Stirlingengineforum.com

Jack - did you see this process table I just posted?

http://www.stirlingengineforum.com/view ... 717#p23717

Tom Booth wrote:What's wrong with the online calculator?

I had a chance to play with those calculators. Tom's is only a single point PV=nRT calculator and will not calculate a two point process.

Matt's has several standard two point processes to choose from, such as isothermal and adiabatic.

Jack's question is for a three point process, or broken into two two point processes.

First one isochronic, constant volume heat addition. Easy to run through Matt's calculator and get the same numbers I got.

Second is adiabatic. Matt's calculator will do it but with some idiosyncrasies. The process will not be calculated from temperature values. It will calculate temperature and pressure from volume.

Jack your English here is very good.

I was able to use Matt's calculator and verified my numbers. Those surprising numbers a few pages back are correct after all. I did put in a P2 that should be P3.

They are generic numbers with no specific units. So they don't relate well to experiment. And assume a zero pressure ambient vacuum of space.

Starting with the following;

Input
T1=300 K
V1=1 m^3
P1=100 Pa

Atmospheric pressure is101 kPa. So reality Q and W will be 1000 time higher in the following:

Calculating the second points using Tom or Matt's website click on isochoric:

Input more:
T2=900 K
V2=1 m^3

So output:
P2=300 Pa <<<
∆U=500.7
Q=500.7
W=0

The adiabatic expansion to get to points 3, needs the calculator Matt linked. Click on adiabatic

Inputs 2's go in as the first point, 3's as the second points. First points next:
T2=900 K
V2=1 m^3
P2=300 Pa <<<

T3=300, It would be desirable to just enter temperature and have it calculate pressure and volume, but it doesn't work with that, an omission by the programer. So guessing a volume or pressure until T3=300 again, works.

Guessing 15, from my calculations gets us very close, guessing 15.59 gets very very close.

So the outcome is:
T3=300 K
V3=15.59 m^3
P3=6.414 Pa
∆U=-500.7 J
W=500.7 J
Q=0 it's always zero for adiabatic processes.

Now for giggles, I did isothermal at 900 K and got:
T3=900 K
V3=3 m^3
P3=100 Pa
∆U=0
W=329.6 J
Q=329.6 J

Back isothermal work:
T3=300 K
V3=3 m^3
P3=33.33333 Pa because heat is stored in the regenerator reducing temperature by 3.
∆U=0
W=-109.86 J
Q=-109.86 J

This would give an efficiency of:
n=(329.6-109.86)/329.6=0.666686893

Or about 66.67% ideal gas law.

Same as (900-300)/900 Carnot

Minimum back work after adiabatic expansion to Tc, followed by isothermal compression back to V1:

W=-274.65
∆U=274.65

n=(500.7-274.65)/500.7=0.4515
Or 45.15%

This is a good mathematical demonstration, proof, of the isothermal cycle being a higher efficiency, and the Carnot limit. Ideally. Real engines will be worse.

Jack wrote: ↑Sat Jun 29, 2024 11:30 pm

Tom Booth wrote: ↑Sat Jun 29, 2024 4:03 pm

air, or any other fluid

You don't want much do you?

All combined you've asked about maybe 100 different questions.

Tell me now. What you can't answer?

You haven't asked a specific identifiable question, just these vague generalizations. You get a response and add more complications.

Anyway Matt's link has several "fluids" to choose from. But not "Any".

You have an incredible talent to trip over one word and forget everything else I said.
"Any", as in "just pick one" as they'll be similar enough in behavior for my purpose.
It doesn't mean "every" or "all". I'm asking to just give an example. And I'm not asking for exact numbers, just general behavior.
Again, my English might not be perfect here and your understanding of my question might be lacking because of it. But I don't think I asked something ridiculous that you can't read into.

Well, friend,

"I" went to some great lengths to try to figure out what in the world you might be wanting to know from your pretty vague request for something.

From the information you provided I tried to fill in the blanks as far as the information you seemed to maybe be looking for and posted that here:

viewtopic.php?p=23705#p23705

Including a link to a calculator where you should be able to find out or calculate in the same way virtually whatever you might actually want to know.

Your two following posts after that appeared to be saying that's was not what you wanted or not what you were looking for.

I don't know, you posted relatively soon after I did, maybe we cross posted???

Did you read that post? Is there something wrong with the calculator? Maybe you missed that post because we cross posted?

Matt found a more elaborate calculator.

Personally I like a calculator that includes moles.

A Stirling engine is a closed system so the moles should not change.

Not all thermodynamic processes are closed systems. Some involve mass flow.

