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

[Fool]

P1V1/T1 = P2•V2/T2

For the given parameters:
P1=P2

Canceling P from both sides:
V1/T1 = V2/T2

From given parameters:
T2=2•T1

Inserting in to the equation:
P1/T1=P2/(2•T1)

Dividing both sides by (2•T1):
P1•(2•T1) / T1 = P2

Cancelling T1 top and bottom and rearranging:
2•P1 = P2

[Tom Booth]

P1V1/T1 = P2•V2/T2

For the given parameters:
P1=P2

Canceling P from both sides:
V1/T1 = V2/T2

From given parameters:
T2=2•T1

Inserting in to the equation:
P1/T1=P2/(2•T1)

Dividing both sides by (2•T1):
P1•(2•T1) / T1 = P2

Cancelling T1 top and bottom and rearranging:
2•P1 = P2

Please show your "given parameters".

As that has changed over time and/or is somewhat ambiguous.

And why does the online example solution cited earlier here:

viewtopic.php?p=22899#p22899

produce P2 = P1/2 ?

[Fool]

Given parameters:

P1=P2
Atmospheric pressure plus constant weight on top, pressure stays constant for entire expansion.

T2=2•T1
300 =T1
600 = T2

600 = 2•300

The example where P2=P1/2?
Is for T1=T2 and V2=2•V1

Taking the same equation:
P1•V1/T1= P2•V2/T2

Since T1=T2 they cancel each other, (multiple both sides by T1 you will get T1/T2=1. One times anything is "anything", so cancelled.) so the equation becomes:
P1•V1= P2•V2

Substituting in V2=2•V1 for V2:
P1•V1= P2•2•V1

Dividing both sides by 2•V1:
P1•V1/(2•V1)=P2

Canceling and rearranging:
P1/2=P2

A result from different given parameters.

T constant.
V doubles.

Instead the first one having
P constant
T doubling

Both are the same expansion.

[Fool]

I suppose I should mention the first is energy/work out, the second requires energy/work input. The first pushes, the second must be pulled. I hope this makes sense.

[Tom Booth]

I suppose I should mention the first is energy/work out, the second requires energy/work input. The first pushes, the second must be pulled. I hope this makes sense.

The online example says nothing about "pulling"

Ideal gas of a volume of 1 m3 at initial pressure of 200 kPa is expanding isothermally to occupy double of its initial volume. Determine the work performed by the gas during expansion, final pressure and the amount of heat supplied to the gas

.

In an isothermal process, all heat Q received by the gas is converted into the work W that it performs (internal energy U does not change). Thus, it holds true:

Q=W.

End result?

Pressure drops 1/2

Piston must return

I "trust" this online example analysis (over "fool" or other anonymous forum posters) because it corresponds with my own experimental results and observations of engine behavior.

The piston returns, fully, even when heat loss to a "sink" is virtually, if not entirely eliminated.

Pressure dropping in half affords a satisfying, and apparently mathematically sound explanation.

[Fool]

The online example says nothing about "pulling"

You would only have to pull on it for the whole stroke if the buffer pressure were at or higher than P1. P2 is even lower. If the buffer pressure where lower than P2 no pulling would be necessary for the whole stroke, but a stop would be needed at V2.

The equation is the same regardless of buffer pressure, only the flow direction of overall energy/work to or from the outside. It customary to equate P1 with buffer or atmospheric pressure, but not required. So you are correct in questioning that. Good pick-up.

[Tom Booth]

The online example says nothing about "pulling"

You would only have to pull on it for the whole stroke if the buffer pressure were at or higher than P1. P2 is even lower. If the buffer pressure where lower than P2 no pulling would be necessary for the whole stroke, but a stop would be needed at V2.

The equation is the same regardless of buffer pressure, only the flow direction of overall energy/work to or from the outside. It customary to equate P1 with buffer or atmospheric pressure, but not required. So you are correct in questioning that. Good pick-up.

I don't question that. I don't question the online example. I question you and Stroller.

Your competence to carry out the mathematics involved is highly questionable.

The online example looks bulletproof.

The sources look impeccable:

https://physicstasks.eu/en

[Fool]

From the link you provided:

The resultant pressure p2 of the air can be determined from the above-mentioned Boyle-Mariotte law as well

p1V1=p2V2.
Therefrom we express the final pressure

p2=p1V1/V2=p1V1/2V1=p1/2.

p2=p1V1/V2=p1V1/2V1=p1/2

Breaking it down:
Dividing both sides by v2:
p2=p1V1/V2

Substituting in v2=v1•2:
p2=p1V1/2V1

Canceling out v1:
p2=p1/2

Same derivation I did.

