How does the centrifugal and coriolis forces affect a centrifugal pump?

In summary, centrifugal pumps require a lot of energy to fight against the coriolis force of the spinning water mass that flows from center toward the circumference. If a 90 degree bend is installed on each ends of the tube, it seems that the pump runs much lighter with less energy input, and also pumps much more water at same RPM.
  • #36
BadBrain said:
OK, I can foresee your next argument, that being that the bend, using only centrifugal force, also uses the rocket effect to increase its rotational velocity.


I would never use terms I regard as ridiculous in an argument. 'Centrifugal force' is as silly as they come. ('Rocket effect' is pretty much there too, but I guess it might be a helpful description to some.)

If you ever catch me relying on non-existant forces in an argument (without paragraphs of caveats to justify their use) like 'centrifugal' and 'Coriolis', then I'll eat my shorts.

You are of a school of thought that I view as throwing away the foundations of Newtonian Mechanics, and as a graduate of that school you are, I believe, prone to erroneous conclusions that will be masked by your misappropriation of the term 'force'.
 
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  • #37
cmb said:
I would never use terms I regard as ridiculous in an argument. 'Centrifugal force' is as silly as they come. ('Rocket effect' is pretty much there too, but I guess it might be a helpful description to some.)

If you ever catch me relying on non-existant forces in an argument (without paragraphs of caveats to justify their use) like 'centrifugal' and 'Coriolis', then I'll eat my shorts.

You are of a school of thought that I view as throwing away the foundations of Newtonian Mechanics, and as a graduate of that school you are, I believe, prone to erroneous conclusions that will be masked by your misappropriation of the term 'force'.

I accept this post of yours as the equivalent of this little ceremony (and, may I remind you of the fact that the chief officiant for this little ceremony just so happens to have attended West Division High School in Milwaukee, Wisconsin, USA, the very same high school from which my own mother was graduated!):



Cheers, mate!
 
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  • #38
cmb said:
I would never use terms I regard as ridiculous in an argument. 'Centrifugal force' is as silly as they come. ('Rocket effect' is pretty much there too, but I guess it might be a helpful description to some.)

If you ever catch me relying on non-existant forces in an argument (without paragraphs of caveats to justify their use) like 'centrifugal' and 'Coriolis', then I'll eat my shorts.

You are of a school of thought that I view as throwing away the foundations of Newtonian Mechanics, and as a graduate of that school you are, I believe, prone to erroneous conclusions that will be masked by your misappropriation of the term 'force'.
Whatever terms we use there is forces - pseudo or not - applied to the mass.
What else than "centrifugal force" would it be to further accelerate water flow when the bottle start to spin by the "rocket effect"? What name do that force have if it isn't "centrifugal force"?

Vidar
 
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  • #39
Vidar:

No need to continue this.

WE'VE WON!

Just as I indicated in my last post to this thread!

Savor victory!

Cheers, mate!
 
  • #40
Low-Q said:
Whatever terms we use there is forces - pseudo or not - applied to the mass.
What else than "centrifugal force" would it be to further accelerate water flow when the bottle start to spin by the "rocket effect"? What name do that force have if it isn't "centrifugal force"?

Vidar

I don't understand what you are asking 'to further accelerate water'?

If you had pipes running out of the bottle which did not kink at the end, then let the water out, which way does the bottle spin?

The acceleration of the water is due to hydrostatic pressure. It is a hydrostatic pressure that causes the water to gain momentum. The rotational acceleration of the assembly is due to the change of momentum of that water at the bend ends of the tube.
 
  • #41
cmb said:
I don't understand what you are asking 'to further accelerate water'?

If you had pipes running out of the bottle which did not kink at the end, then let the water out, which way does the bottle spin?

The acceleration of the water is due to hydrostatic pressure. It is a hydrostatic pressure that causes the water to gain momentum. The rotational acceleration of the assembly is due to the change of momentum of that water at the bend ends of the tube.

That is true when you're describing your gravitational model: Vidar's system is a pump, which elevates the water and INCREASES it's gravitational potential energy.

The application of both centrifugal AND coriolis force is the only thing which explains Vidar's experimentally observed results in his original experiment.

His bathroom experiment does NOT demonstrate his original theory, and, does, in fact, demonstrate a different phenomenon (i.e., the point that you're trying to make).
 
  • #42
cmb said:
I don't understand what you are asking 'to further accelerate water'?

If you had pipes running out of the bottle which did not kink at the end, then let the water out, which way does the bottle spin?

The acceleration of the water is due to hydrostatic pressure. It is a hydrostatic pressure that causes the water to gain momentum. The rotational acceleration of the assembly is due to the change of momentum of that water at the bend ends of the tube.

I understand the thing about the hydrostatic pressure, but what I think you haven't understood is why the bottle emties faster when it starts to rotate. Btw, the bottle will not rotate when the tubes are straight - it refuse to do so due to Coriolis counter torque. This torque will only exist if I try to rotate the bottle by hand - in any direction - doesn't matter.

