From 92cca121f959c6616e3da431c1e2d23c4fa5e886 Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Tue, 7 Apr 2015 15:58:05 +0530 Subject: added books --- Thermodynamics/Chapter14.ipynb | 630 +++++++++++++++++++++++++++++++++++++++++ 1 file changed, 630 insertions(+) create mode 100755 Thermodynamics/Chapter14.ipynb (limited to 'Thermodynamics/Chapter14.ipynb') diff --git a/Thermodynamics/Chapter14.ipynb b/Thermodynamics/Chapter14.ipynb new file mode 100755 index 00000000..4bbbb0eb --- /dev/null +++ b/Thermodynamics/Chapter14.ipynb @@ -0,0 +1,630 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:4c4dfb370e475e99da4c5056464d38bcfd8f5f2d436b23af2eaa8055bd431fec" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 14:CHEMICAL REACTIONS AND COMBUSTION" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.1, Page No:644" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "# From the Table 14.1 \n", + "del_hfHCL=92307; # Enthalpy of Heat in kJ/kmol\n", + "del_hfH2O=-241818; # Enthalpy of Heat kJ/kmol\n", + "\n", + "#Calculation\n", + "del_Ho=4*del_hfHCL-2*del_hfH2O; # The standard heat of reaction from enthalpy equation\n", + "\n", + "#Result\n", + "print \"The standard heat of reaction for the process = \",del_Ho,\"kJ (answer mentioned in the textbook is wrong)\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The standard heat of reaction for the process = 852864 kJ (answer mentioned in the textbook is wrong)\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.2, Page No:645" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "del_Ho=5640000; # Heat released during the process\n", + "# From the Table 14.1 \n", + "del_hfO2=-393509; del_hfH2O=-285830; # Enthalpy of Heat in kJ/kmol\n", + "\n", + "#Calculation\n", + "del_hfsucrose=12*del_hfO2+11*del_hfH2O+del_Ho; # The enthalpy formation of sucrose\n", + "\n", + "#Result\n", + "print \"The enthalpy formation of sucrose = \",del_hfsucrose,\"kJ/kmol (answer mentioned in the textbook is wrong)\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The enthalpy formation of sucrose = -2226238 kJ/kmol (answer mentioned in the textbook is wrong)\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.3, Page No:649" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "# (a).Balancing of chemical equation\n", + "# The unbalanced equation for the process is C8H18 + O2 + N2 \u2192 CO2 + H2O + N2\n", + "x=8; # Carbon balance\n", + "y=9; # Hydrogen balance\n", + "z=12.5; # Oxygen balance in reverse order\n", + "n=z*3.76; # Nitrogen Balance\n", + "\n", + "#Result for (a)\n", + "print \"(a).Balancing of chemical equation\"\n", + "print \" C8H18 + z O2 + n N2 \u2192 x CO2 + y H2O + n1 N2 \\n \"\n", + "print \" z =\",z,\"\\n n =\",n,\"\\n x =\",x,\"\\n y =\",y,\"\\n n1 =\",5*n\n", + "\n", + "#Calculation for (b)\n", + "# (b).The theoretical air-fuel ratio\n", + "a=1; # Mole of C8H18\n", + "AF1=(z+n)/a; #The theoretical air-fuel ratio on mole basis\n", + "ma=28.84; # Molecular mass of air \n", + "mc=114; # Molecular mass of C8H18\n", + "AF2=(AF1*ma)/(a*mc); # The theoretical air-fuel ratio on mass basis\n", + "\n", + "#Result for (b)\n", + "print \"\\n(b).The theoretical air-fuel ratio\",\"\\nThe theoretical