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Power of a fluid pump

  1. Jul 22, 2011 #1
    1. The problem statement, all variables and given/known data

    A pump lifts a liquid of density [itex]\rho[/itex] to a height h and accelerates it from rest to a final velocity v.
    What power P does the pump deliver to the liquid, if the liquid is being pumped through
    a pipe with a cross sectional area A?

    1. Av[itex]\rho[/itex] + [itex]\rho[/itex]gh
    2. Not enough information is provided.
    3. Av[itex]\rho[/itex](gh + 1/2 v2)
    4. 1/2 Av2[itex]\rho[/itex]gh
    5. Av[itex]\rho[/itex](1/2 v2) - gh
    6. Av[itex]\rho[/itex](gh - 1/2 v2)
    7. 1/2 Av3[itex]\rho[/itex]
    8. Av[itex]\rho[/itex]
    9. Zero, of course; a liquid will naturally flow in that way without any pumping.
    10. Av[itex]\rho[/itex]g h

    2. Relevant equations
    Bernoulli's principle: P1 + 1/2 [itex]\rho[/itex]v12 + [itex]\rho[/itex]gh1 = P2 + 1/2 [itex]\rho[/itex]v22 + [itex]\rho[/itex]gh2

    P = W/t = [itex]\Delta[/itex]E/t = F dot v

    W = [itex]\int[/itex]F dh

    v2 = 2ah

    h = 1/2 at2

    3. The attempt at a solution
    I'm actually not sure if kinematics can be applied to a fluid. My attempts have so far eliminated answers 1 and 10, though I came up with 10 using the integral of the force F = PA = [itex]\rho[/itex]ghA. Integration gives W = [itex]\rho[/itex]gh2A, and since h/t = v, P = Av[itex]\rho[/itex]gh.
    I think what I forgot to take into account was the acceleration-- the pump not only did work to lift the liquid, but also accelerated it. To try to incorporate a change in kinetic energy into the work term, I used
    E2 = P2 + 1/2 [itex]\rho[/itex]v2 + [itex]\rho[/itex]gh
    E1 = P1
    [itex]\Delta[/itex]E = P2 - P1 + 1/2 [itex]\rho[/itex]v2 + [itex]\rho[/itex]gh = [itex]\rho[/itex]gh + 1/2 [itex]\rho[/itex]v2 + [itex]\rho[/itex]gh
    Playing around with kinematics I got a = v2/(2h) and t = 2h/v, but all of that together isn't leading me to an answer choice given.
    Any advice is much appreciated!
    Last edited: Jul 22, 2011
  2. jcsd
  3. Jul 22, 2011 #2


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    Staff: Mentor

    If the question is, "what power is delivered to the liquid", does this mean that only the power in the liquid upon delivery is important? That would imply that any work done to get it to the end point is unimportant -- only the end result matters. What's the power in the liquid at the point of delivery?
  4. Jul 22, 2011 #3
    If only the end point matters, I guess that would mean the work done is (1/2 [itex]\rho[/itex]v2 + [itex]\rho[/itex]gh), multiplied by the change in volume = Av. That would be Av[itex]\rho[/itex](1/2 v2 + gh). I'm going to feel a little silly if that's the answer-- it is, after all, the one with a form that suggests an increase in both potential and kinetic energy.
  5. Jul 22, 2011 #4


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    Staff: Mentor

    Perhaps I'm missing the import of the question, but why would the height lifted matter? If you're holding the hose and measuring the power delivered, you don't care if the water was pumped from the well or from the river or shipped UPS from an Antarctic glacier -- you just care about the water velocity and volume at the nozzle as far as power delivered to the water is concerned.
  6. Jul 22, 2011 #5
    I think I see what you're saying... are you reading the problem as stating that the acceleration is the result of a power applied after the fluid has been lifted through whatever distance? As though the liquid was lifted, and after that the power in question was applied? I had assumed it was a simultaneous change-- that some power had been applied resulting in both a lift and acceleration. Sorry if I'm misconstruing your meaning!
  7. Jul 23, 2011 #6


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    Staff: Mentor

    When I read the phrase, "What power P does the pump deliver to the liquid", I interpret it as asking for the power in the liquid as the "end user" would see it emerging from the pipe. The history of the liquid, how it got to the pipe opening, would be irrelevant.

    If the question had said, "What power must the pump consume in order to lift a liquid of density ρ to a height h and accelerate it from rest to a final velocity v through a pipe of cross sectional area A?" then I would agree that the work done lifting the water is to be included in the calculation -- since it applies to the pump power, not the power "delivered to" the liquid.

    But as I say, this is just my interpretation of the what the question is saying. Others may not agree with me :smile:
  8. Jul 23, 2011 #7
    Unfortunately, the answer is indeed Av(rho) (1/2 v2 + gh), but the problems in our database are often worded questionably. Thanks so much for your help!
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