B = F/2 #wl - The reactive force at B\n", + "I_ab = 4 #I The moment of inertia of part AB \n", + "I_bc = 1 #I - The momemt of inertia of part BC\n", + "R_A = F*(l_bc/l_ac) #N- The reaction at joint A\n", + "R_B = F*(l_ab/l_ac) #N- The reaction at joint B\n", + "E = 1 #E youngs modulus\n", + "\n", + "#0<x<4\n", + "x = [0,1,2,3,4]\n", + "M = [0,0,0,0,0]\n", + "y = [0,0,0,0,0]\n", + "for i in range(5):\n", + " M[i] = 4*x[i] #integration of x**n = x**n+1/n+1\n", + " #y_2[i] = 4*x[i]/(E*I_ab) #The \n", + " #y_1[i] = 4*(x[i]**2)/(E*I_ab) -4.8/(E*I_bc) #The constant can be found by conditions y(o) = y(c) = 0\n", + " y[i] = 4*(x[i]**2)/(6*E*I_ab) -4.8*x[i]/(E*I_bc) #elastic curve constant can be found by Y_1(0) = 0 \n", + "\n", + "\n", + "#0<x_1<1\n", + "x_1 = [4,5]\n", + "m = [0,0]\n", + "Y = [0,0]\n", + "for i in range(2):\n", + " m[i] = 16 - 16*x_1[i] #integration of x**n = x**n+1/n+1\n", + " # Y_2 = (16 - 16*x_1[i])/(E*I_ab) \n", + " #Y_1 = (16*x_1[i]-8*(x_1[i]**2) +8 - 4.8)/(E*I_ab)#The constant can be found by conditions y(o) = y(c) = 0\n", + " Y[i] = (8*(x_1[i]**2)-8*(x_1[i]**3)/3 +(8-4.8)*x_1[i] - 4*4.8 )/(E*I_ab) #elastic curve constant can be found by Y_1(0) = 0\n", + "\n", + "#Graphs\n", + "values = y\n", + "y = np.array(values)\n", + "t = np.linspace(0,1,5)\n", + "poly_coeff = np.polyfit(t, y, 2)\n", + "plt.plot(t, y, 'o')\n", + "plt.plot(t, np.poly1d(poly_coeff)(t), '-')\n", + "plt.show()\n", + "print \"b) The shape from x belongs to 0<x<4\"\n", + "values = Y \n", + "y = np.array(values)\n", + "t = np.linspace(0,1,2)\n", + "poly_coeff = np.polyfit(t, y, 2)\n", + "plt.plot(t, y, 'o')\n", + "plt.plot(t, np.poly1d(poly_coeff)(t), '-')\n", + "plt.show()\n", + "print \"b) The shape from x belongs to 4<x<5\"\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.10 page number 529" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The deflection of point D -5.56 mm\n" + ] + } + ], + "source": [ + "k = 24.0*(10**12) #N.mm2 Flexure rigidity\n", + "E = 200.0 #Gpa - Youngs modulus of the string\n", + "l = 5000.0 #mm - The length of the string\n", + "C_A = 300.0 #mm2 - crossection area \n", + "P = 50.0 #KN - The force applies at the end \n", + "a = 2000.0 #mm - The distance C-F\n", + "x = 1#X - let it be a variable X\n", + "y_d = x*(a**3)/(3*k) #Xmm The displacement at D, lets keep the variable in units part\n", + "y_p = -P*(10**3)*(16*(a**3)-12*(a**3)+(a**3))/(k*6) #mm The displacement due to p \n", + "e_rod = l/(C_A*E*(10**3)) #Xmm -deflection, The varible is in units \n", + "e_rod\n", + "X = y_p/(2*e_rod+y_d) # By equating deflections \n", + "y_d_1 = X*(a**3)/(3*k) # the deflection of point D\n", + "print \"The deflection of point D\",round(y_d_1,2),\"mm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.11 page number 530 " + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the horizantal component of deflection 0.246 in\n", + "the vertical component of deflection 0.073 in\n", + "the resultant deflection 0.257 in\n" + ] + } + ], + "source": [ + "import math \n", + "l = 15 #in - The length of the crossection \n", + "b = 33.9 #in - the width of the crossection\n", + "L = 100 #in The length of the cantilever \n", + "E = 29*(10**6) #psi The youngs modulus of the material used \n", + "I_Z = 315 #in4 - the moment of inertia wrt Z axis \n", + "I_y = 8.13 #in4 - the moment of inertia wrt Y axis\n", + "o = 5 # degrees - the angle of acting force \n", + "P = 2000 #k the acting force \n", + "P_h = P*math.sin(math.radians(o)) #k - The horizantal component of P\n", + "P_v = P*math.cos(math.radians(o)) #k - The vertical component of P\n", + "e_h = P_h*(L**3)/(3*E*I_y) # the horizantal component of deflection \n", + "e_v = P_v*(L**3)/(3*E*I_Z ) # the vertical component of deflection\n", + "e = pow((e_h**2 + e_v**2),0.5)\n", + "print \"the horizantal component of deflection\",round(e_h,3) ,\"in\"\n", + "print \"the vertical component of deflection\",round(e_v,3) ,\"in\"\n", + "print \"the resultant deflection\",round(e,3) ,\"in\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.13 page number 533" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) The maximum deflection when the beam is on rigid supports 0.03 mm with impact factor 71.7 b) The maximum deflection when the beam is on spring supports 0.28 mm with impact factor 24.17\n" + ] + } + ], + "source": [ + "#Given \n", + "l = 50.0 #mm - The length of the crossection \n", + "b = 50.0 #mm - the width of the crossection\n", + "m = 15.3 # mass of the falling body\n", + "h = 75.0 #mm - The height of the falling body \n", + "p = m*9.81 #N the force acted due to the body\n", + "L = 1000.0 #mm The length of the cantilever\n", + "E = 200 #Gpa The youngs modulus of the material used \n", + "I = (l**4)/12 #mm - the moment of inertia \n", + "k = 300 #N/mm -the stiffness of the spring \n", + "#Rigid supports \n", + "e = m*9.81*(L**3)*(10**-3)/(48*E*I) #mm - the deflection of beam \n", + "imp_fact_a = 1 +pow((1 +2*h/e),0.5) #no units , impact factor \n", + "#spring supports\n", + "e_spr = h/k #mm the elongation due to spring \n", + "e_total = e_spr + e \n", + "imp_fact_b = 1 +pow((1 +2*h/e_total),0.5) #no units , impact factor\n", + "print \"a) The maximum deflection when the beam is on rigid supports\",round(e,3),\"mm with impact factor\",round(imp_fact_a ,2),\n", + "print \"b) The maximum deflection when the beam is on spring supports\",round(e_total,2),\"mm with impact factor\",round(imp_fact_b,2) ,\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.15 page number 536 " + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the ultimate capacity 120 K-in\n", + "the ultimate curvature 0.000806666666667 in*-1\n" + ] + } + ], + "source": [ + "#Given\n", + "E = 30*(10**3) #ksi - The youngs modulus of the material \n", + "stress_y = 40 #Ksi - yield stress\n", + "stress_max = 24.2 #Ksi - the maximum stress\n", + "l = 2 #in - The length of the crossection \n", + "b = 3 #in - the width of the crossection\n", + "h = 3 #in - the depth of the crossection\n", + "#lets check ultimate capacity for a 2 in deep section \n", + "M_ul = stress_y*b*(l**2)/4 #K-in the ultimate capacity \n", + "curvature = 2*stress_y/(E*(h/2) ) #in*-1 the curvature of the beam \n", + "curvature_max = stress_max/(E*(h/2)) #in*-1 The maximum curvature \n", + "print \"the ultimate capacity\",M_ul,\"K-in\"\n", + "print \"the ultimate curvature\",curvature_max,\"in*-1\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.16 page number 543" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum deflection at tip B -4.11 mm\n", + "The slope at the tip B -0.0 radians\n" + ] + } + ], + "source": [ + "#Given \n", + "l_ad = 1600 #mm - The total length of the beam \n", + "l_ab = 600 #mm - The length of AB\n", + "l_bc = 600 #mm - The length of BC\n", + "e_1 = 0.24 #mm - deflection \n", + "e_2 = 0.48 #mm - deflection\n", + "E = 35 #Gpa\n", + "#Caliculation \n", + "\n", + "A_afe = -(l_ab+l_bc)*e_1*(10**-3)/(2*E)\n", + "A_afe = -(l_ab)*e_2*(10**-3)/(4*E)\n", + "y_1_b = A_afe + A_afe #rad the slope at the tip B\n", + "x_1 = 1200 #com from B\n", + "x_2 = 800 #com from B\n", + "y_b = A_afe*x_1 + A_afe*x_2 #mm The maximum deflection at tip B\n", + "print\"The maximum deflection at tip B\",round(y_b,2),\"mm\"\n", + "print \"The slope at the tip B\",round(y_1_b,2) ,\"radians\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.19 page number 547 " + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The maximum displacement in y direction is -0.0130208333333 W(l**4)/EI \n", + " The maximum deflection occured at 0.5 L\n", + " The maximum deflection is -0.05775 W(l**3)/EI \n" + ] + } + ], + "source": [ + "#Given\n", + "import numpy\n", + "l_ab = 1.0 #L in - The length of the beam\n", + "F_D = 1.0 #W lb/in - The force distribution \n", + "F = F_D*l_ab #WL - The force applied\n", + "#Beause of symmetry the moment caliculations can be neglected\n", + "#F_Y = 0\n", + "R_A = F/2 #wl - The reactive force at A\n", + "R_B = F/2 #wl - The reactive force at B\n", + "#EI - The flxure rigidity is constant and 1/EI =1 # k\n", + "\n", + "#part - A\n", + "#section 1--1\n", + "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.1L distance \n", + "M_1 = [0,0,0,0,0,0,0,0,0,0,0]\n", + "v = [0,0,0,0,0,0,0,0,0,0,0]\n", + "for i in range(10):\n", + " v[i] = R_A - F_D*l_1[i] \n", + " M_1[i] = R_A*l_1[i] - F_D*(l_1[i]**2)/2\n", + "# (EI)y'' = M_1[i] we will integrate M_1[i] twice where variable is l_1[i]\n", + "#(EI)y'- \n", + "\n", + "M_1_intg1 = R_A*(l_1[i]**2)/4 - F_D*(l_1[i]**3)/6 - F_D*(l_ab**3)*l_1[i]/24 #deflection integration of x**n = x**n+1/n+1\n", + "#(EI)y- Using end conditions for caliculating constants \n", + "\n", + "M_1_intg2 = R_A*(l_1[i]**3)/12.0 - F_D*(l_1[i]**4)/24.0 + F_D*(l_ab**3)*l_1[i]/24.0 \n", + "#Equations \n", + "\n", + "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.1L distance \n", + "M_1_intg2 = [0,0,0,0,0,0,0,0,0,0,0]\n", + "Y = [0,0,0,0,0,0,0,0,0,0,0]\n", + "for i in range(10):\n", + " M_1_intg2[i] = (l_1[i]**3)/12.0 - (l_1[i]**4)/24.0 - l_1[i]/24.0 # discluding every term for ruling out float values\n", + " Y[i] = M_1_intg2[i] #W(l**4)/EI k = 1/EI\n", + "#The precision is very less while caliculating through this equation because the least count in X direction is 0.1\n", + "print \" The maximum displacement in y direction is\",min(Y),\"W(l**4)/EI \"\n", + "print \" The maximum deflection occured at\",l_1[Y.index(min(Y))],\"L\"\n", + "\n", + "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.1L distance \n", + "M_1_intg1 = [0,0,0,0,0,0,0,0,0,0,0]\n", + "Y = [0,0,0,0,0,0,0,0,0,0,0]\n", + "for i in range(10):\n", + " M_1_intg1[i] = R_A*(l_1[i]**2)/4 - F_D*(l_1[i]**3)/6 - F_D*(l_ab**3)*l_1[i]/24\n", + "print \" The maximum deflection is\",min(M_1_intg1 ),\"W(l**3)/EI \"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.23 page number 554 " + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The moment at the end is -0.0833333333333 wl**2\n" + ] + } + ], + "source": [ + "import numpy\n", + "l_ab = 1.0 #L in - The length of the beam\n", + "F_D = 1.0 #W lb/in - The force distribution \n", + "F = F_D*l_ab #WL - The force applied\n", + "#Beause of symmetry the moment caliculations can be neglected\n", + "#F_Y = 0\n", + "R_A = F/2 #wl - The reactive force at A\n", + "R_B = F/2 #wl - The reactive force at B\n", + "#EI - The flxure rigidity is constant and 1/EI =1 # k\n", + "#M_A and M_B are applied at the ends\n", + "\n", + "#part - A\n", + "#section 1--1\n", + "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.1L distance \n", + "M = [0,0,0,0,0,0,0,0,0,0,0]\n", + "for i in range(10):\n", + " M[i] = l_1[i]/2.0 - (l_1[i]**2)/2.0 -1.0/12.0 #The moment euation at 1--1 section\n", + "# M_1 = R_A*l_1[i]/2.0 - F_D*(l_1[i]**2)/2.0 -F_D*(l_ab**2)/12.0 #The moment euation at 1--1 section \n", + "# (EI)y'' = M_1[i] we will integrate M_1[i] twice where variable is l_1[i]\n", + "#(EI)y'\n", + "M_1_intg1 = R_A*(l_1[i]**2)/4 - F_D*(l_1[i]**3)/6 - F_D*(l_ab**2)*l_1[i]/12.0 #integration of x**n = x**n+1/n+1\n", + "#(EI)y\n", + " \n", + "print \"The moment at the end is \",M[0],\"wl**2\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.25 pagenumber 556" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0xaf5ca90>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "import matplotlib.pyplot as plt\n", + "%matplotlib inline\n", + "import numpy as np\n", + "# This problem is divided into two parts\n", + "#Part _1\n", + "#Given\n", + "\n", + "l = 1.0 #l - The length of the beam\n", + "p = 1.0 #W - The total load applied\n", + "#since it is triangular distribution \n", + "l_com = 0.66*l #l - The distance of force of action from one end\n", + "#F_Y = 0\n", + "#R_A + R_B = p\n", + "#M_a = 0 Implies that R_B = 2*R_A\n", + "R_A = p/3.0\n", + "R_B = 2.0*p/3\n", + "\n", + "#Taking Many sections \n", + "\n", + "#Section 1----1\n", + "l = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.1L distance \n", + "M = [0,0,0,0,0,0,0,0,0,0,0]\n", + "v = [0,0,0,0,0,0,0,0,0,0,0]\n", + "for i in range(10):\n", + " v[i] = p*(l[i]**2) - p/3.0\n", + " M[i] = p*(l[i]**3)/(3.0)- p*l[i]/3.0\n", + "\n", + "v[10] = R_B #again concluded Because the value is tearing of \n", + "\n", + "\n", + "#Graph\n", + "values = M\n", + "y = np.array(values)\n", + "t = np.linspace(0,1,11)\n", + "poly_coeff = np.polyfit(t, y, 2)\n", + "plt.plot(t, y, 'o')\n", + "plt.plot(t, np.poly1d(poly_coeff)(t), '-')\n", + "plt.show()\n", + "values = v\n", + "y = np.array(values)\n", + "t = np.linspace(0,1,11)\n", + "poly_coeff = np.polyfit(t, y, 2)\n", + "\n", + "plt.plot(t, y, 'o')\n", + "plt.plot(t, np.poly1d(poly_coeff)(t), '-')\n", + "plt.show()\n", + "\n", + "\n", + "#part B\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter11_5.ipynb b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter11_5.ipynb new file mode 100644 index 00000000..411b3c0a --- /dev/null +++ b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter11_5.ipynb @@ -0,0 +1,376 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 11:Stability of Equilibrium: columns " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.2 page number 589" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The length after which the beam starts buckling is 1539.0 mm\n" + ] + } + ], + "source": [ + "#Given \n", + "import math\n", + "h = 60 #mm - the length of the crossection \n", + "b = 100 #mm - the width of hte crossection \n", + "E = 200 #Gpa - The youngs modulus\n", + "stress_cr = 250 #Mpa - The proportionality limit\n", + "#Caliculations \n", + "\n", + "I = b*(h**3)/12 #mm3 The momentof inertia of the crossection\n", + "A = h*b #mm2 - The area of teh crossection \n", + "#From Eulier formula\n", + "r_min = pow((I/A),0.5) #mm - The radius of the gyration \n", + "#(l/r)**2= (pi**2)*E/stress_cr #From Eulier formula\n", + "l = (((math.pi**2)*E*(10**3)/stress_cr)**0.5)*r_min #mm - the length after which the beam starts buckling\n", + "print \"The length after which the beam starts buckling is \",round(l,0),\"mm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.6 page number 613" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a)The following approch is solvable\n", + "a) The allowable stress in this case is 18.9 Kips\n", + "b) The following approch is solvable\n", + "b) The allowable stress in this case is 11.59 Kips\n" + ] + } + ], + "source": [ + "#Given\n", + "import math\n", + "L = 15 #ft - The length of the each rod\n", + "A = 46.7 #in2 - The length of the crossection \n", + "r_min = 4 #in - The radius of gyration\n", + "stress_yp = 36 #Ksi - the yielding point stress\n", + "E = 29*(10**3) #ksi - The youngs modulus\n", + "C_c = ((2*(math.pi**2)*E/stress_yp)**0.5) #Slenderness ratio L/R\n", + "C_s = L*12/r_min # Slenderness ratio L/R of the present situation \n", + "#According to AISC formulas \n", + "if C_s <C_c :\n", + " print \"a)The following approch is solvable\"\n", + "else: \n", + " print \"The caliculation is not possible\"\n", + "F_S = 5.0/3 +3*C_s/(8*C_c) -(3*C_s**3)/(8*C_c**3) #Safety factor \n", + "Stress_all = (1 - (C_s**2)/(2*C_c**2))*stress_yp/F_S #The allowable strees \n", + "print \"a) The allowable stress in this case is\",round(Stress_all,2),\"Kips\" \n", + "#Part - B\n", + "#Given\n", + "L = 40 #ft - The length of the each rod\n", + "A = 46.7 #in2 - The length of the crossection \n", + "r_min = 4 #in - The radius of gyration\n", + "stress_yp = 36 #Ksi - the yielding point stress\n", + "E = 29*(10**3) #ksi - The youngs modulus\n", + "C_c = ((2*(math.pi**2)*E/stress_yp)**0.5) #Slenderness ratio L/R\n", + "C_s = L*12/r_min # Slenderness ratio L/R of the present situation \n", + "#According to AISC formulas \n", + "if C_s <C_c :\n", + " print \"b) The following approch is solvable\"\n", + "else: \n", + " print \"The caliculation is not possible\"\n", + "F_S = 5.0/3 +3*C_s/(8*C_c) -(3*C_s**3)/(8*C_c**3) #Safety factor \n", + "Stress_all = (1 - (C_s**2)/(2*C_c**2))*stress_yp/F_S #The allowable strees \n", + "print \"b) The allowable stress in this case is\",round(Stress_all,2),\"Kips\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.7 page number 614" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a)The following approch is solvable\n", + "The allowable stress in this case is 19.14 Kips\n", + "This stress requires 10.45 in2\n", + "This case is satisfying W8x24 section\n" + ] + } + ], + "source": [ + "#Given\n", + "import math\n", + "L = 15 #ft - The length of the each rod \n", + "p = 200 #Kips The concentric load applied \n", + "r_min = 2.10 #in - The radius of gyration\n", + "stress_yp = 50 #Ksi - the yielding point stress\n", + "E = 29*(10**3) #ksi - The youngs modulus\n", + "C_c = ((2*(math.pi**2)*E/stress_yp)**0.5) #Slenderness ratio L/R\n", + "C_s = L*12/r_min #Slenderness ratio L/R present situation\n", + "if C_s <C_c :\n", + " print \"a)The following approch is solvable\"\n", + "else: \n", + " print \"The caliculation is not possible\"\n", + "F_S = 5.0/3 +3*C_s/(8*C_c) -(3*C_s**3)/(8*C_c**3) #Safety factor \n", + "Stress_all = (1 - (C_s**2)/(2*C_c**2))*stress_yp/F_S #The allowable strees\n", + "a = p/Stress_all #in2 the alloawble area of the beam \n", + "print \"The allowable stress in this case is\",round(Stress_all,2),\"Kips\"\n", + "print \"This stress requires \",round(a,2),\"in2\"\n", + "if a <11.5:\n", + " print \"This case is satisfying W8x24 section\" #From AISC Manual \n", + "else:\n", + " print \"This case is not satisfying W8x24 section\"\n", + " #The ans are quiet varying because of rounding\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.8 pagenumber 614 " + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The following approach is right\n", + "column design compressive strength 1284.51846781 Kips\n" + ] + } + ], + "source": [ + "#Given\n", + "import math\n", + "L = 15.0 #ft - The length of the each rod\n", + "A = 46.7 #in2 - The length of the crossection \n", + "r_min = 4 #in - The radius of gyration\n", + "stress_yp = 36.0 #Ksi - the yielding point stress\n", + "E = 29*(10**3) #ksi - The youngs modulus\n", + "lamda = L*12*((stress_yp/E)**0.5)/(4*(math.pi)) #column slenderness ratio\n", + "if lamda<1.5:\n", + " print \"The following approach is right\"\n", + "else:\n", + " print \"The following approach is wrong\"\n", + "stress_cr = (0.658**(lamda**2))*stress_yp #Ksi - The critical stress \n", + "P_n = stress_cr*A #Kips #Nominal compressive strength \n", + "o = 0.85 #Resistance factor\n", + "p_u = o*P_n #Kips ,column design compressive strength \n", + "print \"column design compressive strength \",p_u,\"Kips\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.9 page number 615" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum allowable stress in case of web width 19.0 Ksi\n", + "The maximum allowable stress in case of flang 17.16 Ksi\n", + "a) The maximum allowable load in case of Buckling 75.58 Kips\n", + "b) The maximum allowable load in case of Buckling 63.14 Kips\n" + ] + } + ], + "source": [ + "#Given \n", + "#FOR FLANGS\n", + "l = 5 #in - The length of the flang\n", + "b = 5 #in - Teh width of the flang\n", + "t = 0.312 #in - the thickness of the flang\n", + "L = 20 #in - Length of the beam, Extracted from AISC manuals\n", + "A = 4.563 #in2 - The area of crossection of the beam\n", + "r = 1.188 #in - radius of the gyration, Extracted from AISC manuals \n", + "#b/t- value of the flang \n", + "k = (5 -t)/(2*t) #b/t ratio \n", + "#AISC, lets check maximum allowable stress for slang\n", + "Stressf_all = 23.1 - 0.79*k #ksi The maximum allowable stress in case of flang,AISC\n", + "\n", + "#web width thickness ratio\n", + "k_2 = (5 -2*t)/(t)\n", + "if k_2<16:\n", + " Stressw_all = 19 #ksi - The allowable stress in case of web width\n", + " \n", + "#a) Overall buckling investment \n", + "k_3 = L/r #slenderness ratio\n", + "Stressb_all = 20.2 - 0.216*k_3#ksi The maximum allowable stress in case of Buckling,AISC\n", + "p_allow = A*Stressb_all #Kips The allowable concentric load \n", + "\n", + "#b) Overall buckling investment\n", + "L_2 = 60 #in \n", + "k_3 = L_2/r #slenderness ratio\n", + "Stressb_all_2 = 20.2 - 0.126*k_3#ksi The maximum allowable stress in case of Buckling,AISC\n", + "p_allow_2 = A*Stressb_all_2 #Kips The allowable concentric load \n", + "\n", + "print \"The maximum allowable stress in case of web width\",round(Stressw_all,2),\"Ksi\"\n", + "print \"The maximum allowable stress in case of flang\",round(Stressf_all,2),\"Ksi\"\n", + "print \"a) The maximum allowable load in case of Buckling\",round(p_allow,2),\"Kips\"\n", + "print \"b) The maximum allowable load in case of Buckling\",round(p_allow_2,2),\"Kips\"\n", + "\n", + "\n", + " \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.11 page number 620 " + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The minimum area is 13.71 in2\n", + "The following approch is solvable\n", + "The following W10x49 section is not satisfying our constraints since f_a/F_a_1 + c_m*f_b*(1-(f_a/F_e))/F_b 1.09 >1\n", + "The following approch is solvable\n", + "The following W10x49 section is satisfying our constraints since f_a/F_a_1 + c_m*f_b*(1-(f_a/F_e))/F_b 0.9 <1\n" + ] + } + ], + "source": [ + "import math\n", + "P = 200.0 #K The force on the beam \n", + "L = 15 #ft - The length of the rod\n", + "F_y = 50.0 #Ksi \n", + "F_a = F_y/(5.0/3) #Ksi -AISC MANUAL ,allowable axial stress if axial force is alone\n", + "F_b = F_a #Allowable compressive bending stress\n", + "M_1 = 600.0 #k-in - The moment acting on the ends of the rod\n", + "M_2 = 800.0 #k-in - the moment acting on the other end of teh rod\n", + "B_x = 0.264 #in - Extracted from AISC manual \n", + "E = 29*(10**3) \n", + "A = P/F_a + M_2*B_x/F_b #in2- The minimum area \n", + "print \"The minimum area is \",round(A,2),\"in2\"\n", + "#we will select W10x49 section \n", + "A_s = 14.4 #in2 - The area of the section \n", + "r_min = 2.54 #in The minimum radius \n", + "r_x = 4.35 #in \n", + "f_a = P/A_s #Ksi- The computed axial stress\n", + "f_b = M_2*B_x/A_s #Computed bending stess\n", + "C_c = ((2*(math.pi**2)*E/F_y)**0.5) #Slenderness ratio L/R\n", + "C_s = L*12/r_min # Slenderness ratio L/R of the present situation\n", + "if C_s <C_c :\n", + " print \"The following approch is solvable\"\n", + "else: \n", + " print \"The caliculation is not possible\"\n", + "F_a_1 = 19.3 #Ksi - AISC lets try this\n", + "c_m = 0.6 - 0.4*(-M_1/M_2) \n", + "F_e = (12*(math.pi**2)*E)/(23*(L*12/r_x)**2) \n", + "k = f_a/F_a_1 + c_m*f_b*(1-(f_a/F_e))/F_b #Condition mentioned in AISC\n", + "if k>1:\n", + " print \"The following W10x49 section is not satisfying our constraints since f_a/F_a_1 + c_m*f_b*(1-(f_a/F_e))/F_b\",round(k,3),\">1\"\n", + "else:\n", + " print \"The following W10x49 section is satisfying our constraints since f_a/F_a_1 + c_m*f_b*(1-(f_a/F_e))/F_b\",k,\"<1\"\n", + " \n", + "#trail - 2\n", + "#Lets take W10 x 60\n", + "A_s = 17.6 #in2 - The area of the section \n", + "r_min = 2.57 #in The minimum radius \n", + "r_x = 4.39 #in \n", + "f_a = P/A_s #Ksi- The computed axial stress\n", + "f_b = M_2*B_x/A_s #Computed bending stess\n", + "C_c = ((2*(math.pi**2)*E/F_y)**0.5) #Slenderness ratio L/R\n", + "C_s = L*12/r_min # Slenderness ratio L/R of the present situation\n", + "if C_s <C_c :\n", + " print \"The following approch is solvable\"\n", + "else: \n", + " print \"The caliculation is not possible\"\n", + "F_a_1 = 19.3 #Ksi - AISC lets try this\n", + "c_m = 0.6 - 0.4*(-M_1/M_2) \n", + "F_e = (12*(math.pi**2)*E)/(23*(L*12/r_x)**2) \n", + "k = f_a/F_a_1 + c_m*f_b*(1-(f_a/F_e))/F_b #Condition mentioned in AISC\n", + "if k>1:\n", + " print \"The following W10x49 section is not satisfying our constraints since f_a/F_a_1 + c_m*f_b*(1-(f_a/F_e))/F_b\",round(k,3),\">1\"\n", + "else:\n", + " print \"The following W10x49 section is satisfying our constraints since f_a/F_a_1 + c_m*f_b*(1-(f_a/F_e))/F_b\",round(k,2),\"<1\"\n", + " \n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter12_5.ipynb b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter12_5.ipynb new file mode 100644 index 00000000..71380130 --- /dev/null +++ b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter12_5.ipynb @@ -0,0 +1,411 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Chapter 12:Energy and Virtual-work Methods" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.1 page number 645 " + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) The deflection is downwards 0.044 in\n", + "b) The deflection is upwards 0.104 in\n", + "c) The deflection is downwards 0.039 in\n" + ] + } + ], + "source": [ + "#Given\n", + "#Virtual loading\n", + "p_ab = -0.833 #lb The recorded virtual loading\n", + "p_bc = + 0.833 #lb The recorded virtual loading\n", + "F_ab = 2500 #lb\n", + "F_bc = -2500 #lb\n", + "l_ab = 60 #in - The length of the rod\n", + "l_bc = 60 #in - The length of the rod\n", + "A_ab = 0.15 #in2 the areaof ab\n", + "A_bc = 0.25 #in2 the areaof bc\n", + "E = 30*(10**6) #psi The youngs modulus of the material\n", + "#Part_a\n", + "e_a =p_ab*l_ab*F_ab/(A_ab*E) + p_bc*l_bc*F_bc/(A_bc*E) #in the deflection\n", + "if e_a<0:\n", + " print \"a) The deflection is downwards\",round(-e_a,3),\"in\"\n", + "else:\n", + " print \"a) The deflection is upwards\",round(e_a,3),\"in\"\n", + "#part-b\n", + "x = 0.125 #Shortening of member Ab\n", + "e_b = p_ab*(-x) + p_bc*0 #in - in\n", + "if e_b<0:\n", + " print \"b) The deflection is downwards\",round(-e_b,3),\"in\"\n", + "else:\n", + " print \"b) The deflection is upwards\",round(e_b,3),\"in\"\n", + "#Part-c\n", + "S = 6.5*(10**-6) #Thermal specific heat\n", + "T = 120 #F - The cahnge in temperature\n", + "e_t = -S*T*l_ab #in - The change in length of member\n", + "e_c = p_bc*e_t #in the deflection\n", + "if e_c<0:\n", + " print \"c) The deflection is downwards\",round(-e_c,3),\"in\"\n", + "else:\n", + " print \"c) The deflection is upwards\",round(e_c,3),\"in\"\n", + "\n", + "\n", + " \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.3 page number 648" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The point C deflects 0.019 mt down\n" + ] + } + ], + "source": [ + "#Given\n", + "#Virtual loading\n", + "#Two parts \n", + "#Part -1 \n", + "p_ab = 5 #KN The recorded virtual loading\n", + "p_bc = -4 #KN The recorded virtual loading\n", + "F_ab = 10 #KN\n", + "F_bc = -8 #KN\n", + "l_ab = 2.5 #mt - The length of the rod\n", + "l_bc = 2 #mt - The length of the rod\n", + "A_ab = 5*(10**-4) #mt2 the areaof ab\n", + "A_bc = 5*(10**-3) #mt2 the areaof bc\n", + "E = 70 #Gpa The youngs modulus of the material\n", + "e_a =(p_ab*l_ab*F_ab/(A_ab*E) + p_bc*l_bc*F_bc/(A_bc*E))*(10**-6) #KN-m\n", + "#Part -2 due to flexure\n", + "I = 60*10**6 #mm4 - the moment of inertia \n", + "#After solving the integration \n", + "e_b = 0.01525 #KN-m\n", + "#Total\n", + "e = (e_a+e_b)*1 #m\n", + "print \"The point C deflects\",round(e,3),\"mt down\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.5 page number 651" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The reaction force at D is 1578.98 lb\n", + "The deflection of nodal point B 0.0211 in\n" + ] + } + ], + "source": [ + "#Given\n", + "#Virtual loading Without f_d\n", + "p_ab = -0.833 #lb The recorded virtual loading\n", + "p_bc = + 0.833 #lb The recorded virtual loading\n", + "F_ab = 2500 #lb\n", + "F_bc = -2500 #lb\n", + "l_ab = 60 #in - The length of the rod\n", + "l_bc = 60 #in - The length of the rod\n", + "A_ab = 0.15 #in2 the areaof ab\n", + "A_bc = 0.25 #in2 the areaof bc\n", + "E = 30*(10**6) #psi The youngs modulus of the material\n", + "#Part_a\n", + "e_a =p_ab*l_ab*F_ab/(A_ab*E) + p_bc*l_bc*F_bc/(A_bc*E) #lb-in the deflection\n", + "#With f_d\n", + "p_bd = 1 #lb The recorded virtual loading \n", + "F_bd = 1 #lb\n", + "l_bd = 40 #in - The length of the rod\n", + "A_bd = 0.1 #in2 the areaof ab\n", + "e_a_1 =p_ab*p_ab*l_ab/(A_ab*E) + p_bc*p_bc*l_bc/(A_bc*E) +p_bd*p_bd*l_bd/(A_bd*E) #lb-in the deflection\n", + "#Since the produced defelection should compensate the other one\n", + "x_d = e_a/e_a_1\n", + "print \"The reaction force at D is\",round(-x_d,2),\"lb\"\n", + "\n", + "#Part - B\n", + "e_b = -x_d*l_bd/(A_bd*E ) #in - The deflection of nodal point B\n", + "print\"The deflection of nodal point B\",round(e_b,4),\"in\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.6 page number 655" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The reaction at A 2.5 k\n", + "The reaction at B -2.5 k\n" + ] + } + ], + "source": [ + "#Given\n", + "#Virtual loading\n", + "import numpy as np\n", + "p_ab = -0.833 #lb The recorded virtual loading\n", + "p_bc = + 0.833 #lb The recorded virtual loading \n", + "l_ab = 60 #in - The length of the rod\n", + "l_bc = 60 #in - The length of the rod\n", + "A_ab = 0.15 #in2 the areaof ab\n", + "A_bc = 0.25 #in2 the areaof bc\n", + "E = 30*(10**6) #psi The youngs modulus of the material\n", + "K_1 = A_ab*E/l_ab #k/in - Stiffness\n", + "K_2 = A_bc*E/l_bc #k/in - Stiffness\n", + "#soving for e_1 and e_2 gives a liner euations to solve\n", + "# 128*e_1 + 24*e_2 = 0\n", + "#24*e_1 + 72*e_2 = -3\n", + "#Solving for e_1,e_2\n", + "a = np.array([[128,24], [24,72]])\n", + "b = np.array([0,-3])\n", + "x = np.linalg.solve(a, b)\n", + "e_1 = x[0] #in\n", + "e_2 = x[1] #in\n", + "u_1 = 0.8*e_1 - 0.6*e_2 #Taking each components\n", + "F_1 = K_1*u_1*(10**-3) #k The reaction at A Force = stiffness x dislacement \n", + "u_2 = 0.8*e_1 + 0.6*e_2 #Taking each components\n", + "F_2 = K_2*u_2*(10**-3) #k The reaction at B Force\n", + "print \"The reaction at A \",F_1,\"k\"\n", + "print \"The reaction at B \",F_2,\"k\"\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.7 page number 655" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The reaction at A 1.18 k\n", + "The reaction at B -1.18 k\n", + "The reaction at D -1.58 k\n" + ] + } + ], + "source": [ + "#Virtual loading\n", + "import numpy as np\n", + "p_ab = -0.833 #lb The recorded virtual loading\n", + "p_bc = + 0.833 #lb The recorded virtual loading \n", + "l_ab = 60 #in - The length of the rod\n", + "l_bc = 60 #in - The length of the rod\n", + "A_ab = 0.15 #in2 the areaof ab\n", + "A_bc = 0.25 #in2 the areaof bc\n", + "E = 30*(10**6) #psi The youngs modulus of the material\n", + "K_1 = A_ab*E/l_ab #k/in - Stiffness\n", + "K_2 = A_bc*E/l_bc #k/in - Stiffness\n", + "p_bd = 1 #lb The recorded virtual loading \n", + "F_bd = 1 #lb\n", + "l_bd = 40 #in - The length of the rod\n", + "A_bd = 0.1 #in2 the areaof ab\n", + "K_3 = A_ab*E/l_ab #k/in - Stiffness\n", + "#soving for e_1 and e_2 gives a liner euations to solve\n", + "# 128*e_1 + 24*e_2 = 0\n", + "#24*e_1 + 72*e_2 = -3\n", + "#Solving for e_1,e_2\n", + "a = np.array([[128,24], [24,147]])\n", + "b = np.array([0,-3])\n", + "x = np.linalg.solve(a, b)\n", + "e_1 = x[0] #in\n", + "e_2 = x[1] #in\n", + "u_1 = 0.8*e_1 - 0.6*e_2 #Taking each components\n", + "F_1 = K_1*u_1*(10**-3) #k The reaction at A Force = stiffness x dislacement \n", + "u_2 = 0.8*e_1 + 0.6*e_2 #Taking each components\n", + "F_2 = K_2*u_2*(10**-3) #k The reaction at B Force\n", + "u_3 = e_2 #Taking each components\n", + "F_3 = K_3*u_3*(10**-3) #k The reaction at D Force\n", + "print \"The reaction at A \",round(F_1,2),\"k\"\n", + "print \"The reaction at B \",round(F_2,2),\"k\"\n", + "print \"The reaction at D \",round(F_3,2),\"k\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.8 page number 659" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "b) The vertical component of the nodal force is [ 12.] \n", + "b) The vertical component of the nodal force is [ 24.] \n", + "a) The components of displacement of point B are 18.75 L/AE and 16.67 L/AE\n" + ] + } + ], + "source": [ + "#Given\n", + "#First we will solve part B\n", + "import numpy as np\n", + "u_1 =5 #L/AE, elastic elongation\n", + "u_2 =25 #L/AE,elastic elongation\n", + "f_1 = u_1#, Units got neutralized , Constitutive relation for elastic bars\n", + "f_2 = u_2# Units got neutralized\n", + "#u_1 = 0.8*e_1 - 0.6*e_2\n", + "#u_2 = 0.8*e_1 + 0.6*e_2\n", + "#u = A*e Matric multiplication \n", + "A = np.array([[0.8,-0.6],[0.8,0.6]]) #The matrix form of A\n", + "F = np.array([[f_1],[f_2]])\n", + "P = np.dot((A.T),F) #Nodal forces matrix\n", + "print \"b) The vertical component of the nodal force is\",P[1],\"\"\n", + "print \"b) The vertical component of the nodal force is\",P[0],\"\"\n", + "#Part A\n", + "#F_1 = (5/8.0)*P_1 - (5/6.0)*p_2 , From statics\n", + "#F_1 = (5/8.0)*P_1 + (5/6.0)*p_2\n", + "#F = BP ,Matric multiplication \n", + "B = np.array([[(5/8.0),-(5/6.0)],[(5/8.0),(5/6.0)]]) #The matrix form of A\n", + "U = np.array([[u_1],[u_2]])\n", + "e = P = np.dot((B.T),U) #L/AE, Nodal forces matrix\n", + "print \"a) The components of displacement of point B are\",round(e[0],2),\"L/AE and\",round(e[1],2),\"L/AE\" \n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.10 page number 667" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The deflection is 0.018 in\n" + ] + } + ], + "source": [ + "#Given\n", + "A_1 = 0.125 #in2 , The area of the crossection of AB\n", + "A_2 = 0.219 #in2 , The area of the crossection of BC\n", + "l_1 = 3*(5**0.5) #in , The length of AB\n", + "l_2 = 6*(2**0.5) #in , The length of BC\n", + "p = 3 #k , Force acting on the system \n", + "E = 10.6*(10**3) #Ksi - youngs modulus of the material\n", + "p_1 = (5**0.5)*p/3 #P, The component of p on AB\n", + "p_2 = -2*(2**0.5)*p/3 #P, The component of p on AB\n", + "\n", + "e = p_1*l_1*p_1/(p*E*A_1) + p_2*l_2*p_2/(p*E*A_2) #in, By virtual deflection method \n", + "print \"The deflection is\",round(e,3),\"in\"" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false + }, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter13_4.ipynb b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter13_4.ipynb new file mode 100644 index 00000000..4d2cad2f --- /dev/null +++ b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter13_4.ipynb @@ -0,0 +1,215 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 13: Statically Indeterminate Problems" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.2 page number 693" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) The maximum displacement in y direction is -0.208333333333 W(l**4)/EI \n", + "a) The maximum deflection occured at 1.0 L\n", + "The reaction at the mid of the bar 1.25 WL\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fad208>" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "b)The above graph is beam displacement graph\n", + "b)The minimum occures in the middle from the above graph \n" + ] + } + ], + "source": [ + "#Given \n", + "#First we will solve without the reaction at middle\n", + "#Given\n", + "import numpy\n", + "import matplotlib.pyplot as plt\n", + "import numpy as np\n", + "l_ab = 1.0 #2L in - The length of the beam\n", + "F_D = 1.0 #W lb/in - The force distribution \n", + "F = F_D*l_ab #WL - The force applied\n", + "#Beause of symmetry the moment caliculations can be neglected\n", + "#F_Y = 0\n", + "R_A = F/2 #wl - The reactive force at A\n", + "R_B = F/2 #wl - The reactive force at B\n", + "#EI - The flxure rigidity is constant and 1/EI =1 # k\n", + "\n", + "#part - A\n", + "#section 1--1\n", + "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.2L distance \n", + "M_1 = [0,0,0,0,0,0,0,0,0,0,0]\n", + "v = [0,0,0,0,0,0,0,0,0,0,0]\n", + "for i in range(10):\n", + " v[i] = R_A - F_D*l_1[i] \n", + " M_1[i] = R_A*l_1[i] - F_D*(l_1[i]**2)/2\n", + "# (EI)y'' = M_1[i] we