air-fuel ratio on mole basis = \",AF1,\"kmol air / kmol C8H18\"\n", + "print \"The theoretical air-fuel ratio on mass basis = \",round(AF2,0),\"kg air / kmol C8H18\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a).Balancing of chemical equation\n", + " C8H18 + z O2 + n N2 \u2192 x CO2 + y H2O + n1 N2 \n", + " \n", + " z = 12.5 \n", + " n = 47.0 \n", + " x = 8 \n", + " y = 9 \n", + " n1 = 235.0\n", + "\n", + "(b).The theoretical air-fuel ratio \n", + "The theoretical air-fuel ratio on mole basis = 59.5 kmol air / kmol C8H18\n", + "The theoretical air-fuel ratio on mass basis = 15.0 kg air / kmol C8H18\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.4, Page No:650" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import numpy as np\n", + "#Variable declaration\n", + "# The combustion equation for C4H10 with 80% theoretical air is C4H10 +5.2(O2 + 3.76 N2) \u2192 a(1)CO + a(2)CO2 + 5H2O + 19.55N2\n", + "# The following matrix shows the balance of C and O\n", + "\n", + "#Calculation\n", + "A = np.array([(1, 1),(1,2)])\n", + "B = np.array([4,5.4])\n", + "m = np.linalg.solve(A,B)\n", + "\n", + "#Result\n", + "print \"The equation for the combustion of butane with 80% theoretical air is \"\n", + "print \"\\nC4H10 +5.2(O2 + 3.76 N2) \u2192\", m.item(0), \"CO + \",m.item(1), \"CO2 + 5H2O + 19.55N2\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The equation for the combustion of butane with 80% theoretical air is \n", + "\n", + "C4H10 +5.2(O2 + 3.76 N2) \u2192 2.6 CO + 1.4 CO2 + 5H2O + 19.55N2\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.5, Page No:650" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "p=101.325; # Atmospheric pressure in kPa\n", + "# The complete combustion equation for actane\n", + " # yC8H18+ x (O2+3.76N2) \u2192 n1 CO2+n2 H2O+n3 O2+n3 N2\n", + "x=12.5*1.5; y=1;\n", + "n1=8; n2=9; n3=6.28; n4=70.5;\n", + "\n", + "#calculation\n", + "n=n1+n2+n3+n4; # Total number of moles of the products\n", + "AFm=(x+x*3.76)/y ;# Air fuel ratio\n", + "m=28.84;\n", + "M=116; # Molecular weight of octane\n", + "AF=AFm*m/M;\n", + "yco2=n1/n; yH2O=n2/n; yO2=n3/n; yN2=n4/n;\n", + "pH2O=p*yH2O; # Partial pressure of water vapour in the products\n", + "Tsat=45.21; # In oC\n", + "\n", + "#Result\n", + "print \"Air fuel ratio = \",round(AF,2),\"kg air/kg octane\"\n", + "print \"Dew point temperature = \",Tsat,\"oC\"\n", + "print \"\\nIf the products are cooled below 25 oC then, the water vapour will condense.\"\n", + "print \"Because the cooled temperature is less than dew point temperature of water vapour i.e., T < Tsat.\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Air fuel ratio = 22.19 kg air/kg octane\n", + "Dew point temperature = 45.21 oC\n", + "\n", + "If the products are cooled below 25 oC then, the water vapour will condense.\n", + "Because the cooled temperature is less than dew point temperature of water vapour i.e., T < Tsat.