{ "metadata": { "name": "", "signature": "sha256:7e02442740cf50f0beae7f1fd9dfdf2fb7a16b9c0848a48a17d80c7a19de3f3d" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 9:Combustion" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex9.1:PG-229" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# initialization of variables\n", "\n", "AFactual=20 # air fuel ratio actual\n", "# The energy balance is done from equation\n", "\n", "# C4H10 + 6.5(O2+3.76N2)-----> 4CO2 + 5H2O + 24.44N2\n", "\n", "P=100 # atmospheic preesure in kPa\n", "mair=6.5*(1+3.76)*29 # mass of air\n", "mfuel=1*58 # mass of fuel\n", "AFth=mair/mfuel # theoritical air-fuel ratio\n", "Pexcessair=(AFactual-AFth)*100/AFth\n", "\n", "print \"The % excess air is\",round(Pexcessair,2),\"% \\n\"\n", "\n", "# NOW THE REACTION IS\n", "# C4H10+ (1+%excessair/100)*6.5*(O2+3.76N2) -----> 4CO2 + 5H2O + 1.903O2 + 31.6N2\n", "\n", "PCO2=4/42.5*100 # VOLUME % OF CO2\n", "\n", "print \"The volume % of CO2 is\",round(PCO2,2),\"% \\n\"\n", "\n", "# NOW WE FIND DEW POINT\n", "Nv=5 # moles of water\n", "N=42.5 # moles of air\n", "Pv=P*(Nv/N) # partial pressure of vapour\n", "Tdp=49# dew point temperature in degree celsius from table C.2\n", "\n", "print \"The Dew point temperature is\",round(Tdp,2),\"degree celsius\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The % excess air is 29.28 % \n", "\n", "The volume % of CO2 is 9.41 % \n", "\n", "The Dew point temperature is 49.0 degree celsius\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex9.2:PG-231" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# initialization of variables\n", "\n", "Pair=0.9 # 90% air is used for combustion\n", "\n", "# THE REACTION IS\n", "# C4H10 + 0.9*6.5*(O2+3.76N2)----> aCO2 + 5H20 + bCO\n", "# a and b are calculated by atomic balance\n", "a=2.7\n", "b=1.3\n", "PCO=b*100/31 # volume % of CO\n", "\n", "print \"The volume % of CO is\",round(PCO,2),\"% \\n\"\n", "\n", "mair=6.5*Pair*4.76*29 # mass of air in kg\n", "mfuel=1*58 # mass of fuel butane in kg\n", "AF=mair/mfuel # air-fuel ratio\n", "\n", "print \"The air to fuel ratio is\",round(AF,2),\"kg air/ kg fuel \"\n", "# THE SOLUTION IS CORRECT BUT THERE ARE SOME PRINTING MISTAKES IN TEXTBOOK\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The volume % of CO is 4.19 % \n", "\n", "The air to fuel ratio is 13.92 kg air/ kg fuel \n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex9.3:PG-231" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# initialization of variables\n", "\n", "# THE REACTION IS\n", "# aC4H10 + b(O2+3.76N2)----> CO2 + 1CO + 3.5H20 + 84.6N2 + cH2O\n", "# a, b and c are calculated by atomic balance\n", "# on balancing the equations we get a=3 b=22.5 c=15\n", "# Now equation becomes\n", "#C4H10 + 7.5(O2+3.76N2)----> 3.67CO2 + 0.33CO + 1.17H20 + 28.17N2 + 5H2O\n", "#MOLES OF AIR in this equation is 7.5 moles\n", "mairactual=7.5 # in moles\n", "#MOLES OF AIR in standard equation of Ex.9 is 6.5\n", "mairtheoritical=6.5\n", "Ptheoriticalair=100*(mairactual/mairtheoritical) \n", "print \"The % theoritical air is\",round(Ptheoriticalair,1),\"% \"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The % theoritical air is 115.4 % \n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex9.4:PG-232" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# initialization of variables\n", "# The reaction equation is \n", "#CaHb + c(O2+3.76N2)---> 10.4CO2 + 