Anyway there are those two calculators and many many more online.

So what additional information are you looking for or needing?

You didn't seem to like when I chose the parameters. Why do I have to choose anyway? I don't know what you want to know.

I'm not going to continue playing a guessing game.

So far you've rejected every offering, every solution offered.

It seems to me you keep adding more restrictions trying to force some specific predetermined answer but didn't like the responses you got.

So what didn't you like about my response here:

viewtopic.php?p=23705#p23705

????

I thought it provided the information you were looking for.

Then you came back wanting to know something else. I'm not a mind reader.

Except maybe with my wife sometimes. More the other way around. Either she hears me thinking sometimes, or I'm talking out loud without realizing it.

Matt found a more elaborate calculator.

No, Matt found a calculator that actually addressed Jack's question.

The fact that you put forth a basic ideal gas law calculator makes me question your actions here, or at least your understanding of the subject.

Sometimes the acceptable right answer is "I don't know".

Jack has been nothing but patient and straightforward with his questions.

VincentG wrote: ↑Sun Jun 30, 2024 11:03 am

Matt found a more elaborate calculator.

No, Matt found a calculator that actually addressed Jack's question.

Which was, exactly what?

So what is the answer?

I'm still trying to figure out the question.

Anyway, either calculator would work as far as I could decipher the question? Whatever it was. And you don't say why it wouldn't and you are again responding for "Jack". Is that your most recent "sock puppet"?

The fact that you put forth a basic ideal gas law calculator makes me question your actions here, or at least your understanding of the subject.

Really? How so?

It seemed he was looking for basic information that could mostly be answered by virtually any ideal or combined gas law calculator. I simply used the first one that popped up on a Google search. It seems simple would be better for a simple question from someone needing to ask in the first place.

Sometimes the acceptable right answer is "I don't know".

Jack has been nothing but patient and straightforward with his questions.

Right, so what is the straightforward answer to this alleged "straightforward question"?

A combined gas law calculator is a straightforward answer to a straightforward question?

What are you so butt hurt about VincentG?

Jacks original question that I saw was:

If a fluid is heated it will keep expanding until it reaches ambient temperatures again?

Originally, a seemingly very simple question with a simple answer.

BTW VincentG, you've been following the conversation closely, apparently, but have offered poor little Jack little in the way of assistance.

You suggest he do research on Wikipedia and not pay any attention to Tom.

So what would your brilliant response be to Jacks straightforward question (that seems to morph and grow after every response), and why have you been holding back?

At least others are trying to help. You just sit back and criticize other people's efforts but offer nothing yourself. A rather strange and disingenuous attitude IMO

Are you just wasting our time with some kind of mind games? Pretending to be this "Jack". Kinda seems like it

Anyway, I suggested a simple calculator that includes moles because I had tried many online ideal or combined gas law calculators that would not keep the moles constant, which is necessary when applying these laws to Stirling engines.

Some calculators "automatically" assume a greater volume means more moles, which in the case of a Stirling engine will produce wrong answers, which I found frustrating and useless.

If you can't specify the number of moles, there is a chance the calculator will "decide" for you.

If a fluid is heated it will keep expanding until it reaches ambient temperatures again?

Originally that appeared to have been a response to something fool said?


viewtopic.php?p=23658#p23658

Seems like he's just trying to get straight what someone else said.

Later expanding to some temperature became a "goal".

The original question, hardly a question at all continued to become more and more convoluted and elaborate from then on.

To check my understanding of this. If a fluid is heated it will keep expanding until it reaches ambient temperatures again?

Not even a question really to begin with, just checking his understanding of something someone else said apparently?

All seems rather strange to me.

Can you explain this VincentG? You seem to be a mind reader and know just exactly what Jack was thinking.

Why should such a simple question require a full blown combined gas law calculator?

The following was Stroller's effort (which I much appreciate) to answer my very similar question to Jack's. An ideal gas law calculator is not enough to answer Jack's question of how much the gas expands and at what temperature it will reach ambient pressure.

Stroller wrote: ↑Fri May 17, 2024 4:13 am

Stroller wrote: ↑Thu May 16, 2024 10:32 pm

VincentG wrote: ↑Thu May 16, 2024 3:31 pm Ok, it may just be me, but I don't have this completely straight.

The joules needed to raise 100cc from 300k to 600k at constant volume = ?

That's a different and more complex case. Since the pressure will be raised as the gas is heated in a constant volume, the rise in pressure will add to the rise in temperature. So less Joules of input heat energy will be required than we calculated with Charles law.

