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Simple fluid problem

  1. Nov 29, 2008 #1
    1. The problem statement, all variables and given/known data
    A container is filled with liquid, density d. The depth of the liquid is h. A narrow, horizontal, tube at the bottom of the container is letting the liquid flow out at a speed v.
    The height of the liquid is kept constant.

    What is the velocity of the liquid flowing out of the tube?

    2. Relevant equations
    Bernoulli's equation:

    p + (1/2)dv^2 + dgh = constant

    3. The attempt at a solution
    I really don't know how to do this! I know the answer is (2gh)^(1/2), but i don't know why that is, as my textbook has no examples. Could someone please explain how to do these kinds of problems?
     
  2. jcsd
  3. Nov 29, 2008 #2
    Ah... this is an example of a special case of the Bernoulli's Equation, known as Torricelli's Theorem.

    I will leave the details of the working out of the equation to you, but I'll point out some important aspects you'll need to consider:

    1) You may make the assumption that the pressure on top = pressure below = atmospheric pressure, since both the top and the bottom are open surfaces
    2) The surface fluid area of the container is very large compared to the area of the bottom. This means you make the approximation that the surface velocity is....
     
  4. Nov 30, 2008 #3
    Thanks for your reply.
    So can we approximate the surface velocity to be zero?

    In which case:

    [tex]
    p_{1} + \frac{1}{2} \rho v_{1}^2 + \rho gh = p_{2} + \frac{1}{2} \rho v_{2}^2
    [/tex]

    But since [tex]p_{1} = p_{2}[/tex] and [tex]v_{1}=0[/tex] then we get [tex]v_{2} = \sqrt{2gh} [/tex].
     
  5. Nov 30, 2008 #4
    Correct, you are absolutely right. Just as an added note, notice that this is also the equation describing the kinematics of the free fall of an object initially at rest.
     
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