Stirlingengineforum.com

What you describe is mechanically impossible.

Basically like a baseball player complaining:

"Ideally, when I'm up to bat, the ball should wait over home plate until I can get ready to swing and hit it.

Aside from that you've only described 1/2 a cycle.

The problem is how does the piston return to TDC after this " ideal" expansion.

PP would remain at TDC until DP reaches BDC. Then PP would move downwards expanding gas while input continues in DP hot space

That's half a cycle. You now have the piston fully extended and the cylinder full of high pressure hot air.

Carried out in a real engine this "ideal" process would leave the engine in a condition where it could no longer move, fully expanded and full of hot high pressure air.

Now you have to somehow get rid of all the heat so the gas can cool and contract.

A real engine runs at about 30 cycles per second. The power piston cannot stop and wait around while the displacer moves and the gas gradually cools back down.

The valve on a steam locomotive allowed the Engineman to set it for a little puff of steam that would expand and drive more efficiently. Or, drive full pressure hot steam onto the piston all the way to the bottom, for more power.

In both the ideal Stirling and Carnot Cycles, heat is added all the way to the bottom from the hot plate.

At that point it becomes necessary to cool the gas for the return stroke. If it isn't cooled the temperature will attempt to rise, being limited isothermally to the hot plate by rejection of heat to the hot plate, rendering work output to zero.

The Stirling and Carnot differ on how they save the body of heat in the gas before compression. Stirling with a regenerator. Carnot with putting work into the system once it has been rendered adiabatic. It is that rendering of a hot cylinder to suddenly an adiabatic, that nobody yet has been able to engineer.

It might be interesting to disconnect a piston from the crankshaft and move the displacer in and out slowly and quickly. There may be a point where just a small displacer movement produces full piston travel. Meaning that it takes very little pressure and temperature change to drive the piston. Be aware the piston may pop out like a cork.

It would be interesting to start with a cold engine and see how long it takes for enough heat to get in. It might then be interesting to heat the cold side to see how long it takes to equalize the temperatures again.

It might be interesting to hold the piston still and feel it's force as the displacer moves in and out. Slowly and or quickly. That should show how fast the gas can absorb and reject heat, as there will be zero volume changes during this test, zero expansion or compression. JFFT. Just Food For Thought.

This video shows the proportionality of displacer position and gas temperature.
https://photos.app.goo.gl/qHLkdf8vsBaq8VjJ6

This video shows the pressure changes without any piston movement at all.
https://photos.app.goo.gl/TjJqvZZSad7dSuaX7

This video shows a heavy steel weight simulating load on the piston, with a simulated higher compression ratio. IIRC the ratio was only still around 1.1:1
https://photos.app.goo.gl/8LxqjKWqiBUMFJpE6

Fool wrote: ↑Wed Aug 09, 2023 9:30 am The valve on a steam locomotive allowed the Engineman to set it for a little puff of steam that would expand and drive more efficiently. Or, drive full pressure hot steam onto the piston all the way to the bottom, for more power.

True.

This is somewhat mirrored I think, in a Stirling engine.

Under load the engine slows down so heat input moves towards "isothermal" allowing more time for heat to be conducted into the engine during expansion.

The external load (real shaft work - output) is compensated for by a simultaneous increase in heat input.

I don't know this for a fact, but I suspect with enough heat input to bring the RPM up to the no-load RPM the PV tracing for the heavy load/high heat input would resemble the no-load PV tracing.

At the higher RPM the pressure changes would tend to flatten.

On the other hand, perhaps the pressure would go higher during expansion and lower during compression. I'd need to experiment.

But at a higher temperature, heat could be input faster as the ∆T is greater, but with a heavier load on the engine work output also takes place within a shorter time frame.

The result- pressure during expansion and contraction should be about the same for high heat input and high work output vs low heat input and low work output at the same RPM

In other words the PV diagram probably does not show any change if there is a simultaneous increase in the load along with an increase in heat input.

The heat in and work out cancel.

So when the engine man turns up the steam as the train approaches a mountain, this is to compensate for the increased work load and the train moves steadily forward.

In an LTD with magnetic displacer, when the engine slows down under load the magnet has more time to lift the displacer so lifts the displacer higher which let's in more heat. With more heat the engine RPM increases. At higher RPM and no-load the magnet has less time to lift the displacer so let's in less heat. Which also parallels the train engineer with his steam valve. Or "stepping on the gas" in an IC engine.

The effect is that these magnetic LTD engines tend to run at a very steady RPM though the displacer movement tends to look rather erratic. The erratic displacer movement is due to always compensating for fluctuations in RPM and power output. Like a hit or miss engine sometimes skips a beat to maintain a steady RPM or hits twice in a row to increase RPM. The ignition seems erratic but it is a means of governing the speed of the engine.

