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# Gravity Torus Heat Engine

Like a Minto Wheel, but better
 (+7) [vote for, against]

Start with a tube of thermally conductive material. The inner diameter of this tube should be relatively small, and the specific heat of the tube material should be low.

Shape the tube into a circle (effectively a torus). The diameter of this circle should be as large as it can conveniently be made.

Fill the circular tube with the working fluid, and seal it. Measured by volume, the fluid should be half liquid, half gas.

Mount the circle on an appropriate rotary bearing, oriented so that each point on the circular tube cycles between the maximum altitude and the minimum altitude.

Looking along the axis, apply a heat source just to the right of the bottommost point, and a heat sink just to the left of the uppermost point.

Since the pressure throughout the system is nearly the same at all points, only a very small temperature differential is needed to boil the liquid fluid at the heat source and condense the gaseous fluid at the heat sink.

Due to the locations of the heat sink and heat source, and due to gravity, and the difference in density between gas and liquid working fluid, the condensate will flow down the left side of the circle, and the boiled fluid will flow up the right side of the circle.

In other words, the fluid will flow counterclockwise through the tube.

Because of the small inner diameter of the tube, the flow of the fluid (both the gas and the liquid) will be restricted. Due to conservation of kinetic energy, this restriction of flow translates into a transfer of momentum from the working fluid to the tube.

Thus, the torus will be made to spin counterclockwise by the fluid flowing through it.

Now, I'm not going to pretend that this is a highly efficient heat engine -- it can't be, since it should work across a very low temperature differential -- but I do think it's probably one of the simplest to build.

After all, it only has one moving part, and a very simple one at that.

 — goldbb, Nov 08 2009

Minto Wheel http://my.voyager.n...all/MintoWheel.html
[goldbb, Nov 08 2009]

Another Minto Wheel Page http://www.genuinei...opicSolarWheel.html
[goldbb, Nov 09 2009]

How Tesla Turbines work. http://www.gyroscop...aTurboMachinery.pdf
"... the rotor efficiency using laminar flow can be very high, even above 95%" [goldbb, Nov 16 2009]

Cool. Tilted slots on the inside of the torus could collect condensation and add a bit more force.
 — 2 fries shy of a happy meal, Nov 08 2009

 For a small diameter tube, slots shouldn't be necessary.

 The surface effects (tension and adhesion between the working fluid and the tube's inner surface) should suffice to transfer momentum from the fluid to the tube.

For a large diameter tube, slots or fins *might* be useful, but I don't see any non-complicated non-costly way of adding them.
 — goldbb, Nov 08 2009

hmmmm, if your thermally conductive material was aluminum you could press the shape you want and fins in a series of crinkles. I don't know how you'd go about sealing it though.
 — 2 fries shy of a happy meal, Nov 09 2009

 In the Minto Wheel, each section is not connected to the adjacent sections -- they're connected to diametrically opposite sections.

 In the Minto Wheel, as the bottommost tank is heated, some of the liquid boils. The gas created pushes some of the liquid upwards, through the crossover pipe to the topmost tank. After the liquid has moved to the topmost tank, gravity pulls that tank downwards, generating torque.

I've just added a link with a better description, and a photo of an actual Minto Wheel, made of glass.
 — goldbb, Nov 09 2009

 The advantages of my design over a Minto Wheel are:

 1) Mechanically less complicated.

 2) Capable of functioning with a smaller temperature difference between heat source and heat sink.

 3) Capable of operating (generating torque) at *much* higher RPM than a Minto Wheel.

 If we assume that 100% of the torque comes from the drag between the liquid and the tube, then torque will at the maximum at 0 RPM, and reach zero when it takes the same amount of time for liquid working fluid to move from the heat sink to the heat source, as it would for that liquid to freefall the same distance.

 However, some portion of the torque will come from drag between the gaseous working fluid and the tube, so in theory, even at that high speed, some torque should be produced.

 In practice, the thermal conductivity and thermal mass of the material of the tube will be the limit on the maximum speed at which torque will be produced.

If the working fluid was water, and the tube material glass, and the heat source a magnetron (heating the water directly), then I don't foresee any upper limit to the device's speed.
 — goldbb, Nov 10 2009

 What's so special about adding baffles / crinkles / slots, that everyone (except me) seems to want to add them to this idea? Sure, they'll transfer kinetic energy from the fluid to the rotor, but they'll also create turbulence, which converts kinetic energy into heat -- mostly in parts of the device where added heat won't improve efficiency.

