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diff --git a/Engineering_Thermodynamics/ch16.ipynb b/Engineering_Thermodynamics/ch16.ipynb new file mode 100755 index 00000000..10fb2321 --- /dev/null +++ b/Engineering_Thermodynamics/ch16.ipynb @@ -0,0 +1,554 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:1aee5040b10cb0c8b88b1f2c29303ab365a0d0e9e802f10ac15f3026e0ae4c4a" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 16 : Reactive Systems" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.2 Page No : 657" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables\n", + "eps_e = 0.27; \n", + "P = 1. ;\n", + "\n", + "# Calculation\n", + "K = (4*eps_e**2*P)/(1-eps_e**2);\n", + "P1 = 100./760; \t\t\t# in Pa\n", + "eps_e_1 = math.sqrt((K/P1)/(4+(K/P1)));\n", + "T1 = 318.; T2 = 298.;\n", + "R = 8.3143; K1 = 0.664; K2 = 0.141;\n", + "dH = 2.30*R*((T1*T2)/(T1-T2))*(math.log(K1/K2));\n", + "\n", + "# Results\n", + "print \"K is\",round(K,4),\"atm\"\n", + "print \"epislon is \",round(eps_e_1,3)\n", + "print \"The heat of reaction is\",round(dH,0),\"kJ/kg mol\"\n", + "\n", + "# note : book answer is wrong. please check." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "K is 0.3145 atm\n", + "epislon is 0.612\n", + "The heat of reaction is 140399.0 kJ/kg mol\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.3 Page No : 659" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "v1 = 1.\n", + "v2 = v1; \n", + "v3 = v2; \n", + "v4 = v2;\n", + "e = 0.56; \t\t\t# Degree of reaction\n", + "P = 1.; \t\t\t# Dummy\n", + "T = 1200.\n", + "R = 8.3143;\n", + "\n", + "# Calculation\n", + "x1 = (1-e)/2; x2 = (1-e)/2;\n", + "x3 = e/2; x4 = e/2;\n", + "K = (((x3**v3)*(x4**v4))/((x1**v1)*(x2**v2)))*P**(v3+v4-v1-v2); \t\t\t# Equillibrium consmath.tant\n", + "dG = -R*T*math.log(K);\n", + "\n", + "# Results\n", + "print \"Equillibrium constant is\",round(K,2)\n", + "print \"Gibbs function change is\",round(dG,1),\"J/gmol\"\n", + "\n", + "# note : rounding off error is there." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equillibrium constant is 1.62\n", + "Gibbs function change is -4812.2 J/gmol\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.5 Page No : 662" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "Veo = 1.777; \t\t\t# Ve/Vo\n", + "e = 1-Veo; \t\t\t# Degree of dissociation\n", + "P = 0.124; \t\t\t# in atm\n", + "\n", + "# Calculation\n", + "K = (4*e**2*P)/(1-e**2);\n", + "\n", + "# Results\n", + "print \"The value of equillibrium constant is\",round(K,3),\"atm\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of equillibrium constant is 0.756 atm\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.6 Page No : 663" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "v1 = 1.; \n", + "v2 = 0.; \n", + "v3 = 1.; v4 = 1./2;\n", + "dH = 250560.; e = 3.2e-03;\n", + "R = 8.3143; T = 1900.;\n", + "\n", + "# Calculation\n", + "Cp = ((dH**2)*(1+e/2)*e*(1+e))/(R*T**2*(v1+v2)*(v3+v4));\n", + "\n", + "# Results\n", + "print \"Cp is\",round(Cp,3),\"j/gmol K\"\n", + "\n", + "# note : rounding off error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Cp is 4.484 j/gmol K\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.7 Page No : 663" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "a = 21.89;\n", + "y = 18.5;\n", + "x = 8.9;\n", + "\n", + "# Calculation\n", + "PC = 100.*(x*12.)