Assumed violation of physics - Heat vs. Work

[Tom Booth]

If you can find a mainstream scientific source which defines a heat pump differently then please provide the reference. Otherwise we will stick with the standard definition:

http://en.wikipedia.org/wiki/Heat_pump
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/heatpump.html
http://www.ecourses.ou.edu/cgi-bin/ebook.cgi?doc=&topic=th&chap_sec=05.5&page=theory
http://www.ohio.edu/people/piccard/phys202/carnot/carnot.html

One of your sources states clearly: "A heat pump is a device which applies external work to extract an amount of heat QC from a cold reservoir and delivers heat QH to a hot reservoir"

Fair enough. But I would argue that that is a definition of its usual function. Not what it actually does or can do.

The heat pump can still operate even if used outside its intended function.

Rather than using a heat pump to extract heat from a cold environment and deliver it to a hot environment, as defined above, you could just as well use it to extract heat from a hot environment and deliver it to a cold one. That doesn't meet the definition of heat pump but it would still work. You can also just run it in one environment which is stable in temperature throughout and make it unstable. Create a temperature difference where there was none.

Perhaps if used in an unusual way it would be better to call it something else. A "generic vapor compressor unit" perhaps. or come up with some other applicable term. A rose by any other name is still a rose.

In my opinion the best definition is simply what is implied by its name. Heat Pump.

It pumps heat from point A to point B Period.

...But Low-Q's arrangement clearly has both the heat pump and the heat engine connected to the same two heat reservoirs.

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.

However, in order to be a heat engine and in order to be a heat pump each must be connected to two reservoirs. They need not be the same reservoirs, but each needs two, by definition.

By definition a computer monitor is not a television set. But the definition is FUNCTIONAL.

Otherwise at heart, they are both simply cathode ray tubes.

I do not agree that a heat pump "MUST be connected to two reservoirs". To meet the usual definition Yes. To actually function as a machine No.

It isn't misleading in the least, it is completely factual. Engineers have been analyzing, designing, and building heat pumps based on this fundamental fact for decades. All heat pumps ever constructed share this feature: the greater the temperature difference between the hot and cold reservoirs the less heat is transferred for the same amount of work input by the formula above*. Your complaint to the contrary is unfounded and goes against both fact and theory.

My argument is not about the factuality of the above statement but with the implication that there MUST BE a HOT and a COLD "reservoir" for a bare bones vapor compressor unit to operate and create a temperature difference.

If you think this is really a necessity or the machine can't work then I would suggest that you strip one down and plug it in and see what happens.

A central air unit (heat pump) sits where ?

Generally it sits outside behind the house in the ambient environment.

You can only get it to deliver heat to the house by using fans and running ductwork but without the ductwork, just sitting outside it will still operate and create a temperature differential.

It doesn't deliver the heat to the "reservoir" inside the house as intended so I guess you can't really call it a heat pump any more and it is no longer "efficient" at heating the house without the ductwork but it is creating heat and cold under the hood just as well as ever. Ductwork or no ductwork. Its actual capability to produce hot and cold from the warm ambient environment has not diminished in the least by removing the ductwork.

 

[Tom Booth]

Tom you seem to have written a lot without saying anything at all. What is the point of your posts? That heat isn't a fluid? That's pretty obvious. Was there some other point you wanted to make? Please elaborate if so. It appears as if you just want to argue what a reservoir is when we already have a perfectly good definition and description of one.

Not what a reservoir IS.

You wrote a few minutes ago:

You cannot create a temperature difference with only one reservoir.

You could argue that it is creating a temperature difference between the input/output sides,

and that would be true

A temperature difference is a temperature difference isn't it ?

The input/output sides are both within the one reservoir. Right ?

It seems to me that you certainly can indeed "create a temperature difference with only one reservoir" I don't know if it is the terminology that needs clarification or what but it appears you contradict yourself. Is it possible or isn't it ? Does a temperature difference exist or not ?

If you can't create a temperature difference with only one reservoir but there is a temperature difference nevertheless then where is the other reservoir ?

 

[Studiot]

A temperature difference is a temperature difference isn't it ?

The input/output sides are both within the one reservoir. Right ?

It seems to me that you certainly can indeed "create a temperature difference with only one reservoir" I don't know if it is the terminology that needs clarification or what but it appears you contradict yourself. Is it possible or isn't it ? Does a temperature difference exist or not ?

If you can't create a temperature difference with only one reservoir but there is a temperature difference nevertheless then where is the other reservoir ?

I can see two different scenarios here.

1) You genuinely want to understand the thermodynamics of heat flow.

2) You just want to argue the toss and and trying to revise standard definitions to suit your case.

I am happy to help in (1) but not (2).

