# Assumed violation of physics - Heat vs. Work

by Low-Q
Tags: assumed, heat, physics, violation, work
P: 21
 Quote by Low-Q Good point. However I am not searching for loopholes in science, just searching for an energysource in the air which a heat pump can harvest in order to power a heat engine. As many links, explanations from you guys has shown, it will not be possible.
I'm disappointed. You give up far to easily.
Mentor
P: 17,545
 Quote by Tom Booth The heat pump can still operate even if used outside its intended function.
Then it is not operating as a heat pump.

Tom, since you have not provided a reference to an alternative definition, I consider the discussion about the definition of a heat pump closed until you can do so.

This forum is not for personal speculation or personal theories. We only deal with mainstream science. In some cases there are multiple mainstream scientific definitions for the same term, in which case it is worth clarifying which definition is being used, but here there appears to be only one definition. Semantic arguments are boring, so I won't continue with this one.

 Quote by Tom Booth That is a result of his haste or enthusiasm to use the diagram before I explained it in detail in the other forum. As he did not clarify the nature of the apparent cold reservoir, I have made an effort to do so.
I missed the clarification amidst all of the definition nonsense. Is the cold reservoir not shared?
P: 21
 Quote by Crazymechanic @Tom , no I don't think we can call those reservoirs...
I don't either.
Mentor
P: 17,545
 Quote by Tom Booth The heat pump in the diagram is maintaining an artificial heat sink within the ambient environment. That is not really a "reservoir". Not in the sense that term is generally used. Naturally, this artificial sink would have to be protected. Insulated. Like an ice box. The only heat entering the ice box would be the waste heat from the engine.
That is exactly what I assumed when I analyzed it above. As I showed above, even for an ideal heat engine and heat pump you would have to use all of the work from the heat engine to run the heat pump. For a real heat engine you will get even less energy and for a real heat pump you will need even more.
 P: 853 Well not being a big mathematician but still understanding physics I think that the problem with excess energy extraction from a heat pump /heat engine cycle is that nowhere in nature atleast not on this earth you have these huge temperature differences from which you could extract alot of energy , the temperature differences are only minimal and that's why I believe it is really not considered a viable source of energy. Ofcourse we have huge temperatures in the mantle down below and there were proposed plans for a steam generation/electricty power plant based on that but I guess the engineering challenge and cost speaks for itself. Not to mention that converting heat to other forms of energy isn't that efficient at all electromagnetism into other forms of energy like mechanical and heat is much more efficient way of converting different forms of energy.So I believe a heat engine is not a very efficient way to run a generator, not to mention the problems associated with running the heat engine in the first place. This whole idea sounds to me like a self running refrigerator, but we all know that if you want to move something from point "a" to point "b" you need to input energy to do that.If you just let the hot reservoir mix with the cold one without keeping them temperature separated then all you come out is a equal temperature at both reservoirs and the heat pump being useless.
PF Gold
P: 244
 Quote by Tom Booth I'm disappointed. You give up far to easily.
I "gave up" because the explanations repeatedly given to me finally sounded reasonable.

I am open to any suggestions that might make this work, but for now (and a few hours to come) I must digest all the information that is already given.

You and me can always discuss this further in another forum.
Equations, reading explanations etc. works not that well for me. I need hands on experience in order to get a good understanding of how things work.
For now it is too expensive to buy a heat pump and a decent heat engine to test this out in practice. So I need some more time to digest all given information, as well as understand where we all disagree or misunderstand eachother.

It is hard to be open minded, and be thinking outside the box, while discussing physics with "rational minded" people (No offence to anyone) who mostly do think inside the box - who is refering to most common textbooks.

I'm pretty open minded, but I understand that energy cannot be created from nowhere, nor be destroied. I try to think outside the box without violating common physics...

