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Work done by a pump

  1. Jan 1, 2016 #1
    1. The problem statement, all variables and given/known data

    Water is pumped steadily out of a flooded basement at
    5.0 m/s through a hose of radius 1.0 cm, passing through a window
    3.0 m above the waterline.What is the pump’s power?

    2. Relevant equations

    Volume and mass flow rates are constant

    [tex]R_v = Av; \quad M_v = \rho A v[/tex]

    Work-energy Theorem

    [tex]W = \Delta E = \Delta U + \Delta K[/tex]

    3. The attempt at a solution

    I've seen one solution leading up to the same answer as in the back of the book, 66 Watts. It goes by imagining a small chunk of water of mass Δm and the work done on it by the pump of a time Δt. Several attempts at that method can be found here: https://www.physicsforums.com/threads/power-of-a-pump.551946/

    However, this seems odd to me. I would think that the water in the hose from the start should be handled differently from the water in the basement that is not (yet) in the hose. The water at the very top of the hose doesn't need to go anywhere, and the water only a little bit below it only needs to go a small distance and then it vanishes. The water at the bottom of the hose needs to be lifted a distance of 3m. The water not yet in the hose just sits there until space is made for it and then only makes it up the hose a certain distance.

    My thought about this was to think of, again, a small chunk of water and the work done on it, but to split the two cases where it's in the hose and where it isn't yet. To model the work done on the water I reason that, in time Δt at a height h from the water level, the work done on the water is [tex]\rho \pi r^2 g(3-h)\Delta h[/tex] as this accounts for the linear variation of the amount of height that the water in the hose has to travel. This would then lead to the integral [tex]\int_0^3 \rho \pi (0.01^2) g (3-h)dh[/tex]

    Then you'd have to do something similar for the water that began not in the hose and add the two calculations of work together.

    Is this a good argument?
     
    Last edited: Jan 1, 2016
  2. jcsd
  3. Jan 1, 2016 #2
    How much work does it take to raise a mass of 1 kg a vertical distance of 3 meters?

    How much work does it take to accelerate a mass of 1 kg from a velocity of 0 m/s to a velocity of 5 m/s?

    What is the mass flow rate through the hose?
     
  4. Jan 1, 2016 #3
    The work to raise 1 kg a vertical distance of 3 m is 9.8*3 N, but my point is that not every quantity of water must be raised 3 m, unless we assume the hose begins empty and ends empty and runs for unit time I suppose.

    I may not be appreciating what you're getting at.
     
  5. Jan 1, 2016 #4

    jbriggs444

    User Avatar
    Science Advisor

    You can ignore the startup and shutdown entirely. Instead, concentrate on the steady state. In this steady state there are many small parcels of water in various stages of their journey simultaneously. We can focus on any parcel. It will be representative of all of the rest.

    How much energy does it take per parcel? How many parcels pass a particular point every second?

    You have calculated the work to raise 1 kg of water 3 meters as 9.8*3 N. You should re-check the units for that result. Then you can proceed with the rest of the calculation.
     
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