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.6, Page No:651" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "# The complete chemical equation is\n", + "#[0.14H2+0.03CH4+0.27CO+0.045CO2+0.01O2+0.505N2]+0.255(O2+3.75N2) \u21920.2H2O+0.345CO2+1.44N2\n", + "a=0.14; # Composition of H2 in air\n", + "b=0.03; # Composition of CH4 in air\n", + "c=0.27; # Composition of CO in air\n", + "d=0.045; # Composition of CO2 in air\n", + "e=0.01; # Composition of O2 in air\n", + "f=0.505; # Composition of N2 in air\n", + "g=(0.265-0.01); # O2 requirement from atmospheric air with 1% O2 already in fuel\n", + "h=3.76; # By nitrogen balance \n", + "i=1; # mole of the air\n", + "\n", + "#Calculation\n", + "#Air fuel ratio on mol (volume) basis\n", + "AFvol=(g+(g*h))/i; # Air fuel ratio (theroretical)\n", + "AFv=1.1*AFvol; # Air fuel ratio on mol (volume) basis\n", + "#Air fuel ratio on mass basis\n", + "M1=2; # Molecular mass of H2\n", + "M2=16; # Molecular mass of CH4\n", + "M3=28; # Molecular mass of CO\n", + "M4=44; # Molecular mass of CO2\n", + "M5=32; # Molecular mass of O2\n", + "M=a*M1+b*M2+c*M3+d*M4+e*M5+f*M3; # Molecular mass of Fuel\n", + "Ma=28.84; # Molecular mass of air\n", + "AFm=AFv*Ma/(i*M); # Air fuel ratio on mass basis\n", + "\n", + "#Results\n", + "print \"Air fuel ratio on mol (volume) basis =\",round(AFv,3),\"kmol actual air/kmol fuel\"\n", + "print \"\\nAir fuel ratio on mass basis = \",round(AFm,2),\"kg air / kg fuel\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Air fuel ratio on mol (volume) basis = 1.335 kmol actual air/kmol fuel\n", + "\n", + "Air fuel ratio on mass basis = 1.56 kg air / kg fuel\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.7, Page No:653" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "#From table 14.2 at 25 oC and 1 atm for C8H8\n", + "del_Ho=-2039.7; # LHV in MJ/kmol\n", + "# Combustion equation is C3H8+ 5O2 +18.8N2 \u2192 3CO2 +4H2O +18.8N2\n", + "# From table 14.3\n", + "h333_C3H8=2751; # h333_h298 of C3H8 in kJ/kmol\n", + "h333_O2=147; # h333_h298 of O2 in kJ/kmol\n", + "h333_N2=145; # h333_h298 of N2 in kJ/kmol\n", + "h1333_CO2=52075; # h1333_h298 of CO2 in kJ/kmol\n", + "h1333_H2O=32644; # h1333_h298 of H2O in kJ/kmol\n", + "h1333_N2=32644; # h1333_h298 of N2 in kJ/kmol\n", + "M=44; # molecular mass of C3H8\n", + "\n", + "#Calculation\n", + "Ha_H1=h333_C3H8+5*h333_C3H8+18.8*h333_N2; # The enthalpy differences\n", + "Hb_H2=3*h1333_CO2+4*h1333_H2O+18.8*h1333_N2; # The enthalpy differences\n", + "Q=(del_Ho+Hb_H2/1000-Ha_H1/1000)/M; # Heat transfer from combustion chamber\n", + "\n", + "#Result\n", + "print \"Heat transfer from combustion chamber =\",round(abs (Q),2),\"MJ/kg C3H8\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transfer from combustion chamber = 26.33 MJ/kg C3H8\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.8, Page No:656" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Ha_H1=6220; # From example 14.7 in kJ/kmol\n", + "del_Ho=-2039.7; # From example 14.7 LHV in MJ/kmol\n", + "\n", + "#Calculation\n", + "Hb_H2=-del_Ho+Ha_H1; # For adiabatic combustion of C3H8\n", + "# Hb_H2=3*h1333_CO2+4*h1333_H2O+18.8*h1333_N2 By iteration process and making use of the values from Table A.3, A.13, A.15\n", + "# we can get the adiabatic flame temperature is\n", + "Tad=2300;# The adiabatic flame temperature in kelvin\n", + "\n", + "#Result\n", + "print \"The adiabatic flame temperature =\",Tad,\"K\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The