if stirling engine is driven as reversed, does it work as cooler?

[Fool]

https://www.e-education.psu.edu/png520/m3_p3.html

figure0306.gif (4.7 KiB) Viewed 6257 times

This shows as volume decreases to the liquid vapor point pressure change per volume increases. Then no pressure change until all gas becomes liquid. Then the pressure changes way more rapidly.

Liquefaction this way only happen for very high compression and specific low temperature. Specific for each specific gas. Not the areas our engines run.

It is interesting to think that during that isothermal compression the latent heat is rejected and work is put into the gas. It takes work to compress gas into a liquid.

This is the process of building a cold hole with heat rejection. Work in, heat pumped, cold to warmer.

For the engines discussed here ideal gas law is appropriate. Zero attractive forces.

However, as described many places, real gasses produce positive pressure in all situations. They don't ever pull a piston.

[Tom Booth]

https://byjus.com/chemistry/deviation-f ... behaviour/

The-Deviation-Of-Real-Gas-From-Ideal-Gas-Behavior-2.png

It looks just the opposite.

The link before the last, is a web page describing the differences between ideal and real gasses. I agree with it. It was posted here for people to read a different wording of my point. It also shows charts depicting a positive pressure for gasses regardless how low the pressure goes. It also shows that expanding a gas leads further away from it becoming a liquid and to an area where it behaves more ideal.

That graph depicts the volume of the gas. That is, the actual space occupied by the gas molecules themselves. It is simply depicting in a very abstract and general way that REAL gas molecules actually takes up some of the volume of the container. "Ideal" gases on the other hand are abstract points with no volume of their own.

What I was talking about is in the paragraphs previous to the graph you cited, which is something altogether different.

[Tom Booth]

https://www.e-education.psu.edu/png520/m3_p3.html



figure0306.gif

This shows as volume decreases to the liquid vapor point pressure change per volume increases. Then no pressure change until all gas becomes liquid. Then the pressure changes way more rapidly.

Liquefaction this way only happen for very high compression and specific low temperature. Specific for each specific gas. Not the areas our engines run.

It is interesting to think that during that isothermal compression the latent heat is rejected and work is put into the gas. It takes work to compress gas into a liquid.

This is the process of building a cold hole with heat rejection. Work in, heat pumped, cold to warmer.

For the engines discussed here ideal gas law is appropriate. Zero attractive forces.

However, as described many places, real gasses produce positive pressure in all situations. They don't ever pull a piston.

Again you are misrepresenting a graph of a pure substance under "ideal" isothermal conditions trying to use that to "disprove" something entirely different.

Like I said, I could really care less if the volume decreases due to "contraction" a "vacuum" or 'atmospheric pressure". Maybe this could be determined experimentally, but I'm not going to bother wasting more time on this kind of infantile nonsense.

Obviously when a gas in a cylinder in an engine cools (or loses "internal energy" due to "work" output resulting in a temperature drop) the piston is pushed, pulled or drawn inward, however you would like to interpret it, the effect is the same.

What is undeniable IMO is that the balance of the forces of molecular attraction and repulsion need to change BEFORE atmospheric pressure can have any effect one way or the other. That's just common sense.

[Tom Booth]

One potentially interesting or useful observation to come out of this:

I just spent some time fooling around with a oversize syringe, a fire piston and some glass tubes, scales and weights.

At "normal" atmospheric pressure and temperature, it certainly appears to be true that a 15 or so pound weight will pretty consistently pull a plunger out of a 1" cylinder regardless of the length.

The reverse is definitely not the case. While pushing the plunger back in the pressure will increase more and more until further compression, (using just my strength, or body weight at least), becomes impossible.

Compressing the air in the syringe downward onto a bathroom scale, the scale will go up to 50 pounds or more with air still in the tube. But with different volumes of air in the tube, 15 pounds will pull the plunger back out every time, without much, if any variation.

