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diff --git a/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch1.ipynb b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch1.ipynb new file mode 100755 index 00000000..04e7e8da --- /dev/null +++ b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch1.ipynb @@ -0,0 +1,212 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:13cb37b7bedc939f2932970f808cc63def43a3ee0a76b1e4d1958377aebef52c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 1 : Introduction" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.1 Page No : 8" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Given \n", + "m = 100 #m is the mass of the object in kg\n", + "a = 10 #a is the acceeleration due to gravity in m/s**2\n", + "\n", + "#To determine the force exerted\n", + "F = m*a #F is the force exerted by the object in kg\n", + "print \"Force exerted by the object= \",\n", + "print \"%.6f\" %F,\n", + "print \"N\"\n", + "F = (1/9.8065)*m*a;#F is the force exerted by the object in kgf\n", + "print \"Force exerted by the object= \",\n", + "print \"%.6f\" %F,\n", + "print \"N\"\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Force exerted by the object= 1000.000000 N\n", + "Force exerted by the object= 101.973181 N\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.4 Page No : 13" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Given\n", + "h = 100 #h is the height of the water fall in m\n", + "n = .855 #n is the efficiency of the turbine\n", + "g = 9.8 #g is the acceleration due to gravity in m/(s**2)\n", + "E = 100*10*3600;#E is the potential enery of water available to the bulb for 10 hours in watt or J/s\n", + "\n", + "#To determine the mass of water required\n", + "m = (E/(g*h*n)) #m is the mass of water required for lighting the bulb for 10 hours in Kg\n", + "print \"Mass of water required for lighting the bulb for 10 hours in Kg= \",\n", + "print \"%.6f\"%m,\n", + "print \"Kg\"\n", + "print \"Mass of water required for lighting the bulb for 10 hours in tonnes= \",\n", + "print \"%.6f\"%(m/907.2),\n", + "print \"Kg\"\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mass of water required for lighting the bulb for 10 hours in Kg= 4296.455424 Kg\n", + "Mass of water required for lighting the bulb for 10 hours in tonnes= 4.735952 Kg\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.5 Page No : 17" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "n = 1. #n is the Kg mole of an ideal gas\n", + "P = 700.*(10**4) #P is the pressure of the system in N/(m**2)\n", + "W = 45. #W is the weight of the mass in Kg\n", + "M = 20. #M is the weight of the piston and piston rod together in Kg\n", + "T = 300. #T is the consmath.tant temperature of the bath in K\n", + "h = .4 #h is the height difference of the piston after expansion in m\n", + "\n", + "#To calculate the work obtained\n", + "a = (10**-4) #a is the cross sectional area of the cylinder in m**2\n", + "V = h*a #V is the volume changed as gas expands in m**3\n", + "\n", + "#(i). If gas alone is the system\n", + "#1Kgf = 9.8065Nm\n", + "P1 = ((W+M)*9.8065)/(10**-4) #P1 is the resisting pressure when the gas confined in the cylinder taken as a system\n", + "W1 = P1*V #W1 is the work done if the gas confined in the cylinder us taken as system\n", + "print \"Work done by the system if the gas confined in the cylinder is taken as a system is \",\n", + "print \"%.6f \"%W1,\n", + "print \"Nm\"\n", + "\n", + "#(ii). If gas + piston + piston rod is a system\n", + "P2 = ((W*9.8065)/(10**-4)) #P2 is the resisting pressure when the gas plus piston plus piston rod is taken as a system\n", + "W2 = P2*V #W2 is the Work done by the system if the gas plus piston plus piston rod is taken as a system\n", + "print \"Work done by the system if the gas plus piston plus piston rod is taken as system is \",\n", + "print \"%.6f\"%W2,\n", + "print \"Nm\"\n", + "\n", + "#(iii). If gas + piston + piston rod +weight is system\n", + "P3 = 0 #P3 is the resisting pressure when the gas plus piston plus piston rod plus weight is taken as a system\n", + "W3 = P3*V #W3 is the work done by the system if the gas plus piston plus piston rod plus weight is taken as a system\n", + "print \"Work done by the system if the gas plus piston plus piston rod plus weight is taken as a system is \",\n", + "print \"%.4f\"%W3,\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Work done by the system if the gas confined in the cylinder is taken as a system is 254.969000 Nm\n", + "Work done by the system if the gas plus piston plus piston rod is taken as system is 176.517000 Nm\n", + "Work done by the system if the gas plus piston plus piston rod plus weight is taken as a system is 0.0000\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.6 Page No : 19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "n = 1 #n is the Kg mole of ideal gas.\n", + "P1 = 700*(10**4) #P1 is the initial pressure of the system in N/(m**2)\n", + "P2 = 638*(10**4) #P2 is the final pressure of the system in N/(m**2)\n", + "T = 300 #T is temperature of the system in K\n", + "R = 8314.4 #R is gas consmath.tant in Nm/Kgmole deg K\n", + "\n", + "#To calculate the work done\n", + "W = n*R*T*math.log(P1/float(P2)) #W is the work done by the system in Nm\n", + "print \"Work done by the system is \",\n", + "print \"%.2e\"%W,\n", + "print \"Nm\"\n", + "#end" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Work done by the system is 2.31e+05 Nm\n" + ] + } + ], + "prompt_number": 7 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch10.ipynb b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch10.ipynb new file mode 100755 index 00000000..4e2aff57 --- /dev/null +++ b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch10.ipynb @@ -0,0 +1,254 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a053f39727ace41fe9dd23fd039f18e2237a3b4848b2ddbeb10075aa5411107c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 10 : Compressor" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.1 Page No : 168" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "V1 = 2.7;#flow rate of CO2 in cubic meter/min\n", + "T1 = 273-51;#temperature in K\n", + "P1 = 1.0;#initial pressure in Kgf/sq cm\n", + "P2 = 10.0;#final pressure in Kgf/sq cm\n", + "y = 1.3;#gamma\n", + "v1 = 0.41;#specific volume in cubic meter/Kg\n", + "H1 = 158.7;# initial enthalpy in Kcal/Kg\n", + "H2 = 188.7;#final enthalpy in Kcal/Kg\n", + "\n", + "#process is isentropic\n", + "#To calculate the horsepower required\n", + "\n", + "#(i)Assuming ideal gas behaviour\n", + "#From equation 10.3 (page no 189)\n", + "W = (y/(y-1))*(P1*1.03*10**4*V1)*(1-(P2/P1)**((y-1)/y));#work in m Kgf/min\n", + "W1 = W/4500.0;\n", + "print \"i)The horsepower required is %f hp\"%(W1);\n", + "\n", + "#(ii)Umath.sing the given data for CO2\n", + "#From equation 10.2 (page no 189)\n", + "W = -(H2 - H1);#work in Kcal/Kg\n", + "M = V1/v1;#Mass rate of gas in Kg/min\n", + "W1 = W*M*(427/4500.0);\n", + "print \" ii)Compressor work is %f hp\"%(W1);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)The horsepower required is -18.779590 hp\n", + " ii)Compressor work is -18.746341 hp\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.2 Page No : 171" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "P1 = 1.0;#Initial pressure in atm\n", + "P2 = 29.0;#Final pressure in atm\n", + "C = 0.05;#Clearance\n", + "y = 1.4;#gamma of air\n", + "\n", + "#To calculate the volumetric efficiency and the maximum possible pressure that can be attained in a math.single stage\n", + "#(i)Calulation of volumetric efficiency\n", + "#From equation 10.11 (page no 194)\n", + "V_E = 1+C-C*(P2/P1)**(1/y);\n", + "print \"i)Volumetric efficiency is %f percent\"%(V_E*100);\n", + "\n", + "#(ii)Calculation of maximum pressure \n", + "V_E = 0;#Minimum efficiency\n", + "P2 = P1*(((1+C-V_E)/C)**y);\n", + "print \" ii)The maximum possible pressure attained is %f atm\"%(P2);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)Volumetric efficiency is 49.596143 percent\n", + " ii)The maximum possible pressure attained is 70.975263 atm\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.3 Page No : 174" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "V_d = 5.15;#print lacement volume in cubic meter/min\n", + "P1 = 1.0;#initial pressure in Kgf/sq cm\n", + "P2 = 8.5;#final pressure in Kgf/sq cm\n", + "C = 0.06;#Clearance\n", + "M_E = 0.8;#Mechenical efficiency\n", + "y = 1.31;#gamma\n", + "\n", + "#To calculate the capacity and the actual horse power of the compressor\n", + "v1 = V_d*(1+C-(C*((P2/P1)**(1/y))));\n", + "print \"The capacity of the copressor is %f cubic meter/min\"%(v1);\n", + "#From equation 10.6 (page no 192)\n", + "W = (y/(y-1))*(P1*1*10**4*v1)*(1-(P2/P1)**((y-1)/y));#work in Kgf/min\n", + "W1 = W/4500.0;#work in hp\n", + "W2 = W1/M_E;\n", + "print \" The actual horse power of the compressor is %f hp\"%(W2);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The capacity of the copressor is 3.876154 cubic meter/min\n", + " The actual horse power of the compressor is -30.000346 hp\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.4 Page No : 177" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "P1 = 1.0;#Initial pressure in Kgf/sq cm\n", + "Pn = 13.0;#Final pressure in Kgf/sq cm\n", + "V1 =27.0;#flow rate of gas in cubic meter/min\n", + "y = 1.6;#gamma of the gas\n", + "n = [1.0,2.0,3.0,4.0,7.0,10.0];#number of stages\n", + "print \"No of stages Horse power in hp\";\n", + "#To Calculate the theoretical horse power required\n", + "W = []\n", + "for i in range(0,6):\n", + " W.append(n[i]*(y/(y-1))*((P1*10**4)/4500)*V1*(1-(Pn/P1)**((y-1)/(n[i]*y))));\n", + " print \" %d\"%(n[i]),\n", + " print \" %f\"%(-1*W[i])\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "No of stages Horse power in hp\n", + " 1 258.647729\n", + " 2 197.623943\n", + " 3 181.430407\n", + " 4 173.977056\n", + " 7 164.971690\n", + " 10 161.541416\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.5 Page No : 180" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "P1 = 1.0;#Initial pressure in Kgf/sq cm\n", + "P4 = 200.0;#Final pressure in Kgf/sq cm\n", + "n = 4.0;#no of stages\n", + "\n", + "#To find out the presure between stages\n", + "r = (P4/P1)**(1/n);#Compression ratio\n", + "P2 = r*P1;\n", + "print \"The pressure after 1st stage is %f Kgf/sq cm\"%(P2);\n", + "P3 = r*P2;\n", + "print \" The pressure after 2nd stage is %f Kgf/sq cm\"%(P3);\n", + "P4 = r*P3;\n", + "print \" The pressure after 3rd stage is %f Kgf/sq cm\"%(P4);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The pressure after 1st stage is 3.760603 Kgf/sq cm\n", + " The pressure after 2nd stage is 14.142136 Kgf/sq cm\n", + " The pressure after 3rd stage is 53.182959 Kgf/sq cm\n" + ] + } + ], + "prompt_number": 6 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch11.ipynb b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch11.ipynb new file mode 100755 index 00000000..e5accd29 --- /dev/null +++ b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch11.ipynb @@ -0,0 +1,160 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:311c74298b5b14a54f4d1278f9c15d5918adb65c1f1cd4717c56b8ba91310447" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 11 : Liquefaction of Gases" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.1 Page No : 195" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Given\n", + "P1 = 8.74;#Initial pressure in Kgf/sq cm\n", + "P2 = 2.41;#Final pressure in Kgf/sq cm\n", + "H1 = 327.13;#Enthalpy of inlet stream in Kcal/Kg\n", + "Hl = 26.8;#Enthalpy of liquid at the final condition in Kcal/Kg\n", + "H2 = H1#Enthalpy of exit stream in Kcal/Kg ,math.since throttling is isenthalpic\n", + "Hg = 340.3;#Enthalpy of gas at the final condition in Kcal/Kg\n", + "vl = 152*10**-5;#Specific volume of liquid at the final condition in cubic meter/Kg\n", + "vg = 0.509;#Specific volume of gas at the final condition in cubic meter/Kg\n", + "v1 = 0.1494;#Initial specific volume in cubic meter/Kg\n", + "\n", + "#To Calculate the dryness fraction of exit stream and the ratio of upstream to downstream diameters\n", + "#(i)Calculation of the dryness fraction of exit stream\n", + "#From equation 3.13(a) (page no 82)\n", + "x = (H2- Hl)/(Hg-Hl);\n", + "print \"i)The dryness fraction of the exit stream is %f\"%(x);\n", + "\n", + "#(ii)Calculation of the ratio of upstream to downstream pipe diameters\n", + "#From equation 3.13(b) (page no 82)\n", + "v2 = (vl*(1-x))+(x*vg);#Total specific volume at the final condition in cubic meter/Kg\n", + "#u1 = u2; math.since KE changes are negligible\n", + "#From continuity equation: A2/A1 = D2**2/D1**2 = v2/v1 ; let required ratio,r = D2/D1;\n", + "r = (v2/v1)**(1/2);\n", + "print \" ii)The ratio of upstream to downstream diameters is %f\"%(r);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)The dryness fraction of the exit stream is 0.957990\n", + " ii)The ratio of upstream to downstream diameters is 1.000000\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.2 Page No : 199" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "P1 = 1000*1.033*10**4;#Initial pressure in Kgf/sq m\n", + "P2 = 1*1.033*10**4;#Final pressure in Kgf/sq m\n", + "T1 = 300.0;#Inital temperature in K\n", + "Cp = 7.0;#Specific heat of the gas in Kcal/Kgmole K\n", + "#Gas obeys the relation: v = (R*T)/P+(b*(T**2))\n", + "b = 5.4392*10**-8;#in cubic meter/Kgmole K**2\n", + "\n", + "#To Calculate the temperature of the throttled gas\n", + "#From equation (a) (page no 212);which we got after integration \n", + "T2 = 1/((1/T1)-((b/Cp)*((P2-P1)/427)));\n", + "print \"The throttled gas is cooled to %f K\"%(T2);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The throttled gas is cooled to 284.000191 K\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.3 Page No : 203" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "#From the figure 11.8 (page no 216) & from figure A.2.7\n", + "H3 = 0.0;\n", + "H7 = -47.0;#in Kcal/Kg\n", + "H6 = -93.0;#in Kcal/Kg\n", + "H8 = 7.0;#in Kcal/Kg\n", + "\n", + "#To Calculate the fraction of air liquified at steady state and temperature of air before throttling\n", + "#(i)Calculation of fraction of air liquified\n", + "#From equation 11.3 (page no 215)\n", + "x = (H8-H3)/(H8-H6);\n", + "print \"The fraction of air liquified is %f\"%(x);\n", + "\n", + "#(ii)Calculation of temperature \n", + "H4 = H3+(H7*(1-x))-(H8*(1-x));#enthalpy of the gas before throttling\n", + "#From figure A.2.7 temperature corresponds to pressure 160 atm and the enthalpy H4 is\n", + "T = -112;\n", + "print \" The temperature of air before throttling is %d deg celsius\"%(T);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The fraction of air liquified is 0.070000\n", + " The temperature of air before throttling is -112 deg celsius\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch12.ipynb b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch12.ipynb new file mode 100755 index 00000000..b466521e --- /dev/null +++ b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch12.ipynb @@ -0,0 +1,218 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:abd9968b31b6fc770d85971aacc78f09c6ba4e5596d9831c5736d23440171297" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 12 : Refrigeration" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.1 Page No : 216" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "m = 500.0;#mass of ice produced per hour in Kg\n", + "T1 = 15.0;#Initial temperature of water\n", + "T2 = -5.0;#Final temperature of ice\n", + "Ci = 0.5;#Specific heat of ice in Kcal/Kg deg celsius\n", + "Cw = 1.0;#Specific heat of water in Kcal/Kg deg celsius\n", + "L_f = 79.71;#Latent heat of fusion in Kcal/Kg\n", + "Tf = 0.0;#Frezzing point of ice in deg celsius\n", + "\n", + "#To Calculate the theoretical horse power required\n", + "Q2 = m*(Cw*(T1-Tf)+L_f+Ci*(Tf-T2));#Heat to be extracted per hour in Kcal\n", + "#From equation 12.1 (page no 220)\n", + "COP = (T2+273)/((T1+273)-(T2+273));\n", + "W = Q2/COP;#Work in Kcal/hr\n", + "W1 = W*(427/(60*4500.0));\n", + "print \"The therotical horse power required is %f hp\"%(W1);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The therotical horse power required is 5.736411 hp\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.2 Page No : 217" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from scipy.integrate import quad\n", + "\n", + "#Given\n", + "Ta = 298.0;#Initial temperature in K\n", + "Tb = 203.0;#Final temperature in k\n", + "T1 = 298.0;#Water temperature in K\n", + "n = 1.0;#Kgmole of CO2\n", + "#Cp = 5.89+0.0112T ; Specific heat of CO2 in Kcal/Kgmole K\n", + "\n", + "#To Calculate the compressor load\n", + "#From equation 12.2a and b (page no 221)\n", + "def f(T):\n", + " y = ((T1-T)/T)*n*(5.89+0.0112*T);\n", + " return y\n", + "W = quad(f,Ta,Tb)[0];\n", + "print \"The compressor load is %f Kcal/Kgmole\"%(W);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The compressor load is -164.797031 Kcal/Kgmole\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.3 Page No : 221" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Given\n", + "#Consider the figure 12.4 (page no 226) \n", + "m = 5.0;#tonnes of refrigeration\n", + "T1 = 273-10.0;#temperature of the saturated vapour in K\n", + "T2 = 273+35.0;#temperature of the super heated vapour in K\n", + "T3 = 273+25.0;#temperature of the saturated liquid in K\n", + "T4 = 273+25.0;#temperature of the wet vapour in K\n", + "H1 = 341.8;#enthalpy of the saturated vapour in Kcal/Kg\n", + "H2 = 409.0;#enthalpy of