Carnot reveal for Tom

[Nobody]

Tom,

I'm going to ignore your bank and water analogies. Easily disproven.
..

Seems like an evasion. If it is so easily disproven why not do so now? I'm listening.

No.

You are the one that keeps bringing up the analogies. You must prove their worth. Q.E.D. (Quite Easily Done)

[Nobody]

../.. 20% ../..

Tom, where is the 20% coming from?

Tom still doesn't have any an answer. 20% of what? 1 watt, 1/10 watt, 1/100 watt. Total Jewels. Good science requires power or energy values.

I gave my nephew a cheap LTD the kind Amazon sells with the gold flywheel. I set it up got it running on a mug of hot water. Showed him how it runs faster/better with an ice cube on the cold side. Cooling works. He was fascinated.

My brother said "micro Watts". Inwardly smiling I said, "right". Our buddy/friend Tom here uses similar engines in his studies. I would guess that my nephew's and Tom's are in the milliwatt range 0.020 Watts or so. Micro-horsepower 0.027 hp. Or so. Not a lot. I think that has a very significant effect on anything noticable.

[stephenz]

@Tom: That's what I figured, thanks for confirming.

It just seems a lot of hate for that limit. haha
Maybe I'm just wired differently but I first saw that formula in school, my first reaction was "Damn this is awesome, what's the catch!?!?". Quite the opposite reaction.

Of course when it's applied in the way you put with roughly 100C of DT, it seems like a hard limit at around 20%, but to me I just found mind blowing that efficiency figures could get dramatically higher under "just" different temperature conditions, particularly on the cold end (you know because Tc is on the denominator). In other words, my personal "thing" for stirling engine was their (amazing) Carnot efficiency, which ultimately is what got me started thinking on possible practical use (pushing Th and Tc as far as possible from each other) like solar concentrators where efficiencies could theoretically be significantly higher.

We live an exciting time, but I also believe we live a time where fundamental discoveries are getting less and less frequent. Once in a while, technological advances will trigger certain domains of Science to somewhat leap forward and things get quiet again. But I don't think we'll see many more decades or centuries of significant science discovery like it happened centuries ago. Don't get me wrong, I don't think that being a scientific research is boring at all, but I do think a lot of today's research is extremely tied to engineering and that without engineering there wouldn't be as much scientific advances.

If you take the ICE as an example, has it received significantly more love than SE's by universities, grants, etc? I personally don't think so, but I am 200% positive that the reason why ICE's are this advanced today is because of the 100 year old car industry which has never stopped improving it and their R/D efforts probably dwarv'ing that of academics by several order of magnitude. I actually believe that one day SE's will have their time to shine.

[stephenz]

My brother said "micro Watts". Inwardly smiling I said, "right". Our buddy/friend Tom here uses similar engines in his studies. I would guess that my nephew's and Tom's are in the milliwatt range 0.020 Watts or so. Micro-horsepower 0.027 hp. Or so. Not a lot. I think that has a very significant effect on anything noticable.

That's the order of magnitude of the LTD studied in that research paper I linked a couple days ago.

[Nobody]

Thanks.

[Nobody]

Show me that some force of this "thermal gravity" exists and you might have a point.

natural convection?

stephenz,
Funny.

Tom,
Gravity, hight, mass, and water, theory of heat are points you bring up. I'm merely pointing out that gravitational strength/value is a better substitute for temperature than height. However it's you that is using it, not I.

[Tom Booth]

Tom,

I'm going to ignore your bank and water analogies. Easily disproven.
..

Seems like an evasion. If it is so easily disproven why not do so now? I'm listening.

No.

You are the one that keeps bringing up the analogies. You must prove their worth. Q.E.D. (Quite Easily Done)

My analogies are just fine and easily understood. Here's another.

If I lift a weight 800 miles above the ground I can recover 100% or so of the potential energy created in lifting it up by lowering the weight back down, turning a crank or whatever.

The Carnot Limit (or it's academic interpretation taught in thermodynamics courses) says that because the center of the Earth is the center of gravity and the 800 miles that the weight was lifted is just 20% of the 4000 mile distance to the center of the earth, the weight can only be lowered 20% of the distance it was lifted up.

The carnot limit claims that if I heat up a pot of water some percentage up the kelvin scale, say 10%, 20%, 30, whatever, supposedly that ratio magically applies to the 100.000 joules or whatever used to heat the quantity of water so that I can now only get back 10, 20 or 30 thousand joules respectively. (10, 20, or 30 percent of the 100, joules used to heat up the water. (Regardless of quantity of water heated)

Can someone explain the reason why the other 90,000 joules of heat I added have been sequestered? Reserved for donation to the "cold reservoir"?

My reasoning has a perfectly good scientific and mathematical basis.

Conservation of energy. Energy can neither be created nor destroyed only converted from one form to another.

What is the basis for this so-called carnot limit?

