Chapter 13 : Thermodynamics of irreversible processes

Example 13.1 pg : 377

In [1]:
			
# Variables
Eab1 = 0.
Eab2 = 5.87 			#mV
T1 = 150.    			#F
T2 = 200. 	    		#F
			
# Calculations
Eab =  -1.12+ 0.035*T1
pi1 = 0.035*(T1+460)
pi2 = 0.035*(T2+460)
			
# Results
print "Thermocouple reading at %d F  =  %.2f mv"%(T1,Eab)
print " Peltier coefficient at %d F  =  %.1f mv"%(T1,pi1)
print " Peltier coefficient at %d F  =  %.1f mv"%(T2,pi2)

Thermocouple reading at 150 F  =  4.13 mv
 Peltier coefficient at 150 F  =  21.4 mv
 Peltier coefficient at 200 F  =  23.1 mv

Example 13.2 pg : 380

In [2]:
			
# Variables
T = 0.   			#C
			
# Calculations
de1 = -72. 			#mV/C
de2 = 500. 			#mv/C
alpha = de1-de2
pi = -(T+273)*alpha
			
# Results
print "Peltier coefficient at %d C  =  %d mv"%(T,pi/1000)

Peltier coefficient at 0 C  =  156 mv