Question

If the velocity distribution of a fluid over a plate is given by u = (3/4)y – y 2 , where u is the velocity

in metre per second at a distance of y metres above the plate, determine the shear stress at

y = 0.15 metre. Take a dynamic viscosity of a fluid as 8.5 x 10 -5 kg.s/m 2 .

Answer 1

Gives

Velocity

$u=\frac{3}{4}y+y^2$

y=0.15 meter

$\eta=8.5\times10^{-5}kg s/m^2$

$\frac{du}{dy}=\frac{3}{4}+2y$

At y=0.15meter

$\frac{du}{dy}=\frac{3}{4}+2(0.15)$

$\frac{du}{dy}=0.75+0.3$

$\frac{du}{dy}=1.05$

Shear stress = $\eta\times\frac{du}{dy}$

Put value

Shear stress =

$$8.5\times10^{-5}\times1.05=8.925\times10^{-05}N/m^2$$

Shear stress=$8.925\times10^{-05}N/M^2$