Heat pump COP theoretical maximum

[mfb]

You do not have a cyclic transformation, you need work as input and get heat flow from cold to warm as output. In other words, just a regular heat pump.

 

[Andrew Mason]

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

 

[Thermolelctric]

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 let's ditch the 4 reservoir idea.

 

[Andrew Mason]

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

 

[Dale]

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.

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.

 

[Thermolelctric]

So now conclusion for the development of the thread:

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.

 

[Dale]

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?

 

[Andrew Mason]

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/FALL/thermodynamics/notes/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

 

[Thermolelctric]

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 Q_h/W=T_h/(T_h-T_c)
since T_h and T_c is starting conditions.

 

[Dale]

Yes indeed but the amount of external energy that is input to the system do not satisfy the formula Q_h/W=T_h/(T_h-T_c)
since T_h and T_c is starting conditions.

That formula still works for an ideal heat pump. In your scenario, all you have is that T_c 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.

 

[Thermolelctric]

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.

 

[Dale]

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: https://www.physicsforums.com/showthread.php?t=667129

 

[RonL]

OK let's 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

 

[Thermolelctric]

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: https://www.physicsforums.com/showthread.php?t=667129

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?

 

[Dale]

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.

 

[russ_watters]

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.

 

[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.

 

[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!

 

[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.

 

[russ_watters]

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.

 

[russ_watters]

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!

 

[Andrew Mason]

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

 

[Andrew Mason]

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

 

[Dale]

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?

 

[Andrew Mason]

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

 

[Thermolelctric]

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?

 

[Andrew Mason]

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

 

[Dale]

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.

 

[Thermolelctric]

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) :)

 

[Dale]

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.

 

[Thermolelctric]

Thanks!

 

[Dale]

You are welcome!

 

[TopiRinkinen]

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