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Heat from an A/C inverter calculation more than energy consumed?

  1. Oct 25, 2014 #1
    Hi guys can anyone please tell me that it is possible with an air conditioner of type inverter to put less energy than what the piece is receiving as heat. So let's say that on the inverter it says output KWh/ input KWh =4.

    Henthalpy= G+TS
    4*Hinput= Houtput inside the house

    So to heat my house I use 4 times less energy if I just used a normal electric heater or this is what they write in product description?
  2. jcsd
  3. Oct 25, 2014 #2


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    If you use a standard electric heater, 100% of the electric energy consumed is converted in heat.

    But, in if a 'heat pump', you use the electric energy to drive a compressor, and the machine converts low-temperature heat, available in the external surroundings, in the external air, mainly, to high temperature heat, that may be used to heat your room. The 'transportation cost' is lower than the energy costs of a resistance heater of the same heating power. How much lower depends on the difference of inside/outside temperatures. The larger the difference, the smaller the advantage with respect to a resistance heater.

    A refrigerator (or heat pump, that is the same) is esentially a thermal engine working in reverse. Instead of taking heat from a high-temperature place, producing useful mechanical work, and discharging that low-temp heat in a low-temperature place, the refrigerator/heat pump takes low-temp heat from a low temperature place, injects mechanical power, and places the resulting high-temp heat in a high temperature environment...

    I believe that using '4 times less energy' may be too optimistic...
    Last edited: Oct 25, 2014
  4. Oct 25, 2014 #3


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    Staff: Mentor

    I guess that what you are describing is more commonly known as a heat pump. If that is the case, then indeed you get more heat out than the electricity input, as additional heat is extracted from the outside air. This is more efficient than direct electric heating.
  5. Oct 25, 2014 #4


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    This two very clear pictures are from 'The Second Law', by Henry Bent:

    15599507266_916eb43343.jpg 15436560709_bc1ee81394.jpg
  6. Oct 26, 2014 #5
    Thanks for your input,
    Just to add to this
    For an ideal heat pump cycle:

    COP = TH/(TL-TH)
    source wikipedia.
  7. Oct 26, 2014 #6
    Thanks for the reply DrClaude, just to add to your thing there is heat pump systems that are air-water, water-water and air-air( air conditioner inverter).
    For an ideal heat pump cycle:

    COP = TH/(TL-TH)
    source wikipedia.
  8. Oct 26, 2014 #7
    How does delta S stay zero in both directions? So the reversed Carnot is the refrigeration cycle, but do these heat pumps use only one cycle, and is it possible that they have some newer technologies so that the efficiency of an air-air heat pump reaches 4 or 5 up to -20 degrees C?
  9. Oct 26, 2014 #8


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    400K reservoir ---> ΔE (cal) = -1200 cal ΔS (cal/K) = -3 cal/K
    300K reservoir ---> ΔE (cal) = +900 cal ΔS (cal/K) = +3 cal/K
    Weight -------------> ΔE (cal) = +300 cal

    Balance -----------> ΔE = 0 cal ΔS = 0 cal/k
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