The Carnot efficiency problem

Discussion on Stirling or "hot air" engines (all types)
Tom Booth
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Re: The Carnot efficiency problem

Post by Tom Booth »

The thing he finds most perplexing I think is represented in this passage, describing the consequence of his new theory of heat that otherwise seems to explain all the new experimental findings:
Heat is then the result of a motion.

Then it is plain that it could be produced by the consumption of motive power, and that it could produce this power.

All the other phenomena composition and de-composition of bodies, passage to the gaseous state, specific heat, equilibrium of heat, its more or less easy transmission, its constancy in experiments with the calorimeter could be explained by this hypothesis.

But it would be difficult to explain why, in the development of motive power by heat, a cold body is necessary ; why, in consuming the heat of a warm body, motion cannot be produced
.

Clearly, IMO he is seriously questioning the necessity of a "cold body". In the absence of Caloric theory, he says, if this new theory is accepted: "it would be difficult to explain why, in the development of motive power by heat, a cold body is necessary...."

Clearly, IMO he did not consider this old idea derived from Caloric theory of a cold sink being necessary compatible with the new, emerging kinetic theory.
Fool
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Re: The Carnot efficiency problem

Post by Fool »

https://m.youtube.com/watch?v=GFfMruoRMGo

At 1:17 they turn on the electric motor by moving the lever to their left CCW. They call that direction forward.

A subtle observation can possibly be seen as the whole rig lurches to the right indicating a CW rotation of the whole setup, a reaction countering a left CCW rotation direction acceleration of the machine, and motor, rotor. Torque to the left produces a lurch to the right.

Regardless they call it forward. Makes sense turn the handle CCW for a rotor rotation to the CCW. The lurch is a CW right equal and opposite reaction to a CCW left acceleration of the rotors inertia.

The "head" gets very cold.

At approximately 2:00 in the video they turn off the motor, by returning the lever to center. They claim "the machine will begin to run but in the reverse direction". I take that to mean spinning to the right CW. It acts as a cold gas engine. The cold head begins to warm.

During both modes the other thermal side is surrounded by a set water bath temperature. Heat coming from the head being rejected to the water bath CCW. Or. Heat coming from the water bath being rejected to the cold head CW.

At about 2:12 the switch is moved to the right CW. They claim, "before the cold gas engine stops the electric motor is switched on again to keep the machine running but in the reverse direction". It makes sense that moving the lever to the right CW makes the machine and motor rotor turn right CW, which they call the reverse direction.

At about 3:07 the electric motor is switched off putting the lever back into the center. They claim "the machine now runs in the forward direction as a hot gas engine". Forward by their definition and observation is again CCW to the left. Or opposite to what it was running before it was switched off.

At 3:19 the machine slows down enough to see the flywheel or coupler moving CCW to the left, which they call forward, verifying that direction and description of the other directions as a consistency.

To me it is blatantly obvious what direction the motor and heat are traveling. It is 100% consistent between their words, the video, their theory, and 200 years of steadily improved thermodynamics, 150 for Phillips.

They claim four modes:

1: Refrigerator. CCW Heat moves out of head and rejected to water bath. Negative work energy output. Power input to drive motor. Heat is pumped out of the cold head/Sink. Rejected to warm water bath Tw=Tm. (Tw temperature of the water. Tm temperature middle same temp as water.)

2: Cold heat engine. CW Heat moves out of water bath and rejected to the cold head. Positive work energy output. Measurable power output to drive windage of unpowered motor, a load. Heat is pumped from water bath to cold head. Tw=Tm

3: Heat pump. CW Heat is pumped out of water bath and into hot head. The same a number 2 except the cold heah is too hot to run as a cold motor so is now driven in same direction to keep the heat flowing in the same direction. Negative work energy output. Heat is pumped from cool Tw=Tm water bath to the hot head.

4: Hot gas engine. CCW Heat flows from hot head to cool water bath. Running as a hot gas motor. Heat is pumped from hot head to cool water bath Tw=Tm.

All four could be called heat pumps. The difference is work energy in or out. Out engine. In heat pump. The gas doesn't know the difference. The gas pumps the same quantity of heat if the speed is the same, loaded, free, or driven. The same amount goes to the same places.

I am sorry. I'm sure that was worse than watching the video and figuring it out for yourself.

That video and a phase PV diagram from a real gas are the bases of thermodynamics. If you don't find those two things reliable you will be lost in thermodynamic theory. I only hope that clarification is helpful.
Tom Booth
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Re: The Carnot efficiency problem

Post by Tom Booth »

What I know is a heat pump or refrigerator, air conditioner, air cycle system etc. all work by compression and expansion of a gas, with or without phase change.

Keeping it simple:

Compression = heat
Expansion = cold

My observation of the little LTD engines I've been working with appear to me to be compressing the working fluid when it is at the hot side and expanding it at the cold side.

