# Example Problem with Complete Solution

6G-1 : Efficiency and Coefficient of Performance of Carnot Cycles 4 pts
A Carnot Cycle operates between thermal reservoirs at 55oC and 560oC. 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: TH 560 oC TC 55 oC
TH 833.15 K TC 328.15 K
Find: h ??? COPR ??? COPHP ???
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 : COPR 0.6498
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 thermal reservoir at 560oC !!
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 : COPHP 1.6498
This is a 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 560oC !!
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% COPR 0.650 COPHP 1.65 