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Thermodynamic Question (refrigerator)

  • Thread starter gotrilum
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  • #1
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I'm having a little trouble in completing this problem. I would appreciate your answer if anyone knows how to solve it. Thanks

An ideal refrigerator removes heat at a rate of 0.15kW from an interior temperature of 1 degree celsius. The exhaust from the refrigerator is 45 degrees celsius. How much electrical power is used.

*Note that this is a refrigerator problem (reverse process from a carnot engine)

I have found the efficiency using Th/Tc - 1 = .29 I do not know how to proceed after this.
 

Answers and Replies

  • #2
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Remember efficieny is always

what you get / what you pay for.

so in this case its
Ql/Win

and for the reversible carnot this comes to:

COP = Ql/(Qh-Ql)

You must remember the for refrigiration cycles its not technically efficiecny, its coefficient of performance we use. This avoids the confision of having efficienys over 100%.

Where you were going wrong was you were taking the efficieny for a carnot cycle and applying it directly.
 
  • #3
Andrew Mason
Science Advisor
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I'm having a little trouble in completing this problem. I would appreciate your answer if anyone knows how to solve it. Thanks

An ideal refrigerator removes heat at a rate of 0.15kW from an interior temperature of 1 degree celsius. The exhaust from the refrigerator is 45 degrees celsius. How much electrical power is used.

*Note that this is a refrigerator problem (reverse process from a carnot engine)

I have found the efficiency using Th/Tc - 1 = .29 I do not know how to proceed after this.
Coefficient of Performance is Qc/W. W = Qh - Qc.

So:

COP = (Qc /(Qh - Qc)

1/COP = Qh/Qc - 1

For a Carnot cycle, this becomes:

1/COP = Th/Tc - 1

You have to determine the COP from the temperatures. Using the value for Qc you can determine W.

AM
 

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