adiabatic flame temperature = 2300 K\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.9, Page No:658" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "# (a).Entropy change per kmol of C\n", + "# From table 14.1 at 298 K and 1 atm\n", + "s_c=5.686; # Absolute entropies of C in kJ/kmol K\n", + "s_o2=205.142; # Absolute entropies of o2 in kJ/kmol K\n", + "s_co2=213.795; # Absolute entropies of CO2 in kJ/kmol K\n", + "\n", + "#Calculation for (a)\n", + "del_s=s_co2-(s_c+s_o2); # The entropy change \n", + "\n", + "#Result for (a)\n", + "print \"(a).The entropy change = \",del_s,\"kJ/K/kmol C\"\n", + "\n", + "#Variable declaration for(b)\n", + "# (b).Entropy change of universe\n", + "Tsurr=298; # Temperature of surroundings in kelvin\n", + "\n", + "#Calculation for (b)\n", + "# From table 14.1 \n", + "del_Ho=-393509; # del_hfco2 in kJ/kmol CO2\n", + "Q=abs (del_Ho);\n", + "del_Ssurr=Q/Tsurr; # Entropy change of surroundings\n", + "del_Suniv=del_s+del_Ssurr; #Entropy change of universe\n", + "\n", + "#Result for (b)\n", + "print \"\\n(b).Entropy change of universe = \",del_Suniv,\"kJ/K\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a).The entropy change = 2.967 kJ/K/kmol C\n", + "\n", + "(b).Entropy change of universe = 1322.967 kJ/K\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.10, Page No:659" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "# (a).The product CO2 is also at 298K\n", + "Pco=2/3; # Paratial pressure of CO in atm \n", + "Po2=1/3; # Paratial pressure of O2 in atm\n", + "Pco2=1; # Paratial pressure of CO2 in atm\n", + "T0=298; # Temperature of surroundings in kelvin\n", + "R_1=8.3143; # Universal gas constant of air in kJ/kmol K\n", + "\n", + "#Calculation for (a)\n", + "# From table 14.1 at 298 K and 1 atm\n", + "s_co2=213.795-R_1*math.log (Pco2); # entropies in kJ/kmol K\n", + "s_co=197.653-(R_1*math.log (Pco)); # entropies in kJ/kmol K\n", + "s_o2=205.03-R_1*math.log (Po2); # entropies in kJ/kmol K\n", + "del_Scv=s_co2-s_co-1/2*s_o2; # Entropy change of comtrol volume \n", + "# From table 14.1\n", + "del_hfco2=-393509; del_hfco=-110525; # Enthalpy of Heat in kJ/kmol\n", + "Q= del_hfco2- del_hfco; # Heat transfer (to surroundings)\n", + "del_Ssurr=abs(Q)/T0; # Entropy change of surroundings\n", + "del_Sgen=del_Scv+del_Ssurr; #Entropy change of universe\n", + "\n", + "#Result for (a)\n", + "print \"(a).The product CO2 is also at 298 K\",\"\\nEntropy change of comtrol volume = \",round(del_Scv,2),\"kJ/K\"\n", + "print \"Entropy change of surroundings = \",round(del_Ssurr,2),\"kJ/K\",\"\\nEntropy change of universe = \",round(del_Sgen,3),\"kJ/K\"\n", + "\n", + "#Calculation for (b)\n", + "# (b).The reaction is adiabatic\n", + "# Let the adiabatic flame temperature be T. Then since\n", + "Q=0;\n", + "C_p=44*0.8414;\n", + "# From table A.16\n", + "T=5057.5; #adiabatic flame temperature in kelvin\n", + "s_CO2=213.795+C_p*math.log (T/T0); # entropies in kJ/kmol K\n", + "del_Scv=s_CO2-s_co-1/2*s_o2; # Entropy change of comtrol volume \n", + "del_Ssurr=abs(Q)/T0; # Entropy change of surroundings\n", + "del_Sgen=del_Scv+del_Ssurr; #Entropy change of universe\n", + "\n", + "#Result for (b)\n", + "print \"\\n(b).The reaction