# Chapter13-Thermo Electric Power¶

## Example 13.2.1-pg707¶

In [1]:
##Ex13.2.1.;Peltier heats absorbed and rejected
##peltier coefficients at these junctions are aplha_p_1-2=alpha_s_1-2*T
##Let A=alpha_s_1-2 at 373 k=55*10^-6 v/degree_k and B=alpha_s_1-2 at 273 k=50*10^-6 v/degree_k
A=(55*10**-6);
B=(50*10**-6);
T1=373.;##k
T2=273.;##k
I=10*10**-3;##current;unit=Ampere
alpha_p_1_2_at_373k=A*T1;
alpha_p_1_2_at_273k=B*T2;
print'%s %.2f %s  %.2f %s'%(" alpha_p_1_2_at_373k=",alpha_p_1_2_at_373k," W/amp"and " \n alpha_p_1_2_at_273k=",alpha_p_1_2_at_273k," W/amp");
##Peltier heats absorned and rejected to be
q2_peltier=alpha_p_1_2_at_373k*I;
q1_peltier=alpha_p_1_2_at_273k*I;
print'%s %.2e %s %.2e %s '%("\n q2_peltier=",q2_peltier," w " and"\n q1_peltier=",q1_peltier," W");
c=q2_peltier-q1_peltier;
print("\n If no other heat transfer were involved,the difference between these vaues,");
print'%s %.2e %s %.2e %s %.2e %s '%("\n ",q2_peltier," "and "",q1_peltier," W"and  "\n,would be supplied as electric power",c,"");

 alpha_p_1_2_at_373k= 0.02
alpha_p_1_2_at_273k=  0.01  W/amp

q2_peltier= 2.05e-04
q1_peltier= 1.36e-04  W

If no other heat transfer were involved,the difference between these vaues,

2.05e-04  1.36e-04
,would be supplied as electric power 6.87e-05


## Example 13.3.2-pg708¶

In [3]:
##Ex.13.3.2.;Find the thomson heat transferred
import  math
import scipy
from scipy import integrate

##Let D=dalpha_s1/dT;
D=5.4*10**-3;##unit=micro V/degree k^2
T1=273;##unit=k
T2=373;##unit=k
I=10*10**-3;##unit=A
##Thomson coefficient sigma,varies with temp.
##sigma_1_of_T=-T*D;unit=V/degree k
##The thomson heat is given by equation
##qth=I*Integration of sigma_1_of_T  w.r.t. T
def fun(T):
y=T
return y
Integration = scipy.integrate.quad(fun,T1,T2);
qth=I*D*Integration[0];
print'%s %.2f %s'%("The THOMSON HEAT=",qth," micro W");

The THOMSON HEAT= 1.74  micro W


## Example 13.4.1-pg715¶

In [3]:
##Ex13.4.1.;Determine the efficiency of the thermoelectric generator.what will be its carnot efficiency
import math
TH=600.;##degree k;##temperature of the hot reservior of source
TC=300.;##degree k;##temperature of the sink
Z=2*(10**-3);##1/degree k;##Figure of merit for the material
M_optimum=(1.+((Z/2.)*(TH+TC)))**0.5;
print'%s %.2f %s'%(" M_optimum=",M_optimum,"");
##Efficiency of the thermoelectric generator is n=(((TH-TC)/TH)*((M_optimum-1)/(M_optimum+(TC/TH)))*100;
a=((TH-TC)/TH);
b=(M_optimum-1)/(M_optimum+(TC/TH));
n=a*b*100;
print'%s %.2f %s'%("\n Efficiency of the thermoelectric generator is n=",n," persent");
##where as efficiency of the carnot cycle (reversible) nc=((TH-TC)/TH)*100
nc=a*100;
print'%s %.2f %s'%("\n Efficiency of the carnot cycle (reversible) nc=",nc," persent");

