Heat pump COP theoretical maximum

[Thermolelctric]

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.

 

[mfb]

The formulas in the wikipedia seems to be wrong

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.

 

[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.

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!

 

[Thermolelctric]

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.

 

[DrClaude]

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.

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

1st law: Q_h = W + Q_c

Therefore \mathrm{COP} = \frac{Q_h}{W} = {Q_h}{Q_h - Q_c} = \frac{1}{1 - Q_c/Q_h} &gt; 1

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

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).

I don't see how any losses from the building come into play in the calculation of the COP of heat pump itself.

 

[Thermolelctric]

Thanks DrClaude.

Q_h: heat pumped into the building

Therefore \mathrm{COP} = \frac{Q_h}{W} = {Q_h}{Q_h - Q_c} = \frac{1}{1 - Q_c/Q_h} &gt; 1

Can you explain more how you get this?

Q_h
I don't see how any losses from the building come into play in the calculation of the COP of heat pump itself.

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.

 

[Dale]

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.

Please show your calculations for this claim.

 

[DrClaude]

Thanks DrClaude.
Can you explain more how you get this?

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

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.

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.

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.

No error, it is just how the COP is defined. Infinite COP will only occur for T_h=T_c, 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 \mathrm{COP} = \eta \frac{T_h}{T_h-T_c} with \eta &lt; 1 and itself a function of T_h and T_c.

 

[Thermolelctric]

No error, it is just how the COP is defined. Infinite COP will only occur for T_h=T_c, 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 \mathrm{COP} = \eta \frac{T_h}{T_h-T_c} with \eta &lt; 1 and itself a function of T_h and T_c.

Thanks.
But what happens to COP when Th closes to infinity?

 

[Dale]

But what happens to COP when Th closes to infinity?

The ideal COP approaches 1 as Th approaches infinity.

 

[Thermolelctric]

Please show your calculations for this claim.

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!

 

[Dale]

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).

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.

 

[DrClaude]

Thanks.
But what happens to COP when Th closes to infinity?

\mathrm{COP} \sim 1

Consider the following hypothetical situation. An ideal (Carnot engine) heat pump installed between a hot sink at T_h and a cold source at T_c, and functions with a constant consumption of electrical energy (W = \mathrm{const.}). Both sources are connected such that heat can flow from hot to cold to maintain the hot source at T_h (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
<br /> T_c = T_{c0} - \frac{Q_c}{C} t + \frac{Q_h}{C} t <br />
where T_{c0} is the initial temperature of the cold source and C its heat capacity. Since Q_h = W+Q_c, we have
<br /> T_c = T_{c0} - \frac{Q_c}{C} t + \frac{W + Q_c}{C} t = T_{c0} + \frac{W}{C} t<br />
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.

 

[Thermolelctric]

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.

This is what I'm trying to prove. That this COP formula is incorrect.

 

[DrClaude]

This is what I'm trying to prove. That this COP formula is incorrect.

But the COP formula is just
<br /> \mathrm{COP} = \frac{\textrm{benefit}}{\textrm{cost}} = \frac{Q_h}{W}<br />

 

[Thermolelctric]

By the COP formula is just
<br /> \mathrm{COP} = \frac{\textrm{benefit}}{\textrm{cost}} = \frac{Q_h}{W}<br />

Not that one!

 

[DrClaude]

Not that!

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

So now we get even more output heat because of increased COP.

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

The COP is increased to infinity.
We have perpetum mobile!

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.

 

[Thermolelctric]

\mathrm{COP} \sim 1

Consider the following hypothetical situation. An ideal (Carnot engine) heat pump installed between a hot sink at T_h and a cold source at T_c, and functions with a constant consumption of electrical energy (W = \mathrm{const.}). Both sources are connected such that heat can flow from hot to cold to maintain the hot source at T_h (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
<br /> T_c = T_{c0} - \frac{Q_c}{C} t + \frac{Q_h}{C} t <br />
where T_{c0} is the initial temperature of the cold source and C its heat capacity. Since Q_h = W+Q_c, we have
<br /> T_c = T_{c0} - \frac{Q_c}{C} t + \frac{W + Q_c}{C} t = T_{c0} + \frac{W}{C} t<br />
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.

