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 airconditioner 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 

Q_{C} 
15 
Btu/s 


Find: 
a.) 
COP_{HP} 
??? 

b.) 
COP_{R} 
??? 




Diagram: 













Assumptions: 
1  
The heat pump and the refrigerator operate at steadystate. 










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 Q_{C} 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 


Q_{C} 
54000 
Btu/h 







COP_{R} 
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 tiefighter diagrams are used. The arrows on the diagrams indicate the direction that heat and work are moving. 





Eqn 3 



Use Eqn 3 to eliminate Q_{H} from Eqn 2 to get : 





Eqn 4 



Now, we can plug in
the numbers in consistent units that we used in part (a). 



COP_{HP} 
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 : 
COP_{R} 
2.12 




COP_{HP} 
3.12 




