the super heated vapour in Kcal/Kg\n", + "H3 = 350.0;#enthalpy of the saturated liquid in Kcal/Kg\n", + "H4 = 71.3;#enthalpy of the wet vapour in Kcal/Kg\n", + "\n", + "#To Calculate the C.O.P, mass of refrigerant required, compressor horse power required and the C.O.P & compressor horse power for a reversed Carnot cycle\n", + "#(i)Calculation of the C.O.P of the compression cycle\n", + "#From equation 12.6 (page no 226)\n", + "COP = (H1-H4)/(H2-H1);\n", + "print \"i)C.O.P of the compression cycle is %f\"%(COP);\n", + "\n", + "#(ii)Calculation of mass of refrigerant required\n", + "#From equation 12.7 (page no 226)\n", + "M = (m*50.4)/(H1-H4);\n", + "print \" ii)The mass of refrigerant required is %f Kg/mt\"%(M);\n", + "\n", + "#(iii)Calculation of the compressor horse power\n", + "#From equation 12.5 (page no 226)\n", + "C_hp = (H2-H1)*M*(427/4500.0);\n", + "print \" iii)The compressor horse power is %f hp\"%(C_hp);\n", + "\n", + "#(iv)Calculation for reversed Carnot cycle\n", + "#From equation 12.1 (page no 220)\n", + "COP = T1/(T3-T1);\n", + "print \" iv)C.O.P for the reversed Carnot cycle is %f\"%(COP);\n", + "Q2 = m*50.4*(427/4500.0);#in hp\n", + "C_hp = Q2/COP\n", + "print \" Compressor horse power for the reversed Carnot cycle is %f hp\"%(C_hp);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)C.O.P of the compression cycle is 4.025298\n", + " ii)The mass of refrigerant required is 0.931608 Kg/mt\n", + " iii)The compressor horse power is 5.940430 hp\n", + " iv)C.O.P for the reversed Carnot cycle is 7.514286\n", + " Compressor horse power for the reversed Carnot cycle is 3.182205 hp\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.4 Page No : 225" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "#Water at 20 deg cel is chilled to 10 deg cel by flash evaporation\n", + "Pv = 0.012;#Vapour pressure of water at 10 deg celsius in Kgf/sq.cm\n", + "H1 = 20.03;#Enthalpy of liquid water at 20 deg cel in Kcal/Kg\n", + "H2 = 10.4;#Enthalpy of liquid water at 10 deg cel in Kcal/Kg\n", + "Hv = 601.6;#Enthalpy of saturated vapour at 10 deg cel in Kcal/kg\n", + "\n", + "#To calculate the pressure in the math.tank and the amount of make up water required\n", + "P = Pv;#pressure in the math.tank = vapour pressure of water\n", + "print \"The pressure in the math.tank is %f Kgf/sq.cm\"%(P);\n", + "#From equation 12.8 (page no 234)\n", + "x = (H1-H2)/(Hv-H2);\n", + "print \" The amount of make up water required is %f Kg\"%(x);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The pressure in the math.tank is 0.012000 Kgf/sq.cm\n", + " The amount of make up water required is 0.016289 Kg\n" + ] + } + ], + "prompt_number": 5 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch13.ipynb b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch13.ipynb new file mode 100755 index 00000000..2d92c849 --- /dev/null +++ b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch13.ipynb @@ -0,0 +1,956 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:98e5c2c383a1a1550ca177061033a45bcfa7bc464bb9def6e21c0e4e3bbd4374" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 13 : Thermodynamics in Phase Equilibria" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.1 Page No : 238" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "#N2 obeys the relation : Z = 1+(2.11*10**-4*P)\n", + "Tc = 126;#Critical temperature in K\n", + "Pc = 33.5;#Critical pressure in atm\n", + "T = 373;#in K\n", + "P = 100;#in atm\n", + "\n", + "#To Calculate the fugacity of N2 at 373K and 100 atm\n", + "#(i)Umath.sing the Z relation given above\n", + "#From equation 13.12 (page no 239)\n", + "phi = math.e**(2.11*10**-4*(P-0));#fugacity coefficient\n", + "f = phi*P;\n", + "print \"i)The fugacity of N2 umath.sing the given Z relation is %f atm\"%(f);\n", + "\n", + "#(ii)Umath.sing the fugacity chart given in figure A.2.9\n", + "Pr = P/Pc;#Reduced pressure in atm\n", + "Tr = T/Tc;#Reduced temperature in K\n", + "#From figure A.2.9,\n", + "phi = 1.04\n", + "f = phi*P;\n", + "print \" ii)The fugacity of N2 umath.sing the fugacity chart is %f atm\"%(f);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)The fugacity of N2 umath.sing the given Z relation is 102.132418 atm\n", + " ii)The fugacity of N2 umath.sing the fugacity chart is 104.000000 atm\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.2 Page No : 241" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "P1 = 50*1.03*10**4;#Initial pressure in Kgf/sq m\n", + "T = 373.0;#Temperature in K\n", + "P2 = 1.03*10**4;#Final pressure in Kgf/sq m\n", + "V = 0.001*18;#Volume in cubic meter\n", + "R = 848.0;#gas consmath.tant in m Kgf/Kgmole K\n", + "\n", + "#To Calculate the fugacity of liquid water\n", + "#From equation 13.13(page no 240)\n", + "del_u = (V/(R*T))*(P2-P1);#del_u = ln(f2/f1); Change in chemical potential\n", + "f1 = P2;#in Kgf/sq m\n", + "f2 = f1*(math.e**del_u);\n", + "print \"The fugacity of the liquid water at 50 atm is %4.2e Kgf/sq m\"%(f2);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The fugacity of the liquid water at 50 atm is 1.00e+04 Kgf/sq m\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.3 Page No : 242" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "import math\n", + "import matplotlib.pyplot as plt\n", + "import numpy\n", + "\n", + "#Given\n", + "x1 = 0.1;#mole fraction of methane\n", + "x2 = 0.9;#mole fraction of propane\n", + "P = [28.1,31.6,35.1];#Pressure in Kgf/sq cm are\n", + "K1 = [5.8,5.10,4.36];#Vapourisation consmath.tants of methane at the corresponding presssures\n", + "K2 = [0.61,0.58,0.56];#Vapourisation consmath.tants of propane at the correspondig pressures\n", + "\n", + "#To Calculate the bubble point pressure of the solution\n", + "#From equation 13.27 (page no 245)\n", + "y1 = []\n", + "y2 = []\n", + "y = []\n", + "for i in range(0,3):\n", + " y1.append(K1[i]*x1);#mole fraction of methane in the vapour phase\n", + " y2.append(K2[i]*x2);#mole fraction of propane in the vapour phase\n", + " y.append(y1[i]+y2[i]);#sum of the mole fraction in the vapour phase\n", + " \n", + "\n", + "plt.plot(P,y)\n", + "plt.title(\"y vs pressure\")\n", + "plt.xlabel(\"P\")\n", + "plt.ylabel(\"y\")\n", + "plt.show()\n", + "P1 = numpy.interp(1,y,P)\n", + "print \"The bubble point pressure of the solution is %f Kgf/sq cm\"%(P1);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "metadata": {}, + "output_type": "display_data", + "png": 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+ "text": [ + "<matplotlib.figure.Figure at 0x10643f210>" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The bubble point pressure of the solution is 35.100000 Kgf/sq cm\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.4 Page No : 244" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "T = [80.6,79.0,77.3,61.4];#Various temperature in deg cel\n", + "x1 = [0.0,15.0,29.0,100.0];#mole fraction of CHCl3 in liquid phase\n", + "y1 = [0.0,20.0,40.0,100.0];#mole fraction of CHCl3 in vapour phase\n", + "P1 = [1370,1310,1230,700];#Vapour pressure of CHCl3 in mm Hg\n", + "P = 760.0;#Total pressure in mm Hg\n", + "\n", + "#To Calculate the equilibrium data i.e y/x and compare with the experimental values\n", + "#From equation 13.27 (page no 245);K = y1/x1 = Pi/P\n", + "print \"Temperature Experimental Calculated\";\n", + "Ke_x = []\n", + "K_c = []\n", + "for i in range(0,4):\n", + " print \" %f\"%(T[i]),\n", + " if x1[i] == 0 :\n", + " print \" Not defined\",\n", + " Ke_x.append(0)\n", + " else:\n", + " Ke_x.append(y1[i]/x1[i]);\n", + " print \" %f\"%(Ke_x[i]),\n", + " K_c.append(P1[i]/P);\n", + " print \" %f\"%(K_c[i])\n", + "\n", + "if Ke_x[i] == K_c[i]:\n", + " print ' The liquid solution is perfect';\n", + "else:\n", + " print \" The liquid solution is imperfect\";\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temperature Experimental Calculated\n", + " 80.600000 Not defined 1.802632\n", + " 79.000000 1.333333 1.723684\n", + " 77.300000 1.379310 1.618421\n", + " 61.400000 1.000000 0.921053\n", + " The liquid solution is imperfect\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.6 Page No : 246" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "x1 = 0.1;#Mole fraction of dichloromethane (CCl2H2)\n", + "x2 = 0.9;#Mole fraction of methyl acetate (C3H6O2)\n", + "M1 = 85.0;#Molecular weight of CCl2H2\n", + "M2 = 74.0;#Molecular weight of C3H602\n", + "D1 = 1.3163;#Density of CCl2H2 in gm/cc\n", + "D2 = 0.9279;#Density of C3H6O2 in gm/cc\n", + "\n", + "#To Calculate the volume of 10% dichloromethane solution\n", + "V1 = M1/D1;#Specific volume of pure CCL2H2 in cc/gmole\n", + "V2 = M2/D2;#Specific volume of C3H6O2 in cc/gmole\n", + "#From equation 13.62(page no 256)& 13.78 (page no 257)\n", + "V_e = x1*x2*(1.2672-0.771*x1);#excess volume in cc/gmole\n", + "V = V1*x1+V2*x2+V_e;\n", + "print \"The volume of 10 percent dichloromethane is %f cc/gmole\"%(V);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The volume of 10 percent dichloromethane is 78.339579 cc/gmole\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.7 Page No : 249" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "import math\n", + "import matplotlib.pyplot as plt\n", + "import numpy\n", + "\n", + "#Given\n", + "x_T = 0.957;#mole fraction of Toluene\n", + "x_D = 0.043;#mole fraction of 1,2-dichloroethane\n", + "t = [90, 100, 110];#temperature in deg cel\n", + "R = 1.98;#gas consmath.tant in Kcal/Kgmole K\n", + "\n", + "#To Calculate the vapour pressure of the solution, bubble point at 686 mm Hg and the vapour composition at equilibrium,\n", + "#compare the experimental value of 91.2% toluene in vapour with the calculated value & calculate the free energy of mixing\n", + "#(1)Calculation of vapour pressure\n", + "print \"1Tempdeg cel P_TmmHg P_DmmHg P_smmHg\";\n", + "P_T = []\n", + "P_D = []\n", + "P_s = []\n", + "for i in range(0,3):\n", + " P_T.append(10**(6.95464-(1344.8/(219.482+t[i]))));#Given as equation(a)(page no 260)\n", + " P_D.append(10**(7.03993-(1274.079/(223+t[i]))));#Given as equation(b)(page no 260)\n", + " P_s.append(x_T*P_T[i]+x_D*P_D[i]);#pressure of the solution in mm Hg\n", + " print ' %f'%(t[i]),\n", + " print ' %f'%(P_T[i]),\n", + " print ' %f'%(P_D[i]),\n", + " print ' %f'%(P_s[i])\n", + "\n", + "#(2)Calculation of bubble point and comparison of values\n", + "plt.plot(t,P_s)\n", + "plt.title(\"t vs P_s\")\n", + "plt.xlabel(\"t\")\n", + "plt.ylabel(\"P_s\")\n", + "plt.show()\n", + "T = numpy.interp(686,P_s,t)\n", + "P = 686.0;#pressure of solution in mm Hg\n", + "y_T_e = 0.912;#experimental value of mole fraction of toluene\n", + "#From the graph we found that the temperature at P = 686 mm Hg is\n", + "#t = 105.3;#in deg cel\n", + "print '2)The bubble point is %f deg cel'%(T);\n", + "#From equation (a)(page no 260)\n", + "P_T = 10**(6.95464-(1344.8/(219.482+T)));#vapour pressure of Toluene in mmHg\n", + "#From equation 13.27 (page no 245)\n", + "y_T_c = (x_T*P_T)/P;\n", + "y_D_c = 1-y_T_c;\n", + "print ' The vapour composition of toluene is %f'%(y_T_c);\n", + "print ' The vapour composition of 1,(2-dichloroethane is %f'%(y_D_c);\n", + "e = ((y_T_e-y_T_c)/y_T_e)*100;\n", + "print ' The percentage error is %f percent'%(e);\n", + "\n", + "#(3)Calculation of free energy\n", + "del_F = R*(T+273)*((x_T*math.log(x_T))+(x_D*math.log(x_D)));\n", + "print '3)The free energy of mixing is %f Kcal/Kgmole'%(del_F);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1Tempdeg cel P_TmmHg P_DmmHg P_smmHg\n", + " 90.000000 406.737847 931.944531 429.321734\n", + " 100.000000 556.321921 1245.698588 585.965118\n", + " 110.000000 746.589152 1636.315096 784.847367\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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+ "text": [ + "<matplotlib.figure.Figure at 0x1057dc4d0>" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "2)The bubble point is 105.029855 deg cel\n", + " The vapour composition of toluene is 0.901898\n", + " The vapour composition of 1,(2-dichloroethane is 0.098102\n", + " The percentage error is 1.107696 percent\n", + "3)The free energy of mixing is -132.756668 Kcal/Kgmole\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.8 Page No : 252" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "#Consider the diagram shown in page no 263\n", + "w1 = 100.0;#weight of LiBr entered as feed in the evaporator per hour in Kg\n", + "x1 = 0.45;#weight fraction of LiBr entered as feed\n", + "x2 = 0.0;#weight fraction of steam in the LiBr soln\n", + "x3 = 0.65;#weight fraction of LiBr formed as product\n", + "H1 = -39.0;#Enthalpy of 45% solution at 25 deg cel in Kcal/Kg\n", + "H3 = -4.15;#Enthalpy of 65% solution at 114.4 deg cel in Kcal/Kg\n", + "H2 = 649.0;#Enthalpy of superheated steam at 100 mmHg and 114.4 deg cel in Kcal/Kg\n", + "\n", + "#To Calculate the heating load required for the process\n", + "#According to material balance\n", + "w3 = (w1*x1)/x3;#weight of LiBr solution formed after evaporation per hour in Kg\n", + "w2 = w1-w3;# weight of steam formed in Kg/hr\n", + "#According to energy balance\n", + "Q = (w2*H2)+(w3*H3)-(w1*H1);\n", + "print 'The heat that has to be supplied for this concentration process is %f Kcal/hr'%(Q);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The heat that has to be supplied for this concentration process is 23581.923077 Kcal/hr\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.12 Page No : 253" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "import math\n", + "import matplotlib.pyplot as plt\n", + "import numpy\n", + "\n", + "#Given\n", + "x_A = [0.0, 0.0435, 0.0942, 0.1711, 0.2403, 0.3380, 0.5981];#mole fraction of acetic acid\n", + "p_A = [0.0, 17.2, 30.5, 46.5, 57.8, 69.3, 95.7];#partial pressure of acetic acid in mmHg\n", + "P_T1 = 202.0;#vapour pressure of toulene in mmHg\n", + "P_T2_ex = 167.3;#experimental partial pressure in mmmHg\n", + "\n", + "#To Calculate the partial pressure of toulene in the solution and check with the experimental value\n", + "#From the equation 13.95,\n", + "#ln(P_T2/P_T1) = -intg(x_A/((1-x_A)*p_A))\n", + "x = []\n", + "for i in range(0,7):\n", + " if (p_A[i] != 0):\n", + " x.append((x_A[i]/((1-x_A[i])*p_A[i]))*10**4)\n", + " else:\n", + " x.append(0)\n", + "\n", + "\n", + "plt.plot(x,p_A)\n", + "plt.title(\" \")\n", + "plt.xlabel(\"(x_A/((1-x_A)*p_A))*10**4\")\n", + "plt.ylabel(\"p_A\")\n", + "plt.show()\n", + " \n", + "#Area of the graph drawn is\n", + "A = -0.138;\n", + "P_T2 = (math.e**A)*P_T1;\n", + "e = ((P_T2-P_T2_ex)*100)/P_T2_ex;\n", + "print 'The partial pressure of toulene is %f mmHg'%(P_T2);\n", + "print ' This deviates %i percent from the reported value'%(e);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "metadata": {}, + "output_type": "display_data", + "png": 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+ "text": [ + "<matplotlib.figure.Figure at 0x10654c750>" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The partial pressure of toulene is 175.961936 mmHg\n", + " This deviates 5 percent from the reported value\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.13 Page No : 254" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "\n", + "import math\n", + "import matplotlib.pyplot as plt\n", + "import numpy\n", + "from numpy.linalg import solve\n", + "\n", + "#Given\n", + "P = 760.0;#pressure at maximum boiling azeotrope of A and B in mmHg\n", + "x_A = 0.6;#mole fraction of A in liquid phase\n", + "x_B = 0.4;#mole fraction of B in liquid phase\n", + "p_A = 600.0;#vapour pressure of A at 90 deg cel\n", + "p_B = 300.0;#vapour pressure of B at 90 deg cel\n", + "\n", + "#To Check whether the activity coefficient of the solution can be represented by the Margules equation\n", + "y_A = P/p_A;#Activity coefficient of A\n", + "y_B = P/p_B;#Activity coefficient of B\n", + "#From the Margules equation or equation (a) & (b)\n", + "U = [[((x_B**2)-(2*(x_B**2)*x_A)), (2*(x_B**2)*x_A)], [(2*(x_A**2)*x_B), ((x_A**2)-(2*(x_A**2)*x_B))]];\n", + "V = [math.log(y_A), math.log(y_B)];\n", + "W = solve(U,V);\n", + "#Now the value of consmath.tants A and B in equations(a)&(b) are given as\n", + "\n", + "A = W[0];\n", + "B = W[1];\n", + "#let us assume \n", + "x_A = [0.0,0.2,0.4,0.6,0.8,1.0];\n", + "x_B = [1.0,0.8,0.6,0.4,0.2,0.0];\n", + "#C = lny_A; D = lny_B; E = ln(y_A/y_B)\n", + "C = []\n", + "D = []\n", + "E = []\n", + "for i in range(6):\n", + " C.append((x_B[i]**2)*(2*(B-A)*x_A[i]+A));\n", + " D.append((x_A[i]**2)*(2*(A-B)*x_B[i]+B));\n", + " E.append(C[i]-D[i]);\n", + " \n", + "plt.plot(x_A,E)\n", + "plt.title(\" \")\n", + "plt.xlabel(\"x_A\")\n", + "plt.ylabel(\"ln(y_A/y_B)\")\n", + "plt.show()\n", + "#Since the graph drawn is approximately symmetrical.Thus it satisfies the Redlich-Kister Test\n", + "print 'The actvity coefficients of the system can be represented by Margules equation';\n", + "#end\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "metadata": {}, + "output_type": "display_data", + "png": 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+ "text": [ + "<matplotlib.figure.Figure at 0x10654cdd0>" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The actvity coefficients of the system can be represented by Margules equation\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.14 Page No : 259" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "import math\n", + "import matplotlib.pyplot as plt\n", + "import numpy\n", + "\n", + "#Given\n", + "P = 760.0#Total pressure of the mixture in mmHg\n", + "T = [80, 90, 95, 100];#Temperature in deg celsius\n", + "P1 = [87.4, 129.0, 162.0, 187.0];#vapour pressure of 1,1,2,2-tetrachloroethane