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{
"metadata": {
"name": "",
"signature": "sha256:7dac7ff4d7c0b0efd562d141ece5753006bca27ad006e31a868bd655b584d707"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter10-Chemical Energy Sources"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.2.8.1-pg585"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"##Ex10.2.8.1;Find Reversible voltage for hydrogen oxygen fuel cell\n",
"del_G=-237.3*10**3;##Joules/gm-mole of H2\n",
"##Reversible voltafe E of a cell is given by =del_Wrev/nF=-del_G/nF\n",
"##since 2 electrons are transferred per molecule of H2.thus\n",
"n=2.;\n",
"F=96500.;##Faraday's constant\n",
"E=-del_G/(n*F);\n",
"print'%s %.2f %s'%(\"Reversible voltage=\",E,\" volts\");\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Reversible voltage= 1.23 volts\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.2.8.2-pg585"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"##Ex10.2.8.2;calculate voltage output of cell,efficiency,electric work output,heat transfer to the surroundings\n",
"import math\n",
"##1] voltage output of cell\n",
"del_G=-237.3*10**3;##Joules/gm-mole of H2\n",
"n=2.;\n",
"F=96500.;##Faraday's constant\n",
"E=-del_G/(n*F);\n",
"print'%s %.2f %s'%(\" E=\",E,\" volts\");\n",
"##2] Efficiency\n",
"##nmax=del_Wmax/-(del_H)25 degree celcuis = -(del_G)T/(-del_H)25\n",
"del_G_at298k=-56690.;##unit=kcal/kg mole\n",
"del_H_at298k=-68317.;##unit=kcal/kg mole\n",
"nmax=del_G_at298k/del_H_at298k\n",
"print'%s %.2f %s'%(\"\\n nmax=\",nmax,\"\")\n",
"##3]Electric work output per mole\n",
"F=(96500/4.184);\n",
"del_Wrever=(n*F*E);\n",
"print'%s %.2f %s'%(\"\\n Electric work output per mole=\",del_Wrever,\" kcal/kg mole\");\n",
"##4] Heat transfer to the surroundings\n",
"##the heat transfer is Q=T*del-s=del_H_at298k-del_G_at298k\n",
"Q=del_H_at298k-del_G_at298k;\n",
"print'%s %.2f %s'%(\"\\n The heat transfer is Q=\",Q,\" kcal/kg mole\");\n",
"##The negative sign indicates that the heat is removed from the cell and transferred to the surrounding\n",
"\n",
"##value of \"Electric work output per mole\" is approximate in the text book to the real calculated value"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" E= 1.23 volts\n",
"\n",
" nmax= 0.83 \n",
"\n",
" Electric work output per mole= 56716.06 kcal/kg mole\n",
"\n",
" The heat transfer is Q= -11627.00 kcal/kg mole\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.2.8.3-pg587"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"##Ex10.2.8.3;The heat transferred to the surrounding\n",
"import math\n",
"del_G_at298k=-237191.;##unit=kJ/kg mole\n",
"del_H_at298k=-285838.;##unit=kJ/kg mole\n",
"ne=2.;\n",
"F=96500.;##Faraday's constant\n",
"E=-del_G_at298k/(ne*F);\n",
"print'%s %.2f %s'%(\" E=\",E,\" volts\");\n",
"nmax=del_G_at298k/del_H_at298k\n",
"print'%s %.2f %s'%(\"\\n nmax=\",nmax,\"\");\n",
"nmax=nmax*100;\n",
"print'%s %.2f %s'%(\"=0\",nmax,\" persent\");\n",
"##Electric work output per mole of the fule is We=-del_G kJ/kg mole\n",
"We=del_G_at298k##kJ/kg mole\n",
"print'%s %.2f %s'%(\"\\n Electric work output per mole of the fule is We=\",We,\" kJ/kg mole\")\n",
"##since there is 1 mol os H2O for each mole of fule,there is also a work output of 237191 kJ/kg mole\n",
"##Heat transferred is Q=T*del-s=del_H_at298k-del_G_at298k\n",
"Q=del_H_at298k-del_G_at298k;\n",
"print'%s %.2f %s'%(\"\\n The heat transfer is Q=\",Q,\" kJ/kg mole\");\n",
"##The negative sign indicates that the heat is removed from the cell and transferred to the surrounding\n",
"\n",
