Answer to Question #142984 in Other for Usman

Question

In a stream of glycerine in motion, at a certain point the velocity gradient is 0.25 metre per sec per metre . The mass density of fluid is 1268.4 kg per cubic metre and kinematic viscosity is 6.30 x 10^-4 square metre per second. Calculate the shear stress at the point.

Answer 1

By the Newton's Law of Viscosity

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

Kinematic viscosity is defined as the ratio between the dynamic viscosity and and density of fluid.

$$\nu=\dfrac{\mu}{\rho}$$

Substitute

$$\tau=\nu\cdot\rho\cdot\dfrac{du}{dy}$$

Given

$\rho=1268.4kg/m^3,$

$\nu=6.3\times10^{-4}m^2/s ,$

$\dfrac{du}{dy}=0.25/s$

$$\tau=1268.4kg/m^3\cdot6.3\times10^{-4}m^2/s\cdot0.25/s=$$ $$=0.199773N/m^2\approx0.2N/m^2$$

The shear stress at the point is $0.2N/m^2.$