When the bottle start to rotate (with the kinked tubes) the hydrostatic pressure at the tubes inlet at the bottom of the bottle will decrease because the water that is already inside the tubes will be thrown out towards the periphery due to the centrifugal force that occours during the rotation. This will accelerate the water in addition to the hydrostatic pressure that is already applied by gravity. So the bottle will get emty faster. If I prevent the bottle from rotating, only the hydrostatic pressure is applied, and the water takes about 50-70% longer time to pour out.

Vidar
 
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  • #43
BadBrain said:
Vidar:

No need to continue this.

WE'VE WON!

Just as I indicated in my last post to this thread!

Savor victory!

Cheers, mate!
I think the subject(s) at hand is not very clear to us... I am not very good in explaining subjects in a foreign language, but I think I have catched some points, figures, and theories that explains to me, in an educational way, how the principle works - with a pump like I have described.

However, the pump does not have to lift water very high. In fact the pump could likely be located under water surface. I am only interesting in the mechanics, the figures that comes into play with such a design. The main questions is (friction is out of the question - only forces, and lifting water is also out of the question):
Does it take energy to rotate a pump like this, if it is not going to lift water?
Does it take energy to rotate a pump like this if the tube is partially clogged by a restrictor, or friction? If the answer is "no" in both these questions, I have a bad feeling that this thread will be banned by the rules of this forum, LOL!

The only question I know for sure the answer to is:
Does it take energy to rotate a pump like this if the tube is totally blocked? No

Vidar
 
  • #44
Low-Q said:
I think the subject(s) at hand is not very clear to us... I am not very good in explaining subjects in a foreign language, but I think I have catched some points, figures, and theories that explains to me, in an educational way, how the principle works - with a pump like I have described.

However, the pump does not have to lift water very high. In fact the pump could likely be located under water surface. I am only interesting in the mechanics, the figures that comes into play with such a design. The main questions is (friction is out of the question - only forces, and lifting water is also out of the question):
Does it take energy to rotate a pump like this, if it is not going to lift water?
Does it take energy to rotate a pump like this if the tube is partially clogged by a restrictor, or friction? If the answer is "no" in both these questions, I have a bad feeling that this thread will be banned by the rules of this forum, LOL!

The only question I know for sure the answer to is:
Does it take energy to rotate a pump like this if the tube is totally blocked? No

Vidar

Yes to both questions, as any motion requires energy, whether the energy input is from a motor rotating the pipe or from gravitational potential energy causing water to run through the pipe.
 
  • #45
You could also use your hydrostatic pressure system as a water wheel to power something else; put a power take-off wheel on the bottom and make something spin.

In my neighboring city of Pawtucket, Rhode Island, there's an open air museum built around the first factory in the US (See:

http://www.slatermill.org/

which operated with a water wheel. At one point, there were so many water wheels in the river that all the useful energy had been extracted and the wheels furthest downstream wouldn't turn anymore, so one man had to run his factory at night when the other wheels had been shut down.
 
  • #46
A silly question (Just to be even more annoying than I already am. I'm really crap in such physics :-)):

I have a tube like the one I have already described - with a bend at the end pointing away from rotation. The tube 1m long from center and out, the circumference would be 6.28m. The tube is rotating at 1 round per second. The circumference will travel 6.28m/s.
If I put a marble of 10grams in the inlet. The velocity of the marble start at zero, but it will increase to 6.28m/s tangential velocity when it reach the end of the tube. The tangential(?) kinetic energy in the marble will be 1/2 * m*v^2 = 1/2 * 0.01 * 6.28^2 = 0.2J. Have I used 0.2J in order to move the marble from center an out, or more?
Now, the marble will change direction to the opposite of the rotation at the bend. Will the marble now loose all of its kinetic energy, -0.2J, and drop vartically to the floor?

There will also "occour" a centripetal force when the marble "hits" the bend. Will this also be a momentum that we are getting back?

The average circumference is at 0.707m from center, if that is important to know...

Maybe I should experiment with this - I have tubes, bends and steel balls...

Vidar
 
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  • #47
Many of the posts here are a complete mystery to me, but this is in part perhaps because I find people's reliance on fictional forces a mystery.

This is terribly simple:

Take a pipe laid radially on a turntable. Put something in it that moves around (water, marble, whathaveyou). If the turntable is caused to move, the object in the pipe will want to remain at rest. However, the side of the pipe is trying to rotate around. Therefore, the side of the pipe applies a force to the object, in a direction instantaneously normal to the pipe. However, the object cannot move normal to the pipe, because the pipe constrains it.

However, the object in the pipe can move towards or away from the centre. The object 'tries' to stay stationary (that is, it will unless forced otherwise). The object therefore reacts against the pipe as the pipe pushes around.