in mmHg\n", + "P2 = [356, 526, 648, 760];#Vapour pressure of water in mmHg\n", + "\n", + "#To Calculate the composition of the vapour evolved\n", + "plt.plot(T,P1,\"green\",T,P2,\"red\")\n", + "plt.title(\" \")\n", + "plt.xlabel(\"Temp in deg cel\")\n", + "plt.ylabel(\"Vapour pressure in mmHg\")\n", + "plt.show()\n", + "#From the graph we conclude that at 93.8 deg cel\n", + "P1 = 155.0;#in mm Hg\n", + "P2 = 605.0;#in mm Hg\n", + "y_1 = P1/P;\n", + "y_2 = P2/P;\n", + "print 'Mole fraction of 1,(1,(2,(2-tetrachloroethane in vapour is %f'%(y_1);\n", + "print ' Mole fraction of water in vapour is %f'%(y_2);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "metadata": {}, + "output_type": "display_data", + "png": 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+ "text": [ + "<matplotlib.figure.Figure at 0x106604950>" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Mole fraction of 1,(1,(2,(2-tetrachloroethane in vapour is 0.203947\n", + " Mole fraction of water in vapour is 0.796053\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.15 Page No : 263" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "import math\n", + "import matplotlib.pyplot as plt\n", + "import numpy\n", + "\n", + "#Given\n", + "T = [146.2, 142.3, 126.1, 115.9, 95.0, 98.0, 100.0];#Temperature in deg cel\n", + "P1 = [760.0, 685.0, 450.3, 313.0];#Vapour pressure of 1,1,2,2-tetrachloroethane at the coressponding temperature in mm Hg\n", + "P2_5 = 648.0;#Vapour pressure of water at 95 deg cel in mm Hg\n", + "P2_6 = 711.0;#Vapour pressure of water at 98 deg cel in mm Hg\n", + "P = 760.0;#Total pressure of mixture in mm Hg\n", + "\n", + "x1 = [0, 0, 0, 0, 0, 0, 0];\n", + "#To plot a graph between temperature and vapour phase composition\n", + "for i in range(0,4):\n", + " x1[i] = P1[i]/P;#mole fraction of 1,1,2,2-tetrachloroethane\n", + "x2_5 = P2_5/P;#mole fraction of water at 95 deg cel\n", + "x2_6 = P2_6/P;#mole fraction of water at 98 deg cel\n", + "x1[4] = 1-x2_5;\n", + "x1[5] = 1-x2_6;\n", + "\n", + "plt.plot(x1,T)\n", + "plt.title(\"\")\n", + "plt.xlabel(\"mole fraction of 1,1,2,2-tetrachloroethane\")\n", + "plt.ylabel(\"Temperature in deg cel\")\n", + "plt.show()\n", + "print 'The required graph has been ploted in the graphic window';\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "metadata": {}, + "output_type": "display_data", + "png": 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+ "text": [ + "<matplotlib.figure.Figure at 0x1064ce750>" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The required graph has been ploted in the graphic window\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.16 Page No : 266" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "#B = -(1.203*10**10)*(T**2.7); second virial coefficient, T is in K\n", + "#math.log P = 6.95464-(1344.8/(219.482+t))...(a);Vapour pressure of toulene\n", + "t = 107.2;#Temperature in deg cel\n", + "T = t+273.16;#in K\n", + "H_ex = 7964.0;#experimental value of heat of vapourisation in Kcal/Kgmole\n", + "d = 800.0;#density of liquid toulene in Kg/cubic meter\n", + "R = 1.98;#gas consmath.tant in Kcal/Kgmole K\n", + "M = 92.14;#molecular weight of toulene\n", + "\n", + "#To Calculate the heat of vapourization of toulene by umath.sing ideal gas law, second virial coefficient but neglecting vl and including vl\n", + "#From equation (a), let K = dmath.logP/dT\n", + "K = 1344.8/(219.482+t)**2;\n", + "#(i)Umath.sing ideal gas behaviour\n", + "#From equation 13.112(page no 286)\n", + "H_c = (2.303*R*(T**2))*K;\n", + "print 'i)The heat of vapourization umath.sing ideal gas behaviour is %f Kcal/Kgmole'%(H_c);\n", + "D = ((H_c-H_ex)/H_c)*100;\n", + "print ' The deviation is %f percent'%(D);\n", + "\n", + "#(ii)Umath.sing second virial coeff but neglecting vl\n", + "#From equation(a)\n", + "P = 10**(6.95464-1344.8/(219.482+t));#in mm Hg\n", + "P1 = P*1.033*10**4/760;#in Kgf/sq m\n", + "B = -((1.203*10**10)/(T**2.7))*10**-3;#in cubic meter/Kgmole\n", + "#From equation 13.111 (page no 286) neglecting vl,\n", + "l = (R*T)+((B*P1)/427);#in Kcal/Kgmole\n", + "H_c = K*2.303*T*l;\n", + "print 'ii)The heat of vapourisation umath.sing second virial coefficient but neglecting vl is %f Kcal/Kgmole'%(H_c);\n", + "D = ((H_c-H_ex)/H_c)*100;\n", + "print ' The deviation in this case is %f percent'%(D);\n", + "\n", + "#(iii)Umath.sing second virial coeff including vl\n", + "vl = M/d;#Liquid specific volume in cubic meter/Kgmole\n", + "n = P1*vl/427;#in Kcal/Kgmole\n", + "H_c = K*2.303*T*(l-n);\n", + "print 'iii)The heat of vapourisation umath.sing second virial coefficient including vl is %f Kcal/Kgmole'%(H_c);\n", + "D = ((H_c-H_ex)/H_c)*100;\n", + "print ' The deviation in this case is %f'%(D);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)The heat of vapourization umath.sing ideal gas behaviour is 8312.967730 Kcal/Kgmole\n", + " The deviation is 4.197872 percent\n", + "ii)The heat of vapourisation umath.sing second virial coefficient but neglecting vl is 7998.487742 Kcal/Kgmole\n", + " The deviation in this case is 0.431178 percent\n", + "iii)The heat of vapourisation umath.sing second virial coefficient including vl is 7970.612948 Kcal/Kgmole\n", + " The deviation in this case is 0.082967\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.17 Page No : 269" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "H_ex = 539.0;#Heat of vapoization of water in Kcal/Kg\n", + "Tc = 647.0;#Critical temperature in K\n", + "Pc = 218.0;#Critical pressure in atm\n", + "Tb = 373.0;#Boiling point of water in K\n", + "t = 100.0;#temperature in deg cel\n", + "M = 18.0;#Molecular weight of water\n", + "P = 1.0;#pressure at boiling point in atm\n", + "P1 = 1.033*10**4;#pressure in Kgf/sq m\n", + "\n", + "#To Calculate the heat of vapourisation of water by Vishwanath and Kuloor method and by Riedel's method and compare with the experimental value\n", + "#(i) Umath.sing Vishwanath and Kuloor method\n", + "H_c = (4.7*Tc*((1-(P/Pc))**0.69)*math.log(P/Pc))/((1-(Tc/Tb))*18);\n", + "print 'i)The heat of vapourisation of water umath.sing Vishwanath and Kuloor method is %f Kcal/Kg'%(H_c);\n", + "D = (H_c-H_ex)*100/H_c;\n", + "print ' The deviation occurs umath.sing this method is %f percent'%(D);\n", + "\n", + "#(ii)Umath.sing Riedel's method\n", + "H_c = (Tb*2.17*(math.log(218)-1))/((0.93-(Tb/Tc))*18);\n", + "print 'ii)The heat of vapourisation of water umath.sing Riedel method is %f Kcal/Kg'%(H_c);\n", + "D = (H_c-H_ex)*100/H_c;\n", + "print ' The deviation occurs umath.sing this method is %f percent'%(D);\n", + "\n", + "#(iii)By umath.sing given vapour equation; math.logP = 8.2157-(2218.8537/(273.16+t)), t is in deg cel\n", + "#From steam table,\n", + "Vv = 1.673;#in cubic meter/Kg\n", + "Vl = 0.001;#in cubic meter/Kg\n", + "H_c = (2218.8/(273.16+t)**2)*(2.3*Tb*P1*(Vv-Vl)/427);\n", + "print 'iii)The heat of vapourisation umath.sing the given vapour equation is %f Kcal/Kg'%(H_c);\n", + "D = (H_c-H_ex)*100/H_c;\n", + "print ' The deviation occurs umath.sing this method is %f percent'%(D);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)The heat of vapourisation of water umath.sing Vishwanath and Kuloor method is 1234.397750 Kcal/Kg\n", + " The deviation occurs umath.sing this method is 56.334982 percent\n", + "ii)The heat of vapourisation of water umath.sing Riedel method is 557.743828 Kcal/Kg\n", + " The deviation occurs umath.sing this method is 3.360652 percent\n", + "iii)The heat of vapourisation umath.sing the given vapour equation is 552.934102 Kcal/Kg\n", + " The deviation occurs umath.sing this method is 2.520029 percent\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.18 Page No : 270" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "T1 = 273-87.0;#temp in K\n", + "T2 = 273.0;#temp in K\n", + "H1 = 115.0;#Latent heat of saturated ethane at 1 atm and -87 deg cel in Kcal/Kg\n", + "H2_ex = 72.44;#Experimental value of latent heat at 0 deg cel in Kcal/Kg\n", + "Tc = 306.0;#Critical temperature in K\n", + "M = 30.0;#Molecular weight of ethane\n", + "\n", + "#To Calculate the latent heat of saturated ethane at 0 deg cel\n", + "Tr1 = T1/Tc;#reduced temp in K\n", + "Tr2 = T2/Tc;#reduced temp in K\n", + "#(i)Umath.sing Waton's method:\n", + "H2_c = H1*((1-Tr2)/(1-Tr1))**0.38;\n", + "print 'i)The latent heat of saturated ethane at 0 deg cel umath.sing Waton method is %f Kcal/Kg'%(H2_c);\n", + "D = (H2_ex-H2_c)*100/H2_ex;\n", + "print ' The deviation occurs umath.sing this method is %f percent'%(D);\n", + "\n", + "#(ii)Umath.sing Vishwanath and Kuloor method\n", + "#From equation 13.117 (page no 289)\n", + "n = (0.00133*(H1*M/T1)+0.8794)**(1/0.1);\n", + "H2_c = H1*((1-Tr2)/(1-Tr1))**n;\n", + "print 'ii)The latent heat of saturated ethane at 0 deg cel umath.sing Vishwanath and Kuloor method is %f Kcal/Kg'%(H2_c);\n", + "D = (H2_ex-H2_c)*100/H2_ex;\n", + "print ' The deviation occurs umath.sing this method is %f percent'%(D);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)The latent heat of saturated ethane at 0 deg cel umath.sing Waton method is 70.411608 Kcal/Kg\n", + " The deviation occurs umath.sing this method is 2.800099 percent\n", + "ii)The latent heat of saturated ethane at 0 deg cel umath.sing Vishwanath and Kuloor method is 71.809846 Kcal/Kg\n", + " The deviation occurs umath.sing this method is 0.869898 percent\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 13.19 Page No : 272" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "H_s_ex = 32.7;#experimental value of latent heat of the solution in KJ/mole\n", + "x1 = 0.536;#mole percent of toulene in the solution\n", + "x2 = 1-0.536;#mole percent of 1,1,1-trichloroethane in the solution\n", + "H1 = 33.34;#Latent heat of toulene in KJ/gmole\n", + "H2 = 29.72;#Latent heat of 1,1,1-trichloroethane in KJ/gmole\n", + "He = 0.0;#excess enthalpy is neglected\n", + "Cp1 = 39.55;#Specific heat of toulene in cal/gmole deg cel\n", + "Cp2 = 24.62;#Specific heat of 1,1,1-trichloroethane in cal/gmole deg cel\n", + "T_D = 100.0;#dew point temperature in deg cel\n", + "T_B = 92.6;#bubble point temperature in deg cel\n", + "\n", + "#To calculate the latent heat of the solution and compare it with the one which calculated from the given vapour pressure equation\n", + "#(i)Calculation of latent heat of the solution\n", + "#From equation 13.118 (page no 291)\n", + "H_s = H1*x1+H2*x2+He+(Cp1*x1+Cp2*x2)*10**-3*4.17*(T_D-T_B);\n", + "print 'i)The latent heat of the solution is %f KJ/gmole'%(H_s);\n", + "D = ((H_s_ex-H_s)*100)/H_s_ex;\n", + "print ' The deviation occurs using this method is %f percent'%(D);\n", + "\n", + "#(ii)Calculation of latent heat from the vapour pressure equation\n", + "#From equation (a) (page no 291)\n", + "K = 1657.599/((273.16+5)**2);\n", + "H_s = (K*2.303*8.314*(273.16+5)**2)*10**-3;\n", + "print 'ii)The latent heat of the solution is %f KJ/gmole'%(H_s);\n", + "D = ((H_s_ex-H_s)*100)/H_s_ex;\n", + "print ' The deviation occurs using this method is %f percent'%(D);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)The latent heat of the solution is 32.666984 KJ/gmole\n", + " The deviation occurs using this method is 0.100965 percent\n", + "ii)The latent heat of the solution is 31.738283 KJ/gmole\n", + " The deviation occurs using this method is 2.941029 percent\n" + ] + } + ], + "prompt_number": 23 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch14.ipynb b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch14.ipynb new file mode 100755 index 00000000..4474f761 --- /dev/null +++ b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch14.ipynb @@ -0,0 +1,1041 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:1e67847320cea04698ca89c9f40ead4304c8c00593843ef51265231f33b2b44d" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 14 : Thermodynamics of Chemical Reactions" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.1 Page No : 282" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "H_f_C2H4 = 12500.0;#Smath.radians(numpy.arcmath.tan(ard heat of formation of ethylene at 25 deg cel in Kcal/Kgmole\n", + "H_f_C2H4O = -11667.0;#Smath.radians(numpy.arcmath.tan(ard heat of formation of ethylene oxide at 25 deg cel in Kcal/Kgmole\n", + "\n", + "#To Calculate the smath.radians(numpy.arcmath.tan(ard heats of reaction at 25 deg celsius\n", + "#The reaction is: C2H4 + (1/2)O2 - C2H4O\n", + "del_H_rxn = H_f_C2H4O-H_f_C2H4;#Since, Smath.radians(numpy.arcmath.tan(ard heat of formation of O2 is zero\n", + "print 'The standard heats of reaction at 25 deg celsius is %d Kcal/Kgmole'%(del_H_rxn);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The standard heats of reaction at 25 deg celsius is -24167 Kcal/Kgmole\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.2 Page No : 284" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "T1 = 25+273.0;#Initial temperature in K\n", + "T2 = 450+273.0;#Final temperature in K\n", + "#Specific heat of sulphur dioxide is given by the relation:\n", + "#Cp = 7.116+9.512*10**-3*T+(3.511*10**-6)*T**2\n", + "\n", + "#To Calculate the sensible heat required\n", + "#Basis: 1 Kgmole of sulphur dioxide\n", + "Q = 7.116*(T2-T1)+(9.512*10**-3*(T2**2-T1**2)/2)+((3.51*10**-6)*(T2**3-T1**3)/3);\n", + "print 'Sensible heat required is %d Kcal/Kgmole'%(Q);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Sensible heat required is 5499 Kcal/Kgmole\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.3 Page No : 289" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "H_C2H4 = 12500;#Smath.radians(numpy.arcmath.tan(ard heat of formation of ethylene at 25 de cel\n", + "H_C2H4O = -12190;#Smath.radians(numpy.arcmath.tan(ard heat of formation of ethylene oxide at 25 deg cel\n", + "H_CO2 = -94052;#Smath.radians(numpy.arcmath.tan(ard heat of formation of CO2 at 25 deg cel\n", + "H_H2O = -57798;#Smath.radians(numpy.arcmath.tan(ard heat of formation of H2O at 25 deg cel\n", + "T1 = 200;#temperature at which mixture entered in deg cel\n", + "Ti = 25;#intermediate temperature in deg cel\n", + "T2 = 260;#product temperature in deg cel\n", + "Cp_air_a = 7;#Mean specific heat for 25 to 200 deg cel in Kcal/Kgmole\n", + "Cp_C2H4_a = 18;#Mean specific heat for 25 to 200 deg cel in Kcal/Kgmole\n", + "#Mean specific heat for 25 to 260 deg cel in Kcal/Kgmole are given as:\n", + "Cp_C2H4_b = 19;\n", + "Cp_C2H4O_b = 21;\n", + "Cp_O2_b = 7.30;\n", + "Cp_N2_b = 7.00;\n", + "Cp_CO2_b = 10.00;\n", + "Cp_H2O_b = 8.25;\n", + "#Basis: 1 Kgmole of ethylene as feed\n", + "n_air = 9;#Kgmoles\n", + "n_C2H4 =1;#Kgmoles\n", + "n_C2H4_1 = 0.6#ethylene consumed while converting in C2H4O in Kgmoles (Rxn 1)\n", + "n_C2H4_2 = 0.3;#ethylene burnt completely to CO2 in Kgmoles (Rxn 2)\n", + "\n", + "#To Calculate the heat must be removed if the product temperature should not exceed 260 deg cel\n", + "n_O2 = n_air*0.21;#Kgmoles of O2 fed\n", + "n_N2 = n_air-n_O2;#Kgmoles of N2 fed\n", + "#Rxn 1: (0.6)C2H4 + (0.3)O2 - (0.6)C2H4O;...(i) 60% of C2H4 is converted to C2H4O\n", + "n_O2_1 = n_C2H4_1/2;#Kgmoles of oxygen consumed\n", + "n_C2H4O_1 = n_C2H4_1;#C2H4O formed in Kgmoles\n", + "#Rxn 2: (0.3)C2H4 + (0.9)O2 - (0.6)CO2+ (0.6)H20;...(ii) 30%conversion\n", + "n_O2_2 = 3*n_C2H4_2;#Kgmoles of O2 reacted\n", + "n_CO2_2 = 2*n_C2H4_2;#Kgmoles of CO2 formed\n", + "n_H2O_2 = 2*n_C2H4_2;#Kgmoles of H2O formed\n", + "n_C2H4_r = n_C2H4 - n_C2H4_1 - n_C2H4_2;#unreacted ethylene in Kgmoles\n", + "n_O2_r = n_O2 - n_O2_1-n_O2_2;#unreacted O2 in Kgmoles\n", + "#The overall rxn is given by (i)&(ii),\n", + "#(0.9)C2H4 + (1.2)O2 - (0.6)C2H4O + (0.6)CO2 + (0.6)H2O...