@DaleSpam: Only dealing with mainstream physics is a huge disadvantage for those more open minded thinkers who want answers to their problem. However, in my opinion, discussions which is all about over unity, perpetual motion, which has nothing to do in this forum, is far beyond being open minded. I think this forum should have enough headroom to allow a discussion take place between pundits and scientists. The main thing to be sure of is that we understand each other's ideas and come to a conclusion that seems rational - and even provable.

Vidar
Mentor
P: 17,545
 Quote by Low-Q To make this work we need a third reservoir. The ground for example. Its temperature will not change as much as the air temperature outside. At very hot or very cold days, the heat engine can be ran by those two reservoires. It powers the heat pump which removes heat from the house at warm days, and supply heat to the house at cold days. Lucky for us, the heat engine will increase its efficiency as the temperature difference is increasing.
So you would use the temperature difference between the air and the ground to drive a heat engine which would power a heat pump to heat or cool a house? I see no reason that couldn't be done.
P: 21
 Quote by DaleSpam That is exactly what I assumed when I analyzed it above. As I showed above, even for an ideal heat engine and heat pump you would have to use all of the work from the heat engine to run the heat pump. For a real heat engine you will get even less energy and for a real heat pump you will need even more.
I still don't have any real confidence that you understand the proposal represented by the diagram. If you really want to discuss it I think it would probably be necessary to go back to the beginning.

A mathematical analysis might prove to be quite interesting and useful but so far you have chosen figures that seem inapplicable to the scenario described, so I'm not sure you really understand the proposal.

It doesn't seem to me that you can really prove that x - y = 7 by showing that q - z = 12.

You may be right but the argument is not convincing when you appear to be solving for a situation that wasn't suggested. It seems like your just throwing numbers around in a broad sweeping dismissal.

For example, after looking at the image you said: "In that image the heat pump and the heat engine share both the same hot reservoir and the same cold reservoir. Everything in my posts 5 and 8 applies"

In #5 you begin by saying: "Assuming that the cold reservoir is a typical 300 K temperature,..."

In 8 you say: "An ideal COP 10 heat pump with a cold reservoir at 300 K could only raise heat to a hot reservoir of 333 K."

It is rather difficult for me to relate this to a scenario that has to do with a heat engine running on ambient heat. The numbers describe a situation that is the opposite of what was being proposed.

You say: "then for 70% efficiency the minimum hot reservoir temperature would be 1000 K."

We need to agree on what these percentages actually represent. Your interpreting the percentages in terms of Carnot efficiency it seems, which was not the intention. Your taking it too literally. The diagram was not intended to be technically accurate just suggestive as originally presented to a lay audience not a scientific community.

Carnot efficiency is on a scale of absolute zero. The percentages on the chart are only meant to represent energy flow. Like if I fill my bathtub to the top it is 100% full. How then can I distribute that 100% for various necessary purposes. The percentages are not intended to plum the depths of the ocean, which by comparison, that is what using Carnot efficiency amounts to.

It seems Low-Q got the answer he was looking for. He started this thread however by quoting me and posting a diagram I drew. Perhaps its redundant and unnecessary at this point, but if anyone thinks it might be worth the bother I wouldn't mind going back and doing some kind of analysis of what was actually proposed in the materials utilized, and maybe put some real numbers on it. Perhaps the conclusion will be the same. i.e. it can't possibly work, but at least the proof would be valid for the actual proposal rather than one turned upside down.
 P: 5,462 In which case you should carry on the process I have been advocating all along. Using real world data. Here is an extract from the manufacturers measured and confirmed capabilities of currently available heat pumps, I could buy off the shelf. Note that the COP values are realistic not pipe dreams. Attached Thumbnails
Thanks
PF Gold
P: 12,269
 Quote by Low-Q Good point. However I am not searching for loopholes in science, just searching for an energysource in the air which a heat pump can harvest in order to power a heat engine. As many links, explanations from you guys has shown, it will not be possible.
This is the crucial misaprension of the thread. There is not an energy source in the air that you can harvest. Afaik, in accepted terminology, you can 'harvest' energy from an energy 'resource' where it is available in some form of chemical, gravitational, thermal, etc. form. Just shuffling the internal energy of a body of air (using an input of energy from some resource) in order to create a temperature difference and then using a heat engine is a pretty pointless exercise. Of course, you can 'power a heat engine' this way but why not use the original energy resource and cut out the (always lossy) middle man? You won't get any more out of the system - as has been ascertained several times.
Is there anything more to the main question of this thread?
P: 21
 Quote by Studiot In which case you should carry on the process I have been advocating all along. Using real world data. Here is an extract from the manufacturers measured and confirmed capabilities of currently available heat pumps, I could buy off the shelf. Note that the COP values are realistic not pipe dreams.
Around 3. What is the application being analyzed ? Heating a home ?

