6G1 :  Efficiency and Coefficient of Performance of Carnot Cycles  4 pts 

A Carnot
Cycle operates between thermal
reservoirs at 55^{o}C and 560^{o}C. Calculate… a.) The thermal efficiency, h, if it is a power cycle b.) The COP if it is a refrigerator c.) The COP if it is a heat pump 

Read :  This is a straightforward application of the definitions of efficiency and coefficient of performance.  
Given:  T_{H}  560  ^{o}C  T_{C}  55  ^{o}C  
T_{H}  833.15  K  T_{C}  328.15  K  
Find:  h  ???  COP_{R}  ???  COP_{HP}  ???  
Diagram:  Not necessary for this problem.  
Assumptions:  None.  
Equations / Data / Solve:  
Part a.)  The thermal efficiency of a Carnot Cycle depends only on the temperatures of the thermal reservoirs with which it interacts. The equation that defines this relationship is :  

Eqn 1  
Just be sure to use absolute temperature in Eqn 1 ! In this case, convert to Kelvin. Temperatures in Rankine will work also.  
h  60.6%  
Part b.)  The coefficient of performance of a Carnot Refrigeration Cycle also depends only on the temperatures of the thermal reservoirs with which it interacts. The equation that defines this relationship is :  

Eqn 2  
Using T in Kelvin yields :  COP_{R}  0.6498  
This is an exceptionally BAD COP_{R} because it is less than 1. This isn't terribly surprising when you consider that the refrigerator must reject heat to a thermal reservoir at 560^{o}C !!  
Part c.)  The coefficient of performance of a Carnot Heat Pump Cycle also depends only on the temperatures of the thermal reservoirs with which it interacts. The equation that defines this relationship is :  

Eqn 3  
Using T in Kelvin yields :  COP_{HP}  1.6498  
This is a BAD COP_{HP} because it is just barely greater than 1. This isn't terribly surprising when you consider that the heat pump must put out heat to a reservoir at 560^{o}C !!  
Notice also that : 

Eqn 4  
This is always true for Carnot Cycles.  
Verify:  No assumptions to verify that were not given in the problem statement.  
Answers :  h  60.6%  COP_{R}  0.650  COP_{HP}  1.65 