(iii)\n", + "del_H = (0.6*H_C2H4O)+(0.6*H_CO2)+(0.6*H_H2O)-(0.9*H_C2H4);#math.since,smath.radians(numpy.arcmath.tan(ard enthalpy of O2 is zero; Smath.radians(numpy.arcmath.tan(ard heat of rxn in Kcal/Kgmole\n", + "Q1 = (n_C2H4*Cp_C2H4_a + n_air*Cp_air_a)*(Ti-T1);#Sensible heat in feed in Kcal\n", + "Q2 = (n_C2H4_r*Cp_C2H4_b + n_C2H4O_1*Cp_C2H4O_b + n_CO2_2*Cp_CO2_b + n_H2O_2*Cp_H2O_b + n_O2_r*Cp_O2_b + n_N2*Cp_N2_b)*(T2-Ti);#Sensible heat in product in Kcal\n", + "Q = Q1+Q2+del_H;\n", + "print 'The heat to be removed is %f Kcal so that the product temperature is 260 deg celsius'%(-Q);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The heat to be removed is 104988.605000 Kcal so that the product temperature is 260 deg celsius\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.4 Page No : 293" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "import math\n", + "\n", + "import matplotlib.pyplot as plt\n", + "import numpy\n", + "\n", + "#Given\n", + "#Basis: 1Kgmole of C0\n", + "#CO + (1/2)O2 - CO2\n", + "#Whem mixture entered,their compositions are given as:\n", + "n_CO = 1.0;#Kgmole\n", + "del_H = -67636.0;#Smath.radians(numpy.arcmath.tan(ard heat of reaction in Kcal/Kgmole at 25 deg cel\n", + "T1 = 95.0;#Temperature at which mixture entered in deg celsius\n", + "T2 = 25.0;#Intermediate temperature in deg celsius\n", + "#Mean specific heat values for the temperature between 25 and 95 deg celsius in Kcal/Kgmole are given as (from figure 14.4)(page no 303)\n", + "Cpm_CO = 6.95;\n", + "Cpm_O2 = 7.1;\n", + "Cpm_N2 = 6.95;\n", + "\n", + "#To Calculate the theoretical flame temperature when both air and CO2 enter at 95 deg celsius\n", + "n_O2 = 1.0;#Kgmole, as 100% excess air is given\n", + "n_N2 = n_O2*(0.79/0.21);#Kgmole\n", + "#After the rxn:\n", + "n_CO2 = n_CO;#Kgmole\n", + "n_O2_r = n_O2-(n_CO/2);#remaining Kgmole of O2\n", + "#In equation 14.18 (page no 307) say: H_2-H_R = Ha, H_P-H_3 = Hb, Hc = del_H+Ha & Ht = Hc+Hb\n", + "Ha = (n_CO*Cpm_CO + n_O2*Cpm_O2 + n_N2*Cpm_N2)*(T2-T1);#in Kcal/Kgmole\n", + "Hc = del_H+Ha;#in Kcal/Kgmole\n", + "#For calculating Hb let us assume the temperature as\n", + "T = [530, 1000, 1650];#in deg celsius\n", + "Cpm_CO2 = [10.85, 12, 12.75];#Mean specific heat of CO2 at the coresspondig temperature (from figure 14.4)\n", + "Cpm_O2 = [7.55, 7.8, 8.3];#Mean specific heat of O2 at the coresspondig temperature (from figure 14.4)\n", + "Cpm_N2 = [7.15, 7.5, 7.85];#Mean specific heat of N2 at the coresspondig temperature (from figure 14.4)\n", + "Hb = []\n", + "Ht = []\n", + "for i in range(0,3):\n", + " Hb.append((n_CO2*Cpm_CO2[i]+n_O2_r*Cpm_O2[i]+n_N2*Cpm_N2[i])*(T[i]-T2));#in Kcal/Kgmole\n", + " Ht.append(Hc+Hb[i]);#in Kcal/Kgmole\n", + "\n", + "plt.plot(T,Ht)\n", + "plt.title(\"\")\n", + "plt.xlabel(\"Temperature, deg celsius\")\n", + "plt.ylabel(\"Ht in Kcal/Kgmole\")\n", + "plt.show()\n", + "Tf = numpy.interp(0,Ht,T) \n", + "print 'The adiabatic temperature is read as %d degree celsius'%(Tf);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "metadata": {}, + "output_type": "display_data", + "png": 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+ "text": [ + "<matplotlib.figure.Figure at 0x10ee76150>" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The adiabatic temperature is read as 1549 degree celsius\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.5 Page No : 294" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Given\n", + "T = 298.0;#Temperature in Kelvin\n", + "#standard enthalpy in Kcal/Kgmole of the following components at 298 K are given as\n", + "H_SO2 = -70960;\n", + "H_SO3 = -94450;\n", + "H_O2 = 0;\n", + "#standard entropy in Kcal/Kgmole K of the following components at 298 K are given as\n", + "S_SO2 = 2.48;\n", + "S_SO3 = -19.7\n", + "S_O2 = 0;\n", + "#Basis: 1 Kgmole of SO2\n", + "#SO2 +(1/2)O2 - SO3\n", + "\n", + "#To Calculate the standard free energy for the reaction\n", + "n_SO2 = 1;#Kgmole of SO2 fed\n", + "n_O2 = (1/2)*2;#Kgmole of O2 fed as 100% excess O2 is given\n", + "n_SO3 = n_SO2;#Kgmole of SO3 formed\n", + "#From equation 14.38 (page no 312)\n", + "del_F = (H_SO3-(T*S_SO3))-(H_SO2-(T*S_SO2))-(H_O2-(T*S_O2));\n", + "print 'The standard free energy for the reaction at 25 degree celsius is %f Kcal/Kgmole'%(del_F);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The standard free energy for the reaction at 25 degree celsius is -16880.360000 Kcal/Kgmole\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.6 Page No : 299" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "import math\n", + "import matplotlib.pyplot as plt\n", + "import numpy\n", + "\n", + "#Given\n", + "#SO2 + (1/2)O2 - SO3\n", + "#Basis: 1 Kgmole of SO2\n", + "n_SO2 = 1.0;# SO2 fed in Kgmole\n", + "#From table 14.1 (page no 301)\n", + "#alpha values for the following components are given as\n", + "a_SO2 = 7.116;\n", + "a_O2 = 6.148;\n", + "a_SO3 = 6.077;\n", + "#beta values for the following components are given as\n", + "b_SO2 = 9.512*10**-3;\n", + "b_O2 = 3.102*10**-3;\n", + "b_SO3 = 25.537*10**-3;\n", + "#standard enthalpy of the following components at 25 deg cel in Kcal/Kgmole are given as\n", + "H_SO2 = -70960.0;\n", + "H_O2 = 0.0;\n", + "H_SO3 = -94450;\n", + "#standard free energy of the following components at 25 deg cel in Kcal/Kgmole K are given as\n", + "F_SO2 = -71680.0;\n", + "F_O2 = 0.0;\n", + "F_SO3 = -88590.08;\n", + "n_O2 = n_SO2;#O2 fed in Kgmole; math.since 50 mole percent mixture of SO2 & O2 is fed\n", + "n_SO3 = n_SO2;#SO3 formed in Kgmole\n", + "n_O2_e = n_O2-(n_O2/2);#Kgmoles of O2 in exit gas\n", + "n_O2_r = n_O2/2;#Kgmoles of O2 reacted\n", + "R = 1.98;#gas consmath.tant in Kcal/Kgmole K\n", + "\n", + "#To show the variation of the standard heats of reaction with temperature and the equilibrium consmath.tant with temperature graphically in the given temperature range\n", + "#(i)Variation of the standard heats of reaction with temperature\n", + "del_H = (n_SO3*H_SO3)-(n_O2_r*H_O2)-(n_SO2*H_SO2);# in Kcal/Kgmole\n", + "del_F = (n_SO3*F_SO3)-(n_O2_r*F_O2)-(n_SO2*F_SO2);# in Kcal/Kgmole\n", + "#From equation 14.10 (page no 301)\n", + "del_a = (n_SO3*a_SO3)-(n_O2_r*a_O2)-(n_SO2*a_SO2);\n", + "del_b = (n_SO3*b_SO3)-(n_O2_r*b_O2)-(n_SO2*b_SO2);\n", + "#In equation 14.11 (page no 302), substituting del_H at\n", + "T = 298;#in deg cel\n", + "I = del_H - del_a*T - (del_b*(T**2)/2);# integrating consmath.tant\n", + "print 'iThe standard heat of reaction at any tempperature can be calculated by the relation:';\n", + "print ' del_Ht = %fT'%del_a,\n", + "print ' + %fT**2 '%(del_b/2),\n", + "print '%f'%I\n", + "\n", + "#(ii)Variation of the equilibrium consmath.tant with temperature\n", + "#K1 = lnKa (say)\n", + "K1 = -del_F/(R*T);\n", + "#From equation 14.42 (page no 316); M1 = M/R (say)\n", + "M1 = K1-(del_a/R)*math.log(T)-(del_b/(2*R))*T+(I/(T*R));\n", + "#Let us assume the temperature in the range 800K to 1500K as\n", + "Ta = [700, 800, 825, 850, 900, 1000, 1100, 1300, 1500];\n", + "Ka = []\n", + "for i in range(0,9):\n", + " Ka.append(math.e**((del_a/R)*math.log(Ta[i])+(del_b*Ta[i]/(2*R))-(I/(Ta[i]*R))+M1));\n", + "\n", + "\n", + "plt.plot(Ta,Ka);\n", + "plt.title(\"\" )\n", + "plt.xlabel(\"Temperatur in K\")\n", + "plt.ylabel(\"equilibrum constant K\")\n", + "plt.show()\n", + " \n", + "print 'ii)From the graph it can be seen that as temperature increases Ka decreases exponentially,(so the reaction is exothermic.';\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "iThe standard heat of reaction at any tempperature can be calculated by the relation:\n", + " del_Ht = -4.113000T + 0.007237T**2 -22907.000548\n" + ] + }, + { + "metadata": {}, + "output_type": 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+ "text": [ + "<matplotlib.figure.Figure at 0x10b2ade10>" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "ii)From the graph it can be seen that as temperature increases Ka decreases exponentially,(so the reaction is exothermic.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.7 Page No : 300" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "import math\n", + "import matplotlib.pyplot as plt\n", + "import numpy\n", + "\n", + "#Given\n", + "#SO2 + (1/2)O2 - SO3\n", + "#Basis: 1 Kgmole of SO2\n", + "n_SO2 = 1.0;# SO2 fed in Kgmole\n", + "n_O2 = n_SO2;#O2 fed in kgmole\n", + "\n", + "#To Calculate the conversion of SO2 to SO3 at 1atm and at various temperature\n", + "#(1)Calculate the conversion of SO2 to SO3 \n", + "P = 1.0;#Pressure in atm\n", + "T = 850.0;#Temperature in K\n", + "m = 1-1-(1/2.0);\n", + "#From example 14.6\n", + "Ta = [700.0, 800.0, 825.0, 850.0, 900.0, 1000.0, 1100.0, 1300.0, 1500.0];\n", + "Ka = [395.40, 52.51, 34.60, 23.44, 11.59, 3.527, 1.48, 0.398, 0.0016];\n", + "\n", + "\n", + "\n", + "plt.plot(Ta,Ka)\n", + "plt.title(\"Equilibrium consmath.tant vs Temperature\")\n", + "plt.xlabel(\"Temperature in K\")\n", + "plt.ylabel(\"Ka\")\n", + "plt.show()\n", + "Ka1 = numpy.interp(850,Ta,Ka)\n", + "\n", + "\n", + "#Let Nc be the moles of SO3 at equilibrium\n", + "Nc = [0.1, 0.2, 0.3, 0.4, 0.5, 0.7, 0.8, 0.9, 0.930, 0.95, 0.98, 0.988, 0.989, 0.9895, 0.9897, 0.9899, 0.9900];\n", + "#From equation 14.49 (page no 320) and umath.sing the given data ,we got equation (b) (page no 323)\n", + "Ka = []\n", + "for i in range (0,17):\n", + " Ka.append((((n_SO2+n_O2-0.5*Nc[i])/(n_O2-0.5*Nc[i]))**(1/2.0))*(Nc[i]/(n_SO2-Nc[i])));\n", + "\n", + " \n", + "\n", + "\n", + "plt.plot(Nc,Ka)\n", + "plt.title(\"Equilibrium consmath.tant vs Kgmoles of SO3\")\n", + "plt.xlabel(\"Kg moles of SO3\")\n", + "plt.ylabel(\"Ka\")\n", + "Nc1 = numpy.interp(Ka1,Ka,Nc)\n", + "\n", + "\n", + "C = Nc1*100.0/n_SO2;\n", + "print '1)The conversion of SO2 to SO3 at 1atm and 850K is %f percent'%C;\n", + "\n", + "#(2)Calculation of conversion at 1 atm and 850 K under the following conditions\n", + "#(i) Given:\n", + "n_N2 = 3.75;#Kgmoles of N2 fed\n", + "#Let Nc be the moles of SO3 at equilibrium\n", + "Nc = [0.85, 0.87, 0.90];\n", + "#From equation 14.49 (page no 320) and umath.sing the given data ,we got equation (c) (page no 324)\n", + "Ka2 = []\n", + "for i in range(0,3):\n", + " Ka2.append((((+n_N2+n_SO2+n_O2-0.5*Nc[i])/(n_O2-0.5*Nc[i]))**(1/2.0))*(Nc[i]/(n_SO2-Nc[i])));\n", + "\n", + "\n", + "\n", + "plt.plot(Nc,Ka2)\n", + "Nc2 = numpy.interp(Ka1,numpy.transpose(Ka2),Nc)\n", + "\n", + " \n", + "C2 = Nc2*100.0/n_SO2;\n", + "print ' 2)i)The conversion of SO2 to SO3 at 1 atm and 850 K when inert gas is also added is %f percent'%C2;\n", + "\n", + "\n", + "\n", + " \n", + "\n", + "#(ii)SO3 is also sent along the original feed\n", + "n_SO3 = 1.0;#Kgmoles of SO3 fed\n", + "#Let Nc be the moles of SO3 at equilibrium\n", + "Nc = [0.80, 0.86, 0.92];\n", + "#From equation 14.49 (page no 320) and umath.sing the given data ,we got equation (d) (page no 326)\n", + "Ka3 = []\n", + "for i in range(0,3):\n", + " Ka3.append((((+n_SO3+n_SO2+n_O2-0.5*Nc[i])/(n_O2-0.5*Nc[i]))**(1/2.0))*((n_SO3+Nc[i])/(n_SO2-Nc[i])));\n", + "\n", + "\n", + "plt.plot(Nc,Ka3)\n", + "Nc3 = numpy.interp(Ka1,Ka3,Nc)\n", + "\n", + "\n", + "C3 = Nc3*100.0/n_SO2;\n", + "print ' ii)The conversion of SO2 to SO3 at 1 atm and 850 K when SO3 is also added along the original feed is %f percent'%C3;\n", + "\n", + "#(iii)Variation of SO2 to O2 ratio:\n", + "#(a)SO2:O2 = 1:1 ; This has been worked out in part 1\n", + "print ' iii)a)The conversion of SO2 to SO3 at 1atm and 850K when SO2:O2 = 1:1 is %f percent'%(C);\n", + "Xc = Nc1/(n_SO2+n_O2-0.5*Nc1);\n", + "\n", + "#(b)SO2:O2 = 1.1:0.5,Now\n", + "n_SO2 = 1.1;#Kgmoles of SO2 fed\n", + "n_O2 = 0.5;#Kgmoles of O2 fed\n", + "#Let Nc be the moles of SO3 at equilibrium\n", + "Nc = [0.9, 0.91, 0.92];\n", + "#From equation 14.49 (page no 320) and umath.sing the given data ,we got equation (e) (page no 327)\n", + "Ka4 = []\n", + "for i in range(0,3):\n", + " Ka4.append((((n_SO2+n_O2-0.5*Nc[i])/(n_O2-0.5*Nc[i]))**(1/2.0))*(Nc[i]/(n_SO2-Nc[i])));\n", + "\n", + "\n", + "plt.plot(Nc,Ka4)\n", + "Nc4 = numpy.interp(Ka1,Ka4,Nc)\n", + "\n", + "\n", + "\n", + "\n", + "C4 = Nc4*100.0/n_SO2;\n", + "print ' iii)b)The conversion of SO2 to SO3 at 1atm and 850K when SO2:O2 = 1.1:0.5 is %f percent'%C4;\n", + "Xc1 = Nc4/(n_SO2+n_O2-0.5*Nc4);\n", + "\n", + "#(c)SO2:O2 = 1:0.5\n", + "n_SO2 = 1.0;#Kgmoles of SO2 fed\n", + "n_O2 = 0.5;#Kgmoles of O2 fed\n", + "#Let Nc be the moles of SO3 at equilibrium\n", + "Nc = [0.8, 0.85, 0.86, 0.87];\n", + "#From equation (a)\n", + "Ka5 = []\n", + "for i in range(0,4):\n", + " Ka5.append((((n_SO2+n_O2-0.5*Nc[i])/(n_O2-0.5*Nc[i]))**(1/2.0))*(Nc[i]/(n_SO2-Nc[i])));\n", + "\n", + "\n", + "\n", + "plt.plot(Nc,Ka5)\n", + "Nc5 = numpy.interp(Ka1,Ka5,Nc)\n", + "\n", + " \n", + "C5 = Nc5*100.0/n_SO2;\n", + "print ' iii)c)The conversion of SO2 to SO3 at 1atm and 850K when SO2:O2 = 1:0.5 is %f percent'%(C5);\n", + "Xc2 = Nc5/(n_SO2+n_O2-0.5*Nc5);\n", + "\n", + "if(Xc2>Xc) and (Xc2>Xc1):\n", + " print ' SO2:O2 = 1:0.5 gives the maximum concentration of SO3 at equilibrium.';\n", + "else:\n", + " if(Xc1>Xc) and (Xc1>Xc2):\n", + " print ' SO2:O2 = 1.1:0.5 gives the maximum concentration of SO3 at equilibrium';\n", + " else:\n", + " if(Xc>Xc1) and (Xc>Xc2):\n", + " print ' SO2:O2 = 1:1 gives the maximum concentration of SO3 at equilibrium';\n", + "\n", + "\n", + "\n", + "#(3)Conversion of SO2 to SO3 at 50 atm and 850 K when SO2:O2 = 1:1\n", + "n_SO2 = 1.0;#Kgmole of SO2 fed\n", + "n_O2 = 1.0;#Kgmoles of O2 fed\n", + "P = 50.0;#Pressure in atm\n", + "#From figure A.2.9\n", + "phi_SO2 = 0.99;\n", + "phi_SO3 = 0.972;\n", + "phi_O2 = 1.0;\n", + "#From equation 14.48 (page no320), Ka = Ky*(P**m)*K_phi\n", + "K_phi = phi_SO3/(phi_SO2*(phi_O2**2));\n", + "#Let Nc be the moles of SO3 at equilibrium\n", + "Nc = [0.99, 0.985, 0.97, 0.96];\n", + "Ka6 = []\n", + "for i in range(0,4):\n", + " Ka6.append(K_phi*(P**m)*((((n_SO2+n_O2-0.5*Nc[i])/(n_O2-0.5*Nc[i]))**(1/2.0))*(Nc[i]/(n_SO2-Nc[i]))));\n", + "\n", + "\n", + "plt.plot(Nc,Ka6)\n", + "plt.legend(['1 part', '2.(i) part', '2.(ii) part', '2.(iii).(b) part', '2.(iii).(c) part', '3 part'])\n", + "plt.show()\n", + "Nc6 = numpy.interp(Ka1,Ka6,Nc)\n", + "\n", + "\n", + "C = Nc6*100.0/n_SO2;\n", + "print ' 3)The conversion of SO2 to SO3 at 50atm and 850K when SO2:O2 = 1:1 is %f percent'%(C);\n", + "#legend(\"1 part\",\"2.(i) part\",\"2.(ii)part\",\"2.(iii).(b)part\",\"2.(iii).(c)part\",\"3 part\");\n", + "\n", + "\n", + "\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "metadata": {}, 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+ "text": [ + "<matplotlib.figure.Figure at 0x10fc16a50>" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1)The conversion of SO2 to SO3 at 1atm and 850K is 93.189481 percent\n", + " 2)i)The conversion of SO2 to SO3 at 1 atm and 850 K when inert gas is also added is 88.179412 percent\n", + " ii)The conversion of SO2 to SO3 at 1 atm and 850 K when SO3 is also added along the original feed is 82.979198 percent\n", + " iii)a)The conversion of SO2 to SO3 at 1atm and 850K when SO2:O2 = 1:1 is 93.189481 percent\n", + " iii)b)The conversion of SO2 to SO3 at 1atm and 850K when SO2:O2 = 1.1:0.5 is 82.473630 percent\n", + " iii)c)The conversion of SO2 to SO3 at 1atm and 850K when SO2:O2 = 1:0.5 is 85.774995 percent\n", + " SO2:O2 = 1:0.5 gives the maximum concentration of SO3 at equilibrium.\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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Qo6bXjxs3jgkTJtCsWTMKCwuZN29e2XJ/HNK0fn274VRZoI2UIUOOClFFtXXI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\nyqaNRiNGo71NCCGEB2ka3Hc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+ "text": [ + "<matplotlib.figure.Figure at 0x10fccd5d0>" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " 3)The conversion of SO2 to SO3 at 50atm and 850K when SO2:O2 = 1:1 is 96.000000 percent\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.9 Page No : 305" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "#2A(g) - B(g)+C(g)\n", + "T = 400.0;#Temperature in Kelvin\n", + "P = 1.0;#Pressure in atm\n", + "R = 1.98;#gas consmath.tant in Kcal/Kgmole K\n", + "del_Fo = 3000.0;#standard free energy of the reaction in Kcal/Kgmole\n", + "n_A = 1.0;#Kgmoles of A\n", + "n_B = 0.1;#Kgmoles of B\n", + "n_C = 0.1;#Kgmoles of C\n", + "\n", + "#To find out the equilibrium has been attained or not. If not then calculate the equilibrium composition and also find out whether the equilibrium composition will change or not if pressure were increased to 3 atm.\n", + "#(1)To find out the equilibrium has been attained or not. If not then calculate the equilibrium composition\n", + "#P_B/P_C = 1 (Given)\n", + "n_T = n_A+n_B+n_C;#Total Kgmoles of the component\n", + "p_A = (n_A/n_T)*P;#Partial pressure of A\n", + "p_B = (n_B/n_T)*P;#Partial pressure of B\n", + "p_C = (n_C/n_T)*P;#Partial pressure of C\n", + "#Umath.sing the relation 14.36 (page no 312)\n", + "del_F = del_Fo + (R*T)*math.log((p_B*p_C)/(p_A**2));\n", + "if del_F == 0:\n", + " print '1.Equilibrium has been attained.';\n", + "else:\n", + " print '1.Equilibrium has not been attained.';\n", + "#Equilibrium composition\n", + "#At equilibrium del_F = 0\n", + "#From equations 14.35(page no 312) and 14.49(page no 320), we got the relations (a),(b)(page no 331) &(c)(page no 332) and\n", + "#ln(p_C**2/(P-(2*p_C**2)))= -del_Fo/(R*T); \n", + "Kp = math.e**(-del_Fo/(R*T))#equilibrium consmath.tant in terms of pressure\n", + "p_C = (Kp**(1/2))/(1+2*(Kp**(1/2)));#Partial pressure of C at equilibrium in atm\n", + "p_B = p_C;#as p_B/p_C = 1\n", + "p_A = P-(2*p_C);#Partial pressure of A at equilibrium in atm\n", + "N_A = p_A*P*100.0;\n", + "N_B = p_B*P*100.0;\n", + "N_C = p_C*P*100.0;\n", + "print ' Equilibrium composition of A,(B and C are %f'%N_A,\n", + "print '(%f and '%N_B,\n", + "print '%f mole percent respectively.'%N_C\n", + "#2.Calculation of composition at 400 K and 3 atm and vapour pressure of C at 400 K is 0.3 atm\n", + "P = 3.0;#Pressure in atm\n", + "P_C = 0.3;#Vapour pressure of C in atm\n", + "#Since m=0,pressure will not have an effect on equilibrium compositions.\n", + "print ' 2.The compositions will be the same as above on increamath.sing the pressure.';\n", + "p_B1 = p_B*P;#Partial pressure of B in atm\n", + "p_C1 = p_B1;#Partial pressure of C in atm\n", + "p_A1 = p_A*P;#Partial pressure of A in atm\n", + "if P_C < p_C1:\n", + " p_A2 = ((p_B1*P_C)/Kp)**(1/2.0);#decreased partial pressure of A in atm\n", + "print ' The partial pressure of A drops from %f'%p_A1,\n", + "print ' to %f to compensate for the amount of C condensed.'%p_A2;\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1.Equilibrium has not been attained.\n", + " Equilibrium composition of A,(B and C are 33.333333 (33.333333 and 33.333333 mole percent respectively.\n", + " 2.The compositions will be the same as above on increamath.sing the pressure.\n", + " The partial pressure of A drops from 1.000000 to 3.639888 to compensate for the amount of C condensed.