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Thread: Heat Pump Question

  1. #61
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    Quote Originally Posted by kzb View Post
    I know you've said it, but just saying it doesn't mean you've proved it. There's lots of things been said on here and they can't all be correct.
    I donít see any major disagreements between other posters on the key points. I see a misunderstanding of COP, but thatís not relevant to your question. I see a debate on whether the air compressor scheme would be a little worse or a little better than resistance heating at heating a house, but it makes little difference to your question, since I donít see anyone claiming it would be a good heat pump or practical to use. I donít see anyone disagreeing on the basic physics or why it would be a poor heat pump.

    The question in your OP has been answered. If you still donít understand it, I donít know what else to say unless you can explain where youíre having trouble. I have no interest in proving something you donít seem to understand.

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  2. #62
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    8A8E858E-F186-47CE-86ED-CB8D14C34CA1.png
    If this image works, it is from the WP page you cited.
    The area under the adiabatic curve is the work done,
    There are two isotherms on the graph, one at the lower pressure.
    Those are at constant temperature which is close enough for your project.

    The area above the lower isotherm and below the adiabatic line is the heat generated.
    The rest of the work done area is the pressure work.
    You are venting away the pressure work.
    sicut vis videre esto
    When we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.
    Originally Posted by Ken G

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    Quote Originally Posted by profloater View Post
    8A8E858E-F186-47CE-86ED-CB8D14C34CA1.png
    If this image works, it is from the WP page you cited.
    The area under the adiabatic curve is the work done,
    There are two isotherms on the graph, one at the lower pressure.
    Those are at constant temperature which is close enough for your project.

    The area above the lower isotherm and below the adiabatic line is the heat generated.
    The rest of the work done area is the pressure work.
    You are venting away the pressure work.
    Yay finally we are getting somewhere ! All we need now is some numbers on the sketch, or some equations which explain the process, or better still a worked example.

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    Quote Originally Posted by Van Rijn View Post
    I don’t see any major disagreements between other posters on the key points. I see a misunderstanding of COP, but that’s not relevant to your question. I see a debate on whether the air compressor scheme would be a little worse or a little better than resistance heating at heating a house, but it makes little difference to your question, since I don’t see anyone claiming it would be a good heat pump or practical to use. I don’t see anyone disagreeing on the basic physics or why it would be a poor heat pump.

    The question in your OP has been answered. If you still don’t understand it, I don’t know what else to say unless you can explain where you’re having trouble. I have no interest in proving something you don’t seem to understand.
    I probably didn't put the question very well. In retrospect I am curious about the equations which would give the heat output of such a system and its thermodynamic efficiency.

  5. #65
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    Quote Originally Posted by kzb View Post
    Yay finally we are getting somewhere ! All we need now is some numbers on the sketch, or some equations which explain the process, or better still a worked example.
    The inevitable COP less than 1, much less than 1 in a small system with no work recovered, is enough of an argument without numbers. You cannot compress the air without heating it but conversely you cannot get that heat without raising pressure.

    Compressors are designed to generate compressed air and shed the “waste heat”. Otherwise the adiabatic rise in temperature will be damaging and dangerous.

    As was said at the beginning, if your COP is less than one, you do better to heat a resistor at 100%.
    sicut vis videre esto
    When we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.
    Originally Posted by Ken G

  6. #66
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    In thermodynamics, we do not find worked examples of your project because it is not a good idea. We work with enthalpy and entropy, usually h and s, in order to include vapours which are important. Real compressors cannot be either adiabatic nor isothermal, somewhere in between. Air above freezing point has some water vapour usually, hence the enthalpy or total heat idea. Vapour carries more heat, because of latent heat, than dry air.

    If you look at compresser thermodynamics, they are complex with engineering compromises. You do not build 10 to 1 ratios in single stages because of that temperature rise. The graphical representation of P and V are replaced with h and s as you move from ideal gases to real air. The graphs are handy because the work is the area between lines and most cycles can be studied in practice from the hs chart rather than numbers in equations because of the many fudge factors of losses in real machines.

    Having tried to explain why your scheme is a non starter as a heat pump, you could delve into real cycles but you should start with enthalpy and entropy as used in engineering thermodynamics. It is not learned in one post!
    sicut vis videre esto
    When we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.
    Originally Posted by Ken G

  7. #67
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    Here is an HS diagram for a multistage compressor to make my point.
    749F7D55-272A-4CB1-8AA6-7595CEE4D75B.png
    sicut vis videre esto
    When we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.
    Originally Posted by Ken G

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    Quote Originally Posted by profloater View Post
    Here is an HS diagram for a multistage compressor to make my point.
    749F7D55-272A-4CB1-8AA6-7595CEE4D75B.png
    OK thanks but without numbers it doesn't really help me.