A force does not work 'around a bend'. (You have to invent a force for that to happen!) The reaction of the object is therefore normal to where the pipe just was. It experiences a force that therefore has an instantaneously radial component. That radial component causes it to accelerate, and the direction of that component is outwards.

The object tries to continue along a straight line. But the rotating pipe constrains it from doing so. Therefore, there is a continuation of the object's reaction force with the pipe. As the direction of the pipe at this reaction point is curved inwards whilst the reaction between the pipe and the object is tangential, so this means the tangential reaction consists of normal and radial parts. The object is accelerated by the raidal component.

It'd be simple to draw with a diagram, but more time consuming to scan it all up. I'll trust that the text is sufficiently explanatory.
 
  • #48
Low-Q said:
A silly question (Just to be even more annoying than I already am. I'm really crap in such physics :-)):

I have a tube like the one I have already described - with a bend at the end pointing away from rotation. The tube 1m long from center and out, the circumference would be 6.28m. The tube is rotating at 1 round per second. The circumference will travel 6.28m/s.
If I put a marble of 10grams in the inlet. The velocity of the marble start at zero, but it will increase to 6.28m/s tangential velocity when it reach the end of the tube. The tangential(?) kinetic energy in the marble will be 1/2 * m*v^2 = 1/2 * 0.01 * 6.28^2 = 0.2J. Have I used 0.2J in order to move the marble from center an out, or more?
Now, the marble will change direction to the opposite of the rotation at the bend. Will the marble now loose all of its kinetic energy, -0.2J, and drop vartically to the floor?

There will also "occour" a centripetal force when the marble "hits" the bend. Will this also be a momentum that we are getting back?

The average circumference is at 0.707m from center, if that is important to know...

Maybe I should experiment with this - I have tubes, bends and steel balls...

Vidar

Not a silly question in the least!

I won't do the math (my health is poor, and I'm currently working on somebody else's problem), but one thing that I can tell you is that I had the idea (no certainty, just an idea) that, with your original pump set-up (assuming the axis of rotation of the pump to be perpendicular to the surface of the Earth), case B would have projected the water outwards from the nozzle, due to the wasted coriolis force operating upon the water to give it unused momentum which would have caused it to continue its movement laterally beyond the nozzle, whereas case C, having fully exploited both centrifugal and coriolis forces to raise the level of the water, would have simply dropped the water from the nozzle vertically to the bottom of the pump casing.

Use my earlier posts to work out the math for your marble question yourself.

I'll be happy to check your results.

And, please, perform the marble experiment and report the results. That's the only way to be certain of anything.
 
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  • #49
I will make a turntable and test this in practice. I can foresee what will happen to the object inside the pipe during rotation, and I can also foresee a counterforce that will prevent the turntable from rotating as the object/mass inside the pipe moves radially outwards.
When the object, with its momentum, hits the curve that gradually turn into tangential direction that points in opposite direction of the turntable rotation, I will soon enough see what will happen to the rotation of the turntable.

I would believe if I by hand pushed the object along the pipe when the turntable is not rotating, the object will roll anong the pipe, and finally cause the turntable to rotate a little when the object follows the 90 degree curve at the end of the pipe.

Vidar
 
  • #50
cmb said:
Many of the posts here are a complete mystery to me, but this is in part perhaps because I find people's reliance on fictional forces a mystery.

This is terribly simple:

Take a pipe laid radially on a turntable. Put something in it that moves around (water, marble, whathaveyou). If the turntable is caused to move, the object in the pipe will want to remain at rest. However, the side of the pipe is trying to rotate around. Therefore, the side of the pipe applies a force to the object, in a direction instantaneously normal to the pipe. However, the object cannot move normal to the pipe, because the pipe constrains it.

However, the object in the pipe can move towards or away from the centre. The object 'tries' to stay stationary (that is, it will unless forced otherwise). The object therefore reacts against the pipe as the pipe pushes around.

A force does not work 'around a bend'. (You have to invent a force for that to happen!) The reaction of the object is therefore normal to where the pipe just was. It experiences a force that therefore has an instantaneously radial component. That radial component causes it to accelerate, and the direction of that component is outwards.

The object tries to continue along a straight line. But the rotating pipe constrains it from doing so. Therefore, there is a continuation of the object's reaction force with the pipe. As the direction of the pipe at this reaction point is curved inwards whilst the reaction between the pipe and the object is tangential, so this means the tangential reaction consists of normal and radial parts. The object is accelerated by the raidal component.

It'd be simple to draw with a diagram, but more time consuming to scan it all up. I'll trust that the text is sufficiently explanatory.

I believe you've just described centrifugal and coriolis forces!

Suppose there is no pipe. Suppose you're playing an old vinyl record on a phonograph and you lay a small coin onto that record. If the record's spinning fast enough, the coin is going to fly off along a vector away from the rotating center. No constraint required, just forces operating upon a massive body exerting inertial resistance.
 
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