Heat pump COP theoretical maximum
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. 
Re: Heat pump COP theoretical maximum
Quote:
A COP>2 is possible, if the temperature difference is not too large. Looks like you have some error in your calculations. 
Re: Heat pump COP theoretical maximum
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=T_{hot}/(T_{hot}T_{cold}) maybe it is for calculating the possible improvment over air to air pump. T_{hot} being the earth and T_{cold} being the air around the building. The common miscontseption being that using T_{cold} = outside temp, and T_{hot} = temp inside building. While should be T_{cold} =outside temp, and T_{hot} = earth (or other heat) source temperature. Huge amount of people are scammed with this false calculation! 
Re: Heat pump COP theoretical maximum
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(T_{cold}) to the hot reservoir(T_{hot}). T_{cold}<T_{hot}. 
Re: Heat pump COP theoretical maximum
Quote:
[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. Quote:

Re: Heat pump COP theoretical maximum
Thanks DrClaude.
Quote:
Quote:
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. 
Re: Heat pump COP theoretical maximum
Quote:

Re: Heat pump COP theoretical maximum
Quote:
Quote:
Quote:

Re: Heat pump COP theoretical maximum
Quote:
But what happens to COP when Th closes to infinity? 
Re: Heat pump COP theoretical maximum
Quote:

Re: Heat pump COP theoretical maximum
Quote:
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! 
Re: Heat pump COP theoretical maximum
Quote:

Re: Heat pump COP theoretical maximum
Quote:
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. 
Re: Heat pump COP theoretical maximum
Quote:

Re: Heat pump COP theoretical maximum
Quote:
[tex] \mathrm{COP} = \frac{\textrm{benefit}}{\textrm{cost}} = \frac{Q_h}{W} [/tex] 
Re: Heat pump COP theoretical maximum
Quote:

Re: Heat pump COP theoretical maximum
Quote:
Quote:
Quote:

Re: Heat pump COP theoretical maximum
Quote:

All times are GMT 5. The time now is 03:11 AM. 
Powered by vBulletin Copyright ©2000  2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums