Talk:Heat pump/Archive 1

Link to Peltier technology?
We might want to link this to the Thermoelectric_cooling article. —Preceding unsigned comment added by 129.2.135.72 (talk) 19:44, 22 August 2008 (UTC)

On efficiency
I am not sure what the potential for bias is here, but it would help us non-specialists to know what the standard for efficiency is; is there an absolute standard or is it entirely relative; is the relative efficiency of heat pumps being compared to other pumps; can we have details on relative efficiency among heat pumps (for example, what is the most efficient refrigerator; what is the least efficient refrigerator; how efficient are average household and inudstrial refrigerators -- can this be included without endorsing any specific product?) Slrubenstein


 * Bias, because I am a great supporter of heat pumps. Now that I have explained the efficiency, I feel the text shows that even more. - Icarus


 * Well, bias or no, I think the more information the article presents, the better it is -- so I appreciate your additions. Nevertheless, they raise more questions.  First, what does it mean that an 100 watt electric heater delivers 100 watts of heat?  I thought watts = volts x amps.  How does this measure "heat?"  Are electric heaters really 100% efficient?  You still haven't explained how efficiency is being measured.  Does an electric heater convert into the electricity into exactly the amount of heat that was originally used to generate the electricity?  Is this really possible?  Also, the article is still vague on how, and under what specific conditions, a heat pump can excede 100% efficiency.  I am not an engineer or a physicist so the answers to these questions are not at all evident to me -- I hope you ro someone else can develop the article so it will be clearer to a mass audience, Slrubenstein


 * Watts are units of power (= energy/time), and yes, you can work out electrical power from voltage x current, but that's not the basic definition of power. Efficiency is generally defined as (energy in)/(_useful_ energy out), however this does not appear to be the definition of efficiency being used in this article - nothing can be 100% efficient by this definiton, let alone more so.  It might be best to quote the efficiency of a heat pump and that of an electric heater by this (widely accepted) definition, and allow comparison.  (This is all complicated by the fact that in most cases the heat is the _useless_ energy, not the useful ...) --  Bth

This article should be written by someone with at least a good knowledge of basic Thermodynamics (Ch415 as we teach it here). I dont really have the time for it now. Yest if the product you want is heat the effciency can well be above 100%. -- 152.1.193.141


 * Okay, the previous two comments just confuse me (nothing can be 100%efficient by one definition, but that isn't the definition here) -- as I said I am not a pysicist or engineer, but I assume the article is being written to help educate people just like me. The point is not to deocument an argument over how efficiency "should" be measured, the point is that the article has to explain clearly what is going on.  I wish it did, but so far it does not, Slrubenstein


 * I suppose if you imagine that the heater is 100% efficient because all of the heat it produces heats the room. Now consider a fireplace consuming an amount of would equal to 100 watts, it might be only 50% efficient because half the heat goes up the flue and only half goes into the room. A heat pump might be 200% because it uses the same 100 watts to move twice as much heat into the room as the heater. I think. -- rmhermen


 * Okay, I know the flu allows for airflow, and also the exhaust of smoke; I know electric heaters do not need airflow and do not produce smoke, so it doesn't need a flu -- is this the main reason why it is so much more efficient? Is it possible for someone to explain all this -- and the physical laws behind it -- in the article in a way that is still accessible to people like me?  I appreciate the answers to my questions, but I also pose them as suggestions, hopeful of improving the article.  As far as I am concerned, the article still does not "explain" efficiency, at least not clearly. That said, I do think the article has improved significantly today.Slrubenstein


 * I would suggest following the link that someone (rmhermen, I assume?) put in to the Carnot heat engine article, and the link to Thermodynamics (which I just wikified).  The theory is explained in those articles (including a definition of efficiency, though perhaps not one you would find very helpful).  Apologies for the lack of clarity in my earlier attempt to explain, I was getting confused myself (shame on me for skipping those Thermo classes). -- Bth

As a device to bring light into a room a standard lightbulb is only 5 percent efficient. 5 percent of the electric energy you put in comes out in the form of light, 95 percent in the form of heat. Efficiency is generally defined as (energy in)/(_useful_ energy out) A 60 Watt bulb requires 60 Watts, and provides (5 percent) 3 Watts of light (useful energy). So it's not very efficient as a light source.

As a heater it is more efficient: A 60 Watt blub still requires 60 Watts, but now it provides (95 percent) 57 Watts of usefull energy, in the form of heat.

An electric heater is 100 percent efficient (not counting the energy lost in transport fom the power plant). No energy is lost in any form but heat. 100 Watts go in, 100 Watts come out, in the form of heat. Don't argue, they do. For example, if the goal is to heat water, it takes 4.18*103 Joules of energy to heat 1 kilogram of water 1 degree Celcius.

So an electric heater of 100 Watts (100 Joules a second) can heat 1 kilogram of water at a rate of 1.43 degrees every minute.

A heat pump is a miraculous device. It does not produce (the bulk) of the heat it 'produces', it gets it from another source, such as groundwater or the air. It uses mechanical devices to do so, which of course require electricity to run.

Depending on the source (temperature) of the heat a heat pump is less or more efficient. It can theoretically 'pump' ten times more energy around than it could ever hope to produce. And as an added bonus:

The machinery also produces heat as a by-product, as do all electrical devices (feel up top your monitor, or around a lightbulb). In fact, if you put 100 Watts of electrical power into the machinery, 100 Watts of power (1.43 degrees of temperature rise a minute) will come out again.

So no, technically nothing can be more than 100 percent efficient. But they can do more with the energy they are given.

Let's say the heat pump has found a loophole.

And let's say it is several times more energy efficient than an electrical heater, if you want to heat say your house.

And let's say an electric heater is 100 percent efficient, (prove me wrong).

Efficiency is generally defined as (_useful_ energy out)/(energy in)

Define efficiency as (useful energy out)/(electrical energy in)

Do the math. -- Icarus


 * Try explaning, not insulting. And all the electric heaters I know also produce light so they are not 100% efficient. --rmhermen


 * I'm not trying to be insulting, I'm just a little shortfiused today, sorry.
 * Well, I appreciate the time that you, rmhermen, and bth are putting into this. The point is not to be argumentative, but to make the article as informative and clear and accesible as possible.   When I have asked questions it wasn't in the hopes of proving anyone wrong, but rather in learning more.  And I am... Slrubenstein

You will find that much of the light emitted by an electric heater is infrared and red light. Have you ever noticed how the sun feels warm on your skin? That's because a lot of the energy emitted by the sun is in the form of infrared light. Technically (IR-)light is not heat, but as soon as it strikes a (black) surface it will be. The IR radiation will be converted into heat, or reflected away. Black absorbs absolutely, other colors do not.

So the light from your electric heater will still become heat, and unless you aim your heater outside, or use mirrors you reflect the radiation outwards, the radiation will always heat your house.

ps: of course some energy will escape, it always does. It'll go out your window, or through cracks underneath doors. But all energy will be(come) heat. - Icarus

A heat pump moves energy from one place to another (i.e. from the air outside to the air inside). If you think of a heat pump as a furnace heating your house, it would appear to have more than 100% efficiency, since 100 Watts used by the heat pump might produce 400 Watts of heat in your home, it appears to have a 400% efficiency. The heat pump did not create energy, this violates the laws of thermodynamics, it merely transported 400W of energy from the air outside to the air inside, and it took 100W of work to transport it. Since you don't pay anything for the thermal energy taken from outside your house, we tend to forget that part of the equation.

It's like free firewood, you don't need to pay for the wood, just to transport it. A cord of wood can contain 20,000,000 BTU of energy, a gallon of regular gas has 126,200 BTU. So you burn 8 gallons (1 MBTU) of gas transporting the wood, and get 20 MBTU of energy to heat your house with... does your truck have 2,000% efficiency?

The efficiency of the pump can never be more than 100%, it never does 100W of useful work per 100W of power, nor does all of the thermal energy it takes in end up transported to the output side. If you ignore the thermal energy in, it can give you more useful energy than you paid for... this is the Coefficient of Performance (COP). - Dan

There is diffrent concepts of Ppump efficiency, and system efficiency, in case od room heating we do not care how that heat comes to us, the fact is we paid for 1kwh electric power, and got 5kwh of heat. so efficiency of your heating system is 500%, someone who used simple heater, paid fro 1kwh and got 1kwh his efficiency is 100%, and yest someone uses gas heater, and he wastes 20% of heat through ehasts system so hie sficiency is only 80% in result we do not care about details. fact is for (what you pay)/ (what USEFULL you get) = efficiency. —Preceding unsigned comment added by 88.118.84.126 (talk) 16:03, 23 January 2009 (UTC)


