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I think that chart is for steady state currents and temperatures and COP's.

At 0.1 times Amax, it will produce 10° of delta T, at a COP of 2.

At 0.2 times Amax, it will produce 20° of delta T, at a COP of .7

Very tiny amounts of power for small steady state COP's, and delta T's, max.

So does that mean the "M" is for Mythical. LOL Don't forget my Unicorn.

It doesn't look like it's supposed to run on ambient energy, as it shows an outside heat source just before that heat exchanger, rendering the heat exchanger useless. Qin #2.

Jack, I offer no disrespect. I think you should try it. We probably don't "know" anything. However, lots of people, some are scientists, have tried lots of things and gotten them to apparently work. Reinventing the wheel, or worse reinventing a failed square wheel, is time wasted. I've lea...

Tom, the Peltier displacer should work just fine. It will produce a temperature difference, hotter and colder inside the engine. The entire displacer chamber could be insulated and it will still work fine, maybe even better. I liked the hollow displacer rod and the wires running up through it. Put a...

Jack, I think you could use a vacuum cleaner and mason jar with a pinwheel like turbine inside for a simple test. A bit more testing could be done with a five gallon bucket and larger turbine. The following is an interesting homemade kinda inside out Tesla turbine: https://m.youtube.com/watch?v=h02V...

... "heat of compression" , a misleading statement. No heat involved. ... Your dissertation does not address Work . Seems to deny a work-heat relationship. "Heat of compression" is a transfer of energy. Work input results in an increase in temperature. Calling that increase in t...

Matt, Thanks for the kudos. I've procrastinated responding to this thread because I've been busy and the second and third posts have questions and comments that require careful wording. The easiest to explain is Jack's point. He is very close to understanding the difference between heat, work, compr...

I think the Phillips Company had nothing slip by. In the mid of century numbered "19", 1900's, they were building Stirling Engines and Cryo Coolers that people contributing to this forum only dream of. Thanks for the attaboys, and thanks for the ah rats too. The attaboys are more fun. LOL ...

In fact I'll do it again. Qh heat added from Tc to get to Th. Uh total energy contained in the gas at Th Uc total energy contained in the gas at Tc Qh=Uh-Uc 100=400-300 Heat equals the change in internal energy. Starting point one: Uc=MCvTc 300=MCv300 Maximum point two: Uh=MCvTh 400=MCv400 MCv= Mass...

Only the ratios are equivalent. Because the differences cancel.

That was shown many times over. The 45 pages of waste is from you not seeing the obvious.

Heat is never hot or cold. Internal energy can be hot or cold. Speaking only thermodynamically, of course. Heat is a description of the transfer of energy from hot to cold. Internal energy is a condition, a state, of how hot something is compared to absolute zero Kelvin. Internal energy is calculate...

The heat doesn't just jump off the bolt and head for the water on its own. Heat transfer in the air is by conduction, convection, and radiation. Which one of these are not the heat "jumping off and heading to the water"? None. For your statement to pass scientific testing, all three must ...

I also don't think that Tom will ever understand how the differences between a state variable and path variable cancel when put into a ratio, and how they are depicted on a PV diagram.

Still he tries to equate, heat a path variable with, internal energy a state variable. He just loses it every time.

Yes, I get it, but take another look at your Q=.................=U and try to justify how any path function can (ever) equal a state function ??? I said I was wrong. Good catch. I don't think Tom will ever understand how a state variable differs from a path variable. The heat doesn't just jump off ...

Tim Booth wrote:But I think we've had this conversation before. If you keep adding wood to a fire it doesn't mean wood is not being added or burned if the fire looks the same after three days.

What the fire looks like is internal energy. The wood added is heat in. The wood burnt is heat out.