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Fool wrote: ↑Sun May 05, 2024 2:53 am

Fool wrote:Pulling a vacuum on a plugged syringe requires an input of work, the gas inside expands doing work for an adiabatic temperature drop. The work input by pulling, plus the work output by the gas, are both absorbed by the atmosphere, changing it negligibly, storing the energy as a spring. The springiness is cause by the difference in pressure inside and out.

After expanding the gas from point two and, Th, V1/V2 and P2 to T3, V3 and P3/Atmospheric/P1, Tom wants to continue expanding adiabatically until the gas reaches Tc. This will be similar to the instructor pulling on a plugged syringe.

This will diminish the pressure below atmospheric, producing an inward force. An extra outward force must be applied as energy into the engine. It could come from the outgoing energy from the momentum, using up that energy. If all the forward stroke gained energy is used, the volume will double at the point where the piston stops from that decelerating force.

Inside pressure 200 will be half the starting pressure 400. Temperature will be lower than Th. And momentum will be zero. The energy earned in the forward stroke, momentum, will be spent and stored in the force pushing inward. Similar to a spring.

The piston now moves inward accelerating. The gas heats up from compression. At the half way point, again the gas will be T3,V3, P3, and the same amount of energy will be stored in momentum going inward.

The atmosphere only provides a free U-turn. Now the momentum is being used to recompress the gas. Pressure inside 400 gets higher than outside3
300. The temperature continues to rise adiabatically. The momentum is used up when back at T2, V2/V1, and P2. The starting point of the expansion. Zero work can be output.

Atmospheric pressure 300, temperature 300 and volume (infinite) remain almost unchanged at all points of the cycle. A negligible change.

Why do you say atmospheric pressure and/or temperature is 300. But "starting pressure" is 400?

Seems to me the engine starts out at equilibrium with the outside temperature and pressure.

Heat is added "isothermally" presumably, in this scenario?

The temperature then should not have gone above the initial "starting temperature" where everything was in thermal equilibrium i.e. 300.

Also you say something about "The energy earned in the forward stroke, momentum, will be spent and stored in the force pushing inward. Similar to a spring."

Previous to that you said:

An extra outward force must be applied as energy into the engine. It could come from the outgoing energy from the momentum, using up that energy. If all the forward stroke gained energy is used, the volume will double at the point where the piston stops from that decelerating force.

Inside pressure 200 will be half the starting pressure 400. Temperature will be lower than Th. And momentum will be zero

All the momentum is gone or "used up", but you also say it is somehow "stored".

You assume a "starting temperature" that is 100 degrees higher than equilibrium, before heat is added.

If inside temperature goes to half after adiabatic expansion that should be half of 300, I think, not 400, which is never achieved with isothermal expansion.

Energy cannot, in most circumstances be both "used up" or "spent" but also stored. In particular not in an "infinite reservoir", as you conclude:

"Atmospheric pressure 300, temperature 300 and volume (infinite) remain almost unchanged at all points of the cycle. A negligible change."

So you have still not explained how or in what way you suppose this "spent" and "used up" energy is "stored" and supposedly returned.

Again, you are still claiming the return of energy that was "used up" and "spent" somehow being "stored" and returned "similar to a spring".

Lots of inconsistencies, contradictions, errors and just general lack of clarity.

I'm still unclear about your use of V1 v2 V3 etc. in statements like:

After expanding the gas from point two and, Th, V1/V2 and P2 to T3, V3 and P3/Atmospheric/P1, Tom wants to continue expanding adiabatically until the gas reaches Tc.

What is "point two"?

You seem to equate P3 with P1 as well as atmosphere? "P3/Atmospheric/P1"

You said:

After expanding the gas from point two and, Th, V1/V2 and P2 to T3, V3 and P3/Atmospheric/P1, Tom wants to continue expanding adiabatically until the gas reaches Tc. This will be similar to the instructor pulling on a plugged syringe.

This will diminish the pressure below atmospheric

Should this "below atmosphere" volume and pressure be represented in some way?

V4 P4 perhaps?

I don't know.

The atmosphere only provides a free U-turn. Now the momentum is being used to recompress the gas. Pressure inside 400 gets higher than outside3
300. The temperature continues to rise adiabatically. The momentum is used up when back at T2, V2/V1, and P2. The starting point of the expansion. Zero work can be output.

Suddenly on the return stroke you allege pressure is quickly back (sic.) to 400. When it was never above 300.