 * Hehe, just thought about that, if you can use say 70 watts of energy to move and collect heat energy that't equall or more than 70 watts (something that a normal heat pump easly can do), than you could easly build a large scale powerplant that compressed air and than used the heat difference to drive somthing like the stirling engine(since its extremely efficient at converting heat into momentum) and create extra power avaible for everybody, kinda geothermal plant you could say.--Nabo0o (talk) 13:56, 13 August 2008 (UTC)

Efficiency omitted from article
Wouldn't it be more conventional to define efficiency as (useful energy out) / (electrical energy in), or more generally (useful energy out) / (useful energy in)? That is, the other way up from the definitions used above. This is particularly important for heat pumps as they are fairly unusual in that for them this number can be greater than one (or 100%, if you prefer). This is because this definition of efficiency ignores the useless energy input. Perhaps the main article should make this point explicit. Ed Davies


 * Yes, you are referring to what is called the Second Law Efficiency in thermodynamics. For some reason, popular use of thermodynamic terms never got beyond the first law (probably because the concept of entropy seems so scary - and has philosophical connotations?).  I am going to think about edits to this article that internalize this distinction in a way that isn't too scary.  As it is, saying it doesn't make sense to talk about the efficiency of heat pumps is a little discouraging. Wes Hermann 05:06, 3 December 2006 (UTC)


 * Okey, its two years since someone wrote here, but I have to add that when one talks about efficiency in something like a heat pump, you don't use efficiency but COP (Coefficient of Perfomence), that way we talk talk about how much energy we can create (or extract) in form of heat based on the ammount of electricity we use. For a normal cheap heat pump the COP might be about 3, which means 300% gain in energy, compared to what you would get if you just used a normal heat element.Nabo0o (talk) 13:49, 13 August 2008 (UTC)

the term "heat"
The use of the term heat in the previous entry is incorrect: A ball contains 100 units of Energy (in the form of kinetic energy that belongs to the individual molecules of gas enclosed in the ball). When the volume of the ball is reduced, the amount of energy per volume (the energy density) increases, and heat is given off to put the energy density of the gas in the ball in equilibrium with that of its surrounding.

--ParisFrog 11:42, 21 July 2005 (UTC)

Yes. It appears the main article was written by someone who was taught that heat is some kind of substance -- caloric theory. I hope we can make the article use the term "heat" more accurately, without making it more difficult to understand. --DavidCary 07:16, 6 January 2006 (UTC)

ParisFrog is also missing something: in a heat pump the compression is (ideally) an adiabatic one, i.e. without opportunity to equilibrate with the surroundings, because this results in the highest possible temperature. The heat energy in the ball after compression is equal to the heat energy in the ball beforehand plus the mechanical work done in compressing the ball. Moreover, only the latter contributes to increasing the *temperature*, as opposed to just the energy density. --Alistair

Football/soccer and ping pong/table tennis
"...say the volume of a football (or soccerball). The air within the volume of the ball has say 100 units of heat. This air is then compressed to the size of ping pong ball (table tennis ball)..."

This is hardly a critical error to fix, but I think it might help a little if the explanation referred to balls that aren't ambiguously named. If ball X is a "football" here and a "soccerball" over there, maybe "basketball" would be a better choice for the analogy, since it's called the same thing everywhere? I'd suggest "basketball" and "golf ball" -- they seem like balls most people would be aware of. CSWarren 13:07, 6 January 2006 (UTC)

Theoretical maximum efficiency?
Is there a theoretical limit of COP? -tom2146


 * I think that the theoretical maximum performance is determined by the number of degrees of absolute temperature over which you're pumping. That is, if the desired hot-side temperature is 300 Kelvins (27 degrees C) and the cool-side temperature is 273 K (0 degrees C), then you're trying to pump heat over 10% of the absolute temperature and the maximum COP you can achieve is 11 (with the electricity supplying the extra 1/10 of the temperature range).


 * But I could definitely be wrong on this and if I'm wrong, I'm sure someone will correct me soon ;-).


 * Atlant 15:20, 7 August 2006 (UTC)

In your case the efficiency would increase if the cool-side temperature would be even lower, which IMO is not correct. The efficiency gets lower (and not higher) with the increasing temperature difference between outside and inside temperature (if you're trying to increase the inside temperature, which is already hotter than the outside temperature). I may be misinterpreting your formula though ;-)


 * No; look at my suggested formula again. If the cold-side temperature decreases, you'd be pumping over a larger portion of the absolute temperature space, say 1/9 instead of 1/10, so I'd sugges that the maximum COP you could achieve in that case would be 10 (with electricity supplying the extra bit).