So, when the piston, or the gas rather, in a Stirling engine is expanding outward against atmosphere, the outside pressure does not ever increase appreciably, since atmosphere is so large it can't be "compressed" so the resistance to expansion is never any higher than 15 psi.

Likely, however, there is a definite limit to how far the gas inside the cylinder can be compressed.

It seems like there should be some way to turn this to some advantage when designing an engine. All else being equal, expansion may be inherently "easier" than compression, at least under "normal" atmosphere at ambient temperature.

I still need to see if cooling the cylinder with a 15 lb weight on the piston, would have the effect of causing the gas inside to "contract", lifting the 15 pound weight hanging upside down from the piston.

In theory (my theory that is), the 15 lb weight should balance or cancel the 15 pounds of atmospheric pressure.

If there is no "contraction" upon cooling of the gas in the cylinder there should be no possibility of the piston lifting the 15 lb weight due to lowering of internal pressure by cooling the cylinder.

If the gas actually in fact contracts or "clumps together" more and more due to molecular attraction as the temperature is lowered further and further then the piston should be drawn inward more and more in spite of the 15 pound weight counterbalancing atmospheric pressure.

Or will the gas only contract some when each constituent gas reaches liquefaction temperature, or not ever contract at all?

Watching a video of a 55 gallon barrel violently "imploding", it's hard to imagine such a force could be counteracted by attaching a 1" cylinder to the barrel with a piston and hanging a 15 pound weight onto the piston. With the 15 lb weight the piston should not be drawn in when the barrel is cooled. Right?

Looking at that twisted metal, I can't help but think something more than 15 psi of atmospheric pressure is involved.

https://youtu.be/WmztG5f12s4

What about a balloon.

If you cool a balloon with ice it will shrink gradually, smaller and smaller the colder it gets.

Is that due only to the elasticity of the rubber?

Suppose we filled a non-elastic plastic bag with a gas instead of a balloon, to eliminate the contracting force from the rubber.

If the gas "never contracts", then the plastic bag should stay inflated and not shrink in size at all until the temperature of liquefaction. Right?

[Tom Booth]

Watching the above video again, what if the experiment, demonstration or whatever were repeated without using any water?

That is, just heat up the air in the steel barrel really really hot so a lot of air escapes and expands, going out the opening.

In that case, there would be no phase change involved to confuse matters.

With a lot of heat all the water vapor can be driven off, so if the remaining air/gas "never contracts" with cooling or loss or energy then the pressure in the barrel (after the lid is put on when still hot) will never go down. The gas still in the can can "only expand". Never contract. Right?

How does an LTD Stirling engine ever operate by a mere 2° ∆T and no phase change involved?

How could an Air Cycle refrigeration system ever work without phase change?

[Tom Booth]

I just filled these two balloons with approximately equal amounts of helium.

Resize_20230820_182831_1892.jpg (112.55 KiB) Viewed 6233 times

Using helium should eliminate confusion due to water vapor or any other gas condensing. Anyway, I've got lots of helium.

I just put the balloon on the left in the freezer.

Helium does not phase change until something like 4 Kelvin so we should not see any gas contraction Right?

There is no contraction of a gas until just at the point liquefaction. Right?

I think that's is what Fool has been saying over and over and over and over...

No such thing as "contraction" of a gas, except with phase change. Gases can only forever expand. Right?

Anyway, in half an hour or so I'll take the balloon out of the freezer. Maybe it will not "contract" at all. Maybe both balloons will stay the same size.

After all, the gas inside the balloon doesn't care about "outside conditions".

After that, maybe repeat the experiment with two zip lock storage bags to rule out the elasticity of the balloons.

[Tom Booth]

Well, it has only been about 15 minutes but I took a peek in the freezer and something really weird is happening.

Resize_20230820_191656_6052.jpg (142.04 KiB) Viewed 6228 times

The helium balloon I put in the freezer, (on the left) appears to have gotten BIGGER!

