Calculating Surface Shear Stress in a Boundary Layer with Given Velocity Profile

In summary, the conversation discusses finding the surface shear stress in a boundary layer given the velocity and distance from the surface. The equation used is τ = μ(du/dy) and the constants a and b are solved for using the given velocity data. The final answer is 1.17 N/m^2.
  • #1
exphaze
4
0

Homework Statement


If the velocity v of the air in a boundary layer having a dynamic viscosity of
18x10-6 kg/ms is given in terms of the distance y from the surface by v = ay + by2

where a and b are constants, calculate the surface shear stress if at 1.5 mm from the surface the velocity is 75 m/s, and at 3.0 mm from the surface it is 105 m/s.

Homework Equations


τ = μ(du/dy)

The Attempt at a Solution

 
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  • #2
What have you figured out so far?
 
  • #3
Chestermiller said:
What have you figured out so far?
I have tried the differences of the velocities and distance (du/dy) . Then multiplied it by the dynamic viscosity which came out as 0.36 N/m^2 but the answer is apparently 1.17 N/m^2.
 
  • #4
exphaze said:
I have tried the differences of the velocities and distance (du/dy) . Then multiplied it by the dynamic viscosity which came out as 0.36 N/m^2 but the answer is apparently 1.17 N/m^2.
You need to solve for a and b from the velocity data; a is the shear rate at the wall.
 
  • #5
Chestermiller said:
You need to solve for a and b from the velocity data; a is the shear rate at the wall.
I have worked out a = 50000 and substituted to τ*shear rate =μ and this came out as 0.9 N/m^2. I am still stuck.
 
Last edited by a moderator:
  • #6
exphaze said:
I have worked out a = 50000 and substituted to τ*shear rate =μ and this came out as 0.9 N/m^2. I am still stuck.
I get a = 65000/sec
 
  • #7
Chestermiller said:
I get a = 65000/sec
I got it now, Thanks for the advice!
 
  • #8
Hello
I have the same issue, but unfortunately I can't solve it.
any hind please?
 

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