 * Atlant 13:13, 30 October 2006 (UTC)

I understand now, what you're trying to say. So the COP in your case would be:

$$COP = \frac{T_{\mathrm{hot}}}{T_{\mathrm{hot}}-T_{\mathrm{cool}}}+1$$

Where T is absolute temperature (in K).

Right?


 * Yes, but keep in mind that while the formula must be something like that, I'm not at all certain it is exactly that. We really need a subject matter expert to tell us the correct answer.


 * Atlant 13:38, 2 November 2006 (UTC)

"Misguided" belief that CFC's cause damage to the ozone layer if released into the atmosphere. (?)
i have no technical expertise in CFC's or ozone, so i didn't edit this article directly, but ...

Under Refrigerants, the current text states, "Its [CFC] manufacture was discontinued in 1995 due to the misguided belief that CFC's cause damage to the ozone layer if released into the atmosphere.''' [my bold]

The current text's statement that belief "that CFC's caused damage to the ozone layer ..." was "misguided" is a controversial claim that is not substantiated with facts in the text nor in the provided references, especially in light of, which is cited within this very article and which asserts the generally accepted positive correlation between CFC's and ozone depletion.

IMHO, either factual substantiation should be provided or the word "misguided" should be deleted in the spirit of neutral point of view

GaroldStone 06:46, 16 December 2006 (UTC)GaroldStone


 * I agree; done.


 * Atlant 17:14, 16 December 2006 (UTC)

Someone who knows what's going on might want to check the "Air or Ground Heat Source/Sink?" section of this article.
I made what I think are relevant grammitical changes to that section, but I had to finish suddenly. There was at least one part of it that I wasn't able to make sure I kept factually correct while changing its structure—that is, the last two sentences of the first paragraph: "Two common types of heat pumps for homes are air-coupled and ground-coupled heat pumps (geothermal heating). The difference between these is whether heat is transferred between the indoor air and the outdoor air or between the indoor air and the ground, respectively."

I'll try to look it over later. I'm just saying—y'know, "FYI" and stuff. -Dan 05:13, 17 December 2006 (UTC)

Mixup between geothermal and ground/rock heat pumps on the "Air or Ground Heat Source/Sink?" section
There's a mixup between the concept of using geothermal heat sources (geothermal heating) and using solar energy that is stored in the ground or in the rock. Most ground heat pumps or rock heat pumps do NOT use a geothermal heat source, but rather solar energy stored in the ground. This is a common mistake. Locations with access to geothermal heat are rare, whereas rock or the ground can be used as a heat source for a heat pump almost anywhere. See this page for some further explanation: http://www.thermia.com/heatpump/heatpump-types.asp (It does not mention geothermal heating, probably because it's more unusual. I'm not representing the company behing the web page in any way.)

/Erik

81.231.246.155 09:20, 29 August 2007 (UTC)

Explanation of physics of heat pumps is very wrong
The gas laws cited in the explanation are ideal gas laws that have no bearing on temperature changes (temperature change on compression/expansion is in fact one of the most obvious deviations from ideal gas laws in common practice). Yes, raising the temperature of an ideal gas will increase the pressure, but raising the pressure will simply result in a smaller volume at the same temperature if there is no heat flow. Real (non-ideal) gases do actually show temperature changes due to forces acting between the gas molecules (in the extreme case of condensation this is the latent heat) - but the actual amount will depend heavily on the actual gas and the conditions - eg at normal temperatures hydrogen actually get warmer on expansion. 220.240.255.193 03:05, 19 January 2007 (UTC)

1) Do not remove posts from the talk page, even if they are your own. 2) The ideal gas laws are exactly the laws that should be used to describe heat pumps.

The Idea Gas Laws state more or less 'Pressure * Volume = n * R * Temperature'. 'If nRt is constant, then with change in pressure, there must be a change in volume.' However, the Ideal gas laws still apply to compression, volume and temperature, because thermal energy does not equal temperature; Specific heat capacity determines how 'hot' a substance becomes with a given amount of energy transfered to it, however, the temperature of said object despite being extremely high would not be able to boil water or burn your hand, while a much cooler object in theory could.

