For two 0.2 m long rotating concentric cylinders, the velocity distribution is given by u(r) = 0.4/r - 1000r m/s. if the diameters are 2 cm and 4 cm, respectively, calculate the fluid viscosity if the torque on the inner cylinder is measured to be 0.0026 n*m.

5.9*10^-3 Pa*s

Explanation:

The fluid will create a tangential force on the surface of the cylinder depending on the first derivative of the speed respect of the radius.

τ = μ * du/dr

u(r) = 0.4/r - 1000*r

The derivative is:

du/dr = -1/r^2 - 1000

On the radius of the inner cylinder this would be

u'(0.02) = -1/0.02^2 - 1000 = -3500

So:

τ = -3500 * μ

We don't care about the sign

τ = 3500 * μ

That is a tangential force per unit of area.

The area of the inner cylinder is:

A = h * π * D

And the torque is

T = F * r

T = τ * A * D/2

T = τ * h * π/2 * D^2

T = 3500 * μ * h * π/2 * D^2

Then:

μ = T / (3500 * h * π/2 * D^2)

μ = 0.0026 / (3500 * 0.2 * π/2 * 0.02^2) = 5.9*10^-3 Pa*s

how is this a question? in order for me to answer it i have to understand the question

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