Question

The velocity distribution for flow over a flat plate is given by u = 3/2 y - y^(3/2) , where u is the point velocity in metre per second at a distance y metre above the plate. Determine the shear stress at y = 9 cm. Assume dynamic viscosity as 8 poise.

Answer 1

Given $u=\dfrac{3}{2}y-y^{3/2}.$ Then

$$\dfrac{du}{dy}=\dfrac{3}{2}-\dfrac{3}{2}y^{1/2}$$

At $y=9 cm=0.09m$

$$\dfrac{du}{dy}=\dfrac{3}{2}-\dfrac{3}{2}(0.09)^{1/2}=1.05(s^{-1})$$

Shear stress is given

$$\tau=\mu\cdot\dfrac{du}{dy}$$

Assume dynamic viscosity as $\mu=8\ \text{poise}=0.8\ N\cdot s/m^2$

Then

$$\tau=0.8\ N\cdot s/m^2\cdot1.05s^{-1}=0.84 N/m^2$$

The shear stress at y = 9 cm is $0.84N/m^2.$