This IMO indicates heat being "pumped" by being taken in at the cold side while the gas is colder and expanded and then the heat is released when at the hot side when the gas is compressed and hotter than the hot side.

These are my own direct observations.

You say things like: "They claim "the machine will begin to run but in the reverse direction". I take that to mean spinning to the right CW"

That kind of supposition is not conclusive IMO.
Fool
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Re: The Carnot efficiency problem

Post by Fool »

Questioning something is not the same as accepting zero cold sink. Especially during the infancy of thermodynamics 200 years ago.

What impresses me is he appears to understand without the proof that came later with Clausius. Seeing the proof intuitively and writing the proof are two different things.

Reading this website gives me great respect for those that can understand this. Your problem is that you've lost your ability to trust any thermodynamics schooler.

I spend too much time here. As professor Feynman said there will be those that get it. Don't worry about those that don't. I can commiserate with his point. From what I can see, he loved teaching.

Compression is not heat.
Adiabatic compression equals temperature rise.
Adiabatic expansion equals temperature drop.

Temperature difference equals heat transfer, from hotter to colder. The only way to have heat, or heat change.

Adiabatic means/equals zero heat change.

Heat at different temperatures has different quantities of energy.
Tom Booth
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Re: The Carnot efficiency problem

Post by Tom Booth »

Fool wrote: Fri Aug 04, 2023 8:13 pm (...)

Adiabatic means/equals zero heat change.
That is not my understanding.

There can be a heat change as measured by temperature reading due to work input or output during an adiabatic expansion or contraction.

Adiabatic simply means that there is no transfer of energy in or out of the system in the form of heat. But energy can change form.
Tom Booth
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Re: The Carnot efficiency problem

Post by Tom Booth »

Take adiabatic compression for example.

A compression ratio of say 10 to 1 does not result in a simple 10 fold increase in pressure.

It might results in something more like a 25 fold increase in pressure

This is because the "work" of compression increases the temperature and therefore the pressure on top of the same energy being forced to occupy a smaller space. True no Joules were added in the FORM of "heat" but Joules were added in the FORM of work.

Similarly with expansion "work" the cooling effect of expansion is multiplied due to the additional energy loss in the form of "work".

You can quibble over the strict meaning of "heat" supposing that if there is a temperature drop because of work output you still need to remove "heat" to have a pressure drop but that IMO is just semantic nonsense.
Tom Booth
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Re: The Carnot efficiency problem

Post by Tom Booth »

In other words, the underlying reality is that an exchange of "heat" or and exchange of "work" are exactly the same thing on a molecular level.

Transfer of motion or "kinetic energy".
matt brown
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Re: The Carnot efficiency problem

Post by matt brown »

Fool wrote: Fri Aug 04, 2023 8:13 pm
Temperature difference equals heat transfer, from hotter to colder. The only way to have heat, or heat change.
OK, so far...
Fool wrote: Fri Aug 04, 2023 8:13 pm
Adiabatic means/equals zero heat change.
Ouch, walked into that one like a big dog. Bad choice of words and I know the feelin'. This is what happens when using the word "heat" which infected the word change.

Tom, what Fool should have have said is...adiabatic means/equals zero heat added. Without attempting to be pedantic, further note that this is only for an adiabatic expansion (since an adiabatic compression has 'heat' added where heat could mean work input energy or temperature increase.
Fool wrote: Fri Aug 04, 2023 8:13 pm
Heat at different temperatures has different quantities of energy.
Awesome fun fact, Fool is redeemed !!!
matt brown
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Re: The Carnot efficiency problem

Post by matt brown »

Tom Booth wrote: Fri Aug 04, 2023 9:11 pm Take adiabatic compression for example.

A compression ratio of say 10 to 1 does not result in a simple 10 fold increase in pressure.

It might result in something more like a 25 fold increase in pressure
Consider
Vr=volume ratio
Tr=thermal ratio
Pr=pressure ratio

For air (and other diatomic gasses) when
Vr=10
Tr=2.512
Pr=25.12

So, good call. Now notice that Vr(Tr)=Pr
Tom Booth
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Re: The Carnot efficiency problem

Post by Tom Booth »

Fool wrote: Fri Aug 04, 2023 8:13 pm
...Your problem is that you've lost your ability to trust any thermodynamics schooler.

...
That's a real interesting criticism, kind of incongruous in the course of a scientific inquiry.

So now we should accept "the second law" or whatever on FAITH?

That's rich.
matt brown
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Re: The Carnot efficiency problem

Post by matt brown »

Here's my LTD test suggestion...

