# Example Problem with Complete Solution

8A-1 : Entropy Generation and Thermal Efficiency in Power Cycles 5 pts
A power cycle exchanges heat with only two thermal reservoirs at 500oR and 2000oR. QH = 4500 Btu/h.
For each of the following, calculate the rate of entropy generation in Btu/oR 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: TH 2000 oR a.) hA 0.83
QH 4500 Btu b.) hB 0.75
TC 500 oR c.) hC 0.44
Diagram: Find: Parts a-c.) Sgen ??? Btu/oR
Internally Reversible ? Internally Irreversible ? Impossible ?
Assumptions: 1 - The system undergoes a power cycle while receiving QH at TH and discharging QC at TC.
2 - The region of the system that receives heat from the hot reservoir remains at a constant temperature of TH = 2000oR.
3 - The region of the system that rejects heat to the cold reservoir remains at a constant temperature of TC = 500oR.
Equations / Data / Solve:
Entropy generation is defined by: Eqn 1
We can solve Eqn 1 for Sgen : Eqn 2
Since we are dealing with a cycle,
DS = 0 and Eqn 2 becomes: Eqn 3
In our process, the system  receives heat, QH, at a constant temperature, TH , and rejects heat, QC, at a constant temperature, TC.  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 +dQH and -dQC 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 QC.  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 QC yields : Eqn 6
Now, we can plug numbers into Eqn 6 to determine QC and then plug QC and the given values of QH, TH and TC into Eqn 4 to complete the solution.
Part (a) QC 765 Btu Sgen -0.72 Btu/oR
Part (b) QC 1125 Btu Sgen 0.00 Btu/oR
Part (c) QC 2520 Btu Sgen 2.79 Btu/oR
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 Sgen = 0 the cycle is internally reversible.
If Sgen > 0 the cycle is internally irreversible.
If Sgen < 0 the cycle is impossible.
Verify: The assumptions made in this solution cannot be verified with the given information.
Answers : Part (a) Sgen -0.72 Btu/oR   Part (a) is impossible.
Part (b) Sgen 0.00 Btu/oR   Part (b) is internally reversible.
Part (c) Sgen 2.79 Btu/oR   Part (c) is internally irreversible. 