What does COP actually represent ?

What, from that chart is relevant to the proposal ?

I have no problem with real world data if it is applicable or is somehow relateable to what was proposed but I think the subject needs to be broadened somewhat beyond Heat Pumps and COP. Basically a "heat pump" is the same thing as a refrigerator, a freezer, an air conditioner,... There are many methods for cooling.

In a simple chart it is impossible to do more than be suggestive, so Heat Pump - COP, relatively understandable at a glance but only one possibility.

I has originally sent a PM to Low-Q suggesting that he might want to read Tesla's original paper on the subject crudely represented by the diagram but he didn't have time and as far as I know still has not read it.

Tesla's method of cooling was put forward simply as "By some means or another". Heat Pumps did not exist at the time. As I say, the diagram was only intended to be suggestive not technically accurate.

I think it might be best to make sure we have a clear idea what the intent was. Then it night be possible to consider what the options might be suggested by "Some means or another".

I know what the abbreviation COP stands for but really only a vague idea what it actually means. My impression is that it is very application dependent. How well does the appliance satisfy the requirements. The COP of a vapor-compression unit could be practically doubled if it is utilized for both heating and cooling rather than just one or the other.

If the kids turn up the thermostat the COP is reduced. etc. etc. It seems a rather wild variable to me and somewhat difficult to pin down. Neither am I an HVAC technician.
 P: 853 Hey guys, let's be real, were fighting thin air here.I can agree to Low-Q that even with all the academic books and all the knowledge we already have we need to stay open minded , but for someone to be really open minded he needs to know the facts and figures that were made before him because without that basic understanding of physics this new generation is starting to play with magnets and thinking of the so called "free energy" basically a misconception from those who either don't know what their doing or the ones who just enjoy a good laugh when their free energy ideas are being realized by people who thought that the earth is flat until yesterday. @Tom without getting into any numbers why don't we talk about the basic concept itself as I believe that all ideas first start off with a thought or concept which is then later if considered realistic , calculated and built if possible. So by concept , you say that yourself that a heat pump is nothing more than a refrigerator, and that's right well as we know for the fridge to be able to function it needs two things a as good as possible thermal isolation and a energy source to run the pump which then in turn will start to move the liquid that will move the heat from A to B. So if you ask me I just don't see a way how you could make a heat pump which heats your house already,+ requires a external energy source that moves the heat in and out ,, yet make a electricity generator no matter what you add to the device would it be a heat engine or what not. To create the heat difference you had to use energy in the first place and now when you have done it you want to get that energy back by dong something further with that moved heat but it just won't do as all devices are lossy and you can never get back as much as you used in the first place.And this basic rule applies to pretty much everything in and around us. Like I said in my earlier posts you have to have a really huge temperature difference for you to be able to use that and extract considerable amounts of power out of it, just like you have to have a really big water height and displacement difference for a hydropowerstation to get you a considerable amount of energy out of that water, no wonder why they try to build dams on either huge rivers with lots of water flow or between canions which have a huge height and also huge potential difference that will be used to power the turbines.With heat it just isn't the same thing because there aren't such a noticable difference and to create one requires energy. So aren't we just playing with magnets and saying that they don't make "free energy" but then turning them around to hope that something will change?...
P: 5,462
 In a simple chart it is impossible to do more than be suggestive, so Heat Pump - COP, relatively understandable at a glance but only one possibility.
 Neither am I an HVAC technician.
Exactly.

