# Example Problem with Complete Solution

8D-1 : Lost Work Associated with Heat Transfer 5 pts
A thermal reservoir at 1550 K transfers 10,000 kJ of heat to a thermal reservoir at 350 K.  The temperature of the surroundings is 298 K.  Determine the lost work for this process based on :
a.) The performance characteristics of Carnot Cycles.
b.) The total entropy generation of this process.

Read : Heat transfer through a finite temperature difference is irreversible. It results in entropy generation and represents a lost opportunity to do work. Part (a) is more challenging because we need to build a hypothetical process out of reversible HE's, Ref's and HP's that accomplishes the same net heat transfer as the real process and then determine how much work we COULD have obtained using these reversible devices. Part (b) is a straightforward application of the definition of entropy generation, the entropy change for an isothermal process such as a thermal reservoir and the relationship between lost work and entropy generation.
Given: Q 10000 kJ TH 1550 K
Tsurr 298 K TC 350 K
Find: Ws,lost ??? kJ
Assumptions: 1 - No shaft work is obtained in the actual process.
2 - Both reservoirs are true thermal reservoirs whose temperature does not change as heat is added or removed.
Diagram: Actual Process Hypothetical Process Equations / Data / Solve:
Part a.) The definition of lost work is : Eqn 1
In our actual process, no shaft work is produced, so : Ws,act 0 kJ
So, we need to evaluate the reversible work in order to determine the lost work.
In order to evaluate the reversible work, we must setup a reversible process that accomplishes the same thing as the actual process.
The hypothetical process must result in a transfer of 10000 kJ of heat from the hot reservoir to the cold reservoir.
We can build the hypothetical process from Carnot Cycles.  The diagram of the hypothetical process I have chosen includes a heat engine and a heat pump, both of which are reversible.
The HE must absorb the 1000 kJ from the hot reservoir and the HP must reject 10000 kJ to the cold reservoir.  In this way, the hypothetical process does indeed accomplish the same thing as the real process.  The work that this hypothetical (reversible) process produces is the reversible work and Eqn 1 tells us that is also the lost work because the actual work is zero.
So, now we need to use the Carnot Efficiency and COP to evaluate the lost work.
The Carnot Efficiency of our HE is : Eqn 2
hR 0.8077
Solving Eqn 2 for the work produced by the HE yields : Eqn 3
WHE 8077 kJ
The COP of a Carnot HP is : Eqn 4
COPR 6.731
Solving Eqn 4 for the work required by the HP yields : Eqn 5
WHP 1486 kJ
The reversible work, and therefore the lost work, is equal to the difference between the work produced by the reversible HE and the work required by the reversible HP.
Ws,rev 6592 kJ Ws,lost 6592 kJ
Part b.) The entropy generation by the real process is equal to the entropy change of the universe resulting from the process.  The entropy change of the universe is made up of the entropy increase of the cold reservoir and the entropy decrease of the hot reservoir because of the transfer of 1000 kJ. Eqn 6
Sgen 22.12 kJ/K
Lost work is related to the total entropy generation by : Eqn 7
Ws,lost 6592 kJ
We can combine Eqns 6 & 7 to obtain a convenient equation for calculating the lost work associated with heat transfer through a finite temperature difference. Eqn 8
Note: In this equation Q is the absolute value of the amount of heat transferred (a positive quantity).
Verify: None of the assumptions made in this problem solution can be verified.
Answers : Parts a & b : Ws,lost 6590 kJ 