Question

The Velocity Distribution for flow over a flat plate is given by u=(2/3)y-y2, Where u is the point velocity in meter per second at a distance y meter above the plate.

Determine the shear stress at y=0 and y=15 cm. Assume dynamic viscosity as 8.63 poises

Answer 1

Given $u=\dfrac{2}{3}y-y^2.$ Then

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

Shear stress is given

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

Assume dynamic viscosity as $\mu=8.63\ \text{poise}=0.863\ N\cdot s/m^2$

At $y=0$

$$\dfrac{du}{dy}=\dfrac{2}{3}-2(0)=\dfrac{2}{3}(s^{-1})$$

Then

$$\tau=0.863\ N\cdot s/m^2\cdot\dfrac{2}{3}s^{-1}=0.575 N/m^2$$

The shear stress at y = 0 is $0.575N/m^2.$

At $y=15 cm=0.15m$

$$\dfrac{du}{dy}=\dfrac{2}{3}-2(0.15)=\dfrac{1.1}{3}(s^{-1})$$

Then

$$\tau=0.863\ N\cdot s/m^2\cdot\dfrac{1.1}{3}s^{-1}=0.316 N/m^2$$

The shear stress at y = 15 cm is $0.316N/m^2.$