LTD magnetic vs gamma

Discussion on Stirling or "hot air" engines (all types)
Tom Booth
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Re: LTD magnetic vs gamma

Post by Tom Booth »

I honestly did not know if this simple motor was really possible, so tried building one.

https://youtu.be/5P6OV7_u8m4

What I don't really understand about it is that; I thought inductive repulsion was due to a CHANGING magnetic field.

Here, it would appear that there is simply a short circuit in the wire from + to - without change or interruption. For example, all the examples I've seen of two wires repelling or attracting each other had to do with AC current. Or so I thought.

I guess I was mostly looking at AC because we are talking about an oscillating circuit.

I think I'm a little confused.
Tom Booth
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Re: LTD magnetic vs gamma

Post by Tom Booth »

There is, of course, the couple of rather powerful neodymium magnets the battery is sitting on.

In one way this avenue of research seems pointless to me, in that it takes almost no energy at all to move a displacer, usually, simple straightforward mechanical energy. So why bother with some kind of circuitry to move the displacer?

The reason I find this a very interesting possibility is that for a long time I've felt that the main advantage an IC engine has over an external combustion engine is a means of adjusting the ignition timing very precisely to maximize performance with changing conditions.

An IC engine has a circuit that controls a spark that determines the moment when heat is generated.

Perhaps if a Stirling engine had a circuit of some sort or other controlling the displacer action a similar result could be achieved and Stirling engine performance might be improved dramatically for certain applications where speeds and loads are variable. Like a lawnmower hitting a tall patch of grass. The timing on a typical small IC engine BTW, like a push mower has fixed timing, which is probably why it bogs down and often stalls and has to be restarted when it hits such a sudden heavy load, like a patch of tall grass.

An automobile, on the other hand, does not stall going up hills. Not so easily anyway. It adjusts to the load.

So if Stirling engines could be controlled electronically, they could also be able to adjust to sudden changing loads.

In theory, the bit of loss that might result from relying on a circuit, instead of direct mechanical control, might be more than compensated for by the general improvement in performance, and it could potentially open up some uses for Stirling engines that may otherwise seem to be out of the question.
omblauman
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Re: LTD magnetic vs gamma

Post by omblauman »

I understand we have a completely different way to approach this design, in itself a stimulating situation.
What i have in mind is an engine which is based on a conventional, Amazon type, LTD , where the displacer is a foam disk which goes up and down, with only the fluid friction to overcome on the mechanical dissipation side. A pendulum version is a very good alternative suggestion, it gets rid of the stability of the displacer issue at a very small efficiency cost.
All other losses are electrical resistance losses which I find easier to predict and measure with fantastic accuracy than to test. It takes nevertheless some work to do it. We know of small, simple minded, LTDs which work quite efficiently on small deltaTs, so forget about the power extraction side, which is conventional and efficient, and concentrate on the EM driven displacer. Apart from the above mentioned friction all the losses are due to at most 2 currents which obey equations simple for me to integrate. The need to integrate a differential eq system arises from the moving displacer, a fixed circuit would not need it.
The advantage of treating the problem theoretically is that it's much more flexible and insightful than experiments. If you do the test and nothing moves nothing is achieved. If we calculate we can easily change parameters until it works and we have learned something. Plus all experimental work can be checked in detail and the difficult to predict mechanical losses will become measurable.
So this is my proposal for the first crucial half of the engine project:
1) design a chamber-displacer unit very similarly to what already exist and works efficiently ( = small delta T) in standard LTDs
2) design an EM scheme, or more than one, which seems promising on paper (if we can't find something similar on the web even better, we will be more motivated to insist)
3) verify EM parameters , EM losses in particular because those are the ones the engine limited generating power has to overcome
4) if, as I believe, the theory predictions are favorable, build the chamber-displacer unit only and verify the predictions. We don't need the power generating side of the engine, the hardest to build, because it's easier to supply the circuitry power with electronic means.
5) at the end of the process one can think of designing a whole engine
Tom Booth
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Re: LTD magnetic vs gamma

Post by Tom Booth »

Right, sounds logical. I know zilch about integral calculus though.

My thinking, however, is, a limiting factor is the current that an LTD Stirling pushing a magnetic linear AC generator can produce.

