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Fluid mechanics flowrate problem

  1. Feb 13, 2012 #1
    1. The problem statement, all variables and given/known data
    A long water trough of triangular cross section is formed from two planks as is show in Fig P3.66. A gap of 0.1 in. remains at the junction of the two planks. If the water depth initially was 2 ft, how long a time does it take for the water depth to reduce to 1 ft?


    2. Relevant equations
    Bernoulli equation

    Conservation of the mass

    3. The attempt at a solution

    Q=water flow
    V=Velocity of the water
    Vol=Volume of water
    L=Length of the planks
    z=height of the water level






    V(t)=20z(t)*z'(t) (EQ1)

    With the bernoulli equation we can find the velocity of the water in fonction of the height of the water level


    V(t)=[itex]\sqrt{\frac{2z(t)\gamma}{\rho}}[/itex] (EQ2)

    By combining EQ1 and EQ2

    [itex]\sqrt{z}[/itex] dz = [itex]\frac{\sqrt{\frac{2\gamma}{\rho}}}{20}[/itex] dt

    By integrate

    -2z3/2/3 = [itex]\frac{\sqrt{\frac{2\gamma}{\rho}}}{20}[/itex]*t+C

    When t=0, z=2 so C=1,89

    if I put z=1, t=3.04s and the answer is 36.5s

    I'm probably out of line, please help me.

  2. jcsd
  3. Feb 14, 2012 #2
    I get your answer of 36.5 seconds. But what is the 3.04s?
  4. Feb 14, 2012 #3
    I am sorry my sentence was wrong I meant:

    when I put z=1, my answer is 3.04s. But the answer in the book is 36.5s.

    I have no idea if my error is at the beginning of the problem or at the end. Please help.

  5. Feb 15, 2012 #4
    What are your units of gamma and rho? Are your terms in your equations uniform in units?

    My solution is similar to yours. First I found an expression for the volume of the trough.

    V = h^2

    I use unit length to unclutter the equations. Then I take the derivative of the above expression.

    I used Bernoulli's eqn. to get another expression for the rate of change of volume.
    It is based on unit length as well.

    I equate the derivatives. It is a variable separable differential equation. One variable is height, the other is time.

    Then I integrate using definite integrals. The time integral goes from 0 to T. The height integral goes from 2 to 1. You wind up with an expression for T and it works out to be 36.5 seconds.
  6. Feb 15, 2012 #5
    I found my error, I used 0.1*L for the area of the falling water but I needed to put it in feet. Thank you very much, you have no idea how much time I passed on that problem.

  7. Feb 15, 2012 #6
    Good for you!
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