The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Discussion on Stirling or "hot air" engines (all types)
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

Leaving out the math as it does not c/p easily, here are a few additional sample problems, just so I can't be accused of "cherry picking"
Question

A heat engine is supplied with 250 kJ/s of heat at a constant fixed temperature of
227°C; the heat is rejected at 27°C, the cycle is reversible, then what amount of heat is rejected?

Solution
The correct option is C 150kJ/s
...
https://byjus.com/question-answer/a-hea ... ant-fixed/

Question
A Carnot engine operates with a source at
500K and sink at 375K. If the engine consumes 600kcal of heat in one cycle, the heat rejected to the sink per cycle is

Solution
The correct option is B 450kcal
https://byjus.com/question-answer/a-car ... -at-375-k/

More examples from this site can be found by scrolling down to "similar questions"


Similar examples from other sources:
advanced physics
questions and answers
example 6.11.

a carnot's engine is operated between two reservoirs at temperatures of 450 k and 350 k. if the engine receives 1000 calories of heat from the source in each cycle, calculate the amount of heat rejected to the sink in each cycle. ...
https://www.chegg.com/homework-help/que ... q119365781

There are, of course many more.

A running engine is obviously taking in some heat to operate.

At a typical 10 or 20% efficiency five or ten times more heat than the engine needs to operate must be "rejected'.

Again, your criticisms above apper disingenuous.

Obviously the heat "rejected" by a running engine supplied with enough heat to operate should at least be measurable, should never be zero, and the temperature of the sink should should certainly not be at or below ambient if the Carnot Limit equation were valid.

Again, if you or anyone thinks my results are flawed, anyone can do their own experiments. If they get different results, wonderful.

I see no point in drawing this out. If you have something to prove, do your own experiments.

Mine are mostly all recorded on video and described in detail easy and relatively inexpensive experiments to replicate.

I have nothing to prove. I just reported my results as objectively as possible, without much personal bias or interpretation on my part. By Video.

Anyone can watch the videos and draw their own conclusions or repeat the experiments and see what results they get themselves.
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

Anyway Mr fool,

I did not make videos of my experiments to prove anything or convince or persuade anyone of anything.

I'm just recording and sharing these experiments for anyone who might be interested.

My focus is on determining for myself how these engines work. I have quite a lot on the line at this point. Just paying the utility bills and property taxes on my workshop is a rather heavy burden. Thousands of dollars. Add to that metal working equipment materials etc. etc. I have a lot invested.

My goal is to have a small cottage business building Stirling engines that are efficient and powerful enough to be practical.

I have no interest in deceiving myself.

My videos are just the tip of the iceberg. In most cases I've repeated these experiments over and over. Whenever or wherever I've discovered some error I've posted a retraction or correction.

I'm not trying to push any agenda, but you certainly seem to be.
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

As far as measurements, I have recorded the values deemed necessary or required by the Carnot efficiency theory itself.

According to the Carnot Limit efficiency is based only on the temperature difference.

I don't have the means to carry out experiments with impossible levels or accuracy using expensive instruments.

I've had to use common sense.

If a metal bottom engine sits over boiling water for hours the bottom is going to get hot.

Air inside the engine will pick up heat.

If the engine is running, obviously it is taking in some heat to run.

You are making what are. IMO wild, irrational suppositions to save the baseless Carnot formula and dismiss experimental results that don't align with your beliefs and preconceptions.

You imagine no heat is getting into the engine. The engine is doing "zero" work etc.

All I can say is do your own experiments. I may do more from time to time out of my own curiosity, but I've done enough.

Enough to draw my own conclusions. You are, of course free to do or think whatever you like. I don't care.
Fool
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Fool »

The temperatures are needed, inside the engine, to calculate the Carnot predicted efficiency. Additional temperatures outside the engine are needed to calculate the heat energy flow. The shaft work measurement is also needed to calculate the total efficiency. The total efficiency can then be compared to the projected Carnot efficiency. Only then can a conclusion be reported about efficiency, and Carnot validity.

An alternate method would be to measure the work by the gas only. Shaft work and total efficiency, will be lower. It can be measured from an indicator diagram. All these can be done and calibrated fairly cheaply.