Now I'm wondering if that might have something to do with the Joule Thomson effect and Helium's unusually low inversion temperature.

https://www.chemeurope.com/en/encyclope ... ffect.html

Kind of like how water expands when it gets cold, so does helium?

[Tom Booth]

To try and check what was going on, I switched balloons and put the other one in the freezer.

This time the one in the freezer got smaller. (Or stayed smaller, though it looked smaller to me in the freezer)(Left).

Resize_20230820_200710_0613.jpg (115.31 KiB) Viewed 6223 times

I left it in the freezer a lot longer while I went out shopping.

Likely the smaller balloon is just not holding air (helium) as well as the other.

[Tom Booth]

Maybe try a different approach. Reading various websites about helium balloons, apparently they can be effected by humidity, but pretty much universally it is said that helium is VERY sensitive to heat and cold and helium balloons could expand and pop in a hot car or shrivel and shrink in cold weather. Seems unlikely since helium at room temperature is miles away from any phase change, and so far, I'm not seeing a dramatic difference.

This Scientific American article suggests an experiment putting balloons into the freezer but marking them with measurements.

https://www.scientificamerican.com/arti ... nd-expand/

To avoid differences in porosity, potential leaks, and possible humidity differences inside or outside the freezer, heating variations between lights in the room and the dark in a refrigerator etc. I'm just using one balloon.

Resize_20230820_211256_6164.jpg (146.32 KiB) Viewed 6216 times

Marked by the inch around the circumference.

Another problem could be the impurities in the dollar store balloon helium which can settle in the canister, so one balloon might potentially have more helium than the other. Filling one after the other could also cause temperature variations. The gas cools when leaving the container, then warms once in the balloon, possibly resulting in differences in inflation due to changes in size after and during filling.

[Tom Booth]

Very odd, but after just a few minutes in the freezer, it appears again, like the helium balloon has EXPANDED slightly!

Resize_20230820_215628_8529.jpg (187.15 KiB) Viewed 6210 times

I'm leaving it in the freezer to see what happens.

How to explain this?

Maybe after being released from the compressed canister into the balloon, the helium continues to gradually expand for several minutes?

The balloon rubber has more give to it the longer it stays inflated?

Helium is supposed to transfer heat better than air, or so I thought.

My marks are not 100% perfect or straight but I don't think they were that far off. At any rate, there certainly does not appear to be any discernable shrinkage so far, though it has only been a few minutes, I was surprised at what looks like an initial expansion after going into tbe freezer though.

[Fool]

Tom,

Obviously when a gas in a cylinder in an engine cools (or loses "internal energy" due to "work" output resulting in a temperature drop) the piston is pushed, pulled or drawn inward, however you would like to interpret it, the effect is the same.

If you mislead yourself into believing that the balance of the forces reverses, by the term "contraction", you will confuse the direction of work that an internal working fluid gas "works". The balance never reverses in the temperature range and gasses we use, especially helium, as it doesn't liquefy until about 7 K, colder if lower than standard atmospheric pressure.

Since gasses always have positive pressure, any expansion, piston moving outwards, is positive work outwards. The gas works.

This is what a PV diagram shows. Pressure and volume of the internal gas.

Conversely, compression, decreasing volume, is always work delivered to the gas. It is work in opposition to the force the gas is providing. Outside forces do work on the gas. This is also depicted on a PV diagram. Matt calls it back work.

You keep bringing outside conditions in for a one sided comparison. The atmosphere pushes on the piston, inwardly and only inwardly, with a constant force F. Pressure multiplied by piston area. It does work in opposition of the piston motion full stroke X. Work = F•X. Positive during compression, negative during expansion. Since the work is equal and opposite the total work contributed to the system for a full cycle is zero, and can be ignored. Just because it compresses the gas in one direction, doesn't mean any total work for the entire cycle. It doesn't add up. It cancels.

