Mass flux through a kidney

  • Thread starter giacomh
  • Start date
  • #1
giacomh
36
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Homework Statement



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?

Homework Equations



m=vpa
q=va

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

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.
 

Answers and Replies

  • #2
zmp3
3
0
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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