Heat pump COP theoretical maximum

  1. Hello everybody.

    I have trouble of proving the theoretical maximum of heat pump Coefficent of Performance.
    The thing I'm trying to calculate is for heat pump pumping heat from colder reservoir to the hotter reservoir:
    COP=Qh/W
    Qh - heat supplied to the hot reservoir(output)
    W - mechanical work consumed by the pump(input)

    The formulas in the wikipedia seems to be wrong and I want to derive the correct formulas from the thermodynamic laws. With current Wikipedia formula it is normal to get COP >2. Search wikipedia:Coefficent of Performance.

    As I calculate the heat balance of the basic air to air heat pump, any COP values above 2 will violate the second law of the thermodynamics, and therefore make perpetual motion machine possible.

    So is what is the correct formula and what is the way to prove it?

    I argued a lot with my friend about that, when he wanted to buy the heat pump to heat his office. Eventually he still bought the pump because I could not provide the calculation and he decided to believe what the sales man told him (COP=3 or something). But now he do not seem very happy with the electricity bill.

    Thanks in advance.
     
  2. jcsd
  3. mfb

    Staff: Mentor

    I don't think so.

    A COP>2 is possible, if the temperature difference is not too large. Looks like you have some error in your calculations.
     
  4. According to my understanding COP >2 is not possible for air to air heat pump. Based on assumption that heat from the building is lost to the same air (cold reservoir) from where it is pumped to inside the building (hot reservoir).

    The COP>2 is possible for land to air pump, when the cold reservoir is hotter than the air aruond the building to where the heat is lost. But not for the air to air.

    The formula COP=Thot/(Thot-Tcold) maybe it is for calculating the possible improvment over air to air pump. Thot being the earth and Tcold being the air around the building. The common miscontseption being that using Tcold = outside temp, and Thot = temp inside building. While should be Tcold =outside temp, and Thot = earth (or other heat) source temperature.

    Huge amount of people are scammed with this false calculation!
     
  5. OK, let me put it that way:

    Can anyone prove that COP>2 is possible, based on thermodynamic laws and other postulates?
    Conditions: pumping heat from cold reservoir(Tcold) to the hot reservoir(Thot). Tcold<Thot.
     
  6. DrClaude

    DrClaude 2,136
    Science Advisor
    Homework Helper

    [itex]Q_h[/itex]: heat pumped into the building
    [itex]Q_c[/itex]: heat taken from outside the building
    [itex]W[/itex]: electrical energy used by the heat pump
    [itex]T_h[/itex]: temperature inside the building
    [itex]T_c[/itex]: temperature outside the building

    1st law: [itex]Q_h = W + Q_c[/itex]

    Therefore [tex]\mathrm{COP} = \frac{Q_h}{W} = {Q_h}{Q_h - Q_c} = \frac{1}{1 - Q_c/Q_h} > 1[/tex]

    2nd law: [tex]\frac{Q_h}{T_h} \ge \frac{Q_c}{T_c}[/tex] therefore [tex]1 - \frac{T_c}{T_h} \le 1 - \frac{Q_c}{Q_h}[/tex] for which we get [tex]\mathrm{COP} \le \frac{1}{1 - T_c/T_h} = \frac{T_h}{T_h - T_c}[/tex]
    There is no fundamental reason why the COP cannot be greater than 2.

    I don't see how any losses from the building come into play in the calculation of the COP of heat pump itself.
     
  7. Thanks DrClaude.
    Can you explain more how you get this?

    For building heating, this is the main goal to determine most economic way to get the energy needed to maintain the temperature inside the building above the ambient. So naturally one wants to compare how much heating energy he gets using different apparatus. Correct understanding of the COP of the heat pump is needed for those calculations.

    There must be error in those calculations, because when simple heat pump has a COP of over 2, it is possible to reheat the input with the energy from the output and with that increase the COP infinetly resulting in infinite COP -- the mashine that pumps heat from cold reservoir to hot reservoir without needing any additional energy to run.
     
  8. DaleSpam

    Staff: Mentor

    Please show your calculations for this claim.
     
  9. DrClaude

    DrClaude 2,136
    Science Advisor
    Homework Helper

    Sorry, a part of the equation didn't come out right. [tex]\mathrm{COP} = \frac{Q_h}{W} = \frac{Q_h}{Q_h-Q_c} = \frac{1}{1-Q_c/Q_h} >1[/tex] I hope it is now clear.

