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Homework Help: Two pipes with equal pressures

  1. Apr 29, 2010 #1
    1. The problem statement, all variables and given/known data

    A liquid is flowing through a horizontal pipe whose radius is 0.0215 m. The pipe bends straight upward through a height of 10.1 m. The pipe then bends back to the horizontal direction with a different radius of 0.0399 m. What volume flow rate will keep the pressures in the two horizontal sections of pipe the same?

    2. Relevant equations

    p1 + (1/2) ρ v1^2 + ρgh1 = p2 + (1/2) ρ v2^2 + ρ g h2

    3. The attempt at a solution

    p1 + (1/2) ρ v1^2 + ρgh1 = p2 + (1/2) ρ v2^2 + ρ g h2
    .
    (1/2) v1^2 + 0 = (1/2) v2^2 + g h2
    .
    or v1^2 = v2^2 + 2*9.80*10.1
    .
    v1^2 = v2^2 + 203.84
    .
    Now... the flow rate through each is speed * area, and area is π r2 so...
    .
    flow rate 1 = flow rate 2 or v1 * π r1^2 = v2 * π r2^2
    .
    v1 * 0.02152^2 = v2 * 0.0399^2
    .
    simplify... v1 = 3.43764 v2
    .
    plug this into the first equation...
    .
    (3.43764 v2)^2 = v2^2 + 203.84 11.81736877 v2^2 = v2^2 + 203.84
    .
    11.81736877 v2^2 = 203.84 v2^2 = 17.2491811

    v2 = 4.153205763 is the speed in the upper pipe.
    .
    flow rate = speed * area = 4.153205763 * π * 0.0399 = 0.5206024 m3 /sec

    please help!! thanks
     
  2. jcsd
  3. Apr 30, 2010 #2
    Does 2*9.8*10.1 really equal 203.84?? (I can see without using a calculator that this number is not going to exceed 200)

    And in your flow rate equation, the radius should be squared. I think your equations are fine, but not the "maths".
     
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