 Furthermore, what's so "optimistic" about relying on drag (not to be confused with friction) to transfer kinetic energy from fluid to rotor?

The turbine / pump invented by Nicola Tesla uses drag in that manner, so I don't see why I can't here.
 — goldbb, Nov 12 2009

should be able to do it with a solid metal wheel: up top it gets colder and contracts while down below it gets warmer and expands... so the top is heavier.
 — FlyingToaster, Nov 12 2009

 I'm not sure it would work. An open ring allows an easy outlet for the fluid to readjust without torquing your ring.

Minto's wheel uses the heat to shift the water in closed vessels so has no outlet for the fluid to readjust.
 — wjt, Nov 13 2009

 You seem to believe that zero kinetic energy will transfer from the working fluid to the torus.

 Here is a proof (by contradiction) that the amount of energy transfered is nonzero.

 First, assume that zero energy is transfered from the working fluid to the torus. The interior of the torus will then have no friction and no drag.

 Logically, the torus won't spin at all.

 Instead, the fluid within will spin... at continuously increasing speeds. Eventually, the material of the torus will fail, due to being unable to supply enough centripital force to keep the working fluid travelling in a circle. Put another way, the centrifigal force of the fluid will burst the torus.

 If the torus were infinitely strong, something else would happen. Since energy (including kinetic energy) has mass, the working fluid (with it's ever increasing speed) will create a space-time distortion, and eventually become a black hole.

 This is all unlikely in the extreme, therefore we can conclude (by contradiction) that a non-trivial amount of kinetic energy will be transfered from the working fluid to the torus. Due to the location and direction of the applied force, this becomes a torque when measured at the axis of the torus.

 ...

Even if the amount of kinetic energy transfered from fluid to torus is small, the engine will still function. The speed of the working fluid will increase until the rate of kinetic energy lost due to drag against the torus equals the rate of energy added to the system.
 — goldbb, Nov 15 2009

What does "won't work" mean?
 — goldbb, Nov 18 2009

 [goldb], please correct me if I'm wrong, but I gather that in this idea, you've got the torus half full of liquid at any time? And you're expecting the drag from the condensed liquid flowing down the one side of the pipe to rotate the entire aparatus? In other words you've got a 50% charge of liquid inside, and you're spinning the torus around the liquid, ie the liquid stays at the bottom and the torus spins around it?

 Are we forgetting about friction of the liquid at the bottom against the torus spinning around it? This will be seen as drag against the torus spinning. It will be higher (roughly proportional to the square of the speed increase) the faster the torus is spinning.

 I'm going out on a limb here, but I'd say that the friction (drag) force of the charge of the liquid against the spinning torus will be higher than anything contributed by the condensing liquid.

Then again, I might just have misunderstood what you're talking about. A sketch, perhaps?
 — Custardguts, Nov 19 2009

 Yes, the torus is half full of liquid at any time.

 Yes, I'm expecting the drag from the condensed liquid flowing down one side of the torus (and the drag of the gas rising up the other side of the torus) to rotate the apparatus around the axle.

 No, I'm not "rotating the torus around the liquid" (mechanically spinning the torus, to spin the liquid). I'm heating and cooling the fluid, to cause the fluid to rotate, and the torus moves as a consequence of drag between the fluid and the torus. The rate of rotation of the torus is slower (at all times) than the rate of rotation of the fluid.

 The portion of the torus filled with liquid isn't the bottom half, but (viewed along the axis) on the left side (from the heat sink that's just to the left of the top, down to the bottommost point, across the very short distance to the heat source).

 The bottom right portion has no liquid in it, only gas, because the liquid boils as it passes through the part of the torus heated by the heat source.

 The liquid's drag doesn't slow the torus, because the liquid is moving faster than the torus is.

 The gaseous portion of the working fluid is rotating in the same direction as both the liquid and the torus, but at a hugely faster rate... it's drag against the torus also rotationally accelerates the torus.

 According to Wikipedia, the equation describing the force of drag of an object through a fully enclosing fluid is:

 F_D = 1/2 * rho * upsilon ^ 2 * C_D * A

 Where F_D is force of drag, rho is the mass density of the fluid, upsilon is the velocity of the fluid, C_D is the coefficient of drag (0.001 in the case of a laminar flow across a flat plate which is parallel to that flow), and A is the area.