\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.10 Page No : 307" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "import math\n", + "import matplotlib.pyplot as plt\n", + "import numpy\n", + "\n", + "#Given\n", + "#SO2(A) + (1/2)O2 (B) - SO3(C)\n", + "#Basis: 1 Kgmole SO2\n", + "n_A = 1.0;#Kgmole of SO2 fed\n", + "n_B = n_A;#Kgmole of O2 fed\n", + "T1 = 273+400.0;#Temperature in K at which reacmath.tants enter\n", + "To = 298.0;#room temperature in K\n", + "del_H = -23490.0;#Smath.radians(numpy.arcmath.tan(ar heat of reaction at 25 deg cel from example 14.6 in Kcal/Kgmole\n", + "\n", + "#At T1,\n", + "C_A_T1 = 11.0;\n", + "C_B_T1 = 7.4;#in Kcal/Kgmole\n", + "#Assume the various temperature be\n", + "T = [913.0, 1073.0, 1373.0, 1573.0];#in K\n", + "#Mean specific heats of the components A,B & C at various temperature are given below in Kcal/Kgmole K\n", + "C_A = [11.6, 11.8, 12.3, 12.5];\n", + "C_B = [7.7, 7.8, 8.0, 8.2];\n", + "C_C = [16.6, 17.2, 18.2, 18.6];\n", + "\n", + "#To Calculate the final temperature for various conversions and the maximum conversion that can be attained in a math.single reactor operating adiabatically\n", + "#In equation 14.18 (page no 307), H2-Hr = K & (Hp-H3)= L(say)\n", + "K = ((n_A*C_A_T1)+(n_B*C_B_T1))*(To-T1);#in Kcal/Kgmole\n", + "print 'Adiabatic reaction temp in K pecentage conversion of SO2';\n", + "n_C = []\n", + "for i in range(0,4):\n", + " n_C.append((-K-(C_A[i]*(T[i]-To))-(C_B[i]*(T[i]-To)))/((T[i]-To)*(C_C[i]-C_A[i]-0.5*C_B[i])+del_H));\n", + " n_C[i] *= 100\n", + " print ' %d'%T[i];\n", + " print ' %f'%(n_C[i]*100);\n", + "\n", + "\n", + "plt.plot(T,n_C)\n", + "\n", + "#Now equilibrium conversion at various temperature taken from figure 14.7 (page no 325) are given as\n", + "Ta = [850.0, 900.0, 1000.0, 1100.0, 1200.0, 1300.0, 1400.0,];\n", + "n_C1 = [93.5, 88.2, 69.0, 49.0, 37.0, 21.5, 6.25];\n", + "\n", + "plt.plot(Ta,n_C1)\n", + "plt.title(\"Temperature vs Percentage Conversion\")\n", + "plt.xlabel(\"Temperature in K\")\n", + "plt.ylabel(\"% Conversion\")\n", + "plt.show()\n", + "\n", + "#From the graph,it can be seen that the curve cut each other approximately at the temp\n", + "T1 = 1140.0;#in Kelvin\n", + "C = numpy.interp(T1,Ta,n_C1)\n", + "\n", + "print ' The maximum conversion that can be attained is %d percent'%C;\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Adiabatic reaction temp in K pecentage conversion of SO2\n", + " 913\n", + " 2181.255555\n", + " 1073\n", + " 3712.910088\n", + " 1373\n", + " 6957.687376\n", + " 1573\n", + " 9308.739255\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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+ "text": [ + "<matplotlib.figure.Figure at 0x10fc6d310>" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The maximum conversion that can be attained is 44 percent\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.11 Page No : 309" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "#Basis: 1 Kgmole of benzene\n", + "#C6H6 (A) + HNO3 (B) - C6H5NO2 (C) + H2O (D)\n", + "T = 298.0;#Temperature in K\n", + "R = 1.98;#gas consmath.tant in Kcal/Kgmole K\n", + "#standard enthalpy in Kcal/Kgmole at 25 deg celsius of the above components are given as\n", + "H_A = 11718.0;\n", + "H_B = -41404.0;\n", + "H_C = -68371.0;\n", + "H_D = 3800.0;\n", + "#standard entropy in Kcal/Kgmole K at 25 deg celsius of the above components are given as\n", + "S_A = 41.30;\n", + "S_B = 37.19;\n", + "S_C = 16.72;\n", + "S_D = 53.60;\n", + "\n", + "#To Calculate the conversion of benzene at 25 degree celsius and 1 atm\n", + "del_F = (H_C+H_D-(T*(S_C+S_D)))-(H_A+H_B-(T*(S_A+S_B)));\n", + "Ka = math.e**(-del_F/(R*T));#Equilibrium consmath.tant\n", + "x = (Ka**(1/2.0)/(1+(Ka**(1/2.0))));\n", + "print 'The conversion is almost %f percent for this reaction.'%(x*100);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The conversion is almost 100.000000 percent for this reaction.\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.12 Page No : 311" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "from sympy import Symbol\n", + "import numpy\n", + "\n", + "#Given\n", + "#C + 2H2 - CH4\n", + "#Basis: 1 Kgmole of C fed\n", + "T = 1000.0;#Temperature in K\n", + "P1 = 2.0;#Pressure in atm\n", + "del_F = 4580.0;#standard free energy in Kcal/Kgmole\n", + "\n", + "\n", + "#To Calculate the maximum CH4 concentration under the condition of 2 atm and the quantity of methane obtained if pressure is 1 atm\n", + "Ka = math.e**(-del_F/(R*T));#Equilibrium consmath.tant\n", + "#In relation (d) (page no 339) p_H2 = p (say)\n", + "p = Symbol('p');\n", + "q = [Ka,1,-P1]#*(p**2)+p-P1;\n", + "r = numpy.roots(q);\n", + "p_H2 = r[1];#partial pressure of H2\n", + "p_CH4 = P1-p_H2;#partial pressure of CH4\n", + "X_H2 = p_H2*100/P1;#mole percent of H2\n", + "X_CH4 = p_CH4*100/P1;#mole percent of CH4\n", + "print 'Under the conditions of 2 atm and 1000 K, the maximum CH4 concentration is %f percent and further increase is not pssible'%(X_CH4);\n", + "#Now.pressure has become\n", + "P2 = 1;#in atm\n", + "q = Ka*(p**2)+p-P2;\n", + "r = [Ka,1,-P2]#numpy.roots(q);\n", + "p_H2 = r[1];#partial pressure of H2\n", + "p_CH4 = P2-p_H2;#partial pressure of CH4\n", + "X_H2 = p_H2*100/P2;#mole percent of H2\n", + "X_CH4 = p_CH4*100/P2;#mole percent of CH4\n", + "print ' Under the conditions of 1 atm and 1000 K,Methane = %f percent and Hydrogen = %f percent'%(X_CH4,X_H2);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Under the conditions of 2 atm and 1000 K, the maximum CH4 concentration is 14.475434 percent and further increase is not pssible\n", + " Under the conditions of 1 atm and 1000 K,Methane = 0.000000 percent and Hydrogen = 100.000000 percent\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.13 Page No : 312" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "T1 = 273+110;#Temperature in K\n", + "T = 298.0;#Room temperature in K\n", + "P = 1.0;#Pressure in atm\n", + "R = 1.98;#gas consmath.tant in Kcal/Kgmole\n", + "#Ag2CO3(s) (A) = Ag2O(s) (B) + CO2(g) (C)\n", + "a = 1.0;#Stoichiometry coefficient of A\n", + "b = 1.0;#Stoichiometry coefficient of B\n", + "c = 1.0;#Stoichiometry coefficient of C\n", + "#standard entropy of the above components in Kcal/Kgmole K at 25 deg cel are given as\n", + "S_A = 40.17;\n", + "S_B = 29.09;\n", + "S_C = 51.08;\n", + "#standard enthalpy of the above components in Kcal/Kgmole at 25 deg cel are given as\n", + "H_A = -119900.0;\n", + "H_B = -6950.0;\n", + "H_C = -94036.0;\n", + "#Specific heat (assumed consmath.tant) of the above components in Kcal/Kgmole K are given as\n", + "C_A = 26.1;\n", + "C_B = 16.5;\n", + "C_C = 9.6;\n", + "\n", + "#To Calculate the partial pressure of CO2 required for decomposition and thus determine whether Ag2CO3 will decomposes or not at the given pressure and temperature\n", + "del_H = H_C+H_B-H_A;#standard heat of reaction at 25 deg cel in Kcal/Kgmole\n", + "del_C = b*C_B+c*C_C-a*C_A;\n", + "#From equation 14.15 (page no 340)\n", + "del_Ht = del_H +del_C*(T1-T);\n", + "del_F = del_H-(T*(S_B+S_C-S_A));#standard free energy in Kcal/Kgmole\n", + "Ka1 = math.e**(-del_F/(R*T));#Equilibrium consmath.tant at temperature T\n", + "#Since del_Ht is consmath.tant(as del_C = 0),Ka2 can be calculated by equation 14.43 (page no 316)\n", + "Ka2 = Ka1* math.e**((del_Ht/R)*(1/T1-1/T));\n", + "p_CO2 = Ka2;\n", + "print 'The partial pressure of CO2 required for decomposition is %4.3e atm'%(p_CO2);\n", + "if p_CO2 < P :\n", + " print ' Silver carbonate will not decompose at 110 deg celsius as the pressure given is 1 atm';\n", + "else:\n", + " print ' Silver carbonate will decompose at 110 deg celsius';\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The partial pressure of CO2 required for decomposition is 8.523e-20 atm\n", + " Silver carbonate will not decompose at 110 deg celsius as the pressure given is 1 atm\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.15 Page No : 314" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import poly1d,roots\n", + "#Given\n", + "#C3H8 = C3H6 + H2... (i)\n", + "#C3H8 = C2H4 +CH4...(ii)\n", + "Kp1 = 7.88;#Equilibrium consmath.tant of equation (i)\n", + "Kp2 = 775.0;#Equilibrium consmath.tant of equation (ii)\n", + "T = 760+273;#Temperature in K\n", + "\n", + "#To Calculate the equilibrium composition of the mixture\n", + "#Basis: 1 mole of C3H8 in feed\n", + "#From the equations (a) &(b) (page no 343); y/x = z (say)\n", + "z = (Kp2/Kp1)**(1/2);\n", + "#Substituting y = z*x in equation(a), we got the equation p:\n", + "\n", + "#x = poly(0,'x')\n", + "p = poly1d([1+Kp1+Kp1*(z**2)+2*Kp1,0,-Kp1])\n", + "q = roots(p);\n", + "\n", + "x = q[0];\n", + "y = z*x;\n", + "print 'Moles of H2, C3H6, C2H4 and CH4 formed at equilibrium are %f %f %f %f respectively.'%(x,x,y,y);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Moles of H2, C3H6, C2H4 and CH4 formed at equilibrium are 0.492252 0.492252 0.492252 0.492252 respectively.\n" + ] + } + ], + "prompt_number": 23 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch15.ipynb b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch15.ipynb new file mode 100755 index 00000000..bb8491c4 --- /dev/null +++ b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch15.ipynb @@ -0,0 +1,56 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a162cef37f1f083e6a9ef4ac7692713ff724aae06050b0ea06458a8bbbc97c25" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 15 : Fuel Cells" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.1 Page No : 321" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "del_F = -56.29;#Smath.radians(numpy.arcmath.tan(ard free energy change in Kcal/Kgmole\n", + "del_H = -68.317;#Smath.radians(numpy.arcmath.tan(ard heat of reaction in Kcal/kgmole\n", + "F = 23.06;#Electro-chemical equivalent in Kcal/volt\n", + "J = 2.0;#Valance for H2\n", + "\n", + "#To Calculate the emf of the cell, cell efficiency and heat to be removed to maintain isothermal conditions\n", + "#Basis: 1 Kgmole of H2\n", + "#From equation 15.4 (page no 355)\n", + "E = -del_F/(F*J);\n", + "print '1.The emf of the cell is %f volt.'%(E);\n", + "n = del_F/del_H*100;\n", + "print ' 2.The cell efficiency is %f percent.'%(n);\n", + "Q = del_H-del_F;\n", + "print ' 3.The heat to be removed is %f Kcal to maintain the temperature at 25 degree celsius.'%(Q);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch2.ipynb b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch2.ipynb new file mode 100755 index 00000000..f0ca2482 --- /dev/null +++ b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch2.ipynb @@ -0,0 +1,631 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:3938cdb77cb1cf91b5da41037c46803e1323bc2d84e6ed7eb8e8d2a220b8bc11" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 2 : P V T Relations" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.1 Page No : 29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "m = 140 #m is the mass of N2 in Kg\n", + "P = 4.052*(10**5) #P is the pressure of the system in Pa\n", + "V = 30 #V is the volume of the system in m**3\n", + "R = 8314.4 # R is the gas consmath.tant\n", + "\n", + "#To determine temperature required\n", + "T = P*V/float(((m/28)*R)) #T is the temperature of the system in K\n", + "print \"Temperature of the system is \",\n", + "print \"%.6f\"%T,\n", + "print \"K\"\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temperature of the system is 292.408352 K\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.5 Page No : 33" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "import numpy\n", + "from numpy.polynomial import Polynomial as poly\n", + "\n", + "\n", + "#Given\n", + "n = 1 #n is Kg moles of methane\n", + "T = 423 #T is the temperatue of the system in kelvin\n", + "P = float(100) #P is the pressure of the system in atm\n", + "Tc = 191 #Tc is the critical temperature of the system in K\n", + "Pc = 45.8 #Pc is the critical pressure of the system in atm\n", + "R = 0.08206 #R is the gas consmath.tant in (m**3 atm/Kg mole K)\n", + "\n", + "#To calculate the volume of methane\n", + "#(i)Umath.sing ideal gas equation\n", + "V1 = (n*R*T)/P #V1 is the volume of the gas in m**3\n", + "print \"i)Volume of the gas umath.sing ideal gas equation is \",\n", + "print \"%.6f\"%V1,\n", + "print \"cubic meter\"\n", + "\n", + "#(ii)Umath.sing Vander Waals' equation\n", + "a = (27*(R**2)*(Tc**2))/(64*Pc) #Vander Waais consmath.tant\n", + "b = (R*Tc)/(8*Pc) #Vander Waais consmath.tant\n", + "q = numpy.poly1d([1,-(((R*T)+(b*P))/P),-((a*b)/P)+(a/P),0])\n", + "r = numpy.roots(q)\n", + "print \" ii)Volume of the gas umath.sing Vander Waals equation is\",\n", + "print \"%.6f\"%r[0],\n", + "print \"cubic meter\"\n", + "\n", + "#(iii)Umath.sing generalized Z chart\n", + "Tr = T/Tc #Tr is the reduced temperatue\n", + "Pr = P/Pc #Pr is the reduced pressure\n", + "#From the figure A.2.2,\n", + "Z = 0.97 #Z is the compressibility factor\n", + "V = (Z*R*T)/P \n", + "print \" iii)Volume of the gas umath.sing Z chart is \",\n", + "print \"%.6f\"%V,\n", + "print \"cubic meter\"\n", + "\n", + "#(iv)Umath.sing molar polarisation method\n", + "#From Table 2.2\n", + "Pmc = 6.82 #Pmc is the molar polarisation for methane\n", + "#From figure A.2.4\n", + "Z0 = .965 \n", + "Z1 = 14.8*(10**-4)\n", + "Z = Z0+(Z1*Pmc)\n", + "V = (Z*R*T)/P\n", + "print \" iv)Volume of the gas umath.sing molar polarisation method is \",\n", + "print \"%.6f\"%V,\n", + "print \"cubic meter\"\n", + "\n", + "\n", + "#(v)From experiment\n", + "#Given\n", + "Z = 0.9848\n", + "V = (0.9848*n*R*T)/P\n", + "print \" v)Volume of the gas calculated by experimental Z value is \",\n", + "print \"%.6f\"%V,\n", + "print \"cubic meter\"\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)Volume of the gas umath.sing ideal gas equation is 0.347114 cubic meter\n", + " ii)Volume of the gas umath.sing Vander Waals equation is" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " 0.322787 cubic meter\n", + " iii)Volume of the gas umath.sing Z chart is 0.336700 cubic meter\n", + " iv)Volume of the gas umath.sing molar polarisation method is 0.338468 cubic meter\n", + " v)Volume of the gas calculated by experimental Z value is 0.341838 cubic meter\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.6 Page No : 35" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "import math\n", + "import matplotlib.pyplot as plt\n", + "import numpy\n", + "\n", + "#Given\n", + "P1 = 266\n", + "T1 = 473.16#Initial temperature in Kelvin\n", + "T2 = 273.16#Final temperature in Kelvin\n", + "V1 = 80\n", + "V2 = 80 #Initial & final volume in litres\n", + "N1 = (14.28/28)\n", + "N2 = (14.28/28) #Initial and final Kg moles are equal\n", + "Tc = 126 #Critical temperature of N2 in K\n", + "Pc = 33.5#Critical pressure of N2 in atm\n", + "\n", + "#To calculate the final pressure achieved\n", + "#(i)Umath.sing ideal gas law\n", + "p2 = (P1*V1*N2*T2)/(V2*N1*T1);\n", + "print \"i)Final pressure of N2 using ideal gas law is \",\n", + "print \"%.6f\"%p2,\n", + "print \"atm\"\n", + "\n", + "#(ii)Umath.sing generalized Z chart\n", + "Tr1 = T1/Tc#reduced initial temp in k\n", + "Pr1 = P1/Pc #reduced initial press in K\n", + "#From the Z-chart compressibility factor coressponding to the above Tr1 &Pr1 is\n", + "Z1 = 1.07\n", + "P2 = [125,135,150]\n", + "Z2 = [0.95, 0.96, 0.98]\n", + "F = [];\n", + "for i in range(0,3):\n", + " F.append((P2[i]/(Z2[i]*T2))-(P1/(Z1*T1)));\n", + "plt.plot(P2,F)\n", + "plt.ylabel(\"F\")\n", + "plt.xlabel(\"P2\")\n", + "plt.title(\"P2 vs F\")\n", + "plt.show()\n", + "P3 = numpy.interp(0,F,P2);\n", + "print \" ii)Final pressure of N2 from Z chart is \",\n", + "print \"%.6f\"%P3,\n", + "print \" atm\"\n", + "#(iii)Umath.sing Pseudo reduced density chart\n", + "R = 0.082 #gas consmath.tant\n", + "v = V1/N1 #Volume per moles of nitrogen in m**3/Kg mole\n", + "Dr = (R*Tc)/(Pc*v)\n", + "Tr2 = T2/Tc #final reduced temp in K\n", + "#From figure A.2.1, reduced pressure coressponding to this Dr and Tr2 is\n", + "Pr2 = 4.1#final reduced pressure in atm\n", + "p2_ = Pr2*Pc\n", + "print \" iii)Final pressure achieved umath.sing Dr chart is \",\n", + "print \"%.6f\"%p2_,\n", + "print \"atm\"\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)Final pressure of N2 using ideal gas law is 153.564460 atm\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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+ "text": [ + "<matplotlib.figure.Figure at 0x1108b2610>" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " ii)Final pressure of N2 from Z chart is 138.489629 atm\n", + " iii)Final pressure achieved umath.sing Dr chart is 137.350000 atm\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.7 Page No : 36" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "n = 1 #n is the Kg mole of methane gas\n", + "T = 298 #T is the consmath.tant temperature in K\n", + "P1 = 1 #P1 is the initial pressure of the system\n", + "P2 = 100 #P2 is the final pressure of the system\n", + "R = 8314.4 #R is the gas consmath.tant in Nm/Kgmole deg K\n", + "\n", + "#To compute the work required\n", + "#(i)Umath.sing ideal gas law\n", + "W = R*T*math.log(P1/float(P2))\n", + "print \"i)Work done by the system if the gas obeys ideal gas law is \",\n", + "print \"%.2e\"%W,\n", + "print \n", + "\n", + "#(ii)Umath.sing Vander Waals' equation\n", + "#Given\n", + "#For methane\n", + "a = 2.32*(10**5) #Vander Wals' consmath.tant a in N/m**2\n", + "b = 0.0428 #Vanderwaals' consmath.tant b in m**3\n", + "#V1 and V2 are evaluated by trial and error umath.sing Vanderwaals' equation as P1 and P2 are known\n", + "V1 = 11.1 #initial volume of the gas in m**3\n", + "V2 = 0.089 #final volume of the gas in m**3\n", + "W = (R*T*math.log((V2-b)/(V1-b)))+(a*((1/V2)-(1/V1)))\n", + "print \" ii)Work done by the system if the gas obeys Vander Waals equation is \",\n", + "print \"%.2e\"%W,\n", + "print \"Nm\"\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)Work done by the system if the gas obeys ideal gas law is -1.14e+07\n", + " ii)Work done by the system if the gas obeys Vander Waals equation is -1.10e+07 Nm\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.8 Page No : 38" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "import math\n", + "import matplotlib.pyplot as plt\n", + "import