A balance is maintained because heat input increases "internal energy" and work output decreased "internal energy". i. e. Heat is "converted to work" by the engine.

Tom Booth wrote: ↑Wed Aug 09, 2023 3:13 am What you describe is mechanically impossible.

For a guy chasing unicorns, this should be a walk in the park. No, it's not impossible, there's several ways to do this.

Tom Booth wrote: ↑Wed Aug 09, 2023 3:13 am
Aside from that you've only described 1/2 a cycle.

The problem is how does the piston return to TDC after this " ideal" expansion.

PP would remain at TDC until DP reaches BDC. Then PP would move downwards expanding gas while input continues in DP hot space

That's half a cycle. You now have the piston fully extended and the cylinder full of high pressure hot air.

Carried out in a real engine this "ideal" process would leave the engine in a condition where it could no longer move, fully expanded and full of hot high pressure air.

Now you have to somehow get rid of all the heat so the gas can cool and contract.

Earth to Tom, this is a Stirling (duh) where in the 2nd 1/2 of the cycle, the DP moves the gas from hot space to cold space whereupon gas drops to P low of cycle before PP compresses gas back to start state.

Geez, that barrel of Infinite Wisdom is taking a toll. Maybe you switch over to Billy Idol's Rebel Yell instead (in the midnight hour...)

matt brown wrote: ↑Thu Aug 10, 2023 12:41 am (...)
Earth to Tom, this is a Stirling (duh) where in the 2nd 1/2 of the cycle, the DP moves the gas from hot space to cold space whereupon gas drops to P low of cycle before PP compresses gas back to start state.

Geez, that barrel of Infinite Wisdom is taking a toll. Maybe you switch over to Billy Idol's Rebel Yell instead (in the midnight hour...)

Like I said, to reiterate the end of my statement which you left off your quote and conveniently ignored:

Carried out in a real engine this "ideal" process would leave the engine in a condition where it could no longer move, fully expanded and full of hot high pressure air.

Now you have to somehow get rid of all the heat so the gas can cool and contract.

A real engine runs at about 30 cycles per second. The power piston cannot stop and wait around while the displacer moves and the gas gradually cools back down

If you continue adding heat all the way to BDC, in the real world revolving crankshafts, pistons, flywheels, pulleys, driveshafts, planar springs, etc. cannot just be placed on pause while you embark upon the slow process of trying to conduct away all the excess heat you put into the engine.

matt brown wrote: ↑Thu Aug 10, 2023 12:41 am

Tom Booth wrote: ↑Wed Aug 09, 2023 3:13 am What you describe is mechanically impossible.

For a guy chasing unicorns, this should be a walk in the park. No, it's not impossible, there's several ways to do this.

OK, so, go ahead and describe or name one. (Or several).

It appears we have already seen how both, adding load and adding heat, appears on a PV diagram.

The load slows down the engine. The PV diagram fattens growing upwards and downwards. Producing a larger output of work per cycle. It is readable as a larger area enclosed by the cycle's path. It also shows more heat going in and out by the paths getting closer to T-hot and T-cold. It shows that slowing down an engine allows more time for heating and cooling, and that heat transfer actually happens by the closer approach of the gas temperature to the hot source and cold sink, for both in and out.

I think you are playing with the idea of adding even more heat than that, enough to keep RPMs constant. Let me pause just for a moment so we can all think about ways of doing that.

Pause. For an ICE when pulling a hill we would just push down on the accelerator to keep our speed constant. That burns more fuel per second.

In a steam locomotive the engineman opens the throttle to maintain speed. This would cause the boiler to cool if the fireman didn't throw more coal onto the fire and open the draft. That burns more fuel per second.

For a Stirling Engine vehicle perhaps one could raise the temperature of the hot plate source heater. The RPMs would be maintained from the higher pressure being developed to push the hill at the same speed. That burns more fuel per second. Pause over.

That higher temperature and resultant pressure would very definitely be apparent on a PV/indicator diagram. It would grow taller from the already fattened path from the load being put on. The cold side would shrink back to the original unloaded position or even higher as more heat is being added at the original RPMs and so more will be rejected.

It would be fun to see a live PV diagram to verify that.

Fool wrote: ↑Sat Aug 12, 2023 10:10 am It appears we have already seen how both, adding load and adding heat, appears on a PV diagram.

The load slows down the engine. The PV diagram fattens growing upwards and downwards. Producing a larger output of work per cycle. It is readable as a larger area enclosed by the cycle's path. It also shows more heat going in and out by the paths getting closer to T-hot and T-cold. It shows that slowing down an engine allows more time for heating and cooling, and that heat transfer actually happens by the closer approach of the gas temperature to the hot source and cold sink, for both in and out.