Your agenda sure seems to be fooling yourself, and slamming: the second law, Carnot, Kelvin, ett all, and promoting obscure Tesla writings. But if the following truly is your goal, I'm merely here to correct your apparent agenda, the following being admirable:
Tom Booth wrote:My goal is to have a small cottage business building Stirling engines that are efficient and powerful enough to be practical.
I wish you the best, as if successful I would entertain the possibility of purchasing one from you. By that time you will, of course, need to measure the efficiency and work output to be able to provide that for your customers. That will also provide the missing information from all your experiments to date, and my recommendations will be moot.

If you make claims that you can beat Carnot, accept money because of those claims, and fail to deliver, you open yourself up to potential wire fraud charges, and suits. Most people that are second law deniers get away with it because they sell nothing and promise nothing. Please be very careful.

I'm not here to stop you, just here to help talk about it safely. IMHO, it is very difficult to do "Good Science" and even harder to talk about it "correctly". I know you have the opposite opinion on that, but your videos, and others, have convinced me otherwise. I am apologizing that this is tough to swallow.

Good luck, and carry on. Please.

P.S., If you are interested in constructing a simple, cheap, and accurate, dynamometer and a PV indicator device, I think I could help. The simplest would be a: paper hinge, ruler, and spring/ bungee. I'd be willing to swap phone numbers. I have lots to do on the farm here, so this is a very gracious offer.
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

You and Matt have quite the double Standard.

The Carnot Limit, according to you two needs no experimental verification.

My simple experiments however must be repeated endlessly, with ever more and more stringent requirements imposed.
I'd be willing to swap phone numbers
You've got to be kidding. No thanks.

I get quite enough of you hounding the hell out of me here. That's the last thing I need. Please leave me alone and mind your own business.
matt brown
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by matt brown »

Tom Booth wrote: Sun Jun 09, 2024 11:23 am You and Matt have quite the double Standard.

The Carnot Limit, according to you two needs no experimental verification.
An ideal Stirling cycle engine will adhere to Carnot Theorem. With all the studies on these buggers which support Carnot (among other stuff) just what type of experiment do you fantasize is required ???

You claim to have been a small engine mechanic. Hmmm, didn't you ever notice how high compression effects both output and efficiency. What I can't figure out is how anyone can fantasize any LTD as a credible engine when it lacks a credible volume ratio. By now, we all get your voodoo cycle with ambient compression, heat input, and expansion wherein Mother Nature supplies 'backwork' at no tax to expansion, and all input becomes work output. Yet, even this voodoo cycle will succumb to a volume ratio requirement AND input will be limited by the heat capacity of the working gas.

You have chosen to disregard most thermo science simply because it doesn't fit your agenda. Your grasp of engines and HVAC is lame, but your grasp of cryo sucks (probably due to poor grasp of PV=nRT). A couple weeks ago, I spent a week of evenings reading your Tesla ambient engine thread which runs for 12 years. Across this thread, you chase free energy before harping on Carnot, but the proof is in the pudding, so where's yours ??? You preach your version on everything and become defiant, defensive or hostile when anyone challenges or corrects you. Gadz, you must be desperate...
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

matt brown wrote: Sun Jun 09, 2024 4:28 pm
Tom Booth wrote: Sun Jun 09, 2024 11:23 am You and Matt have quite the double Standard.

The Carnot Limit, according to you two needs no experimental verification.
An ideal Stirling cycle engine will adhere to Carnot Theorem. With all the studies on these buggers which support Carnot (among other stuff) just what type of experiment do you fantasize is required ???
All the various sample problems make the claim that a typical Stirling engine of 20% or so Carnot efficiency MUST "reject" to the "sink" presumably, (or "cold reservoir") at least 5 times more energy as heat than it produces or converts to mechanical work.

Given all the colleges and universities and high tech research facilities like MIT preaching the Carnot Limit doctrine it seems to me verifying that claim experimentally sometime within the past 200 years shouldn't be, or shouldn't have been that difficult.

Count the RPM, figure out the joules of work the engine is doing to keep the flywheel spinning, maybe add a variable work load. Then measure the joules of heat dumped into the sink. Should be 5X higher at least Right?

I've done that basic experiment in my basement, and at my kitchen table, and at my workshop

Further I figure if all that "waste heat" absolutely MUST be gotten rid of, I would have thought, and so did some other people on the forum, that if the heat rejection was blocked, then the engine wouldn't be able to run.