 

[sophiecentaur]

If a heat pump is put on a table and switched on, it will just develop one hot plat and one cold plate. It still pumps heat - just aimlessly, not achieving anything.

 

[Studiot]

Good morning SophieCentaur

If a heat pump is put on a table and switched on, it will just develop one hot plat and one cold plate.

That's true because the heat pump has created a hot reservoir and a cold reservoir.
It maintains these according to the Second Law by the input of work.

This part of thermodynamics is actually very easy to understand for those who want to ( as you clearly do).

 

[Low-Q]

After reading the last posts I have a strong feeling that the inner workings of a heat pump is often misunderstood.
At least I see myself looking at a heat pump as something that can make a temperature difference between expansion- and condensing parts that is greater than the input work. Naturally I think so when a heat pump (If used as a heat source for the house), can deliver more heat pr. input Watt than a resistive electric heater does.

So if I understand the replies from DaleSpam, I could likely used a regular resistive heater to power that heat engine, and have the same consume/output as if I was using the condenser or expansion part of a heat pump, or even both, to run the heat engine.

The heat engine in a closed loop will take heat from the hot condenser and deliver it to the cold expander. This will prevent the condenser to be too hot to be able to condense the working fluid, and prevent the expander to be too cold to expand the working fluid.
Initially, the pump must do work in order to maintain the pressure difference between high pressure hot condenser and low pressure cold expander. That input work increase as the temperatur difference gets higher - COP drops. If a heat engine is trying to equalize those two temperatures, the pump will have less temperature difference which will reduce input work - COP rise, but also the heat engine will have less temperature difference to work on and then decrease its efficiency. This is how I understand a closed insolated loop.

However, the confusion comes when I look at the power requirements for a resistive heater to heat up a given insolated space (house) versus the power requirements for a heat pump to deliver the same heat into the same space. The heat pump heats up the house on expence of the outside temperature, while the resistive heater does not.
Since the air outdoor has a pretty large volume, it will practically not change its temperature. That reservoire is too large for the heat pump to cool it considerably down.
The same reservoire is also too large for the heat engine heat it up.

This gives me headache

 

[Tom Booth]

A temperature difference is a temperature difference isn't it ?

The input/output sides are both within the one reservoir. Right ?

It seems to me that you certainly can indeed "create a temperature difference with only one reservoir" I don't know if it is the terminology that needs clarification or what but it appears you contradict yourself. Is it possible or isn't it ? Does a temperature difference exist or not ?

If you can't create a temperature difference with only one reservoir but there is a temperature difference nevertheless then where is the other reservoir ?

I can see two different scenarios here.

1) You genuinely want to understand the thermodynamics of heat flow.

2) You just want to argue the toss and and trying to revise standard definitions to suit your case.

I am happy to help in (1) but not (2).

As far as what you quoted of my post, I'm simply trying to get Drakkith to clarify his position when he wrote:

The heat pump, after you throw it on the table, is now in ONE reservoir.

You cannot create a temperature difference with only one reservoir.

You could argue that it is creating a temperature difference between the input/output sides,

and that would be true

To me the statement is contradicting itself. If he agrees that the heat pump is "now in one reservoir" just sitting on a table creating a temperature difference "between the input/output sides" than how can he say in the same breath "You cannot create a temperature difference with only one reservoir." ?

Something doesn't click. I would simply like to reach a mutual understanding or at lest some clarity about whatever it is he's trying to say. Is that too much to ask ?

 

[Drakkith]

A temperature difference is a temperature difference isn't it ?

The input/output sides are both within the one reservoir. Right ?

It seems to me that you certainly can indeed "create a temperature difference with only one reservoir" I don't know if it is the terminology that needs clarification or what but it appears you contradict yourself. Is it possible or isn't it ? Does a temperature difference exist or not ?

If you can't create a temperature difference with only one reservoir but there is a temperature difference nevertheless then where is the other reservoir ?

Oh. You are going to be difficult in this manner. Ok then. Let me be clear. You cannot create a temperature difference between two reservoirs if you don't have 2 reservoirs.

 

[sophiecentaur]

Where is this thread supposed to be going? The definitions are actually quite clear and there is plenty if evidence that there's no paradox or law-breaking. Some if the contributors just need to re-visit the thermodynamics in a rigorous way. They can then stop worrying and imagining they have found a loophole in Science.

 

[Crazymechanic]

Well you can create a temperature difference in the same reservoir or room in two objects but to do that you have to supply constant power to them for them to either stay hot or cold.Basically that is useless work as the temperature will try to reach equilibrium all the time.
That's like building a sand castle in heavy rain and storm you can do that but the rain will constantly wash away your castle and you will have to constantly build it up .

