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Mass flux through a kidney

  1. Sep 30, 2012 #1
    1. The problem statement, all variables and given/known data

    Water enters a kidney through a tube at 10 mL/min. It exits through a 6-mm-diameter tube at 20 mm/s.What is the rate of change of mass of water in the kidney?

    2. Relevant equations

    m=vpa
    q=va

    After conversions to meters
    V=.02
    Q=.000167
    A=(pi/4)*.006^2 = .0000282

    3. The attempt at a solution

    dm/dt-Qp+VAp=0
    dm/dt-(.000167)(999.1)+(.0000282)(.02)(999.1)=0
    dm/dt=.165 kg/s

    Needless to say, thats not the right answer, but I can't figure out where I went wrong.
     
  2. jcsd
  3. Oct 1, 2012 #2
    You are incorrectly finding the positive flux. It is 1.67*10^-4 for the inflow and 5.655*10^-4 for the outflow.

    Inflow 10 mL/min * (min/60 s) * (1 cm^3/mL) * (m^3/1000000 cm^3) = 1.67*10^-7 m^3/s

    Outflow=V*A { 20 mm/s * (m/1000mm) } * { [(6 mm/2)/1000]^2 * pi } = 5.65*10^-7 m^3/s

    Δm = -density*Q(in) + density*Q(out)
    = [-1000*.000000167]+[1000*.000000565]
    = 3.99*10^-4 kg/s
     
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