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Density of a liquid in a horizontal pipe

  1. May 10, 2015 #1
    1) Alright, so the problem reads, "The liquid in a horizontal pipe of diameter d1 before a constriction is at pressure P1 and flowing with the speed v1, and then in the constriction with diameter d2 the gauge measures pressure P2. What is the density of the liquid?"

    2) I'm assuming the equations I'd need would be the equation of continuity, A1v1 = A2v2, and Bernoulli's equation.

    3) So my first step was to figure out what v2 was by using the equation of continuity which gave me
    v2 = v1d12 / d22

    The second step was to set up Bernoulli's equation, P1 + ρv12/2 + ρgh = P2 + ρv22/2 + ρgh. Since the pipe is horizontal ρgh cancels out on both sides leaving P1 + ρv12/2 = P2 + ρv22/2. Then I began the process of isolating ρ.

    ρ(v12/2 - v22/2) = P2 - P1

    ρ = (P2 - P1) / (v12/2 - v22/2)

    ρ = (P2 - P1) / [v12/2 - (v12d14)/ (2d24)]

    ρ = (P2 - P1) / [(v12d24 - v12d14) / 2d24]

    And finally I arrived at my final simplified solution;

    ρ = [2d24 (P2 - P1)] / [v12 (d24 - d14)]

    I'm not all that confident in my math for this particular problem. Any input would be greatly appreciated as an exam approaches and I want to make sure I have a good understanding of this particular type of problem.
     
  2. jcsd
  3. May 10, 2015 #2

    mfb

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    2016 Award

    Staff: Mentor

    Looks right.

    As d2<d1, both numerator and denominator will be negative. You can change that by swapping the sign in both of them, but it does not change the result.
     
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