# Example Problem with Complete Solution

8A-2 : Heat, Work and Entropy Generation 5 pts
Two power cycles operate between the same two thermal reservoirs, as shown below. Cycle R is reversible and cycle I is irreversible. They each absorb the same amount of heat from the hot reservoir, QH, but produce different amounts of work, WR and WI, and reject different amounts of heat to the cold reservoir, QC and Q'C.
a.) Derive an equation for Sgen for the irreversible cycle in terms of WI, WR, and TC only.
b.) Show that WI  WR and Q'C  QC.

Read : Start with the equation for the entropy generated and do an energy balance on both the reversible and irreversible cycles. Put the equations together and simplify to get an equation in the desired terms.
Given: A reversible power cycle, R, and an irreversible power cycle, I, operate between the same two reservoirs.
Find: Part (a) Evaluate Sgen for cycle I in terms of WI, WR, and TC.
Part (b) Show that: WI < WR and Q'C > QC.
Diagram: The diagram in the problem statement is adequate.
Assumptions: 1 - The systems shown undergo power cycles. R is reversible and I is irreversible.
2 - Each system receives QH at a constant temperature region at TH from the hot reservoir and rejects heat, QC, at a constant temperature region at TC to the cold reservoir.
Equations / Data / Solve:
Part a.) Let's begin with the defintion of entropy generation: 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 the irreversible process, the system  receives heat, QH, at a constant temperature, TH , and rejects heat, Q'C, at a constant temperature, TC.  Because the temperatures are constant, they can be pulled out of the integral in Eqn 3 leaving : Eqn 4
The 1st Law for cycles is: Eqn 5
We can apply Eqn 5 to both the reversible and the irreversible cycles, as follows : Eqn 6 Eqn 7
We can combine Eqns 6 & 7 to obtain : Eqn 8
Now, solve Eqn 8 for Q'C : Eqn 9
Next, we can use Eqn 9 to
eliminate Q'C from Eqn 4 to get : Eqn 10
We can rearrange Eqn 10 slightly to
make it more clear how to proceed : Eqn 11
Because R is a reversible cycle and
we use the
Kelvin Temperature Scale : Eqn 12
Eqn 12 can be rearranged to help simplify Eqn 11 : Eqn 13
This yields : Eqn 14
Part b.) Because irreversibilities are present in cycle I : Eqn 15
Rearranging Eqn 15 gives us : Eqn 16
Finally, we can rearrange Eqn 9 to help us determine whether Q'C or QC is larger : Eqn 17
Since Eqn 16 tells us that WR > WI, Eqn 17 tells that : Eqn 18
Verify: The assumptions made in this solution cannot be verified with the given information.
Answers : Part a.) Part b.)   