# Chapter10-Chemical Energy Sources¶

## Example 10.2.8.1-pg585¶

In :
##Ex10.2.8.1;Find Reversible voltage for hydrogen oxygen fuel cell
del_G=-237.3*10**3;##Joules/gm-mole of H2
##Reversible voltafe E of a cell is given by =del_Wrev/nF=-del_G/nF
##since 2 electrons are transferred per molecule of H2.thus
n=2.;
F=96500.;##Faraday's constant
E=-del_G/(n*F);
print'%s %.2f %s'%("Reversible voltage=",E," volts");

Reversible voltage= 1.23  volts


## Example 10.2.8.2-pg585¶

In :
##Ex10.2.8.2;calculate voltage output of cell,efficiency,electric work output,heat transfer to the surroundings
import math
##1] voltage output of cell
del_G=-237.3*10**3;##Joules/gm-mole of H2
n=2.;
F=96500.;##Faraday's constant
E=-del_G/(n*F);
print'%s %.2f %s'%(" E=",E," volts");
##2] Efficiency
##nmax=del_Wmax/-(del_H)25 degree celcuis = -(del_G)T/(-del_H)25
del_G_at298k=-56690.;##unit=kcal/kg mole
del_H_at298k=-68317.;##unit=kcal/kg mole
nmax=del_G_at298k/del_H_at298k
print'%s %.2f %s'%("\n nmax=",nmax,"")
##3]Electric work output per mole
F=(96500/4.184);
del_Wrever=(n*F*E);
print'%s %.2f %s'%("\n Electric work output per mole=",del_Wrever," kcal/kg mole");
##4] Heat transfer to the surroundings
##the heat transfer is Q=T*del-s=del_H_at298k-del_G_at298k
Q=del_H_at298k-del_G_at298k;
print'%s %.2f %s'%("\n The heat transfer is Q=",Q," kcal/kg mole");
##The negative sign indicates that the heat is removed from the cell and transferred to the surrounding

##value of "Electric work output per mole" is approximate in the text book to the real calculated value

 E= 1.23  volts

nmax= 0.83

Electric work output per mole= 56716.06  kcal/kg mole

The heat transfer is Q= -11627.00  kcal/kg mole


## Example 10.2.8.3-pg587¶

In :
##Ex10.2.8.3;The heat transferred to the surrounding
import math
del_G_at298k=-237191.;##unit=kJ/kg mole
del_H_at298k=-285838.;##unit=kJ/kg mole
ne=2.;
F=96500.;##Faraday's constant
E=-del_G_at298k/(ne*F);
print'%s %.2f %s'%(" E=",E," volts");
nmax=del_G_at298k/del_H_at298k
print'%s %.2f %s'%("\n nmax=",nmax,"");
nmax=nmax*100;
print'%s %.2f %s'%("=0",nmax," persent");
##Electric work output per mole of the fule is We=-del_G kJ/kg mole
We=del_G_at298k##kJ/kg mole
print'%s %.2f %s'%("\n Electric work output per mole of the fule is We=",We," kJ/kg mole")
##since there is 1 mol os H2O for each mole of fule,there is also a work output of 237191 kJ/kg mole
##Heat transferred is Q=T*del-s=del_H_at298k-del_G_at298k
Q=del_H_at298k-del_G_at298k;
print'%s %.2f %s'%("\n The heat transfer is Q=",Q," kJ/kg mole");
##The negative sign indicates that the heat is removed from the cell and transferred to the surrounding

##value of "Electric work output per mole" is misprinted in the text book.

 E= 1.23  volts

nmax= 0.83
=0 82.98  persent

Electric work output per mole of the fule is We= -237191.00  kJ/kg mole

The heat transfer is Q= -48647.00  kJ/kg mole


## Example 10.2.8.4-pg587¶

In :
##Ex10.2.8.4;calculate del_G,del_S,del_H;
import math
##We have the relation del_G=-n*F*E
##where,del_G=gibbs free energy of the system at 1 atm and temperature(T)
n=1.;##numbers of electons transferred per molecule of reactant
E=0.0455;##volts ;e.m.f. of the cell
F=96500.;##Faraday's constant
##let X=dE/dT
X=0.000338;
del_G=-n*F*E;
print'%s %.2f %s'%(" del_G=",del_G," joules");
##del_S = Entropy change of the system at temperature T and press p=1 atm in the case
del_S=n*F*(X);##del_S=n*F*(dE/dT)
print'%s %.2f %s'%("\n del_S=",del_S," joules/deg.");
##And entropy change is given by the relation del_H=nF[T(dE/dT)-E]
T=298;
del_H=n*F*((T*X)-E);
print'%s %.2f %s'%("\n del_H=",del_H," joule");

##value are taken approximate in the text book to the real calculated value

 del_G= -4390.75  joules

del_S= 32.62  joules/deg.

del_H= 5329.12  joule


## Example 10.2.8.5-pg588¶

In :
##Ex10.2.8.5;heat transfer rate would be involved under these circumstances
import math
del_G_at25degree_celcius=-195500.;##unit=cal/gm mole
del_H_at25degree_celcius=-212800.;##unit=cal/gm mole
F=(96500/4.184);##since F=96500 coulombs/gm-mole
n=8.
E_at25degree_celcius=-del_G_at25degree_celcius/(n*F);##Joules/coulomb
print'%s %.2f %s'%(" E_at25degree_celcius=",E_at25degree_celcius," volts=1.060 volts");
##Max. efficiency  nmax=del_Wmax/-(del_H)at25 degree celcuis = -(del_G)T/(-del_H)25
nmax=del_G_at25degree_celcius/del_H_at25degree_celcius;
print'%s %.2f %s'%("\n nmax=",nmax,"");
##voltage efficiency nv=on load voltage/open circuit voltage=Operating voltage/Theoretical voltage
Theoretical_voltage=1.060/0.92;
print'%s %.2f %s'%("\n Theoretical_voltage=",Theoretical_voltage," volts");
##power developed=100 kW=100*10^3 W
power_developed=(100*10**3)*0.86;##unit=kcal/hr; since 1 watt=1 joule/sec=0.86 kcal/hr
print'%s %.2f %s'%("\n power_developed=",power_developed," kcal/hr");
del_G=-195500.;
##Required flow rate of Methane
R_F_R_O_M=(power_developed*16.)/del_G;##kg/hr;
##(methane moles)=16
print'%s %.2f %s'%("\n flow rate of Methane=",R_F_R_O_M," kg/hr");
##Heat transfer Q=T8del_s=del_H+del_w=del_H-del_G
Q=del_H_at25degree_celcius-del_G_at25degree_celcius;
print'%s %.2f %s'%("\n The heat transfer is Q=",Q," kcal/kg mole");

##The value are approximate in the text book to the real calculated value
##value of "Required flow rate of methane" is wrong in the text book.
##value of "Heat transfer" is wrong in the text book.

 E_at25degree_celcius= 1.06  volts=1.060 volts

nmax= 0.92

Theoretical_voltage= 1.15  volts

power_developed= 86000.00  kcal/hr

flow rate of Methane= -7.04  kg/hr

The heat transfer is Q= -17300.00  kcal/kg mole