By expanding a gas over a very wide area, the average temperature in the volume drops, because there is no extra thermal energy transfered to the gas. Passing said gas over a plate two degrees warmer than the gas will warm the gas up a little bit, but could add massive amounts of thermal energy because the gas can be spread over a very wide area to collect this energy. When compressed after absorbing thermal energy, the temperature will rise in concert with the reduction of volume. This high temperature gas will pass through a small area that is much cooler than it is, transferring the thermal energy it collected when it was expanded.

All of this is covered in the Ideal Gas Laws and it is correct to use the Ideal Gas Laws to explain heat pumps. --  Orbit  One    [ Talk 19:58, 19 January 2007 (UTC)

First up yes I admit deleting my discussion was a little rough _ I originally put in the dispute because I couldn't be bothered writing a correction my self, but after thinking about it the error bugged my, so I decided to take the high road and fix it myself - it never occurred to my that someone was feeling proprietorial about that section, I assumed it was that way because no one cared. Since the inaccuracy was fixed the discussion point seemed pointless - my changes showed what the error was more clearly than a brief explanation (its a scientist thing - we don't see the point of discussion when we're right sometimes it bites us in the arse). Science is democratic - one scientist is one democracy, n scientists are n independent democracies (hint: the majority has nothing to do with the truth).

Well thats the sermon, I tried science, here comes democracy... and that long explanation I was avoiding. First up what hit my about you explanation is that it did not make it clear that the majority of practical heat pumps (refrigeration, air conditioning ect) use liquid/vapor phase changes to do the majority of the heat transfer, the other effects (Joule-Thomson effect and work cooling are only used where phase change is not practical either dew to operating temperature constraints or mechanical issues. There is a good reason for this - the heat change for the phase change dwarfs all but the biggest expansions - (varies a bit with temperature) 40,000 joules for 1 mole of water, 8,000 even for methane - that's enough to to change the temperature of your ideal gas by > 100K all at high density (liquid) in a nice small temperature range.

Now getting back to those specialized heat pumps that don't use phase changes, but use gas compression/expansion cycles (there are other types such as magnetic and thermoelectric heat pumps, but they are even less used).

First up, believe me when I tell you this (check with common sense, check with you text books, check with you chem lecturer whatever it takes) the ideal gas law was not meant to be used that way - it is strictly useful for static equilibrium situations (so yes given amount, pressure and volume you can find temperature but it will not tell you how to get there - you cannot plug in changes P/V and expect it to predict the temperature change resulting from doing that transformation in some way, although it does happen to at least point in the right direction). In fact the temperature change depends on how you do the change...

Now while the ideal gas law can't be used on its own the underlying ideal gas model is up to the task: lets look at it with a little thought experiment: Temperature is a direct measure of heat energy - the exact relationship depends on the number of ways a substance has to distribute the energy - for an ideal gas the energy is exclusively molecular (translational) kinetic energy - I assume this is not controversial.

So let's take a magic box (an implement very common in thermodynamics) now this box doesn't let any heat in or out and has no heat capacity of its own (we are looking at an adiabatic change). This particular box has a perfectly sealed movable partition since we want to study volume change. Now our gas is on one side of the partition, the other side a vacuum (since that way we don't have to deal with the gas outside, or any sound energy carried away).

Now consider two scenarios where we move the partition to double the volume of the gas: a) we move the partition so fast that none of the molecules hit the partition while it's moving (the gas exerts no pressure on the partition while it moves).

b) we move the partition slowly so that the gas exerts its full pressure on the partition at any given moment

Now let's look at the thermodynamics, now once the molecules spread out evenly: a) the molecules still have all the energy they did before (there's nowhere they can have lost it) i.e. the gas is at the same temperature and half the pressure. (look at the ideal gas description - temperature and energy are related independent of the other conditions).

b) the molecules do work against the moving partition since the pressure, hence force varies with time we have to use calculus to calculate the total work. Now for an incremental movement the work is equal to the force times the distance, but force is pressure times area - re arranging we find the incremental work is pressure times change in volume (delta V = area * delta position) this work is energy that is removed from the kinetic energy of the molecules - reducing the temperature (as it turns out the relationship is total KE = 3nRT/2) so P dV = (3/2)nR dT

ie (nRT/V)dV = (3/2)nR dT

2dV/V = 3dT/T

some calculus magic later (via integral of 1/x = ln(x)):

T2 = T1(V1/V2)^(2/3)

This cycle represents the maximum amount of work that the expanding gas can do - hence the maximum cooling which is still less than the result from the incorrect use of the ideal gas law. For most real gases the cooling is less again since their heat capacity is larger - try 3 or 4 instead of the 1.5 (3/2) for an ideal gas (actual significance of this depends of the temperature difference you are pumping against).

