Isolated cold hole

[VincentG]

Thanks Toms that’s exactly what I was thinking as well. It’s pretty significant in most systems, just look at the low pressure side of you cars ac under the hood.

[Fool]

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Here are two links that provide block diagrams for two different ways of providing a reversible heat pump:

https://hvacrschool.com/how-heat-pump-r ... lve-works/

https://www.nuclear-power.com/nuclear-e ... d-cooling/

Maybe you two can use the block diagrams, in those sites, to show the modifications you are proposing, so I don't have to guess how you two are scheming to do it? After all, it is you two whom are claiming this magical 'free' energy or extra efficiency is possible.

[VincentG]

Maybe you two can use the block diagrams, in those sites, to show the modifications you are proposing, so I don't have to guess how you two are scheming to do it? After all, it is you two whom are claiming this magical 'free' energy or extra efficiency is possible.

Hold on a second here, I think you've understood what I was proposing, Tom certainly has. I'm not sure what magical free energy you mean? It has been stated that the measure of efficiency is the percentage of heat that is converted to work vs the percentage of heat that must be wasted to the sink.

The evaporator is tasked with removing heat energy from the room. What is to stop that energy from doing work along the way? And more importantly, why wouldn't the production of work, from the heat energy in the room, reduce the amount of heat energy sinking to the evaporator?

The room is the hot plate and the cold plate is the low side of the air conditioner. The hot plate could easily be over 100 degrees F if it had direct sunlight from a window and the cold plate could easily be 30 degrees F. Absolute temperature and efficiency of the engine is irrelevant anyway, this is theoretical example to demonstrate the most fundamental concepts being discussed here in the most simplistic and relatable way possible.

[Fool]

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It has been stated that the measure of efficiency is the percentage of heat that is converted to work vs the percentage of heat that must be wasted to the sink.

I think you are missing some of the consequences of the second law. A consequence of the first law is that: 'no more energy will come out out of the vacuum of space than is put into its closed system boundary'. A consequence of the second law is that: 'the Carnot Limit can't be broken unless the first law can also be broken.'. So, to get extra efficiency from a cyclic closed system, you are proposing perpetual motion. So, you are proposing magic energy or efficiency.

The second law prevents you from putting work into a system to cool it so that you can get work out by cooling it. It won't be any more efficient. It will lessen the cooling effect at the additional expense of increased mechanical friction. You can't pull yourself up by your own bootstraps.

The evaporator is tasked with removing heat energy from the room. What is to stop that energy from doing work along the way?

Let me put it this way. It is possible to do what you are proposing, but the system will become less efficient, and less effective. It will cost more energy in, to do the same cooling, and it will require a larger cooler to do the cooling. Pulling up on your bootstraps, while jumping gets you up, but no higher than jumping.

Getting work out doesn't cool any more than the amount of (work in minus the work out) times COP.

And more importantly, why wouldn't the production of work, from the heat energy in the room, reduce the amount of heat energy sinking to the evaporator?

The room is adding around 6000 Watts per square meter of heat energy to the cooler. A low temperature Stirling Engine probability can't absorb anymore heat into its engine than 100 Watt/m^2. Probably way less. That is a reduction of capability by about a factor of 60. If you cover the whole exchanger area, the maximum heat into your engine, will be, one sixtieth of what it was before. Now Carnot kicks in and makes that conversation to work less than 20% so the effective return on that blockage will be 60/0.20 or 300 time less power out than put in to cool the house in the first place. (With the entire exchanger blocked.)


I tried to explain this before by something like the following:

Heat source 100 F, sink 30 F. Your numbers. Forget efficiency, your criteria. The amount of heat pulled from room will be affected by that temperature difference, the heat exchanger, metal tube area, metal tube thickness, and type.

You are proposing to put a Stirling Engine in between those two temperatures. In effect the engine will be blocking that flow of heat.

The engines heat exchanger will have the same limitations on it. The engine must have a high temperature difference for more power output, more cylinder piston push, pressure. So the gas inside must be as close to Th as practical. But wait, to get the heat to transfer through the engines heat exchanger, there must be a temperature difference, the larger the better.

A conundrum.

Let us say for example, 20 F. The temperature outside the engine is 100, the temperature max inside is 80 F. Savvy? So now the heat must travel through a heat exchanger at a much reduced thermal force, temperature difference.

That will require a greatly larger area for the engine's hot exchanger, by 60 times, probably more. The same is true for the cold side, or the heat could be discarded to the 78 F room.

Effectiveness of the cooler reduced by a factor of 60, or more.

Big deal if we can get a more efficient overall system. Right?

Okay, add back efficiency requirements. Since the cooler is providing heat removal to a exchanger that is 60 times larger, (or 10 to 20% smaller than 60 times larger.), it is working harder. How much harder.

Since the engine is producing work that can come out of the system, let's start there. The work into the cooler must equal the work out plus the heat moved divided by the COP.

