Heat pump COP theoretical maximum

by Thermolelctric
Tags: heat, maximum, pump, theoretical
 Share this thread:
Mentor
P: 22,292
 Quote by Thermolelctric I'm still trying to undrestand how we can have so high COP numbers without violating the thermodynamic laws. The heat engine COP is below about 70% never more than 1, but the reverse process of heat pump can have COP=8 or more still we do not have the violation of thermodynamic laws.
Try the calculations. You'll be surprised at just how far "below about 70%" the heat engine efficiency is for a D-T that also gives an 8 COP.
Mentor
P: 22,292
 Quote by Thermolelctric Want to calculate: how much less electrical energy will the house consume, when it is heated with air to air heat pump, versus when it is heated by direct electrical heating. This is not the same value as COP. So there is the catch!
Er, yes it is!
Sci Advisor
HW Helper
P: 6,677
 Quote by russ_watters Er, well, I was way off on that. The federal requirement is 2.3. Heat pumps are much less efficient than air conditioners, which surprises me. Must be an issue of wider D-T in heating mode. Could also be an issue of including the defrost cycle and electric heat backup.
Your original point is well taken, though. It makes little practical sense because if the COP of the heat pump is less than about 3.5 there is not much point in having a heat pump powered by electricity produced from a thermal source.

The maximum efficiency of thermal plants is about 1/3 and the efficiency of an electric motor is about 80%, So with a COP of 3.5, for each 1.25 units of electrical energy (1 unit of output work) you would get 3.5+.25 = 3.75 units of heat (assuming the waste heat from the motor was added to the heat delivered to the heated space). But at 1/3 efficiency it takes 3.75 units of thermal energy to produce that 1.25 units of electricity - not to mention the infrastructure costs etc. of getting that electricity to you. So you are using 3.75 units of thermal energy to deliver 3.75 units of thermal energy.

It would still make some sense to use a heat pump with a COP less than 3.5 using if the heat pump was powered by electricity produced from wind or solar energy, provided that electricity could not be used to replace thermally produced electricity.

AM
Sci Advisor
HW Helper
P: 6,677
 Quote by Thermolelctric I'm still trying to undrestand how we can have so high COP numbers without violating the thermodynamic laws.
COP is not the efficiency of producing thermal energy from a low entropy source such as a chemical or electrical source. That can never exceed 1.

Rather COP is a measure of the advantage in using mechanical work to move thermal energy from a lower temperature reservoir to a slightly higher temperature reservoir over producing all of that thermal energy from a low entropy source (such as electricity). A COP > 1 or even COP > 10 does not necessarily violate the second law of thermodynamics.

AM
Mentor
P: 17,259
 Quote by Thermolelctric Want to calculate: how much less electrical energy will the house consume, when it is heated with air to air heat pump, versus when it is heated by direct electrical heating. This is not the same value as COP.
So if the amount of heat is Q and the electrical heating energy to produce Q is E and the work to pump Q is W then if I am understanding correctly you want E-W? Or perhaps (E-W)/Q?
Sci Advisor
HW Helper
P: 6,677
 Quote by Thermolelctric Thanks for the link. So when we run heat engine inside building and use the mechanical work produced to pump some heat energy from the outside(colder), total we get more heat energy, than when just burning the fuel inside the building with 100% efficiency?
Just to add to what Dalespam has said, the COP of the heat pump has to be ≥ 1 since COP = (Qc+W)/W. If Qc < 0 it is no longer a heat pump.

So long as all the thermal energy that is exhausted from the heat engine is retained inside the building, the heat pump delivering heat to the building from the outside will always give you more heat energy than burning the fuel inside the building with 100% efficiency. So long as the heat pump has positive heat flow from the cold reservoir (ie it is working as a heat pump), the heat flow that is converted by the engine into work on the working fluid of heat pump will always return to the room plus some amount of heat from the outside.

AM
 P: 28 I must admit that there was logical mistake in my COP=5 cycle. The heat pump will not pump any more when the cold reservoir is heated(above the ambient) back from hot reservoir. Thanks everybody for explaining patiently. So there is no other limitation to the COP than COP≤Th/(Th−Tc). The COP can be very high if temperature dT is very small. The practical heat pumps COP is so much below the theoretical maximum only because the various constructional problems. The other interesting question is: theoretically, can we run the heat engine outside the building and use the work from the heat engine to run the heat pump to heat the building from the outside air and get the same amount or more heat, than burning the fuel inside the building with 100% efficiency?
Sci Advisor
HW Helper
P: 6,677
 Quote by Thermolelctric The other interesting question is: theoretically, can we run the heat engine outside the building and use the work from the heat engine to run the heat pump to heat the building from the outside air and get the same amount or more heat, than burning the fuel inside the building with 100% efficiency?
That is exactly how most heat pumps run. The input work is provided by electricity generated by heat engines operating in an outside power plant. As I explained in my earlier post #57, in that case you need a COP of about 3.5 just to break even (ie. to get the same amount of heat than you would from burning the fuel with 100% efficiency). If the COP>3.5 you get more heat than burning the fuel inside at 100% efficiency. And a COP > 3.5 is easily possible. If the temperatures are Th/Tc = 293K/273K, the maximum Carnot COP would be 293/20 ≈ 15

AM
Mentor
P: 17,259
 Quote by Thermolelctric The other interesting question is: theoretically, can we run the heat engine outside the building and use the work from the heat engine to run the heat pump to heat the building from the outside air and get the same amount or more heat, than burning the fuel inside the building with 100% efficiency?
I answered this already in post 50.
P: 28
 Quote by DaleSpam I answered this already in post 50.
This was supposed to be the case when the heat engine is inside the house, you answered in post 50.

The usual case is the is that heat engine is outside the building (in power plant). I have feeling that when the heat engine in the power plant uses same temperature as outside for cooling then no way we can get more heat with the air to air heat pump to heat the building, than was used at the power plant.

I was out gathering some free energy (firewood) :)
Mentor
P: 17,259
 Quote by Thermolelctric This was supposed to be the case when the heat engine is inside the house, you answered in post 50. The usual case is the is that heat engine is outside the building (in power plant). I have feeling that when the heat engine in the power plant uses same temperature as outside for cooling then no way we can get more heat with the air to air heat pump to heat the building, than was used at the power plant. I was out gathering some free energy (firewood) :)
I was assuming that the heat engine and the heat pump were both using the outside air as the cold reservoir in both cases. The difference is that the heat engine uses a hot reservoir heated by the fuel and the heat pump uses the inside of the building as the hot reservoir. The equation I provided applies.
 P: 28 Thanks!
 Mentor P: 17,259 You are welcome!
 P: 8 An engineering way to proceed is to try it in practice ;) Years ago, I deviced LTSPICE models of Peltier thermoelements. Can be found from LTspice yahoo groups: Files > tut > Thermal > tec_fridge.zip. The TEC allows you to set realistic parameters, and also creating ideal (lossless) models. Then you should generate a LTSpice model where you have the hot container (capacitor) and a cold container (another capacitor) and a heat pump (TEC) and the electrical power source (e.g. current source). If you can build a heat pump model (using supplied TEC model, and resistors (R>0) and other passive realistic components) which seems to break laws of thermodynamics, I'll buy you a beer. BR, -Topi

 Related Discussions General Physics 8 Classical Physics 1 Introductory Physics Homework 2 Engineering, Comp Sci, & Technology Homework 3 Classical Physics 3