"##value of \"Electric work output per mole\" is misprinted in the text book.\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" E= 1.23 volts\n",
"\n",
" nmax= 0.83 \n",
"=0 82.98 persent\n",
"\n",
" Electric work output per mole of the fule is We= -237191.00 kJ/kg mole\n",
"\n",
" The heat transfer is Q= -48647.00 kJ/kg mole\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.2.8.4-pg587"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"##Ex10.2.8.4;calculate del_G,del_S,del_H;\n",
"import math\n",
"##We have the relation del_G=-n*F*E\n",
"##where,del_G=gibbs free energy of the system at 1 atm and temperature(T)\n",
"n=1.;##numbers of electons transferred per molecule of reactant\n",
"E=0.0455;##volts ;e.m.f. of the cell\n",
"F=96500.;##Faraday's constant\n",
"##let X=dE/dT\n",
"X=0.000338;\n",
"del_G=-n*F*E;\n",
"print'%s %.2f %s'%(\" del_G=\",del_G,\" joules\");\n",
"##del_S = Entropy change of the system at temperature T and press p=1 atm in the case\n",
"del_S=n*F*(X);##del_S=n*F*(dE/dT)\n",
"print'%s %.2f %s'%(\"\\n del_S=\",del_S,\" joules/deg.\");\n",
"##And entropy change is given by the relation del_H=nF[T(dE/dT)-E]\n",
"T=298;\n",
"del_H=n*F*((T*X)-E);\n",
"print'%s %.2f %s'%(\"\\n del_H=\",del_H,\" joule\");\n",
"\n",
"\n",
"##value are taken approximate in the text book to the real calculated value\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" del_G= -4390.75 joules\n",
"\n",
" del_S= 32.62 joules/deg.\n",
"\n",
" del_H= 5329.12 joule\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10.2.8.5-pg588"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"##Ex10.2.8.5;heat transfer rate would be involved under these circumstances\n",
"import math\n",
"del_G_at25degree_celcius=-195500.;##unit=cal/gm mole\n",
"del_H_at25degree_celcius=-212800.;##unit=cal/gm mole\n",
"F=(96500/4.184);##since F=96500 coulombs/gm-mole\n",
"n=8.\n",
"E_at25degree_celcius=-del_G_at25degree_celcius/(n*F);##Joules/coulomb\n",
"print'%s %.2f %s'%(\" E_at25degree_celcius=\",E_at25degree_celcius,\" volts=1.060 volts\");\n",
"##Max. efficiency nmax=del_Wmax/-(del_H)at25 degree celcuis = -(del_G)T/(-del_H)25\n",
"nmax=del_G_at25degree_celcius/del_H_at25degree_celcius;\n",
"print'%s %.2f %s'%(\"\\n nmax=\",nmax,\"\");\n",
"##voltage efficiency nv=on load voltage/open circuit voltage=Operating voltage/Theoretical voltage\n",
"Theoretical_voltage=1.060/0.92;\n",
"print'%s %.2f %s'%(\"\\n Theoretical_voltage=\",Theoretical_voltage,\" volts\");\n",
"##power developed=100 kW=100*10^3 W\n",
"power_developed=(100*10**3)*0.86;##unit=kcal/hr; since 1 watt=1 joule/sec=0.86 kcal/hr\n",
"print'%s %.2f %s'%(\"\\n power_developed=\",power_developed,\" kcal/hr\");\n",
"del_G=-195500.;\n",
"##Required flow rate of Methane\n",
"R_F_R_O_M=(power_developed*16.)/del_G;##kg/hr;\n",
"##(methane moles)=16\n",
"print'%s %.2f %s'%(\"\\n flow rate of Methane=\",R_F_R_O_M,\" kg/hr\");\n",
"##Heat transfer Q=T8del_s=del_H+del_w=del_H-del_G\n",
"Q=del_H_at25degree_celcius-del_G_at25degree_celcius;\n",
"print'%s %.2f %s'%(\"\\n The heat transfer is Q=\",Q,\" kcal/kg mole\");\n",
"\n",
"##The value are approximate in the text book to the real calculated value\n",
"##value of \"Required flow rate of methane\" is wrong in the text book.\n",
"##value of \"Heat transfer\" is wrong in the text book.\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" E_at25degree_celcius= 1.06 volts=1.060 volts\n",
"\n",
" nmax= 0.92 \n",
"\n",
" Theoretical_voltage= 1.15 volts\n",
"\n",
" power_developed= 86000.00 kcal/hr\n",
"\n",
" flow rate of Methane= -7.04 kg/hr\n",
"\n",
" The heat transfer is Q= -17300.00 kcal/kg mole\n"
]
}
],
"prompt_number": 2
}
],
"metadata": {}
}
]
}
|