(1) take LTD with metal hot and cold plates but plastic case
(2) insulate plastic case if desired
(3) place insulated hot water container securely under LTD
(4) construct insulated cold water container securely over LTD with DP and PP tube extensions thru container similar Tom's past vacuum tank setup
(5) carefully measure and monitor hot and cold water values

Does everyone see how this KISS test works ? With this setup, it doesn't matter if there's a 'heat short' thru the engine. Nope, all that matters is how much energy (temperature x mass) disappears between both water containers (when summed) vs work output. If little work is produced, everything is resolved, and it won't matter if some energy escaped to the...ether (or whatever).
Tom Booth
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Re: The Carnot efficiency problem

Post by Tom Booth »

matt brown wrote: Fri Aug 04, 2023 9:55 pm
Fool wrote: Fri Aug 04, 2023 8:13 pm
Temperature difference equals heat transfer, from hotter to colder. The only way to have heat, or heat change.
OK, so far...
Fool wrote: Fri Aug 04, 2023 8:13 pm
Adiabatic means/equals zero heat change.
Ouch, walked into that one like a big dog. Bad choice of words and I know the feelin'. This is what happens when using the word "heat" which infected the word change.

Tom, what Fool should have have said is...adiabatic means/equals zero heat added. Without attempting to be pedantic, further note that this is only for an adiabatic expansion (since an adiabatic compression has 'heat' added where heat could mean work input energy or temperature increase.
Seems like your making a false distinction.

If you can call work input with adiabatic compression "heat" why not work output witb adiabatic expansion?
Fool wrote: Fri Aug 04, 2023 8:13 pm
Heat at different temperatures has different quantities of energy.
Awesome fun fact, Fool is redeemed !!!
"Heat at different temperatures" does not have "different quantities of heat" necessarily.

Neither term (heat or temperature) has anything much to do with "quantities of heat".

Regardless, heat output or input in Joules has the same effect and the same end result as "work" output or input.

Basically two different ways of conceptualizing the same underlying reality, which is a transfer of kinetic energy.
Tom Booth
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Re: The Carnot efficiency problem

Post by Tom Booth »

matt brown wrote: Fri Aug 04, 2023 10:20 pm Here's my LTD test suggestion...

(1) take LTD with metal hot and cold plates but plastic case
(2) insulate plastic case if desired
(3) place insulated hot water container securely under LTD
(4) construct insulated cold water container securely over LTD with DP and PP tube extensions thru container similar Tom's past vacuum tank setup
(5) carefully measure and monitor hot and cold water values

Does everyone see how this KISS test works ? With this setup, it doesn't matter if there's a 'heat short' thru the engine. Nope, all that matters is how much energy (temperature x mass) disappears between both water containers (when summed) vs work output. If little work is produced, everything is resolved, and it won't matter if some energy escaped to the...ether (or whatever).
Looks good to me.

I'd suggest including a variable load on the engine to see what if any influence this might have on how much energy "disappears".

You say: "...vs work output. If little work is produced..."

Do you propose monitoring work output in some way or is that unnecessary?
matt brown
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Re: The Carnot efficiency problem

Post by matt brown »

Yes, some type of variable load and measurement method would be xlnt. The biggest problem is that all values are tiny for common LTD, so a large "model" would be preferred. This setup only simplifies tracking heat in and out, and any thermal conversion to work will include shaft output AND parasite losses like friction and windage (which would drastically increase relative model size). I'll continue scheming in hopes of finding similar simple way to "see" output variations.
Tom Booth
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Re: The Carnot efficiency problem

Post by Tom Booth »

Personally I'm not particularly concerned about measuring all the various possible avenues of heat input converted to output. You have sound/noise squeezing of the bearings, air resistance on the spinning flywheel, friction, angular momentum, "work", parasitic heat losses... Radiation what else?

Too complicated.

Whatever doesn't arrive at the "sink" (water bath) can be assumed to have escaped "elsewhere".

Nearly all these possible "loses" cannot exist as loses until AFTER the heat has already been converted to motion (work).

Moving wheels squeak. Moving bearings and piston create friction, momentum is a property of a moving object, air resistance is resisting motion, etc.

Potential direct loses of heat from the lower water bath can be controlled with insulation to whatever degree possible but would be difficult to quantify.

What I want to know is, does any heat go into the "sink" or not. If it does, is the quantity anywhere near what is predicted by the "efficiency formula/limit"

Theoretically identifiable loses could be eliminated. NASA has done a pretty good job of that with the frictionless FPSE. Magnetic air bearings, frictionless piston clearances etc.

Squeaky wheels can be oiled.

Heat inescapably traveling through the engine to the "cold reservoir" as a "Law of the universe" cannot be resolved by any means.

The question is, is this so-called "law" anything we should really pay any attention to or is it just a hold over from Caloric theory.

Does the cold water bath actually receive any heat from the hot water bath through the engine while it is converting heat into work or "other forms of energy" (squeaks, friction etc.) that we need not be overly concerned with.
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