/((8.9*12)+(18.5*1));\n", + "PH = 100-PC;\n", + "AFR = ((32*a)+(3.76*a*28))/((12*x)+y);\n", + "EAU = (8.8*32)/((21.89*32)-(8.8*32));\n", + "\n", + "# Results\n", + "print \"carbon = \",round(PC,2),\"% and Hydrogen = \",round(PH,2),\"%\"\n", + "print \"Air fuel ratio is\",round(AFR,0)\n", + "print \"Percentage of excess air used is\",round((EAU*100),2),\"%\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "carbon = 85.24 % and Hydrogen = 14.76 %\n", + "Air fuel ratio is 24.0\n", + "Percentage of excess air used is 67.23 %\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.8 Page No : 664" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "hf_co2 = -393522;\n", + "hf_h20 = -285838;\n", + "hf_ch4 = -74874;\n", + "\n", + "# Calculation\n", + "D = hf_co2 + (2*hf_h20);\n", + "QCV = D - (hf_ch4+1);\n", + "\n", + "# Results\n", + "print \"Heat transfer per kg mol of fuel is\",D,\"kJ\"\n", + "print \"Qcv = %.f kJ\"%QCV\n", + "\n", + "# note : rounding off error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer per kg mol of fuel is -965198 kJ\n", + "Qcv = -890325 kJ\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.9 Page No : 664" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Below values are taken fron table 16.4\n", + "\n", + "# Variables\n", + "Hr = -249952.+(18.7*560)+(70*540);\n", + "Hp = 8.*(-393522.+20288)+9*(-241827+16087)+6.25*14171+70*13491;\n", + "Wcv = 150.; \t\t\t# Energy out put from engine in kW\n", + "Qcv = -205.; \t\t\t# Heat transfer from engine in kW\n", + "\n", + "# Calculation\n", + "n = (Wcv-Qcv)*3600/(Hr-Hp);\n", + "\n", + "# Results\n", + "print \"Fuel consumption rate is\",round((n*114),1),\"kg/h\"\n", + "\n", + "# rounding error is there." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Fuel consumption rate is 38.5 kg/h\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.10 Page No : 665" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import interp\n", + "\n", + "# Variables\n", + "Hr1 = -249952.; \t\t\t# For ocmath.tane\n", + "Hp1 = Hr1;\n", + "# Below values are calculated umath.sing value fron table 16.4\n", + "T2 = 1000.; \n", + "Hp2 = -1226577.\n", + "T3 = 1200.;\n", + "Hp3 = 46537.;\n", + "T4 = 1100.;\n", + "Hp4 = -595964.;\n", + "\n", + "# Calculation\n", + "Hp = [Hp2 ,Hp3, Hp4]\n", + "T = [T2 ,T3, T4]\n", + "T1 = interp(Hp1,Hp , T); \t\t\t# Interpolation to find temperature at Hp1\n", + "\n", + "# Results\n", + "print \"the adeabatic flame temperature is\",T1,\"K\"\n", + "\n", + "# note : answer varies because the method - interp gives some different answer. please check." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the adeabatic flame temperature is 1100.0 K\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.11 Page No : 666" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "T0 = 298.;\n", + "Wrev = -23316-3*(-394374)-4*(-228583);\n", + "Wrev_ = Wrev/44.; \t\t\t# in kJ/kg\n", + "Hr = -103847.;\n", + "T = 980.; \t\t\t# Through trial and error\n", + "\n", + "# Calculation\n", + "Sr = 270.019+20*205.142+75.2*191.611;\n", + "Sp = 3*268.194 + 4*231.849 + 15*242.855 + 75.2*227.485;\n", + "IE = Sp-Sr; \t\t\t# Increase in entropy\n", + "I = T0*3699.67/44;\n", + "Si = Wrev_ - I;\n", + "\n", + "# Results\n", + "print \"Reversible work is\",Wrev_,\"kJ/kg\"\n", + "print \"Increase in entropy during combustion is\",round(Sp-Sr,2),\"kj/kg mol K\"\n", + "print \"Irreversibility of the process\",round(I,0),\"kJ/kg\"\n", + "print \"Availability of products of combustion is\",round(Si,1),\"kJ/kg\"\n", + "\n", + "# note : there are rounding off errors. please check." