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Speed of water leaving trough

  1. May 26, 2008 #1
    **Below is the question. No need to solve it for me. Providing me with an equation would be fine. Thanks**

    A dairy farmer notices that a circular water trough near the barn has become rusty and now has a hole near the base. The hole is 0.14 m below the level of the water that is in the tank.

    The acceleration of gravity is 9.81 m/s^2.

    If the top of the trough is open to the atmosphere, what is the speed of the water as it leaves the hole?

    Assume that the trough is large enough that the velocity of the water at the top is zero. Answer in units of m/s.
     
    Last edited: May 26, 2008
  2. jcsd
  3. May 26, 2008 #2

    Doc Al

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    Staff: Mentor

    Try Bernoulli.
     
  4. May 26, 2008 #3
    So...

    pressure + 1/2 * density * velocity^2 + density * acceleration * elevation = constant


    I believe that density crosses out.

    One question is what is the pressure equal to?
     
  5. May 26, 2008 #4
    ok i think its 100,000 Pascals. When I plug everything in I get 447.2

    Somehow that doesn't seem correct though
     
  6. May 26, 2008 #5

    Doc Al

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    Staff: Mentor

    You need to compare a point at the water surface to one at the hole. Both are open to the atmosphere.
     
  7. May 26, 2008 #6
    Could you explain this further?
     
  8. May 26, 2008 #7
    Could someone possibly walk me through this. Sound a little more difficult than I'm used to.
     
  9. May 27, 2008 #8

    alphysicist

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    Homework Helper

    Hi European Sens,

    Bernoulli's equation indicates that

    [tex]
    P + \frac{1}{2} \rho v^2 + \rho g h
    [/tex]

    is constant along a flow. So the first step is to compute that quantity for the top of the trough, and then separately compute it at the hole. What do you get for those two quantities? Since the quantity is constant, what would you do next?
     
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