Stirlingengineforum.com

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...

Only if some working fluid has been bled from the engine at working temperature near TDC. Otherwise, you're not going to get the partial vacuum at BDC needed for the ambient atmosphere to do any pushing on the return stroke. Negative. We are seeing a strong return stoke with these engines while hav...

I've been avoiding the rock analogy. But to me rolling the same rock up and down a hill seems pointless. How about we ask, if the expansion stroke rolls a rock uphill, can the return stroke roll another rock up the same hill, thus performing more useful work? Not exactly. The point is, in doing wor...

If your asking me, conversion of heat into work is a "refrigeration effect" of a sort. I'm still working out exactly what I am asking so bear with me lol. And I mean thermal efficiency. Air cools down when it expands, and heats when it's compressed. If the air cools down further than it w...

There is a good reason why these equations are turning out with the values of 100 = 400. Because according to the Carnot limit, the 100 joules of heat added is not just in some kind of RATIO. It says, mathematically that the heat supplied actually IS all the heat down to absolute zero because it tri...

I mean, we are only talking about a piston returning to TDC after being pushed "up hill" against 1 atmosphere. Similar to lifting a weight against gravity or winding a spring. If it is "harder" to lift a heavier weight, the equivalent is still stored as "potential energy&quo...

Stroller wrote: ↑Wed Apr 24, 2024 2:57 pm The further the boulder makes it up the hill on the other side of the valley, due to it's momentum; the harder it was to push it up the hill on this side of the valley, due to its inertia.

So?

I mean, in principle. If you store "potential energy" in a bolder, by rolling it up a hill, it can then roll down. I don't think this is any different then driving out a piston "up hill" against atmospheric pressure. It can then return. But what happens when the bolder reaches th...