numpy\n", + "\n", + "#Given\n", + "V = 27*(10**-3) #Volume of the container in m**3\n", + "n = (15/70.91) #n is the Kg moles of chlorine\n", + "T = float(293)#T is the temperature in K\n", + "R = 0.08206\n", + "P = pow(10,(4.39-(1045/293))) #P is the vapour pressure of chlorine\n", + "Pc = 76.1 #Critical pressure of Chlorine\n", + "Tc = float(417) #Critical temperature of Chlorine \n", + "Pr = P/Pc #Reduced pressure of Chlorine\n", + "Tr = T/Tc #Critical temperature of Chlorine\n", + "M = 70.91 #Molecular weight of the Chlorine\n", + "\n", + "#To determine the vapour pressure of chlorine, amount of liquid Cl2 and temperature required\n", + "#(i)Specific volume of liquid Chlorine\n", + "#From figure A.2.2\n", + "Zg = 0.93\n", + "#From figure A.2.6\n", + "Zl = 0.013\n", + "vl = ((Zl*R*T)/P)\n", + "print \"i)Specific volume of liquid Chlorine from compressibility chart is \",\n", + "print \"%.6f\"%vl,\n", + "print \"cubic meter /Kgmole\"\n", + "\n", + "#From Francis relation, taking the consmath.tants from Table 2.3\n", + "D = (1.606-(216*(10**-5)*20)-(28/(200-20)))*10**3 #Density of liq Cl2 in Kg/m**3\n", + "Vl = M/D;\n", + "print \" Specific volume of liquid Chlorine from Francis relation is \",\n", + "print \"%.6f\"%Vl,\n", + "print \"cubic meter /Kgmole\"\n", + "\n", + "#(ii)Amount of liquid Cl2 present in the cylinder\n", + "vg = ((Zg*R*T)/P)\n", + "V1 = V-vg #V1 is the volume of liquid Chlorine\n", + "Vct = 0.027 #volume of the container\n", + "Vg = (0.212-(Vct/vl))/((1/vg)-(1/vl)) #By material balance\n", + "W = ((V-Vg)*70.9)/vl \n", + "print \" ii)Weight of Chlorine at 20deg cel is \",\n", + "print \"%.6f\"%W\n", + "print \"Kg\"\n", + "\n", + "#(iii)Calculation of temperature required to evaporate all the liquid chlorine\n", + "#math.log P' = 4.39 - 1045/T (given)\n", + "#Assume the various temperature\n", + "Ng = 0.212#total Kg moles of gas\n", + "Ta = [413,415,417]\n", + "N = [0,0,0]\n", + "for i in range(0,3):\n", + " Tr = Ta[i]/Tc #reduced temperature in K\n", + " P = pow(10,(4.39-(1045/Ta[i])))\n", + " Pr = P/Pc #reduced pressure in K\n", + "#From the compressibility factor chart,Z values coressponding to the above Tr &Pr are given as\n", + " Z = [0.4,0.328,0.208]\n", + " N[i] = (P*Vct)/(Z[i]*R*Ta[i])\n", + "#end\n", + "\n", + "\n", + "plt.plot(N,Ta)\n", + "plt.ylabel(\"Ta\")\n", + "plt.xlabel(\"N\")\n", + "plt.title(\"Ta vs N\")\n", + "plt.show()\n", + "T1 = numpy.interp(0.212,N,Ta) #in K\n", + "print \" iii)The temperature required to evaporate all the liquid chlorine is \",\n", + "print \"%.6f\"%(T1-273),\n", + "print \"deg celsius\"\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)Specific volume of liquid Chlorine from compressibility chart is 0.012733 cubic meter /Kgmole\n", + " Specific volume of liquid Chlorine from Francis relation is 0.045374 cubic meter /Kgmole\n", + " ii)Weight of Chlorine at 20deg cel is 13.112602\n", + "Kg\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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+ "text": [ + "<matplotlib.figure.Figure at 0x1108b5a50>" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " iii)The temperature required to evaporate all the liquid chlorine is 140.000000 deg celsius\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.9 Page No : 40" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "import numpy\n", + "\n", + "\n", + "#Given\n", + "N1 = 0.7 #Kg mole of CH4\n", + "N2 = 0.3 #Kg mole of N2\n", + "R = 0.08206 #Gas consmath.tant\n", + "T = float(323) #Temperature in Kelvin\n", + "V = 0.04 #Volume in m**3\n", + "a1 = 2.280\n", + "b1 = 0.0428 #Vanderwaals consmath.tants for CH4\n", + "a2 = 1.345\n", + "b2 = 0.0386 #Vanderwaals consmath.tants for N2\n", + "Tc1 = 191\n", + "Pc1 = 45.8 #Critical temperature in K and pressure of CH4 in atm\n", + "Tc2 = 126\n", + "Pc2 = 33.5 #Critical temperature in K and pressure of N2 in atm\n", + "\n", + "#To find Approx Value\n", + "def approx(V,n):\n", + " A = numpy.around([V*pow(10,n)])/pow(10,n)\n", + " return A[0]\n", + "\n", + "#To calculate the pressure exerted by the gas mixture\n", + "#(i)Umath.sing ideal gas law\n", + "P = (N1+N2)*((R*T)/V)\n", + "print \"i) Pressure exerted by the gas mixture umath.sing ideal gas law is \",\n", + "print \"%d\"%P,\n", + "print \"atm\"\n", + "\n", + "#(ii)Umath.sing Vander waal equation\n", + "P1 = ((N1*R*T)/(V-(N1*b1)))-((a1*(N1**2))/(V**2)) #Partial pressure of CH4\n", + "P2 = ((N2*R*T)/(V-(N2*b2)))-((a2*(N2**2))/(V**2)) #Partial pressure of N2\n", + "Pt = P1+P2 \n", + "print \"ii) Pressure exerted by the gas mixture umath.sing Vander waal equation is \",\n", + "print \"%.6f\"%Pt,\n", + "print \"atm\"\n", + "\n", + "#(iii)Umath.sing Zchart and Dalton's law\n", + "Tra = T/Tc1 #reduced temperature of CH4\n", + "Trb = T/Tc2 #reduced temperature of N2\n", + "#Asssume the pressure\n", + "P = [660,732,793,815,831]\n", + "Pa =[]\n", + "Pb = []\n", + "Pra = []\n", + "Prb = []\n", + "for i in range(0,5):\n", + " Pa.append(N1*P[i]) # partial pressure of CH4 for the ith total pressure\n", + " Pb.append(N2*P[i]) # partial pressure of N2 for the ith total pressure\n", + " Pra.append(Pa[i]/Pc1) #reduced pressure of CH4 for the ith total pressure\n", + " Prb.append(Pb[i]/Pc2) #reduced pressure of N2 for the ith total pressure\n", + "#end\n", + "\n", + "#For the above Pr and Tr values compressibility factors from the figure A.2.3 are given as\n", + "Za = [1.154,1.280,1.331,1.370,1.390] #Z values of CH4 \n", + "Zb = [1,1,1,1,1]#Z values of N2\n", + "V3 = 0.0421\n", + "for i in range(0,5):\n", + " Pa[i] = Za[i]*N1*((R*T)/V);#partial pressure of CH4 coressponding to the ith total presure\n", + " Pb[i] = Zb[i]*N2*((R*T)/V);#partial pressure of N2 coressponding to the ith total pressure\n", + " Pt = Pa[i]+Pb[i] #total pressure of the gas mixture\n", + " if Pt-P[i] < 15:\n", + " print \"iii) pressure exerted by the gas mixture umath.sing Z chart and Dalton Law is \",\n", + " print \"%d\"%Pt,\n", + " print \"atm\"\n", + " #end\n", + "#end\n", + "\n", + "#(iv)Umath.sing Amagat's law and Z chart\n", + "P = [1000,1200,1500,1700]\n", + "for i in range(0,4):\n", + " Pra[i] = P[i]/Pc1\n", + " Prb[i] = P[i]/Pc2\n", + "#end\n", + "#For the above Pr and Tr values compressibility factors from the figure A.2.3 are given as\n", + "Za = [1.87,2.14,2.52,2.77]\n", + "Zb = [1.80,2.10,2.37,2.54]\n", + "Va = []\n", + "Vb = []\n", + "V1 = []\n", + "for i in range(0,4):\n", + " Va.append(approx((N1*Za[i]*((R*T)/P[i])),4))\n", + " Vb.append(approx((N2*Zb[i]*((R*T)/P[i])),4))\n", + " V1.append(approx((Va[i]+Vb[i]),4))\n", + " if V1[i]-V <= 0.003:\n", + " print \"iv) Pressure exerted by the gas mixture umath.sing Amagat law and Zchart is \",\n", + " print \"%d\"%P[i],\n", + " print \"atm\"\n", + "#end\n", + "#end\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i) Pressure exerted by the gas mixture umath.sing ideal gas law is 662 atm\n", + "ii) Pressure exerted by the gas mixture umath.sing Vander waal equation is 1353.867785 atm\n", + "iii) pressure exerted by the gas mixture umath.sing Z chart and Dalton Law is 843 atm\n", + "iv) Pressure exerted by the gas mixture umath.sing Amagat law and Zchart is 1700 atm\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.10 Page No : 41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "yN2 = 1.0/4 #mole faction of N2 in the mixture\n", + "yH2 = 3.0/4 #mole fraction of H2 in the mixture\n", + "V = 5.7 #V is the rate at which mixture enters in m**3 in 1 hour\n", + "P = float(600) #P is in atm\n", + "T = float(298) #T is in K\n", + "TcN2 = float(126) #critical temp of N2 in K\n", + "TcH2 = 33.3 #critical temp of H2 in K\n", + "TcNh3 = 406.0 #critical temp of NH3 in K\n", + "PcN2 = 33.5 #critical pressure of N2 in atm\n", + "PcH2 = 12.8 #critical pressure of H2 in atm\n", + "PcNH3 = 111.0 #critical pressure of NH3 in atm\n", + "R = 0.082 #gas consmath.tant\n", + "\n", + "#To calculate the amount of ammonia leaving the reactor and the velocity of gaseous product leaving the reactor\n", + "#(i)Calculation of amount of NH3 leaving the reactor\n", + "Tcm = (TcN2*yN2)+(TcH2*yH2) #critical temperature of the mixture\n", + "Pcm = (PcN2*yN2)+(PcH2*yH2) #critical pressure of the mixture\n", + "Trm = T/Tcm \n", + "Prm = P/Pcm \n", + "#From figure A.2.3\n", + "Zm = 1.57 #compressibility factor of the mixture\n", + "N = (P*V)/(Zm*R*T) #Kg mole of the mixture \n", + "N1 = 0.25*N #Kg mole of N2 in feed\n", + "#N2+3H2 - 2NH3\n", + "W = 2*0.15*N1*17 \n", + "print \"i)Ammonia formed per hour is \",\n", + "print \"%.6f\"%W,\n", + "print \"Kg\"\n", + "\n", + "#(ii)Calculation of velocity\n", + "N1 = 0.25*N-(0.25*N*0.15) #Kg mole of N2 after reactor\n", + "N2 = 0.75*N-(0.75*N*0.15) #Kg mole of H2 after reactor\n", + "N3 = 0.25*N*2*0.15 #Kg mole of NH3 after reactor\n", + "Nt = N1+N2+N3 #total Kg moles after reactor\n", + "y1NH3 = N3/Nt #mole fraction of NH3 after reactor\n", + "y1N2 = N1/Nt #mole fraction of N2 after reactor\n", + "y1H2 = N2/Nt #mole fraction of H2 after reactor\n", + "T1cm = (TcN2*y1N2)+(TcH2*y1H2) \n", + "P1cm = (PcN2*y1N2)+(PcH2*y1H2) \n", + "T1 = 448 #in K\n", + "P1 = 550 #in atm\n", + "T1rm = T1/T1cm \n", + "P1rm = P1/P1cm\n", + "#From Figure A.2.2\n", + "Zm1 = 1.38\n", + "V1 = (Zm1*Nt*R*T1)/P1\n", + "d = 5*(10**-2)#diameter of pipe\n", + "v = V1/((math.pi/4)*(d**2)*3600)\n", + "print \"ii)Velocity in pipe is \",\n", + "print \"%.6f\"%v,\n", + "print \"m/s\"\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)Ammonia formed per hour is 113.659704 Kg\n", + "ii)Velocity in pipe is 1.075261 m/s\n" + ] + } + ], + "prompt_number": 9 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch3.ipynb b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch3.ipynb new file mode 100755 index 00000000..8385d110 --- /dev/null +++ b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch3.ipynb @@ -0,0 +1,562 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:753b89f4a49b587e649e81b025cbd0732ffa221e89b7d46871678d863ea83955" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 3 : First Law of Thermodynamics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.1 Page No : 50" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "W = -((2*745.6*(10**-3)/4.18)*3600) #work added to the system in Kcal/hr\n", + "m = 10.0 #Amount of fluid in math.tank in Kg\n", + "Q = -378.0 #Heat losses from the system in Kcal/hr\n", + "\n", + "#To calculate the change in internal energy\n", + "delE=(Q-W)/m # Change in internal energy in Kcal/hr kg\n", + "print \"Change in Internal energy is \",\n", + "print \"%.6f\"%delE,\n", + "print \"Kcal/hr Kg\"\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Change in Internal energy is 90.628708 Kcal/hr Kg\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.2 Page No : 53" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "n = 1.0 #kg moles of a gas\n", + "Cv = 5.0 #specific heat in Kcal/Kgmole\n", + "delT = 15.0 #increase in temperature in deg celsius\n", + "\n", + "#To calculate the change in internal energy\n", + "Q = n*Cv*delT #heat given to the system in Kcal\n", + "W = 0 #work done\n", + "delE = Q-W #Change in internal energy\n", + "print \"Change in internal energy is \",\n", + "print \"%.6f\"%delE,\n", + "print \"Kcal\"\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Change in internal energy is 75.000000 Kcal\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3 Page No : 55" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "P = 1.0 #consmath.tant pressure throughout the process in atm\n", + "T1 = 273.0 #Initial temperature in K\n", + "T2 = 373.0 #Final temperature in K\n", + "V1 = 0.0#Volume of liquid water or initial volume\n", + "V0 = 22.4 #volume of vapour at smath.radians(numpy.arcmath.tan(ard condition in cubic meter\n", + "Q = 9.7 #Heat of vapourisation in Kcal\n", + "\n", + "#To calculate the work done by the expanding gas and increase in internal energy\n", + "#(i)Calculation of work done\n", + "V2 = 22.4*(T2/T1)*(P)*(10**-3) #Volume of final vapour in cubic meter\n", + "w = P*(V2-V1) #Work done in atm cubic meter\n", + "W = w*(1.03*10**4)/427 #Work done in Kcal\n", + "print \"i)Work done by the expanding gas is \",\n", + "print \"%.6f\"%W,\n", + "print \"Kcal\"\n", + "\n", + "#(ii)Calculation of change in internal energy\n", + "delE = Q-W\n", + "print \" ii)Increase in internal energy is \",\n", + "print \"%.6f\"%delE,\n", + "print \"Kcal\"\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)Work done by the expanding gas is 0.738250 Kcal\n", + " ii)Increase in internal energy is 8.961750 Kcal\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.4 Page No : 58" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "W = 0.0 #work done during the process\n", + "P1 = 1.0 #Initial pressure in atm\n", + "P2 = 10.0 #Final pressure in atm\n", + "#V2 = V1;#Initial & final volume are equal\n", + "Cv = 0.23#specific heat at consmath.tant volume in Kcal/Kg deg K\n", + "#(delQ/delT)=Q\n", + "Q = 1.3 #Rate of heat addition in Kcal/min\n", + "m = 2.5 #Weight of an ideal gas in Kg\n", + "T1 = 298.0 #Initial temperature in Kelvin\n", + "\n", + "#To calculate the time taken for the gas to attain 10 atm\n", + "#Q = m*Cv*(delT/delt)=1.3\n", + "T2 = (P2*T1)/(P1) #Final temperature in Kelvin\n", + "t = ((m*Cv)/1.3)*(T2-T1) #time taken in minutes\n", + "print \"The time taken to attain a pressure of 10 atm is \",\n", + "print \"%.6f\"%(t/60),\n", + "print \"hours\"\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The time taken to attain a pressure of 10 atm is 19.771154 hours\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.6 Page No : 61" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Given\n", + "R = 1.98 #gas consmath.tant in kcal/Kgmole deg K\n", + "T = 293.0 #Temperature in K\n", + "M = 29.0 #Molecular weight of air\n", + "\n", + "#To calculate the flow work per kg of air\n", + "#W=(P*V)=(R*T)\n", + "W = R*T #Flow work in Kcal/Kg mole\n", + "W1 = W/M \n", + "print \"Flow work is %f Kcal/Kg\"%W1\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Flow work is 20.004828 Kcal/Kg\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.7 Page No : 62" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "m = 5000.0 #Amount of steam recived per hour in Kg\n", + "H1 = 666.0 #Specific enthalpy when steam entered in the turbine in Kcal/Kg\n", + "H2 = 540.0 #Specific enthalpy when steam left the turbine in Kcal/Kg\n", + "u1 = 3000/60.0 #velocity at which steam entered in m/sec\n", + "u2 = 600/60.0 #velocity at which steam left in m/sec\n", + "Z1 = 5.0 #height at which steam entered in m\n", + "Z2 = 1.0#height at which steam left in m\n", + "Q = -4000.0 #heat lost in Kcal\n", + "g = 9.81\n", + "\n", + "#To calculate the horsepuwer output of the turbine\n", + "delH = H2-H1#change in enthalpy in Kcal\n", + "delKE = ((u2**2)-(u1**2)/(2*g))/(9.8065*427) #change in kinetic energy in Kcal; 1kgf = 9.8065 N\n", + "delPE = ((Z2-Z1)*g)/(9.8065*427) #change in potential energy in Kcal\n", + "W = -(m*(delH+delKE+delPE))+Q #work delivered in Kcal/hr\n", + "W1 = W*(427/(3600*75.0))#work delivered by turbine in hp\n", + "print \"Work delivered by turbine is %f hp\"%W1\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Work delivered by turbine is 990.133290 hp\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.8 Page No : 63" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "m = 183.0 #rate of water flow in Kg/min\n", + "H1 = 95.0 #enthalpy of storage math.tank 1 in Kcal/Kg\n", + "h = 15.0#height difference between two storage math.tanks in m\n", + "Q = -10100.0 #extraced heat from storage math.tank 1 in a heat exchanger in Kcal/min\n", + "W = -2.0 #work delivered by motor in hp\n", + "\n", + "# To find out the enthalpy of water math.tank2 and the temperature of water in the second math.tank\n", + "delPE = h/427.0 #change in potential energy in Kcal/Kg\n", + "delKE = 0.0 #change in kinetic energy\n", + "W1 = W*(75/427.0) #work delivered by motor in Kcal/sec\n", + "W2 = W1*60.0#work delivered by motor in Kcal/min\n", + "H2 = ((Q+W2)/m)-delKE-delPE+H1#enthalpy of storage math.tank 2 in Kcal/Kg\n", + "print \"The enthalpy of storage tank 2 is %f Kcal/Kg\"%(H2)\n", + "\n", + "#The enthalpy H2=39.66 corresponds to the temperature T according to steam table\n", + "T=40 #Temperature is in deg celsius\n", + "print \" The temperature of water in the second tank is %d deg celsius\"%(T)\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The enthalpy of storage tank 2 is 39.658438 Kcal/Kg\n", + " The temperature of water in the second tank is 40 deg celsius\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.9 Page No : 68" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#To calculate the mass of steam required\n", + "#Given\n", + "m2 = 100.0 #mass of water to be heated\n", + "#From diagram, \n", + "#m3 = m1+m2;..(a)\n", + "#Hs = H1;..(b) math.since throttling is a consmath.tant enthalpy process\n", + "#m3*H3-(m1*H1+m2*H2)=0;..(c) math.since delH=0\n", + "\n", + "#From steam tables, \n", + "Hs = 681.7 #enthalpy of steam at 200 deg cel bleeded at the rate of 5Kgf/(cm**2) in Kcal/Kg\n", + "H2 = 5.03 #enthalpy of liquid water at 5 deg cel\n", + "H3 = 64.98 #enthalpy of liquid water at 65 deg cel\n", + "#from equn (a),(b)&(c);(page no 80)\n", + "m1 = ((H3-H2)/(Hs-H3))*m2 #mass of steam required in Kg (page no 80)\n", + "print \"The mass of steam required to heat 100 Kg of water is %f Kg\"%(m1)\n", + "#end \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The mass of steam required to heat 100 Kg of water is 9.720781 Kg\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.10 Page No : 69" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "V = 0.3 #Volume of the math.tank in m**3\n", + "P1 = 1.0 #Initial pressure of the math.tank in atm\n", + "P2 = 0.0 #Final pressure of the math.tank in atm\n", + "T = 298.0 #Temperature of the math.tank in K\n", + "t = 10.0 #evacuation time in min\n", + "\n", + "#delN=(V/(R*T)*delP)..