For instance in your example of a CO2 extinguisher most of the cooling would be due to latent heat (the highly compressed CO2 is likely to be liquid), and next after that would probably be joule-thomson cooling, since for expansion though a nozzle the main work done is against the air outside.

The physics explanation I put covered the three major heat movement effects (Joule-Thomson, expansion work, and latent heat) and was I'd say 90% accurate compared to the say 30% of yours. Intro chem meet BSc(hons) chem. Cbrettin 08:11, 20 January 2007 (UTC)

Although it is nice that you did put work into this, the process is not as rapid as you would have it (editing the article) when it is in dispute.

Secondly, you pretty much ambushed me with that one. I do not have my PhD. in physics yet, so I need to check with my teacher your explanation. But I do understand science is not a democracy, neither is Wikipedia, but there is still a process to be followed.

Thirdly, I doubt people with less than a few years of university reading physics will understand what you just said. If you can simplify your explanation, it will go a long ways in improving the article. As in mathematics, the simplest formula is the one that is always used.

Lastly, pressure and volume do play a big part in the cycle and can still be explained by Ideal Gas Laws. Let me take a real world example.

If you take an ice box and empty its refrigerants into a pressurized cup with a piston for a lid, it will keep its temperature constant until you lift the piston up. At that point, the liquid will start to boil off, a phase change, and cool off quickly. We changed only the volume and pressure inside the cup, however, the liquid cooled off quickly. What conclusions can we come to from these observations?

However, I will still ask my teacher to read what you have posted and ask him if your explanation is 90% correct or not. I can first do that on Monday.

--  Orbit  One    [ Talk 21:42, 20 January 2007 (UTC)

Ok, I'll check again after Monday.

The Scientist rant was just to try and explain why it hadn't even occurred to me that this was a "real" conflict (in retrospect I should have checked how recently that section had been touched before assuming it was simply neglected).

To introduce some jargon (the names make it easier to look up in a themo text book) scenario a above is called free (or irreversible) expansion and the temperature effects are described by the Joule-Thomson effect - read that article and you will see that (free) expansion will not always cause cooling for gases (and no change for ideal gases).

Scenario b is called reversible expansion (because unlike a it can be symmetrically reversed - often discussed under the heading of "Expansion work") and is much more important industrially since not only does it usually provide good cooling it also does useful work in the process.

So while the ideal gas equation seems to point in the right direction it is not quantitatively true for dynamic changes (especially for ideal gases). For instance in you "expansion" of the liquid refrigerant the pressure over the liquid is the vapor pressure and this will vary with temperature much more dramatically (roughly exponential) than an ideal gas pressure. The exact math of the temperature change is unfortunately rather involved since it must take into account the changing amounts of the gas and liquid and their different heat capacities.

As you can see the real situation is complicated which is why my edit to the article seems a little vague (the idea is in the best encyclopedic traditions the links should help those who want more detail).

I understand that the explanation of these effects is rather above the average reader, which is why in the article edit I simply named and linked the phenomena and summarized their effects - in my talk response I tried to cover the bits that weren't in those links (since I assumed you had looked and found them lacking) - expansion work (Turboexpander doesn't have a lot of explanation (someone should probably fix that ;-) - so as far as I'm concerned the important part for the teacher to look at is my article edit. Cbrettin 23:12, 20 January 2007 (UTC)


 * Having talked with my teacher, he says you are right more or less, but the explanation needs to be worked on and reduced to the simplest words possible.


 * If you want to really impress us, write it as two different levels of understanding. One a junior high school student could understand, a 15 yo, and the full explanation and post that. --  Orbit  One    [ Talk 13:39, 22 January 2007 (UTC)


 * Ideally the edit would integrate with the rest of the operation section, which already includes some explanation of liquid/vapor phase change heat pump. I don't have a lot of free time, but I'll try an find some time in the next couple of weeks to put something together, unless someone beats me to it (well, I can dream). Cbrettin 11:27, 23 January 2007 (UTC)

The gas laws behind heat pumps
The section entitled "The gas laws behind heat pumps" is in a sorry state. Reading the discussion above there seems to be people here who know enough thermodynamics to correct it, but until that is done, I suggest the entire section be removed, as it is better to have too little information in the article than to have inaccuracies.