There can be zero work out from the gas by a shrinking volume also called compression. Hence contraction is a misleading term. Something must force the compression from an outside force pushing inward against the internal gas pressure.

Yes a 55 gallon drum can be imploded by a 15 pound weight, frictionless piston, and an extremely long 1 square inch area cylinder. The volume of the cylinder will need to be larger than the 55 gallon drum's by many many multiples maybe 1000 or more. Very long cylinder. The 1.128 inch diameter piston can be long enough to weigh 15 lbs.

Experiment:
Hang a piston weighing less than 14 lbs, in a 1 square inch area cylinder. A piston of approximately 1.128379167 inches in diameter will have a 1 square inch area. Push it to the top squeezing out all the air, water, E.T.C., zero dead volume. It will not fall. Use a spring scale to pull it one foot, an arbitrary distance. Heat the cylinder and the perfect vacuum. Does it change the tension on the spring? Now try it with some gas or liquid in there. Does it ever have more tension than for the pure vacuum?

A barometer works on that similar principle. Fill a 1000 millimeter long, 1.128 inch inside diameter, glass tube with over 15 pounds of mercury. Invert it into a sufficient cup to hold the mercury. The mercury will drop and stabilize at about 760 mm above the surface in the cup, STP. Heating and cooling will have a minor effect on the density of the mercury and length of glass tube.

Now put some water into the tube. No matter how little is put in, the mercury level will be lower. Heating and cooling the tube will not make the level higher than for the vacuum. It might be close if you cool it below 32° F. The water vapor won't "contract" pulling the mercury level higher than for a pure vacuum.

Your balloon experiments are trying to confuse compression from elastic pressure, with compression from atmospheric pressure.

Try putting ice into a piston and cylinder, then put it into a vacuum chamber. Or start with steam in a cylinder. Put it into a freezer, to freeze the water at atmospheric temperature and compress the steam to ice. Then put it into a vacuum chamber. Does the piston still pop out, move outward? It is possible to do this experiment while cooling the cylinder with liquid nitrogen, however the piston might freeze and stick in the cylinder.

[Tom Booth]

Tom,

Obviously when a gas in a cylinder in an engine cools (or loses "internal energy" due to "work" output resulting in a temperature drop) the piston is pushed, pulled or drawn inward, however you would like to interpret it, the effect is the same.

If you mislead yourself into believing that the balance of the forces reverses, by the term "contraction", you will confuse the direction of work that an internal working fluid gas "works". The balance never reverses in the temperature range and gasses we use, especially helium, as it doesn't liquefy until about 7 K, colder if lower than standard atmospheric pressure.

Since gasses always have positive pressure, any expansion, piston moving outwards, is positive work outwards. The gas works.

This is what a PV diagram shows. Pressure and volume of the internal gas.

Conversely, compression, decreasing volume, is always work delivered to the gas. It is work in opposition to the force the gas is providing. Outside forces do work on the gas. This is also depicted on a PV diagram. Matt calls it back work.

You keep bringing outside conditions in for a one sided comparison. The atmosphere pushes on the piston, inwardly and only inwardly, with a constant force F. Pressure multiplied by piston area. It does work in opposition of the piston motion full stroke X. Work = F•X. Positive during compression, negative during expansion. Since the work is equal and opposite the total work contributed to the system for a full cycle is zero, and can be ignored. Just because it compresses the gas in one direction, doesn't mean any total work for the entire cycle. It doesn't add up. It cancels.

There can be zero work out from the gas by a shrinking volume also called compression. Hence contraction is a misleading term. Something must force the compression from an outside force pushing inward against the internal gas pressure.

(...)

Everybody knows all that.

Again you are just misunderstanding or misinterpreting,

Above I said "work output" as I've said before DURING EXPNSION. Work output is done during the power stroke.

Cooling of the gas takes place by adiabatic expansion during the expansion/power stroke.