Then, the temperature of the gas after free adiabatic expansion(with no additional heat input) from 100cc at 600k to 200cc is = ?

Lower than 600K

The joules needed to heat 100cc at 300k through a free expansion to 200cc at 600k = ?

We did this one already. It's 3.65J

It looks like I goofed my earlier calc by a factor of 10, so it's 36.5J, not 3.65.

In the more complex case of joules needed at constant volume, a reasonable engineering approximation would be to multiply the Cv value of air by the mass of air and the change in Temperature.

0.718 x 0.12g x 300 = 25.85J

This is 9.65J less than the free expansion case. So to answer your other question of what T the air will be once we allow it to expand freely after it reached 600K at constant volume, we just need to work out how much 100cc of freely expanding air will heat up if we put 25.85J of heat energy into it.

Q 25.85J / (Cp value of air is 1.01 x mass of air 0.12g) = Delta T = 213.28K
We add this result to the initial temperature of 300K and we get 513.28K

VincentG wrote: ↑Sun Jun 30, 2024 5:06 pm The following was Stroller's effort (which I much appreciate) to answer my very similar question to Jack's. An ideal gas law calculator is not enough to answer Jack's question of how much the gas expands and at what temperature it will reach ambient pressure.

....

Whatever Strollers response to YOU,

So what?

That was not "Jacks" original question AFAIK.

To check my understanding of this. If a fluid is heated it will keep expanding until it reaches ambient temperatures again?

Jack had asked about temperature originally not pressure.

Without yet having read any more, did not Stroller also later admit his response was not correct?

Is there something you don't like about the correct answer that a proper use of a proper gas law calculator that includes moles provides, as appropriate for a fixed volume of gas

Perhaps you actually prefer an incorrect answer that supports some narrative of yours?

Who knows?

You seem to be up to some kind of shenanigans IMO.

Anyway Vinny, I'm personally more interested in the results of real experiments using real engines.

Enough with all this theoretical BS. It's a waste of time debating such trivial nonsense IMO.

BTW does your much superior calculator produce a different result than what the Online calculator I happened to use spit out? And what's it matter? Supposedly if supposedly not what "Jack" wanted to know anyway?

Why get all huffy about which calculator was used?

Did YOU supply Jack with ANY help or any calculator or any assistance at all?

No.

You told him to go study Wikipedia. LOL, now that's an infallible source of information there. Unfortunately it gets spammed with more disinformation than any resource on the internet.

Good job there Vinny, pointing Jack to a resource.

Oh, I see, that was his later correction.

It looks like I goofed my earlier calc by a factor of 10, so it's 36.5J, not 3.65.

If that can be trusted.

Seems like a bit more complex a question than what Jack was asking.

Bad Tom. He pointed an apparently new guy to a simpler, easier to use resource for a beginner, not a more complicated calculator. Bad bad bad Tom.

Tommy, I have answered his question twice. You appear to not be getting it. And that again doesn't surprise me. You go on and on proving that you don't get it.

For the record it was a simple generic, heat a volume of gas expand it until the temperature is back to the ambient starting pressure T1 Tc. There is zero logic, science, or mathematics, in the whining you are doing now, but there sure is soapboxing, following Jack around, vituperations. Back off.

It seems my initial question was very clear. All I did later was try to clarify it as an isolated system and that I didn't need exact values. This only because everyone was complicating it with values and engine setups.

After that there was a lot of noise and many words that didn't relate to my question. I watch that from a distance, but prefer to stay out of it. It does make me understand what people like in soap series haha.

In the end I'm just here to learn. And for me learning is trying to understand all sides of a story. I thought I'd ask a question to get to, what I thought was, the bottom or the start. From there I can move up.
I've always been a top down learner, so this time I also started with the big picture and I'm now filling in the details.

Anyway, carry on. Thanks for all the help and I'll try to make sense of it when I have some more free brain space.

Jack wrote: ↑Sun Jun 30, 2024 11:29 pm It seems my initial question was very clear. All I did later was try to clarify it as an isolated system and that I didn't need exact values. This only because everyone was complicating it with values and engine setups.

After that there was a lot of noise and many words that didn't relate to my question. I watch that from a distance, but prefer to stay out of it. It does make me understand what people like in soap series haha.

In the end I'm just here to learn. And for me learning is trying to understand all sides of a story. I thought I'd ask a question to get to, what I thought was, the bottom or the start. From there I can move up.
I've always been a top down learner, so this time I also started with the big picture and I'm now filling in the details.

Anyway, carry on. Thanks for all the help and I'll try to make sense of it when I have some more free brain space.

Thanks Jack.