  9. #69
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    Quote Originally Posted by kzb View Post
    OK thanks but without numbers it doesn't really help me.
    What help do you need? An air compressor will not work as a heat pump.
    sicut vis videre esto
    When we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.
    Originally Posted by Ken G

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    Quote Originally Posted by profloater View Post
    The inevitable COP less than 1, much less than 1 in a small system with no work recovered, is enough of an argument without numbers. You cannot compress the air without heating it but conversely you cannot get that heat without raising pressure.

    Compressors are designed to generate compressed air and shed the “waste heat”. Otherwise the adiabatic rise in temperature will be damaging and dangerous.

    As was said at the beginning, if your COP is less than one, you do better to heat a resistor at 100%.
    I still don't think the COP can be less than 100%.

    If the air is rejected to the atmosphere at a temperature below which it was taken in (from that same atmosphere), that means heat has been extracted from it.

    The compressed air, when cooled to ambient and released, will use its compression energy to do work in expanding, and undergo adiabatic cooling as a result. By my understanding, this is how they liquidise permanent gases, like producing liquid nitrogen, by doing several cycles like this.

    Assuming the compressor waste heat is 100% recovered, the COP must exceed 100%. Surely.

  11. #71
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    Quote Originally Posted by kzb View Post
    I still don't think the COP can be less than 100%.

    If the air is rejected to the atmosphere at a temperature below which it was taken in (from that same atmosphere), that means heat has been extracted from it.

    The compressed air, when cooled to ambient and released, will use its compression energy to do work in expanding, and undergo adiabatic cooling as a result. By my understanding, this is how they liquidise permanent gases, like producing liquid nitrogen, by doing several cycles like this.

    Assuming the compressor waste heat is 100% recovered, the COP must exceed 100%. Surely.
    After all this debate. The heat from the outside air is heated, not heat pumped, by work in the compressor. Yes you can get that heat inside by mounting a compressor with its air tank indoors and adding a pipe to the outside.

    For the last time, the main work goes into pressure ending up at indoor temperature. It is hard to get heat out of that pressure but you can get work, like with an air tool.

    If you had a handy vacuum, when you vented the cylinder you would get no cooling, it has to do work against the air. It does do work as useless eddies and in doing the work it cools, although the net result in the atmosphere is heating from those eddies.

    The wasted work is doom for your idea. Even if you could somehow use the work your COP would get higher but always less than one..

    And that is why heat pumps use latent heat and vapours that liquidise under pressure alone.
    sicut vis videre esto
    When we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.
    Originally Posted by Ken G

  12. #72
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    Further notes. Adiabatic processes are reversible but also not possible with real machines. Even in fast cylinders like IC engines, or in this case compressors, heat is lost to the metal. The later venting of air is not adiabatic, it is irreversible. The air does work against an infinite atmosphere and ends up at the ambient temperature .
    The gas energy store is as movement of atoms, ie as temperature, which is why we have to use absolute temperature in calculations. But we can separate pressure out from temperature at all real temperatures.
    The heat rise in compression is inevitable but we can then extract heat to bring the temperature back to where is started, or a little higher in your project.
    In that PV diagram the theoretical adiabatic curve cannot ever be achieved, but it can be used if all the heat is collected indoors.
    The work area under the isothermal curve is less but also not really achievable.
    Once restored to the start temperature, you can only get work out by lowering the temperature below the start point. This is not what we mean by pumping. You cannot get all the energy out as work, which is why you could never get a coefficient of performance of 1. The real COP will be less than 0.5 in practice if you just vent to stored air.
    In a heat pump you do work on the vapour. As you pressurise a vapour it liquifies releases heat. That heat was taken from the environment when the liquid was vapourised at lower pressure. So it is free. Just like an evaporating puddle.
    Without that effect you have no heat pump, only an inefficient heater.
    sicut vis videre esto
    When we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.
    Originally Posted by Ken G

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    Quote Originally Posted by profloater View Post
    In a heat pump you do work on the vapour. As you pressurise a vapour it liquifies releases heat. That heat was taken from the environment when the liquid was vapourised at lower pressure. So it is free. Just like an evaporating puddle.
    Without that effect you have no heat pump, only an inefficient heater.
    Yes, Iíve had trouble understanding how an air compressor could be a heat pump, but I didnít question it because there was no disagreement it would be a very poor heat pump at best. But I think I will question it now.

