From 92cca121f959c6616e3da431c1e2d23c4fa5e886 Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Tue, 7 Apr 2015 15:58:05 +0530 Subject: added books --- .../Chapter19_3.ipynb | 514 +++++++++++++++++++++ 1 file changed, 514 insertions(+) create mode 100755 Thermodynamics:_From_concepts_to_applications/Chapter19_3.ipynb (limited to 'Thermodynamics:_From_concepts_to_applications/Chapter19_3.ipynb') diff --git a/Thermodynamics:_From_concepts_to_applications/Chapter19_3.ipynb b/Thermodynamics:_From_concepts_to_applications/Chapter19_3.ipynb new file mode 100755 index 00000000..95c7b0c1 --- /dev/null +++ b/Thermodynamics:_From_concepts_to_applications/Chapter19_3.ipynb @@ -0,0 +1,514 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:9d79eec6ce58c29eedc2099ff08aa61dc95a4086cf18a82c8aaaa70d00549f3b" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter19-Chemical reactions" + ] + }, + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Example1-pg 528" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#caluclate fuel ratio and excess air and emass air-fuel ratio\n", + "##initialisation of variables\n", + "pN2= 79. ##percent\n", + "VN2= 82.3 ##m^3\n", + "VCO2= 8. ##m^3\n", + "VCO= 0.9 ##m^3\n", + "M= 32. ##gms\n", + "M1= 28. ##gms\n", + "##CALCULATIONS\n", + "P= (pN2/(100-pN2))\n", + "z= VN2/P\n", + "x= VCO2+VCO\n", + "w= VCO2+(VCO/2)+(VCO2/10)\n", + "y= 2*w\n", + "r= y/x\n", + "TO= x+(y/4)\n", + "X= (z/TO)-1\n", + "AF= z*(M+P*M1)/(12*x+y)\n", + "##RESULTS\n", + "print'%s %.3f %s'%('fuel ratio=',r,'')\n", + "print'%s %.3f %s'%('excess air=',X,'')\n", + "print'%s %.2f %s'%('emass air-fuel ratio=',AF,'')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "fuel ratio= 2.079 \n", + "excess air= 0.618 \n", + "emass air-fuel ratio= 23.98 \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example2-pg 534" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#calculate heat interaction\n", + "##initialisation of variables\n", + "m1= 24. ##kg\n", + "M1= 32. ##kg\n", + "m2= 28. ##kg\n", + "M2= 28. ##kg\n", + "e= 0.5\n", + "T3= 1800. ##C\n", + "T0= 25. ##C\n", + "T1= 25. ##C\n", + "T2= 100. ##C\n", + "R= 8.314 ##Jmol K\n", + "cp= 4.57 ##J/mol K\n", + "cp1= 3.5 ##J/mol K\n", + "cp2= 3.5 ##J/mol K\n", + "hCO2= -393522. ##J\n", + "hCO= -110529. ##J\n", + "##CALCULATIONS\n", + "n1= m1/M1\n", + "n2= m2/M2\n", + "N= n1-0.5*e\n", + "N1= n2-e\n", + "N2= e\n", + "N3= N+N1+N2\n", + "y1= N/N3\n", + "Q= ((N*cp+N1*cp1+N2*cp2)*R*(T3-T0)-(n1*cp*(T1-T0)+n2*cp2*(T2-T1))+N*(hCO2-hCO))/60.\n", + "##RESULTS\n", + "print'%s %.f %s'%(' Heat interaction=',Q,'kW ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Heat interaction= -940 kW \n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex3-pg536" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "##initialisation of variables\n", + "hCO2= -393520. ##kJ/kg mol\n", + "hH2O= -285840. ##kJ/kg mol\n", + "hC7H16= -187820. ##kJ/kg mol\n", + "M= 100\n", + "hH2O1= -241830. ##kJkg mol\n", + "##CALCULATIONS\n", + "HHV= -(7*hCO2+8.*hH2O-hC7H16)/M\n", + "LLV= -(7*hCO2+8.