It appears to be just a mangled hold over from the caloric theory.

Tom still doesn't have any an answer. 20% of what? 1 watt, 1/10 watt, 1/100 watt. Total Jewels. Good science requires power or energy values.

Exactly, I agree!

20% of what indeed!

Where is the power or actual energy value in the Carnot limit?

The Carnot limit is calculated on the basis of temperature difference alone and nothing else. No reference to joules, power, energy values, quantity of substance heated, nothing , it's completely meaningless and baseless.

Your complaint is about the carnot limit because that is where the 20% comes from.

[Nobody]

If you lift a weight 800 miles and lower it back down the best 100% that you can hope for is zero zero work total in and or out. Your cycle energy acquired is zero. Work in rasing the weight, minus work out descending the weight will equal zero.

The exact same thing can be said about an adiabatic spring. Push it in and get it he same energy back out. Zero heat involve with either. Adiabatic pairs cancel each other and zero work or heat is obtained or used.

Both processes obey the laws of thermodynamics however neither is a process of heat. Heat works differently because it has to be converted to pressure and volume changes. Those pressure and volume changes are dependent on temperature. Temperature is an absolute scale and depending on how far a temperature is from absolute zero, the pressure and volume changes are affected. Going from 1 to 2 K doubles the volume or pressure. Going from 100 to 101 K doesn't do anything. Negligible. Volume and pressure remain. This is where irreversibility rears it's ugly head. Energy can diffuse into the background. It is also why a temperature difference is needed, so there will be a pressure difference.

[matt brown]

Carnot law is a mathematical construct and provable only by mathematics. Reality can only be used for it's disproof.

I thought Nobody was dropping the ball, until right afterwards he says

Why? Is it so difficult to accept, if an engine absorbs some heat and produces some work it then requires some work/heat-rejection to cycle back to the beginning. The difference between absorbed and rejected heat, becomes the heat that is converted to work.

Carnot limit/s can easily be proven via simple experiments where it should be obvious that it's the backwork ratio of a compression cycle. An ideal 300-600k Stirling cycle has .50 eff simply due that (ideal) compression work equals .50 (ideal) expansion work. Overall, I think this more of a Carnot coincidence than a Carnot consequence.

The whole Carnot buzz only relates to single phase gas cycles, not individual processes, nor multi phase cycles. Imagine any ideal ambient engine with a cold hole to absolute zero and Carnot still wins, it's just 100% efficient. For Carnot to lose, you have to beat him.

If you move beyond PVT buzz and study internal energy (U) things will fall into place. Maybe it's possible to beat Carnot, but not with common regular cycles...due to backwork ratio/s. The only chance I see is some type of irregular/combined cycle, and likely with multiple regeneration.

[Tom Booth]

Carnot law is a mathematical construct and provable only by mathematics. Reality can only be used for it's disproof.

I thought Nobody was dropping the ball, until right afterwards he says

Why? Is it so difficult to accept, if an engine absorbs some heat and produces some work it then requires some work/heat-rejection to cycle back to the beginning. The difference between absorbed and rejected heat, becomes the heat that is converted to work.

Carnot limit/s can easily be proven via simple experiments where it should be obvious that it's the backwork ratio of a compression cycle. An ideal 300-600k Stirling cycle has .50 eff simply due that (ideal) compression work equals .50 (ideal) expansion work. Overall, I think this more of a Carnot coincidence than a Carnot consequence.

The whole Carnot buzz only relates to single phase gas cycles, not individual processes, nor multi phase cycles. Imagine any ideal ambient engine with a cold hole to absolute zero and Carnot still wins, it's just 100% efficient. For Carnot to lose, you have to beat him.

If you move beyond PVT buzz and study internal energy (U) things will fall into place. Maybe it's possible to beat Carnot, but not with common regular cycles...due to backwork ratio/s. The only chance I see is some type of irregular/combined cycle, and likely with multiple regeneration.

What I see as important here, and have no problem agreeing with is this:

The difference between absorbed and rejected heat, becomes the heat that is converted to work.

So in my most recent experiment I made an error and plugged the thermocouple in backwards and maybe jumped to some hasty conclusions, but the numbers are revealing in light of the above (quoted in bold)

stephenz did us the service of calculating the temperature change in joules/second at the cold "sink" which turned out to be a temperature RISE rather than a temperature FALL or refrigeration effect, how embarrassing for me, I didn't know thermocouples were polarity sensitive, but switching polarity does not invalidate the readings, the thermocouple just indicated the degrees of temperature rise as degrees in temperature fall, the degrees of temperature CHANGE are still valid numbers

So what do we know?

The steam generator draws a constant 85 watts which is equivalent to 85 joules/second

stephenz calculations indicated from the thermocouple readings a temperature change of 1.32 joules/second at the sink.

It is debatable how much of the 85 joules/second actually transfered from the steam generator to the engine, but let's just assume for arguments sake as a starting point that all 85 joules per second entered the engine.

