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Integral Form of the Momentum Equation - Reducer Question

  1. Apr 4, 2013 #1
    1. The problem statement, all variables and given/known data

    The internal volume of the reducer is 0.2m^3 and its mass is 25 kg. The fluid being pumped is oil (specific gravity of 0.72).
    Evaluate the total force that must be provided to support the reducer.

    d1 = 0.4m
    d2 = 0.2m
    u1 = 3m/s
    p1 = 58.7 kPa
    p2 = 49kPa (gauge)


    2. Relevant equations


    Qin=Qout

    mdot=ρ*A*u

    A=(∏*d^2)/4


    3. The attempt at a solution

    Tried. Failed. Help.
     
  2. jcsd
  3. Apr 4, 2013 #2
    the answer is apparently F = −3.4xˆ +1.66yˆ kN but I am in need of the working please.
     
  4. Apr 4, 2013 #3

    rude man

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    What does a 'reducer" do?
     
  5. Apr 4, 2013 #4

    SteamKing

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    A reducer couples a larger diameter pipeline to a smaller diameter pipeline, hence the two diameters specified in the OP.
     
  6. Apr 4, 2013 #5

    rude man

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    Thanks SK!

    But - what is meant by its volume? Is it a tapered section of pipe going from the larger to the smaller diameter?
     
  7. Apr 4, 2013 #6

    SteamKing

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    It can be. More often it is a cast fitting, either flanged or suited to welding, with a curved transition between the larger and smaller diameters, so that the overall length of the fitting is kept small. The volume, I believe, is just what it implies, the volume of the internal space of the reducer.
     
  8. Apr 5, 2013 #7
    yea this is right. it can be basically thought of as a converging pipe and the question is based on the internal volume as this helpful fella said. and the mass is the mass of the actual reducer.

    i have obtained the answer but not confident on my methods. anyone else had any luck?
     
  9. Apr 5, 2013 #8

    rude man

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    Thanks to my friend Chestermiller from a similar previous problem, the approach here is to consider the change in momentum per unit time of the oil and equate that to the (longitudinal) force exerted on the reducer.
     
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