Circulating a fluid at a very high speed in a spiral pipe

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  • #26
cjl
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Unfortunately, it still will not work. The fact that the entire system is not coplanar is irrelevant - if you were to analyze the forces on the fluid (and on the pipe) along the entire path, and integrate them to find the net force, it would still be zero. This is a case where your intuition is leading you astray - the same principle as applies in the simple case still applies here, but due to the increased complexity, it is no longer as obvious.
 
  • #27
A.T.
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Next up:

pipemaze.jpg
 
  • #28
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Imbalance

Unfortunately, it still will not work. The fact that the entire system is not coplanar is irrelevant - if you were to analyze the forces on the fluid (and on the pipe) along the entire path, and integrate them to find the net force, it would still be zero. This is a case where your intuition is leading you astray - the same principle as applies in the simple case still applies here, but due to the increased complexity, it is no longer as obvious.
From another point of view, you are wrong, I'll tell you why... Let's see what happens along the entire path with a very simple example:

http://www.2live.ro/demo/ball.bmp

Count the turns in each side and imagine you launched a metal ball inside the pipe. As you can see from the picture, there is an imbalance in one side which contains one more arc. When the ball has reached the final, one force will be generated exactly like in case of a simple u-piece or similar with this:

http://www.2live.ro/demo/Serpentina1.bmp

I think the force will be just down, like shown in the picture, due to momentum change. It is the same force as with a simple U-piece. To find the distribution go to the rotating rest frame of the water in the turn and consider the pressure gradient caused by the centrifugal force.

For the gyroscopic effect the number of windings does matter, because you have more mass and thus more angular momentum.
Now let's see the difference between first picture and this one:

http://www.2live.ro/demo/arc_open.bmp

Where's the imbalance? Disappeared? I need to know which principle says the result is zero in this case.
 
  • #29
A.T.
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  • #30
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:)

The principle is called Conservation of Momentum:
http://en.wikipedia.org/wiki/Momentum#Conservation

The net force on the pipe by the fluid equals the negated net momentum change of the fluid in the pipe.
You said: "It is the same force as with a simple U-piece", about this:

http://www.2live.ro/demo/Serpentina1.bmp

And this one is too complex... Like a maze and because of Conservation of Momentum, everything is zero:

http://www.2live.ro/demo/ball.bmp

If you take out the coils you will have a U-piece.

Closed system - one that does not exchange any matter with the outside and is not acted on by outside forces
Closed loop is not equal with closed system (electromagnetic fields can carry energy and momentum).
It's easy to invoke a rule when you don't have an argument.
 
  • #31
A.T.
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And this one is too complex... Like a maze and because of Conservation of Momentum, everything is zero:

http://www.2live.ro/demo/ball.bmp
No, in the one above the net force is not zero. But in the one below it is zero:

http://www.2live.ro/demo/arc_open.bmp


It's easy to invoke a rule when you don't have an argument.
You asked which principle, I told you the principle. I also gave you my argument: The net force on the pipe by the fluid must equal the negated net momentum change of the fluid in the pipe.
 
  • #32
cjl
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And this one is too complex... Like a maze and because of Conservation of Momentum, everything is zero:

[PLAIN]http://www.2live.ro/demo/ball.bmp[/QUOTE] [Broken]

No - actually, conservation of momentum works perfectly well in this case. All you need to do is look at the flow entering the pipe, and the flow exiting the pipe. In this case, the flow entering the pipe is going upwards, but the exit flow is downwards. Thus, by conservation of momentum, there must be a net upwards force on the pipe to counter the downwards force on the fluid. However, this all changes when you bend the above system:

http://www.2live.ro/demo/arc_open.bmp

Now, the incoming flow (I'll assume the inlet is on the left) is traveling to the left. The outcoming flow is also traveling to the left. Assuming the pipe is constant diameter, and the fluid is incompressible, the flow speed is the same at the inlet and outlet. Therefore, since the flow speed and direction are identical at both the inlet and the outlet, there is no net momentum change in the fluid, so there is no force on the pipe.

I think you're confusing yourself with the spiral pipe. It's true that in the first image above, the net force can be considered to be entirely from the U-shaped section, and the forces in the spirals all cancel out. However, this is no longer the case when you bend the spirals. The force nearly cancels out in the spirals when they are bent (so long as the bend radius is large relative to the spiral radius), but it doesn't entirely cancel out, and if you felt inclined to do the vector sum along the entire path, you would discover that the slight imbalance in the spiral is exactly enough to counteract the force in the U-shaped section. That method is absolutely valid, it simply isn't used very frequently since conservation of momentum allows us to arrive at the same result with much less effort (and it allows us to bypass errors of intuition more easily).
 
