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Fluid mechanics - flow through pipe

  1. Sep 20, 2012 #1
    Please see attached. I'm pretty sure i did the problem right, but I'd like to check with the great minds of physicsforums.

    Summing up flow rates and inlet and outlet gives:
    1500+1200=2700 gal/min

    2700gal/min= area at exit *Velocity at exit
    2700 gal/min * (1ft^3 / 7.48 gal) ( 1min / 60 s) = Area at exit * 30 ft/2
    Area=0.2 ft^3 or 1.5 gallons
     

    Attached Files:

  2. jcsd
  3. Sep 20, 2012 #2
    You are correct but my mind is not a great one.
     
  4. Sep 23, 2012 #3
    I'm second guessing myself. The way I did the problem is assuming that the volume flow rate is conserved, but can I really assume that? What I basically said is that the densities at the 2 inlets and outlet are the same.

    The densities given aren't the same tho.


    I.E


    p1*V1*A1+p2*V2*A2=p3*V3*A3
    where p=density
    V=velocity
    A=area

    That is the conservation of mass flow equation.
    To get from that equation to the
    Q1+Q2=Qout equation, the densities have to cancel. If they cancel, that means the densities are the same for both inlets and outlets, but that is not the case since they give different specific gravities for the 2 inlets, thus different densities.

    It seems the problem contradicts itself by saying "volume flow is conserved" and giving 2 different SGs.
     
    Last edited: Sep 23, 2012
  5. Sep 23, 2012 #4
    UGHHH the more i read this problem the more I get confused.

    Read the 1st sentence. imiscible=not mixable
    How can 2 not mixable liquids be mixed????
    How can there be a "mixture" that exists, if it's not mixable???
     
  6. Sep 24, 2012 #5
    "How can 2 not mixable liquids be mixed????"
    Have you ever mixed salad dressing..oil and vinegar? Vinegar and oil do not mix but each still maintains its unique volume when combined. Each is incompressible. Therefore flow volume per unit time is conserved.
     
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