# Heat pump COP theoretical maximum

by Thermolelctric
Tags: heat, maximum, pump, theoretical
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P: 6,679
 Quote by Thermolelctric It seems good idea to describe the pump as a system of 4 reservoirs: Input reservoir, pump cold reservoir, pump hot reservoir, and output reservoir. And try to calculate heat pumped from input reservoire to the output reservoire, and the work done to pump.
Why is this a good idea?

The whole idea behind a heat pump is to move heat from the cold reservoir to the hot reservoir. By the second law, this can only occur by doing work on the system. The only reason it makes any sense at all is if you get more heat flow out at the hot reservoir than you put in as work. Otherwise, as DrClaude has pointed out, you might as well have the input energy source just provide heat flow (e.g. an electric heater).

AM
P: 28
 Quote by Andrew Mason Why is this a good idea? The whole idea behind a heat pump is to move heat from the cold reservoir to the hot reservoir. By the second law, this can only occur by doing work on the system. The only reason it makes any sense at all is if you get more heat flow out at the hot reservoir than you put in as work. Otherwise, as DrClaude has pointed out, you might as well have the input energy source just provide heat flow (e.g. an electric heater). AM
OK lets ditch the 4 reservoir idea.
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P: 6,679
 Quote by mfb You could define COP with the other possible choice as well. This would give more pumped heat, as you had to add your work input to this quantity. The result would be the same, but the notation would differ a bit.
For refrigerators is COP = Qc/Qh-Qc. Is that the other possible choice you are referring to?

AM
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P: 17,344
 Quote by Thermolelctric Mechanical work is no extracted, heat energy is extracted from pump hot reservoire.
With the pump doing work and requiring energy input. The heat energy extracted from the hot reservoir does no work, so the process only continues as long as external work is supplied. Not perpetually.

 Quote by Thermolelctric A cyclic transformation whose only final result is to transfer heat from a body at a given temperature to a body at a higher temperature is impossible. - German physicist Rudolf Clausius Agree to that?
Agreed, but it doesn't apply to your scenario since there is another final result: work was done on the system requiring external input energy.
P: 28
So now conclusion for the development of the thread:

 Quote by DrClaude No, you reach $T_c = T_h$ 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.
DrClaude got the point there: in the limiting conditions of Tc=Th the formula COP=Th/(Th−Tc) does not hold. When Tc=Th there is no more heat pump. So the formula COP=Th/(Th−Tc) can not be used to determine the limiting values of the heat pump, but some other formulas must be used, whitch we are about to develop yet.
My COP=5 example cycle proves that COP=Th/(Th−Tc) is not correct for that COP value and therefore can not assume it is correct for other values.
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P: 17,344
 Quote by Thermolelctric My COP=5 example cycle proves that COP=Th/(Th−Tc) is not correct for that COP value
No, it doesn't. I can't even figure out why you think it does prove that. There is not even a hint of perpetual motion involved.

No work is extracted from the system and external work needs to be continually supplied to the system. How can you possibly think that is perpetual motion?
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P: 6,679
 Quote by Thermolelctric Mechanical work is no extracted, heat energy is extracted from pump hot reservoire. A cyclic transformation whose only final result is to transfer heat from a body at a given temperature to a body at a higher temperature is impossible. - German physicist Rudolf Clausius Agree to that?
This appears to be the source of confusion here. First of all, you are using a poor translation. Clausius actually stated the second law as:
"Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time."
This usually restated as follows:
"No process is possible whose sole result is the transfer of heat from a cooler to a hotter body. "
see: http://web.mit.edu/16.unified/www/FA...es/node37.html

Neither statement is particularly clear (to make it a little clearer, the term "sole result" excludes a process in which mechanical work is done on the system). A clearer statement of the second law would be to simply say "heat flow cannot occur spontaneously from a body A to a body B if the temperature of body B is higher than that of body A. Heat flow can only occur (ie. from a body A to a body B if the temperature of body B is higher than that of body A) if something else is done (eg. addition of work)".

