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Power and Work

  1. Sep 10, 2010 #1
    1. The problem statement, all variables and given/known data
    Water is pumped from the ocean to a large reservoir on land. A total lift of 140m. At a rate of 60 cubic meters per hour and eject it with a speed of 65 m/s. If one cubic meter of sea water has a mass of 1,025 kg, find:
    a) The work done in lifting the water
    b) the work done in ejecting it
    c) the total work done
    d) the power developed
    e) if the pump is 90% efficient, what horse power engine is needed

    given:
    d = 140m
    r (rate) = 60 cubic meters per hour
    v = 65 m/s
    m = 1,025 kg

    for sure, A (work done in lifting the water) can be solved with W (work) = Fd.
    Since F is not given, W = (ma)d
    since a is not given, a = -v12 / 2d.

    my dad said B (work done in ejecting it) can be solved with KE = 1/2mv2. But Im not sure.

    Total work can be solved by adding A and B.

    I don't know how to solve D and E.

    Are my fomulas for A, B and C correct? What formulas should I use for D and E?
    Please and Thanx! :)
     
  2. jcsd
  3. Sep 10, 2010 #2

    rock.freak667

    User Avatar
    Homework Helper

    It is much easier to use energy indeed.

    you were given ρ=1025 kg/m3, not 'm'.

    If you have the volume flow rate of 60 m3/hr, what do you get if you multiply that by the density (convert the volume flow rate to m3/s and then multiply)?

    If the energy needed to lift the water a distance 'h' is E=mgh and we differentiate w.r.t. t, we will get dE/dt = d/dt(mgh) = gh*dm/dt. We know that power P=dE/dt, so P=gh*dm/dt.
     
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