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Energy consumption of Fridge with faulty light

  1. Sep 24, 2015 #1
    1. The problem statement, all variables and given/known data

    Coefficient of performance of a refrigerator is 1.4

    A faulty 20w bulb remains on when the fridge is closed.

    Under normal circumstances, the bulb turns on when the fridge is open. The door is opened 20 times a day for 30 seconds.

    Calculate the increase in energy consumption in 1 year.



    2. Relevant equations

    Coefficient of performance = heat extracted from cold reservoir Qc/work delivered to pump W

    COP = Qc/W

    3. The attempt at a solution

    This is where I am stuck, the work done by a compressor is not given, a bulb doesnt relate to the COP? Here's my attempt:

    P=20w, under normal operation, T= 30x20x365 s therefor total energy in a year is 20x219x10^3 = 4.38x10^6 J

    Under faulty conditions: P=20w, T= 31.536x10^6 s therefor total energy = 20T = 630.7x10^6 J

    Increase in energy = 630.7x10^6 J - 4.38x10^6 J = 626.3x10^6 J

     
  2. jcsd
  3. Sep 24, 2015 #2

    jbriggs444

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    A 20 watt bulb is glowing in the middle of a closed refrigerator. How much heat is that bulb adding to the interior?
     
  4. Sep 24, 2015 #3
    Thanks for the reply but I am unsure how to calculate the part thermal energy given to the fridge from the light of the bulb
     
  5. Sep 24, 2015 #4

    jbriggs444

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    Try conservation of energy. 20 watts of electrical power goes into the light. How much power comes out?

    Edit: Possibly a little background is called for.

    Light bulbs are traditionally classified based on the electrical power they consume. A 20 watt bulb consumes 20 watts of electrical power. This tradition was standard in the days when incandescent bulbs were ubiquitous and was used for traditional overhead fluorescent lighting, even though the amount of radiated light per input watt was very different for the two lighting types.

    Modern light bulbs are typically compact fluorescent (CFL) or light emitting diode (LED) based. The labelling on these is not based on input power but, for marketting and convenience, is instead based on the power that would be consumed by an equivalent incandescent bulb.

    The person who came up with this problem is fairly clearly assuming a traditional incandescent lamp where the "20 watts" denotes input power consumed.
     
    Last edited: Sep 24, 2015
  6. Sep 24, 2015 #5
    When the door is closed, the fridge is a closed system, light cannot escape so the entire electrical power consumed by the bulb eventually becomes the additional thermal energy of the fridge, 20 J/s
     
  7. Sep 24, 2015 #6
    So an additional 20J/s needs to be extracted from the fridge to keep it cool, reducing the COP?
     
  8. Sep 24, 2015 #7

    jbriggs444

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    The point is that it needs to be extracted. The COP is needed to tell you how much power must be supplied to the compressor motor in order to extract that much heat.
     
  9. Sep 24, 2015 #8
    I can redefine COP as COP= heat to be extracted (Qc)/ Electrical work to operate compressor. We have found the additional heat under faulty conditions, therefor the additional work is Additional heat/COP. This makes sense to me, is this the correct approach?
     
  10. Sep 24, 2015 #9

    jbriggs444

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    Yes. Of course this is the additional work beyond the additional power already required just to operate the lamp for the additional time.
     
  11. Sep 24, 2015 #10
    Thanks for your help, appreciate it
     
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