Why are "back at T2, V2/V1, and P2. The starting point of the expansion"

Whatever you mean by that is not clear.

What is V2/V1?

Are you dividing?

Are you equating the two?

How can T2, V2 and P2 be associated with "the starting point of the expansion"?

"The" expansion means what?

The start of the cycle OR the start of "Tom's" extended adiabatic expansion?

I'm frankly just getting tired of your poorly constructed, carelessly worded, incomprehensible, disconnected from any common sense reality, abstract arguments.

You present this garbage talk as if it were conclusive proof of anything. All it is is your attempt at a wave of the hand dismissal.

Maybe you can clarify this.

Draw a cylinder marking the actual positions that V1, V2 etc are supposed to represent might help.

Sorry but I don't have time to try and unscramble your omelet.

If you can put together some coherent arguments for consideration, fine, but I'm likely not going to respond to any more of this kind of illogical, confused, mixed up contradictory nonsense.

Absolutely correct. The problem lies in explaining it to someone who isn't knowledgeable of integral calculus. It takes a good instructor to enlightened someone on the usefulness of the area under the curve, and above the axis. And a chalkboard too.

Momentum isn't energy, but it contains energy. The same M and V in MV becomes 1/2MV^2.

Tom seems to think that atmospheric "spring" during the return stroke is "extra free" energy, when infact it noting more than a free U-turn. Like a ball bouncing. Ball going down at V, bounces, ball going up at V. Zero energy charge, different direction. MV outward cocks the spring. V slows to zero. Spring snaps accelerates M back up to V in the inward opposite direction at the same position. Extra energy equals zero.
If adiabatic compression continues,V slows to zero back at Top Dead Center. Qhz.

I know you know this. Adiabatic bounce, is a zero heat flow, zero momentum change, and zero energy change, event/process/cycle, except for unavoidable natural losses, like friction and windage. Zero energy in, starting energy is rejected by frictional heating and hysteresis.

The engine starts out at 300, atmospheric. 100 heat is added. The expansion stroke starts at point 2, 400. If all the processes are to be adiabatic after that, the complete cycle will end back at point 2, Qhz, V2, T2, and P2, which is 400.

The only way back to 300 at that point, is to reject 300 units of heat. The same 300 input to get from Qcz to Qhz, or DQh.

If the processes are not adiabatic, that point will be different from 400, worse or better depending on the order and type of processes.

I wish I could sit down with paper and draw while explaining that. It would be so much easier than descriptive words.

Think about an adiabatic line on a PB diagram and what it takes to get off that line. Heat flow in or out is the only answer. Heat-in goes one way from the line, heat-out goes the other.

Think about an isothermal line and what getting off one means. Different temperature.

Think about how the two lines relate to each other. And what happens between the two, if a processes path travels there. Not quite isothermal, not quite adiabatic. Right?

Point 1 starting point Qcz, V1, T1, P1. Each process ends at a new point. Point 2, 3, 4 etc..

Each as it's own Pressure P, Temperature T, and Volume V. Some are the same V1=V2 Volume 1 and 2 are the same, said another way "V1/V2<both the same". Sorry. Again it would be easier with a chart.

Same volume different temperature and pressure.

Fool wrote: ↑Sun May 05, 2024 5:05 pm The engine starts out at 300, atmospheric. 100 heat is added. The expansion stroke starts at point 2, 400 ...

This is already inherently contradictory.

Needless to say, I haven't read any further.

Is there a point 1 ?

How do you get from point 1 to point 2 without expansion?

What do you mean by "point"?

Position? Position of the piston in the cylinder?

Is this supposed to be isochoric heat addition at TDC?

Please try to talk sense.

Fool wrote: ↑Sun May 05, 2024 5:05 pm ...If all the processes are to be adiabatic after that, ...

After what?

Sounds like the piston hasn't gone anywhere yet.

Fool wrote: ↑Sun May 05, 2024 5:05 pm The engine starts out at 300, atmospheric. 100 heat is added. The expansion stroke starts at point 2, 400. If all the processes are to be adiabatic after that, the complete cycle will end back at point 2, Qhz, V2, T2, and P2, which is 400.

The only way back to 300 at that point, is to reject 300 units of heat. ...

Are you out of your mind?

You say the engine starts at 300.

A cycle returns to where it started.

And how do you figure that 400 minus 300 gets you "back to 300" from 400?

Fool wrote:The only way back to 300 at that point, is to reject 300 units of heat. ...