Very simple common sense experiments.

If 5X more heat is sent straight through the engine to the sink, what about an engine running on ice? Shouldn't the ice melt rather than re-freez?

I've done all these experiments and more, simply because in my early research starting more than ten years ago, it seemed nobody had ever bothered. 200 years and not one record of any such basic common sense experiments.

As I said many times, someone should have done all these experiments at least 100 years ago. Why not? Why the resistance now?
Fool wrote: Sat Jun 08, 2024 10:20 pm Do you really want to demand proof beyond proof.

Demanding impossible proof is one of the 5 reasoning errors of science deniers.
Seriously?

"Impossible proof"? Are you kidding me?

I've devised and conducted dozens of such experiments. Certainly any well equipped university could do any of my experiments MUCH much better, with the most accurate instrumentation and controls.

How about ANYTHING???

Any experiment at all ?

I thought maybe someone connected with the Science and Physics forums could follow up on something. Replicate one or more of my experiments.

Nope. They'd rather ban me.

Oh well I guess I'll just have to carry on as best I can with my very limited time and resources.
You claim to have been a small engine mechanic. Hmmm, didn't you ever notice how high compression effects both output and efficiency. What I can't figure out is how anyone can fantasize any LTD as a credible engine when it lacks a credible volume ratio. By now, we all get your voodoo cycle with ambient compression, heat input, and expansion wherein Mother Nature supplies 'backwork' at no tax to expansion, and all input becomes work output. Yet, even this voodoo cycle will succumb to a volume ratio requirement AND input will be limited by the heat capacity of the working gas.

You have chosen to disregard most thermo science simply because it doesn't fit your agenda. Your grasp of engines and HVAC is lame, but your grasp of cryo sucks (probably due to poor grasp of PV=nRT). A couple weeks ago, I spent a week of evenings reading your Tesla ambient engine thread which runs for 12 years. Across this thread, you chase free energy before harping on Carnot, but the proof is in the pudding, so where's yours ??? You preach your version on everything and become defiant, defensive or hostile when anyone challenges or corrects you. Gadz, you must be desperate...
I just put up some videos.

It's you guys that come at me spitting venom.

The Science forums went through all kinds of histrionics claiming my little LTD was a "perpetual motion machine".

No it isn't. I argued. It will stop when the ice melts. But they banned me anyway for "invoking perpetual motion without knowing it".because I couldn't measure much heat output or the ice melted too slow or something. Or god forbid I made mention of Tesla.

Now you are engaging in similar histrionics.

The above rant contains a lot of name calling and snide remarks but nothing to explain why some very simple experiment should be so "impossible" or "proof beyond proof". What?

Anyway, I don't "require" any experiment from anyone. I just think experiments are more valid than ranting and raving.

Nobody else seems to have ever had the motivation, so I'm doing what experiments I can myself.

All you Carnot advocates can't stand it and won't leave me be.
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

matt brown wrote: Sun Jun 09, 2024 4:28 pm ...AND input will be limited by the heat capacity of the working gas.
...
If this is true, and I think we can agree that it is.

Let's say my engine is running at a 10 or 20 % Carnot efficiency.

We also, I think agree, or you have agreed in the past that with an isothermal expansion 100% efficiency is possible, just for the expansion but not the whole cycle.

So, if the expansion stroke were 100% efficient where does the 5X or 10X additional heat that needs to be rejected during the subsequent compression stroke come from exactly ?

I mean, if ALL the input heat has been used up or converted, and there is so little heat that can get into the working fluid in the first place, due to having such a a limited heat capacity.

The temperature is the same or possibly lower, at BDC the pressure is lower, all the input energy has been used up to drive the engine thus far, but...

...on the return stroke now, we have to pull 5X or 10X more than all the total input energy that has already been used up so this "waste heat" can be "rejected" on the return stroke.

Let's say, 1 unit of heat is added before and/or during expansion.

1 unit of heat is converted to work or mechanical motion to drive the engine to BDC. Is there additional adiabatic expansion? I think so, probably.

1-1=0 right?

So the return stroke begins with zero remaining of the original added heat.

Now on the return stroke we are going to see 5 or maybe 10 units of heat "rejected".

Am I getting this straight so far?

No reason to question these assumptions right? This all seems perfectly reasonable to me, doesn't it seem reasonable to you?