1.2CO + 2.8O2 + 85.6N2 + dH2O\n", "\n", "# using atomic balancing the equations become\n", "\n", "# C11.6H37.9 + 21.08(O2+3.76N2)---> 11.6CO2 + 18.95H2O + 79.26N2\n", "Ptheoriticalair=22.8*100/21.08 # theoritical air\n", "excessair=Ptheoriticalair-100\n", "\n", "print\"The excess air is\",round(excessair),\"%\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The excess air is 8.0 %\n" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex9.5:PG-235" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# initialization of variables\n", "# The reaction equation is \n", "#C3H8 + 5(O2+3.76N2)---> 3CO2 + 18.8N2 + 4H2O\n", "\n", "# All the enthalpy of formation values are taken from Table B.5 with units in kJ/mol\n", "hfCO2=-393520 # enthalpy associated with CO2\n", "hfH2O=-285830 # enthalpy associated with H2O(l)\n", "hfC3H8=-103850# ehthalpy associated with C3H8\n", "\n", "# by first law Q= Hproducts - Hreactants\n", "\n", "Qg=3*(hfCO2)+4*(hfH2O)-(hfC3H8) # enthalpy of combustion for gaseous propane\n", "\n", "print \" The enthalpy of combustion for gaseous propane is\",round(Qg),\"kJ\\n\"\n", "\n", "hv=15060 # enthalpy of vaporization for propane\n", "\n", "Ql=3*(hfCO2)+4*(hfH2O)-(hfC3H8-hv) # enthalpy of combustion for liquid propane\n", "\n", "print \" The enthalpy of combustion for liquid propane is\",round(Ql),\"kJ\\n\"\n", "\n", "#The answers are slightly different in textbook as they have approximated the result while in Python results are precise\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The enthalpy of combustion for gaseous propane is -2220030.0 kJ\n", "\n", " The enthalpy of combustion for liquid propane is -2204970.0 kJ\n", "\n" ] } ], "prompt_number": 15 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex9.6:PG-235" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# initialization of variables\n", "# The reaction equation is \n", "#C3H8 + 5(O2+3.76N2)---> 3CO2 + 18.8N2 + 4H2O\n", "\n", "# All the enthalpy of formation values are taken from Table B.5 with units in kJ/mol\n", "hfCO2=-393520 # enthalpy of formation associated with CO2\n", "hbarCO2=22280 #enthalpy associated with CO2 at 600K from table E.4\n", "hdotbarCO2=9364#enthalpy associated with CO2 at 298K from table E.4\n", "\n", "hfH2O=-241820 # enthalpy of formation associated with gaseous H2O\n", "hbarH2O=20402 #enthalpy associated with H20 at 600K from table E.6\n", "hdotbarH2O=9904#enthalpy associated with H20 at 298K from table E.6\n", "\n", "hfC3H8=-103850# ehthalpy of formation associated with C3H8\n", "\n", "hbarN2=17563 #enthalpy associated with N2 at 600K from table E.2\n", "hdotbarN2=8669#enthalpy associated with N2 at 298K from table E.2\n", "# by first law Q= Hproducts - Hreactants\n", "\n", "Qg=3*(hfCO2+hbarCO2-hdotbarCO2)+4*(hfH2O+hbarH2O-hdotbarH2O)+18.8*(hbarN2-hdotbarN2)-(hfC3H8) # enthalpy of combustion\n", "\n", "print\"The heat transfer required is\",round(Qg),\"kJ\\n\"\n", "\n", "#The answer is WRONG textbook as they have made an error in calculating Qg \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The heat transfer required is -1796043.0 kJ\n", "\n" ] } ], "prompt_number": 17 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex9.7:PG-236" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "# initialization of variables\n", "# The reaction equation is \n", "\n", "#C8H18 + 12.5(O2+3.76N2)---> 8CO2 + 47N2 + 9H2O\n", "\n", "# All the enthalpy