If the cold plate only rose in temperature /heat energy 1.32 joules per second where did all the other 83.68 joules per second go???

According to the quote in bold above;

"The difference between absorbed and rejected heat, becomes the heat that is converted to work."

That would mean the missing 83.68 joules/second of missing heat were converted into work.

That indicates an efficiency of 99.62%

Carnot efficiency given the ∆T is around about maybe 20%

So we have a discrepancy of around 80%

[Tom Booth]

I have nothing to prove, yet I am willing to do some experiments for you, just layout your protocol and I will run your experiment with instrumentation that has at least one order of magnitude better accuracy than yours. Send me links to the engine you want me to buy.

How about something like this

Resize_20230712_023439_9883.jpg (181.96 KiB) Viewed 13879 times

The engine in a vacuum chamber.

The connecting rods would need to extend through rigid tubing to the outside atmosphere to provide air to the engine, otherwise it is effectively inside a Dewar(vacuum insulated)

A small heater below putting out a watt or two. Minimal amount of heat to run the engine or variable to test different inputs.

Wires for thermocouples can pass through the chamber walls to monitor temperatures.

Probably a sealed water reservoir with x grams of fluid for a "sink" to monitor heat rejection.

[matt brown]

It is debatable how much of the 85 joules/second actually transferred from the steam generator to the engine, but let's just assume for arguments sake as a starting point that all 85 joules per second entered the engine.

If the cold plate only rose in temperature /heat energy 1.32 joules per second where did all the other 83.68 joules per second go???

Input to gas is limited by heat capacity of gas, so a small LTD can't suck up much heat without high speed. In theory, gas volume x dT x rpm = input, regardless of efficiency.

[matt brown]

I like the vacuum chamber idea, but two in tandem would alleviate ambient extensions. You probably considered this and opted for single unit for more accurate readings.

[stephenz]

stephenz did us the service of calculating the temperature change in joules/second at the cold "sink" which turned out to be a temperature RISE rather than a temperature FALL or refrigeration effect, how embarrassing for me, I didn't know thermocouples were polarity sensitive, but switching polarity does not invalidate the readings, the thermocouple just indicated the degrees of temperature rise as degrees in temperature fall, the degrees of temperature CHANGE are still valid numbers

do not oversimplify things.
1. the reversing of the polarity of the thermocouple - ugh, just redo your measurements once you figured out how reliable those measurements are
2. as i mentioned the losses on your set up are everywhere: your cup, your hot plate, the bolts linking the hot and cold plate, and the convection at the cooler

I only used this equation to quantify the cooling power that appeared from those erroneous measurements.

If you want to know how much heat is being rejected on the plate. Insulate everything, including the top plate, particularly the top plate. Measure the top plate, and properly time or film and see how this temperature increases for a few minutes. Tinitial, Tfinal, and the time it took to from Ti to Tf. Know the material of your plate, know its dimensions. Apply the formula I used. But before you do that:

The main issue is the characterization of your heat input. Using an electrical boiler, boiling water whose vapor condensates at the base of your engine, is difficult to insulate. Calculating the heat going in the engine requires you to evaluate losses that are complex to evaluate. If you want to know how much heat you are putting in the engine, simplify your set up and do this:

install a cartridge heater mounted to the base of your engine. The best way to do that: ditch your current base plate and replace with a thicker aluminum base plate. Drill a 3/16" hole through its thickness, maybe 2" deep. Use a 3/16" cartridge heater - they're cheap and available in a variety of heat loads. They're resistors, find a DC version and use a DC power supply which will allow you to get clear values for the power you are supplying that that plate. that plate will be infinitely easier to insulate with foam which will give you a much better idea of your engine heat source.

[Tom Booth]

I like the vacuum chamber idea, but two in tandem would alleviate ambient extensions. You probably considered this and opted for single unit for more accurate readings.

No, don't really know what you mean by "two in tandem" or how that would be arranged.

Watching the video, using a stopwatch, I count about 8 revolutions/second. At least 7 Translates to 400 RPM conservatively.

Would it be possible for the air to take in 1 joule per revolution?

Anyway, It looks like the heat capacity of air is very low.

I am wondering however if that accounts for simultaneous work output.

That is, take in a joule of heat, do a joule of work. Take in another joule of heat, do another joule of work.

How many joules in work output required to turn the engine one cycle?

Lots of unknowns and difficult questions, hard to answer. I don't like relying on "theoretical".

The only thing of significance for my purposes is how much heat is "rejected" to the sink.

Heat going other places or not reaching the engine at all is irrelevant as far as a "cold hole" engine.

How much heat is actually being dumped into the cold hole is what we need to know.

https://youtu.be/t6cGw1scLvc

After accounting for work output to keep the engine running.

I think 400 RPM is rapid enough to assume mostly adiabatic compression and expansion.