That is why mathematics is so important in science. If done correctly it comes up with the same answer.

If done incorrectly the error is identifiable, and can be corrected.

Matt, Stroller, and myself are capable of good mathematics. Questioning any of us is always a good idea. The source noted in your link probably has had many people review it and question it. That process makes science more reliable, but not perfect. Text books with errors, almost all of them, even the ones that pride themselves of zero errors, have errors. The students that lack dogmatic beliefs will find them, and still learn around those errors. Matt Stroller and I will make errors, but the important points can still be discerned.

Your questioning my point about pulling on the system to get the response asked for, required my response as a qualifier. The point is that even when the expanding gas provides work, it is not enough to overcome the opposing work of a higher pressure outside. The link you sent left out what accomplished the expansion.

[Tom Booth]

...The link you sent left out what accomplished the expansion.

It states clearly: "heat supplied to the gas".

Also:

"work performed by the gas"

Perfectly clear IMO. "Heat supplied to the gas" "work performed by the gas"

Anyway, this is VincentG's question to Stroller.

I've had my say so I'm bowing out.

[Stroller]

The clearest way to compare Tom's online example (OE) with Stroller's example (SE) is to scale up Stroller's example to match it.

OE starts with 1 cubic metre of air whereas SE starts with 100cc. OE's initial volume is 10,000 times bigger.
SE requires 36.5J of energy input Q. Multiply by 10,000 = 365,000J to scale up. But OE uses kiloJoules not Joules, so now we need to divide by 1000 = 365kJ

OE achieves isothermal expansion with 140kJ but scaled up SE uses 2.6 times more at 365kJ (+10kJ to lift the 1 tonne piston). That's because SE is not an isothermal expansion, it's a diabatic expansion balancing atmospheric pressure and lifting a piston.

This is why the SE air needs to be hit with 2.6 times more input energy and ends up at 600K from 300K. It has to be heated up that much more because by the time it reaches double its original volume, it only has half the density of the atmospheric ambient air it has to be in pressure balance with.

Neither example is wrongly calculated, they're just different. My example is more relevant to thinking about DIY Stirling engines, because it starts with ambient pressure and temperature air, and heats it in order to raise a piston against atmospheric pressure.

By the way, now we've taken Vincent's advice about scaling up, my DIY Stirling engine design is now using a heap of coal to raise a barrel sized piston. Eat your heart out Wilcox.

[Stroller]

This is why the SE air needs to be hit with 2.6 times more input energy and ends up at 600K from 300K. It has to be heated up that much more because by the time it reaches double its original volume, it only has half the density of the atmospheric ambient air it has to be in pressure balance with.

This situation simply obeys Charles Law

V1/T1 = V2/T2

1 cubic metre/300K = 2 cubic metres/600K

Temperature doubles, and so does volume. Density halves, pressure remains constant: similar to what you can see in the animated diagram below (the animation doesn't quite double the temperature, but you get the idea).



Now we've got this sorted out, we can think about getting back on topic and looking at the efficiency. We've only done half the Stirling cycle and things are looking bad. We've used 375kJ to get the 1 tonne piston up the cylinder, and dropping it on someone's foot is only going to give us 10kJ back. Let's hope we can improve matters.

[Stroller]

Since Tom has graciously bowed out from his own thread, I'll give it back to him and take the next stage over to my own efficiency thread here:

https://www.stirlingengineforum.com/vie ... php?t=5661

I'm away for the weekend now, so it'll be next week before I get to it. Cheers all.

[Tom Booth]

.... My example is more relevant to thinking about DIY Stirling engines, ...

Your example is crap, obviously.

[Stroller]

Your anticipated response is why I gave you your thread back, obviously.
Kindly stay away from mine.
Thank you.

[VincentG]

Stroller wrote: ↑Tue May 21, 2024 5:24 am
It's the action of the expanding gas "consuming" energy that I am interested in. Is internal energy really reduced when the gas expands, other than by conduction losses to the surrounding colder surfaces of the engine. And if so, how exactly?
The internal energy of the entire expanding volume isn't reduced, but the internal energy per unit volume is, because the total internal energy of the gas is spread out more, into a bigger space.
....
This statement (bold) is, or would be, a violation of conservation of energy.

Your statement is basically that energy can GO OUT from the working fluid as WORK but all the energy is still in the working fluid but just spread out more.

That would double the total amount of energy.

Say 1000 joules gone out as "work" but the same 1000 joules still in the working fluid but just spread out in a greater volume.

So the original 1000 joules has split and turned into 2000, 1000 going out as "work" and 1000 remaining but just spread out in a larger volume.

This is exactly what I would like to focus on.