Theoretical vs. Experimental Single Process Expansion

[VincentG]

Hopefully Fool or others can help with this. This will be based on my 60l drum project so the final numbers may change. https://www.stirlingengineforum.com/vie ... f2af24984b

Let's start with a displacer having a diameter of 13" and a stroke of 3" making internal air volume roughly 6,500cc.

Starting air temperature is 50 degrees F and the final air temperature will be 200 degrees F.

The power piston will have a diameter of 13" and a stroke of .25", displacing 540cc of air.

1: How many joules are required to raise the 6,500cc of air to 200 degrees F, considering the final volume will be 540cc more than the start?

2: How much weight, in lbs. can be raised .25" with that number of joules, considering 100% efficiency?

3: What is the theoretical weight that can be raised .25" with the expanding air given real world efficiency but assuming zero friction loss?

Further discussion:

1: How much heat will actually be converted to work output if heat input is limited to constant volume addition, blocked off, and then allowed to expand 540cc?

2: What differences will monoatomic gas make in this experiment?

3: Any links to such experiments?

[matt brown]

Checking source and sink values, 50F = 283k and 200F = 366.5k whereby ideal Tr = 1.286 (real Tr will be far less)

Checking your basic dimensions, DPvol = 6517cc and PPvol = 543cc with DP/PP = 12 whereby swept Vr = 1.08333

You need to rethink your volumes here...DP volume must exceed DP swept volume by some type of clearance volume that depends upon scheme itself. As far as I can tell, your small heater "reservoir" will be part of total DP volume but not part of DP swept volume. Likewise, you will probably need a small clearance volume between DP and PP. These "dead" volumes must be included in total DP volume as they will effect all real PVT values.

I think I'm getting the layout...(very) short stroke PP at top with a rolling seal between PP and DP case, similar Bumpkin's inner tube idea ??? PP is centrally located and as heavy as possible whereby upstroke is largely all Wneg and downstroke is largely all Wpos. Any 'vacuum' achieved after TDC (during downstroke) becomes...icing on the cake, but a thing apart vs common LTD model. Overall, very similar an early Watt steam engine (inverted config).

[VincentG]

Matt, for this exercise assume the stated temperatures are actually reached by the gas. Assume there are no clearance volume issues, such that displacer piston kisses power piston when dp is up and pp is down(does that make sense?). Also ignore any hot reservoir effects that may exist in the real chamber for now as I can start with a totally cold chamber and then introduce heat and observe the difference. No rolling piston, just a single rubber diaphragm similar to Shane's latest engine.

Here's the basic idea, based on a previous thread I started. Force x 0" (distance)= no work. Force x.001" (distance)= work. So it's about gaming the potential bell curve of when force becomes work, and when work ends and just becomes a force. And on top of that, just to help my dense skull visualize the losses in a heat powered expansion cycle.

[matt brown]

I was letting you off easy with steam engine reference, since the super-Carnot give away here is how much WEIGHT can be raised.

This simple scheme should raise some eyebrows...convert all heat to work in a single expansion process that requires no compression process to complete the 'cycle'. However, there's a slight flaw in your logic and I'm curious how Fool will sort this out.

[VincentG]

Ok Matt, just looked up Watt's engine. Very cool and I see the similarity.

Don't think of a heavy piston for this example. Upstroke is Wpos, raising weight, weight is held up after being raised.

[Fool]

50 F is approximately 283 K
200 F is approximately 366.5 K

T1 = 283 K
T2 = 366.5 K

Volume
V1 = 6,500cc
V2 = 6,500cc + 540cc = 7040cc

P1 = 14.7 psi

P1V1/T1 = P2V2/T2

P2 = P1V1T2/(T1V2) = 14.7•6500•366.5/(283•7040) = 17.577 psi absolute

Piston area
13^2•π/4 = 132.7 in^2

Weight max :
132.7(17.577 - 14.7) = 381.77 lbs max, ideally.

Questions :

1. Get the Cv and convert all numbers to SI, mks. More than I care to do right now on my puny little cell phone calculator. Be sure to use the correct Cv. Moles vs kilograms. J = M•Cv•DT. That won't be exact but volume changes little enough that it should be pretty good.

2. About 382 lbs. See above derivation.

3. As a guess, cut that value in half, or a quarter, or if you can 1/10. Real world you may only get half the Delta T, or worse. As it runs faster it will be worse. Heat conduction into air will be much slower as the volume increases. Plus, that is a maximum weight. For fast action, try to minimize it as low as possible.

Further discussion:

1. Similar to the calculations that I gave in the "Let's Beat Up Carnot" thread. They demonstrated a much larger work output if heated through the entire expansion stroke, and, hence, cooled through the compression stroke.

2. I'm not sure. Look up the Cv, Cp, alpha, and gammas for the gasses you are modeling.

3. Try searching for 'gas calculator', or 'thermodynamic calculator'. The Wikipedia link I gave in your .....Carnot thread has formulas. I put them into a spreadsheet. It took three months, LOL.

Good luck. Awesome project. Bumpkin is working a similar project. Both of you are way ahead of me. Thanks for sharing.

[VincentG]

Thank you for your time Fool. I can and will do the maths myself but I wanted a bit of a peer review going in. I'd rather spend my time in the shop than behind the keyboard. Updates will follow.

[Fool]

Me too. Alas I spend it mowing and hobby farm maintenance. That is fun too. Thanks for letting me share here.