Chapter 8: Basic applications of the second law

Exa 8.1

In [ ]:
#It is proposed to obtain a power from the hot surface water of tropical seas
#using the cold water from the depths as a sink for heat rejection. The surface
#water is 85 F the deep water is at 50 F. In the light of the second law is such
#scheme possible? If so, what is the max. thermal efficiency possible under law?
#initialisation of variables
T1= 85. 					#F
T2= 50. 					#F
#CALCULATIONS
n= (T1-T2)/(T1+460)			#Max. efficiency
n1= n*100 					#percentage
#RESULTS
print '%s %.2f' %('Maximum thermal efficiency (percent) = ',n1)
raw_input('press enter key to exit')

Maximum thermal efficiency (percent) =  6.42

Exa 8.2

In [1]:
#A proposed steam power plant will provide for supplying heat to steam at temp
#upto 1050F. The temperature of heat rejection is about 90 F. It is stated that
#the efficiency of the plant will approach 34%. Does this seem reasonable per law?
#initialisation of variables
T1= 1050. 	 				#F
T2= 90. 					#F
#CALCULATIONS
n= (T1-T2)/(T1+460)			#Efficiency
n1= n*100 					#Percentage
#RESULTS
print '%s %.2f' %('Maximum thermal efficiency (percent) = ',n1)
raw_input('press enter key to exit')

Maximum thermal efficiency (percent) =  63.58
press enter key to exit
Out[1]:
''

Exa 8.3

In [2]:
#How much does te entropy of a pound if air change when the air is heated 
#irreversibly from 1 atm pressure, 50 F to 1 atm, 150 F if the Cp=0.240?
import math
#initialisation of variables
m= 1 											#lb
cp= 0.240 										#btu/lb F
T2= 150 										#F
T1= 50 											#F
#CALCULATIONS
S= m*cp*(math.log(460.+T2)-math.log(460.+T1))	#Entropy
#RESULTS
print '%s %.4f' %('Entropy change (Btu/Fabs) = ',S)
raw_input('press enter key to exit')

Entropy change (Btu/Fabs) =  0.0430
press enter key to exit
Out[2]:
''

Exa 8.4

In [4]:
#If in the previous example, the process is adiabatic with work against friction
#How much the entropy change during this process
import math
#initialisation of variables
m= 1 											#lb
cp= 0.240 										#btu/lb F
T2= 150 										#F
T1= 50 											#F
#CALCULATIONS
S= m*cp*(math.log(460.+T2)-math.log(460.+T1))	#Entropy
#RESULTS
print '%s %.4f' %('Entropy change (Btu/Fabs) = ',S)
#This result is same as the above since change in entropy does not depend on the process involved
# but only on the initial and final states
raw_input('press enter key to exit')

Entropy change (Btu/Fabs) =  0.0430
press enter key to exit
Out[4]:
''

Exa 8.5

In [3]:
#In a steam boiler, hot gases from a fire transfer heat to water which evaporates
#at const. temp. In certain case the gases are cooled from 2000 F to 1000F while
#the water evaportates to 400 F. the Cp=0.24 and L=826. No wastage. How much does
#the entropy increase as a result of the irreversible heat transfer.?
import math
#initialisation of variables
Q= 826 											#Btu/lb
T= 400.											#F
T1= 1000. 										#F
T2= 2000. 										#F
#CALCULATIONS
Sw= Q/(T+460.)									#Entropy change for water
Sg= (Q/T1)*(math.log(T1+460)-math.log(T2+460))	#Entropy change for gas
S= Sw+Sg										#Total entropy change
#RESULTS
print '%s %.3f' %('Total Entropy change (Btu/R) = ',S)
raw_input('press enter key to exit')

Total Entropy change (Btu/R) =  0.530
press enter key to exit
Out[3]:
''

Exa 8.6

In [5]:
#For the above example find the increase in unavailable energy due to the 
#irreversible heat transfer. Assume the temperature of the surroundings is 80 F?
import math
#initialisation of variables
Q= 826.												#Btu/lb
T= 400.												#F
T1= 1000. 											#F
T2= 2000. 											#F
T3= 80.												#F
#CALCULATIONS
Sw= Q/(T+460.)										#Entropy change for water
Sg= (Q/T1)*(math.log(T1+460)-math.log(T2+460))		#Entropy change for gas
S= Sw+Sg 											#Total entropy change
Q1= (T3+460.)*S 									#Heat generated
Q2= Q+(T3+460.)*Sg 									#Total heat
n= Q1/Q2 											#Efficiency
n1= n*100 											#Loss percent
#RESULTS
print '%s %d' %('Loss percent = ',n1)
raw_input('press enter key to exit')

Loss percent =  48
press enter key to exit
Out[5]:
''