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Homework Help: Minimum consumption rate of a heat pump

  1. Dec 26, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-12-26_17-1-11.png

    2. Relevant equations


    3. The attempt at a solution

    Effiency of a heat pump is given as ## \eta = \frac { T_h }{T_h – T_c} = 30 ##

    W * 30/s = 6000*4.2 cal /s

    Minimum consumption rate = W/s = 840 watt

    Is this correct?
     
  2. jcsd
  3. Dec 26, 2017 #2

    rude man

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    This formula is not quite correct. It results in only about a 3% error in this case but it could be a lot more if Th - Tc were a lot larger.

    So actually none of the choices is particularly good. But maybe you picked the closest ...
     
  4. Dec 27, 2017 #3
    The question is asking for minimum consumption rate.
    For this I have to use a heat pump with carnot engine.
    For this case, isn't the followiing right ?
    The textbook by blundell and blundell says,
    upload_2017-12-27_12-16-26.png
     
  5. Dec 27, 2017 #4

    rude man

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    OK I see the problem I think.
    Your textbook defines efficiency as e = Qh/W = (Qc + W)/W for some reason.
    But most textbooks talk about the "coefficient of performance" (COP) which is much more logical since this is heat removed per unit of work.
    So per your textbook e = 30 but for the COP it's 29.
    The word "efficiency" really doesn't apply here IMO since by definition efficiency can never exceed unity.
     
    Last edited: Dec 27, 2017
  6. Dec 27, 2017 #5
    What is meant by IMO?
     
  7. Dec 27, 2017 #6

    rude man

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    " In My Opinion".
    Many such abbreviations were started with texting. This one is extremely widely used.
     
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