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diff --git a/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter10.ipynb b/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter10.ipynb new file mode 100755 index 00000000..edc9f942 --- /dev/null +++ b/Non-conventional_Energy_Sources_by_G._D._Rai/Chapter10.ipynb @@ -0,0 +1,285 @@ +{
+ "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": {}
+ }
+ ]
+}
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