will integrate M_1[i] twice where variable is l_1[i]\n", + "#(EI)y'- \n", + "\n", + "M_1_intg1 = R_A*(l_1[i]**2)/4 - F_D*(l_1[i]**3)/6 - F_D*(l_ab**3)*l_1[i]/24 #integration of x**n = x**n+1/n+1\n", + "#(EI)y- Using end conditions for caliculating constants \n", + "\n", + "M_1_intg2 = R_A*(l_1[i]**3)/12.0 - F_D*(l_1[i]**4)/24.0 + F_D*(l_ab**3)*l_1[i]/24.0 \n", + "#Equations \n", + "\n", + "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.2L distance \n", + "M_1_intg2 = [0,0,0,0,0,0,0,0,0,0,0]\n", + "Y = [0,0,0,0,0,0,0,0,0,0,0]\n", + "for i in range(10):\n", + " M_1_intg2[i] = (l_1[i]**3)/12.0 - (l_1[i]**4)/24.0 - l_1[i]/24.0 # discluding every term for ruling out float values\n", + " Y[i] = M_1_intg2[i] #W(l**4)/EI k = 1/EI\n", + "Y_min = 16*min(Y)\n", + "print \"a) The maximum displacement in y direction is\",16*min(Y),\"W(l**4)/EI \"\n", + "print \"a) The maximum deflection occured at\",2*l_1[Y.index(min(Y))],\"L\"\n", + "f_bb = 2**3/48.0 #l**3/EI - flexibility coefficient\n", + "Reac = - Y_min/f_bb #WL , The reaction at the mid of the bar\n", + "print \"The reaction at the mid of the bar\",Reac ,\"WL\"\n", + "\n", + "#Graphs \n", + "Y.extend(Y) #Because of symmetry\n", + "values = Y \n", + "y = np.array(values)\n", + "t = np.linspace(0,1,22)\n", + "poly_coeff = np.polyfit(t, y, 2)\n", + "\n", + "plt.plot(t, y, 'o')\n", + "plt.plot(t, np.poly1d(poly_coeff)(t), '-')\n", + "plt.show()\n", + "print \"b)The above graph is beam displacement graph\"\n", + "print \"b)The minimum occures in the middle from the above graph \"\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.3 page number 694 " + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The reactive moment at A i.e M_A -0.0714285714286 WL**2\n", + "The reactive force at A i.e R_A -1.14285714286 WL\n" + ] + } + ], + "source": [ + "#Given \n", + "#First we will solve without the reaction at middle\n", + "#Given\n", + "import numpy as np\n", + "l_ab = 1.0 #2L in - The length of the beam\n", + "F_D = 1.0 #W lb/in - The force distribution \n", + "F = F_D*l_ab #WL - The force applied\n", + "#Beause of symmetry the moment caliculations can be neglected\n", + "#F_Y = 0\n", + "R_A = F/2 #wl - The reactive force at A\n", + "R_B = F/2 #wl - The reactive force at B\n", + "#EI - The flxure rigidity is constant and 1/EI =1 # k\n", + "\n", + "#part - A\n", + "#section 1--1\n", + "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.2L distance \n", + "M_1 = [0,0,0,0,0,0,0,0,0,0,0]\n", + "v = [0,0,0,0,0,0,0,0,0,0,0]\n", + "for i in range(10):\n", + " v[i] = R_A - F_D*l_1[i] \n", + " M_1[i] = R_A*l_1[i] - F_D*(l_1[i]**2)/2\n", + "# (EI)y'' = M_1[i] we will integrate M_1[i] twice where variable is l_1[i]\n", + "#(EI)y'- \n", + "\n", + "M_1_intg1 = R_A*(l_1[i]**2)/4 - F_D*(l_1[i]**3)/6 - F_D*(l_ab**3)*l_1[i]/24 #integration of x**n = x**n+1/n+1\n", + "#(EI)y- Using end conditions for caliculating constants \n", + "\n", + "M_1_intg2 = R_A*(l_1[i]**3)/12.0 - F_D*(l_1[i]**4)/24.0 + F_D*(l_ab**3)*l_1[i]/24.0 \n", + "#Equations \n", + "\n", + "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.2L distance \n", + "M_1_intg2 = [0,0,0,0,0,0,0,0,0,0,0]\n", + "Y = [0,0,0,0,0,0,0,0,0,0,0]\n", + "for i in range(10):\n", + " M_1_intg2[i] = (l_1[i]**3)/12.0 - (l_1[i]**4)/24.0 - l_1[i]/24.0 # discluding every term for ruling out float values\n", + " Y[i] = M_1_intg2[i] #W(l**4)/EI k = 1/EI\n", + "e_1 = 16*min(Y) #WL4/EI - The maximum defection \n", + "e_2 = - F_D*((2*l_ab)**3)/24.0 #WL3/EI - The maximum angle\n", + "#Caliculating for momentum and force\n", + "f_ab = ((2*l_ab)**2)/16.0 #L2/EI \n", + "f_bb = ((2*l_ab)**3)/48.0 #L3/EI \n", + "f_aa = 2*l_ab/3.0 #L/EI\n", + "f_ba = ((l_ab)**2)/4.0 #L2/EI\n", + "#F*X = e - Matrix multiplication \n", + "#Solving for X\n", + "a = np.array([[f_aa,f_ba], [f_ba,f_bb]])\n", + "b = np.array([e_2,e_1])\n", + "x = np.linalg.solve(a, b)\n", + "print \"The reactive moment at A i.e M_A\",x[0],\"WL**2\"\n", + "print \"The reactive force at A i.e R_A\",x[1],\"WL\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter2_5.ipynb b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter2_5.ipynb new file mode 100644 index 00000000..0c496644 --- /dev/null +++ b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter2_5.ipynb @@ -0,0 +1,325 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2:Axial strains and Deformations in bars " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.1 page number 77" + ] + }, + { + "cell_type": "code", + "execution_count": 77, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total deflection is : 1.0 mm\n" + ] + } + ], + "source": [ + "l_ob = 2000 #mm - length of rod ob\n", + "l_bc = 1000 #mm - length of rod bc\n", + "l_cd = 1500 #mm - length of rod cd\n", + "p_ob = 100 #kN - Force in rods \n", + "p_bc = -150 #KN\n", + "p_cd = 50 #KN \n", + "A_ob = 1000 #mm2 - Area of rod ob\n", + "A_bc = 2000 #mm2 - Area of rod bc \n", + "A_cd = 1000 #mm2 - Area of rod cd\n", + "E = 200.0 #GPA \n", + "# the total deflection is algebraic sums of `deflection in each module \n", + "e_1 = p_ob*l_ob/(A_ob*E)\n", + "e_2 = p_bc*l_bc/(A_bc*E)\n", + "e_3 = p_cd*l_cd/(A_cd*E)\n", + "#All units are satisfied \n", + "e_total = e_1+ e_2 + e_3\n", + "print \"The total deflection is :\",round(e_total,3) ,\"mm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.4 page number 80" + ] + }, + { + "cell_type": "code", + "execution_count": 36, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "0.0112677358491\n", + "The vertical stiffness of the combination is 113.14 kips/inch\n" + ] + } + ], + "source": [ + "p_app = 3 #Kips - applied force \n", + "P_A = 2.23 #Kips \n", + "p_B = -2.83 #kips - compressive force\n", + "l_ab = 6.71 #inch\n", + "l_bc = 8.29 #inch\n", + "s_ab = 17.8 #ksi - tensile stress\n", + "s_bc = -12.9 #ksi - compressive stress\n", + "E = 10.6 * pow(10,3) #ksi -youngs modulus \n", + "e_ab = s_ab*l_ab/E\n", + "\n", + "e_bc = s_bc*l_bc/E\n", + "x = e_ab/e_bc #the Ratio of cosines of the deflected angles \n", + "# t_1 and t_2 be deflected angles \n", + "#t_2 = 180-45-26.6-t_1 the sum of angles is 360\n", + "#t_1 = 52.2 degress\n", + "import math\n", + "e = e_ab/math.acos(math.radians(52.2)) #inch\n", + "k = p_app/e # kips/in vertical stiffness of the combination\n", + "print \"The vertical stiffness of the combination is\",round(k,3),\"kips/inch\"\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.6 page number 83" + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The lateral strain is: 0.00243 no units\n", + "The longitudinal strain is: 0.00073 no units\n", + "The poissions ratio is: 3.32876712329 no units\n", + "Youngs modulus: 69.8 N/mm2\n" + ] + } + ], + "source": [ + "dia = 50 #mm - diameter of aluminium \n", + "p = 100 # KN - instant force applid\n", + "dia_c = 0.1215 #mm- change in diameter \n", + "l_c = 0.219 #mm - change in length\n", + "l = 300 #mm - length \n", + "strain_dia = dia_c/dia # lateral strain \n", + "strain_l = l_c/l #longitudinal strain \n", + "po = strain_dia/strain_l # poission ratio \n", + "area = 3.14*dia*dia/4 #mm2 area\n", + "E = p*l/(area*l_c) #N/mm2 youngs modulus \n", + "print \"The lateral strain is:\",strain_dia,\"no units\"\n", + "print \"The longitudinal strain is:\",strain_l,\"no units\"\n", + "print \"The poissions ratio is:\",po,\"no units\"\n", + "print \"Youngs modulus:\",round(E,2),\"N/mm2\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.7 page number 86" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The displacement in point B is : 0.00795578950395 in\n" + ] + } + ], + "source": [ + "T = 12.9*pow(10,-6) #/F\n", + "t = 100.00 # F \n", + "l_ab = 6.71 #inch\n", + "l_bc = 8.29 #inch\n", + "e_ab = T*t*l_ab #in-elongation \n", + "e_bc = T*t*l_bc #in-elongation\n", + "k = e_ab/e_bc # ratio of cosines of deflected angles \n", + "# t_1 and t_2 be deflected angles \n", + "#t_2 = 180-45-26.6-t_1 the sum of angles is 360\n", + "t_1 = 26.6\n", + "import math\n", + "e = e_ab/math.acos(math.radians(26.6))\n", + "print \"The displacement in point B is :\",e ,\"in\"\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.11 page number " + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum stress is: 3.59377281766 Mpa\n" + ] + } + ], + "source": [ + "mass = 4 #kg \n", + "dist = 1 #mt freely falling distance\n", + "l = 1500 #mm length of rod\n", + "d = 15 #mm diameter\n", + "l_ab = 6.71 #inch\n", + "l_bc = 8.29 #inch\n", + "E = 200 #GPA youngs modulus \n", + "k = 4.5 # N/mm stiffness costant\n", + "F = mass*9.81# The force applying\n", + "Area = 3.14*(d**2)/4 \n", + "# Two cases \n", + "#youngs modulus \n", + "e_y = F*l/(Area*E*pow(10,3))\n", + "# stiffness\n", + "e_f = F/k \n", + "#total\n", + "e = e_y +e_f\n", + "k = 1+(2/(e*pow(10,-3)))\n", + "stress_max_1 = F*(1+pow(k,0.5))/Area\n", + "print \"The maximum stress is:\",stress_max_1,\"Mpa\"\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.12 page number 103" + ] + }, + { + "cell_type": "code", + "execution_count": 75, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The reactions at bottom is -1.25 p\n" + ] + } + ], + "source": [ + "flex_a = 1#f\n", + "flex_b = 2#f\n", + "#removing lower support and solving FBD\n", + "e = -2 -(2+1)#fp\n", + "#e_1 = (2+1+1)*R\n", + "#e_1 = -e Making the elongations zero since the both ends are fixed\n", + "R = e/(2+1+1.0) #p\n", + "print \"The reactions at bottom is\",R,\"p\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.19 page number 113" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The end deflection is 0.12 in\n" + ] + } + ], + "source": [ + "#Given \n", + "l = 30 #in - The length of the rod\n", + "p_1 = 80 #kips - The Force on the end\n", + "p_2 = 125 #kips - The force on the other end\n", + "A_s = 0.5 #in2 - The crossection of the steel rod\n", + "A_a = 0.5 #in2 - The crossection of the aluminium \n", + "E_a = 10*(10**6) #psi - The youngs modulus of the aluminium \n", + "E_s = 30*(10**6) #psi - The youngs modulus of the steel\n", + "#Internally stastically indeterminant \n", + "p_a = p_1/4 #From solving we get p_s = 3*P_a\n", + "#From material properties point of view \n", + "#stress_steel = stress_aluminium\n", + "e = p_a*l*(10**3)/(A_a*E_a) #The end deflection \n", + "print \"The end deflection is\",e,\"in\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter4_5.ipynb b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter4_5.ipynb new file mode 100644 index 00000000..f687074e --- /dev/null +++ b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter4_5.ipynb @@ -0,0 +1,439 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4:Torsion" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.2 page number 183" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The maximum shear due to torsion is 152.87 Mpa\n", + "stress tensor matrix [[ 0. 152.9 0. ]\n", + " [ 152.9 0. 0. ]\n", + " [ 0. 0. 0. ]]\n" + ] + } + ], + "source": [ + "#Given\n", + "dia = 10 #diameter of shaft(A-C)\n", + "c = dia/2 #mm - Radius\n", + "T = 30 #N/mm -Torque in the shaft \n", + "#Caliculations\n", + "\n", + "J = 3.14*(dia**4)/32 #mm4\n", + "shear_T = T*c*pow(10,3)/J # The torsion shear in the shaft AC\n", + "import numpy as np \n", + "print \"The maximum shear due to torsion is \",round(shear_T,2),\"Mpa\"\n", + "arr_T = np.zeros((3,3))\n", + "arr_T[0][1]=round(shear_T,1) #arranging the elements in array\n", + "arr_T[1][0]=round(shear_T,1)\n", + "print \"stress tensor matrix\",(arr_T),\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.3 page number 184" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum shear due to torsion is 43.15 Mpa\n", + "The minimum shear due to torsion is 34.52 Mpa\n" + ] + } + ], + "source": [ + "#Given \n", + "dia_out = 20 #mm- outer diameter of shaft\n", + "dia_in = 16 #mm- inner diameter of shaft \n", + "c_out = dia_out/2 #mm - outer Radius of shaft \n", + "c_in = dia_in/2 #mm - inner radius of shaft \n", + "T = 40 #N/mm -Torque in the shaft \n", + "#caliculations\n", + "\n", + "J = 3.14*((dia_out**4)- (dia_in**4))/32 #mm4\n", + "shear_T_max = T*c_out*pow(10,3)/J # The maximum torsion shear in the shaft\n", + "shear_T_min = T*c_in*pow(10,3)/J # The maximum torsion shear in the shaft\n", + "print \"The maximum shear due to torsion is \",round(shear_T_max,2),\"Mpa\"\n", + "print \"The minimum shear due to torsion is \",round(shear_T_min,2),\"Mpa\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.4 page number 187" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Diameter of the shaft used is 15.26 mm\n" + ] + } + ], + "source": [ + "#Given\n", + "hp = 10 # horse power of motor \n", + "f = 30 # given \n", + "shear_T = 55 #Mpa - The maximum shearing in the shaft \n", + "#caliculations\n", + "\n", + "T = 119*hp/f # N.m The torsion in the shaft \n", + "#j/c=T/shear_T=K\n", + "k = T*pow(10,3)/shear_T #mm3\n", + "#c3=2K/3.14\n", + "c = pow((2*k/3),0.33) #mm - The radius of the shaft \n", + "diamter = 2*c #mm - The diameter of the shaft\n", + "print \"The Diameter of the shaft used is\",round(diamter,2),\"mm\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.5 page number 188" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Diameter of the shaft1 is 6.87 mm\n", + "The Diameter of the shaft2 is 0.702590481015 mm\n" + ] + } + ], + "source": [ + "#Given \n", + "hp = 200 #Horse power\n", + "stress_sh = 10000 #psi- shear stress\n", + "rpm_1 = 20.0 # The rpm at which this shaft1 operates \n", + "rpm_2 = 20000.0 # The rpm at which this shaft2 operates\n", + "T_1= hp*63000.0/rpm_1 #in-lb Torsion due to rpm1\n", + "T_2= hp*63000/rpm_2 #in-lb Torsion due to rpm1\n", + "#caliculations \n", + "\n", + "#j/c=T/shear_T=K\n", + "k_1= T_1/stress_sh #mm3\n", + "#c3=2K/3.14\n", + "c_1= pow((2*k_1/3),0.33) #mm - The radius of the shaft \n", + "diamter_1 = 2*c_1 #mm - The diameter of the shaft\n", + "print \"The Diameter of the shaft1 is\",round(diamter_1,2),\"mm\"\n", + "\n", + "#j/c=T/shear_T=K\n", + "k_2= T_2/stress_sh #mm3\n", + "#c3=2K/3.14\n", + "c_2= pow((2*k_2/3),0.33) #mm - The radius of the shaft \n", + "diamter_2 = 2*c_2 #mm - The diameter of the shaft\n", + "print \"The Diameter of the shaft2 is\",diamter_2,\"mm\"\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.7 page number 193" + ] + }, + { + "cell_type": "code", + "execution_count": 37, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum angle rotated is 0.0232628450106 radians \n" + ] + } + ], + "source": [ + "#Given \n", + "T_ab = 0 #N.m - torsion in AB \n", + "T_bc = 150 #N.m - torsion in BC\n", + "T_cd = 150 #N.m - torsion in CD\n", + "T_de = 1150 #N.m - torsion in DE\n", + "l_ab = 250 #mm - length of AB\n", + "l_bc = 200 #mm - length of BC\n", + "l_cd = 300 #mm - length of cd \n", + "l_de = 500.0 #mm - length of de\n", + "d_1 = 25 #mm - outer diameter \n", + "d_2 = 50 #mm - inner diameter\n", + "G = 80 #Gpa -shear modulus\n", + "#Caliculations \n", + "\n", + "J_ab = 3.14*(d_1**4)/32 #mm4\n", + "J_bc = 3.14*(d_1**4)/32 #mm4\n", + "J_cd = 3.14*(d_2**4 - d_1**4)/32 #mm4\n", + "J_de = 3.14*(d_2**4 - d_1**4)/32 #mm4\n", + "rad = T_ab*l_ab/(J_ab*G)+ T_bc*l_bc/(J_bc*G)+ T_cd*l_cd/(J_cd*G)+ T_de*l_de/(J_de*G) # adding the maximum radians roteted in each module\n", + "print \"The maximum angle rotated is \",rad,\"radians \" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.9 Pagenumber 196" + ] + }, + { + "cell_type": "code", + "execution_count": 39, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Torsion at rigid end A is -141.72 N-m\n", + "The Torsion at rigid end B is 1291.72 N-m\n" + ] + } + ], + "source": [ + "#given \n", + "#its a statistally indeterminant \n", + "#we will take of one of the support \n", + "#Given \n", + "T_ab = 0 #N.m - torsion in AB \n", + "T_bc = 150 #N.m - torsion in BC\n", + "T_cd = 150 #N.m - torsion in CD\n", + "T_de = 1150 #N.m - torsion in DE\n", + "l_ab = 250 #mm - length of AB\n", + "l_bc = 200 #mm - length of BC\n", + "l_cd = 300 #mm - length of cd \n", + "l_de = 500.0#mm - length of de\n", + "d_1 = 25 #mm - outer diameter \n", + "d_2 = 50 #mm - inner diameter\n", + "#Caliculations \n", + "\n", + "J_ab = 3.14*(d_1**4)/32 #mm4\n", + "J_bc = 3.14*(d_1**4)/32 #mm4\n", + "J_cd = 3.14*(d_2**4 - d_1**4)/32 #mm4\n", + "J_de = 3.14*(d_2**4 - d_1**4)/32 #mm4\n", + "G = 80 #Gpa -shear modulus\n", + "rad = T_ab*l_ab/(J_ab*G)+ T_bc*l_bc/(J_bc*G)+ T_cd*l_cd/(J_cd*G)+ T_de*l_de/(J_de*G) \n", + "#now lets consider T_A then the torsion is only T_A\n", + "# T_A*(l_ab/(J_ab*G)+ l_bc/(J_bc*G)+ l_cd/(J_cd*G)+ l_de/(J_de*G)) +rad = 0\n", + "# since there will be no displacement \n", + "T_A =-rad/(l_ab/(J_ab*G)+ l_bc/(J_bc*G)+ l_cd/(J_cd*G)+ l_de/(J_de*G)) #Torsion at A\n", + "T_B = 1150 - T_A #n-m F_X = 0 torsion at B\n", + "print \"The Torsion at rigid end A is\",round(T_A,2),\"N-m\"\n", + "print \"The Torsion at rigid end B is\",round(T_B,2),\"N-m\"\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.12 Pagenumber 202" + ] + }, + { + "cell_type": "code", + "execution_count": 49, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The allowable torsion on the 8 bolt combination 27129600.0 N-m\n" + ] + } + ], + "source": [ + "#Given\n", + "dai_bc = 240 #mm- daimeter of '8'bolt circle \n", + "dia = dai_bc/8 #Diameter of each bolt\n", + "A = 0.25*(dia**2)*3.14 # Area of a bolt\n", + "S_allow = 40 #Mpa - The maximum allowable allowable shear stress \n", + "P_max = (S_allow)*A #N - The maximum allowable force \n", + "D = 120.0 #mm - the distance from central axis \n", + "T_allow =P_max*D*8 #N-m The allowable torsion on the 8 bolt combination \n", + "print \"The allowable torsion on the 8 bolt combination\",T_allow ,\"N-m\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.15 page number 211" + ] + }, + { + "cell_type": "code", + "execution_count": 57, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the Equivalent Torsion constant is 1.97 in4\n" + ] + } + ], + "source": [ + "#Given \n", + "#AISC MANUALS\n", + "#approximated by three narrow tubes \n", + "#J = Bbt^3\n", + "B = 0.33 # constant mentiones in AISC\n", + "#three rods \n", + "\n", + "#rod_1\n", + "t_1 = 0.605 #inch - Thickness \n", + "b = 12.0 #inches - width \n", + "J_1 = B*b*(t_1**3) #in4 - Torsion constant \n", + "\n", + "#rod_2\n", + "t_2 = 0.605 #inch - Thickness \n", + "b = 12 #inches - width \n", + "J_2 = B*b*(t_2**3) #in4 - Torsion constant \n", + "\n", + "#rod_3\n", + "t_3 = 0.390 #inch - Thickness \n", + "b = 10.91 #inches - width \n", + "J_3 = B*b*(t_3**3) #in4 - Torsion constant \n", + "\n", + "#Equivalent\n", + "J_eq = J_1+J_2+J_3 #in4 - Torsion constant \n", + "print \"the Equivalent Torsion constant is \",round(J_eq,2), \"in4\"\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.16 page number 214" + ] + }, + { + "cell_type": "code", + "execution_count": 58, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum shear due to torsion is 345.23 Mpa\n", + "The minimum shear due to torsion is 276.18 Mpa\n" + ] + } + ], + "source": [ + "#Given \n", + "dia_out = 10 #mm- outer diameter of shaft\n", + "dia_in = 8 #mm- inner diameter of shaft \n", + "c_out = dia_out/2 #mm - outer Radius of shaft \n", + "c_in = dia_in/2 #mm - inner radius of shaft \n", + "T = 40 #N/mm -Torque in the shaft \n", + "#caliculations\n", + "\n", + "J = 3.14*((dia_out**4)- (dia_in**4))/32 #mm4\n", + "shear_T_max = T*c_out*pow(10,3)/J # The maximum torsion shear in the shaft\n", + "shear_T_min = T*c_in*pow(10,3)/J # The maximum torsion shear in the shaft\n", + "print \"The maximum shear due to torsion is \",round(shear_T_max,2),\"Mpa\"\n", + "print \"The minimum shear due to torsion is \",round(shear_T_min,2),\"Mpa\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter5_5.ipynb b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter5_5.ipynb new file mode 100644 index 00000000..d6f0670f --- /dev/null +++ b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter5_5.ipynb @@ -0,0 +1,865 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Chapter 5:Axial force, Shear and Bending moment " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example5.2 pagenumber 231" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The X,Y components of reaction force at A is 0 , -410.0 N\n", + "The X,Y components of reaction force at B is 0 , 670.0 N\n" + ] + } + ], + "source": [ + "#Given \n", + "L_ab = 0.4 #mt The total length of the rod\n", + "M = 200 #N_m - the moment acting on rod\n", + "l_1 = 0.1 #mt -moment acting point the distance from 'a'\n", + "R_1 = 100 #N - The Force acting \n", + "l_2 = 0.2 #mt -R_1 acting point the distance from 'a'\n", + "R_2 = 160 #N The Force acting \n", + "l_3 = 0.3 #mt -R_2 acting point the distance from 'a'\n", + "#caliculations\n", + "\n", + "#F_X = 0 forces in x directions \n", + "R_A_X = 0 # since there are no forces in X-direction \n", + "R_B_X = 0\n", + "#M_A = 0 momentum at point a is zero\n", + "\n", + "# M + R_1*l_2 + R_2*l_3 = R_B*L_ab *the moment for a force is FxL\n", + "R_B_Y = (M + R_1*l_2 + R_2*l_3)/L_ab\n", + "\n", + "#M_B= 0 momentum at point b is zero\n", + "# R_A_Y*L_ab + M - R_1*l_2 - R_2*0.1 = 0\n", + "\n", + "R_A_Y = -(M - R_1*l_2 - R_2*0.1)/L_ab\n", + " \n", + "print \"The X,Y components of reaction force at A is \",R_A_X,\",\",R_A_Y,\"N\"\n", + "print \"The X,Y components of reaction force at B is \",R_B_X,\",\",R_B_Y,\"N\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.2 page number 233" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The X,Y components of reaction force at A is 0 , -9.0 N\n", + "The X,Y components of reaction force at B is 0 , 6.0 N\n" + ] + } + ], + "source": [ + "#Given \n", + "P_Max = 10 #N - the maximum distribution in a triangular distribution\n", + "L = 3 #mt the total length of force distribution \n", + "L_X = 5 #mt - the horizantal length of the rod\n", + "#caliculations \n", + "\n", + "F_y = P_Max*L*0.5 #N - The force due to triangular distribition \n", + "L_com = 2*L /3 #mt - the resultant force acting as a result of distribution acting position \n", + "#F_X = 0 forces in x directions\n", + "R_A_X = 0 # since there are no forces in X-direction\n", + "R_B_X = 0\n", + "#M_A = 0 momentum at point a is zero\n", + "#F_y*L_com - R_B_Y*L_X = 0\n", + "R_B_Y = F_y*L_com/L_X\n", + "\n", + "#M_B= 0 momentum at point b is zero\n", + "#- R_A_Y*L_X = F_y*(L_X-L )\n", + "\n", + "R_A_Y = - F_y*L/L_X \n", + "print \"The X,Y components of reaction force at A is \",R_A_X,\",\",R_A_Y,\"N\"\n", + "print \"The X,Y components of reaction force at B is \",R_B_X,\",\",R_B_Y,\"N\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.3 page number 233 " + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The X,Y components and resultant of reaction force at A is 4 , 3 , 5.0 N\n", + "The X,Y components and resultant of reaction force at B is 1 , 1 , 1.41 N\n" + ] + } + ], + "source": [ + "#given\n", + "F = 5 #K - force acting on the system\n", + "tan = (4/3) # the Tan of the angle of force with x axis\n", + "l_ab = 12 #inch - the total length of ab \n", + "l = 3 # inch - Distance from 'a'\n", + "#caliculation\n", + "F_X = 4 #K\n", + "F_Y = 3 #k\n", + "\n", + "#M_A = 0 momentum at point a is zero\n", + "# F_X*l- R_B_Y*l_ab = 0 \n", + "R_B_Y = F_X*l/l_ab\n", + "\n", + "#M_B= 0 momentum at point b is zero\n", + "# R_A_Y*l_ab - F_X*(l_ab - l)\n", + "R_A_Y = F_X*(l_ab - l)/l_ab\n", + " \n", + "#F_X = 0 forces in x directions\n", + "R_A_X = F_Y + R_B_Y \n", + "R_B_X = R_B_Y # since the angle is 45 degrees\n", + "\n", + "#resultants \n", + "R_A = pow(R_A_X**2 + R_A_Y**2,0.5)\n", + "R_B = pow(R_B_X**2 + R_B_Y**2,0.5)\n", + "\n", + "print \"The X,Y components and resultant of reaction force at A is \",R_A_X,\",\",R_A_Y,\",\",R_A,\"N\"\n", + "print \"The X,Y components and resultant of reaction force at B is \",R_B_X,\",\",R_B_Y,\",\",round(R_B,2),\"N\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.4 page number 239" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The force and moment in section a--a are -2.33 KN , -13.644 KN-m\n", + "The force and moment in section b--b are 6.0 KN , 6.0 KN-m\n" + ] + } + ], + "source": [ + "#Given \n", + "P_Max = 10 #N - the maximum distribution in a triangular distribution\n", + "L = 3 #mt the total length of force distribution \n", + "L_X = 5 #mt - the horizantal length of the rod\n", + "#caliculations \n", + "\n", + "F_y = P_Max*L*0.5 #N - The force due to triangular distribition \n", + "L_com = 2*L /3 #mt - the resultant force acting as a result of distribution acting position \n", + "#F_X = 0 forces in x directions\n", + "R_A_X = 0 # since there are no forces in X-direction\n", + "R_B_X = 0\n", + "#M_A = 0 momentum at point a is zero\n", + "#F_y*L_com - R_B_Y*L_X = 0\n", + "R_B_Y = F_y*L_com/L_X\n", + "\n", + "#M_B= 0 momentum at point b is zero\n", + "#- R_A_Y*L_X = F_y*(L_X-L )\n", + "\n", + "R_A_Y = - F_y*L/L_X\n", + "\n", + "#For a---a section \n", + "l_a = 2 #mt - a---a section from a \n", + "l_com_a = 2*l_a/3\n", + "v_a = R_A_Y + 0.5*l_a*(10.0*2/3) #*(10*2/3) because the maximum moves\n", + "\n", + "M_a = (10.0*0.66)*l_a*(0.33) + R_A_Y*l_a \n", + "\n", + "#For b---b section \n", + "\n", + "v_b = F_y + R_A_Y #equilabrium conditions\n", + "M_b = (F_y + R_A_Y)*(-1)\n", + "\n", + "print \"The force and moment in section a--a are\",round(v_a,2),\"KN ,\",M_a,\"KN-m\"\n", + "print \"The force and moment in section b--b are\",v_b,\"KN ,\",M_b,\"KN-m\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.5 page number 241" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Given problem is for drawing diagram, this diagram is drawn by step by step manner. \n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x41889b0>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#Given \n", + "#Lets divide the section into two sections \n", + "l_ac = 10 # ft -The total length of the rod\n", + "R = 5 #k - The applies force at c\n", + "tan = 4/3 # The tan of the angle of the force \n", + "l_ab = 5 #ft - The distance of applied force from A\n", + "R_y = 4 #k,downwards X- component of the force\n", + "R_X = 3 #k Y- component of the force , lets consider only Y direction since we are concentrating on the Shears \n", + "\n", + "#F_Y = 0 forces in Y directions\n", + "#R_A +R_B = R_y\n", + "#M_c = 0 making the moment zero at point c \n", + "#Caliculations \n", + "# R_A= R_B*(l_ac - l_ab)/(l_ab) so R_A = R_B\n", + "\n", + "R_A = R_y/2 #F_Y = 0\n", + "R_B = R_y/2\n", + "#considering section x--x\n", + "l_x = 2 #ft - length of section from A\n", + "v_x = R_A #k ,F_X = 0 \n", + "M_x = R_A*l_x #k-ft M_c = 0\n", + "\n", + "#considering section at midpoint t--t\n", + "l_t = 2 #ft - length of section from A\n", + "v_t = 0 #k ,F_X = 0 \n", + "M_t = (R_A)*l_t #k-ft M_c = 0\n", + "\n", + "##considering section y---y\n", + "l_y = 2 #ft - length of section from B\n", + "v_y = - R_B #k ,F_X = 0 \n", + "M_y = R_B*l_y #k-ft M_c = 0\n", + "\n", + "#Graph\n", + "%matplotlib inline\n", + "import math \n", + "from matplotlib.pyplot import plot,suptitle,xlabel,ylabel\n", + "#Drawing of shear and bending moment diagram\n", + "print \"Given problem is for drawing diagram, this diagram is drawn by step by step manner. \"\n", + "X = [0,2,4.9999999999999,5,5.00000000000000001,7,10] # For graph precision \n", + "\n", + "V = [R_A,v_x,v_x,v_t,v_y,v_y,-R_B];\t\t\t#Shear matrix\n", + "M = [0,M_x,M_t,M_t,M_t,M_y,0];\t\t\t#Bending moment matrix\n", + "plot(X,V);\t\t\t#Shear diagram\n", + "plot(X,M,'r');\t\t\t#Bending moment diagram\n", + "suptitle( 'Shear and bending moment diagram')\n", + "xlabel('X axis')\n", + "ylabel( 'Y axis') ;" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.6 pagenumber 243 " + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Given problem is for drawing diagram, this diagram is drawn by step by step manner. \n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x8b0bef0>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#Given\n", + "l = 1 #L - Length of the cantilever \n", + "F_app = ((2**0.5))*2 #p - force applies \n", + "tan = 1 # The angle of force applied\n", + "F_app_x = F_app/((2**0.5)) #p The horizantal component of the force , neglected \n", + "F_app_y = F_app/((2**0.5)) #p The Vertical component of the force \n", + "#F_Y = 0 \n", + "R_A = 1 #p\n", + "\n", + "#Considering section 1-----1\n", + "l_1 = 0.5 # The length of the section from one end\n", + "v_1 = R_A #F_Y = 0\n", + "M_1 = -R_A*l_1 #MAking moment at section 1 = 0\n", + "\n", + "#considering end of cantilever\n", + "l_2 = 1 # The length of the section from one end\n", + "v_2 = R_A #F_Y = 0\n", + "M_2 = -R_A*l_2#MAking moment at section 1 = 0\n", + "\n", + "#Graph\n", + "%matplotlib inline\n", + "import math \n", + "from matplotlib.pyplot import plot,suptitle,xlabel,ylabel\n", + "#Drawing of shear and bending moment diagram\n", + "print \"Given problem is for drawing diagram, this diagram is drawn by step by step manner. \"\n", + "X = [0,0.5,1] # For graph precision \n", + "\n", + "V = [R_A,v_1,v_2 ];\t\t\t#Shear matrix\n", + "M = [0,M_1,M_2];\t\t\t#Bending moment matrix\n", + "plot(X,V);\t\t\t #Shear diagram\n", + "plot(X,M,'r');\t\t\t #Bending moment diagram\n", + "suptitle( 'Shear and bending moment diagram')\n", + "xlabel('X axis')\n", + "ylabel( 'Y axis') ;\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.7 page number 243" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Given problem is for drawing diagram, this diagram is drawn by step by step manner. \n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x414ff60>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#Given\n", + "l_ab = 3 #L - The total length lets say '3L'\n", + "R_1 = 1 #p - The force applied at b\n", + "R_2 = 1 #p - The force applied at c\n", + "l_ab = 1 #L\n", + "l_bc = 1 #L \n", + "\n", + "#Logical step \n", + "#Since the system is in symmetry we can avoid moment M = 0 caliculations\n", + "\n", + "#F_Y = 0 \n", + "R_A = (R_1 + R_2)/2\n", + "R_B = (R_1 + R_2)/2\n", + "\n", + "#Lets take '3' sections \n", + "#Considering section 1-----1 at 0.5L\n", + "l_1 = 0.5 #L - distance of the section from the A\n", + "v_1 = R_A #F_Y = 0 \n", + "M_1 = R_A*l_1 #MAking moment at section 1 = 0\n", + "\n", + "#Considering section 2-----2 at 1L\n", + "l_2 = 1 #L - distance of the section from the A\n", + "v_2 = R_A #F_Y = 0 \n", + "M_2 = R_A*l_2 #MAking moment at section 2 = 0\n", + "\n", + "#Considering section 3-----3 at 1.5L\n", + "l_3 = 1.5 #L - distance of the section from the A\n", + "v_3 = 0 #F_Y = 0 \n", + "M_3 = R_A*l_2 #MAking moment at section 2 = 0 and symmetry \n", + "\n", + "#GRAPH\n", + "#Since the symmetry exists the graphs are also symmetry\n", + "%matplotlib inline\n", + "import math \n", + "from matplotlib.pyplot import plot,suptitle,xlabel,ylabel\n", + "#Drawing of shear and bending moment diagram\n", + "print \"Given problem is for drawing diagram, this diagram is drawn by step by step manner. \"\n", + "X = [0,0.5,1,1.000000001,1.5,1.999999999999,2,2.5,3] # For graph precision \n", + "\n", + "V = [R_A,v_1,v_2,v_3,v_3,v_3,-v_2,-v_1,-R_B];\t\t\t#Shear matrix\n", + "M = [0,M_1,M_2,1,1,1,M_2,M_1,0];\t\t\t#Bending moment matrix\n", + "plot(X,V);\t\t\t#Shear diagram\n", + "plot(X,M);\t\t\t#Bending moment diagram\n", + "suptitle( 'Shear and bending moment diagram')\n", + "xlabel('X axis')\n", + "ylabel( 'Y axis') ;" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.8 page nmber 244" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "0.5\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0xab25c50>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#Given \n", + "import numpy as np\n", + "l_ab = 1.0 #L - The length of the beam\n", + "F_D = 1.0 #W - The force distribution \n", + "F = F_D*l_ab #WL - The force applied\n", + "#Beause of symmetry the moment caliculations can be neglected\n", + "#F_Y = 0\n", + "R_A = F/2 #wl - The reactive force at A\n", + "R_B = F/2 #wl - The reactive force at B\n", + "\n", + "#considering many sections \n", + "\n", + "#section 1--1\n", + "l_1 = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.1L distance \n", + "M_1 = [0,0,0,0,0,0,0,0,0,0,0]\n", + "v = [0,0,0,0,0,0,0,0,0,0,0]\n", + "for i in range(10):\n", + " v[i] = R_A - F_D*l_1[i] \n", + " M_1[i] = R_A*l_1[i] - F_D*(l_1[i]**2)/2 #M = 0 in the section\n", + "print R_A\n", + "#Graphs\n", + "import numpy as np\n", + "values = [0.5,0,-0.5]\n", + "y = np.array(values)\n", + "t = np.linspace(0,1,3)\n", + "poly_coeff = np.polyfit(t, y, 2)\n", + "import matplotlib.pyplot as plt\n", + "plt.plot(t, y, 'o')\n", + "plt.plot(t, np.poly1d(poly_coeff)(t), '-')\n", + "plt.show()\n", + "\n", + "import numpy as np\n", + "values = M_1\n", + "y = np.array(values)\n", + "t = np.linspace(0,1,11)\n", + "poly_coeff = np.polyfit(t, y, 2)\n", + "import matplotlib.pyplot as plt\n", + "plt.plot(t, y, 'o')\n", + "plt.plot(t, np.poly1d(poly_coeff)(t), '-')\n", + "plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.9 page number 245" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The force and moment in section a--a are -2.33 KN , -13.644 KN-m\n" + ] + } + ], + "source": [ + "#Given \n", + "P_Max = 10 #N - the maximum distribution in a triangular distribution\n", + "L = 3 #mt the total length of force distribution \n", + "L_X = 5 #mt - the horizantal length of the rod\n", + "#caliculations \n", + "\n", + "F_y = P_Max*L*0.5 #N - The force due to triangular distribition \n", + "L_com = 2*L /3 #mt - the resultant force acting as a result of distribution acting position \n", + "#F_X = 0 forces in x directions\n", + "R_A_X = 0 # since there are no forces in X-direction\n", + "R_B_X = 0\n", + "#M_A = 0 momentum at point a is zero\n", + "#F_y*L_com - R_B_Y*L_X = 0\n", + "R_B_Y = F_y*L_com/L_X\n", + "\n", + "#M_B= 0 momentum at point b is zero\n", + "#- R_A_Y*L_X = F_y*(L_X-L )\n", + "\n", + "R_A_Y = - F_y*L/L_X\n", + "\n", + "#caliculating for some random value\n", + "#For a---a section \n", + "l_a = 2 #mt - a---a section from a \n", + "l_com_a = 2*l_a/3\n", + "v_a = R_A_Y + 0.5*l_a*(10.0*2/3) #*(10*2/3) because the maximum moves\n", + "\n", + "M_a = (10.0*0.66)*l_a*(0.33) + R_A_Y*l_a\n", + "\n", + "print \"The force and moment in section a--a are\",round(v_a,2),\"KN ,\",M_a,\"KN-m\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.13 page number 254" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Given problem is for drawing diagram, this diagram is drawn by step by step manner. \n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x8f3aa58>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#Given\n", + "l_ab = 4 #L - The total length lets say '3L'\n", + "R_1 = 1 #p - The force applied at b\n", + "R_2 = 1 #p - The force applied at c\n", + "l_ab = 1 #L\n", + "l_bc = 3 #L \n", + "\n", + "#Logical step \n", + "#Since the system is in symmetry we can avoid moment M = 0 caliculations\n", + "\n", + "#F_Y = 0 \n", + "R_A = (R_1 + R_2)/2\n", + "R_B = (R_1 + R_2)/2\n", + "\n", + "#Lets take '3' sections \n", + "#Considering section 1-----1 at 0.5L\n", + "l_1 = 0.5 #L - distance of the section from the A\n", + "v_1 = R_A #F_Y = 0 \n", + "M_1 = R_A*l_1 #MAking moment at section 1 = 0\n", + "\n", + "#Considering section 2-----2 at 1L\n", + "l_2 = 1 #L - distance of the section from the A\n", + "v_2 = R_A #F_Y = 0 \n", + "M_2 = R_A*l_2 #MAking moment at section 2 = 0\n", + "\n", + "#Considering section 3-----3 at 1.5L\n", + "l_3 = 3 #L - distance