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Fluid flow question using bernoulli's and continuity equations

  1. Nov 29, 2009 #1
    1. The problem statement, all variables and given/known data

    The pressure in a section of horizontal pipe with a diameter of 2.0 cm is 140 kPa. Water flows through the pipe at 2.80 L/s. Assume laminar nonviscous flow. If the pressure at a certain point is to be reduced to 102 kPa by constricting a section of the pipe, what should the diameter of the constricted section be in cm?

    2. Relevant equations

    Bernoulli's equation for constant elevation: P1 + .5p(v1)^2 = P2 + .5p(v2)^2

    P1 = 140000 Pa
    p = 1000 kg/m^3
    v1 = 2.8 L/s

    P2 = 102000 Pa
    v2 = ?

    Continuity equation = A1*v1 - A2*v2 = 0

    A1 = pi(.01)^2
    v1 = 2.8 L/s

    A2 = pi(r)^2
    v2 = answer to Bernoulli equation above (63.21392252 L/s)

    3. The attempt at a solution

    I plugged in all known variables to Bernoulli's equation above, and got 63.21392252 L/s for the v2 flow speed. I then used the continuity equation to search for A2, and then to find r, and then to convert to d. I found r to be .0021046149 m, doubled that to find d to be .0042092298m, and converted to cm to find the final answer to be 0.42092298 cm. This is incorrect. Can someone see where I might have done something wrong? Maybe it's a simple math error?
  2. jcsd
  3. Nov 29, 2009 #2


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    But the quantity given in L/s is a volume flow rate, not a velocity.
  4. Nov 29, 2009 #3
    Oh, of course. I just googled it, but I can't figure out how to convert from L/s to m/s, can you help me with that?
  5. Nov 29, 2009 #4


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    If you have a cylindrical volume representing the amount of fluid to "flow by" in one second, and you know its cross sectional area, then you know the what the third dimension (the length of the cylinder) has to be in order for the cylinder to be of that volume. Therefore, you know how much distance is covered by the fluid in one second.
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