The COP equals the heat out of the room divided by the work in. Neither has changed, so the COP hasn't changed. The only thing that has changed is the work out, so now, overall work in must change.

The problem you two are having is your lack of a block diagram of the system boundary, and your concentration on one little piece of the system. No block diagram, makes it very difficult to visualize the effects of all parts of the system. That is why I asked for one, and provided a starting point with the two links.

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[VincentG]

I think you are missing some of the consequences of the second law. A consequence of the first law is that: 'no more energy will come out out of the vacuum of space than is put into its closed system boundary'. A consequence of the second law is that: 'the Carnot Limit can't be broken unless the first law can also be broken.'. So, to get extra efficiency from a cyclic closed system, you are proposing perpetual motion. So, you are proposing magic energy or efficiency.

The second law prevents you from putting work into a system to cool it so that you can get work out by cooling it. It won't be any more efficient. It will lessen the cooling effect at the additional expense of increased mechanical friction. You can't pull yourself up by your own bootstraps.

You are misunderstanding what I am proposing. I am not suggesting that this would be a good idea to increase overall system efficiency. You are getting bogged down talking about real world material issues and absolute temperature instead of heat quantity. If we can talk about a Carnot engine idealistically, then we should be able to keep this idealistic as well. The overall system doesn't matter, the heat to work is what matters.

If it helps you visualize this, lets put a 600k piece of steel inside a freezer that is 200k. Ok, now the temperatures are good enough to get some real work out. If the 600k lump of steel is made to do work inside the freezer, will the freezer have less cooling load than if the steel is left to cool without doing work?

There should be a straightforward answer to this, and perhaps the answer is that we don't know. To say that the cooling load is the same would be implying free energy, and so far it seems that would be your answer.

[matt brown]

If it helps you visualize this, lets put a 600k piece of steel inside a freezer that is 200k. Ok, now the temperatures are good enough to get some real work out. If the 600k lump of steel is made to do work inside the freezer, will the freezer have less cooling load than if the steel is left to cool without doing work?

I saw your earlier comment the same way where T spread within a system has a work potential that might be utilized via a subsystem (engine) that benefits total system (shrinks waste heat). Meanwhile, Fool focused on Q limits for subsystem that limits this work potential (due to integrated system).

[Tom Booth]

I think you are missing some of the consequences of the second law. A consequence of the first law is that: 'no more energy will come out out of the vacuum of space than is put into its closed system boundary'. A consequence of the second law is that: 'the Carnot Limit can't be broken unless the first law can also be broken.'. So, to get extra efficiency from a cyclic closed system, you are proposing perpetual motion. So, you are proposing magic energy or efficiency.

The second law prevents you from putting work into a system to cool it so that you can get work out by cooling it. It won't be any more efficient. It will lessen the cooling effect at the additional expense of increased mechanical friction. You can't pull yourself up by your own bootstraps.

You are misunderstanding what I am proposing. I am not suggesting that this would be a good idea to increase overall system efficiency. You are getting bogged down talking about real world material issues and absolute temperature instead of heat quantity. If we can talk about a Carnot engine idealistically, then we should be able to keep this idealistic as well. The overall system doesn't matter, the heat to work is what matters.

If it helps you visualize this, lets put a 600k piece of steel inside a freezer that is 200k. Ok, now the temperatures are good enough to get some real work out. If the 600k lump of steel is made to do work inside the freezer, will the freezer have less cooling load than if the steel is left to cool without doing work?

There should be a straightforward answer to this, and perhaps the answer is that we don't know. To say that the cooling load is the same would be implying free energy, and so far it seems that would be your answer.

An interesting experiment might be to get an actual small Stirling generator, something like:

stirlingkit-enjomor-gamma-hot-air-stirling-engine-movement-is-everything_8_3a473d33-a1e1-4009-87f7-a38b19a26bfc_1600x.jpg (112.63 KiB) Viewed 1190 times

With a long wire to a light bulb. Put the hot end of the engine into a vacuum insulated heat proof dewar with your hot lump of steel and put the box and engine into the freezer or just another insulated box, with a thermometer the wire from the generator running outside both box and freezer to the outside environment.

The heat/energy from the lump of hot steel will theoretically be converted by the engine to electricity and transfered through the wire and emitted as heat and light outside the boxes.

Maybe put the light bulb inside another insulated box and check the temperature rise.

Probably the cold box would warm some from the generator and wires, but generally the engine should run until the lump of steel cools down.

You should be able to determine something. If at least some heat was converted so the cold box does not heat up but the box outside with the lightbulb does.

I mean, I think with a little ingenuity something like that could be more than just a thought experiment, but could be an actual experiment.

[Fool]

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The above has gotten totally off topic. Here is the original question. It is the one I keep trying to answer:

The "waste heat" from the room is sinking to the evaporator, as it would anyway, but if ambient heat is converted to work, the load on the air conditioner is reduced.