Or do you think maybe we should run some kind of actual experiment and see?

Let's compress the cold low pressure expanded gas and see if we can get five or ten times more heat out than what we put in.

The Carnot formula says that's how it is, so it must be true Right?

Still, I think it might make a good high school physics lab demonstration. Show all the students how the Carnot efficiency limit can pull a bunch of heat right out of thin air.

A few cubic centimeters of very low heat capacity thin air inside a Stirling engine to be exact.

May as well add that to the curriculum, don't you think?

Par for the course I guess.
matt brown wrote: Sun Jun 09, 2024 4:28 pm the proof is in the pudding, so where's yours ??? You preach your version on everything and become defiant, defensive or hostile when anyone challenges or corrects you. Gadz, you must be desperate...
Me defensive? Hostile? Desperate?

No not at all.

I think you may just be projecting a wee bit.

Your Carnot nonsense just doesn't really hold water.
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

Talking about experiments. We can go all the way back to my first.

I had just gotten my first Stirling engine kit.

Right away I decided that it would be a good idea to replace the bolts holding it together with plastic bolts to reduce heat transfer between the hot and cold plate through the bolts.

The clear engine displacer chamber was already plastic.

This is the engine:

https://youtu.be/415iIJtfdLA

Then I thought, I'd run it on some hot water in a double wall vacuum insulated cup I bought at Walmart. The vacuum insulated cup would hold in the heat so the engine might run for a long time.

It did. Pretty long. About 3 hours.

https://youtu.be/HQT5JviF-qk

I also ordered a second engine.

I fooled around some more for a while just playing with my new toy.

I ran it on ice for a while.

Then I remembered an old controversy on the forum. Not much of a controversy really.

Everybody basically agreed that if the engine were covered with insulation it would stop because the heat flow would be blocked.

I thought it would be interesting to demonstrate this. I could post the video to the forum.

I decided to record the experiment live. Then I could post the video showing the engine slow down and stop when covered with insulation.

At the time I had no doubt the engine would stop.

I insulated the whole engine, except the top cold plate so there was nowhere else for the best to go but up and out through the top.

Of course, when I also blocked the heat flow out through the top, then the engine would overheat and stop.

https://youtu.be/fFByKkGr5bE

Well, that experiment was a failure. I was flabbergasted.

I checked, rewatching the video several times counting the RPMs to see how much the engine slowed down.

Well, it didn't.

It actually started running faster after the insulation went over the top.

You could try that experiment if you want Matt.

Do you think you'll get a different result ?

I posted the video anyway. Hey, look at this?

People commented. I should add insulation over the power cylinder, I should remove insulation on the bottom. etc.

I followed all the suggestions of other forum members, tried all the variations.

I tried better, thicker, different insulation. Silica Aerogel. I tried a plastic top on the engine. I covered the whole thing with a glass globe to keep out drafts.

Nothing I tried stopped that engine or any other engine.

They never overheated and never stopped. After hours and hours, the tops didn't even feel warm.

So I began to question. Where is all the "waste heat" going?

Do you blame me?

Thermocouples, infrared cameras...

Where is this waste heat.

Finally I began to question if there really was any waste heat.

This was not the results I was expecting. It wasn't the results anyone anticipated.

Nobody, I guess had ever been dumb enough or ignorant enough or so stupid to do something so silly as cover the heat sink with insulation.

Everybody knows without a doubt that the engine will stop running. Right?

Or everybody thought they knew. I thought I knew. I was going to video tape it just for laughs. But it didn't turn out like I was expecting.

Now fool thinks I'll be suspected of wire fraud?

Incredible.
matt brown
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by matt brown »

Tom Booth wrote: Sun Jun 09, 2024 10:32 pm
matt brown wrote: Sun Jun 09, 2024 4:28 pm ...AND input will be limited by the heat capacity of the working gas.
...
If this is true, and I think we can agree that it is.
correct
Tom Booth wrote: Sun Jun 09, 2024 10:32 pm Let's say my engine is running at a 10 or 20 % Carnot efficiency.
Let's use 10%
Tom Booth wrote: Sun Jun 09, 2024 10:32 pm We also, I think agree, or you have agreed in the past that with an isothermal expansion 100% efficiency is possible, just for the expansion but not the whole cycle.
correct
Tom Booth wrote: Sun Jun 09, 2024 10:32 pm So, if the expansion stroke were 100% efficient where does the 5X or 10X additional heat that needs to be rejected during the subsequent compression stroke come from exactly ?

