Centrifugal pump question.

  1. Oct 19, 2011 #1


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    When a centrifugal pump is running, how much energy is required to fight against the coriolis force of the spinning water mass that flows from center toward the circumference?

    I have done some simple experiments with a tube. The tube have a water inlet in the middle of its length, so the open ends of the long tube is following the circumference. When it pumps water, even only 1cm above the water surface, it requires lots of energy to sustain rotation - even if the water is lifted only 1cm. This energy requirement is due to the coriolis counter force. However, If I put on a 90 degree bend, on each ends of the tube, that points away from rotation, it seems the pump runs much lighter with less energy input, and also pumps much more water at same RPM. How can this be possible (Well, it IS, but why?)?

  2. jcsd
  3. Oct 22, 2011 #2
    Excellent question! Great experiment!

    I'll work on it, but the main thing that comes to my mind is a wet dog shaking the water out of his fur. The water flies along sort of a spiraling trajectory. Perhaps the change in trajectory when the water enters the bent tube, being 45°, partly compensates for the coriolis force better than the 90° change in trajectory required by the straight tube.

    Just a guess.


    By the way, the last time I discussed the coriolis force on the net I was talking about attacking planes flying between Tel Aviv and Tehran, and pointed out that the Iranian planes would hit Tel Aviv before the Israeli planes hit Tehran due to the Earth's rotation, and some idiot came up with the idea that the coriolis force would counteract that effect, despite the fact that Tel Aviv-Tehran is a dead East-West trip (i.e., no coriolis effect would exist!).

    EDIT: I've just watched a video of dogs and other animals shaking water out of their fur, and it occurred to me that the initial angular momentum imparted to the water droplets by the torsional motion of the animal is more responsible for the spiral motion of the water droplets than the coriolis force.

    Here's the video:

    Last edited by a moderator: Sep 25, 2014
  4. Oct 22, 2011 #3


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    I do not think the Coriolis effect have anything to do with those planes. Their initial velocity on the ground is the same (as the earth rotation at that point), so the land masses will too. The planes will not be affected by the rotation of the earth.

    Well, back to the centrifugal pump experiment:

    I just tried with a LOOONG tube with a 90 degree bend at the end - in air. I held the tube in my hand, and spun myself around, and so the tube.
    However, on my side of the tube I attached a plastic bag (wrapped around the tube end) that was filled with air - as much air a plastic bag can carry with an open end.
    When the 90 degree bend pointet towards rotation, the air did not escape from the bag - the bag did not deflate or inflate.
    When I aligned the bend towards the floor or ceiling, it took about 4 rounds (1/2 round pr second) to deflate the bag.
    When I pointed the bend away from rotation, the bag deflated in only 2.5 rounds (1/2 round pr. second) to deflate the bag.

    I repeated the experiments many times - spinning in both directions untill I got sick...

    The average result is what I explained above.

    PS! Yes I removed the TV set, the speakers, my wifes many flowers and "stuff", and the kids, before the experiment :eek:

    Last edited: Oct 22, 2011
  5. Oct 23, 2011 #4

    My claim is precisely that air travel along a due east-west axis will involve no coriolis forces, as there's no change in the distance of the plane from the center of the rotating body (earth). That's why I questioned the intelligence of the guy on the other forum who suggested it.

    Assuming a perfectly calm day (i.e., the air is moving at the same velocity as the Earth), the plane flying from Tehran will have benefit of the fact that the Earth's rotation is bringing Tel Aviv closer to him, whereas the plane flying from Tel Aviv to Tehran will experience the opposite effect. Even artillery shells, which travel far shorter distances than airplanes, must have the Earth's rotation accounted for by their gun's guidance system if they are to hit their targets.


    You're partly right! Under the conditions I described, earth's rotation would be cancelled out by a headwind hindering the first plane and a tailwind assisting the second plane, if they were both flying at haircut altitude. But most modern air combat operations are conducted at far higher altitudes, where the thinner air will exert less dynamic pressure against the speeding aircraft than will the thicker air on deck. Thus, the effect I mentioned will be observed.

    Once again, great experiment! You're working out all of the corollaries of your theories. You're doing a thorough job. I like that.
    Last edited: Oct 23, 2011
  6. Oct 23, 2011 #5


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    That is due to the Criolis effect - which is caused by earth rotation. However, if an airplane is pointing its nose towards east, at equator, the plane have a reverse velicoty of about 1600 km/h, while the other plane with its nose pointing to west, have a forward velocity of 1600 km/h. The earth surface also have a velocity of 1600 km/h. So if the air planes can reach a maximum velocity of 1600 km/h, flying towards eachother, the first plane will stand still relative to space, while the other will fly at 3200 km/h relative to space.
    What effect that will be different on those planes is the centrifugal force. The second plane is actually flying in a curve at 3200 km/h, but the first plane is not because the space velocity is zero. That means the second plane must spend energy to keep the nose down in order to not change altitude to a higher level.

