# Heat Pumps and COP

1. Aug 5, 2008

### Gear300

I was reading on the Clausius statement and heat pumps. Apparently, the COP (coefficient of performance) for heat pumps can be greater than unity (1), and it usually is for several systems (depending on the conditions). So the energy transfer due to work done on the system results in a greater transfer of energy from cold to hot by heat. So that implies that the output ends up being greater than the input. I'm not completely new to the concept of higher output than input processes (there are mechanical processes that allow for such, such as pulleys). I'm just wondering how processes like these function in terms of energy.

Last edited: Aug 5, 2008
2. Aug 5, 2008

### alphysicist

Hi Gear300,

I think it's best to compare it with the efficiency of a heat engine. A certain amount of heat is absorbed from the hot reservoir; part of that is transferred to a cold reservoir and part is extracted as work. Since efficiency is (work/heat abosrbed from hot reservoir), it makes no sense to talk about an efficiency greater than 1. (An important point is that the heat flow is from hot to cold, which is the natural direction.)

For a heat pump, a certain amount of heat is forced from the cold reservoir to the hot reservoir. To accomplish this, a certain amount of work has to be done. The heat that shows up on the hot reservoir side is equal to the sum of the heat absorbed from the cold reservoir plus the work done to move the heat from cold to hot. So we would expect the coefficient of performance (heat dumped in hot reservoir/work) to be greater than 1.

(Of course I'm not talking about the real machines, just ideal theoretical models.)

3. Aug 6, 2008

### Gear300

I see where you're going with this. Though in this sense, doesn't it imply that most of what goes to the hot reservoir is from the work applied rather than from the cold reservoir?

4. Aug 6, 2008

### alphysicist

I'm not sure what part you are saying implies that. The numerical value of the coefficient of performance will tell you the ratio of the heat transfers and work.

For example, take the coefficienf of performance as 3. Then, for a particular cycle, for every unit of work input into the system, three units of heat are output to the hot reservoir; which means two units of heat are transferred out of the cold reservoir.

So as long as the COP is above two, the contribution from the heat from the cold reservoir will be greater than the contribution from the input work.

5. Aug 6, 2008

### Gear300

I see, I see...and this input of work is also responsible for the heat from the cold reservoir to the hot reservoir. Supposedly, I was confused about how 1 unit of work could displace, say, 2 units of heat (I thought the ratio should have been 1:1 with 1 unit of work displacing 1 unit of heat). Therefore, can it be said that when inputting energy into a system, the way the system organizes after that does not depend on the amount of energy added but on how the energy is added?

6. Aug 6, 2008

### alphysicist

Right; there are Carnot limits for the COP, but how close you get to those depend on the details of the actual process.

7. Aug 6, 2008

### RonL

From Van Nostrand's Scientific Encyclopedia, Fifth Edition.

After describing the Carnot cycle, the following statement is made,

"It should be noted that the efficiency of the forward cycle is highest when T1 is as high as possible. Since, in practice, T0 will always be fixed by the temperature of the surrounding atmosphere, a high efficiency corresponds to a large difference T1-T0. In contrast, a high coefficient of performance, or a high effectiveness of a heat pump corresponds to a small difference T1-T0.
It would appear that increasing T0 for a power cycle below that of the surrounding atmosphere is advantageous in that the efficiency n is increased. However, it must be realized that this can only be achieved at the expense of work in operating a refrigerator, and no advantage is gained".

In line with the question of the OP, and all the basic laws I read, how is it not possible to design a system, that after it is put in motion and acquires a mass of energy, that a state of imbalance is maintained, and the work of operating the refrigerator kept internal where all losses are continually recycled?

If the high pressure hot side is insulated to prevent heat loss, and the cold side is kept a good bit below atmospheric temperature, and designed to absorb heat at a maximum value, the design mechanical device that separates the two sides, would have to transfer out, a work value equal to the difference of heat loss through insulation, and the heat absorbed by the cold side.

As Russ Watters stated in another thread, the greater the vacuum the less energy that will transfer, that helped me a lot, but I'm still a little hung up on why we can't make a system work that gives us the benefits of a large amount of cold air, and a little bit of work (electric charge) and some amount of condensation based on humidity.

I have never been able to understand why BTUs can't be collected and moved to a higher state of usable condition.

I know I'm not the only one with this mind block, so maybe this thread will clear the air a little.

Thanks

Ron

8. Aug 7, 2008

### Gear300

Sorry, didn't completely get your question. Are you asking why systems tend towards equilibrium or why energy converted from heat to work is never 100% efficient?

9. Aug 7, 2008

### stewartcs

In nature things always tend to balance themselves out - that is just the way it is and has been proven empirically.

Essentially, the hot reservoir would eventually lose energy by heat transfer to the cold reservoir until the energy of the two are balanced. Therefore, the imbalance would not last, and no perpetual motion machine could be created.

Keep in mind that the Carnot Heat Engine is a fictitious one since it is a reversible heat engine and ALL things in nature have irreversibilities that must be accounted for.

CS

10. Aug 7, 2008

### RonL

CS, Gear300,

The question I'm asking, is if the mechanical power transfer device is set in motion (a mass of input energy Stored) and a cycle set into play, where pulses of energy (compressed, high pressure gas, originally transfered from the larger, low pressure section ) are applied to a turbine type device to maintain speed of motion, and transfer some work out (producing a thermal reduction of the gas pulse, in essence a heat pump action) why can this compression cycle, not be sustained, if the thermal energy flowing spontaneously into the large colder section, is greater than the losses through the insulation of the, hot high pressure end?