is adiabatic\",\"\\nEntropy change of comtrol volume = \",round(del_Scv,3),\"kJ/K\"\n", + "print \"Entropy change of surroundings = \",round(del_Ssurr,3),\"kJ/K\",\"\\nEntropy change of universe = \",round(del_Sgen,3),\"kJ/K\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a).The product CO2 is also at 298 K \n", + "Entropy change of comtrol volume = -94.31 kJ/K\n", + "Entropy change of surroundings = 949.61 kJ/K \n", + "Entropy change of universe = 855.299 kJ/K\n", + "\n", + "(b).The reaction is adiabatic \n", + "Entropy change of comtrol volume = 10.517 kJ/K\n", + "Entropy change of surroundings = 0.0 kJ/K \n", + "Entropy change of universe = 10.517 kJ/K\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.11, Page No:661" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "# The Combustion of H2 with Q2 from H2O\n", + "#H2(g)+1/2 O2 (g)\u2192H2O(l)+285830 kJ/kmol H2\n", + "T0=298; # Temperature of surroundings in kelvin\n", + "# From table 14.1 at 298 K\n", + "del_hfH2O=-285830; # Enthalpy of Heat in kJ/kmol\n", + "s_298H2O=69.94; s_298H2=130.684; s_298O2=205.142; # entropies in kJ/kmol K\n", + "\n", + "#Calculation\n", + "GP_GR=del_hfH2O-T0*(s_298H2O-s_298H2-1/2*s_298O2); # Formation of Gibbs function\n", + "GR=0;\n", + "GP=GP_GR-GR; # Standard Gibbs function of formation of liquid H2O\n", + "\n", + "#Result\n", + "print \"Standard Gibbs function of formation of liquid H2O = \",round(GP,0),\"kJ/kmol\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Standard Gibbs function of formation of liquid H2O = -237162.0 kJ/kmol\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.12, Page No:662" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "# the combustion equation\n", + "# n1C3H8+n2O2+n3 N2 \u2192 n4 CO2+ n5 H2O+n6 O2+n7 N2\n", + "T0=298; # Temperature of surroundings in kelvin\n", + "# (a).Product species at 25 oC and 1 atm\n", + "d_gfC3H8=-24290; d_gfCO2=-394359; d_gfH2O=-228570; # in kJ/kmol\n", + "GR=d_gfC3H8;\n", + "\n", + "#Calculation for (a)\n", + "GP=3*d_gfCO2+4*d_gfH2O;\n", + "Wmax=GR-GP; # Maximum possible work output\n", + "M=44;#Molecular weight\n", + "Wmax=Wmax/M;\n", + "\n", + "#Result for (a)\n", + "print \"(a).\",\"\\nMaximum possible work output = \",round(Wmax,3),\"kJ/kg fuel (answer mentioned in the textbook is wrong)\"\n", + "\n", + "#Calculation for (b)\n", + "# (b).The actual partial pressures of products\n", + "n1=1; n2=20; n3=75.2;\n", + "n4=3; n5=4; n6=15; n7=75.2; # refer equation\n", + "SR=19233; SP=19147; # in kJ/K from table\n", + "HR=-104680; # in kJ/kmol fuel\n", + "d_h0fCO2=-393509; d_h0fH2O=-241818; # in kJ/kmol\n", + "HP=3*d_h0fCO2+4*d_h0fH2O;\n", + "Wmax=HR-HP-T0*(SR-SP); # Maximum possible work output\n", + "Wmax=Wmax/M;\n", + "\n", + "#Result for (b)\n", + "print \"\\n(b).\",\"\\nMaximum possible work output = \",round(Wmax,3),\"kJ/kg (round off error)\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a). \n", + "Maximum possible work output = 47115.159 kJ/kg fuel (answer mentioned in the textbook is wrong)\n", + "\n", + "(b). \n", + "Maximum possible work output = 45852.068 kJ/kg (round off error)\n" + ] + } + ], + "prompt_number": 12 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit