Thermodynamic work vs. real work

[matt brown]

Hopefully Matt chimes in as this is somewhat similar to that non compression internal combustion engine that blows a weight up and extracts work from the weight falling, of which I can't remember the name.

Vincent - your idea is a "low end" Otto-Langen which used the 3 legged Lenoir cycle

Otto-Langen.png (322.29 KiB) Viewed 1057 times

Otto-Langen etc.png (9.5 KiB) Viewed 1057 times

Here's an unbundled version where both "DP" pistons and cylinders could be a conventional displacer. I made this open cycle per your post with no compression within engine. Anyone who thinks Carnot sucks for compression cycles will be blown away by the paltry eff of any Lenoir cycle. As Fool points out, the path is paramount, and a few quick calcs will illuminate the problem...the massive isochoric input is lost to internal energy change that never translates into much work output. So, in this case, merely nixing typical compression backwork doesn't bring home more bacon (nice try).

Lower part of graphic exposes that even "constant volume" regen is bogus when cold gas pools. This unideal cold gas compression is often referred to as squeeze or pinch and leads to classic regen "flutter".

[Fool]

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Consider a Gamma chamber at 1bar and 300k. At 600k it is allowed to drive out a piston that does work on a crankshaft or directly raising a weight. Now separate the chamber from the power producing system. and is repressurized by the atmosphere itself. The piston is allowed to return back to TDC without any compression work as pressure on both sides is 1 bar. The two systems (chamber and power) are reconnected. Repeat.

This I think would be considered a repetitive open process open cycle. Work is extracted on expansion at high efficiency and there is zero work negative from compression.

I think the devil is in the details. When the displacer, "chamber returns to 300k ", it has less gas in it. That will store less heat than needed to return to 600 K, when there is a full charge of gas. A need of number running is apparent.

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[VincentG]

Thanks Matt that’s the one. The thing is, I think this can be viewed as a compression cycle.

The chamber starts at 300k with a full 1 bar(or more I would argue if using tuned port intakes)charge. Then, raising the temperature to 600k at constant volume, before driving the power piston, is akin to an adiabatic compression cycle of larger volume compressing into the same chamber before expansion at the same 600k.

Fool, not sure what you mean by less gas in the chamber. When the chamber returns to 300k there is an atmospheric driven compression cycle that fills it back up.

[matt brown]

Thanks Matt that’s the one. The thing is, I think this can be viewed as a compression cycle.

Indeed a compression cycle, but where the compression process occurs outside the engine via an "ambient" compression that only taxes heat input.

The chamber starts at 300k with a full 1 bar(or more I would argue if using tuned port intakes)charge. Then, raising the temperature to 600k at constant volume, before driving the power piston, is akin to an adiabatic compression cycle of larger volume compressing into the same chamber before expansion at the same 600k.

Hmmm, consider this...

300-600k helium Lenoir.png (16.29 KiB) Viewed 1007 times

Fool is right and running some numbers will be a major eye opener. I used helium since it has a faster T drop than air and note 450k 'exhaust' is 1/2 input. If only this meant cycle has .50 eff, but that would require all Wpos at 600k vs 600-450k gradient. CalcTool is calling and allows adiabatic processes, just remember to keep an eye on gas choice.

Most cycles can be gamed as 4 single-acting pistons with 180 phasing. Later, things can be optimized such as Green cycle in lower graphic could easily be 2 double-diameter, double-acting pistons, with 180 phasing.

We're on the same page, again...I've been gaming non-compression schemes over the past few weeks as a sidebar.

Fool, not sure what you mean by less gas in the chamber. When the chamber returns to 300k there is an atmospheric driven compression cycle that fills it back up.

I think he missed your open cycle pitch...

[matt brown]

Here's a graphic of a helium 300-600k Lenoir cycle with regen

Lenoir regen.png (14.82 KiB) Viewed 982 times

Shown as closed cycle with multi-bar P, this could also be open cycle with 1 bar ambient Pmin. AFAIK this is the only cycle that can be gamed this way and should be in every thermo book as an example of an adiabatic process shoehorned into a common cycle. Note Q and W values where isobaric Wneg steals 2/3 Wpos whereby this cycle has ideal eff=.33 so Carnot keeps his crown, but not bad for meager 1.5 volume ratio. However, an xlnt example of common isobaric issues where a hidden tax sinks many schemes, especially when an isobaric process is on the compression half of a cycle.