The high and low pressure portions would be separated by the turbine/compressor. Based on work applied to the turbine, and resisted by the acts of compression, and electrical generation, the gas returning to the cold side should be colder to the same extent as work transfered into the power unit. (compression/generation)

This question is based on a continual motion transfer, where cycle time does not allow for equilibrium to be established, and once the high and low temperatures are reached, a controlled exact balance has to be maintained.

10:1 ratios are common for heat pumps (much greater in some cases).

I think as I learn more about flow volumes, some of my questions will be resolved.
Based on my previous post, the operation of the refrigerator should not be a negative value, if it is inside the system, where any losses feed the thermal flow.

Triple digit temperatures in Texas sure makes this worth thinking about.

Ron

11. Aug 7, 2008

### Andrew Mason

A reservoir has an infinite capacity to absorb or lose heat without changing temperature. The heat engine or refrigerator operates between two reservoirs. In the real world, this is accomplished by various methods of adding or removing thermal energy from the reservoir (eg. the radiator on a car or refrigerator).

If you insulated the hot reservoir, you would increase the temperature of the hot reservoir as you pumped heat from the cold reservoir. This would result in more and more work being required to move heat from the cold reservoir. You have to have some way of keeping constant the temperature of the hot reservoir (the reservoir that absorbs the heat flowing from the cold reservoir). You have to do that, of course, without using the cold reservoir to cool it as that would defeat the whole purpose.

The reason has to do with the second law of thermodynamics. You can produce large amounts of cold air with a little bit of work provided the temperature difference between the cold air (reservoir) and the reservoir to which the heat from the cold air reservoir is delivered is small.

If you could deliver large amounts of heat from a cold reservoir to the hot reservoir over a large temperature difference with a small amount of work, you could build up heat in the hot reservoir and then reverse the cycle to run a heat engine to get a larger amount of work out than you put in. This would violate the second law of thermodynamics (ie. you would create a perpetual motion machine of the second kind).

AM

12. Aug 7, 2008

### stewartcs

I can't seem to visualize your system, however, based on what I can interpret of your description the high pressure gas that is expanded across the turbine to produce work will result in a lower pressure at the outlet of the turbine. Evenutally, since this works on a cycle, the gas will need to be compressed again back to the higher pressure (think of a Rankine cycle). That will require a pump or compressor which requires a work input from somewhere. Hence, no perpetual motion machine.

CS

13. Aug 7, 2008

### RonL

The turbine unit is the barrier between the high and low portions of the system. The high pressure gas is applied at the outer diameter of the turbine, and the compressor portion is at the center of the unit.

The entire body including generator (the quickest and most useful way to extract the proper amount of energy from the system), spins and acts as a flywheel energy storage system. The faster, more mass spins, the greater the compression, and the more efficient the system becomes (within reason).

A cycle time would be established to balance everything (and maintain speed of rotation), based on the amount of thermal transfer into the cold section.

Hope this helps a little bit more.

Ron

14. Aug 7, 2008

### RonL

Andrew,
I'm having a little difficulty understanding the need to control heat in the insulated area, my impression is that the heat/work relationship (based on the Clausius statement) requires more heat to produce more work. The AC unit that cools my house, removes heat from the refrigerant before expansion, as I understand it, this is because there is no application toward a work producing device.

I'm not sure if CS, or others see that the turbine unit is fully enclosed inside the system.

I think that if the turbine discharge is too hot, the cold side will be defeated.

The only thing crossing boundaries would be either thermal, or electric.

It just seems to me that if COP is large enough, and system volume is right, that accumulation of thermal energy can be moved to a higher state.

Ron

15. Aug 7, 2008

### stewartcs

Not really. Perhaps a schematic is in order.

CS

16. Aug 7, 2008

### stewartcs

An ideal vapor-compression refrigeration cycle works like this: Refrigerant enters the compressor as saturated vapor and is compressed isentropically to the condenser pressure. The temperature of the refrigerant increases during the isentropic compression process to well above the temperature of the surrounding medium. The refrigerant then enters the condenser as superheated vapor and leaves as saturated liquid as a result of heat rejection to the surroundings. The temperature of the refrigerant at that point is still above the temperature of the surroundings. The saturated liquid refrigerant is then throttled to the evaporator pressure by passing it through and expansion valve or capillary tube. The temperature of the refrigerant drops below the temperature of the refrigerated space during this process. The refrigerant then enters the evaporator as a low-quality saturated mixture, and it completely evaporates by absorbing heat from the refrigerated space. Finally, the refrigerant leaves the evaporator as saturated vapor and reenters the compressor, thus completing the cycle.

If the throttling device were replaced with an isentropic turbine, the refrigerant would enter the evaporator at a constant entropy. This would result in an increased refrigeration capacity and the net work input would decrease by the amount of work output by the turbine. This is not practical since the benefits do not justify the added complexity and cost.

CS

17. Aug 7, 2008

### RonL

Thanks for the replies, I'll try to work on that a little, and some more study.

I'm a little too close to the line, so I had better back off a little.

Thanks again

Ron

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