I mean, if ALL the input heat has been used up or converted, and there is so little heat that can get into the working fluid in the first place, due to having such a a limited heat capacity.

The temperature is the same or possibly lower, at BDC the pressure is lower, all the input energy has been used up to drive the engine thus far, but...

...on the return stroke now, we have to pull 5X or 10X more than all the total input energy that has already been used up so this "waste heat" can be "rejected" on the return stroke.

Let's say, 1 unit of heat is added before and/or during expansion.

1 unit of heat is converted to work or mechanical motion to drive the engine to BDC. Is there additional adiabatic expansion? I think so, probably.

1-1=0 right?

So the return stroke begins with zero remaining of the original added heat.

Now on the return stroke we are going to see 5 or maybe 10 units of heat "rejected".

Am I getting this straight so far?
Nope, you're messing up with how the ratios play out. Furthermore, there's a major distinction between a cycle adding heat prior expansion vs during expansion.

Let's start with the Stirling cycle and add 10 units of heat during an isothermal expansion (vs your 1 unit). During this isothermal expansion, these 10 units of heat are converted into 10 units of work and the piston is at BDC. Now at BDC, the temperature is the same as it was at TDC, so compressing the gas back to TDC for a Stirling cycle will simply reverse everything with 10 units of work compressing the gas back to original volume...as an isothermal compression...whereby 10 units of heat are 'forced out' of gas to heat sink similar the 10 units of heat forced into gas during expansion. In this manner, Qin=Wout during expansion, Win=Qout during compression, but Qin=Qout and Wout=Win whereby Wnet=Wout-Win=0 aka gas spring.

The only way to achieve Wout during expansion > Win during compression is by a temperature differential that coincides with a pressure differential. The Stirling cycle uses 2 temperatures to achieve a pressure swing across the cycle since dP=dT when constant volume similar dU=dT. Borrowing the overworked 300-400k cycle, if a fixed mass of gas under constant volume is 1 bar at 100k, the same volume is 2 bars at 200k, 3 bars at 300k, and 4 bars at 400k. So, a 300-400k Stirling cycle with a 2:1 volume ratio will have 400k expansion from 4 bar to 2bar, then drop temperature to 300k and 1.5 bar at BDC, then 300k compression from 1.5 bar to 3 bar, then temperature rise back to 400k at TDC.

Since this is a Stirling cycle, both temperature changes between 300k and 400k are via regen. The takeaway here is that the 300k compression required Win=3/4 Wout of expansion. Thus, this 300-400k cycle has eff=.25 simply because the backwork of compression consumed 3/4 the output of expansion. IOW 10 units of heat converted to 10 units of work during expansion (at 400k) but 7.5 units of heat were lost to sink (at 300k) at a cost of 7.5 units of work during compression.
Tom Booth wrote: Sun Jun 09, 2024 10:32 pm
No reason to question these assumptions right? This all seems perfectly reasonable to me, doesn't it seem reasonable to you?

Or do you think maybe we should run some kind of actual experiment and see?

Let's compress the cold low pressure expanded gas and see if we can get five or ten times more heat out than what we put in.

The Carnot formula says that's how it is, so it must be true Right?
Nope, your 5x or 10x heat out is not vs heat input, but vs work output.

If 10 units of heat are input and converted into 10 units of work output, but 5 units of work input are required for compression to complete the cycle then it's 1:1

However, if 10 units of heat are input and converted into 10 units of work output, but 9 units of work input are required for compression to complete the cycle then it's 9x
VincentG
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by VincentG »

Matt that was one of the best overall descriptions of a Stirling cycle to date.
Fool
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Fool »

Tom Booth wrote:My simple experiments however must be repeated endlessly, with ever more and more stringent requirements imposed.
You sure got that right. Extraordinary claims must be backed by extraordinary scientific data. Get ready for the onslaught of all kinds of attempts to rip your theory apart coming from both qualified kind people, and all through the entire spectrum up to and including unqualified and hostile people. From the postings here it very apparent that Matt and I and a few others lean more towards the qualified and kind type. Kudos to all of the kind type. The physics stack seems more towards the qualified and protective type.