of the section from the A\n", + "v_3 = 0 #F_Y = 0 \n", + "M_3 = R_A*l_2 #MAking moment at section 2 = 0 and symmetry \n", + "\n", + "#GRAPH\n", + "#Since the symmetry exists the graphs are also symmetry\n", + "%matplotlib inline\n", + "import math \n", + "from matplotlib.pyplot import plot,suptitle,xlabel,ylabel\n", + "#Drawing of shear and bending moment diagram\n", + "print \"Given problem is for drawing diagram, this diagram is drawn by step by step manner. \"\n", + "X = [0,0.5,1,1.0000001,2,2.9999999999,3,3.5,4] # For graph precision \n", + "\n", + "V = [R_A,v_1,v_2,v_3,v_3,v_3,-v_2,-v_1,-R_B];\t\t\t#Shear matrix\n", + "M = [0,M_1,M_2,M_3,M_3,M_3,M_2,M_1,0];\t\t\t#Bending moment matrix\n", + "plot(X,V);\t\t\t#Shear diagram\n", + "plot(X,M);\t\t\t#Bending moment diagram\n", + "suptitle( 'Shear and bending moment diagram')\n", + "xlabel('X axis')\n", + "ylabel( 'Y axis') ;" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.14 page number 255" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0xac8e208>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#Given\n", + "import numpy as np\n", + "import matplotlib.pyplot as plt\n", + "l = 1.0 #l - The length of the beam\n", + "p = 1.0 #W - The total load applied\n", + "#since it is triangular distribution \n", + "l_com = 0.66*l#l - The distance of force of action from one end\n", + "#F_Y = 0\n", + "#R_A + R_B = p\n", + "#M_a = 0 Implies that R_B = 2*R_A\n", + "R_A = p/3.0\n", + "R_B = 2.0*p/3\n", + "\n", + "#Taking Many sections \n", + "\n", + "#Section 1----1\n", + "l = [0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1] #L taking each section at 0.1L distance \n", + "M = [0,0,0,0,0,0,0,0,0,0,0]\n", + "v = [0,0,0,0,0,0,0,0,0,0,0]\n", + "for i in range(10):\n", + " v[i] = p*(l[i]**2) - p/3.0\n", + " M[i] = p*(l[i]**3)/(3.0)- p*l[i]/3.0\n", + "\n", + "v[10] = R_B #again concluded Because the value is tearing of \n", + "\n", + "\n", + "#Graph\n", + "values = M\n", + "y = np.array(values)\n", + "t = np.linspace(0,1,11)\n", + "poly_coeff = np.polyfit(t, y, 2)\n", + "\n", + "plt.plot(t, y, 'o')\n", + "plt.plot(t, np.poly1d(poly_coeff)(t), '-')\n", + "plt.show()\n", + "%matplotlib inline\n", + "values = v\n", + "y = np.array(values)\n", + "t = np.linspace(0,1,11)\n", + "poly_coeff = np.polyfit(t, y, 2)\n", + "\n", + "plt.plot(t, y, 'o')\n", + "plt.plot(t, np.poly1d(poly_coeff)(t), '-')\n", + "plt.show()\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.16 page number 259" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "0.166666666667\n", + "Given problem is for drawing diagram, this diagram is drawn by step by step manner. \n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0xad0f2e8>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#Given\n", + "import math \n", + "from matplotlib.pyplot import plot,suptitle,xlabel,ylabel\n", + "l = 6.0 #a - length of the rod\n", + "F = 1.0 #p - force applies in x direction \n", + "d = 1.0 #a \n", + "M = 1.0 #pa - torque applies on the rod\n", + "l_ab = 4.0 #a application of torque point from A\n", + "#M = 0 implies that\n", + "R_A = F/6.0 #p - The reaction at A\n", + "R_B = - R_A #F_Y = 0\n", + "\n", + "#Caliculations \n", + "\n", + "#Taking sections \n", + "#Section 1---1\n", + "l_1 = 1 #a - the length of the section \n", + "M_1 = - R_A*l_1 #M = 0\n", + "\n", + "#Section 2---2\n", + "l_2 = 4 #a - the length of the section \n", + "M_2 = - R_A*l_2 #M = 0\n", + "\n", + "l_4 = 2 #a - the length of the section \n", + "M_4 = 1/3.0 #pa #M = 0 '-M' because there is moment couple in between\n", + "\n", + "\n", + "#Section 3---3\n", + "l_3 = 1 #a - the length of the section \n", + "M_3 = 1/6.0#pa M = 0 '-M' because there is moment couple in between\n", + "print R_A\n", + "\n", + "#GRAPH\n", + "#Since the symmetry exists the graphs are also symmetry\n", + "%matplotlib inline\n", + "#Drawing of shear and bending moment diagram\n", + "print \"Given problem is for drawing diagram, this diagram is drawn by step by step manner. \"\n", + "X = [0,1,4,4.00001,5,6] # For graph precision \n", + "M = [0,M_1,M_2,M_4,M_3,0];\t\t\t#Bending moment matrix\n", + "plot(X,M);\t\t\t#Bending moment diagram\n", + "suptitle( 'Shear and bending moment diagram')\n", + "xlabel('X axis')\n", + "ylabel( 'Y axis') ;\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter6_5.ipynb b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter6_5.ipynb new file mode 100644 index 00000000..6b74082c --- /dev/null +++ b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter6_5.ipynb @@ -0,0 +1,570 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 6:Pure Bending and Bending with Axial force " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.3 page number 293" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The moment of inertia of total system is 655000.0 mm4\n" + ] + } + ], + "source": [ + "#Given \n", + "#Entire area - hallow area\n", + "l_e = 60.0 #mm - length of the entire area\n", + "b_e = 40 #mm - width of the entire area\n", + "l_h = 30 #mm - length of the hallow area\n", + "b_h = 20 #mm - width of the hallow area\n", + "A_e = l_e*b_e #mm2 - The entire area\n", + "A_h = -l_h*b_h #mm2 - The hallow area '-' because its hallow\n", + "A_re = A_e + A_h #mm2 resultant area\n", + "y_e = l_e/2 # mm com from bottom \n", + "y_h = 20+l_h/2 #mm com from bottom \n", + "y_com = (A_e*y_e + A_h*y_h)/A_re \n", + "#moment of inertia caliculatins - bh3/12 +ad2\n", + "I_e = b_e*(l_e**3)/12 + A_e*((y_e-y_com)**2) #Parallel axis theorm\n", + "I_h = b_h*(l_h**3)/12 - A_h*((y_h-y_com)**2) #Parallel axis theorm\n", + "I_total = I_e - I_h\n", + "print \"The moment of inertia of total system is \",I_total,\"mm4\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.4 page number 295" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum stress at 2 mt is 4.81 Mpa\n" + ] + } + ], + "source": [ + "#Given \n", + "l = 400 #mm - Length \n", + "b = 300 #mm - breath \n", + "F = 20 #KN _ the force applied on the beam \n", + "F_d = 0.75 #KN-m - The force distribution \n", + "d = 2 #mt - the point of interest from the free end\n", + "#caliculations \n", + "#From moment diagram\n", + "M = F*d - F_d*d*1\n", + "I = b*(l**3)/12 #mm4 - Bending moment diagram \n", + "c = l/2 # the stress max at this C\n", + "S = I/c #The maximum shear stress \n", + "shear_max = M*(10**6)/S #MPA - the maximum stress \n", + "print \"The maximum stress at 2 mt is\",round(shear_max,2),\"Mpa\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.5 pagr number 297" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum tensile stress 16.0 Ksi\n", + "The maximum compressive stress 21.6 Ksi\n" + ] + } + ], + "source": [ + "#Given \n", + "#We will divide this into three parts\n", + "F = 8 #k - force applied\n", + "d = 16 #inch -distance\n", + "l_1 = 1 #in \n", + "l_2 = 3 #in \n", + "b_1 = 4 #in \n", + "b_2 = 1 #in\n", + "A_1 = l_1* b_1 #in2 - area of part_1\n", + "y_1 = 0.5 #in com distance from ab\n", + "A_2 =l_2*b_2 #in2 - area of part_1\n", + "y_2 = 2.5 #in com distance from ab\n", + "A_3 = l_2*b_2 #in2 - area of part_1\n", + "y_3 = 2.5 #in com distance from ab\n", + "\n", + "y_net = (A_1*y_1 +A_2*y_2 + A_3*y_3)/(A_1+A_2+A_3) #in - The com of the whole system\n", + "c_max = (4-y_net) #in - The maximum distace from com to end\n", + "c_min = y_net #in - the minimum distance from com to end\n", + "I_1 = b_1*(l_1**3)/12 + A_1*((y_1-y_net)**2) #Parallel axis theorm\n", + "I_2 = b_2*(l_2**3)/12 + A_2*((y_2-y_net)**2)\n", + "I_3 = b_2*(l_2**3)/12 + A_2*((y_2-y_net)**2)\n", + "I_net = I_1 + I_2 + I_3 #in4 - the total moment of inertia\n", + "M_c = F*d*c_max \n", + "stress_cmax = M_c/I_net #Ksi - The maximum compressive stress\n", + "\n", + "M_t= F*d*c_min \n", + "stress_tmax = M_t/I_net #Ksi - The maximum tensile stress\n", + "print \"The maximum tensile stress\",stress_tmax ,\"Ksi\"\n", + "print \"The maximum compressive stress\",round(stress_cmax,1) ,\"Ksi\"\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.8 page number 303" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum stress in steel 11.49 Mpa\n", + "The maximum stress in wood 97.09 Mpa\n" + ] + } + ], + "source": [ + "#Given \n", + "#Given \n", + "#We will divide this into two parts\n", + "E_w = 10.0 #Gpa - Youngs modulus of wood\n", + "E_s = 200.0 #Gpa - Youngs modulus of steel\n", + "M = 30.0 #K.N-m _ applied bending moment \n", + "n = E_s/E_w \n", + "l_1 = 250 #mm \n", + "l_2 = 10 #mm\n", + "b_1 = 150.0 #mm\n", + "b_2 = 150.0*n #mm\n", + "A_1 = l_1* b_1 #mm2 - area of part_1\n", + "y_1 = 125.0 #mm com distance from top\n", + "A_2 =l_2*b_2 #mm2 - area of part_1\n", + "y_2 = 255.0 #mm com distance from top\n", + "y_net = (A_1*y_1 +A_2*y_2)/(A_1+A_2) #mm - The com of the whole system from top\n", + "I_1 = b_1*(l_1**3)/12.0 + A_1*((y_1-y_net)**2) #Parallel axis theorm\n", + "I_2 = b_2*(l_2**3)/12.0 + A_2*((y_2-y_net)**2)\n", + "I_net = I_1 + I_2 #mm4 - the total moment of inertia\n", + "c_s= y_net # The maximum distance in steel \n", + "stress_steel = M*(10.0**6)*c_s/I_net #Mpa - The maximum stress in steel \n", + "\n", + "c_w= l_1+l_2-y_net # The maximum distance in wood \n", + "stress_wood = n*M*(10.0**6)*c_w/I_net #MPa - The maximum stress in wood \n", + "\n", + "print \"The maximum stress in steel \",round(stress_steel,2) ,\"Mpa\"\n", + "print \"The maximum stress in wood\",round(stress_wood,2) ,\"Mpa\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.9 page number 305" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum stress in concrete 834.07 psi\n", + "The stress in steel 17427.61 psi\n" + ] + } + ], + "source": [ + "#Given \n", + "M = 50000 #ft-lb , positive bending moment applied\n", + "N = 9 # number of steel bars \n", + "n = 15 # The ratio of steel to concrete \n", + "A_s = 30 #in2 area of steel in concrete\n", + "#(10*y)*(y/2) = 30*(20-y)\n", + "#y**2 + 6*y -120\n", + "#solving quadractic equation \n", + "import math\n", + "\n", + "a = 1\n", + "b = 6\n", + "c = -120\n", + "# calculate the discriminant\n", + "d = (b**2) - (4*a*c)\n", + "\n", + "# find two solutions\n", + "sol1 = (-b-math.sqrt(d))/(2*a)\n", + "sol2 = (-b+math.sqrt(d))/(2*a)\n", + "y = sol2 # Nuetral axis is found\n", + "l_1 = y #in- the concrete below nuetral axis is not considered\n", + "b_1 = 10 #in - width\n", + "A_1 = l_1* b_1 #in2 - area of concrete\n", + "y_1 = y/2 #in com of the concrete \n", + "y_2 = 20-y #in com of the transformed steel \n", + "I_1 = b_1*(l_1**3)/12.0 + A_1*((y_1-y)**2) #in4 parallel axis theorm\n", + "I_2 = A_s*((y_2)**2) #in4 first part is neglected\n", + "I_net = I_1 + I_2 #in4 - the total moment of inertia\n", + "c_c= y #in The maximum distance in concrete \n", + "stress_concrete = M*12*c_c/I_net #psi - The maximum stress in concrete \n", + "c_s= 20- y \n", + "stress_steel =n*M*12*c_s/I_net #psi - The maximum stress in concrete \n", + "print \"The maximum stress in concrete \",round(stress_concrete,2) ,\"psi\"\n", + "print \"The stress in steel\",round(stress_steel,2) ,\"psi\"\n", + "\n", + "\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## example 6.10 page number 309" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum stress upward in straight case is 99.984 Mpa\n", + "The maximum stress downward in straight case is -99.984 Mpa\n", + "The maximum stress upward in curved case is 107.093207632 Mpa\n", + "The maximum stress downward in curved case is -93.6813516989 Mpa\n", + "The maximum stress upward in curved case2 is 128.733538525 Mpa\n", + "The maximum stress downward in curved case2 is -81.0307692623 Mpa\n" + ] + } + ], + "source": [ + "#Given \n", + "l = 50.0 #mm - the length of the beam \n", + "b = 50.0 #mm - the width of the beam\n", + "M = 2083 #Nm\n", + "A = l*b #mm2 - The area \n", + "#straight beam \n", + "I = b*(l**3)/12.0 #mm4 - The moment of inertia of the beam\n", + "c_1= l/2 # the distance where the stress is maximum \n", + "c_2 = -l/2 # the distance where the stress is maximum \n", + "s_1 = I/c_1\n", + "s_2 = I/c_2\n", + "stress_max_1 = M*(10**3)/s_1 #Mpa - the maximum strss recorded in the crossection\n", + "stress_max_2 = M*(10**3)/s_2 #Mpa - the maximum strss recorded in the crossection \n", + "print \"The maximum stress upward in straight case is\",stress_max_1,\"Mpa\"\n", + "print \"The maximum stress downward in straight case is\",stress_max_2,\"Mpa\"\n", + "\n", + "#curved beam \n", + "import math\n", + "r = 250.0 #mm Radius of beam curved \n", + "r_0 = r - l/2 # inner radius \n", + "r_1 = r + l/2 # outer radius\n", + "R = l/(math.log(r_1/r_0)) #mm \n", + "e = r - R \n", + "stressr_max_1 = M*(10**3)*(R-r_0)/(r_0*A*e)\n", + "stressr_max_2 = M*(10**3)*(R-r_1)/(r_1*A*e)\n", + "print \"The maximum stress upward in curved case is\",stressr_max_1,\"Mpa\"\n", + "print \"The maximum stress downward in curved case is\",stressr_max_2,\"Mpa\"\n", + "\n", + "#curved beam _2 \n", + "import math\n", + "r = 75.0 #mm Radius of beam curved \n", + "r_0 = r - l/2 # inner radius \n", + "r_1 = r + l/2 # outer radius\n", + "R = l/(math.log(r_1/r_0)) #mm \n", + "e = r - R \n", + "stressr_max_1 = M*(10**3)*(R-r_0)/(r_0*A*e)\n", + "stressr_max_2 = M*(10**3)*(R-r_1)/(r_1*A*e)\n", + "print \"The maximum stress upward in curved case2 is\",stressr_max_1,\"Mpa\"\n", + "print \"The maximum stress downward in curved case2 is\",stressr_max_2,\"Mpa\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page number 6.14 page number 318" + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The plastic moment of the system is 120784.313725 ft-lb\n" + ] + } + ], + "source": [ + "#given \n", + "#from example 6.9\n", + "St_ul = 2500 #psi - ultimate strength\n", + "st_yl = 40000 #psi _ yielding strength \n", + "b = 10 #in - width from example \n", + "A = 2 #in2 The area of the steel\n", + "d = 20 \n", + "t_ul = st_yl*A #ultimate capasity\n", + "y = t_ul/(St_ul*b*0.85) #in 0.85 because its customary\n", + "M_ul = t_ul*(d-y/2)/12 #ft-lb Plastic moment \n", + "print \"The plastic moment of the system is \",M_ul,\"ft-lb\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.15 page number 231" + ] + }, + { + "cell_type": "code", + "execution_count": 50, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The angle at which nuetral axis locates is 0.0226547191205 radians\n" + ] + } + ], + "source": [ + "#Given \n", + "#From example 5.8 \n", + "W = 4.0 #N/m - The force distribution \n", + "L = 3 # m - The length of the force applied\n", + "M = W*L/8.0 # KN.m The moment due to force distribution\n", + "o = 30 # the angle of force applid to horizantal\n", + "l = 150.0 #mm length of the crossection \n", + "b = 100.0 #mm - width of the crossection \n", + "import math \n", + "M_z = M*(math.cos(3.14/6))\n", + "M_y = M*(math.sin(math.pi/6))\n", + "I_z = b*(l**3)/12.0\n", + "I_y = l*(b**3)/12.0\n", + "#tanb = I_z /I_y *tan30\n", + "b = math.atan(math.radians(I_z*math.tan(3.14/6.0)/I_y ))\n", + "print \"The angle at which nuetral axis locates is\",b,\"radians\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.16 pagenumber 323" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum tensile stress -76.1 Mpa\n", + "The maximum compressive stress 67.73 Mpa\n" + ] + } + ], + "source": [ + "import math\n", + "M = 10 #KN.m - The moment applied\n", + "I_max = 23.95*(10**6) #mm4 - I_z The moment of inertia\n", + "I_min = 2.53*(10**6) #mm4 - I_y The moment of inertia\n", + "o = 14.34 # degress the principle axis rotated\n", + "#Coponents of M in Y,Z direction \n", + "M_z = M*(10**6)*math.cos(math.radians(o))\n", + "M_y = M*(10**6)*math.sin(math.radians(o))\n", + "#tanb = I_z /I_y *tan14.34\n", + "b = math.atan((I_max*math.tan(math.radians(o))/I_min ))\n", + "B = math.degrees(b) \n", + "y_p = 122.9 # mm - principle axis Y cordinate\n", + "z_p = -26.95 #mm - principle axis z cordinate\n", + "stress_B = - M_z*y_p/I_max + M_y*z_p/I_min #Mpa - Maximum tensile stress\n", + "y_f = -65.97 # mm - principle axis Y cordinate\n", + "z_f = 41.93 #mm - principle axis z cordinate\n", + "stress_f = - M_z*y_f/I_max + M_y*z_f/I_min #Mpa - Maximum compressive stress\n", + "print \"The maximum tensile stress\",round(stress_B,2) ,\"Mpa\"\n", + "print \"The maximum compressive stress\",round(stress_f,2),\"Mpa\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.18 page number 328" + ] + }, + { + "cell_type": "code", + "execution_count": 51, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum stress in the beam 3.332 Mpa \n" + ] + } + ], + "source": [ + "l = 50 #mm - The length of the beam \n", + "b = 50 #mm - The width of the beam \n", + "A = l*b #mm2 - The area of the beam \n", + "p = 8.33 #KN - The force applied on the beam \n", + "stress_max = p*(10**3)/A #Mpa After cutting section A--b\n", + "print \"The maximum stress in the beam\",stress_max ,\"Mpa \"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.24 page number 339" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "By sketching the line with angle 53.2 degrees The farthest point associated with B and F\n" + ] + } + ], + "source": [ + "import math\n", + "M = 10 #KN.m - The moment applied\n", + "I_max = 23.95*(10**6) #mm4 - I_z The moment of inertia\n", + "I_min = 2.53*(10**6) #mm4 - I_y The moment of inertia\n", + "o = 14.34 # degress the principle axis rotated\n", + "#Coponents of M in Y,Z direction \n", + "M_z = M*(10**6)*math.cos(math.radians(o))\n", + "M_y = M*(10**6)*math.sin(math.radians(o))\n", + "#tanb = I_z /I_y *tan14.34\n", + "b = math.atan((I_max*math.tan(math.radians(o))/I_min ))\n", + "B = math.degrees(b) \n", + "y_p = 122.9 # mm - principle axis Y cordinate\n", + "z_p = -26.95 #mm - principle axis z cordinate\n", + "stress_B = - M_z*y_p/I_max + M_y*z_p/I_min #Mpa - Maximum tensile stress\n", + "y_f = -65.97 # mm - principle axis Y cordinate\n", + "z_f = 41.93 #mm - principle axis z cordinate\n", + "stress_f = - M_z*y_f/I_max + M_y*z_f/I_min #Mpa - Maximum compressive stress\n", + "#location of nuetral axis To show these stresses are max and minimum \n", + "#tanB = MzI_z + MzI_yz/MyI_y +M_YI_yz\n", + "I_z = 22.64 *(10**6) #mm4 moment of inertia in Z direction\n", + "I_y = 3.84 *(10**6) #mm4 moment of inertia in Y direction\n", + "I_yz =5.14 *(10**6) #mm4 moment of inertia in YZ direction \n", + "M_y = M #KN.m bending moment in Y dorection \n", + "M_z = M #KN.m bending moment in Y dorection \n", + "B = math.atan(( M_z*I_yz)/(M_z*I_y )) #radians location on neutral axis\n", + "beta = math.degrees(B)\n", + "print \"By sketching the line with angle\",round(beta,1),\"degrees The farthest point associated with B and F\" " + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter7_5.ipynb b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter7_5.ipynb new file mode 100644 index 00000000..cec54f64 --- /dev/null +++ b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter7_5.ipynb @@ -0,0 +1,257 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# chapter 7:Shear stress in Beams and Related Problems " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.1 page number 365" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The minimal space between the nails 42.0 mm\n" + ] + } + ], + "source": [ + "#Given\n", + "shear_v = 3000 #N - Transmitted vetical shear \n", + "shear_al = 700 #N - The maximum allowable \n", + "#We will divide this into two parts\n", + "l_1 = 50.0 #mm \n", + "l_2 = 200.0 #mm \n", + "b_1 = 200.0 #mm \n", + "b_2 = 50.0 #mm\n", + "A_1 = l_1* b_1 #mm2 - area of part_1\n", + "y_1 = 25.0 #mm com distance \n", + "A_2 =l_2*b_2 #mm2 - area of part_1\n", + "y_2 = 150.0 #in com distance \n", + "y_net = (A_1*y_1 +A_2*y_2)/(A_1+A_2) #mm - The com of the whole system\n", + "c_max = (4-y_net) #mm - The maximum distace from com to end\n", + "c_min = y_net #mm - the minimum distance from com to end\n", + "I_1 = b_1*(l_1**3)/12 + A_1*((y_1-y_net)**2) #Parallel axis theorm\n", + "I_2 = b_2*(l_2**3)/12 + A_2*((y_2-y_net)**2)\n", + "I_net = I_1 + I_2 #mm4 - the total moment of inertia\n", + "Q = A_1*(-y_1+y_net) #mm3\n", + "q = shear_v*Q/I_net #N/mm - Shear flow\n", + "d = shear_al/q # The space between the nails \n", + "print \"The minimal space between the nails \",round(d,0) ,\"mm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.2 pagenumber 365" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The minimal space between the nails 123.0 mm\n" + ] + } + ], + "source": [ + "#Given \n", + "l = 6 #m -length of the beam \n", + "p = 3 #KN-m _ the load applied\n", + "R_a = l*p/2 #KN -The reaction at a, Since the system is symmetry \n", + "R_b = l*p/2 #KN -The reaction at b \n", + "l_s = 10 #mm - The length of the screw \n", + "shear_al = 2 #KN - The maximum load the screw can take \n", + "I = 2.36*(10**9) #mm2 The moment of inertia of the whole system\n", + "#We will divide this into two parts\n", + "l_1 = 50.0 #mm \n", + "l_2 = 50.0 #mm \n", + "b_1 = 100.0 #mm \n", + "b_2 = 200.0 #mm\n", + "A_1 = l_1* b_1 #in2 - area of part_1\n", + "y_1 = 200.0 #mm com distance \n", + "A_2 =l_2*b_2 #mm2 - area of part_1\n", + "y_2 = 225.0 #in com distance\n", + "Q = 2*A_1*y_1 + A_2*y_2 # mm3 For the whole system\n", + "q = R_a*Q*(10**3)/I #N/mm The shear flow \n", + "d = shear_al*(10**3)/q #mm The space between the nails\n", + "print \"The minimal space between the nails \",round(d,0),\"mm\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.6 page number 376" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The shear centre from outside vertical face is 1.825 in\n" + ] + } + ], + "source": [ + "#Given\n", + "#we will divide this into two equal parts and other part\n", + "l = 10.0 # in - The height \n", + "t = 0.1 # in - The width\n", + "b = 5.0 #mm- The width of the above part \n", + "A = t* b #in2 - area of part\n", + "y_net = l/2 # The com of the system \n", + "y_1 = l # The position of teh com of part_2\n", + "I_1 = t*(l**3)/12 #in4 The moment of inertia of part 1\n", + "I_2 = 2*A*((y_1-y_net)**2) #in4 The moment of inertia of part 2 \n", + "I = I_1 + I_2 #in4 The moment of inertia \n", + "e = (b**2)*(l**2)*t/(4*I) #in the formula of channels\n", + "l_sc = e - t/2 #in- The shear centre \n", + "print \"The shear centre from outside vertical face is \",l_sc ,\"in\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.8 page number 387" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The direct maximum stress 4.25 Mpa\n", + "The torsion maximum stress 101.91 Mpa\n", + "The total stress 106.16 Mpa\n" + ] + } + ], + "source": [ + "#Given \n", + "dia = 10.0 #mm - The diameter of the cylinder \n", + "c = dia/2 #mm - the radius of the cylinder \n", + "A = 3.14*(c**2) #mm2 The area of the crossection \n", + "y = 4*c/(3*3.14) #mm The com of cylinder \n", + "I = 3.14*(c**4)/4 #mm4 - The moment of inertia of the cylinder\n", + "j = 3.14*(dia**4)/32 #mm4\n", + "T = 20.0 #N.m - The torque \n", + "V = 250.0 #N - The shear \n", + "M = 25.0 #N-m The bending moment \n", + "Q = A*y/2 #mm\n", + "stress_dmax = 4*V/(3*A) #V*Q/(I*d) #Mpa The direct maximum stress\n", + "stress_tmax = T*c*(10**3)/j #-Mpa The torsion maximum stress\n", + "stress_total = stress_dmax + stress_tmax #Mpa The total stress\n", + "print \"The direct maximum stress\",round(stress_dmax,2),\"Mpa\"\n", + "print \"The torsion maximum stress\",round(stress_tmax,2),\"Mpa\"\n", + "print \"The total stress\",round(stress_total,2),\"Mpa\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.9 page number 393" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum stress in the system 4.84 Mpa\n" + ] + } + ], + "source": [ + "#Given\n", + "dia = 15 #mm - The diameter of the rod\n", + "h = 0.5 #mt - The freely falling height \n", + "A = 3.14*(dia**2)/4 #mm2 The area of the crossection\n", + "E = 200 #Gpa -Youngs modulus\n", + "L = 750 #mm - The total length of the rod\n", + "G = 80 #gpa - Shear modulus \n", + "N = 10 #number of live coils\n", + "d = 5 #mm the diameter of live coil \n", + "m = 3 # the mass of freely falling body\n", + "H = 500 #mm -from mass to spring \n", + "F= m*9.81 #Kg the force due to that mass\n", + "p = 3 #KN-m _ the load applied\n", + "#e = e_rod + e_spr\n", + "#e_rod\n", + "e_rod = p*L*(10**-3)/(A*E) #mm The elongation due to freely falling body\n", + "#e_spr\n", + "e_spr = 64*F*(dia**3)*N*(10**-3)/(G*(d**4)) #mm The elongation due to spring\n", + "e = e_rod + e_spr #mm The total elongation \n", + "p_dyn =F*(1+pow((1+(2*H/e)),0.5))\n", + "Stress_max = p_dyn/A #MPa - The maximum stress in the system \n", + "print \"The maximum stress in the system \",round(Stress_max,2),\"Mpa\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter8_5.ipynb b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter8_5.ipynb new file mode 100644 index 00000000..4066d756 --- /dev/null +++ b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter8_5.ipynb @@ -0,0 +1,345 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 8:Transformation of stress and strain and Yield and Fracture criteria " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.1 page number 405 " + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The stress action in normal direction on AB 1.29 Mpa\n", + "The stress action in tangential direction on AB 2.12 Mpa\n" + ] + } + ], + "source": [ + "#Given \n", + "import math \n", + "from math import radians\n", + "o = 22.5 #degrees , The angle of infetisimal wedge \n", + "A = 1 #mm2 The area of the element \n", + "A_ab = 1*(math.cos(radians(o))) #mm2 - The area corresponds to AB\n", + "A_bc = 1*(math.sin(radians(o))) #mm2 - The area corresponds to BC\n", + "S_1 = 3 #MN The stresses applying on the element \n", + "S_2 = 2 #MN\n", + "S_3 = 2 #MN\n", + "S_4 = 1 #MN \n", + "F_1 = S_1*A_ab # The Forces obtained by multiplying stress by their areas \n", + "F_2 = S_2*A_ab\n", + "F_3 = S_3*A_bc\n", + "F_4 = S_4*A_bc\n", + "#sum of F_N = 0 equilibrim in normal direction \n", + "N = (F_1-F_3)*(math.cos(radians(o))) + (F_4 - F_2)*(math.sin(radians(o)))\n", + "\n", + "#sum of F_s = 0 equilibrim in tangential direction \n", + "\n", + "S = (F_2-F_4)*(math.cos(radians(o))) + (F_1 - F_3)*(math.sin(radians(o)))\n", + "\n", + "Stress_Normal = N/A #Mpa - The stress action in normal direction on AB\n", + "Stress_tan = S/A #Mpa - The stress action in tangential direction on AB\n", + "print \"The stress action in normal direction on AB\",round(Stress_Normal,2),\"Mpa\"\n", + "print \"The stress action in tangential direction on AB\",round(Stress_tan,2),\"Mpa\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.2 page number 413" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a) The stress action in normal direction on AB 4.12 Mpa\n", + "a) The stress action in tangential direction on AB 0.71 Mpa\n", + "b) The principle stress 4.2 Mpa tension\n", + "b) The principle stress -0.06 Mpa compression\n", + "b) The principle plane angles are 32.0 , 122.0 degrees\n", + "c) The maximum shear is -2.24 Mpa\n", + "a) [ 4.2 -0.1 0. ] Mpa\n", + "b) [ 2. -2.24 -2.24 2. ] Mpa\n" + ] + } + ], + "source": [ + "#Given\n", + "o = -22.5 #degrees , The angle of infetisimal wedge \n", + "A = 1 #mm2 The area of the element \n", + "import math \n", + "from math import radians\n", + "from numpy import array\n", + "A_ab = 1*(math.cos(radians(o))) #mm2 - The area corresponds to AB\n", + "A_bc = 1*(math.sin(radians(o))) #mm2 - The area corresponds to BC\n", + "S_1 = 3.0 #MN The stresses applying on the element \n", + "S_2 = 2.0 #MN\n", + "S_3 = 2.0 #MN\n", + "S_4 = 1.0 #MN\n", + "#Caliculations \n", + "\n", + "F_1 = S_1*A_ab # The Forces obtained by multiplying stress by their areas \n", + "F_2 = S_2*A_ab\n", + "F_3 = S_3*A_bc\n", + "F_4 = S_4*A_bc\n", + "#sum of F_N = 0 equilibrim in normal direction \n", + "N = (F_1-F_3)*(math.cos(radians(o))) + (F_4 - F_2)*(math.sin(radians(o)))\n", + "\n", + "#sum of F_s = 0 equilibrim in tangential direction \n", + "\n", + "S = (F_2-F_4)*(math.cos(radians(o))) + (F_1 - F_3)*(math.sin(radians(o)))\n", + "\n", + "Stress_Normal = N/A #Mpa - The stress action in normal direction on AB\n", + "Stress_tan = S/A #Mpa - The stress action in tangential direction on AB\n", + "print \"a) The stress action in normal direction on AB\",round(Stress_Normal,2),\"Mpa\"\n", + "print \"a) The stress action in tangential direction on AB\",round(Stress_tan,2),\"Mpa\"\n", + "\n", + "#Part- b\n", + "\n", + "S_max = (S_4+S_1)/2 + (((((S_4-S_1)/2)**2) + S_3**2)**0.5) #Mpa - The maximum stress\n", + "S_min = (S_4+S_1)/2.0 - (((((S_4-S_1/2))**2) + S_3**2)**0.5) #Mpa - The minumum stress\n", + "k = 0.5*math.atan(S_3/((S_1-S_4)/2)) #radians The angle of principle axis\n", + "k_1 = math.degrees(k)\n", + "k_2 = k_1+90 #The principle plane angles\n", + "print \"b) The principle stress \",round(S_max,1),\"Mpa tension\"\n", + "print \"b) The principle stress \",round(S_min,2),\"Mpa compression\"\n", + "print \"b) The principle plane angles are\",round(k_1,0),\",\",round(k_2,0),\"degrees\"\n", + "\n", + "#part-c\n", + "#The maximum shear stress case\n", + "t_xy = (((((S_4-S_1)/2)**2) + S_3**2)**0.5) #Mpa - The maximum shear stress case\n", + "K = 0.5*math.atan((-(S_1-S_4)/(2*S_3))) #radians The angle of principle axis\n", + "K_0 = math.degrees(K)\n", + "if K_0<0:\n", + " K_1 = K_0+90\n", + "else:\n", + " K_1 = K_0\n", + "K_2 = K_1+90 #PRinciple plain angles\n", + "T_xy = -((S_1-S_4)/2)*(math.sin(radians(2*K_1))) + ((S_4+S_1)/2)*(math.cos(radians(2*K_1))) # Shear stress\n", + "print \"c) The maximum shear is \",round(T_xy,2),\"Mpa\" \n", + "S_mat_a = array([round(S_max,1),round(S_min,1),0]) #MPa maximum stress matrix\n", + "S_mat_b = array([(S_4+S_1)/2,round(T_xy,2),round(T_xy,2),(S_4+S_1)/2]) #MPa maximum stress matrix at maximum shear\n", + "print \"a)\",S_mat_a,\"Mpa\"\n", + "print \"b)\",S_mat_b,\"Mpa\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.3 page number 421" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The principle stresses are 6.0 Mpa -4.0 Mpa\n", + "The maximum shear stress 5.0 Mpa\n" + ] + } + ], + "source": [ + "#Given \n", + "import math \n", + "from math import radians \n", + "S_x = -2 #Mpa _ the noraml stress in x direction\n", + "S_y = 4 #Mpa _ the noraml stress in Y direction\n", + "c = (S_x + S_y)/2 #Mpa - The centre of the mohr circle \n", + "point_x = -2 #The x coordinate of a point on mohr circle\n", + "point_y = 4 #The y coordinate of a point on mohr circle\n", + "Radius = pow((point_x-c)**2 + point_y**2,0.5) # The radius of the mohr circle\n", + "S_1 = Radius +1#MPa The principle stress\n", + "S_2 = -Radius +1 #Mpa The principle stress\n", + "S_xy_max = Radius #Mpa The maximum shear stress\n", + "print \"The principle stresses are\",S_1 ,\"Mpa\",S_2,\"Mpa\"\n", + "print \"The maximum shear stress\",S_xy_max,\"Mpa\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.4 page number 423" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The normal stress on the 221/2 plane 4.82 Mpa\n", + "The tangential stress on the 221/2 plane 1.43 Mpa\n" + ] + } + ], + "source": [ + "#Given\n", + "import math \n", + "S_x = 3.0 #Mpa _ the noraml stress in x direction\n", + "S_y = 1.0 #Mpa _ the noraml stress in Y direction\n", + "c = (S_x + S_y)/2 #Mpa - The centre of the mohr circle \n", + "point_x = 1 #The x coordinate of a point on mohr circle\n", + "point_y = 3 #The y coordinate of a point on mohr circle\n", + "#Caliculations \n", + "\n", + "Radius = pow((point_x-c)**2 + point_y**2,0.5) # The radius of the mohr circle\n", + "#22.5 degrees line is drawn \n", + "o = 22.5 #degrees \n", + "a = 71.5 - 2*o #Degrees, from diagram \n", + "stress_n = c + Radius*math.sin(math.degrees(o)) #Mpa The normal stress on the plane \n", + "stress_t = Radius*math.cos(math.degrees(o)) #Mpa The tangential stress on the plane\n", + "print \"The normal stress on the 221/2 plane \",round(stress_n,2),\"Mpa\"\n", + "print \"The tangential stress on the 221/2 plane \",round(stress_t,2),\"Mpa\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.7 page number 437" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The principle strains are 400 um/m -600 um/m\n", + "The angle of principle plane 18.43 degrees\n" + ] + } + ], + "source": [ + "import math\n", + "e_x = -500 #10-6 m/m The contraction in X direction\n", + "e_y = 300 #10-6 m/m The contraction in Y direction\n", + "e_xy = -600 #10-6 m/m discorted angle\n", + "centre = (e_x + e_y)/2 #10-6 m/m \n", + "point_x = -500 #The x coordinate of a point on mohr circle\n", + "point_y = 300 #The y coordinate of a point on mohr circle\n", + "Radius = 500 #10-6 m/m - from mohr circle\n", + "e_1 = Radius +centre #MPa The principle strain\n", + "e_2 = -Radius +centre #Mpa The principle strain\n", + "k = math.atan(300.0/900) # from geometry\n", + "k_1 = math.degrees(k)\n", + "print \"The principle strains are\",e_1,\"um/m\",e_2,\"um/m\"\n", + "print \"The angle of principle plane\",round(k_1,2) ,\"degrees\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.8 page number 441" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The principle stresses are 48.35 Mpa -105.49 MPa\n" + ] + } + ], + "source": [ + "#Given\n", + "e_0 = -500 #10-6 m/m \n", + "e_45 = 200 #10-6 m/m \n", + "e_90 = 300 #10-6 m/m\n", + "E = 200 #Gpa - youngs modulus of steel \n", + "v = 0.3 # poissions ratio \n", + "#Caliculations \n", + "\n", + "e_xy = 2*e_45 - (e_0 +e_90 ) #10-6 m/m from equation 8-40 in text\n", + "# from example 8.7\n", + "e_x = -500 #10-6 m/m The contraction in X direction\n", + "e_y = 300 #10-6 m/m The contraction in Y direction\n", + "e_xy = -600 #10-6 m/m discorted angle\n", + "centre = (e_x + e_y)/2 #10-6 m/m \n", + "point_x = -500 #The x coordinate of a point on mohr circle\n", + "point_y = 300 #The y coordinate of a point on mohr circle\n", + "Radius = 500 #10-6 m/m - from mohr circle\n", + "e_1 = Radius +centre #MPa The principle strain\n", + "e_2 = -Radius +centre #Mpa The principle strain\n", + "\n", + "stress_1 = E*(10**-3)*(e_1+v*e_2)/(1-v**2) #Mpa the stress in this direction \n", + "stress_2 = E*(10**-3)*(e_2+v*e_1)/(1-v**2) #Mpa the stress in this direction \n", + "print\"The principle stresses are \",round(stress_1,2),\"Mpa\",round(stress_2,2),\"MPa\" " + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter9_5.ipynb b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter9_5.ipynb new file mode 100644 index 00000000..34d5fcb7 --- /dev/null +++ b/Engineering_Mechanics_of_Solids_by_Popov_E_P/chapter9_5.ipynb @@ -0,0 +1,229 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 9:Elastic stress analysis and design" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.4 pagenumber 465" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a)The principle stresses are 16.67 MPa, -16.67 Mpa\n", + "b)The stresses on inclines plane 11.11 Mpa noraml, -7.06 Mpa shear \n" + ] + } + ], + "source": [ + "#Given \n", + "import math \n", + "b = 40.0 #mm - The width of the beam crossection\n", + "h = 300.0 #mm - The length of the beam crossection \n", + "V = 40.0 #KN - The shear stress in teh crossection\n", + "M = 10.0 #KN-m - The bending moment on K----K crossection \n", + "c = h/2 #mm -The position at which maximum stress occurs on the crossection\n", + "I = b*(h**3)/12 #mmm4 - the moment of inertia \n", + "#Caliculations \n", + "\n", + "stress_max_1 = M*c*(10**6)/I #The maximum stress occurs at the end\n", + "stress_max_2 = -M*c*(10**6)/I #The maximum stress occurs at the end\n", + "y = 140 #mm The point of interest, the distance of element from com\n", + "n = y/(c) # The ratio of