Isolated cold hole

[Tom Booth]

I think I would need to update that to reflect recent experimental evidence.

My experiments show much less than 30% "waste heat" from a Stirling engine. More like zero.

[VincentG]

Lol hold on Tom, we’re still trying to have Fool explain the disappearance of 30% or less of total heat as work let alone 100%.

I will submit a block diagram soon.

[Fool]

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No they don't. Otherwise you would measure the data necessary to confirm it.

VincentG, thanks.

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[Tom Booth]

I think I would need to update that to reflect recent experimental evidence.

My experiments show much less than 30% "waste heat" from a Stirling engine. More like zero.

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No they don't. Otherwise you would measure the data necessary to confirm it.

I've seen the readings, felt the cold plate, the measurements (Infrared readings etc.) are video recorded and publicly posted on the internet.

No amount of measurement or data is going to convince a prejudiced mind.
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[VincentG]

I hope this is clear enough for a quick sketch. I scaled the engine to look like all of the heat must pass through the engine to leave the room/freezer or however you want to look at it. It could just as well be that only a partial amount of heat must pass through the engine, and the evaporator be much larger in that case, or engine much smaller. As far as I see it, the principle is the same.


The red arrow to the right is indicating "ambient heat(from room) minus work" going out to the condenser.

isolated cold hole diagram.jpg (462.51 KiB) Viewed 1088 times

[Fool]

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VincentG, the following is your diagram with a system boundary drawn in so the apparent Carnot limit for it, can be more easily seen. It doesn't prove your idea wrong, it just says it will need to be able to break the second law to work. The why it won't break the second law is more difficult to show, but goes along the lines of an earlier post to this thread, and the 'Bottleneck' YouTube video you found for us.

isolated cold hole diagram (1) (2).jpg (571.96 KiB) Viewed 971 times

Building inside temperature Tc 78 F Qc. Outside temperature Th 100 F Qh. W work/electricity going in to power the AC.

COP=Qc/W = (Qh/W)-1
Carnot COP.

Note: All the friction-heat of the LTD Stirling Engine, as you've drawn it, is rejected into the inside of the building. That will also degrade efficiency, but isn't mentioned in the Carnot limited issues.

I don't know if I'll have time to analyze it further. It would need to be a piece by piece temperature and area assignment. At the very best it would get closer to but not pass Carnot efficiency. Most likely the added complications would be less efficient than a simple cooler.

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[VincentG]

VincentG, the following is your diagram with a system boundary drawn in so the apparent Carnot limit for it, can be more easily seen. It doesn't prove your idea wrong, it just says it will need to be able to break the second law to work. The why it won't break the second law is more difficult to show, but goes along the lines of an earlier post to this thread, and the 'Bottleneck' YouTube video you found for us.

Building inside temperature Tc 78 F Qc. Outside temperature Th 100 F Qh. W work/electricity going in to power the AC.

COP=Qc/W = (Qh/W)-1
Carnot COP.

Note: All the friction-heat of the LTD Stirling Engine, as you've drawn it, is rejected into the inside of the building. That will also degrade efficiency, but isn't mentioned in the Carnot limited issues.

I don't know if I'll have time to analyze it further. It would need to be a piece by piece temperature and area assignment. At the very best it would get closer to but not pass Carnot efficiency. Most likely the added complications would be less efficient than a simple cooler.

To accurately represent a real-world scenario, I think Th should be 90F(like an indoor surface in direct sunlight) and Tc should be something like 40F(typical evaporator temperature).

Sure we can say friction is rejected inside the building, I don't think it will be significant.

Rather than breaking the second law, I think the second law is the reason why this would work. This is more like two closed systems working in series rather than one closed system. The evaporator should see a reduced load from less heat due to work output, and the work output could go back into running the heat pump but would likely be better off satisfying other power demands.

This is of course all banking on heat actually being converted to work, instead of just a temperature difference being converted to work.

[Fool]

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You may use any temperature you like, but I for one vote against an 80 F building. Th would need to be closer to 40 F, to cool the building properly. Same T as the cooler. Otherwise the building will be way hotter. Or the engines hot plate will need to be 100 times larger. As I said in an earlier post.