The main problem is that the section seems to say that a gas that always cools down when the pressure drops, which is false. An ideal gas passing through an expansion valve would not change temperature. (A real gas might, see Joule-Thompson effect.)

--PeR 09:25, 9 March 2007 (UTC)

Someone should give a mention of the Stirling cycle. It is used to liquefy gases by pumping the heat out of them. --Wrinkleworm 04:26, 18 September 2007 (UTC)
 * I don't think that's how Stirling engines work, my understanding is that they use a compression expansion cycle to transfer heat between a hot and a cold reservoir. In the process the cycle can be used to extract work. I don't think that there is a phase change involved in the ideal cycle. It's like the carnot cycle, but you can actually implement it, with some effort. User A1 08:07, 18 September 2007 (UTC)

I agree with you. What I meant to say is that one of the current applications of Stirling cycle heat pumps is in the liquefaction of gases by removing the heat energy from the gases. I mentioned this to illustrate that the useful limits of the Stirling cycle compare favorably to those of phase change heat pumps. In addition they do not require specialized gases or high pressures which make them much more durable in application.

Since the article is entitled "Heat Pumps," and not "Phase Change Heat Pumps," and it addresses only phase change heat pumps while ignoring all other types, either the article is incomplete or the title is inappropriate. --Wrinkleworm (talk) 22:21, 12 January 2008 (UTC)

Heat transfer
After the heat has been absorbed from the source (air or ground), the heat is transferred and used in the home or building (for space heating. This is generally done by pipes in the floor, wall or ceiling . The heat pump can also be used to heat water (thus in solar hot water systems), yet this is generally not done as it is much less efficient. Finally, the heat pumps may also be used to cool the house (eg in summer). However, the latter is also less efficient, and therefore less commonly practiced.

KVDP (talk) 10:09, 30 March 2008 (UTC)
 * Thanks for the contributions, Though I think we need to be careful we don't confuse the thermodynamic concept of a heat pump with the applications thereof, such as heating using external heat sources to an internal sink. This is clearly an application of heat pumps and not a property of the heat pump itself. User A1 (talk) 23:07, 30 March 2008 (UTC)

Error in Figure: In the "Operation" section, it correctly says that a heat pump takes in energy at the cold side. However, the figure ("A simple stylized diagram...") erroneously shows energy being given off at the cold side. The figure should be corrected. Redbelly98 (talk) 17:57, 17 February 2009 (UTC)

Types of heat pumps
A number of sources have been used for the heat source for heating private and communal buildings. The two main types of heat pumps are compression heat pumps and absorption heat pumps. Compression heat pumps always operate on mechanical energy (trough electricity), while absorption heat pumps may also run on heat as an energy source (trough electricity or burnable fuels).

Air-source heat pumps
Air-source heat pumps are relatively easy (and inexpensive) to install and have therefore historically been the most widely used heat pump type. However, they suffer limitations due to their use of the outside air as a heat source or sink. The higher temperature differential during periods of extreme cold or heat leads to a lower efficiency, as explained above. In mild weather, COP may be around 3.5, while at temperatures below around −5°C (23°F) an air-source heat pump's COP will drop below 2. The average COP over seasonal variation is typically 2.5-2.8.

The text above is a lot clearer then the current gobledigook: "In mild weather, COP may be around 4.0, while at temperatures below around −8°C (17°F) an air-source heat pump can achieve a COP of 2.5 or better, which is considerably more than the COP that may be achieved by conventional heating systems. The average COP over seasonal variation is typically 2.5-2.8,[9] and high efficiency model in Japan over 6.0(2.8kW) written in the IPCC 4th Working Group III report chapter 6 [10]." I didn't revert as I don't want to interfere with more qualified people's work, but reverting seems required here.

