What is the mass flux through a kidney?

  • Thread starter Thread starter giacomh
  • Start date Start date
  • Tags Tags
    Flux Mass
Click For Summary
SUMMARY

The discussion focuses on calculating the mass flux through a kidney, given an inflow rate of 10 mL/min and an outflow through a 6-mm-diameter tube at 20 mm/s. The correct mass flow rates were determined to be 1.67 x 10^-7 m³/s for inflow and 5.65 x 10^-7 m³/s for outflow. The final calculation for the rate of change of mass in the kidney is 3.99 x 10^-4 kg/s, correcting the initial miscalculation of 0.165 kg/s.

PREREQUISITES
  • Understanding of fluid dynamics principles
  • Familiarity with mass flow rate calculations
  • Knowledge of unit conversions (mL to m³, mm to m)
  • Basic understanding of density and its role in mass calculations
NEXT STEPS
  • Study fluid dynamics equations, specifically mass flow rate equations
  • Learn about unit conversions in physics and engineering contexts
  • Explore the application of the continuity equation in fluid systems
  • Investigate the impact of tube diameter on flow rates and pressure
USEFUL FOR

This discussion is beneficial for students in physics or engineering, particularly those studying fluid dynamics, as well as professionals involved in biomedical engineering and renal physiology.

giacomh
Messages
36
Reaction score
0

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, that's not the right answer, but I can't figure out where I went wrong.
 
Physics news on Phys.org
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
 

Similar threads

Replies
7
Views
5K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
7
Views
2K
  • · Replies 10 ·
Replies
10
Views
6K
  • · Replies 13 ·
Replies
13
Views
16K
  • · Replies 1 ·
Replies
1
Views
11K
  • · Replies 4 ·
Replies
4
Views
9K
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
9K