*hH2O1-hC7H16)/M\n", + "#RESULTS\n", + "print'%s %.2f %s'% (' Higher heating vlue= ',HHV,' kJ/kg mol ')\n", + "print'%s %.2f %s'% (' Lower heating vlue= ',LLV,' kJ/kg mol ')\n", + "#round off error\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Higher heating vlue= 48535.40 kJ/kg mol \n", + " Lower heating vlue= 45014.60 kJ/kg mol \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example4-pg 537" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#calculate adiabatic flame temperature\n", + "##initialisation of variables\n", + "T0= 25. ##C\n", + "T1= 220. ##C\n", + "hCO2= -393520 ##kJ/kg\n", + "hH2O= -241830 ##kJ/kg\n", + "hC3H8= -103850 ##kJ/kg= 1.4\n", + "R= 8.314 ##Jmol K\n", + "k= 1.4\n", + "k1= 1.29\n", + "##CALCULATIONS\n", + "T= T0+((15*(R*(k/(k-1)))*4.762*(T1-T0)-(3*hCO2+4*hH2O-hC3H8))/(R*((3+4)*(k1/(k1-1))+(10+56.43)*(k/(k-1)))))\n", + "##RESULTS\n", + "print'%s %.1f %s'%('adiabatic flame temperature=',T,'C ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "adiabatic flame temperature= 1142.4 C \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example6-pg 415" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#calculate enthalpy formation\n", + "##initialisation of variables\n", + "T= 25. ##C\n", + "hfT= -241820 ##kJ/kmol\n", + "R= 8.314 ##J/mol K\n", + "k= 1.4\n", + "cpH2O= 4.45\n", + "cpO2= 3.5\n", + "T1= 1000. ##C\n", + "##CALCULATIONS\n", + "S= (cpH2O-k*cpO2)\n", + "hfT1= hfT+S*(T1-T)\n", + "##RESULTS\n", + "print'%s %.f %s'%('enthalpy formation=',hfT1,'kJ/kmol ')\n", + "#there is error because of round off error \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "enthalpy formation= -242259 kJ/kmol \n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example7-pg 545" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#calculate equlibrium constant at K and KT1\n", + "##initialisation of variables\n", + "R= 8.314 ##J/mol K\n", + "T= 25. ##C\n", + "gf= 16590. ##kJ/kmol\n", + "T1= 500. ##C\n", + "Cp= 4.157 ##J/mol K\n", + "hf= -46190 ##kJ/kmol\n", + "e=0.5\n", + "##CALCULATIONS\n", + "K=math.pow(math.e,gf/(R*(273.15+T)))\n", + "r= (1-((273.15+T)/(273.15+T1)))*((hf/(R*(273.15+T)))+(R/Cp))-2*math.log((273.15+T1)/(273.15+T))+0.6\n", + "KT1= K*math.pow(math.e,r)\n", + "##RESULTS\n", + "print'%s %.1f %s'%('equilibrium constant=',K,'bar^-1 ')\n", + "print'%s %.5f %s'%('equilibrium constant=',KT1,'bar^-1 ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "equilibrium constant= 806.5 bar^-1 \n", + "equilibrium constant= 0.00797 bar^-1 \n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example8-pg 546" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#what equilibrium constant at T1 and T2\n", + "##initialisation of variables\n", + "uCO2= -394374 ##J/mol\n", + "uCO= -137150 ##J/mol\n", + "uO2= 0.\n", + "R= 8.314 ##J/mol K\n", + "T= 25. ##C\n", + "cpCO2= 4.57 ##J/mol K\n", + "cpCO= 3.5 ##J/mol K\n", + "cpO2= 3.5 ##J/mol K\n", + "T1= 1500. ##C\n", + "hf= -393522 ##kJ/kmol\n", + "gf= -110529 ##kJ/kmol\n", + "T2= 2500. ##C\n", + "##CALCULATIONS\n", + "r= -(uCO2-uCO-0.5*uO2)/(R*(273.15+T))\n", + "s= (cpCO2-cpCO-0.5*cpO2)\n", + "r1= (1-((273.15+T)/(273.15+T1)))*((hf-gf)/(R*(273.15+T))-s)+s*math.log((273.15+T1)/(273.15+T))\n", + "KT1= math.pow(math.e,r+r1)\n", + "r2= (1-((273.15+T)/(273.15+T2)))*((hf-gf)/(R*(273.15+T))-s)+s*math.log((273.15+T2)/(273.15+T))\n", + "KT2= math.pow(math.e,r+r2)\n", + "##RESULTS\n", + "print'%s %.f %s'%('equilibrium constant at T1=',KT1,'C ')\n", + "print'%s %.3f %s'%('equilibrium constant at T2=',KT2,'C ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "equilibrium constant at T1= 3477 C \n", + "equilibrium constant at T2= 2.635 C \n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example9-pg548" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#what