Also, a section about real world implementation issues would be nice. For example, why does the COP decrease when the heat pump's heat output increases? Is it only because the air coming out is hotter (hence there is a greater cold-hot source temperature difference)? In this case, running a (inverter?) 6kW heat pump at 2kW would have the same COP as running a 2KW pump (provided the air flow is that same)? And the COP quoted in commercial brochures for inverter heat pump is for which power output? the maximum one? (which would explain why it decreases as output goes up if the explanation above is correct). Thanks. —Preceding unsigned comment added by 202.154.149.26 (talk) 04:44, 9 May 2009 (UTC)

Geothermal heat pumps
Ground-source heat pumps typically have higher efficiencies than air-source heat pumps. This is because they draw heat from the ground or groundwater which is at a relatively constant temperature all year round below a depth of about eight feet (2.5 m). This means that the temperature differential is lower, leading to higher efficiency. Ground-source heat pumps typically have COPs of 3.5-4.0 with little seasonal variation. The tradeoff for this improved performance is that a ground-source heat pump is more expensive to install due to the need for the digging of wells or trenches in which to place the pipes that carry the heat exchange fluid. When compared versus each other, groundwater heat pumps are generally more efficient than heat pumps using heat from the soil.

Solid state heat pumps
In 1881, the German physicist Emil Warburg put a block of iron into a strong magnetic field and found that it increased very slightly in temperature. Some commercial ventures to implement this technology are underway, claiming to cut energy consumption by 40% compared to current domestic refrigerators. The process works as follows: Powdered gadolinium is moved into a magnetic field, heating the material by 2 to 5 &deg;C. The heat is removed by a circulating fluid. The material is then moved out of the magnetic field, reducing its temperature below its starting temperature.

Heat transfer
After the heat has been absorbed from the source (air or ground), the heat is transferred and used in the home or building (for space heating. This is generally done by pipes in the floor, wall or ceiling . The heat pump can also be used to heat water (thus in solar hot water systems), yet this is generally not done as it is much less efficient. Finally, the heat pumps may also be used to cool the house (eg in summer). However, the latter is also less efficient, and therefore less commonly practiced.

The info is correct, so best not to remove any info. You may check it all by translating the document (cut and paste in google translate, babylon, ...

KVDP (talk) 13:14, 6 April 2008 (UTC)


 * "However, the latter [cooling] is also less efficient, and therefore less commonly practiced." I checked the reference, but it was all Dutch to me.  It may be less efficient, but there are also fewer alternatives, and since just about everyone has a refrigerator I strongly disagree with this statement! Constan69 (talk) 05:04, 18 October 2008 (UTC)


 * I removed the entire section. It is out of place and contains very dubious information. Since the sources are dutch, I can't verify anything, thus it is unverified to me and the millions of other non-dutch speaking english-speakers. Really that information should go in the intro or in a section on "applications" (info on applications of heat pumps are in the intro right now i think).
 * Also, there are vauge statements like "the latter is also less efficient". Less efficient than what?? Also, it *wouldn't* be less efficient. Its effficiency depends entirely on the heat gradients. There is unneccessary information like "this is generally done by pipes in the floor, wall or celiling" - totally unneccessary. And the "solar hot water systems" thing is completely non-sequiter. Heat pumps are nothing like solar water heating. This information needs to be much more thought out before its added again. Fresheneesz (talk) 23:15, 19 October 2008 (UTC)

table of COPs
What exactly is the table of COPS - are these theoretical or (I guess)typivcal values? For instance, the COP of an air source heat pump at 0C is shown as different to the COP of a ground source unit - presumably the 'Carnot' COP is the same?

Without further clarification, I think this table should be deleted. (RR) —Preceding unsigned comment added by 207.245.198.135 (talk) 12:46, 26 February 2009 (UTC)

I just discovered the difference between the COP of heating and the COP of cooling (which I think is a stupid difference to have). Anyways, I was wondering what COPs were in the table in the CoP and Lift section? Are they for heating or cooling? The type should be noted in the table. Fresheneesz (talk) 17:08, 7 October 2008 (UTC)

The description above the table says "The CoP increases as the temperature difference, or "Lift", decreases" but the fourth row "Prototype Transcritical CO2 (R744) Heat Pump" demonstrates that is not true (or else there's a serious mistake in the Steen paper).

Also, there's no mention of SPF (seasonal performance factor) anywhere in the article, which seems like a significant omission. —Preceding unsigned comment added by 131.111.85.79 (talk) 09:48, 18 December 2008 (UTC)


 * I sorry but table in the section on “COP and Lift” has an error. It is not possible that the Lorentz Cycle has a higher COP then the Carnot cycle for the same source and sink temperatures. —Preceding unsigned comment added by 24.251.56.214 (talk) 14:15, 25 August 2009 (UTC)