But I think the situation when two objects can have different temperatures in the same room is just because of air that air cannot make the temperatures equal out as fast because of it's properties while if you would have the same reservoir or room only this time not filled with air but water it would be a different situation the heat difference between something hot and cold would be almoust none.

@Low-Q yes your right that a electrical heater will consume more power to create the same amount of heat than a heat pump as the heat pump is not creating the heat just transporting it from one place to another, just like a cargo train isn't creating it's cargo just transporting it.
And if you have a heat pump heated house there is a temperature difference but then you break your window at come home after one day and there is no more temperature difference as the air outside and inside has reached equilibrium , the efficiency has decreased dramatically as it now is like a hydro dam with no dam at all , all the water is in the same level in our case temperature.

But basically I think the discussion between do we need two reservoirs or not is useless as in all cases there are two or more reservoirs of different temperatures like the one in your house and the one outside and then the one outside and the one in your basement or your fridge.

 

[Tom Booth]

If a heat pump is put on a table and switched on, it will just develop one hot plate and one cold plate. It still pumps heat - just aimlessly, not achieving anything.

Thank you.

But it is achieving something if you have an application for "one hot plate and one cold plate".

 

[Crazymechanic]

Well it has an application , but a pretty useless and inefficient one.
That's why they came up with the idea of putting a door to the fridge. :D Also I believe the same reason is used for windows just instead having empty holes for light to come in.

But actually I believe there is a situation where Tom's case applies , when they use the water cooled or other chemical cooled CPU coolers.
Because basically the inner space of the pc case with all the chipset and main cpu that is being cooled is not isolated from the outside where the radiator or heat exchanger is located so basically it's a little bit similar to pumping out water from a hole and displacing it right next to that hole and then very soon afterwards it goes through the soil and gets back into the hole.
Pretty inefficient as it's like having a fan in a vacuum , you can still spin it but without much use.
That's why I believe the huge server rooms are isolated from the outside and use heat pumps or we could say air conditioners to move the excess heat generated by the server equipment to the outside of the building as just having a small water cooled cpu head that displaces the heat right next to the server board wouldn't go through as the temperature of the server room would rise constantly until reach equilibrium or almoust close with the temperature of the cpu itself and then the local heat pump would have no use anymore. Just like in the scenario with the heat pump in one reservoir on table that you mentioned.

 

[Tom Booth]

That's true because the heat pump has created a hot reservoir and a cold reservoir.

So are we going with the idea that the condenser coil and the expansion coil of a stripped down vapor-compression unit running on the kitchen table constitute hot and cold reservoirs ?

The surrounding reservoir of ambient heat in the kitchen itself we can just ignore.

 

[Low-Q]

Where is this thread supposed to be going? The definitions are actually quite clear and there is plenty if evidence that there's no paradox or law-breaking. Some if the contributors just need to re-visit the thermodynamics in a rigorous way. They can then stop worrying and imagining they have found a loophole in Science.

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.

@ Crazymechanic:
To use the cargo train as an example, that train is supplying the heat engine. I understand now that the heat engine is also a train which moves the same cargo back to its origin. The net movement of the cargo is zero - so no work has been done. The pump/engine system will therfor not work with only two reservoires.
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.

Vidar

 

[Crazymechanic]

@Tom , no I don't think we can call those reservoirs as the classical understanding of a reservoir is something that is a closed room or space and it has to have some isolation from the surrounding.
When a heat pump is left in an open space with no isolation between the two sides it's not a heat pump any more by definition , just like if you have a jail and then you have a city the isolation of these two are the walls and guards of the jail if you take those away then there are no two spaces anymore all is one, the people just merge with all the consequences from that.@Low-Q , well I believe that what you wanted to make sounds similar to a hydro dam the water goes through the turbine gives it's potential to the turbine and then when it passes out of the dam at the lower side you somehow take it up to the high side of the dam again so that it could do the job one more time and repeat this in a loop but the problem is if you have to bring the water up you have to use energy and in the end the energy extracted from the falling water compared to the energy used to pump the fallen water back up would pretty much equal out. That's why a heat pump or a hydro dam can only create excess energy if the temperature/water difference is supplied by an external force like sun and water evaporation or sun and temperature difference between the two sides of the heat pump and then the pumps has only to move that heat but only in one direction.

 

[Low-Q]

@Low-Q , well I believe that what you wanted to make sounds similar to a hydro dam the water goes through the turbine gives it's potential to the turbine and then when it passes out of the dam at the lower side you somehow take it up to the high side of the dam again so that it could do the job one more time and repeat this in a loop but the problem is if you have to bring the water up you have to use energy and in the end the energy extracted from the falling water compared to the energy used to pump the fallen water back up would pretty much equal out. That's why a heat pump or a hydro dam can only create excess energy if the temperature/water difference is supplied by an external force like sun and water evaporation or sun and temperature difference between the two sides of the heat pump and then the pumps has only to move that heat but only in one direction.