Maybe if you have the time and inclination, could you do me a favor please?

Mr "fool" above has stated: "I have answered his question twice".

I think, somehow I must have missed that.

Could you please restate your initial question and what, if any answer you received that you considered a satisfactory answer, from fool, or anyone else for that matter.

Anyway, you said:

It seems my initial question was very clear. All I did later was try to clarify it as an isolated system and that I didn't need exact values. This only because everyone was complicating it with values and engine setups.

Looking back it appears VincentG provided some values?

VincentG wrote: ↑Jack if you heat 1 liter of 300k air to 600k and expand it adiabatically, it will reach 1 bar well before it lowers to 300k.

I'll try to find the real values fir you but maybe Matt will beat me to it.

I had erroneously assumed those had been your values I guess. I also assumed your question related to Stirling engines, given its a Stirling engine forum.

But, aside from requiring the input of some actual values of some kind, the ideal or combined gas laws apply to gases, engine or no engine.

A gas doesn't need an engine to expand.

Was this your initial question or was there something earlier?

Jack wrote: ↑Fri Jun 28, 2024 6:14 am To check my understanding of this. If a fluid is heated it will keep expanding until it reaches ambient temperatures again?
So assume you have two separate liters of fluid in a closed fully insulated space and you heat one up, you expand it until it reaches starting temperature again. Now you have one with a bigger volume than the other, the same temperature and pressure but lower density?

A "closed space", logically, that can then be "expanded", kind of describes an engine.

Alternatively, a balloon maybe?

Above I can see "temperature's not pressure: you expand it until it reaches starting temperature

Did you later change that to pressure? Was that you or someone else?

The entire record is here to look back on.

Next post on the subject was from fool:


I'm adding some highlight of the issues such as temperature or pressure:

Fool wrote: ↑Fri Jun 28, 2024 6:55 am ...

Jack wrote: If a fluid is heated it will keep expanding until it reaches ambient temperatures again?

I calculated that theory in the "Let's beat up Carnot" thread. The expansion stops when P inside equals P outside. The temperature ends up hotter than Tc. The extra temperature holds the larger volume. It doesn't go down to Tc until pressure is well below atmospheric. Again that is a manifestation of internal energy. Work out was way less than heat in.

Gas only expands if the chamber gets larger. The enlargement of the chamber can be from internal or external forces, or a battle between the two.

Jack, did you find that response satisfactorily answered your question?

Your next response was:

Jack wrote: ↑Fri Jun 28, 2024 12:32 pm

Fool wrote: ↑Fri Jun 28, 2024 6:55 am

Jack wrote: If a fluid is heated it will keep expanding until it reaches ambient temperatures again?

I calculated that theory in the "Let's beat up Carnot" thread. The expansion stops when P inside equals P outside. The temperature ends up hotter than Tc. The extra temperature holds the larger volume. It doesn't go down to Tc until pressure is well below atmospheric. Again that is a manifestation of internal energy. Work out was way less than heat in.

Gas only expands if the chamber gets larger. The enlargement of the chamber can be from internal or external forces, or a battle between the two.

I was leaving out any work or anything other than pure hypotheticals.

Let's say I do expand the fluid until Tc is reached. Do I end up with a bigger volume but lower density? And as I understand you saying now, I'll end up with lower pressure as well?

Edit, I guess lower density means lower pressure. It's a bit of a weird phenomenon though. Something seems to get lost somewhere.

Fool introduced pressure.

You bring the conversation back to temperature:

Let's say I do expand the fluid until Tc is reached. Do I end up with a bigger volume but lower density? And as I understand you saying now, I'll end up with lower pressure as well?

The questions start pulling up a bit there.

Generally speaking though, you don't seem to accept the response you received from fool at this point.

I made some apparently irrelevant comments about the inadequacy of fools post and how gas behaves in an engine, to which you responded:

Jack wrote: ↑Sat Jun 29, 2024 1:52 am Not to discard or talk over what you just explained, but I'm trying to get to the bottom first. Without engines in mind yet.

What happens to a fluid when heated up and how does manipulating either temperature, volume or pressure change that.

So if you heat a liter of air, expand it until it's back to starting temperature, what do you end up with? Bigger volume, lower density, same pressure?

Again, the question was about, or focused on temperature.

Then VincentG introduced some real values:

VincentG wrote: ↑Sat Jun 29, 2024 5:09 am Jack if you heat 1 liter of 300k air to 600k and expand it adiabatically, it will reach 1 bar well before it lowers to 300k.

I'll try to find the real values fir you but maybe Matt will beat me to it.