the distances from nuetral axis to the elements\n", + "stress_L_1 = n*stress_max_1 #The normal stress on elements L--L\n", + "stress_L_2 = -n*stress_max_1 #The normal stress on elements L--L\n", + "x = 10 #mm The length of the element\n", + "A = b*x #mm3 The area of the element \n", + "y_1 = y+x/2 # the com of element from com of whole system\n", + "stress_xy = V*A*y_1*(10**3)/(I*b) #Mpa - The shear stress on the element \n", + "#stresses acting in plane 30 degrees \n", + "o = 60 #degrees - the plane angle\n", + "stress_theta = stress_L_1/2 + stress_L_1*(math.cos(math.radians(o)))/2 - stress_xy*(math.sin(math.radians(o))) #Mpa by direct application of equations\n", + "stress_shear = -stress_L_1*(math.sin(math.radians(o)))/2 - stress_xy*(math.cos(math.radians(o))) #Mpa Shear stress\n", + " \n", + "print \"a)The principle stresses are \",round(stress_max_1,2),\"MPa,\",round(stress_max_2,2),\"Mpa\"\n", + "print \"b)The stresses on inclines plane \",round(stress_theta,2),\"Mpa noraml, \",round(stress_shear,2),\"Mpa shear \"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.5 page number 476" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The stress developed 0.4 is in allowable ranges for 30077.85 mm2 area\n", + "The minimum area is 5714.28571429 mm2\n" + ] + } + ], + "source": [ + "#Given\n", + "M = 10 #KN-m moment\n", + "v = 8.0 #KN - shear Stress \n", + "stress_allow = 8 #MPa - The maximum allowable stress\n", + "shear_allow_per = 1.4 #Mpa - The allowable stress perpendicular to grain\n", + "stress_allow_shear = 0.7 #MPa - The maximum allowable shear stress\n", + "#Caliculations \n", + "\n", + "S = M*(10**6)/stress_allow #mm3 \n", + "#lets arbitarly assume h = 2b\n", + "#S = b*(h**2)/6\n", + "h = pow(12*S,0.333) #The depth of the beam\n", + "b = h/2 #mm The width of the beam\n", + "A = h*b #mm2 The area of the crossection , assumption\n", + "stress_shear = 3*v*(10**3)/(2*A) #Mpa The strear stress \n", + "if stress_shear<stress_allow_shear:\n", + " print \"The stress developed \",round(stress_shear,2),\" is in allowable ranges for \",round(A,2),\"mm2 area\"\n", + "else:\n", + " print \"The stress developed\",stress_shear,\" is in non allowable ranges\",A,\"area\"\n", + "Area_allow = v*(10**3)/shear_allow_per #mm - the allowable area\n", + "print \"The minimum area is \",Area_allow ,\"mm2\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.6 page number 478" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "W8x24 gives the allowable ranges of shear stress\n", + "lengths of the bearing at A 0.092 in\n", + "lengths of the bearing at B 2.12 in\n" + ] + } + ], + "source": [ + "#Given\n", + "stress_allow = 24 #ksi - The maximum allowable stress\n", + "stress_allow_shear = 14.5 #ksi- The maximum allowable shear stress\n", + "M_max = 36 #k-ft The maximum moment\n", + "l = 16 #in-The length of the rod\n", + "w = 2 #k/ft - The force distribution on the rod\n", + "A = l*w\n", + "R_A = 6.4 #k - The reaction at A\n", + "R_B = 25.6 #k - the reaction at B\n", + "v_max = R_B-l*w #kips the maximum stress, from diagram\n", + "#W8x24 is used from the appendix table 3 and 4 \n", + "l =0.245 #in - W8x24 crossesction length\n", + "#Caliculations \n", + "\n", + "stress_xy = v_max/A #ksi the approximate shear stress \n", + "if stress_xy < stress_allow_shear:\n", + " print \"W8x24 gives the allowable ranges of shear stress\"\n", + "else:\n", + " print \"W8x24 doesnot gives the allowable ranges of shear stress\"\n", + "k = 7.0/8 #in the distance from the outer face of the flange to the webfillet\n", + "#at+kt should not exceed 0.75 of yeild stress\n", + "#a1t+2kt should not exceed 0.75 of yeild stress\n", + "Stress_yp = 36 #Ksi - The yeild stress\n", + "t = 0.245 #in thickness of the web\n", + "#support a \n", + "a = R_A/(0.75*Stress_yp*t)-k #in lengths of the bearings\n", + "#support b\n", + "a_1 = R_B/(0.75*Stress_yp*t)-2*k #in lengths of the bearings\n", + "print \"lengths of the bearing at A \",round(a,3),\"in\"\n", + "print \"lengths of the bearing at B\",round(a_1,3),\"in\"\n", + " \n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.8 page number 483" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The suggested diameter is 2.66 in\n" + ] + } + ], + "source": [ + "#given \n", + "hp = 63000 #horse power\n", + "T = hp*20*(10**-3)/63 #k-in the torsion implies due to horse power\n", + "stress_allow_shear = 6 #ksi- The maximum allowable shear stress\n", + "M_ver = 6.72/2 #k-in the vertical component of the moment \n", + "M_hor = 9.10 #k-in the horizantal component of the moment \n", + "#Caliculations \n", + "\n", + "M = pow(((M_ver**2)+(M_hor**2)),0.5) #K-in The resultant \n", + "d = pow((16*(((M**2)+(T**2))**0.5)/(stress_allow_shear*3.14)),0.333) #in, The suggested diameter from derivation\n", + "print \"The suggested diameter is\",round(d,2),\"in\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/charpter_3_6.ipynb b/Engineering_Mechanics_of_Solids_by_Popov_E_P/charpter_3_6.ipynb new file mode 100644 index 00000000..898a6e79 --- /dev/null +++ b/Engineering_Mechanics_of_Solids_by_Popov_E_P/charpter_3_6.ipynb @@ -0,0 +1,288 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chaper 3:Generalized hooke's law, Pressure vessels and Thick walled cylinders" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.1 page number 141" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the stiffness constant is 64.0 N/mm\n" + ] + } + ], + "source": [ + "#Given\n", + "from numpy.random import randn\n", + "th = 8 #mm - thickness \n", + "G = 0.64 #N/mm2 -shear modulus\n", + " \n", + "a = 40 #mm length\n", + "b = 20 #mm width \\\n", + "Area = a*b #mm2\n", + "e = randn() # lets say any random vale\n", + "strain = e/th # strain in shearing case\n", + "#caliculations\n", + "\n", + "stress =G*strain\n", + "F = stress*Area\n", + "#stiffness = froce/displacement\n", + "k = F/e #N/mm\n", + "print \"the stiffness constant is \",k,\"N/mm\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.2 page number 149" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The change in the dimension between parallel faces is -0.025 mm\n" + ] + } + ], + "source": [ + "#Given\n", + "a = 50 #mm - length of a cube\n", + "E = 200 # Gpa - the youngs modulus \n", + "v = 0.25 # no units- poissions ratio \n", + "pressure = 200 # Mpa - pressure acting on all sides \n", + "#pressure is a compressive stress \n", + "S_x = -200 # Gpa - The stress in X direction \n", + "S_y = -200 # Gpa - The stress in Y direction\n", + "S_z = -200 # Gpa - The stress in Z direction\n", + "#Caliculations\n", + "\n", + "e = S_x*pow(10,-3)/E - v*S_y*pow(10,-3)/E-v*S_z*pow(10,-3)/E#mm - considering all three directions \n", + "x = e*a #mmThe change in the dimension between parallel faces\n", + "print \"The change in the dimension between parallel faces is \",x,\"mm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.3 page number 154" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Hoop stress is 80.0 mm\n", + "The longitudinal stress is 40.0 mm\n", + "The change in daimeter of the cylinder is 0.35 mm\n" + ] + } + ], + "source": [ + "#Given\n", + "R = 1000 # mm - radius of the cylinder \n", + "t = 10 #mm - thickness of the cylinder\n", + "p_in = 0.80 #Mpa- Internal pressure \n", + "E = 200 #Mpa- youngs modulus \n", + "v = 0.25 # poission ratio\n", + "#caliculations\n", + "\n", + "Stress_1 = p_in*R/t #Mpa -Hoop stress #From derived expressions \n", + "Stress_2 = p_in*R/(2*t) #Mpa- Longitudinal stress \n", + "e = Stress_1*pow(10,-3)/E-v*Stress_2*pow(10,-3)/E\n", + "dia_change = e*R #mm- The change in daimeter of the cylinder \n", + "print \"The Hoop stress is \",Stress_1,\"mm\"\n", + "print \"The longitudinal stress is \",Stress_2, \"mm\"\n", + "print \"The change in daimeter of the cylinder is\",dia_change,\"mm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.4 page number 156" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum stress is 40.0 mm\n", + "The change in daimeter of the cylinder is 0.15 mm\n" + ] + } + ], + "source": [ + "#Given\n", + "R = 1000 #mm - radius of the cylinder\n", + "th = 10 #mm - thickness of the cylinder\n", + "E = 200 #Mpa- youngs modulus \n", + "v = 0.25 # poission ratio\n", + "p_in = 0.80 #Mpa- Internal pressure\n", + "t = 10 #mm - thickness of the cylinder\n", + "#caliculations\n", + "\n", + "Stress_1 = p_in*R/(2*t) #Mpa -Hoop stress #From derived expressions \n", + "Stress_2 = p_in*R/(2*t) #Mpa- Longitudinal stress\n", + "# Hoop stress and Longitudinal stress are same in this case\n", + "e = Stress_1*pow(10,-3)/E-v*Stress_2*pow(10,-3)/E\n", + "dia_change = e*R #mm- The change in daimeter of the cylinder\n", + "print \"The maximum stress is \",Stress_2, \"mm\"\n", + "print \"The change in daimeter of the cylinder is\",dia_change,\"mm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.5 page number 156" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The diameter of each bolt is 1.41985914794 mm\n" + ] + } + ], + "source": [ + "#Given\n", + "p_in = 0.7 #Mpa - internal pressure \n", + "n_bolts = 20 # number of bolts \n", + "dia = 650 #mm - bolt circle diameter \n", + "stress_allow = 125 #mm Maximum alowable stress\n", + "Stress_conc = 2 #stress concentration\n", + "d = 25 #mm \n", + "#caliculations\n", + "\n", + "F = p_in*3.14*pow(((dia-2*d)/2),2)*pow(10,6) #N\n", + "F_each = F/n_bolts #N- force per each Bolt\n", + "A = Stress_conc*F_each/(stress_allow*pow(10,-6)) #mm2 The bolt area \n", + "Bolt_dia = 2*pow((round(A,3)/3.14),0.5)*pow(10,-7) #mm the bolt daimeter\n", + "print \"The diameter of each bolt is\",Bolt_dia,\"mm\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.6 page number 166" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The average stress in case a is 10.0238095238 p\n", + "The average stress in case b is 0.133333333333 p\n" + ] + } + ], + "source": [ + "#Given\n", + "from numpy.random import randn\n", + "r_1 = randn()# let r_1 is a random number \n", + "t = 0.1*r_1 #Thickness\n", + "r_2 = r_1+t #Outer Radius\n", + "#caliculations\n", + "\n", + "stress_1_a= ((r_1**2)/((r_2**2)-(r_1**2)))*(1+((r_2**2)/(r_1**2)))#p -r =r_2\n", + "stress_2_a= ((r_1**2)/((r_2**2)-(r_1**2)))*(1+((r_2**2)/(r_2**2)))#p -r =r_1\n", + "stress_avg_a=(stress_1_a+stress_2_a)/2 #p\n", + "print \"The average stress in case a is\",stress_avg_a,\"p\"\n", + "\n", + "#Case-B\n", + "#Given\n", + "r_1 = randn()# let r_1 is a random number \n", + "r_2_b= 4*r_1 \n", + "#caliculations\n", + "\n", + "stress_1_b = ((r_1**2)/((r_2_b**2)-(r_1**2)))*(1+((r_2_b**2)/(r_1**2)))#p -r =r_2\n", + "stress_2_b= ((r_1**2)/((r_2_b**2)-(r_1**2)))*(1+((r_2_b**2)/(r_2_b**2)))#p -r =r_1\n", + "stress_avg_b=(stress_1_b+stress_2_b)/2 #p\n", + "print \"The average stress in case b is\",stress_2_b,\"p\"\n", + "\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.10" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/screenshots/1_3.PNG b/Engineering_Mechanics_of_Solids_by_Popov_E_P/screenshots/1_3.PNG Binary files differnew file mode 100644 index 00000000..3e77324d --- /dev/null +++ b/Engineering_Mechanics_of_Solids_by_Popov_E_P/screenshots/1_3.PNG diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/screenshots/2_3.PNG b/Engineering_Mechanics_of_Solids_by_Popov_E_P/screenshots/2_3.PNG Binary files differnew file mode 100644 index 00000000..9dea04e5 --- /dev/null +++ b/Engineering_Mechanics_of_Solids_by_Popov_E_P/screenshots/2_3.PNG diff --git a/Engineering_Mechanics_of_Solids_by_Popov_E_P/screenshots/3_3.PNG b/Engineering_Mechanics_of_Solids_by_Popov_E_P/screenshots/3_3.PNG Binary files differnew file mode 100644 index 00000000..4ff04bf3 --- /dev/null +++ b/Engineering_Mechanics_of_Solids_by_Popov_E_P/screenshots/3_3.PNG diff --git a/Thermal_Engineering_by_A._V._Arasu/README.txt b/Thermal_Engineering_by_A._V._Arasu/README.txt new file mode 100644 index 00000000..191378eb --- /dev/null +++ b/Thermal_Engineering_by_A._V._Arasu/README.txt @@ -0,0 +1,10 @@ +Contributed By: Dhruva Shastri +Course: others +College/Institute/Organization: freelancer +Department/Designation: freelancer +Book Title: Thermal Engineering +Author: A. V. Arasu +Publisher: - +Year of publication: - +Isbn: - +Edition: 1
\ No newline at end of file diff --git a/Thermal_Engineering_by_A._V._Arasu/ch1.ipynb b/Thermal_Engineering_by_A._V._Arasu/ch1.ipynb new file mode 100644 index 00000000..16247619 --- /dev/null +++ b/Thermal_Engineering_by_A._V._Arasu/ch1.ipynb @@ -0,0 +1,702 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 1 : Fuels and Combustion" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.1 Page no : 15" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Minimum mass of air per kg of coal is 11.45 kg\n" + ] + } + ], + "source": [ + "\n", + "# Variables\n", + "C = 0.91;\t\t\t#Percentage composition of Carbon\n", + "H = 0.03;\t\t\t#Percentage composition of Hydrogen\n", + "O = 0.02;\t\t\t#Percentage composition of Oxygen\n", + "N = 0.008;\t\t\t#Percentage composition of Nitrogen\n", + "S = 0.008;\t\t\t#Percentage composition of Sulphur\n", + "\n", + "# Calculations\n", + "m = (11.5*C)+(34.5*(H-(O/8)))+(4.3*S);\t\t\t#Mass of air per kg of coal in kg\n", + "\n", + "# Results\n", + "print 'Minimum mass of air per kg of coal is %3.2f kg'%(m)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.2 Page no : 16" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Theoretical volume of air at N.T.P per kg fuel is 10.85 m**3)/kg of fuel\n" + ] + } + ], + "source": [ + "\n", + "# Variables\n", + "C = 0.86;\t\t\t#Percentage composition of Carbon\n", + "H = 0.12;\t\t\t#Percentage composition of Hydrogen\n", + "O = 0.01;\t\t\t#Percentage composition of Oxygen\n", + "S = 0.01;\t\t\t#Percentage composition of Sulphur\n", + "v = 0.773;\t\t\t#Specific volume of air at N.T.P in (m**3)/kg\n", + "\n", + "# Calculations\n", + "m = (11.5*C)+(34.5*(H-(O/8)))+(4.3*S);\t\t\t#Theoretical mass of air per kg of coal in kg\n", + "vth = m*v;\t\t\t#Theoretical volume of air at N.T.P per kg fuel in (m**3)/kg of fuel\n", + "\n", + "# Results\n", + "print 'Theoretical volume of air at N.T.P per kg fuel is %3.2f m**3)/kg of fuel'%(vth)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.3 Page no : 16" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Minimum quantity of air required for complete combustion is 10.83 kg \n", + "Total mass of products of combustion is 11.792 kg\n" + ] + } + ], + "source": [ + "\n", + "\n", + "# Variables\n", + "C = 0.78;\t\t\t#Percentage composition of Carbon\n", + "H = 0.06;\t\t\t#Percentage composition of Hydrogen\n", + "O = 0.078;\t\t\t#Percentage composition of Oxygen\n", + "N = 0.012;\t\t\t#Percentage composition of Nitrogen\n", + "S = 0.03;\t\t\t#Percentage composition of Sulphur\n", + "\n", + "# Calculations\n", + "m = (11.5*C)+(34.5*(H-(O/8)))+(4.3*S);\t\t\t#Minimum quantity of air required in kg\n", + "mt = ((11*C)/3)+(9*H)+(2*S)+(8.32+N);\t\t\t#Total mass of products of combustion in kg\n", + "\n", + "# Results\n", + "print 'Minimum quantity of air required for complete combustion is %3.2f kg \\\n", + "\\nTotal mass of products of combustion is %3.3f kg'%(m,mt)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.4 Page no : 17" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Mass of dry flue gases per kg of coal burnt is 19 kg\n" + ] + } + ], + "source": [ + "\n", + "# Variables\n", + "C = 0.84;\t\t\t#Percentage composition of Carbon\n", + "H = 0.09;\t\t\t#Percentage composition of Hydrogen\n", + "CO2 = 0.0875;\t\t\t#Volumetric composition of CO2\n", + "CO = 0.0225;\t\t\t#Volumetric composition of CO\n", + "O2 = 0.08;\t\t\t#Volumetric composition of Oxygen\n", + "N2 = 0.81;\t\t\t#Volumetric composition of Nitrogen\n", + "M1 = 44.;\t\t\t#Molecular mass of CO2\n", + "M2 = 28.;\t\t\t#Molecular mass of CO\n", + "M3 = 32.;\t\t\t#Molecular mass of O2\n", + "M4 = 28.;\t\t\t#Molecular mass of N2\n", + "\n", + "# Calculations\n", + "c1 = CO2*M1;\t\t\t#Proportional mass of CO2\n", + "c2 = CO*M2;\t \t\t#Proportional mass of CO\n", + "c3 = O2*M3;\t\t \t#Proportional mass of O2\n", + "c4 = N2*M4;\t\t\t #Proportional mass of N2\n", + "c = c1+c2+c3+c4;\t\t\t#Total proportional mass of constituents\n", + "m1 = c1/c;\t\t \t#Mass of CO2 per kg of flue gas in kg\n", + "m2 = c2/c;\t\t \t#Mass of CO per kg of flue gas in kg\n", + "m3 = c3/c;\t\t \t#Mass of O2 per kg of flue gas in kg\n", + "m4 = c4/c;\t\t \t#Mass of N2 per kg of flue gas in kg\n", + "d1 = m1*100;\t\t\t#Mass analysis of CO2\n", + "d2 = m2*100;\t\t\t#Mass analysis of CO\n", + "d3 = m3*100;\t\t\t#Mass analysis of O2\n", + "d4 = m4*100;\t\t\t#Mass analysis of N2\n", + "m = ((3*m1)/11)+((3*m2)/7.);\t\t\t#Mass of carbon in kg\n", + "md = C/m;\t\t\t #Mass of dry flue gas in kg\n", + "\n", + "# Results\n", + "print 'Mass of dry flue gases per kg of coal burnt is %.f kg'%(md)\n", + "\n", + "# note : rounding off error" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.5 Page no : 17" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Minimum air required to burn 1 kg of coal is 8.43 kg \n", + "Mass of air actually supplied per kg of coal is 11.521 kg \n", + "Amount of excess air supplied per kg of coal burnt is 3.090 kg\n" + ] + } + ], + "source": [ + "\n", + "\n", + "# Variables\n", + "C = 0.624;\t\t\t#Percentage composition of Carbon\n", + "H = 0.042;\t\t\t#Percentage composition of Hydrogen\n", + "O = 0.045;\t\t\t#Percentage composition of Oxygen\n", + "CO2 = 0.13;\t\t\t#Volumetric composition of CO2\n", + "CO = 0.003;\t\t\t#Volumetric composition of CO\n", + "O2 = 0.06;\t\t\t#Volumetric composition of Oxygen\n", + "N2 = 0.807;\t\t\t#Volumetric composition of Nitrogen\n", + "M1 = 44;\t\t\t#Molecular mass of CO2\n", + "M2 = 28;\t\t\t#Molecular mass of CO\n", + "M3 = 32;\t\t\t#Molecular mass of O2\n", + "M4 = 28;\t\t\t#Molecular mass of N2\n", + "mw = 0.378;\t\t\t#Mass of H2O in kg\n", + "\n", + "# Calculations\n", + "m = (11.5*C)+(34.5*(H-(O/8)));\t\t\t#Minimum air required in kg\n", + "c1 = CO2*M1;\t\t\t#Proportional mass of CO2\n", + "c2 = CO*M2;\t\t\t#Proportional mass of CO\n", + "c3 = O2*M3;\t\t\t#Proportional mass of O2\n", + "c4 = N2*M4;\t\t\t#Proportional mass of N2\n", + "c = c1+c2+c3+c4;\t\t\t#Total proportional mass of constituents\n", + "m1 = c1/c;\t\t\t#Mass of CO2 per kg of flue gas in kg\n", + "m2 = c2/c;\t\t\t#Mass of CO per kg of flue gas in kg\n", + "m3 = c3/c;\t\t\t#Mass of O2 per kg of flue gas in kg\n", + "m4 = c4/c;\t\t\t#Mass of N2 per kg of flue gas in kg\n", + "d1 = m1*100;\t\t\t#Mass analysis of CO2\n", + "d2 = m2*100;\t\t\t#Mass analysis of CO\n", + "d3 = m3*100;\t\t\t#Mass analysis of O2\n", + "d4 = m4*100;\t\t\t#Mass analysis of N2\n", + "mC = ((3*m1)/11)+((3*m2)/7);\t\t\t#Mass of carbon in kg\n", + "md = C/mC;\t\t\t#Mass of dry flue gas in kg\n", + "mact = (md+mw)-(C+H+O);\t\t\t#Actual air supplied per kg of fuel in kg\n", + "me = mact-m;\t\t\t#Mass of excess air per kg of fuel in kg\n", + "\n", + "# Results\n", + "print 'Minimum air required to burn 1 kg of coal is %3.2f kg \\\n", + "\\nMass of air actually supplied per kg of coal is %3.3f kg \\\n", + "\\nAmount of excess air supplied per kg of coal burnt is %3.3f kg'%(m,mact,me)\n", + "#rounding-off errors" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.6 Page no : 19" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Mass of air to be supplied is 9.92 kg \n", + "Mass of CO2 produced per kg of coal is 2.86 kg \n", + "Mass of H2O produced per kg of coal is 0.27 kg\n", + "Mass of SO2 produced per kg of coal is 0.02 kg \n", + "Mass of excess O2 produced per kg of coal is 0.69 kg \n", + "Mass of N2 produced per kg of coal is 9.90 kg \n" + ] + } + ], + "source": [ + "\n", + "\n", + "# Variables\n", + "C = 0.78;\t\t\t#Percentage composition of Carbon\n", + "H = 0.03;\t\t\t#Percentage composition of Hydrogen\n", + "O = 0.03;\t\t\t#Percentage composition of Oxygen\n", + "S = 0.01;\t\t\t#Percentage composition of Sulphur\n", + "me = 0.3;\t\t\t#Mass of excess air supplied\n", + "\n", + "# Calculations\n", + "m = (11.5*C)+(34.5*(H-(O/8)))+(4.3*S);\t\t\t#Mass of air per kg of coal in kg\n", + "mec = me*m;\t\t\t#Excess air supplied per kg of coal in kg\n", + "mact = m+mec;\t\t\t#Actual mass of air supplied per kg of coal in kg\n", + "mCO2 = (11*C)/3;\t\t\t#Mass of CO2 produced per kg of coal in kg\n", + "mHw = 9*H;\t\t\t#Mass of H2O produced per kg of coal in kg\n", + "mSO2 = 2*S;\t\t\t#Mass of SO2 produced per kg of coal in kg\n", + "mO2 = 0.232*mec;\t\t\t#Mass of excess O2 produced per kg of coal in kg\n", + "mN2 = 0.768*mact;\t\t\t#Mass of N2 produced per kg of coal in kg\n", + "\n", + "# Results\n", + "print 'Mass of air to be supplied is %3.2f kg \\\n", + "\\nMass of CO2 produced per kg of coal is %3.2f kg \\\n", + "\\nMass of H2O produced per kg of coal is %3.2f kg\\\n", + "\\nMass of SO2 produced per kg of coal is %3.2f kg \\\n", + "\\nMass of excess O2 produced per kg of coal is %3.2f kg \\\n", + "\\nMass of N2 produced per kg of coal is %3.2f kg '%(m,mCO2,mHw,mSO2,mO2,mN2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.7 Page no : 20" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Minimum mass of air required is 11.4 kg \n", + "Total mass of dry flue gases per kg of fuel is 17.93 kg \n", + "Percentage composition of CO2 by volume is 12.69 percent \n", + "Percentage composition of SO2 by volume is 0.048 percent \n", + "Percentage composition of O2 by volume is 7.2 percent \n", + "Percentage composition of N2 by volume is 80.08 percent\n" + ] + } + ], + "source": [ + "\n", + "\n", + "# Variables\n", + "C = 0.9;\t\t\t#Percentage composition of Carbon\n", + "H = 0.033;\t\t\t#Percentage composition of Hydrogen\n", + "O = 0.03;\t\t\t#Percentage composition of Oxygen\n", + "N = 0.008;\t\t\t#Percentage composition of Nitrogen\n", + "S = 0.009;\t\t\t#Percentage composition of Sulphur\n", + "M1 = 44;\t\t\t#Molecular mass of CO2\n", + "M2 = 64;\t\t\t#Molecular mass of SO2\n", + "M3 = 32;\t\t\t#Molecular mass of O2\n", + "M4 = 28;\t\t\t#Molecular mass of N2\n", + "\n", + "# Calculations\n", + "m = (11.5*C)+(34.5*(H-(O/8)))+(4.3*S);\t\t\t#Minimum mass of air per kg of coal in kg\n", + "mCO2 = (11*C)/3;\t\t\t#Mass of CO2 produced per kg of coal in kg\n", + "mHw = 9*H;\t\t\t#Mass of H2O produced per kg of coal in kg\n", + "mSO2 = 2*S;\t\t\t#Mass of SO2 produced per kg of coal in kg\n", + "mt = 11.5*1.5;\t\t\t#Total mass of air supplied per kg of coal in kg\n", + "me = mt-m;\t\t\t#Excess air supplied in kg\n", + "mO2 = 0.232*me;\t\t\t#Mass of excess O2 produced per kg of coal in kg\n", + "mN2 = 0.768*mt;\t\t\t#Mass of N2 produced per kg of coal in kg\n", + "mtN2 = mN2+N;\t\t\t#Total mass of Nitrogen in exhaust in kg\n", + "md = mCO2+mSO2+mO2+mtN2;\t\t\t#Total mass of dry flue gases per kg of fuel in kg\n", + "CO2 = (mCO2/md)*100;\t\t\t#Percentage composition of CO2 by mass in percent\n", + "SO2 = (mSO2/md)*100;\t\t\t#Percentage composition of SO2 by mass in percent\n", + "O2 = (mO2/md)*100;\t\t\t#Percentage composition of O2 by mass in percent\n", + "N2 = (mN2/md)*100;\t\t\t#Percentage composition of N2 by mass in percent\n", + "c1 = CO2/M1;\t\t\t#Proportional volume of CO2\n", + "c2 = SO2/M2;\t\t\t#Proportional volume of SO2\n", + "c3 = O2/M3;\t\t\t#Proportional volume of O2\n", + "c4 = N2/M4;\t\t\t#Proportional volume of N2\n", + "c = c1+c2+c3+c4;\t\t\t#Total proportional volume of constituents\n", + "m1 = c1/c;\t\t\t#Volume of CO2 in 1 (m**3) of flue gas\n", + "m2 = c2/c;\t\t\t#Volume of SO2 in 1 (m**3) of flue gas\n", + "m3 = c3/c;\t\t\t#Volume of O2 in 1 (m**3) of flue gas\n", + "m4 = c4/c;\t\t\t#Volume of N2 in 1 (m**3) of flue gas\n", + "d1 = m1*100;\t\t\t#Volume analysis of CO2\n", + "d2 = m2*100;\t\t\t#Volume analysis of SO2\n", + "d3 = m3*100;\t\t\t#Volume analysis of O2\n", + "d4 = m4*100;\t\t\t#Volume analysis of N2\n", + "\n", + "# Results\n", + "print 'Minimum mass of air required is %3.1f kg \\\n", + "\\nTotal mass of dry flue gases per kg of fuel is %3.2f kg \\\n", + "\\nPercentage composition of CO2 by volume is %3.2f percent \\\n", + "\\nPercentage composition of SO2 by volume is %3.3f percent \\\n", + "\\nPercentage composition of O2 by volume is %3.1f percent \\\n", + "\\nPercentage composition of N2 by volume is %3.2f percent'%(m,md,d1,d2,d3,d4)\n", + "\n", + "# note : rounding off error" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.8 Page no : 21" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Mass of air actually supplied per kg of coal is 18.20 kg \n", + "Percentage of excess air is 60 percent\n" + ] + } + ], + "source": [ + "\n", + "\n", + "# Variables\n", + "C = 0.88;\t\t\t#Percentage composition of Carbon\n", + "H = 0.036;\t\t\t#Percentage composition of Hydrogen\n", + "O = 0.048;\t\t\t#Percentage composition of oxygen\n", + "CO2 = 0.109;\t\t\t#Volumetric composition of CO2\n", + "CO = 0.01;\t\t\t#Volumetric composition of CO\n", + "O2 = 0.071;\t\t\t#Volumetric composition of Oxygen\n", + "N2 = 0.81;\t\t\t#Volumetric composition of Nitrogen\n", + "M1 = 44.;\t\t\t#Molecular mass of CO2\n", + "M2 = 28.;\t\t\t#Molecular mass of CO\n", + "M3 = 32.;\t\t\t#Molecular mass of O2\n", + "M4 = 28.;\t\t\t#Molecular mass of N2\n", + "\n", + "# Calculations\n", + "m = (11.5*C)+(34.5*(H-(O/8)));\t\t\t#Theoretical air required in kg\n", + "c1 = CO2*M1;\t\t\t#Proportional mass of CO2\n", + "c2 = CO*M2;\t\t\t#Proportional mass of CO\n", + "c3 = O2*M3;\t\t\t#Proportional mass of O2\n", + "c4 = N2*M4;\t\t\t#Proportional mass of N2\n", + "c = c1+c2+c3+c4;\t\t\t#Total proportional mass of constituents\n", + "m1 = c1/c;\t\t\t#Mass of CO2 per kg of flue gas in kg\n", + "m2 = c2/c;\t\t\t#Mass of CO per kg of flue gas in kg\n", + "m3 = c3/c;\t\t\t#Mass of O2 per kg of flue gas in kg\n", + "m4 = c4/c;\t\t\t#Mass of N2 per kg of flue gas in kg\n", + "mC = ((3*m1)/11)+((3*m2)/7);\t\t\t#Mass of carbon in kg\n", + "md = C/mC;\t\t\t#Mass of dry flue gas in kg\n", + "hc = H*9;\t\t\t#Hydrogen combustion in kg of H2O\n", + "mair = (md+hc)-(C+H+O);\t\t\t#Mass of air supplied per kg of coal in kg\n", + "me = mair-m;\t\t\t#Excess air per kg of coal in kg\n", + "mN2 = m4*md;\t\t\t#Mass of nitrogen per kg of coal in kg\n", + "mact = mN2/0.768;\t\t\t#Actual mass of air per kg of coal in kg\n", + "pe = (me/m)*100;\t\t\t#Perccentage excess air in percent\n", + "\n", + "# Results\n", + "print 'Mass of air actually supplied per kg of coal is %3.2f kg \\\n", + "\\nPercentage of excess air is %.f percent'%(mact,pe)\n", + "\n", + "# note : rounding off error" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.9 Page no : 22" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Mass of excess air supplied per kg of fuel burnt is 6.0 kg/kg of fuel \n", + "Air-fuel ratio is 20.7:1\n" + ] + } + ], + "source": [ + "\n", + "\n", + "# Variables\n", + "C = 0.84;\t\t\t#Percentage composition of Carbon\n", + "H = 0.14;\t\t\t#Percentage composition of Hydrogen\n", + "O = 0.02;\t\t\t#Percentage composition of oxygen\n", + "CO2 = 8.85;\t\t\t#Volumetric composition of CO2\n", + "CO = 1.2;\t\t\t#Volumetric composition of CO\n", + "O2 = 6.8;\t\t\t#Volumetric composition of Oxygen\n", + "N2 = 83.15;\t\t\t#Volumetric composition of Nitrogen\n", + "M1 = 44.;\t\t\t#Molecular mass of CO2\n", + "M2 = 28.;\t\t\t#Molecular mass of CO\n", + "M3 = 32.;\t\t\t#Molecular mass of O2\n", + "M4 = 28.;\t\t\t#Molecular mass of N2\n", + "a = 8/3.;\t\t\t#O2 required per kg C\n", + "b = 8.; \t\t\t#O2 required per kg H2\n", + "mair = 0.23;\t\t\t#Mass of air\n", + "\n", + "# Calculations\n", + "c = C*a;\t\t\t#O2 required per kg of fuel for C\n", + "d = H*b;\t\t\t#O2 required per kg of fuel for H2\n", + "tO2 = c+d+O;\t\t\t#Theoreticcal O2 required in kg/kg of fuel\n", + "tm = tO2/mair;\t\t\t#Theoretical mass of air in kg/kg of fuel\n", + "c1 = CO2*M1;\t\t\t#Proportional mass of CO2 by Volume\n", + "c2 = CO*M2;\t\t\t#Proportional mass of CO by Volume\n", + "c3 = O2*M3;\t\t\t#Proportional mass of O2 by Volume\n", + "c4 = N2*M4;\t\t\t#Proportional mass of N2 by Volume\n", + "c = c1+c2+c3+c4;\t\t\t#Total proportional mass of constituents\n", + "m1 = c1/c;\t\t\t#Mass of CO2 per kg of flue gas in kg\n", + "m2 = c2/c;\t\t\t#Mass of CO per kg of flue gas in kg\n", + "m3 = c3/c;\t\t\t#Mass of O2 per kg of flue gas in kg\n", + "m4 = c4/c;\t\t\t#Mass of N2 per kg of flue gas in kg\n", + "mC = ((m1*12)/M1)+((m2*12)/M2);\t\t\t#Mass of carbon per kg of dry flue gas in kg\n", + "md = C/mC;\t\t\t#Mass of dry flue per kg of fuel in kg\n", + "p = (4*m2)/7;\t\t\t#Oxygen required to burn CO in kg\n", + "meO2 = md*(m3-p);\t\t\t#Mass of excess O2 per kg of fuel in kg\n", + "me = meO2/mair;\t\t\t#Mass of excess air in kg/kg fuel\n", + "mt = tm+me;\t\t\t#Total air required per kg fuel\n", + "\n", + "# Results\n", + "print 'Mass of excess air supplied per kg of fuel burnt is %3.1f kg/kg of fuel \\\n", + "\\nAir-fuel ratio is %3.1f:1'%(me,mt)\n", + "\n", + "# rounding off error" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.10 Page no : 23" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Volume of air required for complete combustion is 1.178 m**3)\n" + ] + } + ], + "source": [ + "\n", + "# Variables\n", + "H2 = 0.27;\t\t\t#Percentage composition of H2 by volume\n", + "CO2 = 0.18;\t\t\t#Percentage composition of CO2 by volume\n", + "CO = 0.125;\t\t\t#Percentage composition of CO by volume\n", + "CH4 = 0.025;\t\t\t#Percentage composition of CH4 by volume\n", + "N2 = 0.4;\t\t\t#Percentage composition of N2 by volume\n", + "\n", + "# Calculations\n", + "v = (2.38*(H2+CO))+(9.52*CH4);\t\t\t#Volume of air required for complete combustion in (m**3)\n", + "\n", + "# Results\n", + "print 'Volume of air required for complete combustion is %3.3f m**3)'%(v)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.11 Page no : 24" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Air-fuel ratio by volume is 5.055\n" + ] + } + ], + "source": [ + "\n", + "# Variables\n", + "H2 = 0.5;\t\t\t#Percentage composition of H2 by volume\n", + "CO2 = 0.1;\t\t\t#Percentage composition of CO2 by volume\n", + "CO = 0.05;\t\t\t#Percentage composition of CO by volume\n", + "CH4 = 0.25;\t\t\t#Percentage composition of CH4 by volume\n", + "N2 = 0.1;\t\t\t#Percentage composition of N2 by volume\n", + "pCO2 = 8;\t\t\t#Percentage volumetric analysis of CO2\n", + "pO2 = 6;\t\t\t#Percentage volumetric analysis of O2\n", + "pN2 = 86;\t\t\t#Percentage volumetric analysis of N2\n", + "\n", + "\n", + "# Calculations\n", + "v = (2.38*(H2+CO))+(9.52*CH4);\t\t\t#Volume of air required for complete combustion in (m**3)\n", + "vN2 = v*0.79;\t\t\t#Volume of nitrogen in the air in m**3\n", + "a = CO+CH4+CO2;\t\t\t#CO2 formed per m**3 of fuel gas burnt\n", + "b = vN2+N2;\t\t\t#N2 formed per m**3 of fuel gas burnt\n", + "vt = a+b;\t\t\t#Total volume of dry flue gas formed in m**3\n", + "ve = (pO2*vt)/(21-pO2);\t\t\t#Excess air supplied in m**3\n", + "V = v+ve;\t\t\t#Total quantity of air supplied in m**3\n", + "afr = V/1\n", + "\n", + "# Results\n", + "print 'Air-fuel ratio by volume is %3.3f'%(afr)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.12 Page no : 24" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Volume of air required for complete combustion is 0.952 m**3) \n", + "Volume of CO2 per m**3 of gas fuel is 0.29 m**3/m**3 of gas fuel \n", + "Volume of N2 per m**3 of gas fuel is 1.603 m**3/m**3 of gas fuel \n", + "Volume of excess O2 per m**3 of gas fuel is 0.08 m**3/m**3 of gas fuel \n", + "Total volume of dry combustion products is 1.973 m**3/m**3 of gas fuel \n", + "Percentage volume of CO2 is 14.7 percent \n", + "Percentage volume of N2 is 81.25 percent \n", + "Percentage volume of O2 is 4.05 percent\n" + ] + } + ], + "source": [ + "\n", + "# Variables\n", + "H2 = 0.14;\t\t\t#Percentage composition of H2 by volume\n", + "CO2 = 0.05;\t\t\t#Percentage composition of CO2 by volume\n", + "CO = 0.22;\t\t\t#Percentage composition of CO by volume\n", + "CH4 = 0.02;\t\t\t#Percentage composition of CH4 by volume\n", + "O2 = 0.02;\t\t\t#Percentage composition of O2 by volume\n", + "N2 = 0.55;\t\t\t#Percentage composition of N2 by volume\n", + "e = 0.4;\t\t\t#Excess air supplied\n", + "# Calculations\n", + "v = (2.38*(H2+CO))+(9.52*CH4)-(4.76*O2);\t\t\t#Volume of air required for complete combustion in (m**3)\n", + "ve = v*e;\t\t\t#Volume of excess air supplied in m**3\n", + "vtN2 = v-(v*0.21);\t\t\t#Volume of N2 in theoretical air in m**3\n", + "veN2 = ve-(ve*0.21);\t\t\t#Volume of N2 in excess air in m**3\n", + "vt = vtN2+veN2;\t\t\t#Total volume of N2 in air supplied in m**3\n", + "vCO2 = CO+CH4+CO2;\t\t\t#CO2 formed per m**3 of fuel gas\n", + "vN2 = vt+N2;\t\t\t#N2 formed per m**3 of fuel gas\n", + "veO2 = ve*0.21;\t\t\t#Volume of excess O2 per m**3 of fuel gas\n", + "vT = vCO2+vN2+veO2;\t\t\t#Total volume of dry combustion products\n", + "pCO2 = (vCO2*100)/vT;\t\t\t#Percentage volume of CO2\n", + "pN2 = (vN2*100)/vT;\t\t\t#Percentage volume of N2\n", + "pO2 = (veO2*100)/vT;\t\t\t#Percentage volume of O2\n", + "\n", + "# Results\n", + "print 'Volume of air required for complete combustion is %3.3f m**3) \\\n", + "\\nVolume of CO2 per m**3 of gas fuel is %3.2f m**3/m**3 of gas fuel \\\n", + "\\nVolume of N2 per m**3 of gas fuel is %3.3f m**3/m**3 of gas fuel \\\n", + "\\nVolume of excess O2 per m**3 of gas fuel is %3.2f m**3/m**3 of gas fuel \\\n", + "\\nTotal volume of dry combustion products is %3.3f m**3/m**3 of gas fuel \\\n", + "\\nPercentage volume of CO2 is %3.1f percent \\\n", + "\\nPercentage volume of N2 is %3.2f percent \\\n", + "\\nPercentage volume of O2 is %3.2f percent'%(v,vCO2,vN2,veO2,vT,pCO2,pN2,pO2)\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Thermal_Engineering_by_A._V._Arasu/ch2.ipynb b/Thermal_Engineering_by_A._V._Arasu/ch2.ipynb new file mode 100644 index 00000000..5ffb6e9f --- /dev/null +++ b/Thermal_Engineering_by_A._V._Arasu/ch2.ipynb @@ -0,0 +1,1161 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2 : Gas Power Cycles" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.1 Page no : 55" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Maximum pressure of the cycle is 9.434 MPa \n", + "Maximum temperature of the cycle is 3632 K \n", + "Cycle efficiency is 56.4 percent \n", + "Mean effective pressure is 1.533 MPa\n" + ] + } + ], + "source": [ + "\n", + "# Variables\n", + "P1 = 0.1;\t\t\t#Pressure of air supplied in MPa\n", + "T1 = 308;\t\t\t#Temperature of air supplied in K\n", + "rv = 8;\t\t\t#Compression ratio\n", + "q1 = 2100;\t\t\t#Heat supplied in kJ/kg\n", + "Cp = 1.005;\t\t\t#Specific heat at constant pressure in kJ/kg-K\n", + "Cv = 0.718;\t\t\t#Specific heat at constant volume in kJ/kg-K\n", + "R = 0.287;\t\t\t#Universal gas constant in kJ/kg-K\n", + "\n", + "# Calculations\n", + "y = Cp/Cv;\t\t\t#Ratio of specific heats\n", + "n = (1-(1/(rv**(y-1))))*100;\t\t\t#Cycle efficiency\n", + "v1 = (R*T1)/(P1*1000);\t\t\t#Specific volume at point 1 in (m**3)/kg\n", + "v2 = v1/rv;\t\t\t#Specific volume at point 2 in (m**3)/kg\n", + "T2 = T1*(rv**(y-1));\t\t\t#Temperature at point 2 in K\n", + "T3 = (q1/Cv)+T2;\t\t\t#Temperature at point 3 in K\n", + "P2 = P1*(rv**y);\t\t\t#Pressure at point 2 in MPa\n", + "P3 = P2*(T3/T2);\t\t\t#Pressure at point 3 in MPa\n", + "wnet = (q1*n)/100;\t\t\t#Net workdone in J/kg\n", + "MEP = (wnet/(v1-v2))/1000;\t\t\t#Mean effective pressure in MPa\n", + "\n", + "# Results\n", + "print 'Maximum pressure of the cycle is %3.3f MPa \\\n", + "\\nMaximum temperature of the cycle is %3.0f K \\\n", + "\\nCycle efficiency is %3.1f percent \\\n", + "\\nMean effective pressure is %3.3f MPa'%(P3,T3,n,MEP)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.2 Page no : 57" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Relative efficiency of the engine is 38.8 percent\n" + ] + } + ], + "source": [ + "\n", + "# Variables\n", + "d = 80;\t\t\t#Bore in mm\n", + "L = 85;\t\t\t#Stroke in mm\n", + "Vc = 0.06;\t\t\t#Clearance volume in litre\n", + "n = 0.22;\t\t\t#Actual thermal efficiency\n", + "y = 1.4;\t\t\t#Ratio of specific heats\n", + "\n", + "# Calculations\n", + "Vs = (3.147/4)*(d**2)*L;\t\t\t#Stroke volume in mm**3\n", + "Vt = Vs+(Vc*(10**6));\t\t\t#Total volume in mm**3\n", + "rv = Vt/(Vc*(10**6));\t\t\t#Compression ratio\n", + "ni = (1-(1/(rv**(y-1))));\t\t\t#Ideal thermal efficiency\n", + "nr = (n/ni)*100;\t\t\t#Relative efficiency\n", + "\n", + "# Results\n", + "print 'Relative efficiency of the engine is %3.1f percent'%(nr)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.3 Page no : 57" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Clearance volume is 14.6 percent of swept volume \n", + "Otto cycle efficiency is 56.15 percent\n" + ] + } + ], + "source": [ + "\n", + "\n", + "# Variables\n", + "d = 0.137;\t\t\t#Bore in m\n", + "L = 0.13;\t\t\t#Stroke in m\n", + "Vc = 280*(10**-6);\t\t\t#Clearance volume in m**3\n", + "y = 1.4;\t\t\t#Ratio of specific heats\n", + "\n", + "# Calculations\n", + "Vs = (3.147/4)*(d**2)*L;\t\t\t#Stroke volume in m**3\n", + "rv = (Vc/Vs)*100;\t\t\t#Compression ratio\n", + "rvf = (Vs+Vc)/Vc;\t\t\t#final compression ratio\n", + "n = (1-(1/rvf**(y-1)))*100;\t\t\t#Cycle efficiency\n", + "\n", + "# Results\n", + "print 'Clearance volume is %3.1f percent of swept volume \\\n", + "\\nOtto cycle efficiency is %3.2f percent'%(rv,n)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.4 Page no : 58" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Maximum pressure of the cycle is 6449.19 kPa \n", + "Maximum temperature of the cycle is 1968.7 K \n", + "Amount of heat transferred is 0.65 kJ \n", + "Thermal efficiency is 59.4 percent \n", + "Mean effective pressure is 718.3 kPa\n" + ] + } + ], + "source": [ + "\n", + "# Variables\n", + "rv = 9.5;\t\t\t#Compression ratio\n", + "P1 = 100.;\t\t\t#Air pressure in kPa\n", + "T1 = 290.;\t\t\t#Air temperature in K\n", + "V1 = 600.*(10**-6);\t\t\t#Volume of air in m**3\n", + "T4 = 800.;\t\t\t#Final temperature in K\n", + "R = 287.;\t\t\t#Universal gas constant in J/kg.K\n", + "Cv = 0.718;\t\t\t#Specific heat at constant volume in kJ/kg.K\n", + "y = 1.4;\t\t\t#Ratio of specific heats\n", + "\n", + "# Calculations\n", + "T3 = T4*(rv**(y-1));\t\t\t#Temperature at the end of constant volume heat addition in K\n", + "P2 = P1*(rv**y);\t\t\t#Pressure at point 2 in kPa\n", + "T2 = T1*(rv**(y-1));\t\t\t#Temperature at point 2 in K\n", + "P3 = P2*(T3/T2);\t\t\t#Pressure at point 3 in kPa\n", + "m = (P1*1000*V1)/(R*T1);\t\t\t#Specific mass in kg/s\n", + "Q = m*Cv*(T3-T2);\t\t\t#Heat transferred in kJ\n", + "n = (1-(1/rv**(y-1)))*100;\t\t\t#Thermal efficiency\n", + "Wnet = (n*Q)/100;\t\t\t#Net workdone in kJ\n", + "MEP = Wnet/(V1*(1-(1/rv)));\t\t\t#Mean effective pressure in kPa\n", + "\n", + "# Results\n", + "print 'Maximum pressure of the cycle is %3.2f kPa \\\n", + "\\nMaximum temperature of the cycle is %3.1f K \\\n", + "\\nAmount of heat transferred is %3.2f kJ \\\n", + "\\nThermal efficiency is %3.1f percent \\\n", + "\\nMean effective pressure is %3.1f kPa'%(P3,T3,Q,n,MEP)\n", + "\n", + "# rounding off error" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.5 Page no : 60" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Pressure at the end of heat addition process is 4392.3 kPa\n", + "Temperature at the end of heat addition process is 1733.8 K\n", + "Net work output is 423.54 kJ/kg\n", + "Thermal efficiency is 56.47 percent\n", + "Mean effective pressure is 534 kPa\n" + ] + } + ], + "source": [ + "\n", + "\n", + "# Variables\n", + "rv = 8.;\t\t\t#Compression ratio\n", + "P1 = 95.;\t\t\t#Pressure at point 1 in kPa\n", + "T1 = 300.;\t\t\t#Temperature at point 1 in K\n", + "q23 = 750.;\t\t\t#Heat transferred during consmath.tant volume heat addition process in kJ/kg\n", + "y = 1.4;\t\t\t#Ratio of specific heats\n", + "Cv = 0.718;\t\t\t#Specific heat at constant volume in kJ/kg-K\n", + "R = 287.;\t\t\t#Universal gas constant in J/kg-K\n", + "\n", + "# Calculations\n", + "T2 = T1*(rv**(y-1));\t\t\t#Temperature at point 2 in K\n", + "P2 = P1*(rv**y);\t\t\t#Pressure at point 2 in kPa\n", + "T3 = (q23/Cv)+T2;\t\t\t#Temperature at point 3 in K\n", + "P3 = P2*(T3/T2);\t\t\t#Pressure at point 3 in kPa\n", + "nth = (1-(1/(rv**(y-1))))*100;\t\t\t#Thermal efficiency\n", + "Wnet = (nth*q23)/100;\t\t\t#Net work output in kJ/kg\n", + "v1 = (R*T1)/(P1*1000);\t\t\t#Speific volume at point 1 in (m**3)/kg\n", + "MEP = Wnet/(v1*(1-(1/rv)));\t\t\t#Mean effective pressure in kPa\n", + "\n", + "# Results\n", + "print 'Pressure at the end of heat addition process is %3.1f kPa'%P3\n", + "print 'Temperature at the end of heat addition process is %3.1f K'%T3\n", + "print 'Net work output is %3.2f kJ/kg'%Wnet\n", + "print 'Thermal efficiency is %3.2f percent'%nth\n", + "print 'Mean effective pressure is %3.0f kPa'%MEP\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.6 Page no : 61" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Air standard efficiency is 60.4 percent\n" + ] + } + ], + "source": [ + "\n", + "# Variables\n", + "rv = 14.;\t\t\t#Compression ratio\n", + "c = 0.06;\t\t\t#Cut-off percentage\n", + "y = 1.4;\t\t\t#Ratio of specific heats\n", + "\n", + "# Calculations\n", + "rc = 1.78;\t\t\t#Cut-off ratio\n", + "nth = (1-(((rc**y)-1)/((rv**(y-1))*y*(rc-1))))*100;\t\t\t#Thermal efficiency\n", + "\n", + "# Results\n", + "print 'Air standard efficiency is %3.1f percent'%(nth)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.7 Page no : 62" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Cut-off ratio is 2.01 \n", + "Heat supplied is 884.4 kJ/kg\n", + "Cycle efficiency is 61.3 percent \n", + "Mean effective pressure is 699.35 kPa\n" + ] + } + ], + "source": [ + "\n", + "\n", + "# Variables\n", + "rv = 16.;\t\t\t#Compression ratio\n", + "P1 = 0.1;\t\t\t#Pressure at point 1 in MPa\n", + "T1 = 288.;\t\t\t#Temperature at point 1 in K\n", + "T3 = 1753.;\t\t\t#Temperature at point 3 in K\n", + "y = 1.4;\t\t\t#Ratio of specific heats\n", + "Cp = 1.005;\t\t\t#Specific heat at constant pressure in kJ/kg-K\n", + "R = 0.287;\t\t\t#Universal gas constant in kJ/kg-K\n", + "\n", + "# Calculations\n", + "T2 = int(T1*(rv**(y-1)));\t\t\t#Temperature at point 2 in K\n", + "rc = round(T3/T2,2);\t\t\t#Cut-off ratio\n", + "q1 = Cp*(T3-T2);\t\t\t#Heat supplied in kJ/kg\n", + "nth = (1-(((rc**y)-1)/((rv**(y-1))*y*(rc-1))))*100;\t\t\t#Cycle efficiency\n", + "wnet = int((q1*nth)/100);\t\t\t#Net work done in kJ/kg\n", + "v1 = round((R*T1)/(P1*1000),3);\t\t\t#Speific volume at point 1 in (m**3)/kg\n", + "v2 = round(v1/rv,3);\t\t\t#Speific volume at point 2 in (m**3)/kg\n", + "MEP = wnet/(v1-v2);\t\t\t#Mean effective pressure in kPa\n", + "\n", + "# Results\n", + "print 'Cut-off ratio is %3.2f \\\n", + "\\nHeat supplied is %3.1f kJ/kg\\\n", + "\\nCycle efficiency is %3.1f percent \\\n", + "\\nMean effective pressure is %3.2f kPa'%(rc,q1,nth,MEP)\n", + "\n", + "# rounding off error" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.8 Page no : 64" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Air standard efficiency at 5 percent cut-off is 59.36 percent\n", + "Air standard efficiency at 8 percent cut-off is 57.40 percent\n", + "Percentage loss in efficiency is 1.95 percent\n" + ] + } + ], + "source": [ + "\n", + "\n", + "# Variables\n", + "d = 0.15;\t\t\t#Bore in m\n", + "L = 0.25;\t\t\t#Stroke in m\n", + "Vc = 400*(10**-6);\t\t\t#Clearance volume in m**3\n", + "V2 = Vc;\t\t\t#Clearance volume in m**3\n", + "c1 = 0.05;\t\t\t#Cut-off percentage 1\n", + "c2 = 0.08;\t\t\t#Cut-off percentage 2\n", + "y = 1.4;\t\t\t#Ratio of specific heats\n", + "\n", + "# Calculations\n", + "Vs = (3.147/4)*(d**2)*L;\t\t\t#Stroke volume in m**3\n", + "V31 = V2+(c1*Vs);\t\t\t#Volume at the point of cut-off in m**3\n", + "rc1 = V31/V2;\t\t\t#Cut-off ratio 1\n", + "rv = (Vc+Vs)/Vc;\t\t\t#Compression ratio\n", + "nth1 = (1-(((rc1**y)-1)/((rv**(y-1))*y*(rc1-1))))*100;\t\t\t#Air standard efficiency 1\n", + "V32 = V2+(c2*Vs);\t\t\t#Volume at the point of cut-off in m**3\n", + "rc2 = V32/V2;\t\t\t#Cut-off ratio 2\n", + "nth2 = (1-(((rc2**y)-1)/((rv**(y-1))*y*(rc2-1))))*100;\t\t\t#Air standard efficiency 2\n", + "pl = nth1-nth2;\t\t\t#Percentage loss in efficiency\n", + "\n", + "# Results\n", + "print 'Air standard efficiency at 5 percent cut-off is %3.2f percent\\\n", + "\\nAir standard efficiency at 8 percent cut-off is %3.2f percent\\\n", + "\\nPercentage loss in efficiency is %3.2f percent'%(nth1,nth2,pl)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.9 Page no : 65" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Maximum temperature attained during the cycle is 1595.4 oC \n", + "Thermal efficiency of the cycle is 60.3 percent\n" + ] + } + ], + "source": [ + "\n", + "\n", + "# Variables\n", + "e = 7.5;\t\t\t#Expansion ratio\n", + "c = 15.;\t\t\t#Compression ratio\n", + "P1 = 98.;\t\t\t#Pressure at point 1 in kN/(m**2)\n", + "P4 = 258.;\t\t\t#Pressure at point 4 in kN/(m**2)\n", + "T1 = 317.;\t\t\t#Temperature at point 1 in K\n", + "y = 1.4;\t\t\t#Ratio of specific heats\n", + "\n", + "# Calculations\n", + "T4 = T1*(P4/P1);\t\t\t#Temperature at point 4 in K\n", + "T3 = T4*(e**(y-1));\t\t\t#Temperature at point 3 in K\n", + "t3 = T3-273;\t\t\t#Temperature at point 3 in oC\n", + "T2 = T1*(c**(y-1));\t\t\t#Temperature at point 2 in K\n", + "n = (1-((T4-T1)/(y*(T3-T2))))*100;\t\t\t#Thermal efficiency\n", + "\n", + "# Results\n", + "print 'Maximum temperature attained during the cycle is %3.1f oC \\\n", + "\\nThermal efficiency of the cycle is %3.1f percent'%(t3,n)\n", + "\n", + "# rounding off error" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.10 Page no : 66" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Thermal efficiency is 63.5 percent \n", + "Mean effective pressure is 933 kPa\n" + ] + } + ], + "source": [ + "\n", + "# Variables\n", + "rv = 20.;\t\t\t#Compression ratio\n", + "P1 = 95.;\t\t\t#Pressure at point 1 in kPa\n", + "T1 = 293.;\t\t\t#Temperature at point 1 in K\n", + "T3 = 2200.;\t\t\t#Temperature at point 3 in K\n", + "y = 1.4;\t\t\t#Ratio of specific heats\n", + "R = 287.;\t\t\t#Universal gas constant in J/kg-K\n", + "Cp = 1.005;\t\t\t#Specific heat at constant pressure in kJ/kg-K\n", + "\n", + "# Calculations\n", + "P2 = P1*(rv**y);\t\t\t#Pressure at point 2 in kPa\n", + "T2 = T1*(rv**(y-1));\t\t\t#Temperature at point 2 in K\n", + "v2 = (R*T2)/(P2*1000);\t\t\t#Specific volume at point 2 in (m**3)/kg\n", + "v3 = v2*(T3/T2);\t\t\t#Specific volume at point 3 in (m**3)/kg\n", + "rc = v3/v2;\t\t\t#Cut-off ratio\n", + "nth = (1-(((rc**y)-1)/((rv**(y-1))*y*(rc-1))))*100;\t\t\t#Thermal efficiency\n", + "q23 = Cp*(T3-T2);\t\t\t#Heat flow between points 2 and 3 in kJ/kg\n", + "wnet = (nth*q23)/100;\t\t\t#Net workdone in kJ/kg\n", + "MEP = wnet/(v2*(rv-1));\t\t\t#Mean effective pressure in kPa\n", + "\n", + "# Results\n", + "print 'Thermal efficiency is %3.1f percent \\\n", + "\\nMean effective pressure is %.f kPa'%(nth,MEP)\n", + "\n", + "# rounding off error" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.11 Page no : 68" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Cut-off ratio is 2 \n", + "Air standard efficiency is 65.36 percent\n" + ] + } + ], + "source": [ + "\n", + "# Variables\n", + "rv = 21.;\t\t\t#Compression ratio\n", + "re = 10.5;\t\t\t#Expansion ratio\n", + "y = 1.4;\t\t\t#Ratio of specific heats\n", + "\n", + "# Calculations\n", + "rc = rv/re;\t\t\t#Cut-off ratio\n", + "nth = (1-(((rc**y)-1)/((rv**(y-1))*y*(rc-1))))*100;\t\t\t#Air standard efficiency\n", + "\n", + "# Results\n", + "print 'Cut-off ratio is %3.0f \\\n", + "\\nAir standard efficiency is %3.2f percent'%(rc,nth)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.12 Page no : 69" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Ideal efficiency of engine is 61.5 percent\n" + ] + } + ], + "source": [ + "\n", + "# Variables\n", + "rv = 16.;\t\t\t#Compression ratio\n", + "rp = 1.5;\t\t\t#Pressure ratio\n", + "y = 1.4;\t\t\t#Ratio of specific heats\n", + "cp = 8;\t\t\t#Cut-off percentage\n", + "\n", + "# Calculations\n", + "rc = 2.2;\t\t\t#Cut-off ratio\n", + "ntd = (1-((rp*(rc**y)-1)/((rv**(y-1)*((rp-1)+(y*rp*(rc-1)))))))*100;\t\t\t#Dual cycle efficiency\n", + "\n", + "# Results\n", + "print 'Ideal efficiency of engine is %3.1f percent'%(ntd)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.13 Page no : 69" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Ideal efficiency of the engine is 62.2 percent\n" + ] + } + ], + "source": [ + "\n", + "# Variables\n", + "d = 0.2;\t\t\t#Bore in m\n", + "L = 0.5;\t\t\t#Stroke in m\n", + "c = 0.06;\t\t\t#Cut-off percentage\n", + "y = 1.4;\t\t\t#Ratio of specific heats\n", + "rv = 15.;\t\t\t#Compression ratio\n", + "rp = 1.4;\t\t\t#Pressure ratio\n", + "\n", + "# Calculations\n", + "Vs = (3.147/4)*(d**2)*L;\t\t\t#Stroke volume in m**3\n", + "DV = c*Vs;\t\t\t#Difference in volumes at points 4 and 3\n", + "V3 = Vs/(rv-1);\t\t\t#Specific volume at point 3 in m**3\n", + "V4 = V3+DV;\t\t\t#Specific volume at point 4 in m**3\n", + "rc = V4/V3;\t\t\t#Cut-off ratio\n", + "ntd = (1-((rp*(rc**y)-1)/((rv**(y-1)*((rp-1)+(y*rp*(rc-1)))))))*100;\t\t\t#Ideal efficiency\n", + "\n", + "# Results\n", + "print 'Ideal efficiency of the engine is %3.1f percent'%(ntd)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.14 Page no : 70" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Amount of heat added is 1742.1 kJ/kg \n", + "Amount of heat rejected is 772.91 kJ/kg \n", + "Workdone per cycle is 12.23 kJ/cycle \n", + "Thermal efficiency is 55.63 percent\n" + ] + } + ], + "source": [ + "\n", + "\n", + "# Variables\n", + "d = 0.2;\t\t\t#Bore in m\n", + "L = 0.3;\t\t\t#Stroke in m\n", + "c = 0.04;\t\t\t#Cut-off percentage\n", + "y = 1.4;\t\t\t#Ratio of specific heats\n", + "rv = 8.;\t\t\t#Compression ratio\n", + "P1 = 1.;\t\t\t#Pressure at point 1 in bar\n", + "P3 = 60.;\t\t\t#Pressure at point 3 in bar\n", + "T1 = 298.;\t\t\t#Temperature at point 1 in K\n", + "R = 287.;\t\t\t#Universal gas constant in J/kg\n", + "Cv = 0.718;\t\t\t#Speific heat at constant volume in kJ/kg-K\n", + "Cp = 1.005;\t\t\t#Speific heat at constant pressure in kJ/kg-K\n", + "\n", + "# Calculations\n", + "Vs = (3.147/4)*(d**2)*L;\t\t\t#Stroke volume in m**3\n", + "V2 = Vs/(rv-1);\t\t\t#Specific volume at point 2 in m**3\n", + "V3 = V2;\t\t\t#Specific volume at point 3 in m**3\n", + "V1 = V2+Vs;\t\t\t#Specific volume at pont 1 in m**3\n", + "V5 = V1;\t\t\t#Specific volume at pont 5 in m**3\n", + "P2 = P1*(rv**y);\t\t\t#Pressure at point 2 in bar\n", + "T2 = T1*(rv**(y-1));\t\t\t#Temperature at point 2 in K\n", + "T3 = T2*(P3/P2);\t\t\t#Temperature at point 3 in K\n", + "V4 = V3+(c*(V1-V2));\t\t\t#Specific volume at point 4 in m**3\n", + "T4 = T3*(V4/V3);\t\t\t#Temperature at point 4 in K\n", + "T5 = T4*((V4/V5)**(y-1));\t\t\t#Temperature at point 5 in K\n", + "q1 = (Cv*(T3-T2))+(Cp*(T4-T3));\t\t\t#Heat added in kJ/kg\n", + "q2 = Cv*(T5-T1);\t\t\t#Heat rejected in kJ/kg\n", + "nth = (1-(q2/q1))*100;\t\t\t#Thermal efficiency\n", + "m = (P1*V1*(10**5))/(R*T1);\t\t\t#Mass of air supplied in kg\n", + "W = m*(q1-q2);\t\t\t#Workdone in kJ/cycle\n", + "\n", + "# Results\n", + "print 'Amount of heat added is %3.1f kJ/kg \\\n", + "\\nAmount of heat rejected is %3.2f kJ/kg \\\n", + "\\nWorkdone per cycle is %3.2f kJ/cycle \\\n", + "\\nThermal efficiency is %3.2f percent'%(q1,q2,W,nth)\n", + "\n", + "# rounding off error" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.15 Page no : 72" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Mean effective pressure is 19.78 bar\n", + "Thermal efficiency is 56.48 percent\n" + ] + } + ], + "source": [ + "\n", + "\n", + "# Variables\n", + "P1 = 1.;\t\t\t#Pressure at point 1 in bar\n", + "P3 = 70.;\t\t\t#Pressure at point 3 in bar\n", + "T1 = 310.;\t\t\t#Temperature at point 1 in K\n", + "rv = 10.;\t\t\t#Compression ratio\n", + "y = 1.4;\t\t\t#Ratio of specific heats\n", + "qin = 2805.;\t\t\t#Heat added in kJ/kg\n", + "m = 1.;\t\t\t#Mass of air in kg\n", + "R = 287.;\t\t\t#Universal gas constant in J/kg\n", + "Cv = 0.718;\t\t\t#Speific heat at constant volume in kJ/kg-K\n", + "Cp = 1.005;\t\t\t#Speific heat at constant pressure in kJ/kg-K\n", + "\n", + "# Calculations\n", + "V1 = (m*R*T1)/(P1*(10**5));\t\t\t#Volume at point 1 in m**3\n", + "T2 = T1*(rv**(y-1));\t\t\t#Temperature at point 2 in K\n", + "P2 = P1*(rv**y);\t\t\t#Pressure at point 2 in K\n", + "T3 = T2*(P3/P2);\t\t\t#Temperature at point 3 in K\n", + "q23 = Cv*(T3-T2);\t\t\t#Heat supplied at constant volume in kJ/kg\n", + "q34 = qin-q23;\t\t\t#Heat supplied at constant pressure in kJ/kg\n", + "T4 = (q34/Cp)+T3;\t\t\t#Temperature at point 4 in K\n", + "V2 = V1/rv;\t\t\t#Volume at point 2 in m**3\n", + "V4 = V2*(T4/T3);\t\t\t#Volume at point 4 in m**3\n", + "V5 = V1;\t\t\t#Volume at point 5 in m**3\n", + "T5 = T4*((V4/V5)**(y-1));\t\t\t#Temperature at point 5 in K\n", + "qout = Cv*(T5-T1);\t\t\t#Heat rejected in kJ/kg\n", + "nth = (1-(qout/qin))*100;\t\t\t#Thermal efficiency\n", + "W = qin-qout;\t\t\t#Workdone in kJ/kg\n", + "Vs = V1*(1-(1/rv));\t\t\t#Swept volume in (m**3)/kg\n", + "MEP = (W/Vs)/100;\t\t\t#Mean effective pressure in bar\n", + "\n", + "# Results\n", + "print 'Mean effective pressure is %3.2f bar\\\n", + "\\nThermal efficiency is %3.2f percent'%(MEP,nth)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.16 Page no : 74" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Cycle efficiency is 26.94 percent \n", + "Heat supplied to air is 517.7 kJ/kg \n", + "Work available at the shaft is 139.47 kJ/kg\n", + "Heat rejected in the cooler is 378.23 kJ/kg \n", + "Turbine exit temperature is 674.34 K\n" + ] + } + ], + "source": [ + "\n", + "\n", + "# Variables\n", + "P1 = 1.;\t\t\t#Pressure at point 1 in bar\n", + "T1 = 298.;\t\t\t#Temperature at point 1 in K\n", + "P2 = 3.;\t\t\t#Pressure at point 2 in bar\n", + "T3 = 923.;\t\t\t#Temperature at point 3 in K\n", + "y = 1.4;\t\t\t#Ratio of specific heats\n", + "Cp = 1.005;\t\t\t#Speific heat at constant pressure in kJ/kg-K\n", + "\n", + "# Calculations\n", + "x = (y-1)/y;\t\t\t#Ratio\n", + "rp = P2/P1;\t\t\t#Pressure ratio\n", + "nth = (1-(1/(rp**x)))*100;\t\t\t#Cycle efficiency\n", + "T2 = T1*(rp**x);\t\t\t#Temperature at point 2 in K\n", + "q1 = Cp*(T3-T2);\t\t\t#Heat supplied in kJ/kg\n", + "Wout = (nth*q1)/100;\t\t\t#Work output in kJ/kg\n", + "q2 = q1-Wout;\t\t\t#Heat rejected in kJ/kg\n", + "T4 = T3*((1/rp)**x);\t\t\t#Temperature at point 4 in K\n", + "\n", + "# Results\n", + "print 'Cycle efficiency is %3.2f percent \\\n", + "\\nHeat supplied to air is %3.1f kJ/kg \\\n", + "\\nWork available at the shaft is %3.2f kJ/kg\\\n", + "\\nHeat rejected in the cooler is %3.2f kJ/kg \\\n", + "\\nTurbine exit temperature is %3.2f K'%(nth,q1,Wout,q2,T4)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.17 Page no : 75" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Optimum pressure ratio is 14.74 \n", + "Maximum net specific work output 401 kJ/kg \n", + "Thermal efficiency 54 percent \n", + "Work ratio is 0.54 \n", + "Carnot efficiency is 79 percent\n" + ] + } + ], + "source": [ + "\n", + "import math\n", + "\n", + "# Variables\n", + "T1 = 283.;\t\t\t#Temperature at point 1 in K\n", + "T3 = 1353.;\t\t\t#Temperature at point 3 in K\n", + "y = 1.41;\t\t\t#Ratio of specific heats\n", + "Cp = 1.007;\t\t\t#Specific heat constant pressure in kJ/kg-K\n", + "\n", + "# Calculations\n", + "x = (y-1)/y;\t \t\t#Ratio\n", + "rpmax = ((T3/T1)**(1/x));\t\t\t#Maximum pressure ratio\n", + "rpopt = math.sqrt(rpmax);\t\t\t#Optimum pressure ratio\n", + "T2 = T1*(rpopt**x);\t \t\t#Temperature at point 2 in K\n", + "T4 = T2;\t\t\t #Maximum temperature at point 4 in K\n", + "Wmax = Cp*((T3-T4)-(T2-T1));\t\t\t#Maximum net specific work output in kJ/kg\n", + "nth = (Wmax/(Cp*(T3-T2)))*100;\t\t\t#Thermal efficiency\n", + "WR = nth/100; \t\t\t#Work ratio\n", + "nc = ((T3-T1)/T3)*100;\t \t\t#Carnot efficiency\n", + "\n", + "# Results\n", + "print 'Optimum pressure ratio is %3.2f \\\n", + "\\nMaximum net specific work output %3.0f kJ/kg \\\n", + "\\nThermal efficiency %3.0f percent \\\n", + "\\nWork ratio is %3.2f \\\n", + "\\nCarnot efficiency is %3.0f percent'%(rpopt,Wmax,nth,WR,nc)\n", + "\n", + "# rounding off error. please check." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.18 Page no : 76" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Maximum work per kg of air is 239.47 kJ/kg \n", + "Cycle efficiency is 47 percent\n", + "Ratio of brayton cycle efficiency to carnot efficieny is 0.654\n" + ] + } + ], + "source": [ + "\n", + "\n", + "# Variables\n", + "Tmin = 300.;\t\t\t#Minimum temperature in K\n", + "Tmax = 1073.;\t\t\t#Maximum temperature in K\n", + "Cp = 1.005;\t\t\t#Specific heat at constant pressure in kJ/kg-K\n", + "\n", + "# Calculations\n", + "Wmax = Cp*((math.sqrt(Tmax)-math.sqrt(Tmin))**2);\t\t\t#Maximum work output in kJ/kg\n", + "nB = (1-math.sqrt(Tmin/Tmax))*100;\t\t\t#Brayton cycle efficiency\n", + "nC = (1-(Tmin/Tmax))*100;\t\t \t#Carnot efficiency\n", + "r = nB/nC;\t \t\t #Ratio of brayton cycle efficiency to carnot efficieny\n", + "\n", + "# Results\n", + "print 'Maximum work per kg of air is %3.2f kJ/kg \\\n", + "\\nCycle efficiency is %3.0f percent\\\n", + "\\nRatio of brayton cycle efficiency to carnot efficieny is %3.3f'%(Wmax,nB,r)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.19 Page no : 77" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Net power output of the turbine is 1014 kW \n", + "Thermal efficiency of the plant is 32 percent\n", + "Work ratio is 0.446\n" + ] + } + ], + "source": [ + "\n", + "\n", + "# Variables\n", + "T1 = 291.;\t\t\t#Temperature at point 1 in K\n", + "P1 = 100.;\t\t\t#Pressure at point 1 in kN/(m**2)\n", + "nC = 0.85;\t\t\t#Isentropic efficiency of compressor\n", + "nT = 0.88;\t\t\t#Isentropic effficiency of turbine\n", + "rp = 8.;\t\t\t#Pressure ratio\n", + "T3 = 1273.;\t\t\t#Temperature at point 3 in K\n", + "m = 4.5;\t\t\t#Mass flow rate of air in kg/s\n", + "y = 1.4;\t\t\t#Ratio of speciifc heats\n", + "Cp = 1.006;\t\t\t#Specific heat at constant pressure in kJ/kg-K\n", + "\n", + "# Calculations\n", + "x = (y-1)/y;\t\t\t#Ratio\n", + "T2s = T1*(rp**x);\t\t\t#Temperature at point 2s in K\n", + "T2 = T1+((T2s-T1)/nC);\t\t\t#Temperature at point 2 in K\n", + "t2 = T2-273;\t\t\t#Temperature at point 2 in oC\n", + "T4s = T3*((1/rp)**x);\t\t\t#Temperature at point 4s in K\n", + "T4 = T3-((T3-T4s)*nT);\t\t\t#Temperature at point 4 in K\n", + "t4 = T4-273;\t\t\t#Temperature at point 4 in oC\n", + "W = m*Cp*((T3-T4)-(T2-T1));\t\t\t#Net power output in kW\n", + "nth = (((T3-T4)-(T2-T1))/(T3-T2))*100;\t\t\t#Thermal efficiency\n", + "WR = W/(m*Cp*(T3-T4));\t\t\t#Work ratio\n", + "\n", + "# Results\n", + "print 'Net power output of the turbine is %3.0f kW \\\n", + "\\nThermal efficiency of the plant is %3.0f percent\\\n", + "\\nWork ratio is %3.3f'%(W,nth,WR)\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.20 Page no : 79" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Percentage increase in the cycle efficiency due to regeneration is 41.41 percent\n" + ] + } + ], + "source": [ + "\n", + "# Variables\n", + "P1 = 0.1;\t\t\t#Pressure at point 1 in MPa\n", + "T1 = 303.;\t\t\t#Temperature at point 1 in K\n", + "T3 = 1173.;\t\t\t#Temperature at point 3 in K\n", + "rp = 6.; \t\t\t#Pressure ratio\n", + "nC = 0.8;\t\t\t#Compressor efficiency\n", + "nT = nC;\t\t\t#Turbine efficiency\n", + "e = 0.75;\t\t\t#Regenerator effectiveness\n", + "y = 1.4;\t\t\t#Ratio of specific heats\n", + "Cp = 1.005;\t\t\t#Specific heat at constant pressure in kJ/kg-K\n", + "\n", + "# Calculations\n", + "x = (y-1)/y; \t\t\t#Ratio\n", + "T2s = T1*(rp**x);\t\t\t#Temperature at point 2s in K\n", + "T4s = T3/(rp**x);\t\t\t#Temperature at point 4s in K\n", + "DTa = (T2s-T1)/nC;\t\t\t#Difference in temperatures at point 2 and 1 in K\n", + "DTb = (T3-T4s)*nT;\t\t\t#Difference in temperatures at point 3 and 4 in K\n", + "wT = Cp*DTb;\t \t\t#Turbine work in kJ/kg\n", + "wC = Cp*DTa;\t\t \t#Compressor work in kJ/kg\n", + "T2 = DTa+T1;\t\t\t #Temperature at point 2 in K\n", + "q1 = Cp*(T3-T2);\t\t\t#Heat supplied in kJ/kg\n", + "nth1 = ((wT-wC)/q1)*100;\t\t\t#Cycle efficiency without regenerator\n", + "T4 = T3-DTb;\t\t \t#Temperature at point 4 in K\n", + "T5 = T2+(e*(T4-T2));\t\t\t#Temperature at point 5 in K\n", + "q2 = Cp*(T3-T5);\t\t\t#Heat supplied with regenerator in kJ/kg\n", + "nth2 = ((wT-wC)/q2)*100;\t\t\t#Cycle efficiency with regenerator\n", + "p = ((nth2-nth1)/nth1)*100;\t\t\t#Percentage increase due to regeneration\n", + "\n", + "# Results\n", + "print 'Percentage increase in the cycle efficiency due to regeneration is %3.2f percent'%(p)\n", + "\n", + "# rounding off error. please check." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.21 Page no : 80" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Velocity of air leaving the nozzle is 712.5 m/s\n" + ] + } + ], + "source": [ + "\n", + "# Variables\n", + "P1 = 1.;\t\t\t#Pressure at point 1 in atm\n", + "P3 = 5.;\t\t\t#Pressure at point 3 in atm\n", + "T1 = 288.;\t\t\t#Temperature at point 1 in K\n", + "T4 = 1143.;\t\t\t#Temperature at point 4 in K\n", + "y = 1.4;\t\t\t#Ratio of specific heats\n", + "Cp = 1.005;\t\t\t#Specific heat at constant pressure in kJ/kg-K\n", + "\n", + "# Calculations\n", + "rp = P3/P1;\t\t\t#Pressure ratio\n", + "x = round((y-1)/y,3);\t\t\t#Ratio\n", + "rpx = round(rp**x,2)\n", + "T3 = round(T1*(rpx));\t\t\t#Temperature at point 3 in K\n", + "T5 = T4-(T3-T1);\t\t\t#Temperature at point 5 in K\n", + "T6 = T4/(rpx);\t\t\t#Temperature at point 6 in K\n", + "C6 = math.sqrt(2000*Cp*(T5-T6));\t\t\t#Velocity of air leaving the nozzle in m/s\n", + "\n", + "\n", + "# Results\n", + "print 'Velocity of air leaving the nozzle is %3.1f m/s'%(C6)\n", + "\n", + "# rounding error. Please check. there is rounding off error in book" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.22 Page no : 81" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Pressure at the turbine exit is 374.2 kPa \n", + "Velocity of exhaust gases are 933.5 m/s \n", + "Propulsive efficiency is 26.9 percent\n" + ] + } + ], + "source": [ + "\n", + "\n", + "# Variables\n", + "C1 = 280.;\t\t\t#Velocity of aircraft in m/s\n", + "P1 = 48.;\t\t\t#Pressure at point 1 kPa\n", + "T1 = 260.;\t\t\t#Temperature at point 1 in K\n", + "rp = 13.;\t\t\t#Pressure ratio\n", + "T4 = 1300.;\t\t\t#Temperature at point 4 in K\n", + "Cp = 1005.;\t\t\t#Specific heat at constant pressure in J/kg\n", + "y = 1.4;\t\t\t#Ratio of specific heats\n", + "\n", + "# Calculations\n", + "x = (y-1)/y;\t\t\t#Ratio\n", + "T2 = T1+((C1**2)/(2*Cp));\t\t\t#Temperature at point 2 in K\n", + "P2 = P1*((T2/T1)**(1/x));\t\t\t#Pressure at point 2 in kPa\n", + "P3 = rp*P2;\t\t\t#Pressure at point 3 in kPa\n", + "P4 = P3;\t\t\t#Pressure at point 4 in kPa\n", + "T3 = T2*(rp**x);\t\t\t#Temperature at point 3 in K\n", + "T5 = T4-T3+T2;\t\t\t#Temperature at point 5 in K\n", + "P5 = P4*((T5/T4)**(1/x));\t\t\t#Pressure at point 5 in kPa\n", + "P6 = P1;\t\t\t#Pressure at point 6 in kPa\n", + "T6 = T5*((P6/P5)**x);\t\t\t#Temperature at point 6 in K\n", + "C6 = math.sqrt(2*Cp*(T5-T6));\t\t\t#Velocity of air at nozzle exit in m/s\n", + "W = (C6-C1)*C1;\t\t\t#Propulsive power in J/kg\n", + "Q = Cp*(T4-T3);\t\t\t#Total heat transfer rate in J/kg\n", + "nP = (W/Q)*100;\t\t\t#Propulsive efficiency\n", + "\n", + "# Results\n", + "print 'Pressure at the turbine exit is %3.1f kPa \\\n", + "\\nVelocity of exhaust gases are %3.1f m/s \\\n", + "\\nPropulsive efficiency is %3.1f percent'%(P5,C6,nP)\n", + "\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Thermal_Engineering_by_A._V._Arasu/ch3.ipynb b/Thermal_Engineering_by_A._V._Arasu/ch3.ipynb new file mode 100644 index 00000000..8b9a3246 --- /dev/null +++ b/Thermal_Engineering_by_A._V._Arasu/ch3.ipynb @@ -0,0 +1,439 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:3e461b166983dbc4e8640a4c60ebf5164b7675ebb01d9440eb0b63f7316d9dde" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 3 :\n", + "Internal Combustion Engines" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.1 Page no : 139" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "d = 200.;\t\t\t#diameter of cylinder in mm\n", + "L = 300.;\t\t\t#stroke of cylinder in mm\n", + "Vc = 1.73;\t\t\t#Clearance volume in litres\n", + "imep = 650.;\t\t\t#indicated mean effective pressure in kN/(m**2)\n", + "g = 6.2;\t\t\t#gas consumption in (m**3)/h\n", + "CV = 38.5;\t\t\t#Calorific value in MJ/(m**3)\n", + "y = 1.4;\t\t\t#Ratio of specific heats\n", + "N = 150.;\t\t\t#No. of firing cycles per minute\n", + "\n", + "# Calculations\n", + "Vs = ((3.1415/4)*(d**2)*L)*(10**-6);\t\t\t#Stroke volume in litres\n", + "Vt = Vs+Vc;\t\t\t#Total volume in litres\n", + "rv = (Vt/Vc);\t\t\t#Compression ratio\n", + "n = (1-(1/rv**(y-1)))*100;\t\t\t#Air standard efficiency\n", + "IP = imep*(Vs*10**-3)*(N/60);\t\t\t#Indicated power in kW\n", + "F = (g*CV*1000)/3600;\t\t\t#Fuel energy input in kW\n", + "nT = (IP/F)*100;\t\t\t#Indicated thermal efficiency\n", + "\n", + "# Results\n", + "# 1st answer is wrong in book\n", + "print 'Air Standard Efficiency is %3.1f percent \\\n", + "\\nIndicated Power is %3.1f kW \\\n", + "\\nIndicated thermal efficiency is %3.0f percent'%(n,IP,nT)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Air Standard Efficiency is 52.5 percent \n", + "Indicated Power is 15.3 kW \n", + "Indicated thermal efficiency is 23 percent\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.2 Page no : 140" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "Vs = 0.0008;\t\t\t#Swept volume in m**3\n", + "Vc = 0.00015;\t\t\t#Clearance volume in m**3\n", + "CV = 38.;\t\t\t#Calorific value in MJ/(m**3)\n", + "v = 0.45;\t\t\t#volume in m**3\n", + "IP = 81.5;\t\t\t#Indicated power in kW\n", + "y = 1.4;\t\t\t#Ratio of specific heats\n", + "\n", + "# Calculations\n", + "rv = (Vs+Vc)/Vc;\t\t\t#Compression ratio\n", + "n = (1-(1/rv**(y-1)));\t\t\t#Air standard efficiency\n", + "Ps = (v*CV*1000.)/60;\t\t\t#Power supplied in kW\n", + "nact = IP/Ps;\t\t\t#Actual efficiency\n", + "nr = (nact/n)*100;\t\t\t#Relative efficiency\n", + "\n", + "\n", + "# Results\n", + "print 'Relative Efficiency is %3.2f percent'%(nr)\n", + "\n", + "# rounding error in book answer. please check." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Relative Efficiency is 54.77 percent\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3 Page no : 141" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "n = 6.;\t\t\t #No. of cylinders\n", + "d = 0.61;\t\t\t#Diameter in m\n", + "L = 1.25;\t\t\t#Stroke in m\n", + "N = 2.;\t\t \t#No.of revolutions per second\n", + "m = 340.;\t\t\t#mass of fuel oil in kg\n", + "CV = 44200.;\t\t#Calorific value in kJ/kg\n", + "T = 108.;\t\t\t#Torque in kN-m\n", + "imep = 775.;\t\t#Indicated mean efective pressure in kN/(m**2)\n", + "\n", + "# Calculations\n", + "IP = (imep*L*3.1415*(d**2)*N)/(8);\t\t\t#Indicated power in kW\n", + "TotalIP = (n*IP);\t\t\t #Total indicated power in kW\n", + "BP = (2*3.1415*N*T);\t\t\t#Brake power in kW\n", + "PI = (m*CV)/3600.;\t\t\t #Power input in kW\n", + "nB = (BP/PI)*100.;\t\t \t#Brake thermal efficiency\n", + "bmep = (BP*8)/(n*L*3.1415*(d**2)*2);\t\t\t#Brake mean effective pressure in kN/(m**2)\n", + "nM = (BP/TotalIP)*100;\t\t\t#Mechanical efficiency\n", + "bsfc = m/BP;\t \t\t#Brake specific fuel consumption in kg/kWh\n", + "\n", + "# Results\n", + "print 'Total Indicated Power is %3.1f kW \\\n", + "\\nBrake Power is %3.1f kW \\\n", + "\\nBrake thermal efficiency is %3.1f percent \\\n", + "\\nBrake mean effective pressure is %3.1f kN/m**2 \\\n", + "\\nMechanical efficiency is %3.1f percent \\\n", + "\\nBrake specific fuel consumption is %3.3f kg/kW.hr'%(TotalIP,BP,nB,bmep,nM,bsfc)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Total Indicated Power is 1698.6 kW \n", + "Brake Power is 1357.1 kW \n", + "Brake thermal efficiency is 32.5 percent \n", + "Brake mean effective pressure is 619.2 kN/m**2 \n", + "Mechanical efficiency is 79.9 percent \n", + "Brake specific fuel consumption is 0.251 kg/kW.hr\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.4 Page no : 142" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "Hm = 21.;\t\t\t#Mean height of indicator diagram in mm\n", + "isn = 27.;\t\t\t#indicator spring number in kN/(m**2)/mm\n", + "Vs = 14.;\t\t\t#Swept volume in litres\n", + "N = 6.6;\t\t\t#Speed of engine in rev/s\n", + "Pe = 77.;\t\t\t#Effective brake load in kg\n", + "Re = 0.7;\t\t\t#Effective vrake radius in m\n", + "mf = 0.002;\t\t\t#fuel consumed in kg/s\n", + "CV = 44000.;\t\t\t#Calorific value of fuel in kJ/kg\n", + "mc = 0.15;\t\t\t#cooling water circulation in kg/s\n", + "Ti = 311.;\t\t\t#cooling water inlet temperature in K\n", + "To = 344.;\t\t\t#cooling water outlet temperature in K\n", + "C = 4.18;\t\t\t#specific heat capacity of water in kJ/kg-K\n", + "Ee = 33.6;\t\t\t#Energy to exhaust gases in kJ/s\n", + "g = 9.81;\t\t\t#Acceleration due to geravity in m/(s**2)\n", + "\n", + "# Calculations\n", + "imep = isn*Hm;\t\t\t#Indicated mean efective pressure in kN/(m**2)\n", + "IP = (imep*Vs*N)/(2000);\t\t\t#Indicated Power in kW\n", + "BP = (2*3.1415*N*g*Pe*Re)/1000;\t\t\t#Brake Power in kW\n", + "nM = (BP/IP)*100;\t\t\t#Mechanical efficiency\n", + "Ef = mf*CV;\t\t\t#Eneergy from fuel in kJ/s\n", + "Ec = mc*C*(To-Ti);\t\t\t#Energy to cooling water in kJ/s\n", + "Es = Ef-(BP+Ec+Ee);\t\t\t#Energy to surroundings in kJ/s\n", + "p = (BP*100)/Ef;\t\t\t#Energy to BP in %\n", + "q = (Ec*100)/Ef;\t\t\t#Energy to coolant in %\n", + "r = (Ee*100)/Ef;\t\t\t#Energy to exhaust in %\n", + "w = (Es*100)/Ef;\t\t\t#Energy to surroundings in %\n", + "\n", + "# Results\n", + "print 'Indicated Power is %3.1f kW \\\n", + "\\nBrake Power is %3.0f kW \\\n", + "\\nMechanical Efficiency is %3.0f percent \\\n", + "\\nENERGY BALANCE kJ/s Percentage \\\n", + "\\nEnergy from fuel %3.0f 100 \\\n", + "\\nEnergy to BP %3.0f %3.0f \\\n", + "\\nEnergy to coolant %3.01f %3.1f \\\n", + "\\nEnergy to exhaust %3.1f %3.1f \\\n", + "\\nEnergy to surroundings, etc %3.1f %3.1f'%(IP,BP,nM,Ef,BP,p,Ec,q,Ee,r,Es,w)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Indicated Power is 26.2 kW \n", + "Brake Power is 22 kW \n", + "Mechanical Efficiency is 84 percent \n", + "ENERGY BALANCE kJ/s Percentage \n", + "Energy from fuel 88 100 \n", + "Energy to BP 22 25 \n", + "Energy to coolant 20.7 23.5 \n", + "Energy to exhaust 33.6 38.2 \n", + "Energy to surroundings, etc 11.8 13.4\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.5 Page no : 143" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "t = 30.;\t\t\t#duration of trial in minutes\n", + "N = 1750.;\t\t\t#speed in rpm\n", + "T = 330.;\t\t\t#brake torque in Nm\n", + "m = 9.35;\t\t\t#mass of fuel in kg\n", + "CV = 42300.;\t\t\t#Calorific value in kJ/kg\n", + "mj = 483.;\t\t\t#jacket cooling water circulation in kg\n", + "Ti = 290.;\t\t\t#inlet temperature in K\n", + "T0 = 350.;\t\t\t#outlet temperature in K\n", + "ma = 182.;\t\t\t#air consumption in kg\n", + "Te = 759.;\t\t\t#exhaust temperature in K\n", + "Ta = 256.;\t\t\t#atmospheric temperature in K\n", + "nM = 0.83;\t\t\t#Mechanical efficiency\n", + "ms = 1.25;\t\t\t#mean specific heat capacity of exhaust gas in kJ/kg-K\n", + "Cw = 4.18;\t\t\t#specific heat capacity of water in kJ/kg-K\n", + "\n", + "# Calculations\n", + "BP = (2*3.1415*T*N)/(60*1000);\t\t\t#Brake power in kW\n", + "sfc = (m*2)/BP;\t\t\t#specific fuel consumption in kg/kWh\n", + "IP = BP/nM;\t\t\t#Indicated power in kW\n", + "nIT = IP/(m*2/3600*CV)*100 \t\t\t#Indicated thermal efficiency\n", + "Ef = (m/t*CV) \t\t\t#Eneergy from fuel in kJ/min\n", + "EBP = BP*60;\t\t\t#Energy to BP in kJ/min\n", + "Ec = (mj*Cw*(T0-Ti))/t;\t\t\t#Energy to cooling water in kJ/min\n", + "Ee = ((ma+m)*ms*(Te-Ti))/30;\t\t\t#Energy to exhaust in kJ/min\n", + "Es = Ef-(EBP+Ec+Ee);\t\t\t#Energy to surroundings in kJ/min\n", + "\n", + "# Results\n", + "print 'Break power is %3.1f kW \\\n", + "\\nSpecific fuel consumption is %3.3f kg/kWh \\\n", + "\\nIndicated thermal efficiency is %3.1f percent \\\n", + "\\nEnergy from fuel is %3.0f kJ/min \\\n", + "\\nEnergy to BP is %3.0f kJ/min \\\n", + "\\nEnergy to cooling water is %3.0f kJ/min \\\n", + "\\nEnergy to exhaust is %3.0f kJ/min \\\n", + "\\nEnergy to surroundings is %d kJ/min'%(BP,sfc,nIT,round(Ef,-2),EBP,Ec,Ee,Es)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Break power is 60.5 kW \n", + "Specific fuel consumption is 0.309 kg/kWh \n", + "Indicated thermal efficiency is 33.2 percent \n", + "Energy from fuel is 13200 kJ/min \n", + "Energy to BP is 3628 kJ/min \n", + "Energy to cooling water is 4038 kJ/min \n", + "Energy to exhaust is 3739 kJ/min \n", + "Energy to surroundings is 1777 kJ/min\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.6 Page no : 144" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "BP0 = 12.;\t\t\t#Brake Power output in kW\n", + "BP1 = 40.5;\t\t\t#Brake Power in trial 1 in kW\n", + "BP2 = 40.2;\t\t\t#Brake Power in trial 2 in kW\n", + "BP3 = 40.1;\t\t\t#Brake Power in trial 3 in kW\n", + "BP4 = 40.6;\t\t\t#Brake Power in trial 4 in kW\n", + "BP5 = 40.7;\t\t\t#Brake Power in trial 5 in kW\n", + "BP6 = 40.0;\t\t\t#Brake Power in trial 6 in kW\n", + "\n", + "# Calculations\n", + "BPALL = BP0+BP6;\t\t\t#Total Brake Power in kW\n", + "IP1 = BPALL-BP1;\t\t\t#Indicated Power in trial 1 in kW\n", + "IP2 = BPALL-BP2;\t\t\t#Indicated Power in trial 2 in kW\n", + "IP3 = BPALL-BP3;\t\t\t#Indicated Power in trial 3 in kW\n", + "IP4 = BPALL-BP4;\t\t\t#Indicated Power in trial 4 in kW\n", + "IP5 = BPALL-BP5;\t\t\t#Indicated Power in trial 5 in kW\n", + "IP6 = BPALL-BP6;\t\t\t#Indicated Power in trial 6 in kW\n", + "IPALL = IP1+IP2+IP3+IP4+IP5+IP6;\t\t\t#Total Indicated Power in kW\n", + "nM = (BPALL/IPALL)*100;\t\t\t#Mechanical efficiency\n", + "\n", + "# Results\n", + "print 'Indicated Power of the engine is %3.1f kW \\\n", + "\\nMechanical efficiency of the engine is %3.1f percent'%(IPALL,nM)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Indicated Power of the engine is 69.9 kW \n", + "Mechanical efficiency of the engine is 74.4 percent\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.7 Page no : 145" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math\n", + "\n", + "# Variables\n", + "n = 2.;\t\t\t#No. of cylinders\n", + "N = 4000.;\t\t\t#speed of engine in rpm\n", + "nV = 0.77;\t\t\t#Volumetric efficiency\n", + "nM = 0.75;\t\t\t#Mechanical efficiency\n", + "m = 10.;\t\t\t#fuel consumed in lit/h\n", + "g = 0.73;\t\t\t#spcific gravity of fuel\n", + "Raf = 18.;\t\t\t#air-fuel ratio\n", + "Np = 600.;\t\t\t#piston speed in m/min\n", + "imep = 5.;\t\t\t#Indicated mean efective pressure in bar\n", + "R = 281.;\t\t\t#Universal gas constant in J/kg-K\n", + "T = 288.;\t\t\t#Standard temperature in K\n", + "P = 1.013;\t\t\t#Standard pressure in bar\n", + "\n", + "\n", + "# Calculations\n", + "L = Np/(2*N);\t\t\t#Piston stroke in m\n", + "mf = m*g;\t\t\t#mass of fuel in kg/h\n", + "ma = mf*Raf;\t\t\t#mass of air required in kg/h\n", + "Va = (ma*R*T)/(P*60*(10**5));\t\t\t#volume of air required in (m**3)/min\n", + "D = math.sqrt((2*Va)/(nV*L*N*3.1415));\t\t\t#Diameter in m\n", + "IP = (2*imep*100*L*3.1415*(D**2)*N)/(4.*60);\t\t\t#Indicated Power in kW\n", + "BP = nV*IP;\t\t\t#Brake Power in kW\n", + "\n", + "# Results\n", + "print 'Piston Stroke is %3.3f m \\\n", + "\\nBore diameter is %3.4f m \\\n", + "\\nBrake power is %3.1f kW'%(L,D,BP)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Piston Stroke is 0.075 m \n", + "Bore diameter is 0.0694 m \n", + "Brake power is 14.6 kW\n" + ] + } + ], + "prompt_number": 5 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Thermal_Engineering_by_A._V._Arasu/ch4.ipynb b/Thermal_Engineering_by_A._V._Arasu/ch4.ipynb new file mode 100644 index 00000000..09c1bcde --- /dev/null +++ b/Thermal_Engineering_by_A._V._Arasu/ch4.ipynb @@ -0,0 +1,1482 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4 : Steam nozzles and Steam turbines" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.1 Page no : 161" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Throat area is 255 mm**2 \n", + "Exit area is 344 mm**2 \n", + "Mach number at exit is 1.49\n" + ] + } + ], + "source": [ + "\n", + "# Variables\n", + "P1 = 3.5;\t\t\t#Pressure at entry in MN/(m**2)\n", + "T1 = 773.;\t\t\t#Temperature at entry in K\n", + "P2 = 0.7;\t\t\t#Pressure at exit in MN/(m**2)\n", + "ma = 1.3;\t\t\t#mass flow rate of air in kg/s\n", + "y = 1.4;\t\t\t#Ratio of specific heats\n", + "R = 0.287;\t\t\t#Universal gas constant in KJ/Kg-K\n", + "\n", + "# Calculations\n", + "c = y/(y-1); \t\t\t#Ratio\n", + "Pt = ((2/(y+1))**c)*P1;\t\t\t#Throat pressure in MN/(m**2)\n", + "v1 = (R*T1)/(P1*1000);\t\t\t#Specific volume at entry in (m**3)/kg\n", + "Ct = ((2*c*P1*v1*(1-((Pt/P1)**(1/c))))**0.5)*1000;\t\t\t#Velocity at throat in m/s\n", + "vt = v1*((P1/Pt)**(1/y));\t\t\t#Specific volume at throat in (m**3)/kg\n", + "At = ((ma*vt)/Ct)*(10**6);\t\t\t#Area of throat in (mm**2)\n", + "C2 = ((2*c*P1*v1*(1-((P2/P1)**(1/c))))**0.5)*1000;\t\t\t#Velocity at exit in m/s\n", + "v2 = v1*((P1/P2)**(1/y));\t\t\t#Specific volume at exit in (m**3)/kg\n", + "A2 = ((ma*v2)/C2)*(10**6);\t\t\t#Area of exit in (mm**2)\n", + "M = C2/Ct;\t\t\t #Mach number at exit\n", + "\n", + "# Results\n", + "print 'Throat area is %3.0f mm**2 \\\n", + "\\nExit area is %3.0f mm**2 \\\n", + "\\nMach number at exit is %3.2f'%(At,A2,M)\n", + "\n", + "# rounding off error" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.2 Page no : 163" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Increase in temperature is 356 K \n", + "Increase in pressure is 2.46 MN/m**2 \n", + "Increase in internal energy is 255 kJ/kg\n" + ] + } + ], + "source": [ + "\n", + "# Variables\n", + "T1 = 273.;\t\t\t#Temperature at section 1 in K\n", + "P1 = 140.;\t\t\t#Pressure at section 1 in KN/(m**2)\n", + "v1 = 900.;\t\t\t#Velocity at section 1 in m/s\n", + "v2 = 300.;\t\t\t#Velocity at section 2 in m/s\n", + "Cp = 1.006;\t\t\t#Specific heat at constant pressure in kJ/kg-K\n", + "Cv = 0.717;\t\t\t#Specific heat at constant volume in kJ/kg-K\n", + "y = 1.4;\t\t\t#Ratio of specific heats\n", + "\n", + "# Calculations\n", + "c = y/(y-1);\t\t\t#Ratio\n", + "R = Cp-Cv;\t\t\t#Universal gas constant in KJ/Kg-K\n", + "T2 = T1-(((v2)**2-(v1)**2)/(2000*c*R));\t\t\t#Temperature at section 2 in K\n", + "DT = T2-T1;\t\t\t#Increase in temperature in K\n", + "P2 = P1*((T2/T1)**c);\t\t\t#Pressure at section 2 in KN/(m**2)\n", + "DP = (P2-P1)/1000;\t\t\t#Increase in pressure in MN/(m**2)\n", + "IE = Cv*(T2-T1);\t\t\t#Increase in internal energy in kJ/kg\n", + "\n", + "# Results\n", + "print 'Increase in temperature is %3.0f K \\\n", + "\\nIncrease in pressure is %3.2f MN/m**2 \\\n", + "\\nIncrease in internal energy is %3.0f kJ/kg'%(DT,DP,IE)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.3 Page no : 163" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Throat area is 2888 mm**2 \n", + "Exit area is 4280 mm**2 \n", + "Degree of undercooling at exit is 10.3 K\n" + ] + } + ], + "source": [ + "\n", + "\n", + "# Variables\n", + "P1 = 2;\t\t\t#Pressure at entry in MN/(m**2)\n", + "T1 = 598;\t\t\t#Temperature at entry in K\n", + "P2 = 0.36;\t\t\t#Pressure at exit in MN/(m**2)\n", + "m = 7.5;\t\t\t#mass flow rate of steam in kg/s\n", + "n = 1.3;\t\t\t#Adiabatic gas constant\n", + "v1 = 0.132;\t\t\t#Volume at entry in (m**3)/kg from steam table\n", + "Ts = 412.9;\t\t\t#Saturation temperature in K\n", + "\n", + "# Calculations\n", + "c = n/(n-1);\t\t\t#Ratio\n", + "Pt = ((2/(n+1))**c)*P1;\t\t\t#Throat pressure in MN/(m**2)\n", + "Ct = ((2*c*P1*v1*(1-((Pt/P1)**(1/c))))**0.5)*1000;\t\t\t#Velocity at throat in m/s\n", + "vt = v1*((P1/Pt)**(1/n));\t\t\t#Specific volume at throat in (m**3)/kg\n", + "At = ((m*vt)/Ct)*(10**6);\t\t\t#Area of throat in (mm**2)\n", + "C2 = ((2*c*P1*v1*(1-((P2/P1)**(1/c))))**0.5)*1000;\t\t\t#Velocity at exit in m/s\n", + "v2 = v1*((P1/P2)**(1/n));\t\t\t#Specific volume at exit in (m**3)/kg\n", + "A2 = ((m*v2)/C2)*(10**6);\t\t\t#Area of exit in (mm**2)\n", + "T2 = T1*((P2/P1)**(1/c));\t\t\t#Temperature at exit in K\n", + "D = Ts-T2;\t\t\t#Degree of undercooling at exit in K\n", + "\n", + "# Results\n", + "print 'Throat area is %3.0f mm**2 \\\n", + "\\nExit area is %3.0f mm**2 \\\n", + "\\nDegree of undercooling at exit is %3.1f K'%(At,round(A2,-1),D)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.4 Page no : 165" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Throat velocity is 548 m/s \n", + "Exit velocity is 800 m/s \n", + "Throat area is 3210 mm**2 \n", + "Exit area is 6050 mm**2 \n" + ] + } + ], + "source": [ + "\n", + "\n", + "# Variables\n", + "P1 = 2.2;\t\t\t#Pressure at entry in MN/(m**2)\n", + "T1 = 533.;\t\t\t#Temperature at entry in K\n", + "P2 = 0.4;\t\t\t#Pressure at exit in MN/(m**2)\n", + "m = 11.;\t\t\t#mass flow rate of steam in kg/s\n", + "n = 0.85;\t\t\t#Efficiency of expansion\n", + "h1 = 2940.;\t\t\t#Enthalpy at entrance in kJ/kg from Moiller chart\n", + "ht = 2790.;\t\t\t#Enthalpy at throat in kJ/kg from Moiller chart\n", + "h2s = 2590.;\t\t\t#Enthalpy below exit level in kJ/kg from Moiller chart\n", + "vt = 0.16;\t\t\t#Throat volume in (m**3)/kg\n", + "v2 = 0.44;\t\t\t#Volume at exit in (m**3)/kg\n", + "\n", + "# Calculations\n", + "Ct = (2000*(h1-ht))**0.5;\t\t\t#Throat velocity in m/s\n", + "h2 = ht-(0.85*(ht-h2s));\t\t\t#Enthalpy at exit in kJ/kg\n", + "C2 = (2000*(h1-h2))**0.5;\t\t\t#Exit velocity in m/s\n", + "At = ((m*vt)/Ct)*(10**6);\t\t\t#Area of throat in (mm**2)\n", + "A2 = ((m*v2)/C2)*(10**6);\t\t\t#Area of exit in (mm**2)\n", + "\n", + "# Results\n", + "print 'Throat velocity is %3.0f m/s \\\n", + "\\nExit velocity is %3.0f m/s \\\n", + "\\nThroat area is %3.0f mm**2 \\\n", + "\\nExit area is %3.0f mm**2 '%(Ct,C2,round(At,-1),A2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.5 Page no : 166" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Cross section of nozzle is 26.7 mm * 8.9 mm \n", + "Degree of undercooling is 35.8 K and Degree of supersaturation is 2.58 \n", + "Loss in available heat drop due to irreversibility is 6.16 kJ/kg \n", + "Increase in entropy is 0.01390 kJ/kg-K \n", + "Ratio of mass flow rate with metastable expansion to the thermal expansion is 1.065\n" + ] + } + ], + "source": [ + "\n", + "\n", + "# Variables\n", + "P1 = 35.;\t\t\t#Pressure at entry in bar\n", + "T1 = 573.;\t\t\t#Temperature at entry in K\n", + "P2 = 8.;\t\t\t#Pressure at exit in bar\n", + "Ts = 443.4;\t\t\t#Saturation temperature in K\n", + "Ps = 3.1;\t\t\t#Saturation pressure in bar\n", + "m = 5.2;\t\t\t#mass flow rate of steam in kg/s\n", + "n = 1.3;\t\t\t#Adiabatic gas consmath.tant\n", + "v1 = 0.06842;\t\t\t#Specific volume at entry in (m**3)/kg from steam table\n", + "v3 = 0.2292;\t\t\t#Specific volume at exit in (m**3)/kg from steam table\n", + "h1 = 2979.;\t\t\t#Enthalpy in kJ/kg from Moiller chart\n", + "h3 = 2673.3;\t\t\t#Enthalpy in kJ/kg from Moiller chart\n", + "\n", + "# Calculations\n", + "c = n/(n-1);\t\t\t#Ratio\n", + "C2 = ((2*c*P1*(10**5)*v1*(1-((P2/P1)**(1/c))))**0.5);\t\t\t#Velocity at exit in m/s\n", + "v2 = v1*((P1/P2)**(1/n));\t\t\t#Specific volume at exit in (m**3)/kg\n", + "A2 = ((m*v2)/C2)*(10**4);\t\t\t#Area of exit in (cm**2)\n", + "a = ((A2/18)**0.5)*10;\t\t\t#Length in mm\n", + "b = 3*a;\t\t\t#Breadth in mm\n", + "T2 = T1*((P2/P1)**(1/c));\t\t\t#Temperature at exit in K\n", + "D = Ts-T2;\t\t\t#Degree of undercooling in K\n", + "Ds = P2/Ps;\t\t\t#Degree of supersaturation\n", + "hI = h1-h3;\t\t\t#Isentropic enthalpy drop in kJ/kg\n", + "ha = (C2**2)/2000;\t\t\t#Actual enthalpy drop in kJ/kg\n", + "QL = hI-ha;\t\t\t#Loss in available heat in kJ/kg\n", + "DS = QL/Ts;\t\t\t#Increase in entropy in kJ/kg-K\n", + "C3 = (2000*(h1-h3))**0.5;\t\t\t#Exit velocity from nozzle\n", + "mf = ((A2*C3*(10**-4))/v3);\t\t\t#Mass flow rate in kg/s\n", + "Rm = m/mf;\t\t\t#Ratio of mass rate\n", + "\n", + "# Results\n", + "print 'Cross section of nozzle is %3.1f mm * %3.1f mm \\\n", + "\\nDegree of undercooling is %3.1f K and Degree of supersaturation is %3.2f \\\n", + "\\nLoss in available heat drop due to irreversibility is %3.2f kJ/kg \\\n", + "\\nIncrease in entropy is %3.5f kJ/kg-K \\\n", + "\\nRatio of mass flow rate with metastable expansion to the thermal expansion is %3.3f'%(b,a,D,Ds,QL,DS,Rm)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.6 Page no : 169" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nozzle efficiency is 88.9 percent \n", + "Exit area is 7000 mm**2 \n", + "Throat velocity is 529 m/s\n" + ] + } + ], + "source": [ + "\n", + "import math\n", + "\n", + "# Variables\n", + "m = 14.;\t\t\t#Mass flow rate of steam in kg/s\n", + "P1 = 3.;\t\t\t#Pressure of Steam in MN/(m**2)\n", + "T1 = 300.;\t\t\t#Steam temperature in oC\n", + "h1 = 2990.;\t\t\t#Enthalpy at point 1 in kJ/kg\n", + "h2s = 2630.;\t\t\t#Enthalpy at point 2s in kJ/kg\n", + "ht = 2850.;\t\t\t#Enthalpy at point t in kJ/kg\n", + "n = 1.3;\t\t\t#Adiabatic gas consmath.tant\n", + "C2 = 800.;\t\t\t#Exit velocity in m/s\n", + "v2 = 0.4;\t\t\t#Specific volume at exit in (m**3)/kg\n", + "\n", + "# Calculations\n", + "x = n/(n-1);\t\t\t#Ratio\n", + "Pt = ((2/(n+1))**x)*P1;\t\t\t#Temperature at point t in MN/(m**2)\n", + "h2 = h1-((C2**2)/2000);\t\t\t#Exit enthalpy in kJ/kg\n", + "nN = ((h1-h2)/(h1-h2s))*100;\t\t\t#Nozzle efficiency\n", + "A2 = ((m*v2)/C2)*(10**6);\t\t\t#Exit area in (mm**2)\n", + "Ct = math.sqrt(2*(h1-ht)*10**3);\t\t\t#Throat velocity in m/s\n", + "\n", + "# Results\n", + "print 'Nozzle efficiency is %3.1f percent \\\n", + "\\nExit area is %3.0f mm**2 \\\n", + "\\nThroat velocity is %3.0f m/s'%(nN,A2,Ct)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.7 Page no : 170" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Throat area is 388 mm**2 \n", + "Exit area is 1275 mm**2 \n", + "Steam quality at exit is 95 percent\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "P1 = 10.;\t\t\t#Pressure at point 1 in bar\n", + "P2 = 0.5;\t\t\t#Pressure at point 2 in bar\n", + "h1 = 3050.;\t\t\t#Enthalpy at point 1 in kJ/kg\n", + "h2s = 2480.;\t\t\t#Enthalpy at point 2s in kJ/kg\n", + "ht = 2910.;\t\t\t#Enthalpy at throat in kJ/kg\n", + "n = 1.3;\t\t\t#Adiabatic gas constant\n", + "r = 0.1;\t\t\t#Total available heat drop\n", + "v1 = 0.258;\t\t\t#Specific volume at point 1 in (m**3)/kg\n", + "h2f = 340.6;\t\t\t#Enthalpy for exit pressure from steam tables in kJ/kg\n", + "hfg = 2305.4;\t\t\t#Enthalpy for exit pressure from steam tables in kJ/kg\n", + "m = 0.5;\t\t\t#Mass flow rate in kg/s\n", + "\n", + "# Calculations\n", + "x = n/(n-1);\t\t\t#Ratio\n", + "Pt = ((2/(n+1))**x)*P1;\t\t\t#Temperature at throat in bar\n", + "h2 = h2s+(r*(h1-h2s));\t\t\t#Enthalpy at point 2 in kJ/kg\n", + "vt = ((P1/Pt)**(1/n))*v1;\t\t\t#Specific volume at throat in (m**3)/kg\n", + "v2 = ((P1/P2)**(1/n))*v1;\t\t\t#Specific volume at point 2 in (m**3)/kg\n", + "Ct = math.sqrt(2000*(h1-ht));\t\t\t#Throat velocity in m/s\n", + "At = ((m*vt)/Ct)*(10**6);\t\t\t#Throat area in (mm**2)\n", + "C2 = math.sqrt(2000*(h1-h2));\t\t\t#Exit velocity in m/s\n", + "A2 = ((m*v2)/C2)*(10**6);\t\t\t#Exit area in (mm**2)\n", + "x2 = ((h2-h2f)/hfg)*100;\t\t\t#Steam quality at exit\n", + "\n", + "# Results\n", + "print 'Throat area is %d mm**2 \\\n", + "\\nExit area is %d mm**2 \\\n", + "\\nSteam quality at exit is %3.0f percent'%(At,A2,x2)\n", + "\n", + "# rounding off error" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.8 Page no : 171" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Maximum discharge is 13.294 kg/min \n", + "Exit area is 493.8 mm**2\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "P1 = 3.5;\t\t\t#Dry saturated steam in bar\n", + "P2 = 1.1;\t\t\t#Exit pressure in bar\n", + "At = 4.4;\t\t\t#Throat area in cm**2\n", + "h1 = 2731.6;\t\t\t#Enthalpy at P1 in kJ/kg\n", + "v1 = 0.52397;\t\t\t#Specific volume at P1 in m**3/kg\n", + "n = 1.135;\t\t\t#Adiabatic gas constant\n", + "ht = 2640.;\t\t\t#Enthalpy at Pt in kJ/kg\n", + "vt = 0.85;\t\t\t#Specific volume at throat in m**3/kg\n", + "h2 = 2520.;\t\t\t#Enthalpy at P2 in kJ/kg\n", + "v2 = 1.45;\t\t\t#Specific volume at P2 in m**3/kg\n", + "\n", + "# Calculations\n", + "x = n/(n-1);\t\t\t#Ratio\n", + "Pt = ((2/(n+1))**x)*P1;\t\t\t#Throat pressure in bar\n", + "Ct = math.sqrt(2000*(h1-ht));\t\t\t#Throat velocity in m/s\n", + "mmax = ((At*Ct*(10**-4))/vt)*60;\t\t\t#Maximum discharge in kg/min\n", + "C2 = math.sqrt(2000*(h1-h2));\t\t\t#Exit velocity in m/s\n", + "A2 = ((mmax*v2)/(C2*60))*(10**6);\t\t\t#Exit area in mm**2\n", + "\n", + "# Results\n", + "print 'Maximum discharge is %3.3f kg/min \\\n", + "\\nExit area is %3.1f mm**2'%(mmax,A2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.9 Page no : 172" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Since throat pressure is greater than exit pressure,nozzle used is convergent-divergent nozzle \n", + "Minimum area of nozzle required is 2.14e-03 m**2\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "P1 = 10.;\t\t\t#Pressure at point 1 in bar\n", + "T1 = 200.;\t\t\t#Temperature at point 1 in oC\n", + "P2 = 5.;\t\t\t#Pressure at point 2 in bar\n", + "n = 1.3;\t\t\t#Adiabatic gas consmath.tant\n", + "h1 = 2830.;\t\t\t#Enthalpy at P1 in kJ/kg\n", + "ht = 2710.;\t\t\t#Enthalpy at point Pt in kJ/kg\n", + "vt = 0.35;\t\t\t#Specific volume at Pt in m**3/kg\n", + "m = 3. \t\t\t#Nozzle flow in kg/s\n", + "\n", + "# Calculations\n", + "x = n/(n-1);\t\t\t#Ratio\n", + "Pt = ((2/(n+1))**x)*P1;\t\t\t#Throat pressure in bar\n", + "Ct = math.sqrt(2000*(h1-ht));\t\t\t#Throat velocity in m/s\n", + "At = (m*vt)/Ct;\t\t\t#Throat area in m**2\n", + "\n", + "# Results\n", + "print 'Since throat pressure is greater than exit pressure,nozzle used is\\\n", + " convergent-divergent nozzle \\\n", + " \\nMinimum area of nozzle required is %.2e m**2'%(At)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.10 Page no : 173" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Throat velocity is 443.27 m/s \n", + "Mass flow rate of steam is 1549.90 kg/m**2\n" + ] + } + ], + "source": [ + "import math \n", + "\n", + "# Variables\n", + "P1 = 10.5;\t\t\t#Pressure at point 1 in bar\n", + "x1 = 0.95;\t\t\t#Dryness fraction\n", + "n = 1.135;\t\t\t#Adiabatic gas constant\n", + "P2 = 0.85;\t\t\t#Pressure at point 2 in bar\n", + "vg = 0.185;\t\t\t#Specific volume in m**3/kg\n", + "\n", + "\n", + "# Calculations\n", + "c = n/(n-1);\t\t\t#Ratio\n", + "Pt = round(((2/(n+1))**c)*P1,2);\t\t\t#Throat pressure in MN/(m**2)\n", + "v1 = round(x1*vg,3);\t\t\t#Specific volume at point 1 in m**3/kg\n", + "Ct = round(math.sqrt((2*n*P1*v1*(10**5)/(n+1))),2);\t\t\t#Velocity at throat in m/s\n", + "vt = round(((P1/Pt)*(v1**n))**(1/1.135),3);\t\t\t#Specific volume at throat in m**3/kg\n", + "m = Ct/vt;\t\t\t#Mass flow rate per unit throat area in kg/(m**2)\n", + "\n", + "# Results\n", + "print 'Throat velocity is %3.2f m/s \\\n", + "\\nMass flow rate of steam is %3.2f kg/m**2'%(Ct,m)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.11 Page no : 174" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Degree of supersaturation is 4.98 \n", + "Degree of undercooling 50 C\n" + ] + } + ], + "source": [ + "\n", + "# Variables\n", + "P1 = 10.;\t\t\t#Pressure at point 1 in bar\n", + "T1 = 452.9;\t\t\t#Temperature at point 1 in K\n", + "P2 = 4.;\t\t\t#Pressure at point 2 in bar\n", + "n = 1.3;\t\t\t#Adiabatic gas constant\n", + "Ps = 0.803;\t\t\t#Saturation pressure at T2 in bar\n", + "Ts = 143.6;\t\t\t#Saturation temperature at P2 in oC\n", + "# Calculations\n", + "x = (n-1)/n;\t\t\t#Ratio\n", + "T2 = ((P2/P1)**x)*T1;\t\t\t#Temperature at point 2 in K\n", + "Ds = P2/Ps;\t\t\t#Degree of supersaturation\n", + "Du = Ts-(T2-273);\t\t\t#Degree of undercooling\n", + "\n", + "# Results\n", + "print 'Degree of supersaturation is %3.2f \\\n", + "\\nDegree of undercooling %3.0f C'%(Ds,Du)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.12 Page no : 174" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Quantity of steam used per second is 0.012 kg/s \n", + "Exit velocity of steam is 816.09 m/s\n" + ] + } + ], + "source": [ + "\n", + "import math \n", + "\n", + "# Variables\n", + "P1 = 9.;\t\t\t#Pressure at point 1 in bar\n", + "P2 = 1.;\t\t\t#Pressure at point 2 in bar\n", + "Dt = 0.0025;\t\t\t#Throat diameter in m\n", + "nN = 0.9;\t\t\t#Nozzle efficiency\n", + "n = 1.135;\t\t\t#Adiabatic gas constant\n", + "h1 = 2770.;\t\t\t#Enthalpy at point 1 in kJ/kg\n", + "ht = 2670.;\t\t\t#Throat enthlapy in kJ/kg\n", + "h3 = 2400.;\t\t\t#Enthlapy at point 2 in kJ/kg\n", + "x2 = 0.96;\t\t\t#Dryness fraction 2\n", + "vg2 = 0.361;\t\t\t#Specific volume in m**3/kg\n", + "\n", + "# Calculations\n", + "x = n/(n-1);\t\t\t#Ratio\n", + "Pt = ((2/(n+1))**x)*P1;\t\t\t#Throat pressure in bar\n", + "Ct = math.sqrt(2000*(h1-ht)*nN);\t\t\t#Throat velocity in m/s\n", + "At = (3.147*2*(Dt**2))/4;\t\t\t#Throat area in m**2\n", + "vt = x2*vg2;\t\t\t#Specific volume at throat in m**3/kg\n", + "m = (At*Ct)/vt;\t\t\t#Mass flow rate of steam in kg/s\n", + "hact = nN*(h1-h3);\t\t\t#Actual enthalpy drop in kJ/kg\n", + "C2 = math.sqrt(2000*hact);\t\t\t#Exit velocity of steam in m/s\n", + "\n", + "# Results\n", + "print 'Quantity of steam used per second is %3.3f kg/s \\\n", + "\\nExit velocity of steam is %3.2f m/s'%(m,C2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.13 Page no : 202" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Blade angles are 33 degrees, 33 degrees \n", + "Tangential force on blades is 840 N \n", + "Axial thrust is 0 \n", + "Diagram power is 336 kW \n", + "Diagram efficiency 89.6 percent\n" + ] + } + ], + "source": [ + "\n", + "# Variables\n", + "C1 = 1000.;\t\t\t#Steam velocity in m/s\n", + "a1 = 20.;\t\t\t#Nozzle angle in degrees\n", + "U = 400.;\t\t\t#Mean blade speed in m/s\n", + "m = 0.75;\t\t\t#Mass flow rate of steam in kg/s\n", + "b1 = 33.;\t\t\t#Blade angle at inlet from the velocity triangle in degrees\n", + "b2 = b1;\t\t\t#Blade angle at exit from the velocity triangle in degrees\n", + "Cx = 1120.;\t\t\t#Change in whirl velocity from the velocity triangle in m/s\n", + "Ca = 0;\t\t \t#Change in axial velocity from the velocity triangle in m/s\n", + "\n", + "# Calculations\n", + "Fx = m*Cx;\t\t \t #Tangential force on blades in N\n", + "Fy = m*Ca;\t\t\t #Axial thrust in N\n", + "W = (m*Cx*U)/1000;\t\t\t#Diagram power in kW\n", + "ndia = ((2*U*Cx)/(C1**2))*100;\t\t\t#Diagram efficiency\n", + "\n", + "# Results\n", + "print 'Blade angles are %3.0f degrees, %3.0f degrees \\\n", + "\\nTangential force on blades is %3.0f N \\\n", + "\\nAxial thrust is %3.0f \\\n", + "\\nDiagram power is %3.0f kW \\\n", + "\\nDiagram efficiency %3.1f percent'%(b1,b2,Fx,Fy,W,ndia)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.14 Page no : 203" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Power developed is 3800 kW \n", + "Blade efficiency is 78.7 percent \n", + "Steam consumed is 9.46 kg/kWh\n" + ] + } + ], + "source": [ + "# Variables\n", + "D = 2.5;\t\t\t#Mean diameter of blade ring in m\n", + "N = 3000.;\t\t\t#Speed in rpm\n", + "a1 = 20.;\t\t\t#Nozzle angle in degrees\n", + "r = 0.4;\t\t\t#Ratio blade velocity to steam velocity\n", + "Wr = 0.8;\t\t\t#Blade friction factor\n", + "m = 10.;\t\t\t#Steam flow in kg/s\n", + "x = 3.;\t \t\t#Sum in blade angles in degrees\n", + "b1 = 32.5;\t\t\t#Blade angle at inlet from the velocity triangle in degrees\n", + "W1 = 626.7;\t\t\t#Relative velocity at inlet from the velocity triangle in m/s\n", + "Cx = 967.;\t\t\t#Change in whirl velocity from the velocity triangle in m/s\n", + "\n", + "# Calculations\n", + "U = (3.147*D*N)/60;\t\t\t#Blade velocity in m/s\n", + "C1 = U/r;\t\t\t#Steam velocity in m/s\n", + "b2 = b1-x;\t\t\t#Blade angle at exit in degrees\n", + "W2 = Wr*W1;\t\t\t#Relative velocity at outlet from the velocity triangle in m/s\n", + "W = (m*Cx*U)/1000;\t\t\t#Power developed in kW\n", + "ndia = ((2*U*Cx)/(C1**2))*100;\t\t\t#Blade efficiency\n", + "sc = (m*3600)/W;\t\t\t#Steam consumption in kg/kWh\n", + "\n", + "# Results\n", + "print 'Power developed is %3.0f kW \\\n", + "\\nBlade efficiency is %3.1f percent \\\n", + "\\nSteam consumed is %3.2f kg/kWh'%(round(W,-1),ndia,sc)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.15 Page no : 204" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Blading efficiency is 68.3 percent \n", + "Blade velocity co-efficient is 0.49\n" + ] + } + ], + "source": [ + "\n", + "# Variables\n", + "m = 3.;\t \t\t#Mass flow rate of steam in kg/s\n", + "C1 = 425.;\t\t\t#Steam velocity in m/s\n", + "r = 0.4;\t\t\t#Ratio of blade speed to jet speed\n", + "W = 170.;\t\t\t#Stage output in kW\n", + "IL = 15.;\t\t\t#Internal losses in kW\n", + "a1 = 16.;\t\t\t#Nozzle angle in degrees\n", + "b2 = 17.;\t\t\t#Blade angle at exit in degrees\n", + "W1 = 265.;\t\t\t#Relative velocity at inlet from the velocity triangle in m/s\n", + "W2 = 130.;\t\t\t#Relative velocity at outlet from the velocity triangle in m/s\n", + "\n", + "# Calculations\n", + "U = C1*r;\t\t\t#Blade speed in m/s\n", + "P = (W+IL)*1000;\t\t\t#Total power developed in W\n", + "Cx = P/(m*W);\t\t\t#Change in whirl velocity in m/s\n", + "ndia = ((2*U*Cx)/(C1**2))*100;\t\t\t#Blading efficiency\n", + "Wr = W2/W1;\t\t\t#Blade velocity co-efficient\n", + "\n", + "# Results\n", + "print 'Blading efficiency is %3.1f percent \\\n", + "\\nBlade velocity co-efficient is %3.2f'%(ndia,Wr)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.16 Page no : 205" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Blade angles assumed are 34 degrees, 41 degrees \n", + "Power developed by turbine is 52.8 kW\n" + ] + } + ], + "source": [ + "\n", + "# Variables\n", + "C1 = 375.;\t\t\t#Steam velocity in m/s\n", + "a1 = 20.;\t\t\t#Nozzle angle\n", + "U = 165.;\t\t\t#Blade speed in m/s\n", + "m = 1.;\t\t\t#Mass flow rate of steam in kg/s\n", + "Wr = 0.85;\t\t\t#Blade friction factor\n", + "Ca1 = 130.;\t\t\t#Axial velocity at inlet from the velocity triangle in m/s\n", + "Ca2 = Ca1;\t\t\t#Axial velocity at outlet in m/s\n", + "W1 = 230.;\t\t\t#Relative velocity at inlet from the velocity triangle in m/s\n", + "Cx = 320.;\t\t\t#Change in whirl velocity from the velocity triangle in m/s\n", + "\n", + "# Calculations\n", + "b2 = 41;\t\t\t#Blade angle at exit from the velocity triangle in degrees\n", + "b1 = 34;\t\t\t#Blade angle at exit from the velocity triangle in degrees\n", + "W = (m*Cx*U)/1000;\t\t\t#Power developed by turbine in kW\n", + "\n", + "# Results\n", + "print 'Blade angles assumed are %3.0f degrees, %3.0f degrees \\\n", + "\\nPower developed by turbine is %3.1f kW'%(b1,b2,W)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.17 Page no : 206" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nozzle angle is 19 degrees \n", + "Blade angles are 33 degrees, 36 degrees\n" + ] + } + ], + "source": [ + "\n", + "# Variables\n", + "m = 2.;\t\t\t#Mass flow rate of steam in kg/s\n", + "W = 130.;\t\t\t#Turbine power in kW\n", + "U = 175.;\t\t\t#Blade velocity in m/s\n", + "C1 = 400.;\t\t\t#Steam velocity in m/s\n", + "Wr = 0.9;\t\t\t#Blade friction factor\n", + "W1 = 240.;\t\t\t#Realtive velocity at inlet from the velocity triangle in m/s\n", + "\n", + "# Calculations\n", + "Cx1 = (W*1000)/(m*U);\t\t\t#Whirl velocity at inlet in m/s\n", + "W2 = Wr*W1;\t\t\t#Realtive velocity at outlet from the velocity triangle in m/s\n", + "a1 = 19;\t\t\t#Nozzle angle from the velocity triangle in degrees\n", + "b1 = 33;\t\t\t#Blade angle at inlet from the velocity triangle in degrees\n", + "b2 = 36;\t\t\t#Blade angle at outlet from the velocity triangle in degrees\n", + "\n", + "# Results\n", + "print 'Nozzle angle is %3.0f degrees \\\n", + "\\nBlade angles are %3.0f degrees, %3.0f degrees'%(a1,b1,b2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.18 Page no : 207" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Diagram efficiency is 76.2 percent\n" + ] + } + ], + "source": [ + "# find Diagram efficiency\n", + "\n", + "# Variables\n", + "U = 150.;\t\t\t#Blade speed in m/s\n", + "m = 3.;\t\t\t#Mass flow rate of steam in kg/s\n", + "P = 10.5;\t\t\t#Pressure in bar\n", + "r = 0.21;\t\t\t#Ratio blade velocity to steam velocity\n", + "a1 = 16.;\t\t\t#Nozzle angle in first stage in degrees\n", + "b2 = 20.;\t\t\t#Blade angle at exit in first stage in degrees\n", + "a3 = 24.;\t\t\t#Nozzle angle in second stage in degrees\n", + "b4 = 32.;\t\t\t#Blade angle at exit in second stage in degrees\n", + "Wr = 0.79;\t\t\t#Blade friction factor for first stage\n", + "Wr2 = 0.88;\t\t\t#Blade friction factor for second stage\n", + "Cr = 0.83;\t\t\t#Blade velocity coefficient\n", + "W1 = 570.;\t\t\t#Relative velocity at inlet from the velocity triangle for first stage in m/s\n", + "C2 = 375.;\t\t\t#Velocity in m/s\n", + "W3 = 185.;\t\t\t#Relative velocity at inlet from the velocity triangle for second stage in m/s\n", + "\n", + "# Calculations\n", + "C1 = U/r;\t\t\t#Steam speed at exit in m/s\n", + "W2 = Wr*W1;\t\t\t#Relative velocity at outlet for first stage in m/s\n", + "C3 = Cr*C2;\t\t\t#Steam velocity at inlet for second stage in m/s\n", + "W4 = Wr2*W3;\t\t\t#Relative velocity at exit for second stage in m/s\n", + "DW1 = W1+W2;\t\t\t#Change in relative velocity for first stage in m/s\n", + "DW2 = 275;\t\t\t#Change in relative velocity from the velocity triangle for second stage in m/s\n", + "ndia = ((2*U*(DW1+DW2))/(C1**2))*100;\t\t\t#Diagram efficiency\n", + "\n", + "# Results\n", + "print 'Diagram efficiency is %3.1f percent'%(ndia)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.19 Page no : 208" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Blade speed is 124.7 m/s \n", + "Blade tip angles of the fixed blade are 17 degrees and 43 degrees \n", + "Diagram efficiency is 79.5 percent\n" + ] + } + ], + "source": [ + "import math\n", + "# Variables\n", + "b1 = 30.;\t\t\t#Blade angle at inlet in first stage in degrees\n", + "b2 = 30.;\t\t\t#Blade angle at exit in first stage in degrees\n", + "b3 = 30.;\t\t\t#Blade angle at inlet in second stage in degrees\n", + "b4 = 30.;\t\t\t#Blade angle at exit in second stage in degrees\n", + "t1 = 240.;\t\t\t#Temperature at entry in oC\n", + "P1 = 11.5;\t\t\t#Pressure at entry in bar\n", + "P2 = 5.;\t\t\t#Pressure in wheel chamber in bar\n", + "vl = 10.;\t\t\t#Loss in velocity in percent\n", + "h = 155.;\t\t\t#Enthalpy at P2 in kJ/kg\n", + "W4 = 17.3;\t\t\t#Relative velocity at exit from the velocity triangle for second stage in m/s\n", + "a4 = 90.;\t\t\t#Nozzle angle in second stage in degrees\n", + "C3 = 33.;\t\t\t#Steam velocity at inlet from the velocity triangle for second stage in m/s\n", + "W2 = 49.;\t\t\t#Relative velocity at outlet from the velocity triangle for first stage in m/s\n", + "x = 15.;\t\t\t#Length of AB assumed for drawing velocity triangle in mm\n", + "y = 67.;\t\t\t#Length of BC from the velocity triangle in mm\n", + "\n", + "# Calculations\n", + "C1 = math.sqrt(2000*h);\t\t\t#Velocity of steam in m/s\n", + "W3 = W4/0.9;\t\t\t#Relative velocity at inlet for second stage in m/s\n", + "C2 = C3/0.9;\t\t\t#Velocity in m/s\n", + "W1 = W2/0.9;\t\t\t#Relative velocity at inlet for first stage in m/s\n", + "C1n = C1/y;\t\t\t#Velocity of steam in m/s\n", + "U = x*C1n;\t\t\t#Blade speed in m/s\n", + "a3 = 17.;\t\t\t#Nozzle angle in second stage from the velocity triangle in degrees\n", + "a2 = 43.;\t\t\t#Nozzle angle from the velocity triangle in degrees\n", + "DW1 = 731.5;\t\t\t#Change in relative velocity from the velocity triangle for first stage in m/s\n", + "DW2 = 257.5;\t\t\t#Change in relative velocity from the velocity triangle for second stage in m/s\n", + "ndia = ((2*U*(DW1+DW2))/(C1**2))*100;\t\t\t#Diagram efficiency\n", + "\n", + "# Results\n", + "print 'Blade speed is %3.1f m/s \\\n", + "\\nBlade tip angles of the fixed blade are %3.0f degrees and %3.0f degrees \\\n", + "\\nDiagram efficiency is %3.1f percent'%(U,a3,a2,ndia)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.20 Page no : 210" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Blade speed is 160.5 m/s \n", + "Power developed by the turbine is 530.66 kW\n" + ] + } + ], + "source": [ + "\n", + "# Variables\n", + "C1 = 600.;\t\t\t#Steam velocity in m/s\n", + "b1 = 30.;\t\t\t#Blade angle at inlet in first stage in degrees\n", + "b2 = 30.;\t\t\t#Blade angle at exit in first stage in degrees\n", + "b3 = 30.;\t\t\t#Blade angle at inlet in second stage in degrees\n", + "b4 = 30.;\t\t\t#Blade angle at exit in second stage in degrees\n", + "a4 = 90.;\t\t\t#Nozzle angle in second stage in degrees\n", + "m = 3.;\t\t\t#Mass of steam in kg/s\n", + "x = 15.;\t\t\t#Length for drawing velocity triangle in mm\n", + "y = 56.;\t\t\t#Length of BC from the velocity triangle in mm\n", + "\n", + "# Calculations\n", + "C1n = round(C1/y,1);\t\t\t#Velocity of steam in m/s\n", + "U = round(x*C1n,1);\t\t\t#Blade speed in m/s\n", + "l = 103.;\t\t\t#Length from velocity triangle in mm\n", + "P = (m*l*C1n*U)/1000;\t\t\t#Power developed in kW\n", + "\n", + "# Results\n", + "print 'Blade speed is %3.1f m/s \\\n", + "\\nPower developed by the turbine is %3.2f kW'%(U,P)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.21 Page no : 211" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Mean diameter of drum is 963 mm \n", + "Volume of steam flowing per second is 8.18 m**3/s\n" + ] + } + ], + "source": [ + "import math\n", + "# Variables\n", + "N = 400.;\t\t\t#Speed in rpm\n", + "m = 8.33;\t\t\t#Mass of steam in kg/s\n", + "P = 1.6;\t\t\t#Pressure of steam in bar\n", + "x = 0.9;\t\t\t#Dryness fraction\n", + "W = 10.;\t\t\t#Stage power in kW\n", + "r = 0.75;\t\t\t#Ratio of axial flow velocity to blade velocity\n", + "a1 = 20.;\t\t\t#Nozzle angle at inlet in degrees\n", + "a2 = 35.;\t\t\t#Nozzle angle at exit in degrees\n", + "b1 = a2;\t\t\t#Blade tip angle at exit in degrees\n", + "b2 = a1;\t\t\t#Blade tip angle at inlet in degrees\n", + "a = 25.;\t\t\t#Length of AB from velocity triangle in mm\n", + "vg = 1.091;\t\t\t#Specific volume of steam from steam tables in (m**3)/kg\n", + "\n", + "# Calculations\n", + "Cx = 73.5;\t\t\t#Change in whirl velocity from the velocity triangle by measurement in mm\n", + "y = Cx/a;\t\t\t#Ratio of change in whirl velocity to blade speed\n", + "U = math.sqrt((W*1000)/(m*y));\t\t\t#Blade speed in m/s\n", + "D = ((U*60)/(3.147*N))*1000;\t\t\t#Mean diameter of drum in mm\n", + "v = m*x*vg;\t\t\t#Volume flow rate of steam in (m**3)/s\n", + "\n", + "# Results\n", + "print 'Mean diameter of drum is %3.0f mm \\\n", + "\\nVolume of steam flowing per second is %3.2f m**3/s'%(D,v)\n", + "\n", + "# rounding off error" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.22 Page no : 212" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Drum diameter is 1.030 m \n", + "Blade height is 78 mm\n" + ] + } + ], + "source": [ + "\n", + "import math\n", + "\n", + "# Variables\n", + "N = 300.;\t\t\t#Speed in rpm\n", + "m = 4.28;\t\t\t#Mass of steam in kg/s\n", + "P = 1.9;\t\t\t#Pressure of steam in bar\n", + "x = 0.93;\t\t\t#Dryness fraction\n", + "W = 3.5;\t\t\t#Stage power in kW\n", + "r = 0.72;\t\t\t#Ratio of axial flow velocity to blade velocity\n", + "a1 = 20.;\t\t\t#Nozzle angle at inlet in degrees\n", + "b2 = a1;\t\t\t#Blade tip angle at inlet in degrees\n", + "l = 0.08;\t\t\t#Tip leakage steam\n", + "vg = 0.929;\t\t\t#Specific volume of steam from steam tables in (m**3)/kg\n", + "\n", + "# Calculations\n", + "mact = m-(m*l);\t\t\t#Actual mass of steam in kg/s\n", + "a = (3.147*N)/60;\t\t\t#Ratio of blade velocity to mean dia\n", + "b = r*a;\t\t\t#Ratio of axial velocity to mean dia\n", + "c = 46;\t\t\t#Ratio of change in whirl velocity to mean dia\n", + "D = math.sqrt((W*1000)/(mact*c*a));\t\t\t#Mean dia in m\n", + "Ca = b*D;\t\t\t#Axial velocity in m/s\n", + "h = ((mact*x*vg)/(3.147*D*Ca))*1000;\t\t\t#Blade height in mm\n", + "D1 = D-(h/1000);\t\t\t#Drum dia in m\n", + "\n", + "# Results\n", + "print 'Drum diameter is %3.3f m \\\n", + "\\nBlade height is %3.0f mm'%(D1,h)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.23 Page no : 214" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rotor blade angles are 58.56 degrees and 58.56 degrees \n", + "Flow coefficient is 0.611 \n", + "Blade loading coefficient is 2 \n", + "Power developed is 13.8 MW\n" + ] + } + ], + "source": [ + "import math\n", + "# Variables\n", + "P0 = 800.;\t\t\t#Steam pressure in kPa\n", + "P2 = 100.;\t\t\t#Pressure at point 2 in kPa\n", + "T0 = 973.;\t\t\t#Steam temperature in K\n", + "a1 = 73.;\t\t\t#Nozzle angle in degrees\n", + "ns = 0.9;\t\t\t#Steam efficiency\n", + "m = 35.;\t\t\t#Mass flow rate in kg/s\n", + "Cp = 1.005;\t\t\t#Specific heat at constant pressure in kJ/kg-K\n", + "y = 1.4;\t\t\t#Ratio of specific heats\n", + "\n", + "# Calculations\n", + "tanb1 = math.tan(math.radians(a1))/2;\t\t\t#Blade angle at inlet in degrees\n", + "b1 = math.degrees(math.atan(tanb1))\n", + "b2 = b1;\t\t\t#Blade angle at exit in degrees\n", + "p = 2/math.tan(math.radians(a1));\t\t\t#Flow coefficient\n", + "s = p*(math.tan(math.radians(b1))+math.tan(math.radians(b2)));\t\t\t#Blade loading coefficient\n", + "Dh = ns*Cp*T0*(1-((P2/P0)**((y-1)/y)));\t\t\t#Difference in enthalpies in kJ/kg\n", + "W = (m*Dh)/1000;\t\t\t#Power developed in MW\n", + "\n", + "# Results\n", + "print 'Rotor blade angles are %3.2f degrees and %3.2f degrees \\\n", + "\\nFlow coefficient is %3.3f \\\n", + "\\nBlade loading coefficient is %3.0f \\\n", + "\\nPower developed is %3.1f MW'%(b1,b2,p,s,W)\n", + "\n", + "# answer in book is wrong for W. please check." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.24 Page no : 215" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rotor blade angles for first stage are 53.95 degrees and 53.95 degrees \n", + "Rotor blade angles for second stage are 53.95 degrees and 53.95 degrees \n", + "Power developed is 9.90 MW \n", + "Final state of steam at first stage is 3306.52 kJ/kg \n", + "Final state of steam at second stage is 3257.00 kJ/kg \n", + "Blade height at first stage is 0.0114 m \n", + "Blade height at second stage is 0.0139 m\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "P0 = 100.;\t\t\t#Steam pressure in bar\n", + "T0 = 773.;\t\t\t#Steam temperature in K\n", + "a1 = 70.;\t\t\t#Nozzle angle in degrees\n", + "ns = 0.78;\t\t\t#Steam efficiency\n", + "m = 100.;\t\t\t#Mass flow rate of steam in kg/s\n", + "D = 1.;\t\t\t#Turbine diameter in m\n", + "N = 3000.;\t\t\t#Turbine speed in rpm\n", + "h0 = 3370.;\t\t\t#Steam enthalpy from Moiller chart in kJ/kg\n", + "v2 = 0.041;\t\t\t#Specific volume at P2 from steam tables in (m**3)/kg\n", + "v4 = 0.05;\t\t\t#Specific volume at P4 from steam tables in (m**3)/kg\n", + "\n", + "# Calculations\n", + "U = (3.147*D*N)/60;\t\t\t#Blade speed in m/s\n", + "C1 = (2*U)/math.sin(math.radians(a1));\t\t\t#Steam speed in m/s\n", + "b1 = math.tan(math.radians(a1))/2;\t\t\t#Blade angle at inlet for first stage in degrees\n", + "b1 = math.degrees(math.atan(b1))\n", + "b2 = b1;\t\t\t#Blade angle at exit for first stage in degrees\n", + "b3 = b1;\t\t\t#Blade angle at inlet for second stage in degrees\n", + "b4 = b2;\t\t\t#Blade angle at exit for second stage in degrees\n", + "Wt = (4*m*(U**2))/(10**6);\t\t\t#Total workdone in MW\n", + "Dh = (2*(U**2))/1000;\t\t\t#Difference in enthalpies in kJ/kg\n", + "Dhs = Dh/ns;\t\t\t#Difference in enthalpies in kJ/kg\n", + "h2 = h0-Dh;\t\t\t#Enthalpy at point 2 in kJ/kg\n", + "h2s = h0-Dhs;\t\t\t#Enthalpy at point 2s in kJ/kg\n", + "Dh2 = (2*(U**2))/1000;\t\t\t#Difference in enthalpies in kJ/kg\n", + "Dh2s = Dh2/ns;\t\t\t#Difference in enthalpies in kJ/kg\n", + "h4 = h2-Dh2;\t\t\t#Enthalpy at point 4 in kJ/kg\n", + "h4s = h2-Dh2s;\t\t\t#Enthalpy at point 4s in kJ/kg\n", + "Ca = C1*math.cos(math.radians(a1));\t\t\t#Axial velocity in m/s\n", + "hI = (m*v2)/(math.pi*D*Ca);\t\t\t#Blade height at first stage in m/s\n", + "hII = (m*v4)/(math.pi*D*Ca);\t\t\t#Blade height at second stage in m/s\n", + "\n", + "# Results\n", + "print 'Rotor blade angles for first stage are %3.2f degrees and %3.2f degrees \\\n", + "\\nRotor blade angles for second stage are %3.2f degrees and %3.2f degrees \\\n", + "\\nPower developed is %3.2f MW \\\n", + "\\nFinal state of steam at first stage is %3.2f kJ/kg \\\n", + "\\nFinal state of steam at second stage is %3.2f kJ/kg \\\n", + "\\nBlade height at first stage is %3.4f m \\\n", + "\\nBlade height at second stage is %3.4f m'%(b1,b2,b3,b4,Wt,h2s,h4s,hI,hII)\n", + "\n", + "# rounding off error" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.25 Page no : 218" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rotor blade angles for first stage are 64.11 degrees and 64.11 degrees \n", + "Rotor blade angles for second stage are 34.48 degrees and 34.48 degrees \n", + "Power developed is 19.81 MW \n", + "Final state of steam at first stage is 3171.9 kJ/kg \n", + "Final state of steam at second stage is 3065.27 kJ/kg \n", + "Rotor blade height is 0.0146 m\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "P0 = 100.;\t\t\t#Steam pressure in bar\n", + "T0 = 773.;\t\t\t#Steam temperature in K\n", + "a1 = 70.;\t\t\t#Nozzle angle in degrees\n", + "ns = 0.78;\t\t\t#Steam efficiency\n", + "m = 100.;\t\t\t#Mass flow rate of steam in kg/s\n", + "D = 1.;\t\t\t#Turbine diameter in m\n", + "N = 3000.;\t\t\t#Turbine speed in rpm\n", + "h0 = 3370.;\t\t\t#Steam enthalpy from Moiller chart in kJ/kg\n", + "P4 = 27.;\t\t\t#Pressure at point 4 in bar\n", + "T4 = 638.;\t\t\t#Temperature at point 4 in K\n", + "v4 = 0.105;\t\t\t#Specific volume at P4 from mollier chart in (m**3)/kg\n", + "ns = 0.65;\t\t\t#Stages efficiency\n", + "\n", + "# Calculations\n", + "U = (3.147*D*N)/60;\t\t\t#Blade speed in m/s\n", + "C1 = (4*U)/math.sin(math.radians(a1));\t\t\t#Steam speed in m/s\n", + "Ca = C1*math.cos(math.radians(a1));\t\t\t#Axial velocity in m/s\n", + "tanb1 = (3*U)/Ca;\t\t\t#Blade angle at inlet for first stage in degrees\n", + "b1 = math.degrees(math.atan(tanb1))\n", + "b2 = b1;\t\t\t#Blade angle at exit for first stage in degrees\n", + "b4 = math.degrees(math.atan(U/Ca));\t\t\t#Blade angle at exit for second stage in degrees\n", + "b3 = b4;\t\t\t#Blade angle at inlet for second stage in degrees\n", + "WI = m*6*(U**2);\t\t\t#Power developed in first stage in MW\n", + "WII = m*2*(U**2);\t\t\t#Power developed in second stage in MW\n", + "W = (WI+WII)/(10**6);\t\t\t#Total power developed in MW\n", + "Dh = (W*1000)/100;\t\t\t#Difference in enthalpies in kJ/kg\n", + "Dhs = (W*1000)/(ns*100);\t\t\t#Difference in enthalpies in kJ/kg\n", + "h4 = h0-Dh;\t\t\t#Enthalpy at point 4 in kJ/kg\n", + "h4s = h0-Dhs;\t\t\t#Enthalpy at point 4s in kJ/kg\n", + "h = (m*v4)/(3.147*D*Ca);\t\t\t#Rotor blade height in m\n", + "\n", + "\n", + "# Results\n", + "print 'Rotor blade angles for first stage are %3.2f degrees and %3.2f degrees \\\n", + "\\nRotor blade angles for second stage are %3.2f degrees and %3.2f degrees \\\n", + "\\nPower developed is %3.2f MW \\\n", + "\\nFinal state of steam at first stage is %3.1f kJ/kg \\\n", + "\\nFinal state of steam at second stage is %3.2f kJ/kg \\\n", + "\\nRotor blade height is %3.4f m'%(b1,b2,b3,b4,W,h4,h4s,h)\n", + "\n", + "# rounding off error" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.26 Page no : 221" + ] + }, + { + "cell_type": "code", + "execution_count": 40, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Blade angle at inlet is 10 degrees \n", + "Blade angle at exit is 60 degrees\n" + ] + } + ], + "source": [ + "\n", + "import math \n", + "\n", + "# Variables\n", + "a1 = 30.;\t\t\t#Nozzle angle in degrees\n", + "Ca = 180.;\t\t\t#Axial velocity in m/s\n", + "U = 280.;\t\t\t#Rotor blade speed in m/s\n", + "R = 0.5;\t\t\t#Degree of reaction\n", + "\n", + "# Calculations\n", + "a1n = 90-a1;\t\t\t#Nozzle angle measured from axial direction in degrees\n", + "Cx1 = Ca*math.tan(math.radians(a1n));\t\t\t#Whirl velocity in m/s\n", + "b1 = math.degrees(math.atan((Cx1-U)/Ca));\t\t\t#Blade angle at inlet in degrees\n", + "b2 = a1n;\t\t\t#Blade angle at exit in degrees\n", + "\n", + "# Results\n", + "print 'Blade angle at inlet is %3.0f degrees \\\n", + "\\nBlade angle at exit is %3.0f degrees'%(b1,b2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.27 Page no : 222" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rotor blade angles are 0 degrees and 70 degrees \n", + "Power developed is 1.92 MW \n", + "Isentropic enthalpy drop is 30.12 kJ/kg\n" + ] + } + ], + "source": [ + "\n", + "import math \n", + "\n", + "# Variables\n", + "P0 = 800.;\t\t\t#Steam pressure in kPa\n", + "T0 = 900.;\t\t\t#Steam temperature in K\n", + "a1 = 70.;\t\t\t#Nozzle angle in degrees\n", + "ns = 0.85;\t\t\t#Steam efficiency\n", + "m = 75.;\t\t\t#Mass flow rate of steam in kg/s\n", + "R = 0.5;\t\t\t#Degree of reaction\n", + "U = 160.;\t\t\t#Blade speed in m/s\n", + "\n", + "# Calculations\n", + "C1 = U/math.sin(a1);\t\t\t#Steam speed in m/s\n", + "Ca = C1*math.cos(a1);\t\t\t#Axial velocity in m/s\n", + "b1 = 0;\t\t\t #Blade angle at inlet from velocity triangle in degrees\n", + "b2 = a1; \t\t\t#Blade angle at exit in degrees\n", + "a2 = b1;\t\t\t #Nozzle angle in degrees\n", + "W = (m*(U**2))/(10**6);\t\t\t#Power developed in MW\n", + "Dhs = (W*1000)/(ns*m);\t\t\t#Isentropic enthalpy drop in kJ/kg\n", + "\n", + "# Results\n", + "print 'Rotor blade angles are %3.0f degrees and %3.0f degrees \\\n", + "\\nPower developed is %3.2f MW \\\n", + "\\nIsentropic enthalpy drop is %3.2f kJ/kg'%(b1,b2,W,Dhs)\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Thermal_Engineering_by_A._V._Arasu/ch5.ipynb b/Thermal_Engineering_by_A._V._Arasu/ch5.ipynb new file mode 100644 index 00000000..1c3b2dbf --- /dev/null +++ b/Thermal_Engineering_by_A._V._Arasu/ch5.ipynb @@ -0,0 +1,1100 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:2b6bc93922bd7b11c4334e4b77fa7e0b05d2efd84a162a89a3c4553815d1a094" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 5 :\n", + "Air Compressors" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.1 Page no : 250" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "D = 0.2;\t\t\t#Cylinder diameter in m\n", + "L = 0.3;\t\t\t#Cylinder Stroke in m\n", + "P1 = 1.;\t\t\t#Pressure at entry in bar\n", + "T1 = 300.;\t\t\t#Temperature at entry in K\n", + "P2 = 8.;\t\t\t#Pressure at exit in bar\n", + "n = 1.25;\t\t\t#Adiabatic gas constant\n", + "N = 100.;\t\t\t#Speed in rpm\n", + "R = 287.;\t\t\t#Universal gas constant in J/kg-K\n", + "\n", + "# Calculations\n", + "x = (n-1)/n;\t\t\t#Ratio\n", + "V1 = (3.147*L*(D**2))/4;\t\t\t#Volume of cylinder in m**3/cycle\n", + "W = (P1*(10**5)*V1*(((P2/P1)**x)-1))/x;\t\t\t#Work done in J/cycle\n", + "Pc = (W*100)/(60*1000);\t\t\t#Indicated power of compressor in kW\n", + "m = (P1*(10**5)*V1)/(R*T1);\t\t\t#Mass of air delivered in kg/cycle\n", + "md = m*N;\t\t\t#Mass delivered per minute in kg\n", + "T2 = T1*((P2/P1)**x);\t\t\t#Temperature of air delivered in K\n", + "\n", + "# Results\n", + "print 'Indicated power of compressor is %3.2f kW \\\n", + "\\nMass of air delivered by compressor per minute is %3.2f kg \\\n", + "\\nTemperature of air delivered is %3.1fK'%(Pc,md,T2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Indicated power of compressor is 4.06 kW \n", + "Mass of air delivered by compressor per minute is 1.10 kg \n", + "Temperature of air delivered is 454.7K\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.2 Page no : 251" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# Variables\n", + "IP = 37.;\t\t\t#Indicated power in kW\n", + "P1 = 0.98;\t\t\t#Pressure at entry in bar\n", + "T1 = 288.;\t\t\t#Temperature at entry in K\n", + "P2 = 5.8;\t\t\t#Pressure at exit in bar\n", + "n = 1.2;\t\t\t#Adiabatic gas constant\n", + "N = 100.;\t\t\t#Speed in rpm\n", + "Ps = 151.5;\t\t\t#Piston speed in m/min\n", + "a = 2.;\t\t\t#For double acting compressor\n", + "\n", + "# Calculations\n", + "L = Ps/(2*N);\t\t\t#Stroke length in m\n", + "x = (n-1)/n;\t\t\t#Ratio\n", + "r = (3.147*L)/4;\t\t\t#Ratio of volume to bore\n", + "D = math.sqrt((IP*1000*60*x)/(N*a*r*P1*(10**5)*(((P2/P1)**x)-1)));\t\t\t#Cylinder diameter in m\n", + "\n", + "# Results\n", + "print 'Stroke length of cylinder is %3.4f m \\\n", + "\\nCylinder diameter is %3.4f m'%(L,D)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Stroke length of cylinder is 0.7575 m \n", + "Cylinder diameter is 0.3030 m\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.3 Page no : 251" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables\n", + "IP = 11.;\t\t\t#Indicated power in kW\n", + "P1 = 1.;\t\t\t#Pressure at entry in bar\n", + "P2 = 7.;\t\t\t#Pressure at exit in bar\n", + "n = 1.2;\t\t\t#Adiabatic gas consmath.tant\n", + "Ps = 150.;\t\t\t#Piston speed in m/s\n", + "a = 2.; \t\t\t#For double acting compressor\n", + "r = 1.5;\t\t\t#Storke to bore ratio\n", + "\n", + "# Calculations\n", + "x = (n-1)/n;\t\t\t#Ratio\n", + "y = 3.147/(4*(r**2));\t\t\t#Ratio of volume to the cube of stroke\n", + "z = (P1*(10**2)*y*(((P2/P1)**x)-1))/x;\t\t\t#Ratio of workdone to the cube of stroke\n", + "L = (math.sqrt(IP/(z*Ps)))*1000;\t\t\t#Stroke in mm\n", + "D = (L/r);\t\t\t#Bore in mm\n", + "\n", + "# Results\n", + "print 'Stroke length of cylinder is %3.0f mm \\\n", + "\\nBore diameter of cylinder is %3.0f mm'%(L,D)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Stroke length of cylinder is 30 mm \n", + "Bore diameter of cylinder is 20 mm\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.4 Page no : 252" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "x = 0.05 # ratio\n", + "P1 = 1.;\t\t\t#Pressure at point 1 in bar\n", + "T1 = 310.;\t\t\t#Temperature at point 1 in K\n", + "n = 1.2;\t\t\t#Adiabatic gas constant\n", + "P2 = 7.;\t\t\t#Pressure at point 2 in bar\n", + "Pa = 1.01325;\t\t\t#Atmospheric pressure in bar\n", + "Ta = 288.;\t\t\t#Atmospheric temperature in K\n", + "\n", + "# Calculations\n", + "V1 = 1+x;\t\t\t#Ratio of volume of air sucked to stroke volume\n", + "V4 = ((P2/P1)**(1/n))/20;\t\t\t#Ratio of volume delivered to stroke volume\n", + "DV = V1-V4;\t\t\t#Difference in volumes\n", + "nv1 = DV*100;\t\t\t#Volumetric efficiency\n", + "V = (P1*DV*Ta)/(T1*Pa);\t\t\t#Ratio of volumes referred to atmospheric conditions\n", + "nv2 = V*100;\t\t\t#Volumetric efficiency referred to atmospheric conditions\n", + "W = (n*0.287*T1*((P2/P1)**((n-1)/n)-1))/(n-1);\t\t\t#Work required in kJ/kg\n", + "\n", + "# Results\n", + "print 'Volumetric efficiency is %3.1f percent \\\n", + "\\nVolumetric efficiency referred to atmospheric conditions is %3.1f percent \\\n", + "\\nWork required is %3.1f kJ/kg'%(nv1,nv2,W)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Volumetric efficiency is 79.7 percent \n", + "Volumetric efficiency referred to atmospheric conditions is 73.1 percent \n", + "Work required is 204.5 kJ/kg\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.5 Page no : 253" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "D = 0.2;\t\t\t#Bore in m\n", + "L = 0.3;\t\t\t#Stroke in m\n", + "P1 = 1.;\t\t\t#Pressure at point 1 in bar\n", + "P2 = 7.;\t\t\t#Pressure at point 2 in bar\n", + "n = 1.25;\t\t\t#Adiabatic gas constant\n", + "lc = 0.015\n", + "\n", + "# Calculations\n", + "V3 = (3.147*(D**2)*lc)/4.;\t\t\t#Clearance volume in m**3\n", + "Vs = (3.147*(D**2)*L)/4.;\t\t\t#Stoke volume in m**3\n", + "C = V3/Vs;\t\t\t#Clearance ratio\n", + "nv = (1+C-(C*((P2/P1)**(1/n))))*100;\t\t\t#Volumetric efficiency\n", + "DV = (nv*Vs)/100.;\t\t\t#Volume of air taken in (m**3)/stroke\n", + "\n", + "# Results\n", + "print 'Theoretical volume of air taken in per stroke is %.2e m**3/stroke'%(DV)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Theoretical volume of air taken in per stroke is 7.67e-03 m**3/stroke\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.6 Page no : 254" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# Variables\n", + "D = 0.2;\t\t\t#Bore in m\n", + "L = 0.3;\t\t\t#Stroke in m\n", + "P1 = 1.;\t\t\t#Pressure at point 1 in bar\n", + "r = 0.05 # ratio\n", + "T1 = 293.;\t\t\t#Temperature at point 1 in K\n", + "P2 = 5.5;\t\t\t#Pressure at point 2 in bar\n", + "n = 1.3;\t\t\t#Adiabatic gas constant\n", + "N = 500.;\t\t\t#Speed of compressor in rpm\n", + "\n", + "# Calculations\n", + "x = (n-1)/n;\t\t\t#Ratio\n", + "Vs = (3.147*L*(D**2))/4;\t\t\t#Stroke volume in m**3\n", + "Vc = r*Vs;\t\t\t#Clearance volume in m**3\n", + "V1 = Vc+Vs;\t\t\t#Volume at point 1 in m**3\n", + "V4 = Vc*((P2/P1)**(1/n));\t\t\t#Volume at point 4 in m**3\n", + "EVs = V1-V4;\t\t\t#Effective swept volume in m**3\n", + "W = (P1*(10**5)*EVs*(((P2/P1)**x)-1))/x;\t\t\t#Work done in J/cycle\n", + "MEP = (W/Vs)/(10**5);\t\t\t#Mean effective pressure in bar\n", + "P = (W*N)/(60*1000);\t\t\t#Power required in kW\n", + "\n", + "# Results\n", + "print 'Mean effective pressure is %3.2f bar \\\n", + "\\nPower required is %3.2f kW'%(MEP,P)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mean effective pressure is 1.81 bar \n", + "Power required is 14.21 kW\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.7 Page no : 255" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables\n", + "D = 0.2;\t\t\t#Bore in m\n", + "L = 0.3;\t\t\t#Stroke in m\n", + "P1 = 97.;\t\t\t#Pressure at entry in kN/(m**2)\n", + "P4 = P1;\t\t\t#Pressure at point 4 in kN/(m**2)\n", + "T1 = 293.;\t\t\t#Temperature at point 1 in K\n", + "P2 = 550.;\t\t\t#Compression Pressure in kN/(m**2)\n", + "P3 = P2;\t\t\t#Pressure at point 3 in kN/(m**2)\n", + "n = 1.3;\t\t\t#Adiabatic gas constant\n", + "N = 500.;\t\t\t#Speed of compressor in rpm\n", + "Pa = 101.325;\t\t\t#Air pressure in kN/(m**2)\n", + "Ta = 288.;\t\t\t#Air temperature in K\n", + "\n", + "# Calculations\n", + "x = (n-1)/n;\t\t\t#Ratio\n", + "DV = (3.147*L*(D**2))/4;\t\t\t#Difference in volumes in m**3\n", + "V3 = r*DV;\t\t\t#Clearance volume in m**3\n", + "V1 = V3+DV;\t\t\t#Volume at point 1 in m**3\n", + "V4 = V3*((P3/P4)**(1/n));\t\t\t#Volume at point 4 in m**3\n", + "Vs = V1-V4;\t\t\t#Effective swept volume in m**3\n", + "EVs = Vs*N;\t\t\t#Effective swept volume per min\n", + "Va = (P1*EVs*Ta)/(Pa*T1);\t\t\t#Free air delivered in (m**3)/min\n", + "nV = ((V1-V4)/(V1-V3))*100;\t\t\t#Volumetric effciency\n", + "T2 = T1*((P2/P1)**x);\t\t\t#Air delivery temperature in K\n", + "t2 = T2-273;\t\t\t#Air delivery temperature in oC\n", + "W = (n*P1*(V1-V4)*(((P2/P1)**x)-1))*N/((n-1)*60);\t\t\t#Cycle power in kW\n", + "Wiso = P1*V1*(math.log(P2/P1));\t\t\t#Isothermal workdone\n", + "niso = (Wiso/(4.33*0.493))*100;\t\t\t#Isothermal efficiency\n", + "\n", + "# Results\n", + "print 'Free air delivered is %3.3f m**3/min \\\n", + "\\nVolumetric efficiency is %3.0f percent \\\n", + "\\nAir delivery temperature is %3.1f oC \\\n", + "\\nCycle power is %3.0f kW \\\n", + "\\nIsothermal efficiency is %3.1f percent'%(Va,nV,t2,W,round(niso,-1))\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Free air delivered is 3.820 m**3/min \n", + "Volumetric efficiency is 86 percent \n", + "Air delivery temperature is 164.3 oC \n", + "Cycle power is 14 kW \n", + "Isothermal efficiency is 80.0 percent\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.8 Page no : 257" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# Variables\n", + "Ve = 30.;\t\t\t#Volume of air entering compressor per hour in m**3\n", + "P1 = 1.;\t\t\t#Presure of air entering compressor in bar\n", + "N = 450.;\t\t\t#Speed in rpm\n", + "P2 = 6.5;\t\t\t#Pressure at point 2 in bar\n", + "nm = 0.8;\t\t\t#Mechanical efficiency\n", + "nv = 0.75;\t\t\t#Volumetric efficiency\n", + "niso = 0.76;\t\t\t#Isothermal efficiency\n", + "\n", + "# Calculations\n", + "Vs = Ve/(nv*3600);\t\t\t#Swept volume per sec in (m**3)/s\n", + "V = (Vs*60)/N;\t\t\t#Swept volume per cycle in m**3\n", + "V1 = (Ve*60)/(3600*N);\t\t\t#Volume at point 1 in m**3\n", + "Wiso = P1*100*V1*math.log(P2/P1);\t\t\t#Isothermal workdone per cycle\n", + "Wact = Wiso/niso;\t\t\t#Actual workdone per cycle on air\n", + "MEP = (Wact/V)/100;\t\t\t#Mean effective pressure in bar\n", + "IP = (Wact*N)/60;\t\t\t#Indicated power in kW\n", + "BP = IP/nm;\t\t\t#Brake power in kW\n", + "\n", + "# Results\n", + "print 'Mean effective pressure is %3.3f bar \\\n", + "\\nBrake power is %3.2f kW'%(MEP,BP)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mean effective pressure is 1.847 bar \n", + "Brake power is 2.57 kW\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.9 Page no : 258" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# Variables\n", + "Va = 15.;\t\t\t#Volume of air in (m**3)/min\n", + "Pa = 1.01325;\t\t\t#Pressure of air in bar\n", + "Ta = 302.;\t\t\t#Air temperature in K\n", + "P1 = 0.985;\t\t\t#Pressure at point 1 in bar\n", + "r = 0.04 # ratio\n", + "T1 = 313.;\t\t\t#Temperature at point 1 in K\n", + "y = 1.3;\t\t\t#Ratio of stroke to bore diameter\n", + "N = 300.;\t\t\t#Speed in rpm\n", + "n = 1.3;\t\t\t#Adiabatic gas constant\n", + "P2 = 7.5;\t\t\t#Pressure at point 2 in bar\n", + "\n", + "# Calculations\n", + "x=((P2/P1)**(1./n))-1;\n", + "a = x*r;\t\t\t#Ratio of volume at point 4 to swept volume\n", + "nv = 1-a;\t\t\t#Volumetric efficiency\n", + "V1 = (Pa*Va*T1)/(Ta*P1);\t\t\t#Volume at point 1 in (m**3)/min\n", + "Vs = V1/(nv*N*2);\t\t\t#Swept volume in m**3\n", + "D = ((Vs*4)/(math.pi*y))**(1./3);\t\t\t#Bore in m\n", + "L = y*D;\t\t\t#Stroke in m\n", + "\n", + "# Results\n", + "print 'Cylinder bore in %3.3f m \\\n", + "\\nCylinder stroke %3.3f m'%(D,L)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Cylinder bore in 0.313 m \n", + "Cylinder stroke 0.407 m\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.10 Page no : 259" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# Variables\n", + "P1 = 0.98;\t\t\t#Pressure at point 1 in bar\n", + "P4 = P1;\t\t\t#Pressure at point 4 in bar\n", + "P2 = 7.;\t\t\t#Pressure at point 2 in bar\n", + "P3 = P2;\t\t\t#Pressure at point 3 in bar\n", + "n = 1.3;\t\t\t#Adiabatic gas consmath.tant\n", + "Ta = 300.;\t\t\t#Air temperature in K\n", + "Pa = 1.013;\t\t\t#Air pressure in bar\n", + "T1 = 313.;\t\t\t#Temperature at point 1 in K\n", + "Va = 15.;\t\t\t#Volume of air delivered in m**3\n", + "R = 0.287;\t\t\t#Universal gas constant in kJ/kg-K\n", + "c = 0.04\n", + "\n", + "# Calculations\n", + "x = (n-1)/n;\t\t\t#Ratio\n", + "r = (P2/P1)**(1/n);\t\t\t#Ratio of volumes\n", + "a = r*c;\t\t\t#Ratio of volume at point 4 to swept volume\n", + "DV = 1+c-a;\t\t\t#Difference in volumes\n", + "V = (P1*DV*Ta)/(T1*Pa);\t\t\t#Volume of air delivered per cycle\n", + "nv = V*100;\t\t\t#Volumetric efficiency\n", + "DV1 = (Pa*Va*T1)/(Ta*P1);\t\t\t#Difference in volumes\n", + "T2 = T1*((P2/P1)**x);\t\t\t#Temperature at point 2 in K\n", + "ma = (Pa*100*Va)/(R*Ta);\t\t\t#Mass of air delivered in kg/min\n", + "IP = (ma*R*(T2-T1))/(x*60);\t\t\t#Indicated power in kW\n", + "Piso = (ma*R*T1*math.log(P2/P1))/60;\t\t\t#Isothermal indicated power in kW\n", + "niso = (Piso/IP)*100;\t\t\t#Isothermal efficiency\n", + "\n", + "# Results\n", + "print 'Volumetric efficiency is %3.1f percent \\\n", + "\\nIndicated power is %3.2f kW \\\n", + "\\nIsothermal efficiency is %3.0f percent'%(nv,IP,niso)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Volumetric efficiency is 79.6 percent \n", + "Indicated power is 65.74 kW \n", + "Isothermal efficiency is 79 percent\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.11 Page no : 261" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "V1 = 7.*(10**-3);\t\t\t#Volume of air in (m**3)/s\n", + "P1 = 1.013;\t\t\t#Pressure of air in bar\n", + "T1 = 288.;\t\t\t#Air temperature in K\n", + "P2 = 14.;\t\t\t#Pressure at point 2 in bar\n", + "n = 1.3;\t\t\t#Adiabatic gas constant\n", + "nm = 0.82;\t\t\t#Mechanical efficiency\n", + "\n", + "# Calculations\n", + "x = (n-1)/n;\t\t\t#Ratio\n", + "W = (P1*100*V1*(((P2/P1)**x)-1))/x;\t\t\t#Work done by compressor in kW\n", + "P = W/nm;\t\t\t#Power requred to drive compressor in kW\n", + "\n", + "# Results\n", + "print 'Power requred to drive compressor is %3.2f kW'%(P)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Power requred to drive compressor is 3.12 kW\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.12 Page no : 261" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "L = 0.15;\t\t\t#Stroke in mm\n", + "D = 0.15;\t\t\t#Bore in mm\n", + "N = 8.;\t\t\t#Speed in rps\n", + "P1 = 100.;\t\t\t#Pressure at point 1 in kN/(m**2)\n", + "P2 = 550.;\t\t\t#Pressure at point 2 in kN/(m**2)\n", + "n = 1.32;\t\t\t#Adiabatic gas constant\n", + "C = 0.06 # RATIO\n", + "\n", + "# Calculations\n", + "x = (n-1)/n;\t\t\t#Ratio\n", + "nv = (1+C-(C*((P2/P1)**(1/n))))*100;\t\t\t#Volumetric efficiency\n", + "DV = (3.147*(D**2)*L)/4;\t\t\t#Difference in volumes at points 1 and 3\n", + "DV1 = (nv*DV)/100;\t\t\t#Difference in volumes at points 1 and 4\n", + "V2 = DV1*((P1/P2)**(1/n))*N;\t\t\t#Volume of air delivered per second\n", + "W = (P1*DV1*(((P2/P1)**x)-1))*N/x;\t\t\t#Power of compressor in kW\n", + "\n", + "# Results\n", + "print 'Theoretical volume efficiency is %3.1f percent \\\n", + "\\nVolume of air delivered is %3.5f m**3/s \\\n", + "\\nPower of compressor is %3.3f kW'%(nv,V2,W)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Theoretical volume efficiency is 84.2 percent \n", + "Volume of air delivered is 0.00491 m**3/s \n", + "Power of compressor is 3.774 kW\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.13 Page no : 262" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# Variables\n", + "V = 16.;\t\t\t#Volume of air compresssed in m**3\n", + "P1 = 1.;\t\t\t#Pressure at point 1 in bar\n", + "P3 = 10.5;\t\t\t#Pressure at point 3 in bar\n", + "T1 = 294.;\t\t\t#Temperature at point 1 in K\n", + "Tc = 25.;\t\t\t#Temperature of cooling water in oC\n", + "n = 1.35;\t\t\t#Adiabatics gas constant\n", + "R = 0.287;\t\t\t#Universal gas constant in kJ/kg-K\n", + "Cp = 1.005;\t\t\t#Specific heat at constant pressure in kJ/kg-K\n", + "Cw = 4.187;\t\t\t#Specific heat of water in kJ/kg-K\n", + "\n", + "# Calculations\n", + "x = (n-1)/n;\t\t\t#Ratio\n", + "P2 = math.sqrt(P1*P3);\t\t\t#Pressure at point 2 in bar\n", + "W1 = (2*P1*100*V*(((P2/P1)**x)-1))/(x*60);\t\t\t#Indicated power of compressor from P1 to P2 in kW\n", + "W2 = (P1*100*V*(((P3/P1)**x)-1))/(x*60);\t\t\t#Indicated power of compressor from P1 to P3 in kW\n", + "T4 = T1*((P2/P1)**x);\t\t\t#Maximum temperature for two stage compression in K\n", + "T2 = T1*((P3/P1)**x);\t\t\t#Maximum temperature for single stage compression in K\n", + "m = (P1*100*V)/(R*T1);\t\t\t#Mass of air compressed in kg/min\n", + "Q = m*Cp*(T4-T1);\t\t\t#Heat rejected by air in kJ/min\n", + "mc = Q/(Cw*Tc);\t\t\t#Mass of cooling water in kg/min\n", + "\n", + "# Results\n", + "print 'Minimum indicated power required for 2 stage compression is %3.1f kW \\\n", + "\\nPower required for single stage compression is 18 percent more than that for \\\n", + "two stage compression with perfect intercooling \\\n", + "\\nMaximum temperature for two stage compression is %3.1f K \\\n", + "\\nMaximum temperature for single stage compression is %3.1f K \\\n", + "\\nHeat rejected by air is %3.1f kJ/min \\\n", + "\\nMass of cooling water required is %3.1f kg/min'%(W1,T4,T2,Q,mc)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum indicated power required for 2 stage compression is 73.3 kW \n", + "Power required for single stage compression is 18 percent more than that for two stage compression with perfect intercooling \n", + "Maximum temperature for two stage compression is 398.8 K \n", + "Maximum temperature for single stage compression is 540.9 K \n", + "Heat rejected by air is 1996.6 kJ/min \n", + "Mass of cooling water required is 19.1 kg/min\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.14 Page no : 264" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# Variables\n", + "V = 0.2;\t\t\t#Air flow rate in (m**3)/s\n", + "P1 = 0.1;\t\t\t#Intake pressure in MN/(m**2)\n", + "P3 = 0.7;\t\t\t#Final pressure in MN/(m**2)\n", + "T1 = 289.;\t\t\t#Intake temperature in K\n", + "n = 1.25;\t\t\t#Adiabatic gas constant\n", + "N = 10.;\t\t\t#Compressor speed in rps\n", + "\n", + "# Calculations\n", + "x = (n-1)/n;\t\t\t#Ratio\n", + "P2 = math.sqrt(P1*P3);\t\t\t#Intermediate pressure in MN/(m**2)\n", + "V1 = (V/N)*1000;\t\t\t#Total volume of LP cylinder in litres\n", + "V2 = ((P1*V1)/P2);\t\t\t#Total volume of HP cylinder in litres\n", + "W = ((2*P1*V*(((P2/P1)**x)-1))/x)*1000;\t\t\t#Cycle power in kW\n", + "\n", + "# Results\n", + "print 'Intermediate pressure is %3.3f MN/m**2 \\\n", + "\\nTotal volume of LP cylinder is %3.0f litres \\\n", + "\\nTotal volume of HP cylinder is %3.1f litres \\\n", + "\\nCycle power is %3.0f kW'%(P2,V1,V2,W)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Intermediate pressure is 0.265 MN/m**2 \n", + "Total volume of LP cylinder is 20 litres \n", + "Total volume of HP cylinder is 7.6 litres \n", + "Cycle power is 43 kW\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.15 Page no : 265" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "P1 = 1.;\t\t\t#Pressure at point 1 in bar\n", + "T1 = 290.;\t\t\t#Temperature at point 1 in K\n", + "P3 = 60.;\t\t\t#Pressure at point 3 in bar\n", + "P2 = 8.;\t\t\t#Pressure at point 2 in bar\n", + "T2 = 310.;\t\t\t#Temperature at point 2 in K\n", + "L = 0.2;\t\t\t#Stroke in m\n", + "D = 0.15;\t\t\t#Bore in m\n", + "n = 1.35;\t\t\t#Adiabatic gas constant\n", + "N = 200.;\t\t\t#Speed in rpm\n", + "\n", + "# Calculations\n", + "x = (n-1)/n;\t \t\t#Ratio\n", + "V1 = (3.147*(D**2)*L)/4;\t\t\t#Volume at point 1 in m**3\n", + "V2 = (P1*V1*T2)/(T1*P2);\t\t\t#Volume of air entering LP cylinder in m**3\n", + "W = ((P1*(10**5)*V1*(((P2/P1)**x)-1))/x)+((P2*(10**5)*V2*(((P3/P2)**x)-1))/x);\t\t\t#Workdone by compressor per cycle in J\n", + "P = (W*N)/(60*1000);\t\t \t#Power of compressor in kW\n", + "\n", + "# Results\n", + "print 'Power of compressor is %3.2f kW'%(P)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Power of compressor is 6.59 kW\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.16 Page no : 265" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables\n", + "N = 220.;\t\t\t#Speed of compressor in rpm\n", + "P1 = 1.;\t\t\t#Pressure entering LP cylinder in bar\n", + "T1 = 300.;\t\t\t#Temperature at point 1 in K\n", + "Dlp = 0.36;\t\t\t#Bore of LP cylinder in m\n", + "Llp = 0.4;\t\t\t#Stroke of LP cylinder in m\n", + "Lhp = 0.4;\t\t\t#Stoke of HP cylinder in m\n", + "P2 = 4.;\t\t\t#Pressure leaving LP cylinder in bar\n", + "P5 = 3.8;\t\t\t#Pressure entering HP cylinder in bar\n", + "T3 = 300.;\t\t\t#Temperature entering HP cylinder in K\n", + "P6 = 15.2;\t\t\t#Dicharge pressure in bar\n", + "n = 1.3;\t\t\t#Adiabatic gas constant\n", + "Cp = 1.0035;\t\t\t#Specific heat at constant pressure in kJ/kg-K\n", + "R = 0.287;\t\t\t#Universal gas constant in kJ/kg-K\n", + "T5 = T1;\t\t\t#Temperature at point 5 in K\n", + "C = 0.04\n", + "# Calculations\n", + "x = (n-1)/n;\t\t\t#Ratio\n", + "Vslp = round((math.pi*(Dlp**2)*Llp*N*2)/4,2);\t\t\t#Swept volume of LP cylinder in m**3/min\n", + "nv = round(1+C-(C*((P2/P1)**(1/n))),4);\t\t\t#Volumetric efficiency\n", + "V1 = nv*Vslp;\t\t\t#Volume of air drawn at point 1 in (m**3)/min\n", + "m = round((P1*100*V1)/(R*T1),2);\t\t\t#Mass of air in kg/min\n", + "T2 = round(T1*((P2/P1)**x));\t\t\t#Temperature at point 2 in K\n", + "QR = m*Cp*(T2-T5);\t\t\t#Heat rejected in kJ/min\n", + "V5 = (m*R*T5)/(P5*100);\t\t\t#Volume of air drawn in HP cylinder M**3/min\n", + "Plp = P2/P1;\t\t\t#Pressure ratio of LP cylinder\n", + "Php = P6/P5;\t\t\t#Pressure ratio of HP cylinder\n", + "Vshp = V5/nv;\t\t\t#Swept volume of HP cylinder in m**3/min\n", + "Dhp = math.sqrt((Vshp*4)/(3.147*Lhp*N*2));\t\t\t#Bore of HP cylinder in m\n", + "P = (m*R*(T2-T1))/(x*60);\t\t\t#Power required for HP cylinder in kW\n", + "\n", + "print V5,Plp,Php,Vshp,Dhp,P\n", + "# Results\n", + "print 'Heat rejected in intercooler is %3.1f kJ/min \\\n", + "\\nDiameter of HP cylinder is %3.4f m \\\n", + "\\nPower required for HP cylinder is %3.0f kW'%(QR,Dhp,P)\n", + "\n", + "# rounding off error. please check\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "4.35484736842 4.0 4.0 4.71405863652 0.184511219993 45.0178314444\n", + "Heat rejected in intercooler is 2179.5 kJ/min \n", + "Diameter of HP cylinder is 0.1845 m \n", + "Power required for HP cylinder is 45 kW\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.17 Page no : 267" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables\n", + "P1 = 1.;\t\t\t#Pressure at point 1 in bar\n", + "P3 = 30.;\t\t\t#Pressure at point 3 in bar\n", + "T1 = 300.;\t\t\t#Temperature at point 1 in K\n", + "n = 1.3;\t\t\t#Adiabatics gas constant\n", + "\n", + "# Calculations\n", + "P2 = math.sqrt(P1*P3);\t\t\t#Intermediate pressure in bar\n", + "rD = math.sqrt(P2/P1);\t\t\t#Ratio of cylinder diameters\n", + "\n", + "# Results\n", + "print 'Ratio of cylinder diameters is %3.2f'%(rD)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ratio of cylinder diameters is 2.34\n" + ] + } + ], + "prompt_number": 41 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.18 Page no : 268" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "P1 = 1.013;\t\t\t#Pressure at point 1 in bar\n", + "T1 = 288.;\t\t\t#Temperaturea at point 1 in K\n", + "v1 = 8.4;\t\t\t#free air delivered by compressor in m**3\n", + "P4 = 70.;\t\t\t#Pressure