Short answer, yes the engine will run. But. Whatever power out of it will need the cooler to have the same amount of power, or more, added to what it is already using before adding the engine. Power out will require, power in. No free lunch.

Yes an engine will run and output work from any temperature difference. However, the real question that needs to be asked is where that power is coming from. The answer in the proposed scenario is from the cooler. It will be from the power input to create the cold hole.

You are warming the cold hole up to 80 F, by blocking it with an engine. The cooler will need to make up for that blockage, by requiring more input power and a bigger heat exchanger.

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[VincentG]

Short answer, yes the engine will run. But. Whatever power out of it will need the cooler to have the same amount of power, or more, added to what it is already using before adding the engine. Power out will require, power in. No free lunch.

Yes an engine will run and output work from any temperature difference. However, the real question that needs to be asked is where that power is coming from. The answer in the proposed scenario is from the cooler. It will be from the power input to create the cold hole.

You are warming the cold hole up to 80 F, by blocking it with an engine. The cooler will need to make up for that blockage, by requiring more input power and a bigger heat exchanger.

I don't think anyone ever questioned whether or not the engine would run. For now I won't comment on what seems to be your complete misunderstanding of how an air conditioner operates, as this was a poor example on my part. I should have realized that the conversation might get steered towards the air conditioner instead of the engine. The hot lump of steel inside the freezer may be a better example.

The fundamental question that you seem to be dancing around is whether or not the heat energy that is directly used to perform mechanical work is actually sinked to the cold plate or dissipated to ambient in any way.

If 1000 joules go into raising the temperature of the steel from 200k to 600k, and then 300 joules of mechanical work is performed inside the freezer, how many joules does the freezer need to remove to get back to its starting temperature?

[Tom Booth]

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You are warming the cold hole up to 80 F, by blocking it with an engine. The cooler will need to make up for that blockage, by requiring more input power and a bigger heat exchanger.

As VincentG said, you obviously have a: "complete misunderstanding of how an air conditioner operates" or don't understand, or are pretending not to understand VincentG's proposal.

Without the help from the engine the heat pump "cooler" would have to deal with the full 100°F heat in the room.

At worst, the engine cools the room air down to 80°F by converting some of that heat to work output.

The engine is not "warming the cold hole up to 80°F" it's reducing the heat load down from 100°F to 80°F

Whatever the engine "converts" the heat pump is relieved from removing from the room.

[Fool]

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The hot lump of steel inside the freezer may be a better example.

This is a completely different scenario than the opening question. It show why I asked for a block diagram.

Yes outputting work will reduce the heat going into the ice box, making the coolers work less. But it is a one time deal the hot object has a set ∆Q. There is no time constraint on getting that heat out. Even though putting an engine in-between the hot and cold would slow the process down, it would eventually get done, ideally.

But the original question involved an assumed set influx of heat that was being removed by a set size of cooler. Putting an engine in would block and slow down that process, effectively destroying the effort.

Even in the real world, if the goal is to cool off the object in a freezer, putting an engine in-between will certainly slow the effort.

Still, overall the COP of the cooler will not be greater than the Carnot equation.

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[Tom Booth]

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The hot lump of steel inside the freezer may be a better example.

This is a completely different scenario than the opening question. It show why I asked for a block diagram.

Yes outputting work will reduce the heat going into the ice box, making the coolers work less. But it is a one time deal the hot object has a set ∆Q. There is no time constraint on getting that heat out. Even though putting an engine in-between the hot and cold would slow the process down, it would eventually get done, ideally.

But the original question involved an assumed set influx of heat that was being removed by a set size of cooler. Putting an engine in would block and slow down that process, effectively destroying the effort.

Even in the real world, if the goal is to cool off the object in a freezer, putting an engine in-between will certainly slow the effort.

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"But the original question involved an assumed set influx of heat that was being removed by a set size of cooler. Putting an engine in would block and slow down that process, effectively destroying the effort"

I think maybe you need to reread the opening post:

viewtopic.php?p=24231#p24231

Possibly this could have been worded more clearly:

"The "waste heat" from the room is sinking to the evaporator, as it would anyway, but if ambient heat is converted to work, the load on the air conditioner is reduced"

It might have been better to say the waste heat from the engine...

But seems clear enough to me.

The engine takes in heat from the room and some of that is converted to "work" output and the rest (less heat) ends up as "waste heat" going to the air conditioner.

The air conditioner has less heat to remove from the room as a result of the engine converting some of that heat into work before the left over "waste heat" goes to the evaporator.

The engine is not "blocking" anything, it's delivering a reduced amount of heat to the evaporator.

As a result, the air conditioner can cool more efficiently AND this will increase the ∆T.

The two machines assist each other.

[Fool]

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Imagine air conditioning a building.

Opening line. Assumption, real world building and AC.

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[Tom Booth]

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Imagine air conditioning a building.

Opening line. Assumption, real world building and AC.

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So?

[Fool]

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So!

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