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Bernoulli Equation and flow loss

  1. Oct 30, 2012 #1
    1. The problem statement, all variables and given/known data

    Liquid, specific density 0.8, flows with velocity 4 m/s
    in a pipe that has a downward slope of 1:50. At a
    certain point in the pipe, a pressure gauge shows a
    pressure of 80 kPa. Determine the pressure at a
    point 200 m downstream of the gauge if:

    flow losses are ignored;
    and,there is a flow loss equal to 10% of the total
    initial head.
    2. Relevant equations

    Bernoulli equation

    3. The attempt at a solution
    Solved the first part and got 111,360, the second part is where I'm going wrong.

    So for the second part, first I find the total initial head
    P1/pg +V1^(2)/2g +z = H
    (80*10^3)/(800*9.8) + (4^2)/(2*9.8) + 0
    I get 11.10, so 0.1* 11.1 = 1.11
    therefore pgh = P = 8000*9.8 *1.11 = 8702.4,
    then simply subtract for P2, 11360 -8702.4 = 102657.6 Pa

    However this answer is wrong, what am I doing wrong?

    Thanks in advance
     
    Last edited: Oct 30, 2012
  2. jcsd
  3. Oct 31, 2012 #2
    I don't think it is that simple.
    It is not necessary to include the velocity head since the pipe diameter is (assumed) to be constant. The flow loss means that
    [itex]0.9\frac{p_{upper}}{\rho g}=\frac{p_{lower}}{\rho g}-200[/itex]
     
    Last edited: Oct 31, 2012
  4. Oct 31, 2012 #3
    But solving that equation for Plower results in the wrong answer also.
     
  5. Oct 31, 2012 #4
    By the way i think you got your Z2 mixed up it should be 4, not 200 since the slope is 1:50, either way you're answer is still wrong.
     
    Last edited: Oct 31, 2012
  6. Oct 31, 2012 #5
    How did you get at 111,360 Pa?
     
  7. Oct 31, 2012 #6
    p1/pg = 80*10^3/800*9.8
    v1/2g =0 (because it cancels)
    z1=0

    v2^2/2g = 0 (cancels)
    z2=-4

    so p2 = (80*10^(3))/(800*9.8 +4)*800*9.8 = P2
     
    Last edited: Oct 31, 2012
  8. Oct 31, 2012 #7
    I think we have to assume that the velocity head is also going to be reduced, not just the pressure head.
     
  9. Oct 31, 2012 #8
    The equation then might be:
    [itex]\frac{p_{upper}}{\rho g}=-4+\frac{p_{lower}}{\rho g}+1.11[/itex]
     
  10. Oct 31, 2012 #9
    Hi Basic_Physics thanks for your input thus far, however your equation still results in the wrong answer.
     
  11. Nov 1, 2012 #10
    I know the steps to the solution but I don't understand why those steps are used, can anyone offer some input?
     
  12. Nov 1, 2012 #11
    As far as I can find out this type of problem is usually done via:
    hupper - hlower = hloss
    but the problem is that both the pressure and velocity head will be changed due to the loss.
     
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