The following is another experiment you won't do because it will prove you wrong:

You claim to have completely cut off the heat from escaping the cold side.

That will also mean ambient heat can't get in the cold side. It is completely cut off from the ambient temperature. It is the proverbial "cold hole". It is similar to a thermos bottle. It can be put in a hot room or car and will stay cold all day, especially if loaded with ice.

You claim this will allow an engine to run on ambient heat indefinitely, until the ice melts from ambient heat getting to it, not from heat rejection. Something your tests haven't proven because your heat source shuts down.

So again: Cut off the heat with insulation on the cold side. Get it running on a tumbler of hot water. Then put it into a hot room, car, or oven at, say the temperature of the hot water, 150° F, warm enough to run but not to melt the engine. Could be 120 to be safe.

Time how long the engine runs before the cold side heats up to ambient. This will give an indication of how heat conductive the insulation is. If your theory is true it should run forever/weeks/all day. Ambient heat is cutoff, right? Try it with the engine not running.

Use that information to calculate what the temperature rise Should be if the engine is running back in ambient from one joule rejected heat. You may be astounded how many hours it would take for even one degree.

Alternately, just measure the temperature of an insulated block of steel. It will tell you how "cutoff" the heat energy is.
Matt Brown wrote:Nope, your 5x or 10x heat out is not vs heat input, but vs work output.
Many people seem to get confused with the whole efficiency verses COP inversion. LOL

The heat rejected and or absorbed is related to the work performed, but not in the way common thought or intuition on the surface, appears. Mathematics is only way to view it, and fairly advanced mathematics. I'm glad you and others here understand that and the mathematics.

It doesn't surprise me that it is hard to grasp. If the engine is outputting very very little power, the amount of heat entering the engine may just be very little. That just means the running engine is a good insulator. The whole purpose of the displacer is to insulate the hot from the cold. Tiny little engines are tiny little heat flow devices, otherwise known as insulation. The only way to know this is the combination of the dynamometer with and indicator diagram. I think it was James Watt that discovered that. And people today still reject that. Hostile? Yes. It's always hostile to reject/deny science.
Fool
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Fool »

Tom Booth wrote:Now fool thinks I'll be suspected of wire fraud?
You are safe saying whatever science denial you choose, right to an opinion, right to be wrong, or correct. It's your choice. And you are protected up until you take money with a promise to produce, and then you are protected if you do produce.

Many people have made similar claims, promises, taken money, had beliefs and good intentions, were even warned, and failed to produce. No one claiming similarly has produced. There have been many claimers that suffered because of lack of production after extraordinary claims. The followers are sufferers too. Be very careful. Going safely into a business with extraordinary claims, requires extraordinary working products. Your claims, rejecting accepted science, are extraordinary. Your products, so far aren't. Be careful.
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

matt brown wrote: Mon Jun 10, 2024 12:32 am
Tom Booth wrote: Sun Jun 09, 2024 10:32 pm
matt brown wrote: Sun Jun 09, 2024 4:28 pm ...AND input will be limited by the heat capacity of the working gas.
...
If this is true, and I think we can agree that it is.
correct
Tom Booth wrote: Sun Jun 09, 2024 10:32 pm Let's say my engine is running at a 10 or 20 % Carnot efficiency.
Let's use 10%
Tom Booth wrote: Sun Jun 09, 2024 10:32 pm We also, I think agree, or you have agreed in the past that with an isothermal expansion 100% efficiency is possible, just for the expansion but not the whole cycle.
correct
Tom Booth wrote: Sun Jun 09, 2024 10:32 pm So, if the expansion stroke were 100% efficient where does the 5X or 10X additional heat that needs to be rejected during the subsequent compression stroke come from exactly ?

I mean, if ALL the input heat has been used up or converted, and there is so little heat that can get into the working fluid in the first place, due to having such a a limited heat capacity.

The temperature is the same or possibly lower, at BDC the pressure is lower, all the input energy has been used up to drive the engine thus far, but...

...on the return stroke now, we have to pull 5X or 10X more than all the total input energy that has already been used up so this "waste heat" can be "rejected" on the return stroke.

Let's say, 1 unit of heat is added before and/or during expansion.

1 unit of heat is converted to work or mechanical motion to drive the engine to BDC. Is there additional adiabatic expansion? I think so, probably.

1-1=0 right?