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  • #33
KHMOU YOUSSEF
I propose the following :
Consider the geometry you described in Cartesian base (X,Y,Z), so you have incompressible flow with constant radius of the pipe, with Z changing, you can apply the Bernoulli law :
V^2 / 2 + g*Z+P/rho= constant

You can compute the force by deriving the pressure from the above equation
 
  • #34
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Thank you

No - actually, conservation of momentum works perfectly well in this case. All you need to do is look at the flow entering the pipe, and the flow exiting the pipe. In this case, the flow entering the pipe is going upwards, but the exit flow is downwards. Thus, by conservation of momentum, there must be a net upwards force on the pipe to counter the downwards force on the fluid. However, this all changes when you bend the above system:

Now, the incoming flow (I'll assume the inlet is on the left) is traveling to the left. The outcoming flow is also traveling to the left. Assuming the pipe is constant diameter, and the fluid is incompressible, the flow speed is the same at the inlet and outlet. Therefore, since the flow speed and direction are identical at both the inlet and the outlet, there is no net momentum change in the fluid, so there is no force on the pipe.
Thank you for your answer, now I understand quite well how conservation of momentum works. My knowledge of physics are limited, this is why I sought clarification in this forum.
I made ​​few tests in a rudimentary manner as shown in video in previous posts where I used a diaphragm pump and a flexible/elastic pipe shaped as shown in pictures. Forces represented in the pictures are observed in the experiments, results of experiments generating my confusion.
I assume that these forces have emerged due to differences arising between inlet and the outlet. Diaphragm pump is generating a pulsating flow, in combination with elastic pipe, attenuation occurs on pulsating flow so that the resulting output flow is approximately laminar flow. I assume that between inlet and outlet there was a momentum change in the fluid.

I assume that in this case, if correct, manipulating fluid flow in certain areas of the pipe can lead to the generation of desired forces.
 
  • #35
A.T.
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I made ​​few tests in a rudimentary manner as shown in video in previous posts where I used a diaphragm pump and a flexible/elastic pipe shaped as shown in pictures. Forces represented in the pictures are observed in the experiments, results of experiments generating my confusion.
I assume that these forces have emerged due to differences arising between inlet and the outlet. Diaphragm pump is generating a pulsating flow, in combination with elastic pipe, attenuation occurs on pulsating flow so that the resulting output flow is approximately laminar flow. I assume that between inlet and outlet there was a momentum change in the fluid.
I explained that to you in post #9

I assume that in this case, if correct, manipulating fluid flow in certain areas of the pipe can lead to the generation of desired forces.
What are the forces you desire. What is the goal here?
 
  • #36
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What are the forces you desire. What is the goal here?
To find a way to produce more thrust than this one: Ion Thruster, using only electric current.
 
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  • #37
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What if?

What if the system is splitted in 3 sections: blue, yellow and red?

http://www.2live.ro/demo/sections.bmp

Blue section - fluid flow in this section is constant preserving the velocity vector, the mass in this case comes into pipe and leaves the pipe with the same momentum.
Yellow section - this section is composed by a liquid slug generator (gas/liquid) and a liquid separator.
Red section - the role of this section is to accelerate the liquid slugs. The high momentum of the liquid slugs create considerable force. We have compressibility and pressure wave propagation, so there must be instantaneous net forces in red section.
 
  • #38
A.T.
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What if the system is splitted in 3 sections: blue, yellow and red?

http://www.2live.ro/demo/sections.bmp

Blue section - fluid flow in this section is constant preserving the velocity vector, the mass in this case comes into pipe and leaves the pipe with the same momentum.
No. It leaves with the opposite momentum. The velocity vector is flipped in the blue section.
 
  • #40
CWatters
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In these rudimentary constructions I acquired a superior thrust force...
Thrust or torque? I don't see any evidence in either vid that you have made a reactionless drive. You are rotating the earth the other way.

No doubt the moderators will be along soon to close the thread.
 
  • #41
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No doubt the moderators will be along soon to close the thread.
At least it's not magnets again.
 
  • #42
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If the water circulates one way, the platform will turn the other way. That is just conservation of angular momentum. In the first case the tubes can also transfer momentum.
 
  • #43
CWatters
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Just in case he doesn't get it...
Torque.jpg
 
  • #44
A.T.
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Just in case he doesn't get it...
https://www.physicsforums.com/data/attachments/56/56407-7e82b012611c54eaf0a473a00967f957.jpg [Broken]
In that first case the floating platform reverses the linear momentum of the water, so it's pulled out, and the elastic tubes (which also straighten under pressure) make it swing.

The angular momentum conservation explanation was mainly referring to the second video, with the hanging platform.

Tasks for iridiu:

1) Replace your magic spiral with a straight piece of tube on the hanging platform. Does it change the result?

2) Stop switching the pump on and off. Just let it run until an equilibrium is established, against a significant counter torque of the strings. For this remove the straight piece of the string, and just attach the 4 strings directly to the ceiling. Is the equilibrium orientation with pump running different from the equilibrium orientation with pump off?
 
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  • #45
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I know we do not have the same point of view, but I think there are more forces that manifest themselves simultaneously in the "magic spiral", one is explained here: http://www.google.com/patents/US6321783
Elasticity of the tube is very important, because there is vibration and mechanical resonance.
 
  • #46
A.T.
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I know we do not have the same point of view,
It's not a matter of opinion, but of experiment. Why don't you replace the spiral with a straight piece? Any honest experimenter would have done this control test on his own.
 

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