(...).

I'm not entirely sure, as your wording is a little ambiguous and possibly open to interpretation, or could use some clarification but I don't think I can agree with the part(s) I've quoted in BOLD.

It reads as though you are saying that as the working fluid temperature when hot, gets close to the heat source temperature heat transfer improves and when the working gas temperature, when cold, approaches the cold side T-cold temperature, heat transfer out is better, faster or more effective.

The wording is: "heat transfer actually happens by the closer approach of the gas temperature to the hot source and cold sink, for both in and out"

I don't want to put words in your mouth but BY, seems to imply because of or by means of or as a result of.


Rather than saying " It also shows more heat going in and out by the paths getting closer to T-hot and T-cold" I would say "energy".

OR

It also shows more energy going in as heat and more energy going out as work by the paths getting closer to T-hot and T-cold


Heat input per cycle may increase with increased load (work) but heat input per time interval when and if the engine slows down, is likely less. The engine has to slow down to allow the slow heat input to catch up with work output.

If the temperature of the heat input is increased then heat transfer is improved, takes place faster, RPM increases.

I have a feeling the PV tracing would tend to level off as the heat input and work output returned to a balance but I'm really not certain.

It certainly would be interesting to see an actual real time PV reading with variable load and heat input over time.

My assumption is that work out is indistinguishable from heat transfer out by conduction as both will result in a drop in temperature.

Pressure I'm not so sure about. But I really don't think "b]heat transfer actually happens by the closer approach of the gas temperature to the hot source and cold sink, for both in and out.[/b]

I think you may be seeing a correlation but have causation reversed or something.

What maybe LOOKS LIKE heat rejection to the sink because of apparent "better cooling" because the gas temperature is closer to the sink temperature cannot actually be because the gas temperature is dropping down as cold as the sink.

IMO the gas temperature drops down more because of the increased energy output as work, not because of more effective energy output as heat lost to the sink.

My assertion is that it seems impossible to distinguish a temperature drop from actual heat loss to the sink from a temperature drop due to work output just by looking at a PV diagram.

Possibly you mean:

heat transfer actually happens AS EVIDENCED BY the closer approach of the gas temperature to the hot source and cold sink

???

Anyway, bottom line is I do not think you can automatically attribute the closer approach of the gas to the sink temperature to heat transfer out to the sink without ruling out energy transfer as work, and increasing the load has obviously increased work output which you seem to be neglecting as a probable cause of the increased cooling.

Can you accept that, on a PV diagram, motion to the right, increasing volume, is work output?

Can you also accept the reverse, that motion to the left, decreasing volume, is work input?

For both, we are limiting description to the working gas. The gas is doing work > output. Or work is being input to the gas.

Hopefully you accept those, so, we can derive that a constant volume line, vertical line, is zero work, in or out.

Do you accept

Work = P•(V2--V1) = P•∆V

Do accept what that looks like on a PV diagram?

A change in V equates to an area under the path equivalent to work.

Fool wrote: ↑Sun Aug 13, 2023 3:10 am Can you accept that, on a PV diagram, motion to the right, increasing volume, is work output?
...

Not necessarily.

Suppose I have a Stirling engine at TDC and turn it upside down so the weight of the piston is basically "hanging", held up only by atmospheric pressure.

That is, if the piston were to move towards BDC, pulled down by gravity, a partial vacuum would form inside the engine and the piston would be prevented from falling out of the cylinder, held up only by atmospheric pressure.

Now apply some heat to the engine.

To eliminate mechanical friction and such, perhaps disconnect the piston from the crankshaft and hang a small weight from the piston that compensates for any friction between the piston and cylinder.

Would you not agree that when heat is applied the volume will increase?

If yes, does the increase in volume involve any external work output?

Certainly, in my opinion anyway, unless there is some resistance to overcome, no real work is done. The gas is not expending energy, and an engine is designed to be as low friction as possible so the piston can move very easily without much resistance.

So really running right side up does not involve much more work than running upside down.

So no, not really.

Expansion or movement to the right represents heat input. "Ideally" the piston and cylinder are frictionless.

Tom Booth wrote: ↑Mon Aug 14, 2023 1:50 am no real work is done.

What is your definition of "Work" then?

Classically any movement of a physical object means that work was done.

MikeB, thanks. I agree totally.

Tom, if a horizontal cylinder piston combination at atmospheric temperature and pressure is turned upside down such that the mass of the piston "hangs", the inside working gas will expand pushing the piston outwardly. The gas is outputting work.

The atmospheric pressure is countering this work plus the work caused by the weight of the piston (mgh) until the system stabilizes. The inside working gas doesn't care about the outside system, it only sees a wall expanding so work comes out of it. Pressure drops. Briefly temperature drops. Since the temperature drops, ambient heat enters returning the temperature, the piston drops a little more as the working gas warms.