If Such a small model engine can't actually generate enough current through any kind of wire or coil to charge any kind of capacitor or influence a scrap piece of aluminum foil, then there's not much hope.

I have a half dozen LTD Stirling engines kicking around, some of which have already been taken apart and modified and experimented with rather extensively.

Adding a magnet and coil and such to one of these seems like the most direct avenue.

With some form of variable capacitor and magnet wire, magnets etc. in hand, dozens of configurations could be tested in a relatively short time.

While waiting for the flashlights to arrive, 've been looking at different ways of making variable capacitors:

https://youtu.be/bOYNWXOKmA8

https://youtu.be/w3iLnN3SOXo

https://youtu.be/JOaVpnG_pQ4

https://youtu.be/3CnmuIhgn4g

Or, maybe join a bunch together if just one doesn't manage to move anything, like a marx generator?

https://youtu.be/dje7uhyW23o

Anyway, fun times.
omblauman
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Re: LTD magnetic vs gamma

Post by omblauman »

"If Such a small model engine can't actually generate enough current through any kind of wire or coil to charge any kind of capacitor or influence a scrap piece of aluminum foil, then there's not much hope."

What matters is power not current, in electricity one can always transform power very efficiently to the current-voltage form one needs, so forget about current. The capacitors we are thinking about are real high energy density, every day commercial capacitors, nobody is thinking of building one, especially for the ultra low frequencies we need. The issue here is the efficiency (mechanical energy out / EM energy in) of the displacer actuator.
Tom Booth
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Re: LTD magnetic vs gamma

Post by Tom Booth »

Power-voltage-current-watts, whatever.

A typical LTD model engine from Amazon or wherever will begin to struggle if the flywheel is brushed with a feather. The slightest friction/resistance will bring it to a screeching halt. The basic challenge is often to manage to put it together in a way, such that it can overcome its own friction enough to run, so getting one to put out any useable electrical power at all would be a major hurdle however one looks at it.

I am not able to find any examples online of anyone having successfully done so. That is, getting an off the shelf LTD to so much as power an LED or something.
omblauman
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Re: LTD magnetic vs gamma

Post by omblauman »

Tom, I didn't understand you want to make a generator.
My interest would have been to design an engine where the the inertia to complete the S cycle would have been provided by an RC oscillator and not by a flywheel. Since dissipation in electrical devices can only be ohmic (in absence of chemical reactions and EM radiation) if one can keep ohmic losses down enough it would run, stop. With superconductors it would certainly work.
Did I put a limit on delt T? Even if it would require a one kW kettle heater to run instead of the warmth of a hand it would be a success, it would be a starting point and a demonstration. Especially if consistent with a theory model with simple assumptions.
Actually the use of a well insulated kettle water heater is a good idea so one can accurately measure the input power.
My interest is just to learn something interesting, I don't think the world will decarbonize with stirling engines.
And if it's not already on the web, one more reasons to do it.
omblauman
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Re: LTD magnetic vs gamma

Post by omblauman »

The engine power balance is as follows:
Pd + Po = Pg where d stands for displacer, o for oscillator and g for generator
rewritten in terms of "per cycle" energies
Wd * EFFd + Eo/Q = Ws EFFg
where Wd is the mechanical energy expended by the displacer, EFFd is the efficiency of the electrical device which drives the displacer, Eo is the electrical oscillator energy, Q is the inverse of the per cycle losses, Ws is the Stirling engine energy out, and EFFg the generator efficiency.
Ws is for example the area of the figure in https://www.youtube.com/watch?v=dvomod6 ... ex=26&t=1s
I introduced Q because it's customary to write oscillator losses in this way and it's already calculated in all books, Q= 1/R *SQRt(L/C) .
EFFg is the generator efficiency and must come down to somewhere in between 20% and 80%
Wd is a fraction of your LTD feather energy. As you can see the only real question mark is EFFd, which depends on the design of the only original component in this deal. It is what I wanted to calculate with the spreadsheet, to measure with the displacer-all-by-itself experiment and which would tell us if the LTD with the electric flywheel would work or not.
This quantitatively summarizes the logic behind the 1 to 5 points.
Tom Booth
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Re: LTD magnetic vs gamma