(a) change in moles as V and T are consmath.tant\n", + "#delW=delN*R*T*lnP..(b)pump work required\n", + "#From (a)&(b),delW=V*delP*lnP\n", + "\n", + "#To calculate the pump work required\n", + "#On doing integration of dW we will get\n", + "\n", + "W = V*(P1-P2);#pump work done in J/sec\n", + "W1=(W*(1.033*10**4))/(75*600.0);\n", + "print \"The pump work required is %f hp\"%(W1);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The pump work required is 0.068867 hp\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.11 Page No : 71" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "H1 = 680.6;#Enthalpy of entering steam at 6Kgf/cm**2 &200 deg cel in Kcal/Kg\n", + "u1 = 60.0;#velocity at which steam entered the nozzle in m/sec\n", + "u2 = 600.0;#velocity at which steam left the nozzle in m/sec\n", + "g = 9.8;\n", + "Hg = 642.8; Hlq = 110.2;#Enthalpy of saturated vapour & saturated liquid at 1.46 Kgf/cm**2 respectively\n", + "\n", + "#To calculate the quality of exit steam\n", + "H2 = H1+((u1**2)-(u2**2))/(2*g*427);#enthalpy of leaving steam in Kcal/Kg\n", + "x = (H2-Hlq)/(Hg-Hlq);\n", + "print \"The quality of exit steam is %f percent\"%(x*100);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The quality of exit steam is 99.101631 percent\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.12 Page No : 73" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "W = 0.0;#pump work\n", + "Mi = 0.0;#chamber is initially evacuated\n", + "M2 = 0.0;#no exist stream\n", + "H1 = 684.2;#enthalpy of steam at 200 deg cel & 3 Kgf/cm**2\n", + "\n", + "#To calculate the internal energy of the steam in the chamber\n", + "#Q=150*m1;.. (a) heat lost from the chamber in Kcal/Kg\n", + "#m1=mf;..(b) mass of steam added from large pipe is equal to steam in chamber\n", + " #H1*M1-Q=Mf*Ef; umath.sing (a)&(b)\n", + "Ef = H1-150;\n", + "print \"The internal energy of steam in chamber is %f Kcal\"%(Ef);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The internal energy of steam in chamber is 534.200000 Kcal\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.13 Page No : 76" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "import numpy\n", + "\n", + "#Given\n", + "#Q=W=delPE=delKE=0;\n", + "#M2=0; no exit stream\n", + "Ti = 288.0;#initial temperature in K\n", + "H = 7*Ti;#enthalpy of air in Kcal/Kgmole\n", + "Ei = 5*Ti;# initial internal energy of air in Kcal/Kgmole\n", + "#Ef=5*Tf;Final internal energy of air in Kcal/Kgmole\n", + "Pi = 0.3;#initial pressure in atm\n", + "V = 0.57;#volume of the math.tank in m**3\n", + "R = 848.0;#gas consmath.tant in mKgf/Kg mole K\n", + "Pf = 1.0;#final prssure in atm\n", + "\n", + "#To calculate the final weight and the final temperature of the air in the math.tank\n", + "Mi = (Pi*V*1.03*10**4)/(R*Ti);#initial quantity of air in math.tank in Kg mole\n", + "#Tf=(Pf*V*1.033*10**4)/(Mf*R)..(a) final temperature,Mf=final quantity of air in math.tank in Kg mole\n", + "#M1=Mf-Mi..(b) M1 is mass of steam added in Kg mole\n", + "#H*M1=(Ef*Mf)-(Ei*Mi)\n", + "#H*M1=((5*Pf*V*1.033*10**4)/(Tf*R))*Tf-(Ei*Mi)...(c)\n", + "A = [[1,-1],[0,-H]];\n", + "B = [Mi,((Ei*Mi)-((5*Pf*V*1.03*10**4)/R))];\n", + "x = numpy.divide(A,B)\n", + "\n", + "Mf = x[0][0];\n", + "print \"The final weight of air in the tank is %f Kg\"%Mf;\n", + "\n", + "Tf = (Pf*V*1.03*10**4)/(Mf*R);\n", + "print \" The final temperature of air in the tank is %f K\"%(Tf);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The final weight of air in the tank is 138.661216 Kg\n", + " The final temperature of air in the tank is 0.049930 K\n" + ] + } + ], + "prompt_number": 17 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch4.ipynb b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch4.ipynb new file mode 100755 index 00000000..0e734b53 --- /dev/null +++ b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch4.ipynb @@ -0,0 +1,415 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:cba11d6ad27a555b5f3aecc639d6f99831950fba74d42eb631521eded4620d06" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 4 : Second Law of Thermodynamics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.1 Page No : 84" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "Q1 = 250.0;#Heat absorbed in Kcal\n", + "T1 = (260+273.0);#Temperature at which engine absorbs heat\n", + "T0 = (40+273.0);#Temperature at which engine discards heat\n", + "#To Calculate work output, heat rejected, entropy change of system,surronding & total change in entropy and the efficiency of the heat engine\n", + "\n", + "#(i)Calculation of work output\n", + "W = (Q1*((T1-T0)/T1));#Work done umath.sing equations 4.7 & 4.9 given on page no 98\n", + "print \"i)The work output of the heat engine is %f Kcal\"%(W);\n", + "\n", + "#(ii)Calculation of heat rejected\n", + "Q2 = (Q1*T0)/T1;\n", + "print \" ii)The heat rejected is %f Kcal\"%(Q2);\n", + "\n", + "#(iii)Calculation of entropy\n", + "del_S1 = -(Q1/T1);#Change in the entropy of source in Kcal/Kg K\n", + "del_S2 = Q2/T0;#Change in the entropy of math.sink in Kcal/Kg K\n", + "del_St = del_S1+del_S2;#Total change in entropy in Kcal/Kg K\n", + "print \" iii)Total change in entropy is %d confirming that the process is reversible\"%(del_St);\n", + "\n", + "#(iv)Calculation of efficiency\n", + "n = (W/Q1)*100;\n", + "print \" iv)The efficiency of the heat engine is %f percent\"%(n);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)The work output of the heat engine is 103.189493 Kcal\n", + " ii)The heat rejected is 146.810507 Kcal\n", + " iii)Total change in entropy is 0 confirming that the process is reversible\n", + " iv)The efficiency of the heat engine is 41.275797 percent\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.2 Page No : 89" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "T1 = 373.0;#Temperature of the saturated steam in K\n", + "T2 = 298.0;#Temperature of the saturated water in K\n", + "#To calculate the total change in entropy and hence determine the reversibility of the process\n", + "\n", + "#del_H = del_Q+(V*del_P)\n", + "#del_H =del_Q; math.since it is a consmath.tant pressure process\n", + "\n", + "#From steam table,\n", + "#enthalpy of saturated steam at 373K is\n", + "H1 = 6348.5;# in Kcal/Kg\n", + "#enthalpy of saturated liquid water at 373K is\n", + "H2 = 99.15;#in Kcal/Kg\n", + "Q = H2-H1;#heat rejected in Kcal/Kg\n", + "del_S1 = Q/T1;#change in entropy of the system in Kcal/Kg K\n", + "del_S2 = Q/T2;#change in entropy of the surronding in Kcal/Kg K\n", + "del_St = del_S1+del_S2;#total change in the entropy in Kcal/Kg K\n", + "if(del_St == 0):\n", + " print \"Process is reversible\";\n", + "else:\n", + " print \"Process is irreversible\";\n", + "#end\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Process is irreversible\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.3 Page No : 91" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "Cp = 0.09;#specific heat of metal block in Kcal/Kg K\n", + "m = 10.0;#mass of metal block in Kg\n", + "T1 = 323.0;#initial temperature of the block in K\n", + "T2 = 298.0;#final temperature of the block in K\n", + "#consmath.tant pressure process\n", + "#To find out entropy change of block,air and total entropy change\n", + "\n", + "#(i)To calculate the entropy change of block\n", + "del_S1 = m*Cp*math.log(T2/T1);\n", + "print \"i)Entropy change of block is %f Kcal/Kg K\"%(del_S1);\n", + "\n", + "#(ii)To calculate the entropy change of air\n", + "Q = m*Cp*(T1-T2);#heat absorbed by air = heat rejected by block in Kcal\n", + "del_S2 = (Q/T2);\n", + "print \" ii)Entropy change of air is %f Kcal/Kg K\"%(del_S2);\n", + "\n", + "#(iii)To calculate the total entropy change\n", + "del_St = del_S1+del_S2;\n", + "print \" iii)Total entropy change is %f Kcal/Kg K\"%(del_St);\n", + "if(del_St == 0):\n", + " print \" Process is reversible\";\n", + "else:\n", + " print \" Process is irreversible\";\n", + "#end\n", + "#end \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)Entropy change of block is -0.072503 Kcal/Kg K\n", + " ii)Entropy change of air is 0.075503 Kcal/Kg K\n", + " iii)Total entropy change is 0.003000 Kcal/Kg K\n", + " Process is irreversible\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.4 Page No : 94" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "m1 = 10 #mass of metal block in Kg\n", + "m2 = 50 #mass of water in Kg\n", + "Cp1 = 0.09 #Specific heat of metal block in Kcal/Kg K\n", + "Cp2 = 1 #Specific heat of water in Kcal/Kg K\n", + "T1 = 50 #Initial temperature of block in deg celsius\n", + "T2 = 25 #Final temperature of block in deg celsius\n", + "\n", + "#To calculate the total change in entropy\n", + "#Heat lost by block = Heat gained by water\n", + "Tf = ((m1*Cp1*T1)+(m2*Cp2*T2))/((m1*Cp1)+(m2*Cp2)) #final temperature of water in deg celsius\n", + "Tf1 = Tf+273.16 #final temperature in K\n", + "del_S1 = m1*Cp1*math.log(Tf1/(T1+273)) #change in entropy of the block in Kcal/K\n", + "del_S2 = m2*Cp2*math.log(Tf1/(T2+273)) #change in entropy of the block in Kcal/K\n", + "del_St = del_S1+del_S2\n", + "print \"The total change entropy is \",\n", + "print \"%.6f\" %del_St,\n", + "print \"Kcal/K\"\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The total change entropy is 0.030226 Kcal/K\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.5 Page No : 96" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "#Air at 20 deg celsius\n", + "#P1 = 250;initial pressure in atm\n", + "#P2 = 10;final pressure after throttling in atm\n", + "\n", + "#To calculate the entropy change\n", + "#According to the given conditions from figure4.5(page no 103)\n", + "S1 = -0.38;#initial entropy in Kcal/Kg K\n", + "S2 = -0.15;#final entroy in Kcal/Kg K\n", + "del_S = S2-S1;\n", + "print \"Change in entropy for the throttling process is %f Kcal/Kg K\"%(del_S);\n", + "#From figure 4.6(page no 104), the final temperature is -10 deg celsius\n", + "#end \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Change in entropy for the throttling process is 0.230000 Kcal/Kg K\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.7 Page No : 101" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "#Basis: 1 hour\n", + "m = 10.0;#mass of air in Kg\n", + "T = 293.0;#Consmath.tant temperature throughout the process in K\n", + "#P1 = 1;#Initial pressure in atm\n", + "#P2 = 30;#Final pressure in atm\n", + "#According to the given data and umath.sing the graph or figure A.2.7 given in page no 105\n", + "S1 = 0.02;#Initial entropy in Kcal/Kg\n", + "S2 = -0.23;#Final entropy in Kcal/Kg\n", + "H1 = 5.0;#Initial enthalpy in Kcal/Kg\n", + "H2 = 3.0;#Final enthalpy in Kcal/Kg\n", + "\n", + "W = -((H2-H1)+T*(S2-S1))*m*(427/(3600*75.0));\n", + "print \"The horse power of the compressor is %f hp\"%(W);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The horse power of the compressor is 1.190065 hp\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.8 Page No : 104" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "#Basis: 1 Kg of steam\n", + "#P1 = 30;Intial pressure in Kgf/cm**2\n", + "#P2 = 3;Final pressure in Kgf/cm**2\n", + "#T = 300;#Operating temperature\n", + "#From figure A.2.8, \n", + "H1 = 715.0;#Initial enthalpy of steam in Kcal/Kg\n", + "H2 = 625.0;#Final enthalpy of steam in Kcal/Kg\n", + "S1 = 1.56;#Initial entropy of steam in Kcal/Kg K\n", + "S2 = 1.61;#Final entropy of steam in Kcal/Kg K\n", + "Q = -1.0;#heat loss in Kcal/Kg\n", + "To = 298;#The lowest surronding temperature in K\n", + "\n", + "#To calculate the effectiveness of the process\n", + "W = (-(H2-H1)+Q);#Actual work output by the turbine in Kcal\n", + "#The maximum or available work can be calculated from equation 4.14\n", + "del_B = -((H2-H1)-(To*(S2-S1)));# Maximum work that can be obtained in Kcal\n", + "E = (W/del_B)*100.0;\n", + "print \"The effectiveness of the process is %f percent\"%(E);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The effectiveness of the process is 84.842707 percent\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.9 Page No : 108" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "m = 1.0;#mass of liquid water in Kg\n", + "T1 = 1350.0;#initial temperature in deg celsius\n", + "T2 = 400.0;#final temperature in deg celsius\n", + "Cp = 1.0;#Specific heat of water in Kcal/Kg K\n", + "Cpg = 0.2;#Specific heat of combustion gases in Kcal/Kg K\n", + "Hv = 468.35;#Heat of vapourisation at 14 Kgf/cm**2 and 194.16 deg celsius in Kcak/Kg\n", + "To = 298.0;#Surronding temperature\n", + "Tb = 194.16+273;#Boiling point of liquid water\n", + "\n", + "#To Calculate the maximum work obtained and the entropy change\n", + "#(i)Calculation of maximum work\n", + "#Q = del_H = m*Cp*(T2-T1); gas can be assumed to cool at consmath.tant pressure\n", + "#From equation 4.14 (page no 110)\n", + "del_B = -((m*Cpg*(T2-T1))-(To*m*Cp*math.log((T2+273)/(T1+273))));\n", + "print \"i)The maximum work that can be obtained is %f Kcal/Kg of gas\"%(del_B);\n", + "\n", + "#(ii)To Calculate the change in entropy\n", + "del_S =(m*Cp*math.log(Tb/To))+((m*Hv)/Tb);\n", + "print \"ii)The entropy change per Kg of water is %f\"%(del_S);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)The maximum work that can be obtained is -72.325299 Kcal/Kg of gas\n", + "ii)The entropy change per Kg of water is 1.452126\n" + ] + } + ], + "prompt_number": 7 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch5.ipynb b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch5.ipynb new file mode 100755 index 00000000..d3517ed7 --- /dev/null +++ b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch5.ipynb @@ -0,0 +1,82 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:3e5d8c28d726b79a2c86b3cf54d87d3c39d0c9ab36d48d833fe93313ec8fd001" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 5 : Thermodynamic Potentials and Maxwell Relation" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.1 Page No : 119" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "T = 293.0;#Consmath.tant temperature in K\n", + "w_NH3 = 20/100.0;#weight of NH3 in an aqueous solution in Kg\n", + "w_H2O = 80/100.0;#weight of H2O in an aqueous solution in Kg\n", + "V = 40.0;#feed rate in Kg/min\n", + "M_NH3 = 17.0;#Molecular weight of NH3\n", + "M_H2O = 18.0;#Molecular weight of H20\n", + "R = 1.98;#gas consmath.tant in Kcl/Kg mole K\n", + "V_s = 62.0;#Rate of heating steam in Kg/min\n", + "P1_H2O = 11.6;#Vapour pressure of water in feed in mm Hg\n", + "P2_H2O = 17.5;#Vapour pressure of pure water in mm Hg\n", + "P1_NH3 = 227.0;#Vapour pressure of NH3 in feed in mm Hg\n", + "P2_NH3 = 6350.0;#Vapor pressure of pure NH3 in mm Hg\n", + "#From steam tables:\n", + "Hs = 666.4;#Enthalpy of steam at 160 deg celsius & 2Kgf/cm**2 in Kcal /Kg \n", + "Ss = 1.75;#Entropy of steam at 160 deg celsius & 2Kgf/cm**2V in Kcal/Kg K\n", + "Hl = 20.03;#Enthalpy of liquid water at 20 deg celsius in Kcal/Kg\n", + "Sl = 0.0612;#Entropy of liquid water at 20 deg celsius in Kcal/Kg K\n", + "\n", + "#To Calculate the efficiency of the separation process \n", + "#Material Balance:\n", + "n_NH3 = (V*w_NH3)/M_NH3;#Kg moles of NH3 in feed(tops)\n", + "n_H2O = (V*w_H2O)/M_H2O;#Kg moles of H20 in feed(bottoms)\n", + "#del_F = del_F_NH3 +del_F_H2O;\n", + "del_F = (R*T*n_NH3*math.log(P2_NH3/P1_NH3))+(R*T*n_H2O*math.log(P2_H2O/P1_H2O));#Theoretical minimum work done in Kcal\n", + "#The available energy of the steam can be calculated from equation 4.14(page no 110)\n", + "del_B = -V_s*((Hl-Hs)-T*(Sl-Ss));#Available energy of the steam in Kcal\n", + "E = (del_F/del_B)*100;\n", + "print \"The efficiency of the separation process is %f percent\"%(E);\n", + "#end\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The efficiency of the separation process is 14.192424 percent\n" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch7.ipynb b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch7.ipynb new file mode 100755 index 00000000..d75a5a28 --- /dev/null +++ b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch7.ipynb @@ -0,0 +1,291 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:53f38691736e17bfc88c47ccfeb3fb1c72de31c9b0bf6a81cf8dd2ed8dca8405" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 7 : Ideal Gases" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.3 Page No : 125" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "from scipy.integrate import quad\n", + "\n", + "#Given\n", + "P1 = 15.0;#initial pressure in Kgf/cm**2\n", + "P2 = 1.0;#final pressure in Kgf/cm**2\n", + "V1 = 0.012;#initial volume in m**3\n", + "V2 = 0.06;#final volume in m**3\n", + "T1 = 420.0;#initial temperature in K\n", + "M = 28.0;#molecular weight of the gas\n", + "Cp = 0.25;#specific heat at consmath.tant pressure in Kcal/Kg K\n", + "R = 1.98;#gas consmath.tant in Kcal/Kg mole K\n", + "R2 = 848.0;#gas consmath.tant in mKgf/Kgmole K\n", + "#Cv = a+0.0005*T1; Specific heat at consmath.tant volume\n", + "\n", + "#To Calculate the final temperature of the ideal gas, work done in an open and closed system,internal energy change for the process\n", + "#(a)Calculation of final temperature\n", + "#Umath.sing ideal gas law:(P*V)/(R*T)\n", + "T2 = (P2*V2*T1)/(P1*V1);\n", + "print \"a)The final temperature is %d K\"%(T2);\n", + "\n", + "#(b)Calculation of work in an open and closed system\n", + "#From equation 7.22(page no 147): P1*(V1**n)=P2*(V2**n)\n", + "n = (math.log(P2/P1))/(math.log(V1/V2));\n", + "#From equation 7.25(page no 149)\n", + "W = ((P1*V1)-(P2*V2))/(n-1)*10**4;#work in mKgf\n", + "W1 = W/427;#Work in Kcal\n", + "print \" b)The work in a closed system is %f Kcal\"%(W1);\n", + "Ws = n*W1;#from equation 7.28(page no 149)\n", + "print \" The work in an open system is %f Kcal\"%(Ws);\n", + "\n", + "#(c)Calculation of internal energy change\n", + "R1 = R/M;#gas consmath.tant in Kcal/Kg\n", + "Cv = Cp-R1;#specific heat at consmath.tant volume in Kcal/Kg K\n", + "a = Cv-(0.0005*T1);\n", + "m = (P1*10**4*V1*M)/(R2*T1);#mass of gas in Kg\n", + "def y(T):\n", + " val = m*(a+(0.0005*T));\n", + " return val\n", + "del_E = quad(y,T1,T2)[0];#internal energy change in Kcal/Kg\n", + "del_E1 = M*del_E;#internal energy change in Kcal/Kgmole\n", + "print \" c)The internal energy change for the process is %f Kcal/Kgmole\"%(del_E1);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a)The final temperature is 140 K\n", + " b)The work in a closed system is 4.117022 Kcal\n", + " The work in an open system is 6.927326 Kcal\n", + " c)The internal energy change for the process is -121.245283 Kcal/Kgmole\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.5 Page No : 126" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "import numpy\n", + "\n", + "#Given\n", + "P1 = 1.0;#Initial pressure of air in atm\n", + "T1 = 15+273.0;#Initial temperature in K\n", + "P2 = 5.0;#Final pressure of air in atm\n", + "T2 = 15+273.0;#Final temperature in K\n", + "Cv = 5.0;#specific heat of air at consmath.tant volume in Kcal/Kgmole K\n", + "Cp = 7.0;#specific heat of air at consmath.tant pressure in Kcal/Kgmole K\n", + "R = 0.082;#gas consmath.tant in atm-m**3/Kgmole K\n", + "R1 = 2.0;#gas consmath.tant in Kcal/Kgmole K\n", + "#From the P-V diagram given in page no 155:\n", + "#Line 12 represents Isothermal process\n", + "#Line b2,c2 & 1a represent Isometric process\n", + "#Line a2 & 1c represent Isobaric process\n", + "#Line 1b reprsent Adiabatic process\n", + "\n", + "#To find Approx Value\n", + "def approx(V,n):\n", + " A=numpy.around([V*10**n])[0]/10**n;\n", + " return A\n", + "\n", + "#To Compute del_H, del_E, Q, W, del_S for the processes given above.