of formation values are taken from Table B.5 with units in kJ/mol\n", "hfCO2=-393520 # enthalpy of formation associated with CO2\n", "hbarCO2=42769 #enthalpy associated with CO2 at 1000K from table E.4\n", "hdotbarCO2=9364#enthalpy associated with CO2 at 298K from table E.4\n", "\n", "hfH2O=-241820 # enthalpy of formation associated with gaseous H2O\n", "hbarH2O=35882 #enthalpy associated with H20 at 1000K from table E.6\n", "hdotbarH2O=9904#enthalpy associated with H20 at 298K from table E.6\n", "hfC3H8=-103850# ehthalpy of formation associated with C3H8\n", "\n", "hbarN2p=(30784+29476)/2 #enthalpy associated with N2 at 1000K from table E.2 by averaging enthalpy at 1020K and 980K for product\n", "hbarN2r=17563 #enthalpy associated with N2 at 600K from table E.2 for reactant\n", "hdotbarN2=8669#enthalpy associated with N2 at 298K from table E.2\n", "\n", "hfC8H18=-249910 # enthalpy of formation associated with octane taken from internet as not provided in textbook\n", "\n", "hbarO2=17929 # enthalpy associated with O2 at 600K table E.3\n", "hdotbarO2=8682#enthalpy associated with O2 at 298K table E.3\n", "\n", "# using first law and including kinetic energy change\n", "# 0=Hp-Hr+Mp*(V^2)/2\n", "\n", "Hp=8*(hfCO2+hbarCO2-hdotbarCO2)+9*(hfH2O+hbarH2O-hdotbarH2O)+47*(hbarN2p-hdotbarN2)\n", "# enthalpy of products\n", "\n", "Hr=(hfC8H18)+12.5*(hbarO2-hdotbarO2)+47*(hbarN2r-hdotbarN2)\n", "# enthalpy of reactants\n", "\n", "Mp=8*44+9*18+47*28 #(mass of products by multiplying molecular mass to number of moles)\n", "\n", "V=math.sqrt(2*1000*(Hr-Hp)/Mp)# exit velocity using energy balance\n", "\n", "print \"The exit velocity is\",round(V),\"m/s\"\n", "\n", "#The answers are slightly different in textbook as they have approximated the values while in Python results are precise\n", "\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The exit velocity is 2116.0 m/s\n" ] } ], "prompt_number": 20 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex9.8:PG-237" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# initialization of variables\n", "\n", "# The reaction equation with theoritical air is \n", "# C8H18 + 12.5(O2+3.76N2)---> 8CO2 + 47N2 + 9H2O\n", "\n", "# for 400% theoritical air reaction is\n", "\n", "# C8H18 + 50(O2+3.76N2)---> 8CO2 + 188N2 + 9H2O + 37.5O2\n", "\n", "# All the enthalpy of formation values are taken from Table B.5 with units in kJ/mol\n", "hfCO2=-393520 # enthalpy of formation associated with CO2\n", "hbarCO2=42769 #enthalpy associated with CO2 at 1000K from table E.4\n", "hdotbarCO2=9364#enthalpy associated with CO2 at 298K from table E.4\n", "hfH2O=-241820 # enthalpy of formation associated with gaseous H2O\n", "hbarH2O=35882 #enthalpy associated with H20 at 1000K from table E.6\n", "hdotbarH2O=9904#enthalpy associated with H20 at 298K from table E.6\n", "hbarN2p=(30784+29476)/2 #enthalpy associated with N2 at 1000K from table E.2 by averaging enthalpy at 1020K and 980K \n", "hdotbarN2=8669#enthalpy associated with N2 at 298K from table E.2\n", "\n", "hfC8H18=-249910 # enthalpy associated with octane taken from internet as not provided in textbook\n", "hbarO2=31389 # enthalpy associated with O2 at 1000K table E.3\n", "hdotbarO2=8682#enthalpy associated with O2 at 298K table E.3\n", "\n", "Hp=8*(hfCO2+hbarCO2-hdotbarCO2)+9*(hfH2O+hbarH2O-hdotbarH2O)+37.5*(hbarO2-hdotbarO2)+188*(hbarN2p-hdotbarN2)# enthalpy of products\n", "\n", "Hr=(hfC8H18)\n", "# enthalpy