 M_optimum= 1.38

Efficiency of the thermoelectric generator is n= 10.07  persent

Efficiency of the carnot cycle (reversible) nc= 50.00  persent


## Example 13.4.2-pg716¶

In [3]:
##Ex13.4.12.;Calculare maximum generator efficiency and the efficiency for maximum power,power output
import math
##seedbeck coefficient(alpha_s);unit=volts/degree celcius
alpha_s1=-190.*10**-6;##n-type
alpha_s2=190.*10**-6;##p-type
##Specific resistivity(p);unit=Ohm-cm
p1=1.45*10**-3;##n-type
p2=1.8*10**-3;##p-type
##Figure of merit(Z);unit=degree k**-1
Z1=2.*10**-3;##n-type
Z2=1.7*10**-3;##p-type

##conductivity (n-type),
k1=(alpha_s1**2.)/(p1*Z1);
##similarly
k2=(alpha_s2**2.)/(p2*Z2);
print'%s %.2f %s %.2f %s'%(" Conductivity k1=",k1," W/cm degree celcius" and " \n Conductivity k2=",k2," W/cm degree celcius");
##Z_opt=((alpha_s1-alpha_s2)**2)/[(p1*k1)**2+(p2*k2)**2];
##let
a=(alpha_s1-alpha_s2)
b=(p1*k1)
c=(p2*k2)
A=math.sqrt(b)
B=math.sqrt(c)
C=(A+B);
##/therefore
Z_opt=(a/C)**2.;
print'%s %.4f %s'%("\n Z_opt=",Z_opt," degree k");
##Thermal conductance
A1=2.3;##cm**2
A2=1.303;##cm**2
l1=1.5;##cm
l2=0.653;##cm
K=((k1*A1)/l1)+((k2*A2)/l2)
print'%s %.2f %s'%("\n Thermal conductance K=",K," W/degree celcius");
##R=Resistance of the generator=R1+R2
R=((p1*l1)/A1)+((p2*l2)/A2);
print'%s %.2f %s'%("\n Resistance of the generator R=",R," ohm");
TH=923.;##unit=k
TC=323.;##unit=k
M_opt=(1.+((Z_opt/2.)*(TH+TC)))**0.5;
print'%s %.2f %s'%("\n M_opt=",M_opt," ohm");
RL=M_opt*R;
print'%s %.4f %s'%("\n RL=",RL," ohms");
##Optimum efficiency n_opt=(((TH-TC)/TH)*((M_opt-1)/(M_opt+(TC/TH)))*100;
aa=((TH-TC)/TH);
##taking M_opt=1.43
b=(1.43-1.)/(1.43+(TC/TH));
n_opt=aa*b*100.;
print'%s %.2f %s'%("\n Optimum efficiency n_opt=",n_opt," persent");
##efficiency for max. power output n= (TH-TC)/TH)*m/[((1+m)**2/TH)*(KR/alpha_s_12**2)+(1+m)-(TH-TC)/2TH)]
##Efficiency power output
##RL=R i.e. m=1
## let ab=(1+m)**2/TH;ac=(KR/alpha_s_12**2);ad=(TH-TC)/2TH
m=1.;
ab=4./TH;
ac=1./Z_opt;
ad=aa/2.;
n_max=(aa/(ab*ac+2.-ad))*100.;
print'%s %.2f %s'%("\n max. power output n_max ",n_max," persent")
##Power output P_opt=I**2*RL=alpha_s12**2(TH-TC)*RL/(R+RL)**2=alpha_s12**2(TH-TC)/(1+M_opt)**2*RL
##let at=alpha_s12**2(TH-TC);mi=(1+M_opt)**2*RL
at=a*a*(TH-TC)*(TH-TC);
ml=(1.+1.43)*(1.+1.43)*2.63*10**-3
P_opt=at/ml;
print'%s %.2f %s'%("\n Power output P_opt=",P_opt," watts");
##for max. power P_max (RL=R)
##P_max=alpha_s12**2(TH-TC)*RL/(r+RL)**2=alpha_s12**2(TH-TC)RL*4RL
P_max=at/(4.*1.84*10**-3);
print'%s %.2f %s'%("\n max. power P_max=",P_max," watts");

##Many calcuating mistak are there in a following example,which is corrected in program.