is maximum work of given variable \n", + "##initialisation of variables\n", + "Wc= 12. ##kg\n", + "hf= -393520 ##kJ/kmol\n", + "gf= -394360 ##kJ/kmol\n", + "##CALCULATIONS\n", + "Wmax= -gf/Wc\n", + "##RESULTS\n", + "print'%s %.f %s'%('maximum work=',Wmax,'kJ/kg of carbon ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "maximum work= 32863 kJ/kg of carbon \n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example10-pg549" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#calculate the outlet temperature and energy of formation and energy out let and energy of the products\n", + "##initialisation of variables\n", + "T= 25 ##C\n", + "R= 8.314 ##Jmol K\n", + "k= 1.27\n", + "k1= 1.34\n", + "hf= -393520 ##kJ/kmol\n", + "M= 28 ##gms\n", + "gf= -394360 ##kJ/kmol\n", + "M= 12 ##gms\n", + "##CALCULATIONS\n", + "T1= T+(-hf/((R)*((k/(k-1))+(0.2+4.5144)*(k1/(k1-1)))))\n", + "Bin= 0\n", + "dh= (k1*R/(k1-1))*(T1-T)\n", + "dh1= (k1*R/(k1-1))*math.log((273.15+T1)/(273.15+T))\n", + "H= dh-(273.15+T)*dh1\n", + "h= (k*R/(k-1))*(T1-T)+hf\n", + "h1= (k*R/(k-1))*math.log((273.15+T1)/(273.15+T))+((hf-gf)/(273.15+T))\n", + "h2= h-(273.15+T)*h1\n", + "Bout= (h2+(0.2+4.5144)*H)/M\n", + "##RESULTS\n", + "print'%s %.2f %s'%('outlet temperature=',T1,'C')\n", + "print'%s %.f %s'%('energy of formation=',Bin,'J')\n", + "print'%s %.f %s'%('energy at outlet=',H,'kJ/kmol')\n", + "print'%s %.f %s'%('energy of the products=',Bout,'k')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "outlet temperature= 2057.82 C\n", + "energy of formation= 0 J\n", + "energy at outlet= 46519 kJ/kmol\n", + "energy of the products= -9961 k\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example11-pg553" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#calculate the change in energy and amount of air and gas and netchange in energy and percent change in energy\n", + "##initialisation of variables\n", + "b= 1475.30 ##kJ/kg\n", + "b0= 144.44 ##kJ/kg\n", + "h2= 3448.6 ##kJkg\n", + "h1= 860.5 ##kJ/kg\n", + "k= 1.27 \n", + "k1= 1.34\n", + "R= 8.314 ##J/mol K\n", + "hf= -393520 ##kJ/kmol\n", + "hg= 72596 ##kJ/kmol\n", + "Mc= 12 ##kg\n", + "n= 1.2 ##moles\n", + "n1= 3.76 ##moles\n", + "M= 32. ##gms\n", + "M1= 28. ##gms\n", + "M2= 44. ##gms\n", + "n2= 0.2 ##moles\n", + "n3= 4.512 ##moles\n", + "B1= 25592. ##kJ/kmol C\n", + "B2= 394360. ##kJ/kmol C\n", + "e= 0.008065\n", + "##CALCULATIONS\n", + "B= b-b0\n", + "Q= h2-h1\n", + "CpCO2= k*R/(k-1)\n", + "CpO2= k1*R/(k1-1)\n", + "Qcoal= (hg+hf)/Mc\n", + "mcoal= Q/(-Qcoal)\n", + "ncoal= mcoal/Mc\n", + "r= (n*M+n1*M1)/Mc\n", + "r1= (M2+n2*M+n3*M1)/Mc\n", + "mair= r*mcoal\n", + "mgas= r1*mcoal\n", + "Bfuel= (B1-B2)*e\n", + "Bnet= Bfuel+B\n", + "p= B*100/(-Bfuel)\n", + "##RESULTS\n", + "print'%s %.2f %s'% ('change in energy=',B,'kJ/kg ')\n", + "print'%s %.3f %s'%('amount of air=',mair,'kg/kg ')\n", + "print'%s %.3f %s'%('amount of gas=',mgas,'kg/kg ')\n", + "print'%s %.3f %s'%('net change in energy=',Bnet,'kg/kg steam ')\n", + "print'%s %.2f %s'%('percent energy in original fuel=',p,'percent ')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "change in energy= 1330.86 kJ/kg \n", + "amount of air= 1.159 kg/kg \n", + "amount of gas= 1.425 kg/kg \n", + "net change in energy= -1643.254 kg/kg steam \n", + "percent energy in original fuel= 44.75 percent \n" + ] + } + ], + "prompt_number": 13 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit