# Example Problem with Complete Solution

6B-2 : Coefficient of Performance of a Heat Pump and a Refrigerator 4 pts
A geothermal heat pump absorbs 15 Btu/s of heat from the Earth 50 ft below a house. This heat pump uses a 10 hP compressor.
a.) Calculate the COP of the heat pump.
b.) In the summer, the cycle is reversed to cool the house. Calculate the COP of the cycle when it is operated as an air-conditioner assuming the working fluid rejects 15 Btu/s to the Earth.

Read : Here we must apply the definition of COP for both refrigerators and heat pumps.
Given: W 10 hP QC 15 Btu/s
Find: a.) COPHP ??? b.) COPR ???
Diagram:  Assumptions: 1 - The heat pump and the refrigerator operate at steady-state.
Equations / Data / Solve:
Part a.) If the purpose is to cool the groundwater, then the device is a refrigerator.
So, let's begin with the definition of the coefficient of performance for a refrigerator. Eqn 1
We are given the values of both QC and W, so all we need to do is make the units consistent and then plug values into Eqn 1.
Conversion Factors : 1 hP = 2545 Btu/h W 25450 Btu/h
QC 54000 Btu/h
COPR 2.122
Part b.) If the purpose is to heat a building, then the device is a heat pump.
So, let's begin with the definition of the coefficient of performance for a heat pump. Eqn 2
Next, we apply the 1st Law to the heat pump cycle, keeping in mind that we traditionally do not use our sign convention when tie-fighter diagrams are used.  The arrows on the diagrams indicate the direction that heat and work are moving. Eqn 3
Use Eqn 3 to eliminate QH from Eqn 2 to get : Eqn 4
Now, we can plug in the numbers in consistent units that we used in part (a).
COPHP 3.122
Notice that : Eqn 5
This is always true for Carnot Cycles.
Verify: No assumptions to verify that were not given in the problem statement.
Answers : COPR 2.12 COPHP 3.12 