    The claim was that if the air returning to the outside, after expansion was cooler than inside it would be a heat pump, but would that actually be the case?

    First, would the air returning outside necessarily be cooler than when it came in? Yes, itís compressed and a heat exchanger transfers heat to the house and cools it, but would that remove more heat energy than was put into it through compression?

    The air then further cools through expansion but does it just end up back to where it started, or cooler? And as you said, there is energy added to the air in the form of increased pressure, which will be lost to the outside as it expands and end up as heat. So I am very skeptical it could act as a heat pump just by compressing air.

    I do see one possibility: Water vapor in the air could turn to liquid as the pressure is increased and release latent heat. So perhaps it could act as a heat pump to the extent there is water vapor in the air. Which I expect would make little difference in heating effectiveness, but still.

    Honestly, I think the best way to do this is to have both the air intake and exhaust in the house. That way it would work about as well as resistance heating, it would just be more expensive hardware with more moving parts.

    "The problem with quotes on the Internet is that it is hard to verify their authenticity." ó Abraham Lincoln

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  14. #74
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    Quote Originally Posted by Van Rijn View Post
    Yes, I’ve had trouble understanding how an air compressor could be a heat pump, but I didn’t question it because there was no disagreement it would be a very poor heat pump at best. But I think I will question it now.

    The claim was that if the air returning to the outside, after expansion was cooler than inside it would be a heat pump, but would that actually be the case?

    First, would the air returning outside necessarily be cooler than when it came in? Yes, it’s compressed and a heat exchanger transfers heat to the house and cools it, but would that remove more heat energy than was put into it through compression?

    The air then further cools through expansion but does it just end up back to where it started, or cooler? And as you said, there is energy added to the air in the form of increased pressure, which will be lost to the outside as it expands and end up as heat. So I am very skeptical it could act as a heat pump just by compressing air.

    I do see one possibility: Water vapor in the air could turn to liquid as the pressure is increased and release latent heat. So perhaps it could act as a heat pump to the extent there is water vapor in the air. Which I expect would make little difference in heating effectiveness, but still.

    Honestly, I think the best way to do this is to have both the air intake and exhaust in the house. That way it would work about as well as resistance heating, it would just be more expensive hardware with more moving parts.
    I guess that was the whole confusion. If you vented the compressed air direct to outside it would be hotter. But first you let it cool, but in cooling it retains pressure, otherwise, not much use as a normal compressor.
    Now when you release the air outside, it does work on the outside air, pushing it about and actually heating it but the air flow itself cools. It cools doing the work. It can cool lower than the outside air.

    But you do not compare the temperature of the air stream to judge a heat pump. You measure the heat gained versus the work you did, which is an energy comparison. The magic of heat pumps is that you extract energy from a very big source in boiling a well chosen liquid. That part of the design of heat pumps is essential.

    The cooling effect of expanding air is paradoxical and non intuitive. It is only when the expanding air is doing work that it cools. If you think of the pressure as heat, fast moving atoms, then the second law states that you cannot turn all that heat into work. You have to expel heat at a lower temperature to get work and the amount of work depends on the start and end temperatures. If perfectly vented to vacuum there is no cooling, which is more intuitively correct. The atoms fly off at their velocity as they perfectly escaped.

    Because we ignore the stream of eddies created by the vent in the outside air, we do not realise work is being done. That work was originally provided by the compressor. Wasted.
    sicut vis videre esto
    When we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.
    Originally Posted by Ken G

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    Quote Originally Posted by Van Rijn View Post
    I do see one possibility: Water vapor in the air could turn to liquid as the pressure is increased and release latent heat. So perhaps it could act as a heat pump to the extent there is water vapor in the air. .
    That is where enthalpy comes in. Normal compressors do collect water in their reservoirs that way, and you have to remove it from time to time. Humid air would produce hotter, wet air, but the losses are still very large and will never get that COP near 1.
    sicut vis videre esto
    When we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.
    Originally Posted by Ken G

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    The setup as I described it is nothing to do with achieving work.

    You are thinking of a heat engine where we are usually interested in getting it to do work as efficiently as we can. In this case we are not.