 Conductivity k1= 0.01
Conductivity k2= 0.01  W/cm degree celcius

Z_opt= 0.0018  degree k

Thermal conductance K= 0.04  W/degree celcius

Resistance of the generator R= 0.00  ohm

M_opt= 1.47  ohm

RL= 0.0027  ohms

Optimum efficiency n_opt= 15.70  persent

max. power output n_max  16.13  persent

Power output P_opt= 3.35  watts

max. power P_max= 7.06  watts


## Example 13.4.3-pg718¶

In [1]:
##Ex.13.4.3;maximum efficiency,no. of thermocouple in series,open ckt voltage,heat i/p and reject at full load.
import  math
kA=0.02;##unit=watt/cm degree kelvin
kB=0.03;##unit=watt/cm degree kelvin
pA=0.01;##unit=ohm cm
pB=0.012;##unit=ohm cm
TH=1500.;##unit=degree kelvin
TC=1000;##unit=degree kelvin
AA=43.5;##unit=cm**2
AB=48.6;##unit=cm**2
LA=0.49;##unit=cm
LB=0.49;##unit=cm
I=20.*48.6;##Current density in the element limited to,I=20 amp/cm**2
output=100.;##unit=kW
##alpha_SAB at 1250 degree kelvin=0.0012 volt/degree kelvin=alpha_SA-alpha_SB
alpha_SAB=0.0012;##unit=volt/degree kelvin
##let
b=(pA*kA);
c=(pB*kB);
A=math.sqrt(b);
B=math.sqrt(c);
C=(A+B);
##figure of merit
Z=(alpha_SAB/C)**2.;
print'%s %.4f %s'%(" Z=",Z," degree k^-1");
M=(1+((Z/2.)*(TH+TC)))**0.5;
print'%s %.2f %s'%("\n M=",M,"");
##let
aa=((TH-TC)/TH);
bb=(M-1)/(M+(TC/TH));
##1] MAx. efficiency of a thermoelectric converter is given by n_max=((TH-TC)/TH)*[(M-1)/(M+(TC/TH))]*100;
n_max=aa*bb*100.;
print'%s %.2f %s'%("\n Maximum efficiency n_max=",n_max," persent");
##2] No. of thermocouple in series
V=alpha_SAB*(TH-TC);
print'%s %.2f %s'%("\n V=",V," volt");
R=((pA*LA)/AA)+((pB*LB)/AB);##since  R=RA+RB=((pA*LA)/AA)+((pB*LB)/AB);
print'%s %.4f %s'%("\n R=",R," ohm");
VL=V-(R*I);
print'%s %.2f %s'%("\n VL=",VL," volt");
##NTCS=total voltage required/voltage required by one couple
NTCS=115./VL;
print'%s %.2f %s'%("\n No. of thermocouple in series=",NTCS,"");
##3] Open circuit voltage
OCV=V*309.;
print'%s %.2f %s'%("\n Open circuit voltage=",OCV," volt")
##4] Heat input and reject at full load.
##Heat input at full load.=output/efficency=100/0.091
HIFL=output/n_max;
print'%s %.2f %s'%("\n Heat input at full load=",HIFL," kW")
## Heat reject at full load. =Heat input-Work output
HRFL=HIFL-output;
print'%s %.2f %s'%("\n Heat reject at full load=",HRFL,"kW")

##The value of "pB" is misprinted
##The values are taken in the text book is approximately equal to calculated values

 Z= 0.0013  degree k^-1

M= 1.63

Maximum efficiency n_max= 9.09  persent

V= 0.60  volt

R= 0.0002  ohm

VL= 0.37  volt

No. of thermocouple in series= 308.39

Open circuit voltage= 185.40  volt

Heat input at full load= 11.00  kW

Heat reject at full load= -89.00 kW