    Yes, but this can be made independent of the COP of the pump, by assuming that you have a desired indoor temperature, for a given outdoor temperature. Of course, if your goal is to maintain an indoor temperature relative to the outside, it becomes more complicated. Note also that actual COPs are usually given for a standard pair of indoor and outdoor temperatures, and are mostly useful in comparing one pump with another. If you want to calculate the actual energy needed over a year, including variations of the outside temperature, you would need measured COPs as a function of inside/outside temperatures.

    No error, it is just how the COP is defined. Infinite COP will only occur for [itex]T_h=T_c[/itex], in which case the pump is not doing anything, so indeed it doesn't need any energy to run! Event for finite but huge COPs, they correspond to extremely small differences in temperatures that are not realistic in real world applications. And it is just an inequality: for real heat pumps [tex]\mathrm{COP} = \eta \frac{T_h}{T_h-T_c}[/tex] with [itex]\eta < 1[/itex] and itself a function of [itex]T_h[/itex] and [itex]T_c[/itex].
     
  10. Thanks.
    But what happens to COP when Th closes to infinity?
     
  11. DaleSpam

    Staff: Mentor

    The ideal COP approaches 1 as Th approaches infinity.
     
  12. This is what I am working on.

    We have cold reservoir, and hot reservoir. We are pumping from cold to hot.
    Lets assume we have a COP=5
    Mehanical work into the pump=1kW
    Heat output to the hot reservoir=5kW

    So now we let 2kw go back to heat the cold reservoir.
    This leaves us still 3kW output to hot reservoir, this heats the hot reservoir, but we do not let the hot reservoir temperature increase, the exess heat is conducted away. So Th=const.
    But added 2kW to cold reservoir will increase the cold reservoir temperature. And this increases the COP by formula COP=Th/(Th−Tc). So now we get even more output heat because of increased COP. And the cycle repeats. The COP is increased to infinity.
    We have perpetum mobile!
     
  13. DaleSpam

    Staff: Mentor

    Even with the increased COP it still takes work to move the heat back up. You seem to think that the heat that leaks into the cold reservoir is somehow free to move back to the hot reservoir.
     
  14. DrClaude

    DrClaude 2,136
    Science Advisor
    Homework Helper

    [itex]\mathrm{COP} \sim 1[/itex]

    Consider the following hypothetical situation. An ideal (Carnot engine) heat pump installed between a hot sink at [itex]T_h[/itex] and a cold source at [itex]T_c[/itex], and functions with a constant consumption of electrical energy ([itex]W = \mathrm{const.}[/itex]). Both sources are connected such that heat can flow from hot to cold to maintain the hot source at [itex]T_h[/itex] (imagine that we have a demon that can do that so that we don't need to worry about entropy here). The temperature of the cold source will vary according to
    [tex]
    T_c = T_{c0} - \frac{Q_c}{C} t + \frac{Q_h}{C} t
    [/tex]
    where [itex]T_{c0}[/itex] is the initial temperature of the cold source and [itex]C[/itex] its heat capacity. Since [itex]Q_h = W+Q_c[/itex], we have
    [tex]
    T_c = T_{c0} - \frac{Q_c}{C} t + \frac{W + Q_c}{C} t = T_{c0} + \frac{W}{C} t
    [/tex]
    so the heat pump is basically heating up the cold source as would a perfect electrical heater. No "over unity" or other violation of the laws of TD.

    Note that the initial value of the COP is of no relevance. This will work even if it is initially < 2. The value of the COP grows towards infinity during the process, but so what?

    Hope this helps.
     
  15. This is what I'm trying to prove. That this COP formula is incorrect.
     
  16. DrClaude

    DrClaude 2,136
    Science Advisor
    Homework Helper

    But the COP formula is just
    [tex]
    \mathrm{COP} = \frac{\textrm{benefit}}{\textrm{cost}} = \frac{Q_h}{W}
    [/tex]
     
    Last edited: Feb 14, 2013
  17. Not that one!
     
  18. DrClaude

    DrClaude 2,136
    Science Advisor
    Homework Helper

    I derived the COP in terms of temperatures above. If you can't find any fault with the derivation, then both are equivalent.

    Or you just need less [itex]W[/itex] for the same [itex]Q_h[/itex], which makes sense: the heat pump requires less work if the sink and the source are close to the same temperature.

    No, you reach [itex]T_c = T_h[/itex] and you don't have a heat pump anymore! Or, see my previous post, you just continue heating the cold source as would a electrical heater.
     
  19. Thanks, we are getting closer to the solution -- proving the maximum efficiency of the air to air, building heating, heat pump.
     
  20. DaleSpam

    Staff: Mentor

    It is correct, your perpetual motion argument is wrong. You should post your argument in full.
     
  21. What do you mean wrong. Is the formula COP=Th/(Th−Tc) wrong, or is there something else wrong?
     
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