A PV diagram represents the condition of the internal working gas. The outside system's only effect on that gas is piston motion, volume changes, and heat added or rejected.

Would you not agree that when heat is applied the volume will increase?

Only if the piston moves.

Visualize a closed cylinder tank, like an O2 tank, heating the gas inside doesn't change the volume. A vertical line on the PV diagram. Pressure goes up.

Now cool it. It still has the same volume, the same vertical line. The pressure is now lower. All with zero work realized, with all heat rejected. Efficiency of zero. Total gain of entropy. Zero entropy converted to work. All internal energy dissipated to the cold sink as rejected heat.

If we were inside pushing on the piston and nothing happened, we would feel as if we'd done nothing, equates to zero work, zero volume change. Distance traveled zero. Pressure maintained.

If the wall then moved in the direction we were pushing we would feel as if we did positive work. We really would not know why the wall moved, just that it moved in the direction we were pushing and wanting it to move. This we would think was positive work, increasing volume, work output. Moving to the right on a PV diagram.

Conversely, if the wall were to move in the opposite direction from the way we were pushing, we would call it negative work, decreasing volume, work input. Moving to the left on a PV diagram.

Gasses never contract during an expansion, even when cooling past their, stp condensation point. This can be observed by subjecting water to a vacuum chamber. Sure at first air cools below the dew point and a cloud forms, but continual pumping removes all air and cloud. The water then begins to cold boil. Why? Because the vapor pressure of the water is higher than the gas pressure. If the pump is stopped the water stopps boiling as the pressure increases to the vapour pressure of the water. There is still positive pressure in the chamber.

This will continue even ice forming in the water. It will still have a positive pressure.

Eventually there will only be ice and vapour. The ice will sublime. Ice has a much lower vapor pressure because it it colder. It will continue to have positive pressure all the way to zero Kelvin. Ice has a vapor pressure. If you put an ice cube in a cylinder in a vacuum, adiabatic, it will push the piston out.

Gas never contracts, it is pushed into a smaller area. If gas is cooled under constant volume or constant pressure, it will condensed and liquefy. That requires heat rejection, and it's not from adiabatic work-output expansion.

I think we can look at this another way.

If I hang a weight on the inverted piston, then it creates a vacuum.

Gas expanding into a vacuum does not do any work. The weight on the piston creates intermolecular "free space" between the gas molecules.

When heat is introduced, the gas expands to fill that vacuum.

Gravity is then able to draw the weighted piston down a little more, again leaving behind a vacuum which the gas can fill etc. etc.

Gravity is then the force doing the "work" of moving the piston to increase the volume, not the gas.

If you have an explosive concentration of heat at TDC and the piston is jettisoned down the cylinder, the situation is similar, the piston leaves a vacuum behind itself and the gas moves into the vacuum. So what actually did the "work" if heat is added (mostly) isochorically at or near TDC?

The "internal energy" of the gas increases at TDC, then there is a moment of rapid/explosive pressure increase, where that "internal energy" of the gas is transfered to the piston. The gas immediately cools (due to the transfer of energy to the piston).

The piston is now moving ahead of the "expanding" gas, if the gas could be said to be expanding, which is questionable.

In short then, it seems that the volume increase on the PV graph actually represents, if anything, heat that was added but only expanded the gas into a vacuum without that gas doing any work.

If there was actual work output by the gas, that is, a real transfer of "internal energy" from the gas to the piston, that does not translate into an increase in volume of the gas. Why would the gas expand when it's internal energy is reduced? That energy was transfered to the piston so cannot be used by the gas for expansion.

In other words:

If the gas retains the added heat, then the gas expands, retaining that added energy, and increasing the volume of the cylinder.

If the gas does "work" pushing the piston, then that "internal energy" is transfered to the piston, so tbe gas does not expand and the volume of the cylinder does not increase. At least, the volume does not increase due to the gas expanding as a result of retaining the heat as "internal energy". The momentum of the piston drawing a vacuum behind itself results in an increase in volume down to BDC.

The gas would fill that vacuum only if it took in heat but did not do any work to move the piston.

If the gas took in heat, increasing its internal energy, then transfered that energy to the piston, it no longer has the internal energy required for expansion.

Fool wrote: ↑Tue Aug 15, 2023 9:18 am (...)

The atmospheric pressure is countering this work plus the work caused by the weight of the piston (mgh) until the system stabilizes. The inside working gas doesn't care about the outside system, it only sees a wall expanding so work comes out of it. Pressure drops. Briefly temperature drops. Since the temperature drops, ambient heat enters returning the temperature, the piston drops a little more as the working gas warms.

(...)

That is not a plausible explanation given the extremely brief time interval involved.