Experimentally confirming the conversion of heat energy to work

[Tom Booth]

Maybe for further discussion I should define the room as 10'x10'x10'.

And completely air tight I assume? A virtual compressed air tank then?

Now any continuous air expansion can't take advantage of an infinite buffer space to expand into,

Indeed.

causing the pressure in the room to rise like Fool suggested.

Unless you run the compressor in this 10x10 virtual air tank, now a vacuum chamber. I'm pretty confident the air can expand back out into the vacuum created. But not much air to work with, and the compressor would have to work harder to pull a vacuum.

I know Tom will take issue with this and say I am adding more and more restrictions, and it may seem that way, but this was the original thought of the experiment, a small room.

Then heat energy is supplied to the engine so that measurements of Qin, Qout, and work are taken.

Any experiment is better than no experiment.

You would, of course, be diminishing the expansion potential.

That should not matter, as I said, so long as conditions are the same for both the main experiment (doing work) and the "control" not doing work, which is the one variable under examination.

Other than that your proposal of a 10x10x10 presumably hermetically sealed, perfectly insulated pressure chamber presents some obvious practical challenges.

IMO you obviously have no intention of actually carrying out any such experiment, your just yanking our chain.

[Tom Booth]

...

Then heat energy is supplied to the engine so that measurements of Qin, Qout, and work are taken.

So you plan on adding additional heat to the expansion engine? The engine you want to measure the cooling rate of?

Or are you just talking about the heat generated by the compressor.

Either way, whatever is done, should be done in exactly the same way for both the experiment and the control. Except for the one variable of the gas doing or not doing work.

Of course if you make expansion virtually impossible by putting one air tank inside another slightly bigger air tank there is no experiment.

It would probably not be long before the pressure inside the two tanks equalize especially if the compressor is run outside the room. so without much expansion it may be difficult to get any meaningful results, especially if you are going to start adding additional variables such as adding additional heat.

Pretty obviously you are not being realistic and have zero intention of actually doing any experiment.

[Tom Booth]

BTW, just out of curiosity, is your epoxy engine now operational, have you had it running.

The last report I could find mentioned finding some leak during pressure testing.

[VincentG]

The room was just a thought experiment that has been over complicated by the mention of a continuous air expansion process.

Quite obviously I am not going to build an air tight and insulated room any time soon.

But I do plan on using the fixed volume of hot water in my 60l drum Gamma chamber heat exchanger to try and measure work done inside an insulated box, recording the rise in temperature with and without work.

I am open to suggestions on how to set up an experiment that can accurately record heat input, heat rejection, and work.

[VincentG]

The epoxy chamber is not operational as an engine and remains a test bed for displacer chamber performance.

I got discouraged after realizing that a diaphragm piston needs to be made to get real performance out if it.

[Tom Booth]

...
I am open to suggestions on how to set up an experiment that can accurately record heat input, heat rejection, and work.

I think your overcomplicating things.

The only issue is: is heat being converted into work.

Input heat doesn't matter as long as it's THE SAME in both experiment and control. What is the input heat anyway? The heat already in the compressed air.

Maybe the air flow rate (cfm) should be kept the same also.

There is no "work" in the control. Theoretically any work at all should produce measurable cooling.

There is no "heat rejection". Cooling is by conversion to work not "heat rejection".

As a simple test, I could take my crappy eBay Tesla turbine and my shop compressor.

Using the same PSI for both runs, run the turbine without any load, then couple the things to a small generator with some lights or something and run it again.

The air leaving the turbine should be colder when the turbine is powering the lights.

Maybe I could do that sometime soon, but first I think I need to take it apart and see what's rattling around inside.

Maybe use this:

Compress_20240619_130558_8977.jpg (13.43 KiB) Viewed 1531 times

Or this:

Compress_20240619_130558_8886.jpg (9.44 KiB) Viewed 1531 times

Both available on eBay.


Given my experience with the previous Tesla turbine I ordered I'm reluctant to spend any more money on another.

The little steam turbine/generator might be a better option for $125 and should run on compressed air.

Just run it with a load on the generator and without a load and measure the temperature drop at the exhaust.

If you want to get more fancy and measure additional parameters like voltage do that too, but no point in complicating things too much, it's a relatively simple experiment

[Tom Booth]

Actually, an ordinary shop air tool should work.

https://youtu.be/F5OF0dxbTCE

This one has a pipe attached to the exhaust.

Run it taking out bolts or whatever and measure the temperature at the exhaust.

[VincentG]

As far as I'm concerned, the compressed air expansion is gaming the system.

You have high preasure at ambient heat levels, of course expanding it lowers temperature. Thats a simple gas law calculation.

[Tom Booth]

I've told the story in here before how I first became aware of this topic.

When I moved to Arizona years ago, the first job I got working in an engine repair shop was to fill in for a guy who I was told had been using an impact wrench to remove long head bolts from a big engine.

When he went to put the tool down his finger broke off.

He had his finger over the exhaust port and did not realize it had gotten cryogenically frozen from the cold exhaust.

Simple expansion of air does not get that cold it was the work of removing all those long rusty head bolts.

"Gaming the system". What a lame excuse for not doing a very easy experiment.

You just don't want to prove your pet theory wrong. You could prove me wrong. Do the experiment and show there is no increase in cooling from any extra work

[VincentG]

"Gaming the system". What a lame excuse for not doing a very easy experiment.

You just don't want to prove your pet theory wrong. You could prove me wrong. Do the experiment and show there is no increase in cooling from any extra work

I'm perfectly happy being wrong, but how you don't see that expanding compressed air is entirely different than a heat powered expansion cycle is beyond me.

[Tom Booth]

"Gaming the system". What a lame excuse for not doing a very easy experiment.

You just don't want to prove your pet theory wrong. You could prove me wrong. Do the experiment and show there is no increase in cooling from any extra work

I'm perfectly happy being wrong, but how you don't see that expanding compressed air is entirely different than a heat powered expansion cycle is beyond me.

Ever here of a gas refrigerator? Maybe a vullieumere heat pump perhaps? Solar heating causing air to rise and expand and cool. All "heat powered expansion".

Anyway the question is does a gas expanding and doing work loose energy and cool, FROM DOING WORK.

A basic well recognized principle proven by all kinds of actual experiments by scientist, researchers and industrialists over decades, any number of which you could do yourself to demonstrate the principle.

Your excuses are absolutely lame IMO.

Put cold air in a tank and heat up the tank, you get a tank of compressed air.

[VincentG]

The internal energy of the entire expanding volume isn't reduced, but the internal energy per unit volume is, because the total internal energy of the gas is spread out more, into a bigger space.

By the same token, the temperature of individual molecules of air doesn't change, but the bulk temperature of the gas does reduce, because the thermometer, or finger end, isn't getting hit so often because the molecules are more spread out or rarified in the expanded volume; the pressure falls.

This is why Charles Law holds experimentally (ignoring or minimising conduction losses with insulation)

T1/V1 = T2/V2

As the volume expands, the (bulk) temperature drops, along with the pressure, as we can see from the combined gas law

(P1V1)/T1 = (P2V2)/T2

Tom, you obviously don't understand, or are not willing yourself to address my original question.

Perhaps you could start by addressing this description by Stroller.

[Tom Booth]

I believe Stroller has already retracted his statement after I pointed out it only applies in the case of a "free expansion" into a vacuum doing nearly zero work.

The gas still does some work trying to get out of its own way or pushing itself, because the moment you have one molecule of gas in the vacuum it's no longer a total vacuum, which is why you get Joule Thomson cooling even with expansion into a vacuum. There is still a small amount of work involved. So in reality all expansion of a gas involves some work, if only the gas pushing itself while in the process of expanding. Have it do MORE work and it cools more.

Easily proven by any number of experiments.

Where is Fool with that list about how to recognize "science deniers" now?

5. Setting impossible expectations of proof for others

You could prove the principle to yourself very easily, but your choosing to make that impossible.

No point in further discussion. You're being completely unreasonable IMO.

If you come up with some experiment that meets your strict criteria let us know, you've received a number of good suggestions IMO.

This whole topic is a non-issue.

[Tom Booth]

Just to be clear, those equations, Charles Law, and the combined gas law are generally applicable in most ordinary circumstances where the gas is not doing any appreciable work.

As soon as you add in WORK it's a whole 'nother ball game.

It seems to me you're the one who cannot get that concept through your head no matter how many times it is stated.

When a gas does 1000 joules or work doing something like driving a piston engine it is the equivalent of removing 1000 joules of heat.

Hopefully you can understand that and how it differs from heat simply being conducted away or gas cooling by expanding into a larger area without doing any work such as driving motors or turbines. Cooling by conversion to work would be in addition to cooling by ordinary expansion.

How is that not clear?

What Stroller is citing there does not account for outside WORK.

Without doing work there is obviously no "conversion of heat energy to work".

[VincentG]

As soon as you add in WORK it's a whole 'nother ball game.

It seems to me you're the one who cannot get that concept through your head no matter how many times it is stated.

When a gas does 1000 joules or work doing something like driving a piston engine it is the equivalent of removing 1000 joules of heat.

If you are so sure of this, why are you always asking for experimental confirmation of a Carnot engine?

Can you provide a paper that describes, in detail, the conversion you are so sure of?

You argue the other extreme as I do, so why be so dismissive of another idea?