Question

Find the kinematic viscosity of an oil having density 980 kg/m^2 when at a certain point in the oil, the shear stress is 0.25 N/m^2 and velocity gradient is 0.3/s .

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.

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

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

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

Given $\rho=980kg/m^3,\tau=0.25N/m^2 , \dfrac{du}{dy}=0.3/s$

$$\nu=\dfrac{0.25N/m^2}{980kg/m^3\cdot0.3s^{-1}}\approx0.00085m^2/s=8.5St$$

Kinematic viscosity is 8.5 St.