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It would mean more to people if they took the time to do the math themselves.

I’m certain this perpetual motion machine won’t work.   I think I see a simple physics mistake that caused you to think it might, Jim. Though one would intuitively think, as you do, that the pressure

The height in both would be equal. I think Craig did well explaining why.   To find the height divide the pressure inside the container (at the top) by the pressure outside. Subtract that result fro

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It would mean more to people if they took the time to do the math themselves.

 

Then let me help to fill you in a little. If you click that link up in the top right corner that says "Science Forums Rules" and scroll down to the part that says "How should I behave" you'll find that you are expected to back up your claims.

 

I'll also fill you in on another secret. I've spent over 30 years as a hydraulic systems specialist using math to design and diagnose hydraulic systems and I know for a fact that pipe size has got nothing to do with the static head. The pressure at the bottom of a vertical 1" pipe is the same as it is in a 10" pipe in a static state. Add flow to the equation, a dynamic state, and pipe size then becomes part of the equation.

 

Now, you've claimed that volume is a variable in the static head equation and I expect you to back that up. Craig and I have both provided links to support the claim that volume is not a variable in the equation. Now it's your turn.

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The device has two reservoirs and two columns. Both columns will have the same static head. However, since one reservoir starts at a higher elevation, the static head ends up taller (bottom to top will still be the same length). The device will work for a while. But once the two reservoirs have the same height or the higher one empties, it will stop.

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As one person mis-stated, volume has everything to do with the static head. This would include the diameter of the pipe or tube.
Lets consider a vertical stretch of tube in which the area of the cross section is constant (or at least reasonably so you can consider a single value for it). The weight of liquid is proportional to the area, right? How do you compute the pressure difference from top to bottom of that stretch? Isn't it the weight-area ratio? If so, the pressure difference is directly and inversely proportional to the area and hence doesn't depend on it, it cancels out.

 

Twice the weight, spread over twice the area, gives the same pressure.

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  • 2 months later...
Here is a some what simple design that would have the potential of working...
Won't work - as has been pointed out the head on both sides is the same.

 

Try it, then you'll have more education and experience and will find that we're right. I hold a degree in mechanical engineering and have 20 years experience in fluid systems; I assure you that not only will it not work, it can't work.

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  • 2 years later...
Guest Aemilius

There may be a way, in principle at least, to cause the "Self Siphoning of Water". The fluid employed may change though so I'll define the whole thing as continuous circulation of a fluid in a closed system consisting of a single resevoir under normal atmospheric pressure along with a single U shaped tube. I may use more than one resevoir initially for a couple of the foundational example diagrams. I'll just make one post after another to describe it until an error becomes apparent. Here we go....

 

Here's a siphon (already filled with water). The ends of the inverted U-shaped tube terminate beneath the surface of the water in the resevoir. In this situation nothing will happen.... the height of the two water columns (as measured from the surface where they emerge from the water) created by the tube are equal, and gravity acting on the water is uniform (in the tube and the resevoir). Normal atmospheric pressure coming to bear on the surface of the water is exerting equal pressure at the base of each column of water keeping the water from just falling back into the resevoir (as it would in a vacuum). The sum of all the equal and opposite forces acting on the system is zero.... static equilibrium.

 

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Guest Aemilius

Here's a siphon (already filled with water) where the two ends of the inverted U-shaped tube terminate beneath the surface of the water in two separate resevoirs with two distinct water levels. In this situation water will move up the tube from the resevoir on the right (the higher level) and down the tube to the resevoir on the left (the lower level) until the height of the two water columns created by the tube (as measured from the surface where they emerge from the water in each of the resevoirs) are equal. At that point, the height of the two water columns created by the tube will be equal, and gravity acting on the water will be uniform (in the tube and the resevoirs). Normal atmospheric pressure coming to bear on the surface of the water in both resevoirs will be exerting equal pressure at the base of each column of water keeping the water from just falling back into the resevoirs (as it would in a vacuum). The sum of all the equal and opposite forces acting on the system will be zero.... static equilibrium.