 The gaseous working fluid will have a lower mass density, but a much higher velocity, than the liquid working fluid -- so I can't say whether the gas's drag on the torus will be more than or less than that of the liquid's drag.

 If no mechanical energy is extracted from the torus, and it's axis is has sufficiently low friction, then the torus will *eventually* spin as fast as, or faster than, the liquid portion of the fluid spinning within it; when this occurs, drag caused by the fluid on the torus will slow the torus.

If (at this high speed) the drag of the gaseous working fluid on the torus is greater than the drag of the liquid working fluid on the torus, then the torus will continue to gain rotational speed; otherwise the torus will cease to gain rotational speed. I don't know which would occur, but I suspect that the speed will continue to increase.
 — goldbb, Nov 19 2009

 //The portion of the torus filled with liquid isn't the bottom half, but (viewed along the axis) on the left side (from the heat sink that's just to the left of the top, down to the bottommost point, across the very short distance to the heat source).//

 Okay I need a sketch now. The way I see it, the body of liquid is going to want to settle out under gravity effects, and you've got no way of stopping it doing so. Therefore, you're proposing to boil it off at the same rate that it's trying to flow under gravity. Depending on the size of the vessel, this will be at some fantastic rate. you're going to have problems just getting a torus material with sufficient thermal conductivity for this. Maybe diamond.

 //According to Wikipedia, the equation describing the force of drag of an object through a fully enclosing fluid is: F_D = 1/2 * rho * upsilon ^ 2 * C_D * A Where F_D is force of drag, rho is the mass density of the fluid, upsilon is the velocity of the fluid, C_D is the coefficient of drag (0.001 in the case of a laminar flow across a flat plate which is parallel to that flow), and A is the area//

Whoa there nelly, I smell a horribly simplified model. At the very least, we need some measure for viscosity of the liquid. Secondly, it's the exact opposite to "an object through a fully enclosing fluid" - I'd be looking to modify the existing and very well understood pipe flow equations. There's a lot of work been done for steam piping, which includes gaseous, liquid, and inter-phase fluids. I very much doubt you'll be dealing with laminar flow.
 — Custardguts, Nov 19 2009

''But, sure", he says. Barman gets out a firecracker. Puts it through the middle of the Lifesaver. Lights it and puts both in a glass which he hands back to the guy.
 — wjt, Nov 20 2009

 /Okay I need a sketch now./

 Sadly, I lack the skills to produce a computer sketch -- however, the paragraph you wrote immediately following that sentence accurately describes the process.

 /Therefore, you're proposing to boil it off at the same rate that it's trying to flow under gravity. Depending on the size of the vessel, this will be at some fantastic rate./

 Assuming that water/steam is the working fluid, then I want the inner diameter of the tube to be the same as the inter-disk spacing that a Tesla Turbine would use, for water or steam. In Wikipedia's description of the Tesla Turbine, 0.4mm (0.016 in) is indicated.

 A tube this small size will have a very large surface to volume ratio, and should be easy to heat, even if it's copper.

 /You're going to have problems just getting a torus material with sufficient thermal conductivity for this. Maybe diamond./

 Don't forget, the tube material has thermal mass, as well as conductivity. The heat source heats the tube, then the heat that's been added to the tube heats the liquid. Even if each bit of the tube is only in very brief contact with the heat source, the liquid will be in contact with the tube for a longer period of time.

 As long as the heat is conducted from outside to inside a *reasonable* amount of time, it should work.

 By "reasonable," I mean that if the torus is turning counterclockwise, and heat is applied at the bottom, and the location where the liquid boils is a few degrees counterclockwise of the bottom, then that's ok.

 If the location where the liquid boils is up near the top, or worse, past the top, then that's not ok.

 /I very much doubt you'll be dealing with laminar flow./

 Given the completely smooth inner surface of the tube, and the very small inner diameter of the tube what is going to cause the flow to be non-laminar?

 //Guy walks into a bar//

Surely the only lifesaver one would expect to have in a bar, would be the type made from melon liqueur, pineapple juice, and rum... I don't think it contains enough alcohol to explode.
 — goldbb, Nov 22 2009

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