Post by Tom Booth »

omblauman wrote: Fri Apr 23, 2021 1:28 pm Tom, I didn't understand you want to make a generator.
My interest would have been to design an engine where the the inertia to complete the S cycle would have been provided by an RC oscillator and not by a flywheel. Since dissipation in electrical devices can only be ohmic (in absence of chemical reactions and EM radiation) if one can keep ohmic losses down enough it would run, stop. With superconductors it would certainly work.
Did I put a limit on delt T? Even if it would require a one kW kettle heater to run instead of the warmth of a hand it would be a success, it would be a starting point and a demonstration. Especially if consistent with a theory model with simple assumptions.
Actually the use of a well insulated kettle water heater is a good idea so one can accurately measure the input power.
My interest is just to learn something interesting, I don't think the world will decarbonize with stirling engines.
And if it's not already on the web, one more reasons to do it.
I'm afraid I'm, perhaps not following you,. Or understanding what exactly you are proposing to begin with. You say "I didn't understand you want to make a generator."

Well, what powers the RC circuit? I would assume depending on superconductors for this project is out of the question, so resistance in a real circuit is, I would think, inevitable and resistance would, from what I've been able to glean on the subject, bring the oscillations, however they got going, to a stop in some fractions of a second. So, presumably, there would need to be some electrical input into the circuit to maintain the oscillation. No?

Or are you suggesting that an RC circuit might lift a displacer, repeatedly, without loses or need for recharging. I'm fairly certain that would be impossible, even with superconductors.

Theoretically, I suppose, starting with a charged capacitor, the displacer could lift up, and possibly the energy to do the lifting could be put back into the circuit in some way as the displacer falls back down, but again, some loses are inevitable. For the engine to continue to run, something needs to be making up for the loses.

Generally, an oscillating circuit is sustained by some AC power source, judging by the information on the subject I've seen so far. So, logic would seem to dictate some portion of the engine power output would have to be devoted to providing the AC supply.

In other words, I'm presupposing that the RC oscillator requires a supply current that would have to come from a generator. So that the Stirling engine has to include a generator for this scheme to work, seemed to me to be a given.

Just like the engine has to put energy into a flywheel, it would also have to put energy into the circuit for the circuit to function as a replacement for a flywheel.

Somehow the engine has to be able to convert heat into an electric charge to power the circuit. How else could it do that other than with some sort of generator?

The heat differential can be as high as anyone likes, in theory, but if the heat is not converted to electrical AC with some form of generator, then what is to power the circuit?
Tom Booth
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Re: LTD magnetic vs gamma

Post by Tom Booth »

Also, you do mention: "dissipation in electrical devices can only be ohmic (in absence of chemical reactions and EM radiation) "

Is not what is being proposed, to lift the displacer with an electro-motive force.produced by the current flowing through a loop of wire or coil in the circuit?

I think that takes power out of the circuit which would need to be continually replaced to maintain the oscillation.

With a superconductor, theoretically, energy could be stored indefinitely or without limit, like a flywheel of infinite size or whatever, but. if that energy is used, taken out, by lifting something, it will still be diminished.

The heat generated, which from what I've seen previously in examples of inductive levitation, can be considerable, is also a form of EM radiation in the infrared band.

https://youtu.be/tjD9I95RAbw

In the above video (crystal radio), the antenna coil acts as a generator to maintain the oscillations.

In the proposed Stirling engine, the oscillations., The cycles per second in the circuit, would have to match the RPM (or, in a linear engine, the frequency) of the engine (that is, the coil acting as a generator surrounding the piston, or, associated in some way with the movement of the piston/magnet)

You may have something altogether different in mind, I don't know.
Tom Booth
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Re: LTD magnetic vs gamma

Post by Tom Booth »

The "generator" I'm imagining or thinking of, would be, similar to the crystal radio circuit in that video, something to make up for loses (instead of the radio waves).

Something like this:
Screenshot_20210424-062943_crop_38_resize_46.jpg
Screenshot_20210424-062943_crop_38_resize_46.jpg (68.57 KiB) Viewed 5045 times
I've taken a screenshot from the video. Instead of radio waves supplying power to the antenna coil I imagine a magnetic piston supplying the power as it moves through a coil.