So, if this cycle had 1 bar ambient Pmin then 2/3 of expansion Wpos would be wasted merely pushing ambient pressure, whereby 2/3 of heater input would be wasted during expansion, thereby only 1/3 of heater input would be "available" for useful work output (aka Wnet). This is the sniglet that Tom continually dismisses, insisting that ambient compression is some type of free lunch.

BTW the previous 300-600k Green cycle has eff=.266 with a volume ratio=3 which demonstrates just how crazy chasing these values can become.

[Fool]

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I think he missed your open cycle pitch...

After PP expansion, the DC has less gas than at the beginning of expansion. Some moved into PC.

Valves close.

The DP pushes gas back to the cold side returning heat to the regenerator.

Valves close.

Less gas gets pushed through the regenerator on trip to cold side.

Any heat in PC is exhausted to the atmosphere.

Atmosphere rushes in filling chamber, then DP pushes gas back to hot side.

More gas now goes through the regenerator and the gas doesn't reach TH.

Extra heat must be added to make up for that heat lost.

If a hot piston expansion, isothermal, the leftover heat in the PC is exhausted into the atmosphere. Heat rejection.

If adiabatic expansion, less work will be done and some heat will be exhausted into the atmosphere.

If adiabatic expansion until equal temperature, even less work will be done. And none will be available to fill the regenerator.

I thought all this was demonstrated in the "Let's Beat Up Carnot" thread.

Fool, not sure what you mean by less gas in the chamber. When the chamber returns to 300k there is an atmospheric driven compression cycle that fills it back up.

Yes, but it comes in cold. There is now more gas mass than that which just flowed through the regenerator. More gas by mass, not by volume.

The regenerator didn't get warm enough, so can't heat up the incoming cold air.

Guys, it is harder to show why Carnot can't be beat than come with a scheme, and falsely hope it beats Carnot unless someone else can proves it wrong. The burden of proof is on you schemers. Please look for the errors I bring up.

Carnot has been proven over and over again to be correct. It's best just to accept it, like the Japanese did in the translation of that giant LTD Stirling Engine. And make something practical. If you'd like to keep chasing unicorns and rainbows, please carry on. Just don't accuse me of anything, LOL.

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[matt brown]

Guys, it is harder to show why Carnot can't be beat than come with a scheme, and falsely hope it beats Carnot unless someone else can proves it wrong. The burden of proof is on you schemers. Please look for the errors I bring up.

Common compression cycles prove Carnot via Uhigh vs Ulow cycle constraints that translate into typical Thigh vs Tlow limit buzz, but moving away from common cycles alters these constraints and might beat Carnot.

I've been studying thermo for decades and always favored the ingenuity of designers over the knowledge of academics. Scroll back to my recent Lenoir regen cycle and (1) try to make another such cycle then (2) consider why this is the only such cycle possible. Both cycle and graphic are crude, but they provide an insight into grasping an abstract PVT concept (and yeah, I slightly fudged the adiabatic values for clarity).

I doubt many guys are trying to beat Carnot directly, it's just that 5-10% eff from low thermal ratio pushes interest there indirectly. No one's expecting 100% conversion, just hoping for 50% when Carnot says 20%.

[Fool]

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I am more a designer than a academia. I consider the ingenuity of designers to be comparable to brainstorming. I consider Academia a process of setting the process straight. Nothing kills a project faster than scientific/mathematic fact. If it can't be done, it can't be done. Putting numbers in, allows early comprehension of why it will be a failure. Pursuing things that can be mathematically possible has born more fruit than random hacking and cobbing.together of junk. It wasn't until the Wright Brother's proved with their theories that flight was possible, that they were able to fly. Many people before them cobbed junk together and failed.

To me, design work is like chasing rainbows, theory is like guiding that work towards a bank or realistic goal. What is the the difference between a wild ass guess and a SWAG. Data and theory are used to guide the SWAG. (Scientific Wild Ass Guess.)

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[VincentG]

Guys, it is harder to show why Carnot can't be beat than come with a scheme, and falsely hope it beats Carnot unless someone else can proves it wrong. The burden of proof is on you schemers. Please look for the errors I bring up.

Carnot has been proven over and over again to be correct. It's best just to accept it, like the Japanese did in the translation of that giant LTD Stirling Engine. And make something practical. If you'd like to keep chasing unicorns and rainbows, please carry on. Just don't accuse me of anything, LOL.

I'm reaching the limit of what I can accurately explain with text. The point of this really is to deeply understand the nature of thermodynamic work vs. real work and its relationship to efficiency.

[Fool]

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I still don't get the distinction between "real work" and "thermodynamic work". Can you give an equation? Or method of measuring the two separately?

Work is the transfer of energy.

Energy is stored in different forms. It is transfered by different interactions. It is measured many ways. It is quantified many ways, such as Joules. It is conserved.

When energy is converted to a different form, it often scatters, to more than the expected form. Usually some comes out as heat. Or infrared.

Man has discovered many rules regarding how energy transfer works. Many equations break those rules, from a need to simplify the scientific predictions to within acceptable tolerances. Some of those tolerances are plus-minus, some are just plus, and some are just minus.

The Carnot limit tolerance allows no plus. All real engine will be below it. Minuses.

No one's expecting 100% conversion, just hoping for 50% when Carnot says 20%.

I know you know better than that. That is just wishful thinking. It just as wrong as 100%. Even 20% when Carnot says %20 is out of the realm of reality and reasonability. If you get 10% I would applaud you. 15% would be stellar.

The odd thing I see, that most seekers of high efficiency LTD.Engines, is that typically the fuel is free, sun, scrap wood, etc. so efficiency is moot. Power to cost, isn't.

The Japanese translation showed how they need to be very large for power out. Figuring out how to bring that cost down, and lifespan up is the goal possibly worth exploring.

Carnot made the efficiency a done deal, and severely put a cramp in power to size for LTD heat engines. Sorry, it's just math.


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[VincentG]

I still don't get the distinction between "real work" and "thermodynamic work". Can you give an equation? Or method of measuring the two separately?

An example of pure thermodynamic work is heating a Gamma chamber at constant volume.

An example of real work to thermodynamic work is a fire piston.

An example of real work to real work through a closed gas system with 100 percent conversion is as follows;

Two separate 100lb weights are sitting on two separate pistons that are compressing into the same chamber, let’s say creating an internal pressure of 50psi. They are at the same height. A mechanism is used to lift one piston up 10 feet.

The opposing piston has done real work through the closed internal thermodynamic system to help lift the other piston that was acted upon by external work. But there was no internal thermodynamic work done on the gas.

[Fool]

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I see, now what you are struggling with.

An example of pure thermodynamic work is heating a Gamma chamber at constant volume.

Work originally was defined as force times theú distance moved. W=J=F•d=N•m

It is not the same as heat Q, internal energy U, enthalpy H, nor entropy S. However they all are quantified as energy Joules J.

What you have defined as thermodynamic work is called heat in classical thermodynamics. Heat is the energy transferred by a temperature difference. An equation describing heating a gama chamber is (Matt please correct me if need be.):

∆Q=M•Cv•∆T

An example of real work to thermodynamic work is a fire piston.

Again in classical thermodynamics, that is adiabatic temperature rise. Not heat. At least until the fuel catches fire, then it is an oxygen fuel chemical exothermic release of energy. Adiabatic process for ideal gas, is governed by the following equation:

W=-alpha•M•R•T1(((V2/V1)^(1-gamma))-1)

https://en.m.wikipedia.org/wiki/Adiabatic_process

The opposing piston has done real work through the closed internal thermodynamic system to help lift the other piston that was acted upon by external work. But there was no internal thermodynamic work done on the gas.

Overall system work is zero. The center of mass has not changed. Work done on or by the gas is zero, zero delta Volume. Constant volume. It's similar to the two weights being connected by a rope, or rocking beam. The rope or beam does no work.

If you want the overall system can be broken down to the addition of two systems. One for each weight.

Thermodynamic energy transfer is called heat, not work. A cylinder and piston does work if it pushes with a force, for a distance. We call it pressure times volume change. Or P•∆V=W this is the same as F•d=W

I'm falling asleep. Am I making any sense? Do you want more?


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[matt brown]

Two separate 100lb weights are sitting on two separate pistons that are compressing into the same chamber, let’s say creating an internal pressure of 50psi. They are at the same height. A mechanism is used to lift one piston up 10 feet.

Vincent - I'm assuming you mean something like this...

res w 2 pistons.png (4.41 KiB) Viewed 895 times

[Fool]

That's what I think. Or just a U-tube. Similar to, two blocks (pulleys) and a line (rope) from above.

[VincentG]

Yes Matt just like that. You’re always pulling a rabbit out.