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Homework Help: Fluid cylindrical bucket Question

  1. Sep 25, 2007 #1
    A cylindrical bucket, open at the top has height 27.0 cm and diameter 13.0 cm. A circular hole with a cross-sectional area 1.49 cm^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 2.28×10−4 m^3/s

    my work
    let 1 denote the top and 2 denote the bottom

    Q = V1A1
    V1 = Q/A = 0.01718 m/s

    Q = V2A2
    V2 = 0.01530 m/s

    using bernoulli's
    top bottom
    P + pgh + 1/2pv^2 = pgh + 1/2pv^2 + P
    P's are equal and cancels out
    rho's cancel out
    there's no pgh on bottom
    we are left with
    h = [ 1/2(v2)^2-1/2(v1)^2 ] /g

    the answer i get is 0.00179m which is 0.1179 cm but that's not the answer...can anyone see where i went wrong? thank you very much.
  2. jcsd
  3. Sep 25, 2007 #2


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    you're probably forgetting to convert something from cm to m or some such thing... I found

    h=0.11725 meters

    actually, only good to 2 sig fig?

    h=0.12 meters
  4. Sep 25, 2007 #3
    really? ? any idea what i converted wrong?
    did you get 0.01718m/s and 0.01530m/s for velocities? ... hm
  5. Sep 25, 2007 #4


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    no idea. sorry.

    I don't remember off hand, if I can find my scratch-paper I'll let you know.

    good luck.
  6. Sep 27, 2007 #5
    1cm^2 = .0001m^2 so 2.28x10^-4/1.49x10^-4 gives a velocity of 1.53m/s

    I just plugged it in to toricelli's equation (v1=sqrt(2g(h2-h1)) and solved for h2 (h1 is 0), which gave the answer .119m = 11.9cm.
  7. Sep 27, 2007 #6


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    that neglects the velocity of the water at the top of the bucket... valid if A_top >> A_bottom only.
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