of reactants\n", "\n", "Q=Hp-Hr # using first law2\n", "\n", "print \" The heat transfer is\",round(Q),\"kJ\"\n", "\n", "#The answers are slightly different in textbook as they have approximated the values while in Python results are precise\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The heat transfer is 312593.0 kJ\n" ] } ], "prompt_number": 22 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex9.9:PG-237" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# initialization of variables\n", "\n", "# The reaction equation is \n", "#C3H8 + 5O2---> 8CO2 + 4H2O\n", "\n", "# All the enthalpy of formation values are taken from Table B.5 with units in kJ/mol\n", "hfCO2=-393520 # enthalpy associated with CO2\n", "hfH2O=-241820 # enthalpy associated with gaseous H2O\n", "hfC3H8=103850# enthalpy of formation associated with C3H8\n", "hfgC3H8=15060# enthalpy of vapourization associated with C3H8\n", "T=20+273 # temperature in kelvin\n", "Rbar=8.314 # universal gas constant\n", "Nr=6 # number of moles of reactants\n", "Np=7 # number of moles of products\n", "Hp=3*(hfCO2)+4*(hfH2O) # enthalpy of products\n", "\n", "Hr=hfC3H8+hfgC3H8 # enthalpy of reactants\n", "\n", "Q=(Hp-Hr-(Nr-Np)*Rbar*T)*10**(-3) # heat transfer from first law\n", "\n", "print \" The heat transfer is\",round(Q),\"MJ\"\n", "\n", "#The answers are slightly different in textbook as they have approximated the values while in Python results are precise\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The heat transfer is -2264.0 MJ\n" ] } ], "prompt_number": 24 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex9.10:PG-239" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# initialization of variables\n", "\n", "# The reaction equation for theoritical air is \n", "#C3H8 + 5(O2 + 3.76N2) ---> 3CO2 + 4H2O + 18.8N2\n", "\n", "# for 250% theoritical air reaction becomes\n", "#C3H8 + 12.5(O2 + 3.76N2) ---> 3CO2 + 4H2O + 47N2 + 7.5O2\n", "\n", "# All the enthalpy of formation values are taken from Table B.5 with units in kJ/mol\n", "\n", "Np=47+7.5+4+3 # number of moles of product\n", "hfCO2=-393520 # enthalpy of formation associated with CO2\n", "hbarCO2=(62963+65271)/2 #enthalpy associated with CO2 at 1380 K from table E.4\n", "hbarCO2dash=(58381+60666)/2 #enthalpy associated with CO2 at 1300 K by average from table E.4\n", "hdotbarCO2=9364#enthalpy associated with CO2 at 298K from table E.4\n", "\n", "hfC3H8=-103850# ehthalpy of formation associated with C3H8\n", "\n", "hfH2O=-241820 # enthalpy of formation associated with gaseous H2O\n", "hbarH2O=(51521+53351)/2 #enthalpy associated with H20 at 1380 K by taking average from table E.6\n", "hbarH2Odash=48807 #enthalpy associated with H20 at 1300 K from table E.6\n", "hdotbarH2O=9904#enthalpy associated with H20 at 298K from table E.6\n", "\n", "hbarN2=42920 #enthalpy associated with N2 at 1380K from table E.2 by interpolating enthalpy between 1020K and 980K \n", "hbarN2dash=40170 #enthalpy associated with N2 at 1300 K from table E.2 \n", "hdotbarN2=8669#enthalpy associated with N2 at 298K from table E.2\n", "\n", "hfO2=(44198+45648)/2 # enthalpy associated with O2 at 1380 Kby taking average from table E.3\n", "hfO2dash=48807 # enthalpy associated with O2 at 1380 Kby taking average from table E.3\n", "hdotbarO2=8682#enthalpy associated with O2 at 298K table E.3\n", "\n", "# for adiabatic flame temperature first