AM
P: 28
 Quote by DaleSpam With the pump doing work and requiring energy input. The heat energy extracted from the hot reservoir does no work, so the process only continues as long as external work is supplied. Not perpetually. Agreed, but it doesn't apply to your scenario since there is another final result: work was done on the system requiring external input energy.
Yes indeed but the amount of external energy that is input to the system do not satisfy the formula Qh/W=Th/(Th-Tc)
since Th and Tc is starting conditions.
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P: 17,344
 Quote by Thermolelctric Yes indeed but the amount of external energy that is input to the system do not satisfy the formula Qh/W=Th/(Th-Tc) since Th and Tc is starting conditions.
That formula still works for an ideal heat pump. In your scenario, all you have is that Tc is a function of time due to the leakage.

In any case, even if the function were wrong (which it isn't) that still doesn't make it perpetual motion.
 P: 28 And we could use the heat from the process output to do the work of the pump. It is possible to convert heat difference back to the work, with known coefficent of efficiency. Then it is possible to use iterative method to calculate the maximum COP before perpetum mobile kicks in. There must be some other way with differencial equation, but I must admit I am not very good with differential equations.
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P: 17,344
 Quote by Thermolelctric And we could use the heat from the process output to do the work of the pump.
Such a device is called a heat engine. What do you think the maximum efficiency of a heat engine is?

See here for a discussion: http://www.physicsforums.com/showthread.php?t=667129
PF Gold
P: 707
 Quote by Thermolelctric OK lets ditch the 4 reservoir idea.
Don't give up the 4 (or more) reservoir thinking, this is key to what you are trying to work out....the main reservoir is the coldest, with all others configured inside of it.

Air brought in and compressed gives the heat source that drives the internal heat engine(s).
A two part system (at least) with all heat exchange, work functions taking place inside the cold reservoir where all losses are continually recirculated.

The sum of heat extracted from the air flow through the system, has to have an exactly equal amount of work energy moved out of the cold reservoir (simplest is electric).

If cold air is not the main objective, then it will be an example of how to best wear out a lot of equipment for a small return.

I'm limited in time and ability with words, so just hope this might stimulate your thoughts a little, a little more detail can be found in my post (scroll compressor) I don't know how to link very well.
I tried to explain how the internal functions take place.

Good luck

Ron
P: 28
 Quote by DaleSpam Such a device is called a heat engine. What do you think the maximum efficiency of a heat engine is? See here for a discussion: http://www.physicsforums.com/showthread.php?t=667129

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?
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P: 17,344
 Quote by Thermolelctric 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?
Yes, provided:
1) the heat engine is ideal (maximum efficiency)
2) the heat pump is ideal (maximum COP)
3) the fuel combustion temperature is higher than the temperature to which you want to heat the building.

If those three conditions are met then the heat energy provided is greater than just burning it by a factor of:

$$\frac{1-T_C/T_F}{1-T_C/T_H}$$

Where Tc is the cold outside temperature, Th is the warm inside temperature, and Tf is the combustion temperature of the fuel.
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P: 22,305
 Quote by Thermolelctric 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.
Pick any major air source heat pump manufacturer and look at the ratings. I guarantee you won't find one rated below 3:1. In fact if I had to guess, the legally required minimum is probably 3.5 or 4.

I didn't read the whole thread but since this is just a slight twist on a common misunderstanding, perhaps I can cut to the chase:
1. A heat pump that recycles its own heat to increase temperature has no output, so its COP is zero. No violation.

2. A heat pump that runs a heat engine produces no violation because their efficiency curves are inverses and never sum to greater than 1.
 P: 28 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.
 P: 28 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!
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P: 22,305
 Quote by russ_watters Pick any major air source heat pump manufacturer and look at the ratings. I guarantee you won't find one rated below 3:1. In fact if I had to guess, the legally required minimum is probably 3.5 or 4.
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.

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