As a result of the expansion work the gas temperature drops and the working fluid, for want of a better term "contracts" or is pushed back in by atmosphere due to the lower interior pressure which is commonly called a "vacuum".

Lower pressure inside the balloon and it gets smaller. You can say the helium in the balloon "contracts" or you can say atmosphere pushes it in. I don't care, but part of the balance in REAL gases involves molecular forces of attraction and repulsion.

Real gases DO attract and repel each other. Not just when they liquify.

Resize_20230821_090813_3285.jpg (108.09 KiB) Viewed 6198 times

It looks like there has been significant, measurable "shrinkage" after several hours. But in deference to the vocabulary police I'll call it "crushing in by atmospheric pressure". Or perhaps "elastic contraction" on the part of the balloon.

Or maybe some helium leaked out.

So now I'll leave the balloon out to warm up and see if the atmospheric pressure backs off now. We will wait and see if the atmosphere withdraws it's compressive force from the balloon.

[Tom Booth]

Well, that didn't take long.

Resize_20230821_100642_2518.jpg (171.52 KiB) Viewed 6193 times

It seems a gas will respond to an application of heat much more quickly than to cold. It took hours to cause the balloon to shrink in the freezer but after just a few minutes out of the freezer it has returned to normal.

Oh, sorry "Fool" I meant to say.

After some hours in the freezer the external barometric pressure increased, crushing in on the balloon.

After only a few minutes out of the freezer though, a low pressure weather system moved in and the atmosphere withdrew from the balloon providing additional space for the balloon to fill. Or something like that.

[Tom Booth]

I put the balloon back in the freezer. I'd like it to be a sunny day today.

Resize_20230821_103306_6215.jpg (222.4 KiB) Viewed 6188 times

That should induce a high pressure weather system to form.

[Fool]

Atmospheric pressure doesn't change. The energy, volume, and pressure of the gas changes. Here is an acceptable formula,

PV =nRT

Tom,

That is, just heat up the air in the steel barrel really really hot so a lot of air escapes and expands, going out the opening.

If you heat it to 600 K, cap it, and then cool it to 300 K the pressure will be about 7 psi, minimum crush volume will be 1/2.

Heat 900 cool 300 pressure about 5 psi crush volume 1/3.

Heat 1200 K red hot, cool 300, psi 3. Volume 1/4

How does an LTD Stirling engine ever operate by a mere 2° ∆T and no phase change involved?

Poorly with very little useable output power per size, and low efficiency.

How could an Air Cycle refrigeration system ever work without phase change?

By heat absorbsion at low pressure and heat rejection at high pressure, plus a great deal of work input.

There is no contraction of a gas until just at the point liquefaction. Right?

The scientific way of stating the observed phenomena is, gas when heated pushes out. If at a constant volume the temperature and pressure increase. If at constant pressure. The temperature and volume increase.

If cooled, the gas pushes out, if at constant volume, the temperature and pressure drop. At liquid transition, temperature stops decreasing, pressure continues to drop. Volume remains the same. pressure continues to drop, until most of the gas is liquified. Then pressure and temperature will continue to drop, never quite reaching zero for either.

If at constant pressure, temperature and volume will reduce, gas being compressed by the constant hydraulic, pneumatic, or mechanic pressure/force. As the liquid transition proceeds, the temperature remains relatively constant, and the volume continues to drop. After most of the gas is liquefied, volume reduction slows to relatively almost zero temperature continues to decrease. Volume approaches a minimum but significant value. Vo liquid/Solid at Zero Kelvin.


Except in the Latin Language.

Gases can only forever expand. Right?

I would say gasses forever push. To get a set volume to completely liquefy they need to be under pressure and at a low temperature. Below the critical temperature and above the critical pressure. If an attempt is made to liquefy gas by adiabatic expansion with work, constant entropy, it is impossible to liquefy all the gas. Some will remain as gas. What remains will completely fill the volume and push outward.


To liquefy steam completely, pressure and heat rejection are both needed.