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 



T_{H} 
1550 
K 

T_{surr} 
298 
K 



T_{C} 
350 
K 










Find: 
W_{s,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 : 

W_{s,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 

















h_{R} 
0.8077 












Solving Eqn 2 for the work produced by the HE
yields : 


Eqn 3 

















W_{HE} 
8077 
kJ 











The COP of a Carnot
HP is : 



Eqn 4 

















COP_{R} 
6.731 












Solving Eqn 4 for the work required by the HP
yields : 


Eqn 5 

















W_{HP} 
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. 











W_{s,rev} 
6592 
kJ 



W_{s,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 

















S_{gen} 
22.12 
kJ/K 











Lost
work is related to the total entropy generation by : 





















Eqn 7 







W_{s,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 : 
W_{s,lost} 
6590 
kJ 