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    Quote Originally Posted by kzb View Post
    The setup as I described it is nothing to do with achieving work.

    You are thinking of a heat engine where we are usually interested in getting it to do work as efficiently as we can. In this case we are not.
    No, you need to learn that a compressor does work.
    sicut vis videre esto
    When we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.
    Originally Posted by Ken G

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    Quote Originally Posted by kzb View Post
    The setup as I described it is nothing to do with achieving work.

    You are thinking of a heat engine where we are usually interested in getting it to do work as efficiently as we can. In this case we are not.
    Ok, here we go; energy can be expressed as heat or mechanical work, both in Joules. It was Carnot working with steam, who realised heat had to “flow” from high temperature to lower in order to do work, in a steam engine. His name is given to the theoretical limit of any heat engine, based on two temperatures. Electricity is another energy form and can make work quite well, as in a motor for a compressor, although not as efficient as just heating a resistor.

    You can convert work into heat, indeed mostly it ends up as heat. But you cannot turn heat into work completely unless the sink temperature were to be absolute zero.

    We covered how a compressor must increase temperature while pumping. That is work. In engineering thermodynamics we separate temperature and pressure, but pressure is a kind of measure of temperature. It must be zero at absolute zero too.

    In an ideal gas, and dry air is close to ideal, we can in theory pump it (work on it) with no heat loss, adiabatically , or isothermally with all the heat extracted as we pump.

    If we pump it and then extract the heat at the start temperature, or near in absolute Kelvin terms, we have isothermal pumping.

    We cannot take any more heat out of it unless we find a lower temperature sink.
    But we can make that pressure work by venting back into the atmosphere, pushing against atmos. Pressure until it is all released. To do that work, the air must go to a lower temperature, the molecules slow down doing mechanical, invisible work.

    Even if we make the pressure turn an engine, it has an efficiency, but in that case we might recover most of the pressure work done by the compressor by making the engine produce heat, by stirring the air perhaps. But we never can get all that electrical energy as heat. Whereas in a resistor we can.

    So a compressor on ideal gas is never a heat pump, just a poor kind of heater that also makes air pressure.

    A heat engine is stuck with Carnot’s efficiency and is another subject.
    sicut vis videre esto
    When we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.
    Originally Posted by Ken G

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    OK I realise the cooled compressed air represents stored energy that is lost by simply releasing it outside, without recovering that energy.

    My idea now is that this lost energy is the work done by compressing the air isothermally. It should be possible to calculate this, apparently the work done in isothermal compression of a perfect gas is:

    W= nRTln(V2/V1), where n is the number of moles.

    At T=298K and 10:1 compression, W is 3988J/mole or 0.001108kWh/mole.

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    Quote Originally Posted by kzb View Post
    OK I realise the cooled compressed air represents stored energy that is lost by simply releasing it outside, without recovering that energy.

    My idea now is that this lost energy is the work done by compressing the air isothermally. It should be possible to calculate this, apparently the work done in isothermal compression of a perfect gas is:

    W= nRTln(V2/V1), where n is the number of moles.

    At T=298K and 10:1 compression, W is 3988J/mole or 0.001108kWh/mole.
    Great, a simplification is to replace nR with the specific R for dry air which is 287 joules per kg per degree K (or Cfor changes)
    With that R you can calculate work in joules for a temperature difference. I used that way back in the thread.
    sicut vis videre esto
    When we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.
    Originally Posted by Ken G

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    Quote Originally Posted by profloater View Post
    Great, a simplification is to replace nR with the specific R for dry air which is 287 joules per kg per degree K (or Cfor changes)
    With that R you can calculate work in joules for a temperature difference. I used that way back in the thread.
    I had an epiphany the other day.

    The answer to the problem is surely:

    Heat given out = work done in adiabatic compression - work done in isothermal compression.

    The work done in isothermal compression is lost to the outside unless some of it can be recovered.

    I made an error in the calculation in post #79, the result should have been 5705 Joules.

    Calculating the work done in adiabatic compression is slightly more complicated, but I found a good site:

    https://phys.libretexts.org/Bookshel...r_an_Ideal_Gas

    Going through this the work done in compressing the molar volume (24.47 L at 298K)) of perfect gas adiabatically at a volume ratio 10:1 is 9249 Joules.


    So 9249 - 5705 = 3544 Joules of heat available to heat the building, not including the compressor waste heat. So it is only 38% efficient.