Thanks, we are getting closer to the solution -- proving the maximum efficiency of the air to air, building heating, heat pump.

 

[Dale]

This is what I'm trying to prove. That this COP formula is incorrect.

It is correct, your perpetual motion argument is wrong. You should post your argument in full.

 

[Thermolelctric]

It is correct, your perpetual motion argument is wrong.

What do you mean wrong. Is the formula COP=Th/(Th−Tc) wrong, or is there something else wrong?

 

[Dale]

What do you mean wrong. Is the formula COP=Th/(Th−Tc) wrong, or is there something else wrong?

The formula COP≤Th/(Th−Tc) is correct. Your argument that COP>2 → perpetual motion is wrong.

 

[Thermolelctric]

The formula COP=Th/(Th−Tc) is correct. Your argument that COP>2 → perpetual motion is wrong.

Can you point out what exactly is wrong?

 

[Dale]

Can you point out what exactly is wrong?

No, not until you post the whole argument in detail. Please show in detail why you think COP>2 implies perpetual motion.

 

[Thermolelctric]

No, not until you post the whole argument in detail. Please show in detail why you think COP>2 implies perpetual motion.

Lets analyse the example I posted about COP=5 first.

 

[Thermolelctric]

\mathrm{COP} \sim 1

Consider the following hypothetical situation. An ideal (Carnot engine) heat pump installed between a hot sink at T_h and a cold source at T_c, and functions with a constant consumption of electrical energy (W = \mathrm{const.}). Both sources are connected such that heat can flow from hot to cold to maintain the hot source at T_h (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
<br /> T_c = T_{c0} - \frac{Q_c}{C} t + \frac{Q_h}{C} t <br />
where T_{c0} is the initial temperature of the cold source and C its heat capacity. Since Q_h = W+Q_c, we have
<br /> T_c = T_{c0} - \frac{Q_c}{C} t + \frac{W + Q_c}{C} t = T_{c0} + \frac{W}{C} t<br />
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.


Hope this helps.

[STRIKE]How would be the good way to describe the point when the pump begins to heat the cold source instead of hot reservoire as intended? This is the point from where where Qh/W≠Th/(Th−Tc). Simply saying the heat is not pumped any more to the hot reservoire but instead goes to the cold reservoire[/STRIKE] . Sorry, stupid question. Let's just stick to analysing the COP=5 example, for now.

 

[DrClaude]

How would be the good way to describe the point when the pump begins to heat the cold source instead of hot reservoire as intended? This is the point from where where Qh/W≠Th/(Th−Tc). Simply saying the heat is not pumped any more to the hot reservoire but instead goes to the cold reservoire .

Well, that is what the "demon" has been doing all along. By moving the heat from Th to Tc, you are basically just heating Tc.

If you want to imagine a more realistic situation, take a heat pump connected to two finite reservoirs, which can also exchange heat directly between each other. This direct exchange will depend on a thermal conductivity between the two and the difference in their temperature. As the heat pump works, heat from the cold reservoir will flow to the hot one, but a part will come back to the cold one. Both will slowly increase in temperature, up to a point where Th = Tc = Tf. At this point, you can calculate that the total energy consumed by the pump will be equal to the amount of energy needed to bring Th to Tf and Tc to Tf (conservation of energy). In a real situation, the heat pump itself stops working here because Tc = Th, but you could imagine that compressors are still working, etc., such that both Th and Tc will increase beyond Tf as the electrical energy continues to be converted to heat.
[Edit: I am assuming here that at all times the heat going back to the colr reservoir is > Qc]

 

[mfb]

@Thermolelctric: All equations for heat pumps assume two reservoirs - heat is pumped from the cold reservoir to the hot one, and the required work goes to the hot one as well (this is just a convention). Neglecting engineering issues like thermal conductivity and heat capacity of the materials, there is no difference between outside air and outside ground.

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!

You pump 4kW of power out of the cold reservoir, and let 2kW flow back. You will cool the cold reservoir, so COP will go down.
If you would heat the cold reservoir in some way, COP would go up. So what?

 

[Thermolelctric]

Well, that is what the "demon" has been doing all along. By moving the heat from Th to Tc, you are basically just heating Tc.