Sounds reasonable. You pretty much nailed it

 

[Tom Booth]

Oh. You are going to be difficult in this manner. Ok then. Let me be clear. You cannot create a temperature difference between two reservoirs if you don't have 2 reservoirs.

OK, that seems clearer.

Comparing that to this:

The heat pump, after you throw it on the table, is now in ONE reservoir.

By "ONE reservoir" I assume you mean the ambient heat in the room the table is in. Yes?

You cannot create a temperature difference with only one reservoir.

By that then you actually mean as modified: "You cannot create a temperature difference between two reservoirs if you don't have 2 reservoirs." Yes ?

The source of my confusion was that I was interpreting your statement to mean: "You cannot use a heat pump to create a temperature difference from just the one reservoir it is in"

After all, it followed directly after your statement "the heat pump... is now in one reservoir"

I hope you can understand the reason for the misinterpretation.

You could argue that it is creating a temperature difference between the input/output sides,

and that would be true

I assume the last statement can stand. I'm content with considering the room the engine is in "one reservoir" and refer to the input and output sides as just that. I assume by "input and output sides" you mean the condenser coils and expansion coils of the heat pump. You do not however consider these to be hot and cold reservoirs. Am I correct ?

Studiot, however, just posted in regard to the same scinario:

If a heat pump is put on a table and switched on, it will just develop one hot plate and one cold plate.

That's true because the heat pump has created a hot reservoir and a cold reservoir.

 

[Tom Booth]

Where is this thread supposed to be going?

I believe Low-Q would like to use the temperature difference created by a heat pump to run a Stirling Engine, or some form of heat engine directly from the temperature difference and have the engine in turn run the heat pump with excess energy left over to spare to be used for some useful purpose like generating electricity.

That was my impression at the start.

 

[Studiot]

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.

Of course it is possible.

But it is just that you can't have a closed system driving itself.
But you could power a heat engine of sufficiently lower power.

The originators of thermodynamics called this a 'self acting machine'

A popular goal was once the idea of extracting heat from the very large heat content of the ocean and converting it to mechanical work to drive the ship. Of course this was proved to be less efficient.

"It is impossible for a self acting machine, unaided by any external agancy, to convey heat from a cold body to a hot one"

There is a clear misconception as to the functioning of refrigeration plant in this thread, which in my opinion has lead to the difficulties with the Tom/Low-Q diagram.

No practical heat pump works by reversing the ideal machine cycle.
There is indeed an 'intermediate' temperature. I will come to that.
I think this is where the misunderstanding has arisen.

Here is a calculation for a perfect refrigerator formed from a perfect heat engine driving a perfect heat pump.

A perfect heat engine takes in heat, Q, at T and rejects heat at T₁

Hence
$$W = \frac{Q}{T}\left( {T - {T_1}} \right)$$

The perfect heat pump extracts Q₂ from the cold body at low temperature T₂. It also delivers heat to temperature T₁.

Since it is perfect and therefore reversible

$$W = \frac{{{Q_2}}}{{{T_2}}}\left( {{T_1} - {T_2}} \right)$$

Where W is the heat equivalent of work given by the heat engine and also the the work used in driving the heat pump.

Hence

$$\begin{array}{l} \frac{Q}{T}\left( {T - {T_2}} \right) = \frac{{{Q_2}}}{{{T_2}}}\left( {{T_1} - {T_2}} \right) \\ \frac{{{Q_2}}}{Q} = \frac{{{T_2}}}{T}\frac{{\left( {T - {T_2}} \right)}}{{\left( {{T_1} - {T_2}} \right)}} \\ \end{array}$$

Where $\frac{{{Q_2}}}{Q}$ is the ratio of the heat extracted from the colder body to the heat supplied to the machine reservoir.


Note that the heats (Q₂) used in the calculation of the heat pump and the heat engine use heats at different reservoirs, which differ by the work performed.
Using the wrong Q is a common error by beginners.

 

[Tom Booth]

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.

 

[Dale]

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.

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?

 

[Tom Booth]

@Tom , no I don't think we can call those reservoirs...

I don't either.

 

[Dale]

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.

 

[Crazymechanic]

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 a lot 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.

 

[Low-Q]

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 each other.

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 referring 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

 

[Dale]

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.

 

[Tom Booth]

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.

 

[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.

 

[sophiecentaur]

Your response to my post:

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?

 

[Tom Booth]

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.

 

[Crazymechanic]

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?...

 

[Studiot]

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.

 

[Dale]

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.

 

[Studiot]

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.

 

[Dale]

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.