 

Edited by Aemilius
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Here's a siphon (already filled with water) where the two ends of the inverted U-shaped tube terminate beneath the surface of the water in two separate resevoirs with two distinct water levels. In this situation water will move up the tube from the resevoir on the right (the higher level) and down the tube to the resevoir on the left (the lower level) until the height of the two water columns created by the tube (as measured from the surface where they emerge from the water in each of the resevoirs) are equal. At that point, the height of the two water columns created by the tube will be equal, and gravity acting on the water will be uniform (in the tube and the resevoirs). Normal atmospheric pressure coming to bear on the surface of the water in both resevoirs will be exerting equal pressure at the base of each column of water keeping the water from just falling back into the resevoirs (as it would in a vacuum). The sum of all the equal and opposite forces acting on the system is zero.... static equilibrium.

 

 

How long does it run?

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Guest Aemilius

The water (in this diagram) will flow from the resevoir on the right through the tube to the resevoir on the left until the height of the two water columns created by the tube (as measured from the surface where they emerge from the water in each of the resevoirs) are equal.... It will "run" for as long as it takes for that to happen.

 

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Guest Aemilius

So now, I want to begin exploring/describing how the behaviour of the fluid in the system might be affected using magnetism. Since water is only weakly diamagnetic, for the rest of this thought experiment, in order to enhance the magnetic susceptibility of the weakly diamagnetic water, we'll use a (saturated?) solution of bismuth chloride and water (henceforth referred to as "fluid"). Below is depicted a straight tube open at both ends filled with fluid. The lower end of the tube terminates beneath the surface of the fluid in the resevoir and the upper end is exposed to normal atmospheric pressure (seen in the diagram at the moment of exposure) with no externally applied magnetic field. In this situation, under the force of gravity, the fluid will flow downward into the resevoir until the level of fluid inside of the tube and the level of fluid in the resevoir are the same, and air under normal atmospheric pressure will flow into the upper end of the tube to occupy the volume formerly occupied by the fluid. The sum of all the equal and opposite forces acting on the system will be zero.... static equilibrium.

 

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Guest Aemilius

Moontanman "Shouldn't you start your own thread for this?"

 

If you study the thread title very, very carefully, you'll notice the topic is "Self Siphoning Water" (otherwise known as "fluid"). I'm in the process of exploring a scenario consistent with that so.... No, I don't see any need to start another thread about the same topic.

 

Is there anything else I can help you with?

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I'd originally thought the highest permanent magnet field strength currently available was 1 Tesla, but looking around I see there's actually a 5 Tesla permanent magnet under development.... so for this part of the thought experiment I'll use a cylindrical 5 Tesla permanent magnet with a hole through it (for the tube to pass through).... one pole at the top and the other pole at the bottom. Below is depicted the same straight tube open at both ends filled with fluid. The lower end of the tube terminates beneath the surface of the fluid in the resevoir and the upper end is exposed to normal atmospheric pressure (seen in the diagram at the moment of exposure) as before but now with the 5 Tesla permanent magnet installed around it at about the middle of the tube providing an externally applied magnetic field. In this situation, as the fluid flows downward into the resevoir under the force of gravity, its advance will be diamagnetically resisted as it enters the externally applied magnetic field from above, and it's retreat will be diamagnetically aided as it exits the externally applied magnetic field below on it's way to the resevoir with a net gain of zero in either upward or downward force. Essentially it will behave just as if the magnet was never installed, with the same result as before.... the fluid will flow downward into the resevoir until the level of fluid inside of the tube and the level of fluid in the resevoir are the same, and air under normal atmospheric pressure will flow into the upper end of the tube to occupy the volume formerly occupied by the fluid. The sum of all the equal and opposite forces acting on the system will be zero.... static equilibrium.

 

Edited by Aemilius
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