at point 4 in bar\n", + "n = 1.2;\t\t\t#Adiabatic gas constant\n", + "Cp = 1.0035;\t\t\t#Specific heat at constant pressure in kJ/kg-K\n", + "\n", + "# Calculations\n", + "x = (n-1)/n;\t\t\t#Ratio\n", + "P2 = P1*((P4/P1)**(1./3));\t\t\t#LP cylinder delivery pressure in bar\n", + "P3 = P2*((P4/P1)**(1./3));\t\t\t#IP cylinder delivery pressure in bar\n", + "r = P2/P1;\t\t\t#Ratio of cylinder volumes\n", + "r1 = P3/P2;\t\t\t#Ratio of cylinder volumes\n", + "r2 = r*r1;\t\t\t#Ratio of cylinder volumes\n", + "V3 = 1;\t\t\t#Volume at point 3 in m**3\n", + "T4 = T1*((P2/P1)**x);\t\t\t#Three stage outlet temperature in K\n", + "QR = Cp*(T4-T1);\t\t\t#Heat rejected in intercooler in kJ/kg of air\n", + "W = ((3*P1*100*v1*(((P4/P1)**(x/3))-1))/(x*60));\t\t\t#Total indiacted power in kW\n", + "\n", + "# Results\n", + "print 'LP cylinder delivery pressure is %3.3f bar \\\n", + "\\nIP cylinder delivery pressure is %3.2f bar \\\n", + "\\nRatio of cylinder volumes is %3.2f:%3.1f:%3.0f \\\n", + "\\nTemperature at end of each stage is %3.2f K \\\n", + "\\nHeat rejected in each intercooler is %3.1f kJ/kg of air \\\n", + "\\nTotal indicated power is %3.2f kW'%(P2,P3,r2,r1,V3,T4,QR,W)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "LP cylinder delivery pressure is 4.157 bar \n", + "IP cylinder delivery pressure is 17.06 bar \n", + "Ratio of cylinder volumes is 16.84:4.1: 1 \n", + "Temperature at end of each stage is 364.41 K \n", + "Heat rejected in each intercooler is 76.7 kJ/kg of air \n", + "Total indicated power is 67.72 kW\n" + ] + } + ], + "prompt_number": 42 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.19 Page no : 269" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "D = 0.45;\t\t\t#Bore in m\n", + "L = 0.3;\t\t\t#Stroke in m\n", + "P1 = 1.;\t\t\t#Pressure at point 1 inn bar\n", + "T1 = 291.;\t\t\t#Temperature at point 1 in K\n", + "P4 = 15.;\t\t\t#Pressure at point 4 in bar\n", + "n = 1.3;\t\t\t#Adiabatic gas constant\n", + "R = 0.29;\t\t\t#Universal gas constant in kJ/kg-K\n", + "\n", + "# Calculations\n", + "x = (n-1)/n;\t\t\t#Ratio\n", + "k = (P4/P1)**(1./3);\t\t\t#Pressure ratio\n", + "P2 = k*P1;\t\t\t#Pressure at point 2 in bar\n", + "P3 = k*P2;\t\t\t#Pressure at point 1 in bar\n", + "Vslp = (3.147*(D**2)*L)/4;\t\t\t#Swept volume of LP cylinder\n", + "V7 = C*Vslp;\t\t\t#Volume at point 7 in m**3\n", + "V1 = Vslp+V7;\t\t\t#Volume at point 1 in m**3\n", + "V8 = V7*(k**(1/n));\t\t\t#Volume at point 8 in m**3\n", + "EVs = (V1-V8)*1000;\t\t\t#Effective swept volume in litres\n", + "T4 = T1*(k**x);\t\t\t#Temperature at point 4 in K\n", + "t4 = T4-273;\t\t\t#Delivery temperature in oC\n", + "DV = ((P1*T4*(V1-V8))/(P4*T1))*1000;\t\t\t#Delivery volume per stroke in litres\n", + "W = (3*R*T1*((k**x)-1))/x;\t\t\t#Workdone per kg of air in kJ\n", + "\n", + "# Results\n", + "print 'Intermediate pressures are %3.3f bar and %3.3f bar \\\n", + "\\nEffective swept volume of LP cylinder is %3.2f litres \\\n", + "\\nTemperature of air delivered per stroke is %3.1f oC \\\n", + "\\nVolume of air delivered per stroke is %3.2f litres \\\n", + "\\nWork done per kg of air is %3.1f kJ'%(P2,P3,EVs,t4,DV,W)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Intermediate pressures are 2.466 bar and 6.082 bar \n", + "Effective swept volume of LP cylinder is 44.92 litres \n", + "Temperature of air delivered per stroke is 85.4 oC \n", + "Volume of air delivered per stroke is 3.69 litres \n", + "Work done per kg of air is 254.1 kJ\n" + ] + } + ], + "prompt_number": 43 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.20 Page no : 271" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math \n", + "\n", + "# Variables\n", + "P1 = 1.;\t\t\t#Pressure at point 1 in bar\n", + "Pns = 100.;\t\t\t#Maximum pressure in bar\n", + "p = 4.; \t\t\t#Pressure ratio\n", + "\n", + "# Calculations\n", + "Ns = math.log(Pns)/math.log(p);\t\t\t#Number of stages\n", + "y = math.ceil(Ns);\t \t\t#Rounding off to next higher integer\n", + "ps = (Pns/P1)**(1/y);\t\t\t #Exact stage pressure ratio\n", + "P2 = ps*P1;\t\t\t#Pressure at point 2 in bar\n", + "P3 = ps*P2;\t\t\t#Pressure at point 3 in bar\n", + "P4 = ps*P3;\t\t\t#Pressure at point 4 in bar\n", + "\n", + "# Results\n", + "print 'Number of stages are %3.2f \\\n", + "\\nExact stage pressure ratio is %3.3f \\\n", + "\\nIntermediate pressures are %3.3f bar, %3.2f bar, %3.2f bar'%(y,ps,P2,P3,P4)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of stages are 4.00 \n", + "Exact stage pressure ratio is 3.162 \n", + "Intermediate pressures are 3.162 bar, 10.00 bar, 31.62 bar\n" + ] + } + ], + "prompt_number": 51 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Thermal_Engineering_by_A._V._Arasu/ch6.ipynb b/Thermal_Engineering_by_A._V._Arasu/ch6.ipynb new file mode 100644 index 00000000..f0f6f415 --- /dev/null +++ b/Thermal_Engineering_by_A._V._Arasu/ch6.ipynb @@ -0,0 +1,901 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:4962285b4f62f2bf376e81ac1782d3fcaba245abd75f93d7b849812ce5c45ab3" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 6 :\n", + "Refrigeration Cycles" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.1 Page no : 308" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "COP = 8.5;\t\t\t#Co-efficient of performance\n", + "T1 = 300.;\t\t\t#Room temperature in K\n", + "T2 = 267.;\t\t\t#Refrigeration temperature in K\n", + "\n", + "# Calculations\n", + "COPmax = T2/(T1-T2);\t\t\t#Maximum COP possible\n", + "\n", + "# Results\n", + "print 'Maximum COP possible is %3.2f \\\n", + "\\nSince the COP claimed by the inventor is more than the maximum possible COP\\\n", + " his claim is not correct'%(COPmax)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum COP possible is 8.09 \n", + "Since the COP claimed by the inventor is more than the maximum possible COP his claim is not correct\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.2 Page no : 309" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "TL = 268.;\t\t\t#Low temperature in K\n", + "TH = 293.;\t\t\t#High temperature in K\n", + "t = 24.;\t\t\t#time in hrs\n", + "C = 2100.;\t\t\t#Capacity of refrigerator in kJ/s\n", + "Tw = 10.;\t\t\t#Water temperature in oC\n", + "L = 335.;\t\t\t#Latent heat of ice in kJ/kg\n", + "\n", + "# Calculations\n", + "COP = TL/(TH-TL);\t\t\t#Co-efficient of performance\n", + "Pmin = C/COP;\t\t\t#Minimum power required in kW\n", + "Qr = (4.187*(Tw-0))+L;\t\t\t#Heat removed from water in kJ/kg\n", + "m = C/Qr;\t\t\t#mass of ice formed in kg/s\n", + "W = (m*t*3600)/1000;\t\t\t#Weight of ice formed in tons\n", + "\n", + "# Results\n", + "print 'Minimum power required is %3.2f kW \\\n", + "\\nWeight of ice formed in 24 hours is %3.2f tons'%(Pmin,W)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum power required is 195.90 kW \n", + "Weight of ice formed in 24 hours is 481.44 tons\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.3 Page no : 309" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "TL = -10.;\t\t\t#Temperature of brine in oC\n", + "TH = 20.;\t\t\t#Temperature of water in oC\n", + "L = 335.;\t\t\t#Latent heat of ice in kJ/kg\n", + "\n", + "# Calculations\n", + "Qr = (4.187*(TH-0))+L;\t\t\t#Heat removed from water in kJ/kg\n", + "COP = (TL+273)/(TH-TL);\t\t\t#Co-efficient of performance\n", + "mi = (COP*3600)/Qr;\t\t\t#mass of ice formed per kWh in kg\n", + "\n", + "# Results\n", + "print 'Mass of ice formed per kWh is %3.1f kg'%(mi)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mass of ice formed per kWh is 75.4 kg\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.4 Page no : 310" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "P1 = 1.2;\t\t\t#Pressure at point 1 in bar\n", + "P2 = 7.;\t\t\t#Pressure at point 2 in bar\n", + "m = 0.05;\t\t\t#mass flow rate of refrigerant in kg/s\n", + "h1 = 340.1;\t\t\t#Enthalpy at point 1 from refrigerant-12 tables in kJ/kg\n", + "s1 = 1.57135;\t\t\t#Entropy at point 1 from refrigerant-12 tables in kJ/kg-K\n", + "s2 = 1.57135;\t\t\t#Entropy at point 2 from refrigerant-12 tables in kJ/kg-K\n", + "h2 = 372.;\t\t\t#Enthalpy at point 2 from refrigerant-12 tables in kJ/kg\n", + "h3 = 226.575;\t\t\t#Enthalpy at point 3 from refrigerant-12 tables in kJ/kg\n", + "h4 = 226.575;\t\t\t#Enthalpy at point 4 from refrigerant-12 tables in kJ/kg\n", + "\n", + "# Calculations\n", + "Q2 = m*(h1-h4);\t\t\t#Rate of heat removed from the refrigerated space in kW\n", + "W = m*(h2-h1);\t\t\t#Power input to the compressor in kW\n", + "Q1 = m*(h2-h3);\t\t\t#Rate of heat rejection to the environment in kW\n", + "COP = Q2/W;\t\t\t#Co-efficient of performance\n", + "\n", + "# Results\n", + "print 'Rate of heat removed from the refrigerated space is %3.2f kW \\\n", + "\\nPower input to the compressor is %3.3f kW \\\n", + "\\nRate of heat rejection to the environment is %3.2f kW \\\n", + "\\nCo-efficient of performance is %3.2f'%(Q2,W,Q1,COP)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rate of heat removed from the refrigerated space is 5.68 kW \n", + "Power input to the compressor is 1.595 kW \n", + "Rate of heat rejection to the environment is 7.27 kW \n", + "Co-efficient of performance is 3.56\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.5 Page no : 311" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "T2 = 40.;\t\t\t#Temperature at point 2 in oC\n", + "T1 = -10.;\t\t\t#Temperature at point 1 in oC\n", + "h2 = 367.155;\t\t\t#Enthalpy at point 2 from refrigerant-12 tables in kJ/kg\n", + "s2 = 1.54057;\t\t\t#Entropy at point 2 from refrigerant-12 tables in kJ/kg-K\n", + "s1 = 1.54057;\t\t\t#Entropy at point 1 from refrigerant-12 tables in kJ/kg-K\n", + "sg = 1.56004;\t\t\t#Entropy from refrigerant-12 tables in kJ/kg-K\n", + "sf = 0.96601;\t\t\t#Entropy from refrigerant-12 tables in kJ/kg-K\n", + "hf = 190.822;\t\t\t#Enthalpy from refrigerant-12 tables in kJ/kg-K\n", + "hfg = 156.319;\t\t\t#Enthalpy from refrigerant-12 tables in kJ/kg-K\n", + "h3 = 238.533;\t\t\t#Enthalpy at point 3 from refrigerant-12 tables in kJ/kg-K\n", + "h4 = h3;\t\t\t#Enthalpy at point 4 from refrigerant-12 tables in kJ/kg-K\n", + "\n", + "# Calculations\n", + "x1 = (s1-sf)/(sg-sf);\t\t\t#Quality factor\n", + "h1 = hf+(x1*hfg);\t\t\t#Enthalpy at point 1 from refrigerant-12 tables in kJ/kg\n", + "COP = (h1-h4)/(h2-h1);\t\t\t#Co-efficient of performance\n", + "\n", + "# Results\n", + "print 'COP of the system is %3.2f'%(COP)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "COP of the system is 4.12\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6 Page no : 311" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "Tc = 35.;\t\t\t#Temperature of condenser in oC\n", + "Te = -15.;\t\t\t#Temperature of evaporator in oC\n", + "m = 10.;\t\t\t#Mass of ice per day in tons\n", + "Tw = 30.;\t\t\t#Temperature of water in oC\n", + "Ti = -5.;\t\t\t#Temperature of ice in oC\n", + "nv = 0.65;\t\t\t#Volumetric efficiency\n", + "N = 1200.;\t\t\t#Speed in rpm\n", + "x = 1.2;\t\t\t#Stroke to bore ratio\n", + "na = 0.85;\t\t\t#Adiabatic efficiency\n", + "nm = 0.95;\t\t\t#Mechanical efficiency\n", + "S = 4.187;\t\t\t#Specific heat of water in kJ/kg\n", + "L = 335.;\t\t\t#Latent heat of ice in kJ/kg\n", + "h1 = 1667.24;\t\t\t#Enthalpy at Te from Ammonia chart in kJ/kg\n", + "h2 = 1925.;\t\t\t#Enthalpy at Te from Ammonia chart in kJ/kg\n", + "h4 = 586.41;\t\t\t#Enthalpy at Tc from Ammonia chart in kJ/kg\n", + "v1 = 0.508;\t\t\t#Specific humidity at Te from Ammonia chart in (m**3)/kg\n", + "\n", + "# Calculations\n", + "Qr = (((m*1000)/24)*((S*(Tw-0))+L+(1.94*(0-Ti))))/3600;\t\t\t#Refrigerating capacity in kW\n", + "mr = Qr/(h1-h4);\t\t\t#Refrigerant mass flow rate in kg/s\n", + "T2 = 112;\t\t\t#Discharge temperature in oC\n", + "D = ((mr*v1*4*60)/(nv*3.14*x*N))**(1./3);\t\t\t#Cylinder diameter in m\n", + "L = x*D;\t\t\t#Stroke length in m\n", + "W = (mr*(h2-h1))/(na*nm);\t\t\t#Compressor motor power in kW\n", + "COPth = (h1-h4)/(h2-h1);\t\t\t#Theoretical COP\n", + "COPact = Qr/W;\t\t\t#Actual COP\n", + "\n", + "# Results\n", + "print 'Refrigerating capacity of plant is %3.2f kW \\\n", + "\\nRefrigerant mass flow rate is %3.4f kg/s \\\n", + "\\nDischarge temperature is %3.0f oC \\\n", + "\\nCylinder diameter is %3.3f m \\\n", + "\\nStroke length is %3.3f m \\\n", + "\\nCompressor motor power is %3.2f kW \\\n", + "\\nTheoretical COP is %3.2f \\\n", + "\\nActual COP is %3.2f'%(Qr,mr,T2,D,L,W,COPth,COPact)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Refrigerating capacity of plant is 54.43 kW \n", + "Refrigerant mass flow rate is 0.0504 kg/s \n", + "Discharge temperature is 112 oC \n", + "Cylinder diameter is 0.128 m \n", + "Stroke length is 0.153 m \n", + "Compressor motor power is 16.08 kW \n", + "Theoretical COP is 4.19 \n", + "Actual COP is 3.39\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.7 Page no : 313" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "T1 = -5.;\t\t\t#Temperature at point 1 in oC\n", + "T2 = 30.;\t\t\t#Temperature at point 2 in oC\n", + "m = 13500.;\t\t\t#mass of ice per day in kg\n", + "Tw = 20.;\t\t\t#Temperature of water in oC\n", + "COP = 0.6;\t\t\t#Co-efficient of performance\n", + "h2 = 1709.33;\t\t\t#Enthalpy at point 2 in kJ/kg\n", + "s2 = 6.16259;\t\t\t#Entropy at point 2 in kJ/kg-K\n", + "s1 = 6.16259;\t\t\t#Entropy at point 1 in kJ/kg-K\n", + "sf = 1.8182;\t\t\t#Entropy in kJ/kg-K\n", + "sg = 6.58542;\t\t\t#Entropy in kJ/kg-K\n", + "hf = 400.98;\t\t\t#Enthalpy in kJ/kg\n", + "hfg = 1278.35;\t\t\t#Enthalpy in kJ/kg\n", + "h4 = 562.75;\t\t\t#Enthalpy at point 4 in kJ/kg\n", + "S = 4.187;\t\t\t#Specific heat of water in kJ/kg\n", + "L = 336.;\t\t\t#Latent heat of ice in kJ/kg\n", + "\n", + "# Calculations\n", + "x1 = (s1-sf)/(sg-sf);\t\t\t#Quality factor\n", + "h1 = hf+(x1*hfg);\t\t\t#Enthalpy at point 1 from refrigerant-12 tables in kJ/kg\n", + "COPi = (h1-h4)/(h2-h1);\t\t\t#Ideal COP\n", + "COPact = COP*COPi;\t\t\t#Actual COP\n", + "Qr = ((m*S*(Tw-0))+(m*L))/(24*3600);\t\t\t#Total amount of heat removed in kJ/s\n", + "mr = Qr/(h1-h4);\t\t\t#Circulation rate of ammonia in kg/s\n", + "W = mr*(h2-h1);\t\t\t#Power required in kW\n", + "\n", + "# Results\n", + "print 'Circulation rate of ammonia is %3.3f kg/s \\\n", + "\\nPower required is %3.3f kW \\\n", + "\\nCOP is %3.3f'%(mr,W,COPact)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Circulation rate of ammonia is 0.065 kg/s \n", + "Power required is 9.374 kW \n", + "COP is 4.198\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.8 Page no : 314" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "# Variables\n", + "Tc = 20.;\t\t\t#Temperature of condenser in oC\n", + "Te = -25.;\t\t\t#Temperature of evaporator in oC\n", + "m = 15.;\t\t\t#Mass of ice per day in tons\n", + "Ts = 5.;\t\t\t#Subcooled temperature in oC\n", + "Tsh = 10.;\t\t\t#Superheated temperature in oC\n", + "n = 6.;\t\t\t#No. of cylinders\n", + "N = 950.;\t\t\t#Speed of compressor in rpm\n", + "x = 1.;\t\t\t#Stroke to bore ratio\n", + "h1 = 402.;\t\t\t#Enthalpy at point 1 from R-22 tables in kJ/kg\n", + "h2 = 442.;\t\t\t#Enthalpy at point 2 from R-22 tables in kJ/kg\n", + "h3 = 216.;\t\t\t#Enthalpy at point 3 from R-22 tables in kJ/kg\n", + "h4 = 216.;\t\t\t#Enthalpy at point 4 from R-22 tables in kJ/kg\n", + "v1 = 2.258;\t\t\t#Specific volume at point 1 in (m**3)/min\n", + "\n", + "# Calculations\n", + "Re = h1-h4; \t\t\t#Refrigerating effect in kJ/kg\n", + "mr = (m*14000)/(Re*60);\t\t\t#Mass flow of refrigerant in kg/min\n", + "Pth = (mr*(h2-h1))/60;\t\t\t#Theoretical power in kW\n", + "COP = (h1-h4)/(h2-h1);\t\t\t#Co-efficient of performance\n", + "Dth = v1/n;\t\t\t #Theoretical print lacement per cylinder\n", + "D = (((Dth*4)/(3.147*N))**(1./3))*1000;\t\t\t#Theoretical bore of compressor in mm\n", + "L = D; \t\t\t#Theoretical stroke of compressor in mm\n", + "\n", + "# Results\n", + "print 'Refrigerating effect is %3.0f kJ/kg \\\n", + "\\nMass flow of refrigerant per minute is %3.2f kg/min \\\n", + "\\nTheoretical input power is %3.2f kW COP is %3.2f \\\n", + "\\nTheoretical bore of compressor is %3.2f mm \\\n", + "\\nTheoretical stroke of compressor is %3.2f mm'%(Re,mr,Pth,COP,D,L)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Refrigerating effect is 186 kJ/kg \n", + "Mass flow of refrigerant per minute is 18.82 kg/min \n", + "Theoretical input power is 12.54 kW COP is 4.65 \n", + "Theoretical bore of compressor is 79.56 mm \n", + "Theoretical stroke of compressor is 79.56 mm\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.9 Page no : 316" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "T2 = 40.;\t\t\t#Temperature at point 2 in oC\n", + "T1 = -5.;\t\t\t#Temperature at point 1 in oC\n", + "h2 = 367.155;\t\t\t#Enthalpy at point 2 from F-12 tables in kJ/kg\n", + "sg = 1.55717;\t\t\t#Entropy from F-12 tables in kJ/kg-K\n", + "s1 = 1.54057;\t\t\t#Entropy at point 1 from F-12 tables in kJ/kg-K\n", + "sf = 0.98311;\t\t\t#Entropy from F-12 tables in kJ/kg-K\n", + "hf = 195.394;\t\t\t#Enthalpy from F-12 tables in kJ/kg\n", + "hfg = 153.934;\t\t\t#Enthalpy from F-12 tables in kJ/kg\n", + "h4 = 238.533;\t\t\t#Enthalpy at point 4 from F-12 tables in kJ/kg\n", + "h4s = 218;\t\t\t#Enthalpy at point 4 with subcooling from F-12 tables in kJ/kg\n", + "\n", + "# Calculations\n", + "x1 = (s1-sf)/(sg-sf);\t\t\t#Quality factor\n", + "h1 = hf+(x1*hfg);\t\t\t#Enthalpy at point 1 from refrigerant-12 tables in kJ/kg\n", + "COPns = (h1-h4)/(h2-h1);\t\t\t#Co-efficient of performance with no subcooling\n", + "COPs = (h1-h4s)/(h2-h1);\t\t\t#Co-efficient of performance with subcooling\n", + "\n", + "# Results\n", + "print 'COP with no subcooling is %3.3f \\\n", + "\\nCOP with subcooling is %3.3f'%(COPns,COPs)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "COP with no subcooling is 4.773 \n", + "COP with subcooling is 5.695\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.10 Page no : 309" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "Tg = 470.;\t\t\t#Heating temperature in K\n", + "T0 = 290.;\t\t\t#Cooling temperature in K\n", + "TL = 270.;\t\t\t#Refrigeration temperature in K\n", + "\n", + "# Calculations\n", + "COP = ((Tg-T0)/Tg)*(TL/(T0-TL));\t\t\t#Ideal COP of absorption refrigeration system\n", + "\n", + "# Results\n", + "print 'Ideal COP of absorption refrigeration system is %3.2f'%(COP)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ideal COP of absorption refrigeration system is 5.17\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.11 Page no : 317" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "T1 = -18.;\t\t\t#Temperature at point 1 in oC\n", + "T3 = 27.;\t\t\t#Temperature at point 3 in oC\n", + "rp = 4.;\t\t\t#Pressure ratio\n", + "m = 0.045;\t\t\t#mass flow rate in kg/s\n", + "y = 1.4;\t\t\t#Ratio of specific heats\n", + "Cp = 1.005;\t\t\t#Specific heat at constant pressure in kJ/kg-K\n", + "\n", + "# Calculations\n", + "x = (y-1)/y;\t\t\t#Ratio\n", + "T2 = (rp**x)*(273+T1);\t\t\t#Temperature at point 2 in K\n", + "Tmax = T2-273;\t\t\t#Maximum temperature in oC\n", + "T4 = ((1/rp)**x)*(273+T3);\t\t\t#Temperature at point 4 in K\n", + "Tmin = T4-273;\t\t\t#Minimum temperature in oC\n", + "qL = Cp*(T1-Tmin);\t\t\t#Heat rejected\n", + "Wcin = Cp*(Tmax-T1);\t\t\t#Compressor work\n", + "Wtout = Cp*(T3-Tmin);\t\t\t#Turbine work\n", + "Wnet = Wcin-Wtout;\t\t\t#Net work done\n", + "COP = qL/Wnet;\t\t\t#Co-efficient of performance\n", + "Qref = m*qL;\t\t\t#Rate of refrigeration in kW\n", + "\n", + "# Results\n", + "print 'Maximum temperature in the cycle is %3.0f oC \\\n", + "\\nMinimum temperature in the cycle is %3.0f oC \\\n", + "\\nCOP is %3.2f \\\n", + "\\nRate of refrigeration is %3.2f kW'%(Tmax,Tmin,COP,Qref)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum temperature in the cycle is 106 oC \n", + "Minimum temperature in the cycle is -71 oC \n", + "COP is 2.06 \n", + "Rate of refrigeration is 2.40 kW\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.12 Page no : 318" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "P1 = 1.;\t\t\t#Pressure at point 1 in bar\n", + "T1 = 268.;\t\t\t#Temperature at point 1 in K\n", + "P2 = 5.;\t\t\t#Pressure at point 2 in bar\n", + "T3 = 288.;\t\t\t#Temperature at point 3 in K\n", + "n = 1.3;\t\t\t#Adiabatic gas constant\n", + "Cp = 1.005;\t\t\t#Specific heat at constant pressure in kJ/kg-K\n", + "\n", + "# Calculations\n", + "x = (n-1)/n;\t\t\t#Ratio\n", + "T2 = ((P2/P1)**x)*T1;\t\t\t#Temperature at point 2 in K\n", + "T4 = ((P1/P2)**x)*T3;\t\t\t#Temperature at point 4 in K\n", + "W = Cp*(T3-T4);\t\t\t#Work developed per kg of air in kJ/kg\n", + "Re = Cp*(T1-T4);\t\t\t#Refrigerating effect per kg of air in kJ/kg\n", + "Wnet = Cp*((T2-T1)-(T3-T4));\t\t\t#Net work output in kJ/kg\n", + "COP = Re/Wnet;\t\t\t#Co-efficient of performance\n", + "\n", + "# Results\n", + "print 'Work developed per kg of air is %3.3f kJ/kg \\\n", + "\\nRefrigerating effect per kg of air is %3.3f kJ/kg \\\n", + "\\nCOP of the cycle is %3.2f'%(W,Re,COP)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Work developed per kg of air is 89.795 kJ/kg \n", + "Refrigerating effect per kg of air is 69.695 kJ/kg \n", + "COP of the cycle is 2.22\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.13 Page no : 319" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "T1 = 277.;\t\t\t#Temperature at point 1 in K\n", + "T3 = 328.;\t\t\t#Temperature at point 3 in K\n", + "P1 = 0.1;\t\t\t#Pressure at point 1 in MPa\n", + "P2 = 0.3;\t\t\t#Pressure at point 2 in MPa\n", + "nc = 0.72;\t\t\t#Isentropic efficiency of compressor\n", + "nt = 0.78;\t\t\t#Isentropic efficiency of turbine\n", + "y = 1.4;\t\t\t#Adiabatic gas constant\n", + "Cp = 1.005;\t\t\t#Specific heat at constant pressure in kJ/kg-K\n", + "m = 3.;\t\t\t#Cooling load in tonnes\n", + "\n", + "# Calculations\n", + "x = (y-1)/y;\t\t\t#Ratio\n", + "T2s = T1*((P2/P1)**x);\t\t\t#Temperature at point 2s in K\n", + "T2 = ((T2s-T1)/nc)+T1;\t\t\t#Temerature at point 2 in K\n", + "T4s = T3*((P1/P2)**x);\t\t\t#Temperature at point 4s in K\n", + "T4 = T3-((T3-T4s)*nt);\t\t\t#Temperature at point 4 in K\n", + "Re = Cp*(T1-T4);\t\t\t#Refrigerating effect in kJ/kg\n", + "Wnet = Cp*((T2-T1)-(T3-T4));\t\t\t#Net work output in kJ/kg\n", + "COP = Re/Wnet;\t\t\t#Co-efficient of performance\n", + "P = (m*3.52)/COP;\t\t\t#Driving power required in kW\n", + "ma = (m*3.52)/Re;\t\t\t#Mass flow rate of air in kg/s\n", + "\n", + "# Results\n", + "print 'COP of refrigerator is %3.2f \\\n", + "\\nDriving power required is %3.0f kW \\\n", + "\\nMass flow rate of air is %3.2f kg/s'%(COP,P,ma)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "COP of refrigerator is 0.25 \n", + "Driving power required is 43 kW \n", + "Mass flow rate of air is 0.59 kg/s\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.14 Page no : 321" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "P1 = 2.5;\t\t\t#Pressure at point 1 in bar\n", + "P3 = 9.;\t\t\t#Pressure at point 3 in bar\n", + "COPr = 0.65;\t\t\t#Ratio of actual COP to the theoretical COP\n", + "m = 5.;\t\t\t#Refrigerant flow in kg/min\n", + "T1 = 309;\t\t\t#Temperature at point 1 in K\n", + "T2s = 300;\t\t\t#Temperature at point 2s in K\n", + "h1 = 570.3;\t\t\t#Enthalpy at P1 from the given tables in kJ/kg\n", + "h4 = 456.4;\t\t\t#Enthalpy at P3 from the given tables in kJ/kg\n", + "h2g = 585.3;\t\t\t#Enthalpy at P3 from the given tables in kJ/kg\n", + "s2 = 4.76;\t\t\t#Entropy at P1 from the given tables in kJ/kg-K\n", + "s2g = 4.74;\t\t\t#Entropy at P3 from the given tables in kJ/kg-K\n", + "Cp = 0.67;\t\t\t#Specific heat at P3 in kJ/kg-K\n", + "\n", + "# Calculations\n", + "T2 = (2.718**((s2-s2g)/Cp))*T2s;\t\t\t#Temperature at point 2 in K\n", + "h2 = h2g+(Cp*(T2-T2s));\t\t\t#Enthalpy at point 2 in kJ/kg\n", + "COPR = (h1-h4)/(h2-h1);\t\t\t#Refrigerant COP\n", + "COPact = COPr*COPR;\t\t\t#Actual COP\n", + "qL = COPact*(h2-h1);\t\t\t#Heat rejected in kJ/kg\n", + "QL = ((m*qL*60)/3600)/3.516;\t\t\t#Cooling produced per kg of refrigerant in tonnes of refrigeration\n", + "\n", + "# Results\n", + "print 'Theoretical COP is %3.2f \\\n", + "\\nNet cooling produced per hour is %3.2f TR'%(COPR,QL)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Theoretical COP is 5.40 \n", + "Net cooling produced per hour is 1.75 TR\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.15 Page no : 322" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "T2 = 298.;\t\t\t#Temperature at point 2 in K\n", + "T1 = 268.;\t\t\t#Temperature at point 1 in K\n", + "hf1 = -7.54;\t\t\t#Liquid Enthalpy at T1 in kJ/kg\n", + "x1 = 0.6;\t\t\t#Quality factor 1\n", + "hfg1 = 245.3;\t\t\t#Latent heat at T1 in kJ/kg\n", + "sf1 = 0.251;\t\t\t#Liquid Entropy at T1 in kJ/kg-K\n", + "s1 = 0.507;\t\t\t#Entropy at point 1 in kJ/kg-K\n", + "hfg2 = 121.4;\t\t\t#Latent heat at T2 in kJ/kg\n", + "hf2 = 81.3;\t\t\t#Liquid Enthalpy at T2 in kJ/kg\n", + "h4 = hf2;\t\t\t#Enthalpy at point 4 in kJ/kg\n", + "\n", + "# Calculations\n", + "h1 = hf1+(x1*hfg1);\t\t\t#Enthalpy at point 1 in kJ/kg\n", + "x2 = ((s1-sf1)*T2)/hfg2;\t\t\t#Quality factor 2\n", + "h2 = hf2+(x2*hfg2);\t\t\t#Enthalpy at point 2 in kJ/kg\n", + "COP = (h1-h4)/(h2-h1);\t\t\t#COP of the machine\n", + "\n", + "# Results\n", + "print 'COP of the machine is %3.2f'%(COP)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "COP of the machine is 3.25\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.16 Page no : 323" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "# Variables\n", + "P1 = 25.;\t\t\t#Pressure at point 1 in bar\n", + "P2 = 60.;\t\t\t#Pressure at point 2 in bar\n", + "h2 = 208.1;\t\t\t#Vapour enthalpy at P2 in kJ/kg\n", + "h3 = 61.9;\t\t\t#Liquid enthalpy at P2 in kJ/kg\n", + "h4 = h3;\t\t\t#Liquid enthalpy at P2 in kJ/kg\n", + "s2 = 0.703;\t\t\t#Vapour entropy at P2 in kJ/kg-K\n", + "sf1 = -0.075;\t\t\t#Liquid entropy at P1 in kJ/kg-K\n", + "sfg1 = 0.971;\t\t\t#Entropy in kJ/kg-K\n", + "hf1 = -18.4;\t\t\t#Liquid Enthalpy at P1 in kJ/kg\n", + "hfg1 = 252.9;\t\t\t#Latent heat at P1 in kJ/kg\n", + "m = 5.;\t\t\t#Refrigerant flow in kg/min\n", + "\n", + "# Calculations\n", + "x1 = (s2-sf1)/sfg1;\t\t\t#Quality factor 1\n", + "h1 = hf1+(x1*hfg1);\t\t\t#Enthalpy at point 1 in kJ/kg\n", + "COP = (h1-h4)/(h2-h1);\t\t\t#Co-efficient of performance\n", + "QL = (m*(h1-h4))/60;\t\t\t#Capacity of the refrigerator in kW\n", + "\n", + "# Results\n", + "print 'COP of refrigerator is %3.2f \\\n", + "\\nCapacity of refrigerator is %3.2f kW'%(COP,QL)\n", + "\n", + "# rounding off error" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "COP of refrigerator is 5.13 \n", + "Capacity of refrigerator is 10.19 kW\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.17 Page no : 324" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "T1 = 271.;\t\t\t#Temperature at point 1 in K\n", + "T = 265.;\t\t\t#Temperature at point 1' in K\n", + "Ta = 303.;\t\t\t#Temperature at point 2' in K\n", + "Cpv = 0.733;\t\t\t#Specific heat of vapour in kJ/kg\n", + "Cpl = 1.235;\t\t\t#Specific heat of liquid in kJ/kg\n", + "h = 184.07;\t\t\t#Liquid enthalpy at T in kJ/kg\n", + "s = 0.7;\t\t\t#Entropy at point 1' in kJ/kg-K\n", + "sa = 0.685;\t\t\t#Vapour entropy at Ta in kJ/kg-K\n", + "ha = 199.62;\t\t\t#Enthalpy at point 2' in kJ/kg\n", + "hfb = 64.59;\t\t\t#Liquid enthalpy at Ta in kJ/kg\n", + "DT3 = 5.;\t\t\t#Temperature difference in oC\n", + "Q = 2532.;\t\t\t#Refrigeration capacity in kJ/min\n", + "\n", + "# Calculations\n", + "s2 = s+(Cpv*((math.log(T1/T))/(math.log(2.718))));\t\t\t#Entropy at point 1 in kJ/kg-K\n", + "h1 = h+(Cpv*(T1-T));\t\t\t#Enthalpy at point 1 in kJ/kg-K\n", + "T2 = (2.718**((s2-sa)/Cpv))*Ta;\t\t\t#Temperature at point 2 in K\n", + "h2 = ha+(Cpv*(T2-Ta));\t\t\t#Enthalpy at point 2 in kJ/kg\n", + "h4 = hfb-(Cpl*DT3);\t\t\t#Enthalpy at point 4 in kJ/kg\n", + "COP = (h1-h4)/(h2-h1);\t\t\t#Co-efficient of performance\n", + "m = Q/(h1-h4);\t\t\t#Mass flow rate of refrigerant in kJ/min\n", + "P = (m*(h2-h1))/(60*12);\t\t\t#Power required in kW/TR\n", + "\n", + "# Results\n", + "print 'COP is %3.2f \\\n", + "\\nTheoretical power required per tonne of refrigeration is %3.3f kW/TR'%(COP,P)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "COP is 6.23 \n", + "Theoretical power required per tonne of refrigeration is 0.564 kW/TR\n" + ] + } + ], + "prompt_number": 17 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Thermal_Engineering_by_A._V._Arasu/ch7.ipynb b/Thermal_Engineering_by_A._V._Arasu/ch7.ipynb new file mode 100644 index 00000000..c00c04cd --- /dev/null +++ b/Thermal_Engineering_by_A._V._Arasu/ch7.ipynb @@ -0,0 +1,212 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 7 : Air Conditioning" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.1 Page no : 345" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heating capacity of coil is 5.40 kW \n", + "Surface temperature of coil is 35 C \n", + "Capacity of humidifier is 3.33 kg/hr\n" + ] + } + ], + "source": [ + "\n", + "# Variables\n", + "DBTo = 10.;\t\t\t#Out door Dry bulb temperature in oC\n", + "WBTo = 8.;\t\t\t#Out door Wet bulb temperature in oC\n", + "DBTi = 20.;\t\t\t#In door Dry bulb temperature in oC\n", + "RH = 0.6;\t\t\t#Re-Heat factor\n", + "a = 0.3;\t\t\t#amount of air circulated in (m**3)/min/person\n", + "S = 50.;\t\t\t#Seating capacity of office\n", + "BPF = 0.32;\t\t\t#ByPass factor\n", + "ha = 25.;\t\t\t#Enthalpy at point a from Psychrometric chart shown in Page 346 in kJ/kg\n", + "hb = 42.5;\t\t\t#Enthalpy at point b from Psychrometric chart shown in Page 346 in kJ/kg\n", + "hc = 42.5;\t\t\t#Enthalpy at point c from Psychrometric chart shown in Page 346 in kJ/kg\n", + "Wa = 0.006;\t\t\t#Specific humidity at point a from Psychrometric chart shown in Page 346 in kg/kg dry air\n", + "Wc = 0.009;\t\t\t#Specific humidity at point c from Psychrometric chart shown in Page 346 in kg/kg dry air\n", + "Tb = 27.;\t\t\t#Temperature at point b in oC\n", + "na = 0.81;\t\t\t#Specific Volume from Psychrometric chart shown in page 346 in (m**3)/kg\n", + "\n", + "# Calculations\n", + "ma = (a*S)/(na*60);\t\t\t#mass of air circulated per second in kg/s\n", + "Hc = ma*(hb-ha);\t\t\t#Heating capacity of coil in kW\n", + "Ts = (Tb-(BPF*DBTo))/(1-BPF);\t\t\t#Heating coil surface temperature in oC\n", + "C = (ma*3600)*(Wc-Wa);\t\t\t#Capacity of humidifier in kg/hr\n", + "\n", + "# Results\n", + "print 'Heating capacity of coil is %3.2f kW \\\n", + "\\nSurface temperature of coil is %3.0f C \\\n", + "\\nCapacity of humidifier is %3.2f kg/hr'%(Hc,Ts,C)\n", + "\n", + "# rounding off error" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.2 Page no : 346" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Capacity of cooling coil is 6.16 tonnes \n", + "Capacity of heating coil is 3.6 kW \n", + "Amount of water vapour removed per hour is 17.80 kg/hr \n", + "Bypass factor is 0.385\n" + ] + } + ], + "source": [ + "# Variables\n", + "S = 60.;\t\t\t#No. of staff\n", + "DBTo = 30.;\t\t\t#Out door Dry bulb temperature in oC\n", + "RHo = 0.7;\t\t\t#Re-Heat factor at out-door\n", + "a = 0.4;\t\t\t#amount of air circulated in (m**3)/min/person\n", + "DBTi = 20.;\t\t\t#In door Dry bulb temperature in oC\n", + "RHi = 0.6;\t\t\t#Re-Heat factor at indoor\n", + "Td = 25.;\t\t\t#Heating coil surface temperature in oC\n", + "ha = 82.5;\t\t\t#Enthalpy at point a from Psychrometric chart shown in Page 347 in kJ/kg\n", + "hb = 34.5;\t\t\t#Enthalpy at point b from Psychrometric chart shown in Page 347 in kJ/kg\n", + "hc = 42.5;\t\t\t#Enthalpy at point c from Psychrometric chart shown in Page 347 in kJ/kg\n", + "Wa = 0.020;\t\t\t#Specific humidity at point a from Psychrometric chart shown in Page 347 in kg/kg dry air\n", + "Wb = 0.009;\t\t\t#Specific humidity at point b from Psychrometric chart shown in Page 347 in kg/kg dry air\n", + "Tb = 12.;\t\t\t#Temperature at point b in oC\n", + "na = 0.89;\t\t\t#Specific Volume from Psychrometric chart shown in page 346 in (m**3)/kg\n", + "\n", + "# Calculations\n", + "ma = (a*S)/(na*60);\t\t\t#mass of air circulated per second in kg/s\n", + "Hc = (ma*(ha-hb))/3.5;\t\t\t#Heating capacity of cooling coil in tonnes\n", + "Hh = ma*(hc-hb);\t\t\t#Heating capacity of heating coil in kW\n", + "W = (ma*3600)*(Wa-Wb);\t\t\t#Amount of water vapour removed per hour in kg/hr\n", + "BPF = (Td-DBTi)/(Td-Tb);\t\t\t#By-Pass factor\n", + "\n", + "# Results\n", + "print 'Capacity of cooling coil is %3.2f tonnes \\\n", + "\\nCapacity of heating coil is %3.1f kW \\\n", + "\\nAmount of water vapour removed per hour is %3.2f kg/hr \\\n", + "\\nBypass factor is %3.3f'%(Hc,Hh,W,BPF)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.3 Page no : 347" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Supply air condition to the room is 2.74 kg/s \n", + "Refrigeration load due to reheat is 4.93 ton \n", + "Total refrigerating capacity is 16.28 ton \n", + "Quantity of fresh air supplied is 0.365 m**3/s\n" + ] + } + ], + "source": [ + "\n", + "\n", + "# Variables\n", + "RSH = 10.;\t\t\t#Room sensible heat in kW\n", + "RLH = 10.;\t\t\t#Room latent heat in kW\n", + "td1 = 25.;\t\t\t#Inside temperature in oC\n", + "RH1 = 0.5;\t\t\t#Inside Re-Heat factor\n", + "h1 = 50.4;\t\t\t#Enthalpy at point 1 in kJ/kg\n", + "td2 = 35.;\t\t\t#Out door Dry bulb temperature in oC\n", + "tw2 = 28.;\t\t\t#Out door Wet bulb temperature in oC\n", + "CR = 4.;\t\t\t#Cooling coil ratio\n", + "BPF = 0.1;\t\t\t#Cooling coil bypass factor\n", + "tADP = 10;\t\t\t#Apparatus dew point temperature in oC\n", + "RH3 = 0.55;\t\t\t#Re-Heat factor at point 3\n", + "h3 = 58.2;\t\t\t#Enthalpy at point 3 in kJ/kg\n", + "RH4 = 0.95;\t\t\t#Re-Heat factor at point 4\n", + "h4 = 32.2;\t\t\t#Enthalpy at point 4 in kJ/kg\n", + "RH5 = 0.81;\t\t\t#Re-Heat factor at point 5\n", + "h5 = 36.8;\t\t\t#Enthalpy at point 5 in kJ/kg\n", + "RH6 = 0.54;\t\t\t#Re-Heat factor at point 6\n", + "h6 = 43.1;\t\t\t#Enthalpy at point 5 in kJ/kg\n", + "td6 = 22.;\t\t\t#Temperature at point 6 in oC\n", + "\n", + "# Calculations\n", + "td3 = ((td2-td1)/5)+td1;\t\t\t#Temperature at point 3 from Psychrometric chart shown in Page 348 in oC\n", + "td4 = (BPF*(td3-tADP))+tADP;\t\t\t#Temperature at point 4 from Psychrometric chart shown in Page 348 in oC\n", + "td5 = td4+((td1-td4)/5);\t\t\t#Temperature at point 5 from Psychrometric chart shown in Page 348 in oC\n", + "RSHF = RSH/(RSH+RLH);\t\t\t#Room Sensible Heat Factor\n", + "QR = h1-h6;\t\t\t#Total heat removed in kJ/kg\n", + "S = (RSH+RLH)/QR;\t\t\t#Supply air quantity in kg/s\n", + "R = (S*(h6-h5))/3.5;\t\t\t#Refrigeration load due to reheat in ton\n", + "D = (S*4)/5;\t\t\t#Dehumidified air quantity in kg/s\n", + "T = (D*(h3-h4))/3.5;\t\t\t#Total refrigerating capacity in ton\n", + "Q = (D/5)/1.2;\t\t\t#Quantity of fresh air supplied in (m**3)/s\n", + "\n", + "# Results\n", + "print 'Supply air condition to the room is %3.2f kg/s \\\n", + "\\nRefrigeration load due to reheat is %3.2f ton \\\n", + "\\nTotal refrigerating capacity is %3.2f ton \\\n", + "\\nQuantity of fresh air supplied is %3.3f m**3/s'%(S,R,T,Q)\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Thermal_Engineering_by_A._V._Arasu/screenshots/1.png b/Thermal_Engineering_by_A._V._Arasu/screenshots/1.png Binary files differnew file mode 100644 index 00000000..978ee35f --- /dev/null +++ b/Thermal_Engineering_by_A._V._Arasu/screenshots/1.png diff --git 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