Q. Find the kinematic viscosity of an oil having density 981 kg / (m ^ 3) . The shear stress at a point in oil is 0.2452N / (m ^ 2) and velocity gradient at that point is 0.2 per second.
SOLUTION:
Mass density,rho = 981kg / (m³)
Shear stress, τ = 0.2452N / (m²)
Velocity gradient, du/dy = 0.2s
Using the equation, τ = μ * du/dy (or) 0.2452 = μ * 0.2
μ = 0.245/0.200 = 1.226Ns / m²
Kinematic viscosity u is given by,
v = μ/p = 1.226/981 =.125*10^ -2 m² /sec
= 0.125 * 10-² * 10⁴cm²/ s
= 0.125 * 10² cm² / s
= 12.5 cm² / s
=12.5 stoke (cm²/s = stoke)
Q. Determine the specific gravity of a fluid having viscosity 0.05 poise and kinematic viscosity 0.035 stokes.
SOLUTION:
Given:
Viscosity, µ = 0.05 poise = 0.05/10 Ns/m²
Kinematic viscosity, v = 0.035 stokes = 0.035 cm²/s
= 0.035 * 10-⁴ m²/ s
Using the relation v=µ/p
We get 0.035 * 10-⁴ = 0.05/10 * 1/P
P = 0.05/10 * 1/(0.035 * 10-⁴)
= 1428.5 kg/m³
Sp. Gravity of liquid = Density of liquid/Density of water
=1428.5/1000
= 1.4285 = 1.43 Ans
