# Example Problem with Complete Solution

4F-3 : Coefficient of Performance of a Refrigeration Cycle 3 pts
An industrial refrigerator rejects heat at a rate of 24,750 kJ/min to the surroundings. If the refrigeration cycle has a COP of b = 3.3, determine QC and Wcycle, each in kJ/min.

Read : This one is a straightforward application of the definition of the the 1st Law and the COP of a refrigeration cycle. Diagram : Given: COP = b 3.3 Find: QC ??? kJ/min
QH 24,750 kJ/min Wcycle ??? kJ/min
Assumption: - The cycle only exchanges heat with the two thermal reservoirs.
Equations / Data / Solve:
1st Law applied to the refrigerator: Eqn 1
Definition of COP for a refrigerator : Eqn 2
Degree of freedom analysis: 2 eqns in 2 unknowns: QC and Wcycle.
Solve Eqn 2 for QC and use the result to eliminate QC from Eqn 1 : Eqn 3 Eqn 4
Next, solve Eqn 4 for Wcycle in terms of the known quantities QH and b. Eqn 5
Plug numbers into Eqn 5 : Wcycle 5755.8 kJ/min
Now, use this value for Wcycle and the given value of b in Eqn 3 to evaluate QC :
QC 18994.2 kJ/min
Verify: The only assumption cannot be verified.
Answers : QC = 19000 kJ/min Wcycle = 5760 kJ/min 