Example Problem with Complete Solution

6G-1 : Efficiency and Coefficient of Performance of Carnot Cycles
4 pts
A Carnot Cycle operates between 200°C and 1200°C. Calculate…
                   
a.) …its thermal efficiency if it operates as a power cycle
b.) …its COP if it operates as a refrigerator
c.) …its COP if it operates as a heat pump
Read : This is a straightforward application of the definitions of efficiency and coefficient of performance.
Given : TH
1200
°C TC
200
°C
TH
1473.15
K TC
473.15
K
Find : η
???
COPR
???
COPHP ???
Diagram : Not necessary for this problem.
Equations / Data / Solve :
Part a.) The thermal efficiency of a Carnot Cycle depends only on the temperatures of the reservoirs with which it interacts.  The equation that defines this relationship is :
 
Equation. Eqn 1
Just be sure to use absolute temperature in Eqn 1 !  In this case, convert to Kelvin.  Temperatures in Rankine will work also.
η 67.9%
part b.) The coefficient of performance of a Carnot Refrigeration Cycle also depends only on the temperatures of the reservoirs with which it interacts.  The equation that defines this relationship is :
Equation. Eqn 2
Using T in Kelvin yields : COPR 0.4732
This is an exceptionally BAD COPR because it is less than 1.  This isn't terribly surprising when you consider that the refrigerator must reject heat to a reservoir at 1200°C !!
Part c.) The coefficient of performance of a Carnot Heat Pump Cycle also depends only on the temperatures of the reservoirs with which it interacts.  The equation that defines this relationship is :
Equation. Eqn 3
Using T in Kelvin yields : COPHP 1.4732
This is an BAD COPHP 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 1200°C !!
Notice also that : Equation. Eqn 4
This is always true for Carnot Cycles.
Verify : No assumptions to verify that were not given in the problem statement.
Answers : η 67.9% COPR 0.473 COPHP 1.47
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