A power
cycle exchanges heat with only two thermal reservoirs at 500^{o}R and 2000^{o}R. Q_{H} = 4500 Btu/h. 
For each of the following, calculate the rate of entropy generation in Btu/^{o}R and state whether the power cycle is internally reversible, internally irreversible or impossible.
a.) h = 83%,
b.) h = 75%,
c.) h = 44% 











Read : 
The key equations for
this problem are the defintion
of entropy generation
and thermal efficiency. Because the process operates in a cycle, ΔS
= 0. The
other key point is that we are only asked to decide whether the process is internally reversible. That means we don't have to worry about whether heat exchange with the reservoirs is reversible. The main point is that the temperatures of the regions that exchange heat with the reservoirs must remain constant. This allows us to directly evaluate the integral of dQ/T
and evaluate the entropy generation. 










Given: 
T_{H} 
2000 
^{o}R 


a.) 
h_{A} 
0.83 


Q_{H} 
4500 
Btu 


b.) 
h_{B} 
0.75 


T_{C} 
500 
^{o}R 


c.) 
h_{C} 
0.44 










Diagram: 










Find: 
Parts ac.) 
S_{gen} 
??? 
Btu/^{o}R 

















Internally Reversible ? 
Internally Irreversible ? 
Impossible ? 












Assumptions: 
1  
The system undergoes a
power cycle while receiving Q_{H} at T_{H} and discharging Q_{C} at T_{C}. 


2  
The region of the system that receives heat from the hot reservoir remains at a constant temperature of T_{H} = 2000^{o}R. 


3  
The region of the system that rejects heat to the cold reservoir remains at a constant temperature of T_{C} = 500^{o}R. 










Equations
/ Data / Solve: 


















Entropy
generation is defined by: 


Eqn 1 











We can solve Eqn 1 for S_{gen} : 


Eqn 2 











Since we are dealing
with a cycle,
DS = 0 and Eqn 2
becomes: 



Eqn 3 











In our process, the system receives heat, Q_{H}, at a constant temperature, T_{H}_{ }, and rejects heat, Q_{C}, at a constant temperature, T_{C}. Because the temperatures are constant, they can be pulled out of the integral in Eqn 3 leaving : 















Eqn 4 











In Eqn 4,
notice that dQ becomes +dQ_{H} and dQ_{C} because of the sign convention that heat transfer into
the system is positive. 












The only variable in Eqn 4 that we don't already know is
Q_{C}. But we are given the
value of the thermal efficiency of the power cycle in each
part of this problem. 











The definition of thermal efficiency is: 



Eqn 5 











Rearranging Eqn 5 to solve for Q_{C} yields : 



Eqn 6 











Now, we can plug
numbers into Eqn 6 to
determine Q_{C} and then plug Q_{C} and the given values
of Q_{H}, T_{H} and T_{C} into Eqn 4 to complete the solution. 











Part (a) 
Q_{C} 
765 
Btu 


S_{gen} 
0.72 
Btu/^{o}R 

Part (b) 
Q_{C} 
1125 
Btu 


S_{gen} 
0.00 
Btu/^{o}R 

Part (c) 
Q_{C} 
2520 
Btu 


S_{gen} 
2.79 
Btu/^{o}R 











We can now determine whether the cycle in each part of
the problem is internally reversible, reversible or impossible using the following rules based on the definition of entropy generation. 












If S_{gen} = 0 
the cycle is internally
reversible. 






If S_{gen} > 0 
the cycle is internally
irreversible. 






If S_{gen} < 0 
the cycle is impossible. 














Verify: 
The assumptions made
in this solution cannot be verified with the given information. 










Answers : 
Part (a) 
S_{gen} 
0.72 
Btu/^{o}R 

Part
(a) is impossible. 


Part (b) 
S_{gen} 
0.00 
Btu/^{o}R 

Part (b) is internally reversible. 


Part (c) 
S_{gen} 
2.79 
Btu/^{o}R 

Part
(c) is internally
irreversible. 