    Thanks for the credits, but do you have any idea why the pump runs lighter and faster in the last experiment where the bends points away from rotation?

    Is the velocity of the water mass counteracting the coriolis force when the water mass change direction and "push" the circumference of the tubes?

    How much is this counteracting compared to the Coriolis force?

  7. Oct 23, 2011 #6
    OK, OK, I get it now!

    I don't think coriolis force has any measurable effect on your experiment. We're talking pure centrifugal force here!

    You say it took 4 turns to empty out the bag with the bend pointed towards the floor or the ceiling (i.e., at a 90° angle from the axis of rotation), whereas it only took 2.5 turns to empty the bag when the bend faced away from rotation.

    The ratio of your experimental results is thus 4/2.5 = 1.6, the radian measure of a 90° angle is ∏/2 = 3.14159/2 = 1.570795.

    In other words, this is centrifugal force as a function of angular velocity.

    When the bend faced away from the axis of rotation, centrifugal force kept the bag from deflating.

    The reason the 90° bend helped in your first experiment is because, relative to the axis of application of force, this is actually two 45° angles, which is the usual axis of application of a force: that's the principle underlying the corbel arch and the flying buttress! I can't believe I forgot that!

    With the straight tube, the action of centrifugal force would have directed the water against the side of the tube opposite the action of rotation, which would have hindered, rather than assisted, the flow of water out of the ends of the tube.
    Last edited: Oct 23, 2011
  8. Oct 23, 2011 #7


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    Here is the experiment drawn:

    Attached Files:

  9. Oct 23, 2011 #8
    I had a very different vision of your experiment.

    I thought you were talking about a "V"-shaped tube, with the water access at the bottom of the "V".

    Seeing your diagram, situation "A" would seem to represent a situation in which centrifugal force would be counteracted by the angular inertia of the water in the tip of the tube (the very definition of coriolis force!), whereas situation "B" would represent centrifugal force operating without the assistance of coriolis force, whereas situation "C" would represent centrifugal force operating with the assistance of coriolis force.

    I obviously continue to stand by my mathematical analysis in my post #5, although I now see that it applies to different forces in different directions.
    Last edited: Oct 23, 2011
  10. Oct 23, 2011 #9


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    I had in mind that you might have a different view, that's why I show you a drawing:smile:

    So, maybe I can ask you:
    What is the theoretical energy difference required to run the pump in "C" compared to "B"?
    Is it EC = PI / (EB x 2)

    In "A" it should not require energy to run the pump - because there is no pump function, no mass displacement in the tube.
    In "B" it obviously require energy to run the pump - because of coriolis effect due to radial mass displacement
    In "C" I am not sure - If the radial mass displacement that creates the Coriolis force in the first place, is "gained" back at the bend, it means it will not require energy to displace mass (???)

    Last edited: Oct 23, 2011
  11. Oct 23, 2011 #10
    No, I believe it's actually the reciprocal of your equation, such that EC = (EB x 2)/ ∏.

    Your equation actually requires GREATER energy input for case C than for case B, which is contrary to your experimental results.
  12. Oct 23, 2011 #11


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    OK. If EB/SUB] = PI/2, it shouldn't matter (?).
    Maybe you can help me with this:

    My tube is 1m long. The cross section of the tube is 10cm^2. That would be 9.82N of water flowing at all times.
    The tube is rotating at 1 RPS - the circumference is traveling at 3.14m/s.

    What is the mass displacement rate in case B and C?
    What is the Coriolis countertorque in case B and C?
    What is the total energy required to run the pump in case B and C?

  13. Oct 23, 2011 #12
    What is this?

    Are you trying to trick me into doing a homework assignment for you?

    That drawing was a professionally printed illustration.

    What's going on here?
  14. Oct 24, 2011 #13


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    1. I am 39 years old, and quit school about 20 years ago to start to work. No. It's not homework.

    2. I do proffesional graphic design at work. I make digital designs, engineering speakers etc for a living. The drawing was just a scetch done in 2 minutes - not very proffesional. I have worked with Adobe software for 15 years - learned it by experience as a hobby. No education in graphic design.

    3. I am just looking for a spesific mathematical result, and hoped it would help with some figures to work from.

    I am a very curious person. Things I do not understand, or partialy "think" I understand, is things I want to learn more about. This is just a hobby - among many technical related hobbies. Look at my facebook profile (Vidar Øierås) - I am really not a schoolboy anymore :approve:
  15. Oct 24, 2011 #14

    Let's think this through logically. Your second question should actually be your first, as we need the answer to that to answer your first question.

    The difference between case B and case C is that case C makes use of both centrifugal force and coriolis force, whereas case B uses only centrifugal force and wastes coriolis force. Therefor the coriolis counter torque = ((Centrifugal Force)∏/2) -1 (to subtract out centrifugal force) = 0.57(Centrifugal Force).

    Now, to figure coriolis counter torque at the tip of each arm:

    The volume of the tube is the cross-sectional area times its length = 10cm^2 X 100cm = 1000cm^3. It's velocity at the tip along the coriolis vector is 3.14m/s. Each arm = 0.5m in length and contains 500cm^3. 500cm^3 of water has mass of 500g. Therefor, Coriolis countertorque at the tip = F = mV = 3.14(500g) = 1570g per arm.

    Centrifugal force per arm is 1570g /0.57 = 2754.386g.


    On to question 1. If centrifugal force at the tip is 2754.386g per arm, then that's the mass of water that's coming out of each arm in case B. Therefor the total mass displacement rate of the pump = 5508.77g/s in case B.

    Case C uses centrifugal and coriolis forces, so the answer there is the sum of the forces = 2754.386g + 1570g = 4324.386g/s per arm = 8648.772g/s for the whole pump.


    On question 3. On to question 3. Work = energy expended in Joules:

    J = kg(m^2)/s^2.

    Seeing as we're talking about one meter's worth of motion, and we already have the rate of mass displacement, we need only know the velocity of the water in seconds. Seeing as the velocity along the coriolis vector = 3.14m/s, the velocity along the centrifugal vector = 3.14m/s / 0.57 = 5.5m/s. Therefor, time for one evacuation of the tube in case B in seconds = 1/5.5 = 0.18s. 0.18^2 = 0.0324.

    Energy in Joules in case B = 5.508772kg/0.0324s^2 = 1700.2382J

    Case C uses both forces, so the the velocity of the water at the tip is 5.5m/s + 3.14m/s = 8.64m/s. Time for one evacuation in case C = 1/8.64 = 0.115741s. 0.115741^2 = 0.0134s^2

    Energy in Joules in case C = 8.648772kg/0.0134s^2 = 645.6283J


    This is the energy required per evacuation of the tube. For energy to run the pump per second, divide by seconds per evacuation.

    This comports with your experimental results, as 1700/545.6 = 2.633, the square root of which is 1.62.
    Last edited: Oct 24, 2011
  16. Oct 24, 2011 #15


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    I don't think there is any need to wrangle your way through the 'Coriolis' force.

    Just look at this simply and imagine what happens to the pumped fluid after it has exited the system. What you are after is for the pumped fluid to have no rotational velocity.

    If it has rotational velocity when you don't really want it, then it takes energy to make that happen. This is where your energy is going.

    In most air-flow processes, you have static reaction elements that push back against the air so that it exits the flow-modifyer parts with as little rotational inertia as possible. 'Swirl reducers' if you like. You'll find such parts in erverything from vacuum cleaner blowers through to jet engines; static parts that act to cause process air to exit with minimal rotational inertia.
  17. Oct 24, 2011 #16
    This whole experiment is just a comparison of the efficiency of a system which expends energy to create coriolis force which it then wastes with another which uses this coriolis force that you've expended energy to create.
    Last edited: Oct 24, 2011
  18. Oct 24, 2011 #17


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    Is it?

    What, exactly, experiences a 'Coriolis force'?

    As I see it; fluid in the pipe is merely gaining angular momentum as it travels away from the centre of rotation, and at the end of the pipe there are 3 options - it gains yet more, it gains no more, or it loses angular momentum (back to the pipe, thus the fluid in it). The latter is clearly going to be less power-demanding scenario. No need to conjour up extra concepts here. Call it what you like, but this is a very simple scenario, easily explained.
  19. Oct 24, 2011 #18


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    Thanks for the explanations, and the figures.

    Initially I did expect the rotational velocity of the mass would end up in zero at the circumference of the pump (observed outside the pump), and that is what I wanted. I was further thinking that the kinetic energy in the mass was gained back to the system by the 90 degree bend - translated into an electric car where the motor consumes energy to accelerate, and gain back the energy when it deaccelerate (charging the battery) - total energy to move the mass horizontally from one place to another is zero + loss.

    What happens if the pump is clogged by mud or objects that resist the flow? Say the flow rate is reduced to 50% at the same RPM. Will the energy input to sustain rotation increase, be the same, or reduced?
    If it is completely clogged, then no mass flow at all, should not require energy to sustain rotation, right?

    Please be patient with me. I'm just very curious:smile:

  20. Oct 24, 2011 #19


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    Other than the mechanical rotational losses of the moving parts, yes, that's right.

    Just think what happens to your vacuum cleaner motor when you block the air by putting your hand over the end of the pipe. It will wind up to its maximum (usually an electronically limited speed in modern motors) because there are no fluid-pumping loads on it any more.
  21. Oct 24, 2011 #20
    You're welcome!

    Actually, the kinetic energy that's being recaptured by the system is the equal and opposite reaction to the kinetic energy of the water in the tube along the rotational vector, which equal and opposite reaction is the inertial resistance of the water to the acceleration it undergoes as it moves down the pipe; this is the definition of coriolis force.

    I think you're thinking things are more complicated than they really are, and that you're confusing yourself. Try breaking problems down to their essential parts, and analyzing them in smaller pieces.
    Last edited: Oct 24, 2011
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