Chapter 7 : Reversibility and The Thermodynamic Temperature Scale

Example 7.1 Page No : 174

print "Part a";
			
# Variables
Q1 = 500. 			#kJ 			#Heat transfer from reservoir at t1
Q2 = 187.5 			#kJ 			#Heat transfer from reservoir at t2

			
# Calculations and Results
nr = 1-(Q2/Q1) 			#Efficiency
W = nr*Q1 			#kJ 			#Work output of the engine
print "Work output of the engine = %.1f kJ"%(W);

			#Part(b)
print "Part b";
			
# Variables
Q1 = 500. 			#kJ 			#Heat transfer from reservoir at t1
			
# Calculations and Results
COP_hp = 1/nr
W = Q1/COP_hp 			#kJ 			#Work input to heat pump
Q2 = Q1-W 			#kJ 			#heat ineraction with reservoir at t2
print "Heat ineraction with reservoir at t2 = %.1f kJ"%(Q2);
print "Work input to the heat pump = %.1f kJ"%(W);

Part a
Work output of the engine = 312.5 kJ
Part b
Heat ineraction with reservoir at t2 = 187.5 kJ
Work input to the heat pump = 312.5 kJ

Example 7.2 Page No : 178

			
# Variables
nr = 0.7 			#maximum efficiency
W = 80. 			#kJ 			#Work
Q1 = 100. 			#kJ 			#heat transfered
			
# Calculations and Results
nx = W/Q1 			#claimed efficiency
if nx>nr :
    print "Engine X is not a viable proposition because the claimed efficiency %.2f is greater than maximum efficiency %.2f"%(nx,nr)
else:
    print "Engine X is a viable proposition because the claimed efficiency %.2f is less than maximum efficiency %.2f"%(nx,nr)   

Engine X is not a viable proposition because the claimed efficiency 0.80 is greater than maximum efficiency 0.70

Example 7.4 Page No : 191

			
# Variables
T1 = (527.+273) 			#K
T2 = (27.+273) 			#K

# calculations and results
#Part (a)
print "Part a";
nr = 1 - (T2/T1) 			#reversible efficiency
print "Reversible efficiency = %.3f "%(nr);

#Part(b)
print "Part b";
print "As the values of at least any two from among Q1,Q2 and W \
are not given the efficiency of the engine cannot be evaluated.However\
according to Carnots statement, the efficiency will be less than %.3f"%(nr)

Part a
Reversible efficiency = 0.625 
Part b
As the values of at least any two from among Q1,Q2 and W are not given the efficiency of the engine cannot be evaluated.Howeveraccording to Carnots statement, the efficiency will be less than 0.625

Example 7.5 Page No : 192

			
# Variables
T1 = 273. + 37 			#K 
T2 = 273. - 13 			#K

			#Part(a)
print "Part a";
COP_ref = T2/(T1-T2) 			#COP of reversible heat engine as refrigerator.
print "COP of reversible heat engine as refrigerator = %.1f"%(COP_ref)
			#Part(b)
print "Part b";
COP_hp = T1/(T1-T2) 			#COP of reversible heat engine as heat pump.
print "COP of reversible heat engine as heat pump = %.1f"%(COP_hp)

Part a
COP of reversible heat engine as refrigerator = 5.2
Part b
COP of reversible heat engine as heat pump = 6.2