\n", + "#To indicate the quantities that are state functions \n", + "#To verify that the work required in an isothermal process is less than that in an adiabatic process\n", + "\n", + "#Basis:1 Kgmole of air\n", + "V1 = (R*T1)/P1;#Initial volume in cubic meter\n", + "V2 = (R*T2)/P2;#Final volume in cubic meter\n", + "\n", + "#(i)Isothermal path 12\n", + "#Equations 7.7 to 7.9 will be used (page no 145)\n", + "del_E_12 = Cv*(T2-T1);\n", + "del_H_12 = Cp*(T2-T1);\n", + "W_12 = R1*T1*math.log(P1/P2);\n", + "Q_12 = W_12;\n", + "del_S_12 = approx((R1*math.log(P1/P2)),4);\n", + "print \"i)For isothermal process or path 12 change in internal energy is %f\"%(del_E_12);\n", + "print \" For isothermal process or path 12 change in enthalpy is %f\"%(del_H_12);\n", + "print \" For isothermal process or path 12 heat released is %f Kcal\"%(Q_12);\n", + "print \" For isothermal process or path 12 work is %f Kcal\"%(W_12);\n", + "print \" For isothermal process or path 12 change in entropy is %f Kcal/Kgmole K\"%(del_S_12);\n", + "\n", + "#(ii)Path 1a2 = 1a(isometric)+a2(isobaric)\n", + "#Equation 7.1 to 7.6 will be used (page no 144 & 145)\n", + "Pa = P2;\n", + "Ta = (Pa/P1)*T1;#in K\n", + "Q_1a = Cv*(Ta-T1);\n", + "del_E_1a = Q_1a;\n", + "del_H_1a = Cp*(Ta-T1);\n", + "W_1a = 0;# Consmath.tant volume process\n", + "del_E_a2 = Cv*(T2-Ta);\n", + "del_H_a2 = Cp*(T2-Ta);\n", + "Q_a2 = del_H_a2;\n", + "W_a2 = P2*(V2-V1)*((10**4*1.03)/427);\n", + "del_H_1a2 = del_H_1a+del_H_a2;\n", + "del_E_1a2 = del_E_1a+del_E_a2;\n", + "Q_1a2 = Q_1a+Q_a2;\n", + "W_1a2 = W_1a+W_a2;\n", + "del_S_1a = Cv*math.log(Ta/T1);\n", + "del_S_a2 = Cp*math.log(T2/Ta);\n", + "del_S_1a2 = approx((del_S_1a+del_S_a2),4);\n", + "print \"ii)For path 1a2 change in internal energy is %f\"%(del_E_1a2);\n", + "print \" For path 1a2 change in enthalpy is %f\"%(del_H_1a2);\n", + "print \" For path 1a2 heat released is %f Kcal\"%(Q_1a2);\n", + "print \" For path 1a2 work is %f Kcal\"%(W_1a2);\n", + "print \" For path 1a2 change in entropy is %f Kcal/Kgmole K\"%(del_S_1a2);\n", + "\n", + "#(iii)Path 1b2 = 1b(adiabatic)+b2(isometric)\n", + "#From equation 7.11 (page no 146)\n", + "y = Cp/Cv;\n", + "Tb = T1*((V1/V2))**(y-1);\n", + "#From equation 7.1 to 7.3,7.10 & 7.21,(page no 144,146,147)\n", + "Q_1b = 0;#adiabatic process\n", + "del_E_1b = Cv*(Tb-T1);\n", + "del_H_1b = Cp*(Tb-T1);\n", + "W_1b = -1*del_E_1b;\n", + "Q_b2 = Cv*(T1-Tb);\n", + "del_H_b2 = Cp*(T1-Tb);\n", + "W_b2 = 0;#consmath.tant volume prcess\n", + "del_E_b2 = Cv*(T2-Tb);\n", + "del_H_1b2 = del_H_1b+del_H_b2;\n", + "del_E_1b2 = del_E_1b+del_E_b2;\n", + "W_1b2 = W_1b+W_b2;\n", + "Q_1b2 = Q_1b+Q_b2;\n", + "del_S_1b2 = approx((Cv*math.log(T1/Tb)),4);\n", + "print \"iii)For path 1b2 change in internal energy is %f\"%(del_E_1b2);\n", + "print \" For path 1b2 change in enthalpy is %f\"%(del_H_1b2);\n", + "print \" For path 1b2 heat released is %f Kcal\"%(Q_1b2);\n", + "print \" For path 1b2 work is %f Kcal\"%(W_1b2);\n", + "print \" For path 1b2 change in entropy is %f Kcal/Kgmole K\"%(del_S_1b2);\n", + "\n", + "#(iv)Path 1c2 = 1c(isobaric)+c2(isometric);\n", + "Pc = P1;\n", + "Vc = V2;\n", + "Tc = (Pc/P2)*T2;\n", + "del_E_1c = Cv*(Tc-T1);\n", + "Q_1c = Cp*(Tc-T1);\n", + "del_H_1c = Q_1c;\n", + "W_1c = P1*(Vc-V1)*((10**4*1.03)/427);\n", + "del_E_c2 = Cv*(T2-Tc);\n", + "Q_c2 = del_E_c2;\n", + "del_H_c2 = Cp*(T2-Tc);\n", + "W_c2 = 0.0;#consmath.tant volume process\n", + "del_E_1c2 = del_E_1c+del_E_c2;\n", + "del_H_1c2 = del_H_1c+del_H_c2;\n", + "Q_1c2 = Q_1c+Q_c2;\n", + "W_1c2 = W_1c+W_c2;\n", + "del_S_1c = Cp*math.log(Tc/T1);\n", + "del_S_c2 = Cv*math.log(T2/Tc);\n", + "del_S_1c2 = approx((del_S_1c+del_S_c2),4);\n", + "print \"iv)For path 1c2 change in internal energy is %f\"%(del_E_1c2);\n", + "print \" For path 1c2 change in enthalpy is %f\"%(del_H_1c2);\n", + "print \" For path 1c2 heat released is %f Kcal\"%(Q_1c2);\n", + "print \" For path 1c2 work is %f Kcal\"%(W_1c2);\n", + "print \" For path 1c2 change in entropy is %f Kcal/Kgmole K\"%(del_S_1c2);\n", + "\n", + "#Determination of state & path functions\n", + "if((del_E_12 == del_E_1a2)&(del_E_12 == del_E_1b2)&(del_E_12 == del_E_1c2)):\n", + " print \" del_E is a state function\";\n", + "else:\n", + " print \" del_E is a path function\";\n", + "if((del_H_12 == del_H_1a2)&(del_H_12 == del_H_1b2)&(del_H_12 == del_H_1c2)):\n", + " print \" del_H is a state function\";\n", + "else:\n", + " print \" del_H is a path function\";\n", + "if(del_S_12 == del_S_1a2)&(del_S_12 == del_S_1b2)&(del_S_12 == del_S_1c2):\n", + " print \" del_S is a state function\";\n", + "else:\n", + " print \" del_S is a path function\";\n", + "if((Q_12 == Q_1a2)&(Q_12 == Q_1b2)&(Q_12 == Q_1c2)):\n", + " print \" Q is a state function\";\n", + "else:\n", + " print \" Q is a path function\";\n", + "if((W_12 == W_1a2)&(W_12 == W_1b2)&(W_12 == W_1c2)):\n", + " print \" W is a state function\";\n", + "else:\n", + " print \" W is a path function\";\n", + "\n", + "#Comparison of work required by isothermal & adiabatic process\n", + "if(-(W_12)<-(W_1b2)):\n", + " print \" Work required by isothermal process is less than the work required by an adiabatic process\";\n", + "else:\n", + " print \" Statement is incorrect\";\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)For isothermal process or path 12 change in internal energy is 0.000000\n", + " For isothermal process or path 12 change in enthalpy is 0.000000\n", + " For isothermal process or path 12 heat released is -927.036238 Kcal\n", + " For isothermal process or path 12 work is -927.036238 Kcal\n", + " For isothermal process or path 12 change in entropy is -3.218900 Kcal/Kgmole K\n", + "ii)For path 1a2 change in internal energy is 0.000000\n", + " For path 1a2 change in enthalpy is 0.000000\n", + " For path 1a2 heat released is -2304.000000 Kcal\n", + " For path 1a2 work is -2278.639813 Kcal\n", + " For path 1a2 change in entropy is -3.218900 Kcal/Kgmole K\n", + "iii)For path 1b2 change in internal energy is 0.000000\n", + " For path 1b2 change in enthalpy is 0.000000\n", + " For path 1b2 heat released is -1301.261672 Kcal\n", + " For path 1b2 work is -1301.261672 Kcal\n", + " For path 1b2 change in entropy is -3.218900 Kcal/Kgmole K\n", + "iv)For path 1c2 change in internal energy is 0.000000\n", + " For path 1c2 change in enthalpy is 0.000000\n", + " For path 1c2 heat released is -460.800000 Kcal\n", + " For path 1c2 work is -455.727963 Kcal\n", + " For path 1c2 change in entropy is -3.218900 Kcal/Kgmole K\n", + " del_E is a state function\n", + " del_H is a state function\n", + " del_S is a state function\n", + " Q is a path function\n", + " W is a path function\n", + " Work required by isothermal process is less than the work required by an adiabatic process\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch8.ipynb b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch8.ipynb new file mode 100755 index 00000000..1c35275d --- /dev/null +++ b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch8.ipynb @@ -0,0 +1,86 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:234a7cd4eb239b4fead8bceff4c3350a039b4673e5cc1a175a4862ee9b56966f" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 8 : Third Law of Thermodynamics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.1 Page No : 138" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "\n", + "import math\n", + "import matplotlib.pyplot as plt\n", + "import numpy\n", + "\n", + "#Given\n", + "C_ps = 0.1;#Molal heat capacity of copper at 20 K\n", + "Ti = 0.0;#Initial temperature in K\n", + "Tf = 20.0;#melting point in K\n", + "Tb = 300.0;#boiling point in K\n", + "\n", + "#To calculate the absolute entropy of copper at 300 K\n", + "#From equation 8.4(page no 164)\n", + "a = C_ps/(Tf**3);# a is the charateristic consmath.tant\n", + "C_p = [0.1,0.80,1.94,3.0,3.9,5.0];\n", + "#T1 = math.log(T);\n", + "T1 = [1.301,1.6021,1.7782,1.9031,2.000,2.1761];\n", + "plt.plot(T1,C_p)\n", + "plt.title(\"C_p vs T1\")\n", + "plt.xlabel(\"T1\")\n", + "plt.ylabel(\"Cp\")\n", + "plt.show()\n", + "# Area under the curve is given as\n", + "A = 7.82;\n", + "#From equation 8.5(page no 164)\n", + "S = (a*((Tf**3)/3))+A;\n", + "print \"The absolute entropy of copper is %f Kcal/Kgmole\"%(S);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "metadata": {}, + "output_type": "display_data", + "png": 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+ "text": [ + "<matplotlib.figure.Figure at 0x102f267d0>" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The absolute entropy of copper is 7.853333 Kcal/Kgmole\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch9.ipynb b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch9.ipynb new file mode 100755 index 00000000..cea268a9 --- /dev/null +++ b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch9.ipynb @@ -0,0 +1,290 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:afa9dce93689170d63e47c0994dac86a6544b2fa2de4c3b3095b5b07cd969902" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 9 : Fluid Flow in Pipes and Nozzles" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.1 Page No : 147" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "R = 848.0;#gas consmath.tant in m Kgf/Kgmole K\n", + "M = 29.0;#molecular weight of air\n", + "g = 9.81;\n", + "T1 = 90+273.0;#initial temperature in K\n", + "y = 1.4;#gamma = Cp/Cv\n", + "W = 800/3600.0;#Mass rate of air in Kg/sec\n", + "P1 = 3.5;#initial pressure in atm\n", + "d = 2.5;#diameter of the pipe in cm\n", + "\n", + "#To find out the pressure at the final point\n", + "v1 = (R*T1)/(M*P1*1.033*10**4);#specific volume in cubic meter/Kg\n", + "u1 = (W*v1)/(math.pi*(d**2*(10**-4))/4);#inital velocity in m/sec\n", + "#Assume final temperature as\n", + "T2 = [300,310];\n", + "#Assume specific heat capacity in J/KgK corresponding to the above temperature as\n", + "Cp = [2987.56,2983.56];\n", + "us = []\n", + "u2 = []\n", + "for i in range(0,2):\n", + " us.append((g*y*R*T2[i]/M)**(1/2));#sonic velocity attained in m/sec\n", + " u2.append(((u1**2)-((2*g*Cp[i]/M)*(T2[i]-T1)))**(1/2));#From equation 9.18 & 9.19 (page no 170)\n", + "if us[i]-u2[i] <= 1: \n", + " u2 = u2[i];\n", + " Tnew2 = T2[i];\n", + "v2 = u2*(math.pi/4)*(d**2/10**4)*(1/W);\n", + "P2 = (P1*v1*Tnew2)/(T1*v2);\n", + "print \"The pressure at the final point is %f atm\"%(P2);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The pressure at the final point is 397.261807 atm\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.2 Page No : 151" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "A1 = 0.002 #inlet area in sq meter\n", + "A2 = 0.0005 #throat area in sq meter\n", + "P1 = 1.3*10**4 #inlet pressure in Kgf/sq m\n", + "P2 = 0.7*10**4 #throat pressure in Kgf/sq m\n", + "g = 9.81\n", + "v = 12*10**(-4) #specific volume in cubic m /Kg\n", + "\n", + "#To find out the mass rate of alcohol\n", + "u2 = ((v*2*g*(P1-P2))/(1-((A2/A1)**2)))**(0.5) #throat velocity in m/sec\n", + "W = (u2*A2)/v\n", + "print \"The mass rate of alcohol is \",\n", + "print \"%.4f\" %W,\n", + "print \" Kg/sec\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The mass rate of alcohol is 5.1147 Kg/sec\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.3 Page No : 153" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "import math\n", + "import matplotlib.pyplot as plt\n", + "import numpy\n", + "\n", + "#Given\n", + "P1 = 50.0;#initial pressure in Kgf/sq m\n", + "T1 = 45+273.0;#initial temperature in K\n", + "g = 9.81;\n", + "y = 1.35;#gamma\n", + "R = 848.0;#gas consmath.tant in m Kgf/Kgmole K\n", + "M = 29.0;#molecular weight of air\n", + "d = 1.0;#pipe diameter in cm\n", + "\n", + "#(i)To plot velocity,specific volume,mass velocity against P2/P1\n", + "#(ii)To calculate the critical pressure,critical mass velocity and mass rate of flow\n", + "#(i)Plotting of graph\n", + "V1 = (R*T1)/(M*P1*1.033*10**4);#initial volume of the gas in cubic m/Kg\n", + "#P3 = P2/P1 (say)\n", + "#Assume P3 values as\n", + "P3 = [1.0,0.8,0.6,0.4,0.2,0.1,0.05];\n", + "G = [0,0,0,0,0,0,0];\n", + "u2 = []\n", + "v2 = []\n", + "G = []\n", + "for i in range(0,7):\n", + " u2.append((((2*g*y*R*T1)/((y-1)*M))*(1-(P3[i]**((y-1)/y))))**(1/2.0));#final velocity in m/sec\n", + "for i in range(0,7):\n", + " v2.append(V1/(P3[i]**(1/y)));#final specific volume in cubic meter/Kg\n", + "for i in range(0,7):\n", + " G.append(u2[i]/v2[i]);#Mass velocity in Kg/sq m sec\n", + "\n", + "plt.plot(P3,u2,\"o-\")\n", + "plt.title(\"Velocity vs P2/P1\")\n", + "plt.xlabel(\"P2/P1\")\n", + "plt.ylabel(\"Velocity\")\n", + "plt.show()\n", + "\n", + "plt.plot(P3,G,\"+-\")\n", + "plt.title(\"Mass velocity vs P2/P1\")\n", + "plt.xlabel(\"P2/P1\")\n", + "plt.ylabel(\"Mass Velocity\")\n", + "plt.show()\n", + "\n", + "P_3 = [1.0,0.8,0.6,0.4,0.2,0.1,0];\n", + "plt.plot(P_3,v2,\"*-\")\n", + "plt.title(\"Sp. volume vs P2/P1\")\n", + "plt.xlabel(\"P2/P1\")\n", + "plt.ylabel(\"Specific Volume\")\n", + "plt.show()\n", + "\n", + "#(ii)Calculation of critical pressure,critical mass velocity and mass rate of flow\n", + "#From equation 9.37(page no 181)\n", + "P2 = P1*(2/(y+1))**(y/(y-1));\n", + "print \"The critical pressure is %f atm\"%(P2);\n", + "#From equation a (page no 183)\n", + "u2 = (((2*g*y*R*T1)/((y-1)*M))*(1-((P2/P1)**((y-1)/y))))**(1/2.0);\n", + "print \" The critical velocity is %f m/sec\"%(u2);\n", + "#From equation b (page no 183)\n", + "v2 = ((R*T1)/(M*P1*1.033*10**4))/((P2/P1)**(1.0/y));\n", + "print \" The critical specific volume is %f cubic meter/Kg\"%(v2);\n", + "#From relation c (page no 183)\n", + "Gnew = float(u2)/v2;\n", + "print \" The critical mass velocity is %f Kg/sq meter sec\"%(Gnew);\n", + "W = Gnew*(math.pi/4.0)*(d/(100.0))**2;\n", + "print \" Mass rate of flow through nozzle is %f Kg/sec\"%(W);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ 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+ "text": [ + "<matplotlib.figure.Figure at 0x1045fb150>" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The critical pressure is 26.842561 atm\n", + " The critical velocity is 323.738941 m/sec\n", + " The critical specific volume is 0.028541 cubic meter/Kg\n", + " The critical mass velocity is 11343.108938 Kg/sq meter sec\n", + " Mass rate of flow through nozzle is 0.890886 Kg/sec\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.4 Page No : 156" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#Given\n", + "A1 = 0.1;#Inlet area in sq meter\n", + "u1 = 60.0;#inlet velocity in m/sec\n", + "v1 = 0.185;#inlet specific volume in cubic meter/Kg\n", + "H1 = 715.0;#inlet enthalpy in Kcal/Kg\n", + "H2 = 660.0;#exit enthalpy in Kcal/Kg\n", + "v2 = 0.495;#exit specific volume in cubic meter/Kg\n", + "g = 9.81\n", + "\n", + "#To calculate the area at exit of nozzle and hence decide the type of the nozzle\n", + "#From the first law\n", + "u2 = ((u1**2)-(2*g*(H2-H1)*427))**(1/2);\n", + "W = (u1*A1)/v1;#Mass rate of gas in Kg/sec\n", + "A2 = (W*v2)/u2;#Area at exit of nozzle\n", + "if(A2 < A1):\n", + " print \"The nozzle is convergent\";\n", + "else:\n", + " print \"The nozzle is divergent\";\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The nozzle is divergent\n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +}
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