After that fool responded again on the subject with a very long involve post with a lot of mathematics:

You responded:

Jack wrote: ↑Sat Jun 29, 2024 6:48 am You lost me with the math stuff. I really struggle reading any of that, especially when it's not in it's proper format. Just my math dyslexia.
I guess I was fortunate to be able to avoid it as an engineer. But school was hell because of it haha.
...

It does not appear that your question was answered at this point, or if it was, by fool, you didn't understand it.

After some additional debate between me, fool and VincentG you again respond:

Jack wrote: ↑Sat Jun 29, 2024 9:10 am I don't think you'd need any more info or conditions for the question I asked.

I merely want to know the behaviour of a fluid. Not exact volumes or densities.

My question in short was what happens to the pressure, volume and density ratio when I heat up a fluid and then expand it to reach its starting temperature.
I'm interested in finding out what happens on a molecular level.

Again, you put some emphasis on the gas expanding until it reaches starting temperature.

In your next post you added:

Jack wrote: ↑Sat Jun 29, 2024 9:17 am The answer seemed to be, bigger volume, lower pressure and density.

Well, you heated the gas, then expanded the gas to the original temperature

Taking a balloon, put it in the sun, it will heat up and expand. The volume will increase. The temperature and pressure will also have increased.

Have you arrived at a satisfactory answer here anyway?

"The answer seemed to be, bigger volume, lower pressure and density."

After trying to clarify what you actually wanted to know, I suggested you could get an objective answer yourself by using an online ideal gas calculator.

Plug in some hypothetical values. No opinions.

The ideal gas law shows that when a certain number of moles of a gas are at a certain temperature they will have a certain volume and pressure.

It matters not how much you heated it or how it arrived back at the original temperature.

Are we doing these experiments in outer space, in a vacuum?

If not, then a certain number of moles of gas at a certain temperature, on earth, under atmospheric pressure will return to the original volume. That is a scientific fact. That is the ideal gas law.

You responded with what seemed like a rather hostile rant:

Jack wrote: ↑Sat Jun 29, 2024 1:27 pm I'm trying to get an understanding of what air, or any other fluid, does when heated up. How that turns into internal energy and how that manifests itself.

The theoretical example I thought of would give me an insight into that. And I don't think it's that complicated.
But it seems nobody can give me a straight simple answer. I'm not blaming anybody here, but in stead of flailing around saying my question doesn't make sense, tell me what you need to know then to answer it.

For the sake of imagination, yeah let's say I have a cylinder with a liter of air. 1 bar pressure. I heat it up by 300k (from 300k to 600k) The pressure rises.

Now I expand the cylinder until the air is back to 300k, its starting temperature.
What pressure do I end up with?

Before I was asking this in a very general way because I don't need an exact answer to my question. I just wanted to know if the pressure is higher, lower or the same in the end. That's all.
If this is not answerable, let me know what else you need to know.

“If you can't explain it to a six year old, you don't understand it yourself.”
Einstein (apparently)

Now the question is pressure, apparently.

But the previous ideal gas law calculation gave that.

Assuming this is on earth at 1 atmosphere of external pressure.

You then say:

Jack wrote: ↑Sat Jun 29, 2024 1:31 pm For the sake of clarity. Assume no friction losses or work output. Also no temperature loss through conduction.

The gas could not cool back down to the original temperature and remain expanded.

The hot expanded balloon in the sun will still be hot.

For it to return to the original temperature the internal energy (temperature) would need to be reduced, either by conduction or work output.

Expanding into a vacuum without doing any work and without taking out any heat by conduction does not reduce the temperature.

Such experiments were carried out. A surprising result perhaps but true.

The expanded container would need to be rigid though.

Some search results:
If you look into that in more depth, you will find that some gases actually increase in temperature when expanded. It depends on the type of gas and the inversion temperature. That is the Joule Thomson effect:
The actual setup for these experiments was a container with two chambers.

If you heat the gas in the first chamber, then let it expand into the second (a vacuum). Generally speaking, the gas will not cool down but will remain hot. (Ideal gas law)

Real gases might heat up or cool slightly (Joule Thomson effect). But the effect is negligible.

https://demonstrations.wolfram.com/Joul ... Expansion/

The bottom line is, Jack.

Your question describes a scenario that would not happen in reality.

If a gas expands without doing any work and without loosing any internal energy through heat conduction, then the temperature will stay the same. So the result you describe, the gas first being heated, then expanded is actually a description of Joules experiment.

If the volume is doubled however, I believe the pressure would be halved, but temperature is independent of pressure and volume. So the temperature would not return to the original temperature.