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Bernoulli's Equation and liquid density

  1. Aug 17, 2011 #1
    1. The density of the liquid flowing through the horizontal pipe in the drawing is 1200 kg/m3. The speed of the fluid at point A is 7.5 m/s while at point B it is 11 m/s. What is the difference in pressure, PB – PA, between points B and A?



    2. Bernoulli's Equation P1 + 1/2 pv1^2 + pgy1 = P2 + 1/2 pv2^2



    3. ok to find P is it possible to make P1 = 1/2 pv1^2 + pg (y1 can be neglected) then do the same with P2 then subtract the two values to work out the pressure ? or am i way off ?
     
  2. jcsd
  3. Aug 17, 2011 #2

    NascentOxygen

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    Staff: Mentor

    While you would arrive at the correct final answer, your working would not be quite right.
    You would be saying that P1=0.5 p v12 but that's not right. You are disregarding the constant in Bernoulli's Equation; in effect, equating it to zero.

    It is better to rearrange Bernoulli's Equation to give the pressure difference, viz., P2 - P1 and that way the constant vanishes and your working will be correct all the way through and no more complicated.
     
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