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Fluid Dynamics again

  1. Oct 1, 2007 #1
    A cylindrical bucket, open at the top has height 30.0cm and diameter 10.0cm. A circular hole with a cross-sectional area 1.35cm^2 is cut in the center of the bottom of the bucket. Water flows into the bucket from a tube above it at the rate of 2.20×10−4 m^3/s

    How high will the water in the bucket rise?
    Take the free fall acceleration to be = 9.80

    What I've done is
    A_1*v_1 (bottom) = A_2*v_2 (top)

    So Q_2 = 2.2*10^-4 = A_2*v_2

    (0.0135)v_1 = 2.2*10^-4

    v_1 = 0.016296m/s
    A_1 = 0.0135m^2
    v_2 = 0.028011m/s
    A_2 = 0.007853m^2

    Then I did the bernoulli's principle, cancelling out some stuff
    P_1 + 0.5ρv_1^2 + ρgh = P2 + 0.5ρv_2^2 +ρgh

    It was here where I got confused, but here's what I did..
    (0.5)(1000)(0.016296^2) = (0.5)(1000)(0.028011^2) + (1000)(9.8)h

    Is this the correct way to do it? Because I'm totally unsure.. so thanks for your time
     
  2. jcsd
  3. Oct 1, 2007 #2

    Astronuc

    User Avatar

    Staff: Mentor

    The idea is that the pressure head is related to the flow rate out of the bottom, and the height increases until the flowrate out matches the flowrate in. The use of Bernoulli's equation is not correct as applied.

    Certainly the continuity equation applies (flow in = flow out in steady state, i.e. dh/dt = 0), but one must find the mass flow rate out in terms of the pressure head.
     
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