assume products composed only of nitrogen and Q=0 as adiabatic\n", "hp=(hfC3H8-3*(hfCO2)-4*(hfH2O))/Np +hdotbarN2\n", "# using hp we assume temp=1380 K\n", "# then energy for 1380 k is\n", "H1=3*(hfCO2+hbarCO2-hdotbarCO2)+4*(hfH2O+hbarH2O-hdotbarH2O)+7.5*(hfO2-hdotbarO2)+47*(hbarN2-hdotbarN2) # energy assuming temperature to be 1380 K\n", "\n", "#this is very large \n", "\n", "# now at 1300 K adiabatic temperature\n", "H2=3*(hfCO2+hbarCO2dash-hdotbarCO2)+4*(hfH2O+hbarH2Odash-hdotbarH2O)+7.5*(hfO2dash-hdotbarO2)+47*(hbarN2dash-hdotbarN2) # energy assuming temperature to be 1300 K\n", " \n", " # now interpolation between these two temperatures\n", "Tp=1300-((hp+H2)/(H1-H2))*(1380-1300) # adiabatic temperature by interpolation\n", "print \"The adiabatic flame temperature is\",round(Tp),\"K\"\n", "\n", "#The answers is different in textbook as they have printed the value of hfCO2 with positive sign while calculating H2\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The adiabatic flame temperature is 1311.0 K\n" ] } ], "prompt_number": 26 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex9.11:PG-240" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# initialization of variables\n", "\n", "# The reaction equation for theoritical air is \n", "#C3H8 + 5(O2 + 3.76N2) ---> 3CO2 + 4H2O + 18.8N2\n", "\n", "# All the enthalpy of formation values are taken from Table B.5 with units in kJ/mol\n", "\n", "Np=18.8+4+3 # number of moles of product\n", "hfCO2=-393520 # enthalpy associated with CO2\n", "hbarCO2=137400 #enthalpy associated with CO2 at 2600 K from table E.4 by interpolation\n", "hbarCO2dash=125152 #enthalpy associated with CO2 at 2400 K from table E.4\n", "hdotbarCO2=9364#enthalpy associated with CO2 at 298K from table E.4\n", "\n", "hfC3H8=-103850# ehthalpy associated with C3H8\n", "\n", "hfH2O=-241820 # enthalpy associated with gaseous H2O\n", "hbarH2O=114273 #enthalpy associated with H20 at 2600 K from table E.6\n", "hbarH2Odash=103508 #enthalpy associated with H20 at 2400 K from table E.6\n", "hdotbarH2O=9904#enthalpy associated with H20 at 298K from table E.6\n", "\n", "hbarN2=86600 #enthalpy associated with N2 at 2600 K from table E.2 by interpolation\n", "hbarN2dash=79320 #enthalpy associated with N2 at 2400 K from table E.2 \n", "hdotbarN2=8669#enthalpy associated with N2 at 298K from table E.2\n", "\n", "# for adiabatic flame temperature first assume products composed only of nitrogen and Q=0 as adiabatic\n", "hp=(hfC3H8-3*(hfCO2)-4*(hfH2O))/Np +hdotbarN2 \n", "\n", "# using hp we assume temp=2600 K\n", "# then energy for 2600 k is\n", "H1=3*(hfCO2+hbarCO2-hdotbarCO2)+4*(hfH2O+hbarH2O-hdotbarH2O)+18.8*(hbarN2-hdotbarN2) # energy assuming temperature to be 2600 K\n", "\n", "# now at 2400 K adiabatic temperature\n", "H2=3*(hfCO2+hbarCO2dash-hdotbarCO2)+4*(hfH2O+hbarH2Odash-hdotbarH2O)+18.8*(hbarN2dash-hdotbarN2) # energy assuming temperature to be 2400 K\n", " \n", " # now interpolation between these two temperatures\n", "Tp=2400-((hp+H2)/(H1-H2))*(2600-2400) # adiabatic temperature by interpolation\n", "print \"The adiabatic flame temperature is\",round(Tp),\"K\"\n", "\n", "#The answers are slightly different in textbook as they have approximated the values while in Python results are precise\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The adiabatic flame temperature is 2409.0 K\n" ] } ], "prompt_number": 27 } ], "metadata": {} } ] }