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Hose filling up a bucket

  1. Feb 18, 2012 #1
    1. The problem statement, all variables and given/known data
    A cylindrical bucket, open at the top, is 28.0 cm high and 10.0cm in diameter. A circular hole with a cross-sectional area 1.73cm2 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 2.11×10−4 m3/s. How high will the water in the bucket go (in centimeters)?


    2. Relevant equations
    [itex]P_1+1/2\rho v_1^{2}+\rho gy_1 = P_2+1/2\rho v_2^{2}+\rho gy_2[/itex]

    [itex]A_1v_1=A_2v_2[/itex]

    density of water is 1000kg/m3


    3. The attempt at a solution
    I let 1 be the top of the water, 2 be the hose flowing into the bucket and 3 be right above the hose but in the larger bucket. I know the pressure at 3 must equal the pressure at 2.

    The area of bucket is π(r)(r) = π(0.05)(0.05) = 0.00785

    [itex]P_1+1/2\rho v_1^{2}+\rho gy_1 = P_2+1/2\rho v_2^{2}+\rho gy_2[/itex]

    [itex]P_1+1/2\rho v_1^{2}+\rho gy_1 = P_2+1/2\rho v_2^{2}[/itex]

    [itex]P_3+\rho gh+1/2\rho \frac{A_2v_2}{A_1}^{2}+\rho gy_1 = P_2+1/2\rho v_2^{2}[/itex]

    [itex]1/2\rho \frac{A_2v_2}{A_1}^2+2\rho gy_1 = 1/2\rho v_2^2[/itex]

    [itex]1/2 \frac{A_2v_2}{A_1}^2+2 gh = 1/2v_2^2[/itex]

    [itex]2gh = 1/2v_2^2-1/2 \frac{A_2v_2}{A_1}^2[/itex]

    [itex]2gh = 1/2 (2.11*10^{-4})^2 - 1/2 \frac{(0.000173)(2.11*10^{-4})}{0.00785}^2[/itex]

    [itex]2gh = 1.054999*10^{-4}[/itex]

    [itex]h = 5.377*10^{-6} m = 5.377*10^{-4}cm[/itex]
     
  2. jcsd
  3. Feb 18, 2012 #2

    gneill

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    Staff: Mentor

    I think you should ignore the pressure, velocity, etc., of the hose and just use the given value for the rate at which it delivers fluid. Presumably you want the rate at which fluid leaves the bottom of the container to equal this input rate, that way there will be an equilibrium.

    Bernoulli's equation boils down to Torricelli's law for the velocity of fluid from an aperture:

    ## v = \sqrt{2 g h} ##

    where h is the "head" of fluid above the aperture. You should have enough information to turn that velocity into a flow rate...
     
  4. Feb 19, 2012 #3
    Oh, thank you! I had thought that water was flowing into the bucket from the hole in the bottom ... oops
     
  5. Feb 19, 2012 #4
    So to find the velocity that the water is rising in the bucket, I know that volume = velocity * area
    [itex]v_1=\frac{2.11*10^{-4}}{π(0.05)^{2}} = 0.02686 m/s[/itex]

    [itex]A_1v_1=A_2v_2[/itex]
    [itex]v_2 = \frac{A_1v_1}{A_2} = \frac{(π)(0.05)^2(0.02686)}{1.73*10^{-4}} = 1.21965 m/s[/itex]

    [itex]pv_1^{2} + 2gh = v_2^2[/itex]

    [itex]h=\frac{v_2^2-v_1^2}{2g} = \frac{(1.21965)^2-(0.02686)^2}{(2)(9.81)} = 0.07578m = 7.578 cm[/itex]

    Is this now correct?
     
  6. Feb 19, 2012 #5

    gneill

    User Avatar

    Staff: Mentor

    The above equation does not look dimensionally correct. You've got a pressure x velocity2 term on the left and a velocity2 term on the right.

    Instead, recognize that once you've calculated v2 above, it will be equal to the velocity as given by Torricelli's law. That is,
    [tex] v_2 = \sqrt{2 g h} [/tex]
    and you can solve for h.
     
  7. Feb 19, 2012 #6
    Thank you!
     
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