You can't take a building that needs a 40 F cooler and put an engine in between it and cool the building with an 80 F plate. It won't cool. The 80 F plate needs to be 40 F for the same cooling area.

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[VincentG]

You can't take a building that needs a 40 F cooler and put an engine in between it and cool the building with an 80 F plate. It won't cool. The 80 F plate needs to be 40 F for the same cooling area.

I think this is a misunderstanding of how these engines work. The gas is shuffled between Th and Tc. The work is extracted from Th at the hot plate(temperature of the building), then the gas is cooled down to Tc(at BDC, max volume), then more heat is pumped out as the heat of compression is sinked to the cold plate(until TDC min volume).

The displacer chamber is just intercepting the flow of energy from the building to the evaporator. True, if the cycle rate is too low cooling will be slowed down, but as we have learned from engines, power is not directly linked to efficiency.

[Fool]

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That is not how heat transfer works. There needs to be a temperature difference for heat transfer.

The ideal Carnot assumes Th and isothermal expansion because of an assumed quazi static process. Colloquially,
cycle rates that take a week.

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[VincentG]

That is not how heat transfer works. There needs to be a temperature difference for heat transfer.

The ideal Carnot assumes Th and isothermal expansion because of an assumed quazi static process. Colloquially,
cycle rates that take a week.

The hot plate can be sized to maintain Th, just as the cold plate can be made to maintain Tc. The temperature difference is between the two.

A perfect isothermal cycle may require a quazi-static process but a quazi-isothermal cycle can be quite rapid.

[Fool]

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Okay. Quazi- isothermal. That to me means a temperature difference, but a static temperature difference. Room 80 degrees, plate 70 degrees, internal gas 60 degrees.

So how much bigger does a heat exchanger need to be to absorb the same Wattage of heat from, 80 degrees room to 40 degrees cooler, to 80 to engine 70 degrees then 60 degrees inside?

4 times?

Now how much bigger if a one degree difference? Or zero degrees different?

Area size of heat exchanger is conditional on Wattage and temperature difference. Goes to infinity at zero temperature difference.

[VincentG]

Okay. Quazi- isothermal. That to me means a temperature difference, but a static temperature difference. Room 80 degrees, plate 70 degrees, internal gas 60 degrees.

Not sure the reason to get hung up on an isothermal process. The gas, when heated at TDC can come close to Th at near Cv. The difference between and true isothermal and more realistic semi-adiabatic/isothermal expansion from there is not that great.

Area size of heat exchanger is conditional on Wattage and temperature difference.

I agree, but neither of these concerns are reason to not discuss the possibility of the cycle as we could substitute much more ideal temperatures to test the theory.

In any event the only purpose of this thread was to further examine the conversion of heat to work. From your quote in this thread viewtopic.php?t=5580&start=135 it seems you don't think that work is actually a result of heat but rather a consequence of expansion.

Pressure, volume, and temperature, are all interrelated, and the equations have nothing to do with load on the axel.

[Tom Booth]

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That is not how heat transfer works. There needs to be a temperature difference for heat transfer.
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Unless it's the ambient side of one of my experimental engines you mean, don't you?

Then the waste heat somehow just magically transfers to the "cold reservoir", without any ∆T because, well, "fool" says so.

Anyway, even if the room is 75°F and the evaporator, whatever cold, we are saying 40°? The working fluid cools to near Tc then it's back over sucking heat out of the hot plate.

"Fool" will just lie and make up whatever BS he can dream up. Just a symptom of the chronic negativity of his "it's impossible" mentality. Aside from having no idea what he's talking about when it comes to either heat pumps or heat engines.

[matt brown]

In any event the only purpose of this thread was to further examine the conversion of heat to work. From your quote in this thread viewtopic.php?t=5580&start=135 it seems you don't think that work is actually a result of heat but rather a consequence of expansion.

Here's a graph I made up many years ago that should be in every thermo textbook:

TV graph.png (293.23 KiB) Viewed 861 times

The blue sweeps are isobars and the red sweeps are diatomic adiabats.

Does everyone know what those squares represent ???