Could have sworn I wrote the following, but I didn't...

The only way back to 300 at that point, is to reject 100 units of heat. ...

Tom Booth wrote:Is this supposed to be isochoric heat addition at TDC?

Yes. V1 V2, point 1 to point 2, isochoric heat addition. V1=V2. You got it. Isochoric.

Tom Booth wrote:Is there a point 1 ?

Yes. Qcz, V1, T1, P1. Engine cycle starting point.

Point 2. Qhz Beginning of isentropic expansion stroke. End of isentropic compression stroke. Begining of isentropic processes. And end of isentropic cycle.

Fool wrote: ↑Sun May 05, 2024 6:56 pm

Tom Booth wrote:Is this supposed to be isochoric heat addition at TDC?

Yes. V1 V2, point 1 to point 2, isochoric heat addition. V1=V2. You got it. Isochoric.

Tom Booth wrote:Is there a point 1 ?

Yes. Qcz, V1, T1, P1. Engine cycle starting point.

Point 2. Qhz Beginning of isentropic expansion stroke. End of isentropic compression stroke. Begining of isentropic processes. And end of isentropic cycle.

That makes no sense.

You just skip 1 and go straight to 2?

(On the return stroke that is)

Then backwards to 1 by removing 100 joules?

"You just skip 1 and go straight to 2?"

You are proposing a series of adiabatic processes that start at point 2. The only adiabatic path available is the one that starts at point two, where you switch to adiabatic processes, follows an adiabatic expansion line forward until all motion is exhausted. Then snaps backwards in a reverse stroke that ends at point 2 with all motion stopped.

I'm not suggesting this, it is the results of your description.

To get to point 2, from point 1, 100 J must have been added at TDC.

Fool wrote: ↑Mon May 06, 2024 1:38 am "You just skip 1 and go straight to 2?"

You are proposing a series of adiabatic processes that start at point 2. The only adiabatic path available is the one that starts at point two, where you switch to adiabatic processes, follows an adiabatic expansion line forward until all motion is exhausted. Then snaps backwards in a reverse stroke that ends at point 2 with all motion stopped.

I'm not suggesting this, it is the results of your description.

To get to point 2, from point 1, 100 J must have been added at TDC.

Don't try blaming me for your illogical bs.

How about you draw a PV graph of whatever it is your trying to say because you aren't making any sense.

Top left of adiabatic is point 2.

Isentropic expansion follows down thick green line, and does positive work equal to the shaded green area under the curve.

Compression returns along the same line back to point 2. Requires work input equal to the shaded green area under the thick green line.

Since starting and finishing at the same point, this is considered a full cycle. It has zero area because work output equals work input. Therefore net work output is zero. It ends with the same internal energy it started with.

One more note: To get off the thick green line, heat must flow.

Fool wrote: ↑Mon May 06, 2024 6:12 am Adiabatic.svg.png




Top left of adiabatic is point 2.

Isentropic expansion follows down thick green line, and does positive work equal to the shaded green area under the curve.

Compression returns along the same line back to point 2. Requires work input equal to the shaded green area under the thick green line.

Since starting and finishing at the same point, this is considered a full cycle. It has zero area because work output equals work input. Therefore net work output is zero. It ends with the same internal energy it started with.

One more note: To get off the thick green line, heat must flow.

As I said, a PV diagram does not show all dynamics. It only shows "boundary work', it does not show shaft work, it does not account for heat converted to work, momentum, velocity, passage of time outside atmospheric pressure, inertia, gas compressibility, molecular attraction etc.

It's a simple visual aid for students to conceptualize a cycle. It has little if any practical value for real world engineering.

At any rate, YOUR described cycle has 1,2,3, etc. seemingly more than two points, so how does this relate?

Oh, I can't explain myself. You say. The "good students" will get it. For those who don't, who cares?

A convenient way for you to talk complete nonsense and put your own foot in your mouth spouting buckets of verbal diarrhea on a regular basis and blame it on others failure to "get it".

"Good students" have to swallow irrational crap without question and regurgitate it to pass tests and graduate. Doesn't make the nonsense true or real.

No I'm not a "good student". Sorry to disappoint

Here you can see actual, physical real world data. The results of actual experiment. Solid empirical data. Not charts, not graphs not hypothetical, theoretical "modeling", not questionable mathematical number juggling.

If you want to dismiss it offhand, that's your prerogative.

I have engines to build. REAL engines.

I need REAL data.