    I suspect this could be improved by increasing the compression ratio beyond 10:1, because the work done in isothermal compression is proportional to natural log of compression ratio. But likely it can't be a heat pump unless the isothermal compression energy can be used or recovered with high efficiency, as you said.

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    Duplicate
    Last edited by kzb; 2021-Apr-06 at 11:47 PM. Reason: duplicate post

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    Quote Originally Posted by kzb View Post
    I had an epiphany the other day.

    The answer to the problem is surely:

    Heat given out = work done in adiabatic compression - work done in isothermal compression.

    The work done in isothermal compression is lost to the outside unless some of it can be recovered.

    I made an error in the calculation in post #79, the result should have been 5705 Joules.

    Calculating the work done in adiabatic compression is slightly more complicated, but I found a good site:

    https://phys.libretexts.org/Bookshel...r_an_Ideal_Gas

    Going through this the work done in compressing the molar volume (24.47 L at 298K)) of perfect gas adiabatically at a volume ratio 10:1 is 9249 Joules.


    So 9249 - 5705 = 3544 Joules of heat available to heat the building, not including the compressor waste heat. So it is only 38% efficient.

    I suspect this could be improved by increasing the compression ratio beyond 10:1, because the work done in isothermal compression is proportional to natural log of compression ratio. But likely it can't be a heat pump unless the isothermal compression energy can be used or recovered with high efficiency, as you said.
    Yes. Well done.
    But if you accept the normal definition of COP, a compressor can never be a heat pump until you employ vapours and latent heat.
    sicut vis videre esto
    When we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.
    Originally Posted by Ken G

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    People.

    What Kzb describes already exists, it is generically called a gas cycle or air cycle heat pump. They are not viable for comfort heating/cooling of buildings because:
    1. air piping is bulkier than refrigerant piping
    2. without phase change, practical machines cannot approximate ideal isothermal conditions in the heat transfer phases of the cycle (and are therefore less efficient)

    They are viable for other applications, such cryogenic air distillation and when you have lots of compressed air handy already and weight is a bigger factor than efficiency (for example, the ACM in jet aircraft).

    Basically, for a given cost and size of domestic air heat pump, you can build a better heat pump by using phase change. But if you want to go buy a model Stirling engine and hook it up to a motor, it will function just fine. The Stirling cycle does not employ phase change and an ideal Stirling heat pump performs at Carnot (highest possible) efficiency. Phase change is not necessary for a heat pump/aircon to function.

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    Quote Originally Posted by VQkr View Post
    People.

    What Kzb describes already exists, it is generically called a gas cycle or air cycle heat pump. They are not viable for comfort heating/cooling of buildings because:
    1. air piping is bulkier than refrigerant piping
    2. without phase change, practical machines cannot approximate ideal isothermal conditions in the heat transfer phases of the cycle (and are therefore less efficient)

    They are viable for other applications, such cryogenic air distillation and when you have lots of compressed air handy already and weight is a bigger factor than efficiency (for example, the ACM in jet aircraft).

    Basically, for a given cost and size of domestic air heat pump, you can build a better heat pump by using phase change. But if you want to go buy a model Stirling engine and hook it up to a motor, it will function just fine. The Stirling cycle does not employ phase change and an ideal Stirling heat pump performs at Carnot (highest possible) efficiency. Phase change is not necessary for a heat pump/aircon to function.
    Yes sure, but as you say, not effective for home heating. A Stirling engine is heat made to work at near Carnot, , not work to heat.
    If you have compressed air, yes, you can get work or indeed cryogenic cooling. These examples are heaters or coolers but not heat pumps. Heat pump is a very particular description with COP over 1, otherwise you can get heating with COP equals 1.
    in the compressor example we assume an electric motor and if you have that, you can do other electrically powered tasks. If your primary source is a fuel, you might use a compressor for convenience rather than burning for heat. You can make a heat pump work fired by gas for example, but it is still using latent heat.
    sicut vis videre esto
    When we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.
    Originally Posted by Ken G

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    Cooling is a different matter because usually there is no lower temperature sink to pass heat to, so you have to use expansion cooling, Peltier, or a vapour cycle. That is why liquid air and air con are more recent inventions than heating. When Fahrenheit invented his scale he used chemical cooling of salt in ice to get what seemed at the time the coldest possible state.
    sicut vis videre esto
    When we realize that patterns don't exist in the universe, they are a template that we hold to the universe to make sense of it, it all makes a lot more sense.
    Originally Posted by Ken G