Same principle, or essentially, same kind of circuit but with a different source of energy to make up for loses and maintain the oscillation, that keeps the displacer moving which keeps the engine running.
Tom Booth
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Re: LTD magnetic vs gamma

Post by Tom Booth »

omblauman wrote: Fri Apr 23, 2021 2:21 pm The engine power balance is as follows:
Pd + Po = Pg where d stands for displacer, o for oscillator and g for generator
rewritten in terms of "per cycle" energies
Wd * EFFd + Eo/Q = Ws EFFg
where Wd is the mechanical energy expended by the displacer, EFFd is the efficiency of the electrical device which drives the displacer, Eo is the electrical oscillator energy, Q is the inverse of the per cycle losses, Ws is the Stirling engine energy out, and EFFg the generator efficiency.
Ws is for example the area of the figure in https://www.youtube.com/watch?v=dvomod6 ... ex=26&t=1s
I introduced Q because it's customary to write oscillator losses in this way and it's already calculated in all books, Q= 1/R *SQRt(L/C) .
EFFg is the generator efficiency and must come down to somewhere in between 20% and 80%
Wd is a fraction of your LTD feather energy. As you can see the only real question mark is EFFd, which depends on the design of the only original component in this deal. It is what I wanted to calculate with the spreadsheet, to measure with the displacer-all-by-itself experiment and which would tell us if the LTD with the electric flywheel would work or not.
This quantitatively summarizes the logic behind the 1 to 5 points.
I'm not sure we have the same idea of what the goal is here.

Taking the first part: "The engine power balance is as follows: Pd + Po = Pg where d stands for displacer..."

P, I assume, stands for power? So if d is displacer, is this suppose to represent the power output of the displacer? Power consumption?

"o for oscillator and g for generator"

So is Po the power input to the oscillator, the power stored in the oscillator, or what exactly?

"rewritten in terms of "per cycle" energies
Wd * EFFd + Eo/Q = Ws EFFg
where Wd is the mechanical energy expended by the displacer,"

So then Pd = Wd?

"EFFd is the efficiency of the electrical device which drives the displacer"

What device is that? Isn't that supposed to be the oscillator? Or the inductor loop or coil through which the oscillation travels?

"Eo is the electrical oscillator energy" The same as EFFd?

"Q is the inverse of the per cycle losses" No real idea what you mean there.

"Ws is the Stirling engine energy out"

As far as I'm aware, we are mainly considering a small model engine that has no effective output. No load.

" and EFFg the generator efficiency."

I was thinking, as far as any generator, just a coil and magnet somewhere synchronized with the oscillator or actually generating the electrical oscillation in the circuit.

"As you can see the only real question mark is EFFd,.."

I think there are a number of additional question marks, from my perspective.

"which depends on the design of the only original component in this deal. It is what I wanted to calculate with the spreadsheet, to measure with the displacer-all-by-itself experiment and which would tell us if the LTD with the electric flywheel would work or not."

I don't have any real clue or idea what exactly you're after here.

What do you mean by "the displacer all by itself".

A displacer could be virtually anything. A disk of cardboard, or styrofoam or balsa wood. A tin can, some steel wool, a paper and tape box, or in this proposed device, any of the above, perhaps lined with aluminum foil, like foil faced styrofoam insulation perhaps.

If that is determined. What is needed to make this calculation?

The weight of the displacer?

That too could be anything or virtually nothing, in the case of magnetic suspension, or hanging the displacer, or otherwise balancing or counterbalancing it in some way.

A displacer could also be a diaphragm. Or in some engines the displacer might not exist at all.

In my mind the actual displacer is a wild variable as far as size, weight, material, porosity, air flow, inertia, power consumption (if any) etc. Not really a basis for any kind of calculation.

In general a displacer, as I mentioned before, takes, or consumes a negligible amount of energy, or, theoretically, no energy at all.

It is also an unlikely producer of any power, even if lifted and allowed to fall, say, through a coil, as it took almost no energy to lift in the first place, it has no real energy to give back in falling down. It just kind of floats.

Or take this rotary displacer, revolving on a pin point.

https://youtu.be/WWEEqvnjbfI

The energy required to set it in motion and to keep it going is so small as to be entirely negligible, or virtually non-existent.

As far as I can figure, taking the first equation above:

"The engine power balance is as follows:
Pd + Po = Pg where d stands for displacer, o for oscillator and g for generator"

Pd is, if anything, negative.

Po is also negative. The oscillator requires power input to make up for loses in the circuit.

So Pg is also negative.

But this appears to neglect Ph, or however one might wish to designate heat/power input, which ultimately has to supply the power for the other three.
Tom Booth
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Re: LTD magnetic vs gamma

Post by Tom Booth »

Anyway, the various "Shake" flashlights I sent away for are beginning to arrive.
IMG_20210429_153212534.jpg
IMG_20210429_153212534.jpg (61.24 KiB) Viewed 5019 times
When the rest come in, I may do some sort of video review of their relative worth and suitability or usefulness, or lack thereof, for this sort of project.

I can say, avoid this one:
Screenshot_20210429-154141.png
Screenshot_20210429-154141.png (448.51 KiB) Viewed 5020 times
designated "seachoice 8151" or 08151. from various vendors on Amazon.

The illustration may be different. Black, for instance.
IMG_20210429_160003562.jpg
IMG_20210429_160003562.jpg (267.17 KiB) Viewed 5020 times
Such as I actually received, but which was neither blue, nor any sort of "shake" powered light at all. (The black one is also called a "shake flashlight" that never needs batteries, but may also state "batteries included".

Though the advertising may state otherwise "order now, only 15 left in stock" etc., and the horrendous price may cause one to think it might be better than the others, in spite of the fact that it appears to be a typical "fake-light" that can be purchased in bulk for about 50¢ each, it is actually, or so I was told "out of stock" and I received the ordinary battery operated flashlight pictured above instead, which I'll likely be returning.

I would probably not bother mentioning this, except that I have alerted Amazon of all this, but regardless, this "out of stock" fake-light continues to be advertised. Therefore a warning to others seems more than justified.

Ironically, the replacement flashlight is likely a better value, if someone wants an ordinary battery powered flashlight, but serves no purpose here.
Tom Booth
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Re: LTD magnetic vs gamma

Post by Tom Booth »

In line at the checkout at Walmart, I saw this wireless phone charger:
IMG_20210429_163919946.jpg
IMG_20210429_163919946.jpg (218.17 KiB) Viewed 5018 times
Which, as I suspected, apparently contains what I assume to be some sort of low voltage induction coil:
IMG_20210429_163919946_20210429164230446.jpg
IMG_20210429_163919946_20210429164230446.jpg (143.43 KiB) Viewed 5018 times
That I thought might be useful for this project.
omblauman
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Re: LTD magnetic vs gamma

Post by omblauman »

P, I assume, stands for power? So if d is displacer, is this suppose to represent the power output of the displacer? Power consumption?
The equation represents the engine power balance, left side the losses, right side the gains

rewritten in terms of "per cycle" energies
Wd * EFFd + Eo/Q = Ws EFFg
where Wd is the mechanical energy expended by the displacer,
So then Pd = Wd?


one is the average power, the other is the energy per cycle, a power can not be equal to an energy

EFFd is the efficiency of the electrical device which drives the displacer
What device is that? Isn't that supposed to be the oscillator? Or the inductor loop or coil through which the oscillation travels?


The efficiencies in this context are always the ratios of mechanical to electrical power, or energy per cycle. Electrical energy in the displacer to mechanical energy out, or vice versa for the generator. There are 3 parts to the system: the displacer, the oscillator, the generator, all with electrical and mechanical characteristics. Well no, the oscillator is just electrical.

"Q is the inverse of the per cycle losses" No real idea what you mean there.
That's why I define it and give its expression in term electrical circuit parameters.
Wikipedia has more, this circuit is as simple as it gets for electrical oscillators but one needs to study it.

"Ws is the Stirling engine energy out"
As far as I'm aware, we are mainly considering a small model engine that has no effective output. No load.

In our case all the power goes to drive other parts, like the displacer and the currents in the circuits.

