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Thermodynamics, heat, work, refrigerator

  1. Jul 12, 2013 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Hello guys,
    I've attempted a problem but I get a result that I find too low to make sense. Plus, I have a doubt. So here it comes:
    I've put in red the part I'm unsure of. As far as I know the efficiency of a refrigerator could be any real number from 0 to infinity (1 is not the limit!). So what exactly is 15% of the "ideal" coefficient performance?! The ideal would be an efficiency of infinity?!

    2. Relevant equations
    Not many.


    3. The attempt at a solution
    Per day there are 750 kcal =3139.5 kJ entering the refrigerator.
    1kW h =##3.6 \times 10^ 6 J## which corresponds to a price of 0.025$.
    So if by efficiency of 15% of the ideal one they mean that for each calory that enters the refrigerator I must pay as if I remove 1/0.15 =6.667 calories out of it, then I reach that I must pay as if I must remove 20930 kJ per day. This corresponds to approximately 0.145$ per day. So in a month, 4.36$ only.
    The result seems too low to me and I don't understand why they give the data of "2°C" nor exactly what they mean by 15% of the ideal coefficient performance.
    Thanks for any help.
     
  2. jcsd
  3. Jul 12, 2013 #2

    Andrew Mason

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    I think you have to assume that the room it is in is kept at 20°C or 293K.

    What would the COP be if the refrigerator was operating with maximum possible thermodynamic efficiency? Just make the COP of .15COPmax. How much work would have to be done to move 15 x 50 kcal from 275K to 293K? How much would that cost? Multiply by the number of days in whatever month you choose or use 365.25/12

    AM
     
  4. Jul 14, 2013 #3

    fluidistic

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    Thanks for helping me Andrew.
    Ah I see... I didn't know I have to assume an extra datum, namely the temperature of the environment but that makes sense. I've assumed 300K which is somehow warmer than 20 Celcius. The ideal COP in this case would be 11 (##\frac{300}{300-275}##) so the "real COP" would be 1.65.
    I didn't do any arithmetics but it's an efficiency greater than the one I had assumed in my first post (which was 0.15) so I should reach a price even smaller than the one I had found... Am I right?
     
  5. Jul 16, 2013 #4

    fluidistic

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    Ok I don't know what's wrong here...
    The COP is ##\frac{T_h}{T_h-T_c}## where ##T_c=275 K## and I assume ##T_h=300K##. In this case the COP is worth 12. However the problem states that the refrigerator operates at 15% of that value, which implies a COP of 1.8.
    This means that for each joule of work that I pay, I can extract 1.8 joule from (cold part of) the refrigerator.
    In a day I need to withdraw 750 kcal =3135.5 kJ from the refrigerator.
    For 1$ I can withdraw 1/0.025=40 kW h. That's 144000 kJ.
    So per day I need to pay an amount of 3135.5/144000 ##\approx 0.2## $.
    So for month, roughly 0.65$.

    This doesn't seem to make any sense to me. The value seems way too low.
     
  6. Jul 16, 2013 #5

    rude man

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    COP = Tc/(Th-Tc) = 11, not 12.
    15% of 11 = 1.65. You get 1.65 J of heat removed for every J of energy you put in.

    In 1 month you remove 50kcal x 15/day x 30 days = 22,500 kcal.
    How many joules in 225,000 kcal? http://www.rapidtables.com/convert/energy/Calorie_to_Joule.htm
    Remember to use thermochemical, not food, calories.

    How many kw-hrs per joule? Remember, 1 joule = 1 watt-sec.
    Then multiply the number of kw-hrs by $0.025.
    This is just keeping accurate track of units conversion!
     
    Last edited: Jul 16, 2013
  7. Jul 16, 2013 #6

    fluidistic

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    Thanks rudeman.
    The COP is indeed 11 (as I had found in post 3) although I made a mistake in my last post where I took the efficiency of a heat pump instead of refrigerator. Anyway changing an 11 for a 12 shouldn't change the result by more than 10% (in fact much less than this).
    So as you said, for each joule of work that I pay, I can withdraw 1.65 joule from the refrigerator.
    As for the conversion calory to joule, I used 1 cal =4.186 joule which is pretty close to the one in the website you provide (1 cal=4.184 joule), again the result should not change much.
    For the sake of it, I'll take your units conversion. So I must remove 50*15=750 kcal per day, which are worth 3138 kJ.
    1 kW h= ##\frac{1000J}{s} \cdot 3600s =3.6 \times 10 ^6 J##. To extract this amount of energy costs me 0.025$.
    For 1$ I can extract 40 kW h =##144 \times 10 ^6 J=144000kJ##.
    Per day the cost is thus 3138/144000 $ ##\approx 0.0218##$. So per month this is approximately 0.65$.

    That's exactly the same result I obtained in my last post, even with taking an ideal COP of 12 instead of 11.

    I still don't see where I went wrong...
     
  8. Jul 16, 2013 #7

    rude man

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    You didn't factor in the COP of 1.65. That makes the monthly bill even lower!

    Other than that your computations check out.

    Today's typical rate is 15c per kW-hr so that would make your final answer 6 times as high, divided by the COP.
    I also think removing 750 kcal/day might be a bit low in reality.
     
  9. Jul 16, 2013 #8

    fluidistic

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    You're right, I forgot the factor...
    And ok about the real price. So the way of solving the problem was basically ok, I just wasn't careful enough with the algebra.
    Thanks.
     
  10. Jul 16, 2013 #9

    rude man

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    I think you went about it in a roundaboug way, but that's OK.
    (I would have calculated the number of joules in a month, equated it to kW-hrs, and multiplied by the price per kW-hr. No need to compute the daily costs etc.)
     
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