What is the Critical Radius in Fluid Dynamics?

[kpx001]

Homework Statement

http://img571.imageshack.us/i/reaction.png/

Homework Equations

Velocity = (R^2-r^2) * dP/(4nL)
VMax = R^2 * dP/(4nL)
Average velocity = R^2 * (dP/(L*8n))

dP= change in pressure
n = viscosity

The Attempt at a Solution

Average velocity = R^2 * (dP/(L*8n))
.05 = (.015)^2 * (dP/((10*8*8.4*10^-4))
dP = 14.4 Pa

Velocity = Vcrit
(R^2-r^2) * dP/(4nL) = .07
(.015^2-r^2) * 14.4/(4*8.4*10^-4*10) = .07
r = .00785

so @ r = .00785, it is the Vcritical?

I don't know what to do from here.

 

[KataKoniK]

Your calculations seem to be on the right track, but there are a few things that need to be clarified. First, it would be helpful to define the variables R and r in your equations. I assume R refers to the radius of the cylinder, but what does r represent?

Secondly, you have correctly calculated the pressure change (dP) required to achieve a velocity of 0.05 at the critical radius, but it would also be helpful to include the units for dP (in this case, it would be in Pascals, as you have calculated).

Finally, to determine the critical radius, you need to set the average velocity equation equal to the maximum velocity equation (since the velocity will be at its maximum at the critical radius). This will give you an equation to solve for r. Once you have the value of r, you can plug it into the equations to find the corresponding values for dP and the maximum velocity.

I hope this helps guide you towards finding the solution. Good luck with your calculations!