" and EFFg the generator efficiency."
I was thinking, as far as any generator, just a coil and magnet somewhere synchronized with the oscillator or actually generating the electrical oscillation in the circuit.


like any generator has an efficiency which depends on how well one designs and executes it.

"As you can see the only real question mark is EFFd,.."
I think there are a number of additional question marks, from my perspective.

I don't think so as far as the ability of the engine to run, if the power balance is satisfied it runs

"which depends on the design of the only original component in this deal. It is what I wanted to calculate with the spreadsheet, to measure with the displacer-all-by-itself experiment and which would tell us if the LTD with the electric flywheel would work or not."
I don't have any real clue or idea what exactly you're after here.

What I am saying is that the real question mark in this engine, the least predictable component, is the displacer and its electrical driver.
So this is really the part of the engine one should experiment with first, in the simplest and most diagnosable way. And we can do it: we can build a displacer with its circuitry and electrically power it with an electrical oscillator, which would be very easy to acquire and easy to measure its characteristics

What do you mean by "the displacer all by itself".
A displacer could be virtually anything. A disk of cardboard, or styrofoam or balsa wood. A tin can, some steel wool, a paper and tape box, or in this proposed device, any of the above, perhaps lined with aluminum foil, like foil faced styrofoam insulation perhaps.

Exactly, any would probably do but I would start with a simple Styrofoam disc, like most LTDs, with all the coils, the ones inside and outside the vessel. Or magnets, depends on the design one chooses. Even a mechanical AC generator, a brushless motor connected as a generator and driven by another motor would work instead of an electronic amplifier. Either way we can easily produce all the power we need.

If that is determined. What is needed to make this calculation?
Nothing is needed to make the calculation but being an original calculation needs to be checked with the mentioned experiment.

The weight of the displacer?
That too could be anything or virtually nothing, in the case of magnetic suspension, or hanging the displacer, or otherwise balancing or counterbalancing it in some way.

The displacer should be the lightest possible, all other characteristics should be determined to make it as efficient as possible, at least in theory.

A displacer could also be a diaphragm. Or in some engines the displacer might not exist at all.
it can't be a diaphragm, it needs to let air go around it. To demonstrate that it works I wouldn't go to less efficient design than a traditional displacer.

In my mind the actual displacer is a wild variable as far as size, weight, material, porosity, air flow, inertia, power consumption (if any) etc. Not really a basis for any kind of calculation.
As I said I would start with something which works, a disk of Styrofoam so that we know size, material, weight, porosity, air flow, inertia, plus the circuitry and measure power consumption. Then we might not like the results but we would know how much electrical power the displacer needs.

In general a displacer, as I mentioned before, takes, or consumes a negligible amount of energy, or, theoretically, no energy at all. some mechanical losses but we are really after EFFd.

It is also an unlikely producer of any power, even if lifted and allowed to fall, say, through a coil, as it took almost no energy to lift in the first place, it has no real energy to give back in falling down. It just kind of floats.
that's right, in fact it only appears at the left of the power balance equation.


As far as I can figure, taking the first equation above:

"The engine power balance is as follows:
Pd + Po = Pg where d stands for displacer, o for oscillator and g for generator"

Pd is, if anything, negative.

Po is also negative. The oscillator requires power input to make up for loses in the circuit.

So Pg is also negative.

But this appears to neglect Ph, or however one might wish to designate heat/power input, which ultimately has to supply the power for the other three.

you can decide to write the power balance eq with terms at either side, as I did, or with terms of either sign but on the same side of the equal sign.
-Pd-Po+Pg=0 Up to you, they are equivalent.
I don't neglect Ph, it appears in the energy per cycle eq as Ws, the area in the PV diagram of the youtube example.
deltaP * deltaV is an energy and is equal to Ws. If you are not familiar with this I suggest you try to understand well the youtube example, it's really fundamental to quantify and understand how a thermal energy works. And it's a really beautiful experiment, maybe we should replicate that one first since the diagnostic apparatus needed in our experiment is virtually identical. There is another equivalent one on youtube https://www.youtube.com/watch?v=VcULP9Y ... ex=71&t=9s. but the first one I mentioned is simpler and more elegant, both are worth understanding thoroughly.
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