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Fluid Flow

  1. Nov 29, 2006 #1
    1. The problem statement, all variables and given/known data
    Figure P9.81 shows a water tank with a valve at the bottom. If this valve is opened, what is the maximum height attained by the water stream coming out of the right side of the tank? Assume that h = 8.0 m, L = 1.50 m, and = 30° and that the cross-sectional area at A is very large compared with that at B.

    2. Relevant equations
    Bernouli's equation
    P1 + 1/2(density)(v1)^2 + density(g)(y)= P2 + 1/2(density)(v2)^2 + density(g)(y)

    3. The attempt at a solution

    0+0+density(g)(y)= 0+1/2(density)(v2)^2 + density(g)(Lsin30)

    I tried solving for v2

    I am stuck now at this step

    used that v2 as v initial to find y by

    vfinal^2= vinitial^2-2g(y final-y initial)

    I am not sure but would v initial be the v2 I found? And is v final 0? Also what is y final and what is y initial?

  2. jcsd
  3. Nov 29, 2006 #2


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    Why do you think P1 and P2 are zero?. The water is being launched at an angle. Only the vertical component enters into the height calculation.
  4. Nov 30, 2006 #3
    so the vertical component, which is Lsin30, is that the initial height? I am confused, I tried this again and again but I cant egt it right. And the gauge pressure is 0 since the air is touching the water, air's pressure is 1 atm
  5. Nov 30, 2006 #4


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    I assume they want the maximum height measured from the same place as the 8m height, but maybe they want it measured from where it leaves the spout. The second is easier, but the first can be obtained by just adding Lsinθ. You are OK on the pressure. I was misinterpreting. The height changes from 8m to Lsinθ to the height relative to the end of the spout. Using that (as I think you have done) you can solve for v at the exit of the spout. The vertical component of that velocity is v*sinθ. You can compute the height relative to the end of the spout by finding the point where the vertical velocity is zero. You can either do that by computing the time it takes to reach that point and plugging that into the y(t) equation, or you can use the equation that relates the change in the velocity squared to acceration and distance.
    Last edited: Nov 30, 2006
  6. Nov 30, 2006 #5
    Hey, I solved it myself and get 1.8125 metres (I presume you already know the answer and need the method). Your method is flawed.

    The Bernoulli equation tells us that the mechanical energy per unit mass, let's call it E is:

    E = gz + v2/2 + (p-p0)/rho.

    E = gravitational potential energy + kinetic energy + pressure energy.

    When you open the valve you just say that the mechanical energy at A is the same as that at B. Because the pressure energy is the same at these points they cancel from both sides. Furthermore, because the cross-sectional at A is much larger than that at B you can ignore the kinetic energy at A. You end up with a fairly simple equation that you have to solve for velocity. Once you've got that velocity you need to find how high it gets which will involve a little projectile mechanics, if I could do it I'm sure you can.
  7. Nov 30, 2006 #6
    thank u so much guys, I understand the method better now
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