So the return stroke begins with zero remaining of the original added heat.

Now on the return stroke we are going to see 5 or maybe 10 units of heat "rejected".

Am I getting this straight so far?
Nope, you're messing up with how the ratios play out. Furthermore, there's a major distinction between a cycle adding heat prior expansion vs during expansion.

Let's start with the Stirling cycle and add 10 units of heat during an isothermal expansion (vs your 1 unit). During this isothermal expansion, these 10 units of heat are converted into 10 units of work and the piston is at BDC. Now at BDC, the temperature is the same as it was at TDC, so compressing the gas back to TDC for a Stirling cycle will simply reverse everything with 10 units of work compressing the gas back to original volume...as an isothermal compression...whereby 10 units of heat are 'forced out' of gas to heat sink similar the 10 units of heat forced into gas during expansion. In this manner, Qin=Wout during expansion, Win=Qout during compression, but Qin=Qout and Wout=Win whereby Wnet=Wout-Win=0 aka gas spring.

The only way to achieve Wout during expansion > Win during compression is by a temperature differential that coincides with a pressure differential. The Stirling cycle uses 2 temperatures to achieve a pressure swing across the cycle since dP=dT when constant volume similar dU=dT. Borrowing the overworked 300-400k cycle, if a fixed mass of gas under constant volume is 1 bar at 100k, the same volume is 2 bars at 200k, 3 bars at 300k, and 4 bars at 400k. So, a 300-400k Stirling cycle with a 2:1 volume ratio will have 400k expansion from 4 bar to 2bar, then drop temperature to 300k and 1.5 bar at BDC, then 300k compression from 1.5 bar to 3 bar, then temperature rise back to 400k at TDC.

Since this is a Stirling cycle, both temperature changes between 300k and 400k are via regen. The takeaway here is that the 300k compression required Win=3/4 Wout of expansion. Thus, this 300-400k cycle has eff=.25 simply because the backwork of compression consumed 3/4 the output of expansion. IOW 10 units of heat converted to 10 units of work during expansion (at 400k) but 7.5 units of heat were lost to sink (at 300k) at a cost of 7.5 units of work during compression.
Tom Booth wrote: Sun Jun 09, 2024 10:32 pm
No reason to question these assumptions right? This all seems perfectly reasonable to me, doesn't it seem reasonable to you?

Or do you think maybe we should run some kind of actual experiment and see?

Let's compress the cold low pressure expanded gas and see if we can get five or ten times more heat out than what we put in.

The Carnot formula says that's how it is, so it must be true Right?
Nope, your 5x or 10x heat out is not vs heat input, but vs work output.

If 10 units of heat are input and converted into 10 units of work output, but 5 units of work input are required for compression to complete the cycle then it's 1:1

However, if 10 units of heat are input and converted into 10 units of work output, but 9 units of work input are required for compression to complete the cycle then it's 9x
Your whole analysis contains a number of ambiguous statements making it impossible to actually know what you might be trying to to say.

Let's just take the last paragraph:
However, if 10 units of heat are input and converted into 10 units of work output, but 9 units of work input are required for compression to complete the cycle then it's 9x
What does "it's 9x" referring to?

"It" ?

Maybe go back one more paragraph:
If 10 units of heat are input and converted into 10 units of work output, but 5 units of work input are required for compression to complete the cycle then it's 1:1
Again 'It's 1:1'

What's 1:1 ????

According to the Carnot limit, we must "reject", in the example you've chosen: "Let's use 10%" you said. That is referring to the Carnot efficiency of the engine. Fine with me.

10% efficiency, according to the Carnot theory means 9 times more heat must be "rejected" than the amount of heat converted to work.

10% converted to work
100 - 10 = 90% waste heat.

So we put 10 units of heat in.

These 10 are all consumed or used up, (converted into work) during the expansion stroke.

You say above: "...but 5 units of work input are required for compression to complete the cycle then it's 1:1"

We need to "reject" 90 units of "waste heat". 90 units of heat being rejected to conform with the Carnot limit. 10% converted 90% "rejected".

The "Carnot limit" contradicts reality.

The input heat is all gone at BDC.

You say 5 units of work, provided by atmospheric pressure presumably? are used for compression so add heat of compression on the return stroke.

That doesn't even make up for the ten units already subtracted due to all the input heat having been converted to work.