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Reversible work is 47139.5 kJ/kg\n", + "Increase in entropy during combustion is 3699.67 kj/kg mol K\n", + "Irreversibility of the process 25057.0 kJ/kg\n", + "Availability of products of combustion is 22082.6 kJ/kg\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.12 Page No : 667" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables\n", + "T0 = 298.15; P0 = 1; R = 8.3143;\n", + "xn2 = 0.7567\n", + "xo2 = 0.2035\n", + "xh2o = 0.0312\n", + "xco2 = 0.0003;\n", + "\n", + "# Calculation and Results\n", + "# Part (a)\n", + "g_o2 = 0; g_c = 0; g_co2 = -394380; \n", + "A = -g_co2 + R*T0*math.log(xo2/xco2);\n", + "print \"The chemical energy of carbon is\",round(A,0),\"kJ/k mol\"\n", + "\n", + "# Part (b)\n", + "g_h2 = 0; g_h2o_g = -228590;\n", + "B = g_h2 + g_o2/2 - g_h2o_g + R*T0*math.log(xo2**0.5/xh2o);\n", + "print \"The chemical energy of hydrogen is\",round(B,0),\"kJ/k mol\"\n", + "\n", + "# Part (c)\n", + "g_ch4 = -50790;\n", + "C = g_ch4 + 2*g_o2 - g_co2 - 2*g_h2o_g + R*T0*math.log((xo2**2)/(xco2*xh2o));\n", + "print \"The chemical energy of methane is\",round(C,0),\"kJ/k mol\"\n", + "\n", + "# Part (d)\n", + "g_co = -137150;\n", + "D = g_co + g_o2/2 - g_co2 + R*T0*math.log((xo2**0.5)/xco2);\n", + "print \"The chemical energy of Carbonmonoxide is\",round(D,0),\"kJ/k mol\"\n", + "\n", + "# Part (e)\n", + "g_ch3oh = -166240;\n", + "E = g_ch3oh + 1.5*g_o2 - g_co2 - 2*g_h2o_g + R*T0*math.log((xo2**1.5)/(xco2*(xh2o**2)))\n", + "print \"The chemical energy of methanol is\",round(E,0),\"kJ/k mol\"\n", + "\n", + "# Part (f)\n", + "F = R*T0*math.log(1./xn2);\n", + "print \"The chemical energy of nitrogen is\",round(F,1),\"kJ/k mol\"\n", + "\n", + "# Part (g)\n", + "G = R*T0*math.log(1./xo2);\n", + "print \"The chemical energy of Oxygen is\",round(G,0),\"kJ/k mol\"\n", + "\n", + "# Part (h)\n", + "H = R*T0*math.log(1./xco2);\n", + "print \"The chemical energy of carbondioxide is\",round(H,0),\"kJ/k mol\"\n", + "\n", + "# Part (i)\n", + "g_h2o_l = -237180;\n", + "I = g_h2o_l - g_h2o_g + R*T0*math.log(1./xh2o);\n", + "print \"The chemical energy of water is\",round(I,1),\"kJ/k mol\"\n", + "\n", + "\n", + "# note : rounding off error is there. please check." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The chemical energy of carbon is 410542.0 kJ/k mol\n", + "The chemical energy of hydrogen is 235212.0 kJ/k mol\n", + "The chemical energy of methane is 821580.0 kJ/k mol\n", + "The chemical energy of Carbonmonoxide is 275365.0 kJ/k mol\n", + "The chemical energy of methanol is 716699.0 kJ/k mol\n", + "The chemical energy of nitrogen is 691.1 kJ/k mol\n", + "The chemical energy of Oxygen is 3947.0 kJ/k mol\n", + "The chemical energy of carbondioxide is 20108.0 kJ/k mol\n", + "The chemical energy of water is 5.2 kJ/k mol\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.13 Page No : 669" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "b = 8./(0.114+0.029); \t\t\t# By carbon balance\n", + "C = 18./2 \t\t\t# By hydrogen balance\n", + "a = b*0.114 + (b/2)*0.029 + b*0.016 + C/2 ; \t\t\t# By oxygen balance\n", + "Wcv = 1. \t\t\t# Power developed by engine in kW \n", + "\n", + "# Calculation and Results\n", + "n_fuel = (0.57*1)/(3600*114.22);\n", + "Qcv = Wcv-n_fuel*3845872; \t\t\t# 5.33 \n", + "print \"The rate of heat transfer from the engine is\",round(Qcv,2),\"kW\"\n", + "# Part (b)\n", + "ach = 5407843.; \t\t\t# chemical energy of liquid ocmath.tane\n", + "n2 = Wcv/(n_fuel*ach);\n", + "print \"The second law efficiency is\",round((n2*100),1),\"%\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The rate of heat transfer from the engine is -4.33 kW\n", + "The second law efficiency is 13.3 %\n" + ] + } + ], + "prompt_number": 25 + } + ], + "metadata": {} + } + ] +}
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