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Begining of derivation: Taking the above 'noted' equation "Qhz = Qcz + DQh" and subtracting Qcz from both The term 'n' applies to the engine regardless of 'Qcz' base heat amount or reference point. It will have the same efficiency burning the same energy at the same temperatures but with ...

Tom, why would the balloon inflate? The gas heats up, expands, pushes the piston outward causing some inflation. The engine then runs. Any leakage by the piston would make the piston run closer to the closed end. The gas going by the piston would cool and return to it's original volume. Hence it wo...

But maybe gas can "stretch" like a rubber band. IDK. But I can see that the piston in a Stirling engine does infact return. Without a flywheel. Without a displacer. With no apparent "sink" or viable heat outlet. I've done the experiments myself, so I can't really just dismiss wh...

Begining of derivation: Taking the above 'noted' equation "Qhz = Qcz + DQh" and subtracting Qcz from both The term 'n' applies to the engine regardless of 'Qcz' base heat amount or reference point. It will have the same efficiency burning the same energy at the same temperatures but with ...

a2+b2=C2 would be pretty useless if it only worked for equilateral triangles. Pythagoras is spinning in his grave. :laugh: There is of course, no such thing as an equilateral right triangle. Which is the point. There is no such thing as a "perfect" Carnot engine either An algebraic formul...

Thot = 600 Tcold = 300 Qhot = 150 joules You forgot the units, 'Kelvin', for the temperatures. You put temperatures into an energy equation, and mixed them with heat. The reason it worked in the previous example is that DQh and DT were both 100. It had a one to one relationship between energy and t...

Thot = 600 Tcold = 300 Qhot = 150 joules You forgot the units, 'Kelvin', for the temperatures. You put temperatures into an energy equation, and mixed them with heat. The reason it worked in the previous example is that DQh and DT were both 100. It had a one to one relationship between energy and t...

Suppose we use my original values ? Thot = Qhz = 600 (but also 450 ?) Tcold = 300 = Qcz Qhot = 150 joules = DQh DQh = Qhz - Qhz(1-n) 150 = 600 - 600(1 - .25) 150 = 600 - 600(0.75) 150 = 600 - 450 150 = 150 Hey, worked out! But elsewhere Qcz + DQh = Qhz Qcz + DQh = Qhz 300 + 600 = 900 So Qhz can equa...

Can we try some other more "realistic" (in your opinion) values? How about Thot = 600 Tcold = 300 Qhot = 150 joules Anything wrong with those numbers? Equations are supposed to balance right? Isn't that what the equal sign is for? Isn't that what mathematical precision is all about? We ass...

2+x=4 What does x equal? X=5 2+5=4 7=4 The equation must be wrong. Give me a break. An equation isn't valid for any numbers you want to put in. An equation is only valid if use properly. It should at least be valid for the potential range of application, don't you think? Or are you presupposing ONL...

But by doing work and loosing energy and getting cold, the gas contracts. By contracting, this allows outside atmospheric pressure to do the work of returning the piston to TDC. The gas by pushing to expand, by pushing, the volume increases, because it was heated to a higher temperature, using that...

From that article I thought this was interesting. Something I haven't seen before: So for 𝑇 > 𝑇inv, an expansion at constant enthalpy increases temperature as the work done by the repulsive interactions of the gas is dominant, and so the change in kinetic energy is positive. But for 𝑇 < 𝑇inv, expans...

..., I would imagine the biggest reduction in temperature would be from expanding gas into a vacuum where the gas is able to use all it's kinetic energy to expand unrestricted. Actually it's just the opposite. When expanding into a vacuum the gas has nothing to do work against so doesn't loose ener...

the internal gas pressure is still contributing, still doing work and still actually losing internal energy so still dropping in pressure and temperature due to work output. This is the part I'm having trouble with. The working fluid can only still be doing work while it's above ambient pressure, b...

Some late night experimenting, but no great results so far. https://youtu.be/AaEfauPTqYo?si=0oBcnqy-1X2BbVqA If the flywheel assembly is removed, which can be accomplished rather quickly by just loosening the big hose clamp, the engine will run, apparently quite strongly. Put the flywheel back on an...