If you want to imagine a more realistic situation, take a heat pump connected to two finite reservoirs, which can also exchange heat directly between each other. This direct exchange will depend on a thermal conductivity between the two and the difference in their temperature. As the heat pump works, heat from the cold reservoir will flow to the hot one, but a part will come back to the cold one. Both will slowly increase in temperature, up to a point where Th = Tc = Tf. At this point, you can calculate that the total energy consumed by the pump will be equal to the amount of energy needed to bring Th to Tf and Tc to Tf (conservation of energy). In a real situation, the heat pump itself stops working here because Tc = Th, but you could imagine that compressors are still working, etc., such that both Th and Tc will increase beyond Tf as the electrical energy continues to be converted to heat.

Both will slowly come to the temperature Tf, near the Tf the COP will be very high. But this do not reflect the real Coefficent of Performance, because the heat goes to heating the low temp resevoire, we need to subtract the heat that goes to heating the cold side from Qh. Could this be the difficulty of calculating the COP and current cause of difficulty of thinking?

 

[DrClaude]

Both will slowly come to the temperature Tf, near the Tf the COP will be very high. But this do not reflect the real Coefficent of Performance, because the heat goes to heating the low temp resevoire, we need to subtract the heat that goes to heating the cold side from Qh. Could this be the difficulty of calculating the COP and current cause of difficulty of thinking?

I don't get why the COP being high bugs you so much. What is the problem with a device that, by using 1 W of electricity, can transfer 1000 W from a cold reservoir to a hot reservoir? All the 2nd law tells you is that, to do that, you will need both reservoirs to be very close in temperature.

 

[Thermolelctric]

You pump 4kW of power out of the cold reservoir, and let 2kW flow back. You will cool the cold reservoir, so COP will go down.
If you would heat the cold reservoir in some way, COP would go up. So what?

This is good observation. As long as the cool reservoir is below the ambient(outside temperature) it can absorb the heat from the ambient. So Tc < ambient temp. This brings in new reservoire.
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.

 

[Thermolelctric]

I don't get why the COP being high bugs you so much. What is the problem with a device that, by using 1 W of electricity, can transfer 1000 W from a cold reservoir to a hot reservoir? All the 2nd law tells you is that, to do that, you will need both reservoirs to be very close in temperature.

Problem is that I tried to do that, but it does not work. I can not get the heat energy promised by the heat pump sales man. And instead there are some fundamental laws of nature that prohibit from getting such a good results from heat pumps.
That would be no problem if the pumps really could work at such a high efficiency as promised bu salesman:)

 

[Dale]

Lets analyse the example I posted about COP=5 first.

That's fine. In that example I didn't see ANY work being extracted at all, let alone perpetual motion. So please use that example and demonstrate that it implies perpetual motion.

 

[Andrew Mason]

@Thermolelctric: All equations for heat pumps assume two reservoirs - heat is pumped from the cold reservoir to the hot one, and the required work goes to the hot one as well (this is just a convention).

It is more than a convention!
During the cycle, the work goes into the internal energy of the working fluid (by compression). If the heat flow from the working fluid then went back to the cold reservoir (there are only two reservoirs) you would not have a heat pump!

AM

 

[mfb]

Problem is that I tried to do that, but it does not work. I can not get the heat energy promised by the heat pump sales man.

That is just an engineering issue then - the efficiency might need special conditions to be reached.

And instead there are some fundamental laws of nature that prohibit from getting such a good results from heat pumps.

No, there are not.

It seems good idea to describe the pump as a system of 4 reservoirs

This is a bad idea for a theoretic consideration, it adds complexity but nothing interesting.
For a real heat pump, it can be relevant - if you stress it too much, the imperfect temperature exchange with the environment can become relevant - the temperature of the cool side drops below the air temperature outside (as air flow is not quick enough) and the efficiency reduces. The same can happen at the hot side - to heat the room with non-zero power, it has to be (a bit) hotter than the room in some way.

It is more than a convention!
During the cycle, the work goes into the internal energy of the working fluid (by compression). If the heat flow from the working fluid then went back to the cold reservoir (there are only two reservoirs) you would not have a heat pump!

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.

 

[Thermolelctric]

That's fine. In that example I didn't see ANY work being extracted at all, let alone perpetual motion. So please use that example and demonstrate that it implies perpetual motion.

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?

 

[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.