I have tried to show you, by direct calculation, that the COP and efficiency refer to different ratios, and are not just inverses.

In fact there are two different measures and definitions of COP one for refrigerators and one for heat pumps, which are thermodynamically different.

In a refrigerator we extract as much heat as possible for a given amount of work from the evaporator.

In heat pump we want to maximise the heat obtained from the condenser for a given amount of work.

This leads to different values of Q being used from different points in the cycle to define their COP = Q/W.

This is the last time I will attempt to introduce some reality and reason into this thread.

If you are truly concerned to understand more about the thermodynamics of refrigeration and heat pump cycles read chapter 13 of the book that has been the engineering standard for many years

Engineering Thermodynamics, Work and Heat Transfer by Rogers & Mayhew.
Mentor
P: 17,545
 Quote by Tom Booth A mathematical analysis might prove to be quite interesting and useful but so far you have chosen figures that seem inapplicable to the scenario described, so I'm not sure you really understand the proposal.
That is a reasonable objection. Rather than getting a specific set of numbers to exactly match, we can solve for the general system and just leave it in terms of variables that you can plug in.

Using the same notation as in post 14, the efficiency of an ideal heat engine is again:
$$\frac{W_E}{Q_{HE}}=1-\frac{T_{CE}}{T_{HE}}$$

The COP of an ideal heat pump used in heating mode is:
$$\frac{Q_{CP}+W_P}{W_P}=\frac{T_{HP}}{T_{HP}-T_{CP}}$$

Because they share both reservoirs we have $T_{CE}=T_{CP}$ and $T_{HE}=T_{HP}$, and to keep the cold side from heating up we need $Q_{CP}=Q_{HE}-W_E$. Plugging all of these in and simplifying we obtain:
$$W_E=W_P$$

In other words, assuming a perfectly ideal heat engine and a perfectly ideal heat pump the heat pump requires all of the work that the heat engine outputs. It doesn't matter what temperatures you use for your input and output sides, they all cancel out and all of the work from the engine is used by the heat pump.

Of course, for real engines and pumps it will be even worse. The real engine is not as efficient as the ideal engine, so it will generate less work, and a real heat pump has a lower COP so it will require more work.
 P: 5,462 Good evening Dale, One matter absent from your analysis is the small matter of heat transfer efficiency into and out of the reservoirs. Your perfect machine analysis is silent in this respect.
 Mentor P: 17,545 Hi Studiot, I believe that you are talking about the rate of transfer to the reservoir, not efficiency. Correct? The transfer is passive, so it doesn't use or generate work, so I don't see where efficiency would come in.
 P: 5,462 No I'm not talking about rate of transfer. Let us say you want to use the heat pump to heat a room, obtaining the heat from the outside atmosphere ( a pretty typical practical application). 1)In order to extract heat from the air to your working fluid you have to pass the air over some sort of heat exhanger. Probably you require to expend energy in a fan. 2)Having picked up the heat, you have to pump your working fluid around in heat pump and perform the compression and expansion cycles. 3)You now know what I'm going to say. Having pumped your hot fluid into the room you need a second heat exchanger/fan to remove and distribute the heat. I was intending to come round to explaining that the manufacturers only quote the the COP of the heat exchanger using (2) as the work budget. This makes advertising headlines but poor engineering.
Mentor
P: 17,545
 Quote by Studiot I was intending to come round to explaining that the manufacturers only quote the the COP of the heat exchanger using (2) as the work budget. This makes advertising headlines but poor engineering.
Oh, I didn't realize that. I thought that the COP was for the whole system, which of course includes all of those other fans and parts that you mention for the work.

Well, all that does is reduce the real efficiency further from the ideal efficiency. Since even an ideal heat engine and heat pump could not gain energy this way, what you mention is just one more way for a real system to perform even worse. Worse than 0 is pretty bad.

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