Regardless, "heat of compression" though in reality IMO, only partially compensating for the heat already consumed, so the piston returns by partial vacuum. Even if this return by vacuum generated 5 units of heat that is somehow ALL "rejected" we actually need 85 additional units of "waste heat" to satisfy Carnot.

85

85 units of heat we have to somehow pull out of thin air and then "reject".
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

Fool wrote: Mon Jun 10, 2024 5:49 am
Tom Booth wrote:My simple experiments however must be repeated endlessly, with ever more and more stringent requirements imposed.
You sure got that right. Extraordinary claims must be backed by extraordinary scientific data. Get ready for the onslaught of all kinds of attempts to rip your theory apart coming from both qualified kind people, and all through the entire spectrum up to and including unqualified and hostile people. From the postings here it very apparent that Matt and I and a few others lean more towards the qualified and kind type. Kudos to all of the kind type. The physics stack seems more towards the qualified and protective type.

The following is another experiment you won't do because it will prove you wrong:

You claim to have completely cut off the heat from escaping the cold side.

That will also mean ambient heat can't get in the cold side. It is completely cut off from the ambient temperature. It is the proverbial "cold hole". It is similar to a thermos bottle. It can be put in a hot room or car and will stay cold all day, especially if loaded with ice.

You claim this will allow an engine to run on ambient heat indefinitely, until the ice melts from ambient heat getting to it, not from heat rejection. Something your tests haven't proven because your heat source shuts down.

So again: Cut off the heat with insulation on the cold side. Get it running on a tumbler of hot water. Then put it into a hot room, car, or oven at, say the temperature of the hot water, 150° F, warm enough to run but not to melt the engine. Could be 120 to be safe.

Time how long the engine runs before the cold side heats up to ambient. This will give an indication of how heat conductive the insulation is. If your theory is true it should run forever/weeks/all day. Ambient heat is cutoff, right? Try it with the engine not running.

Use that information to calculate what the temperature rise Should be if the engine is running back in ambient from one joule rejected heat. You may be astounded how many hours it would take for even one degree.

Alternately, just measure the temperature of an insulated block of steel. It will tell you how "cutoff" the heat energy is.
Matt Brown wrote:Nope, your 5x or 10x heat out is not vs heat input, but vs work output.
Many people seem to get confused with the whole efficiency verses COP inversion. LOL

The heat rejected and or absorbed is related to the work performed, but not in the way common thought or intuition on the surface, appears. Mathematics is only way to view it, and fairly advanced mathematics. I'm glad you and others here understand that and the mathematics.

It doesn't surprise me that it is hard to grasp. If the engine is outputting very very little power, the amount of heat entering the engine may just be very little. That just means the running engine is a good insulator. The whole purpose of the displacer is to insulate the hot from the cold. Tiny little engines are tiny little heat flow devices, otherwise known as insulation. The only way to know this is the combination of the dynamometer with and indicator diagram. I think it was James Watt that discovered that. And people today still reject that. Hostile? Yes. It's always hostile to reject/deny science.
You seem to be mixing/confusing different experiments. An engine running on ice with the ice insulated mixed up with an engine running on heat with the "waste heat" out blocked/insulated.

As far as "hostile", I disagree. It is not "hostile" to question "established science". Again, see Popper on falsification.

It is scientific for science to continually question and examine its own postulates. That is how science progresses.

The only thing I'm rejecting is this desperate clinging and glorification of a completely untested, unverified 200 year old theory.

Not that I have anything against 200 year old theories, if tested empirically, by actual experiment, and verified, or at least having strong empirical evidence in support of the theory.

If I was hostile to science I would not have willingly subjected my experiments to scientific criticism on the Science and Physics forums.

THEY pulled the plug on that, not me. They did not like what they could not support being challenged. They chose to ignore and ban ACTUAL EXPERIMENT

Since when are experiments "hostile to science" ?

Since when is suggesting some simple experiments should be conducted "impossible".?

If your suggestion for an experiment were specific enough and coherent enough to actually understand and follow I might certainly consider it, but to be honest I don't understand your proposal. I'm not sure you do either. You seem to just be throwing spaghetti against the wall or rather, in my face, as an attempt at debunking, though you have no actual coherent proposal that could be carried out, so you can complain about my "refusal" or "rejection" of your "science". Which is actually just jabberwocky.
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