Constant volume compression/expansion-displacer chamber analysis-heat powered mechanical amplifier

[VincentG]

The displacer 'pumps' the air back and forth, some adiabatic temperature changes will occur as a result. I'm curious if it is enough to notice, or negligible. I'm also curious, will starting with a delta T effect that. The assumption is not. But I'm curious. Also of effects from cycle speed. Faster more change?

I see. I don't think there is enough pumping effect to change temperature from starting at room temperature.

Starting with a temperature differential, I do think that the gas briefly goes above and below Tmax and Tmin while cycling the displacer. As of now I don't know how to take advantage of this, other than the hope that in a well designed and built engine with distinct events and thermal separation, this will manifest itself in the form of less energy consumption.

[Fool]

If it goes above Th and Tc. It should produce that as a slow buildup from very small Th and Tc. And maybe maintain Th and Tc in room temperature.

I'm just curious. My assumption is that it will return and stay at room temperature. But there may be a subtle temperature change, difference, Or???

I don't think it will be beneficial. Hysteresis maybe. Or nothing. Work is going in to cycle it.

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[VincentG]

If it goes above Th and Tc. It should produce that as a slow buildup from very small Th and Tc. And maybe maintain Th and Tc in room temperature.

My thought is that the heat value is small enough, and the mass of the plates much too large to get results that way. If anything, it would just help to increase effective Tmax and Tmin.

[Fool]

So you are saying it, at best, will be negligible?

[VincentG]

So you are saying it, at best, will be negligible?

These engines run on temperature delta. So if you consider an increase in effective temperature delta negative, then sure.

[Fool]

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Negligible means insignificant. Wether that is positive or negative is to be seen. I'd just like to see if it is noticeable. How much and where, or absent.

Putting work into the displacer gas should have a thermodynamic effect.

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[matt brown]

...this title could go on and on...

This video shows my ltd engine being spun without the power piston connected, operating on boiling water and ice. https://photos.app.goo.gl/17PjLX2A34X3s5AN8

We can view a pressure swing just as if the engine was running normally, only in this case we have constant volume compression and expansion. It does not much seem to matter how fast it is spun, the pressure swings are close to the same. This seems to tell me the displacer chamber is rather efficient in quickly changing the temperature of the gas.

I've demonstrated, at least to myself in this next video, that there is a constant flow of heat energy between the hot and cold sinks based on displacer position that we use to manipulate air pressure. https://photos.app.goo.gl/Xa9e8qaf5dHpkrb89

The best we can do is use interrupted and partial bites of this energy. When powering a piston this is easier said than done. However in this configuration I believe that the most efficient use of this energy will be found at the highest cycle rate in which there is still a full(or maybe nearly full) pressure swing.

Power out is much greater than power in (displacer movement). The major benefit here is the potential to heat and cool a huge volume of air without the need for a large piston and cylinder.

The hurdle is how to utilize this energy. I think the answer is in recovering the resonant energy via a very large vibration harvesting generator similar to a loudspeaker, or any number of methods described here. https://orbray.com/magazine_en/archives/1261

For me the key to this is unlike a free piston, we can directly control the cycle rate regardless of load. This should make it much easier to generate consistent power without the need for high end electronics.

Constant displacer motion could come from something as simple as a gravity powered counterweight driving the crankshaft, or a small electric motor.

What do you guys think?

Everyone needs to review this thread from above OP and pay attention to time stamps.

Whoa...it appears Vincent discovered the Proell effect 1-1/2 years ago, long before any of us knew it as such. Here's some posts...

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CvW4.png (36.64 KiB) Viewed 1943 times

[Fool]

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Matt, thanks for reopening and reviewing this thread. I wanted to comment on the opening post, but didn't get in until it had been run off topic by the end of the Forth out of now six pages. By that time all I could do is provide some data and research questions.

The second post on this thread, Matt's first one, pointed the following out correctly:

No such critters as constant volume compression and expansion, but I get what you're saying.

He understands the phenomenon but by a different description. Good point.

Compression is defined as volume reduction of a closed system.
Expansion is defined as volume increase in a closed system.
A closed system is defined as a system with constant mass.

A long cylinder with one hot left end, and one cold right end is actually two open systems. Mass transfers towards the right cold end during heating of the left end to become hot.
Yes pressure and consequently temperature go up at both ends.

So looking at PV=nRT, needs to treat each end separately. Constant volume left VL, and right VR. Pressure left and right PL and PR. Temperature left and right TL and TR. n left and right nL and nR. It will be assumed that the changes to R are negligible, so only one R.

There is also a state one and two. Before and after heating. So all those will have a one state and a two state subscript such as nL1, nL2 and nR1, nR2.

Since it is a closed system it will be constrained by totals. Example: nL1 + nR1 = nT
nT is the total number of molecules, or mass in the system. nT stays constant.

Now it's important to see that PV=nRT pressure will change if any of the other factors change. So system left even if modeled as a constant volume, (zero compression), can increase in pressure if it has an inflow of mass. This is not compression. It is an increase in pressure from an increase in mass. Also called an increase in density.

Density total is constant = dL+dR.

It can also be cut up with constant mass elements, where each has a non constant volume. And each has its density increase resulting in a volume reduction. Hence a local compression at one end, and local expansion at the other hot end. Total volume constant. Local volumes not. This is not constant volumes left and right. Each chunk is a closed system, by definition of constant mass per element.

If negligible power is put into a displacer at constant total volume, my guess is that the over and under temperatures will be equal so total heat in or out will be zero. It isn't until the piston forces, an increased, and a decreased volume at appropriate times that the system develops work output and heat input, or work input and heat output.

