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Second Law of Thermodynamics - heat pumps and air conditioners

  • Thread starter elcaptain
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  • #1
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Homework Statement


A heat pump operates as an air conditioner, drawing 20000 Btu per hr from a building. It is operating between 21 degrees Celsius and 41 degrees Celsius.

a) If the COP (coefficient of performance) is 35% that of a Carnot air conditioner, what is the value of the effective COP (coefficient of performance)?

b) What is the value of the power (Watts) required for the compressor motor?


Homework Equations


QL = heat removed from the low temperature area (inside the building)
QH = heat removed from the high temperature area (outside the building)
W = work done to remove the heat
TL = temperature of system at lower temperature (inside the building)
TH = temperature of system at higher temperature (outside the building)

COP = QL/W

COP = QL/(QH - QL)

COPideal = TL / (TH - TL)

The Attempt at a Solution


I first converted Btu (British thermal unit) to joules: 1Btu = 1055.056 J. Then I tried finding the ideal COP by using the above equation with the given temperatures and got 1.05, but I'm pretty sure the ideal COP is not calculated this way. I tried multiplying 1.05 by 0.35 but got the wrong answer for part a. Thanks for any help.
 

Answers and Replies

  • #2
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anything?
 

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