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    Quote Originally Posted by profloater View Post
    Yes sure, but as you say, not effective for home heating. A Stirling engine is heat made to work at near Carnot, , not work to heat.
    If you have compressed air, yes, you can get work or indeed cryogenic cooling. These examples are heaters or coolers but not heat pumps. Heat pump is a very particular description with COP over 1, otherwise you can get heating with COP equals 1.
    in the compressor example we assume an electric motor and if you have that, you can do other electrically powered tasks. If your primary source is a fuel, you might use a compressor for convenience rather than burning for heat. You can make a heat pump work fired by gas for example, but it is still using latent heat.

    Cooling is a different matter...
    The Stirling cycle is reversible. It works equally well for work to heat transfer, and one as a heat pump indeed would have a COP over 1. It would work as an air conditioner, too: you compress, reject the heat, expand, and the refrigerant (air in this case) is below ambient temperature. You could run two in series if your hot sink is warmer than the compressed air. The key takeaway that you still appear to be confused about is that phase change (aka latent heat) is not a prerequisite for a heat pump.

  28. #88
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    Quote Originally Posted by profloater View Post
    Yes. Well done.
    But if you accept the normal definition of COP, a compressor can never be a heat pump until you employ vapours and latent heat.
    Actually 20 minutes after I wrote that, I realised that in that equation the only energy is coming from the compressor. If that equation was correct, there could be no heat pumping. Getting back to expanding the compressed cooled air will cool it below the temperature it started with, that still seems to imply heat must be pumped somehow.

    Also I don't know if phase change is essential. You still have to do work on the vapour to turn it into a liquid.

  29. #89
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    Quote Originally Posted by VQkr View Post
    People.

    What Kzb describes already exists, it is generically called a gas cycle or air cycle heat pump. They are not viable for comfort heating/cooling of buildings because:
    1. air piping is bulkier than refrigerant piping
    2. without phase change, practical machines cannot approximate ideal isothermal conditions in the heat transfer phases of the cycle (and are therefore less efficient)

    They are viable for other applications, such cryogenic air distillation and when you have lots of compressed air handy already and weight is a bigger factor than efficiency (for example, the ACM in jet aircraft).

    Basically, for a given cost and size of domestic air heat pump, you can build a better heat pump by using phase change. But if you want to go buy a model Stirling engine and hook it up to a motor, it will function just fine. The Stirling cycle does not employ phase change and an ideal Stirling heat pump performs at Carnot (highest possible) efficiency. Phase change is not necessary for a heat pump/aircon to function.
    I kind of knew that if it was as good to do without a refrigerant, they would already be doing it.

    What set me off was seeing the miserable temperature increase afforded by the heat pumps they want us to buy. Your hot side is 30-40 degrees C. This radiator temperature is no good in a cold winter I don't care how much insulation is in the loft. I recalled being told that when they compress gases into the 230 bar gas bottles they get red hot. So I thought that sounds more like it.

    Anyhow I will Google air cycle heat pump when I get time. I think this thread has shown me I still don't get thermodynamics any better now than when I was at university !

  30. #90
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    Quote Originally Posted by kzb View Post
    What set me off was seeing the miserable temperature increase afforded by the heat pumps they want us to buy.
    That's an engineering tradeoff; it's not the direct fault of the fact that the heat pump takes advantage of a phase change. The maximum possible COP of a heat pump is a function of the temperature of the hot and cold reservoirs: COPmax=Th/(Th-Tc).

    So if Tc is fixed at 5įC as your design OSA temperature, a 30įC hot side (COPmax=12.1) is potentially 2.5x more efficient than a 80įC hot side (COPmax=4.7). 30įC is definitely on the cool side for space heating, but not impossible - that's a good temperature for hydronic slab heating systems, for example. You just need lots of heat transfer area because the deltaT between your heating element and design temperature are low. So much heat transfer area, in that case, that the "radiator" has to be an architectural element like a radiant wall or heated slab. Neither one is likely to be attractive for residential applications.

    40įC is still a cool fluid by traditional service heating standards, but getting up to the range where a fin-tube heat exchanger, with forced convection, is becoming practical - you've more than doubled your temperature difference from the typical comfort heating set point (say, 22įC). The general trend in commercial and industrial heating systems is to use cooler service heating fluids because of the increased efficiency as well as the ability to recover heat from processes such as air compression.

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