The Vuilleumier Heat Pump uses both a displacer and piston to get it's mechanical power and heat pumping action. It is a combination heat engine and pump.

I hope this helps understand Matt's comment and how to better model this kind of system. The necessity of using finite element analysis, and hundreds, if not thousands of elements, is why it is so difficult to model and understand these engines. An indicator measurement diagram is a much faster and cheaper way to gain similar averaged data.

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[Fool]

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PV=nRT there is a battle between the left side and right side when one side is heated and mass moves to the other side. Matt are there any equations to calculate that I know it will revolve around PL and PR being equal, V1 and V2 constant, and nL+nR=nT so it becomes the solving of the following equations:

P1•VL=nL1•R•TL1
P1•VR=nR1•R•TR1

P2•VL=nL2•R•TL2
P2•VR=nR2•R•TR2

nL1+nR1=nT
nL2+nR2=nT

TL1=TR1=Tc=300
TL2=Th=600

Other assigned values:
P1L=P1R=atmospheric
VL=VR
R

Pre calculated values:
nL1, nR1, nT

Unknowns that must be solved for:
P2, nL2, nR2, T2

First, second, and fifth equations are dominant from assigned values. It leaves 3 equations and four unknowns. The fourth equation would be the ideal gas equation for an adiabatic process. Right?

Start by solving for nL2 or nR2 and plug it back into the others, until there is one equation that can be plugged into the adiabatic equation or calculator.

Of course, it can be diced up into more non mixing elements than the two left and right. As you've already done. Gets ugly fast. Need to use Linear Matrix Algebra. LOL.




One more thing. Density can be added to PV=nRT by dividing both sides by V:

P=(n/V)RT

n/V is density d.

Or

P=dRt.

(Pee dirt? Pay dirt?) Yuck!

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[Tom Booth]

What a load of crap!!

You obviously don't know what your talking about, "fool".

[VincentG]

Thanks to all and especially Matt Brown for helping to figure this stuff out with me. I have a lot to add when I get some time.

For now, keep the arguing and personal attacks out of this thread please.

[Tom Booth]

Thanks to all and especially Matt Brown for helping to figure this stuff out with me. I have a lot to add when I get some time.

For now, keep the arguing and personal attacks out of this thread please.

That was not a "personal" attack. "Fool" has completely botched and misrepresented the subject, especially in regard to the Vuilleumier heat pump.

His description is just ignorant nonsense and people should be aware that this "fool" is just spreading misinformation so perhaps they will read the patent themselves and not rely on the ignorant Troll "fool" for any information.

https://patents.google.com/patent/US1275507A/en

[Tom Booth]

Thanks to all and especially Matt Brown ...

I like to at least try and think of you as one of the good people here VincentG, but your frequent backing up and support of these two turd balls who have been targeting me and spamming the forum and derailing topics since their arrival has me wondering if you're not all three working for the same shill mill.

[VincentG]

Tom, everyone here brings something to the table. That is why I can bear your multiple accusations against me(aka Jack). Fool may provide constant resistance to innovation but he has the math(like Arby's has the meat). You may disagree with the math, but it is just you(us) against the established math. So unless you are willing to work with some other people, you will be left bloviating till the end of your days.

Fool, if any new equation can be derived I think it must relate the adiabatic work potential to the thermodynamic work potential.

For instance if 100cc of 300k gas is rapidly made to be 100cc of 600k gas, it can adiabatically expand to "x" volume against 1 bar(no real work). But for out purposes this workless expansion constitutes the energy that may be used to further raise the thermodynamic potential of the gas at a constant volume of 100cc. Maybe this is what you meant by the following?

First, second, and fifth equations are dominant from assigned values. It leaves 3 equations and four unknowns. The fourth equation would be the ideal gas equation for an adiabatic process. Right?

[Fool]

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Vincent G, Thanks. I can't see why anyone here would have any disrespect for you. You are enthusiastic about things here. You are challenging the classical theory as everyone has and should, and you are respectful of others and patient with their errors. Thanks again.

I don't think of myself as being resistant to innovation, I've recommended experiments and materials that night help, or not. I like challenging establishment. However, I think nature resists fantasy. I also think that classical theory, reflects nature. When people use classical theory incorrectly, I try to correct it.

For example:

For instance if 100cc of 300k gas is rapidly made to be 100cc of 600k gas, it can adiabatically expand to "x" volume against 1 bar(no real work). But for out purposes this workless expansion constitutes ...

Any time there is a delta Volume ∆V there is real work involved. W=P•∆V. Force × distance. Adiabatic path puts an extra constraint on the process. It is now possible to calculate the other two variables from a single value and path. A point on the line.


The Proell Effect has two different real work processes on one side increased density, on the other decreased density. The two may tend to cancel each other. The two are not constant volume nor constant mass.

First, second, and fifth equations are dominant from assigned values. It leaves 3 equations and four unknowns. The fourth equation would be the ideal gas equation for an adiabatic process. Right?

Should have read, correction in red:

"First, second, and fifth equations are dependent from assigned values. It leaves 3 independent equations and four unknowns. The fourth equation would be the ideal gas equation for an adiabatic process. Right?"

Linearly dependent equations can't be used to solve for extra variables. It a law of linear mathematics. They must be linearly independent. It just means, in this case, that only three of the equations can be used to determine the four unknown variables. Unfortunately four are necessary. Fortunately the adiabatic path equation is available to be the fourth. No I haven't worked it out so may find a problem. But so far it, to me, looks possible.


It was left for Matt to ponder, mostly.

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