SOLVED EXAMPLE-1 | VISCOSITY

Example-1:

If the velocity distribution over a plate is given by u = (2/3)y - y² in which u is the velocity in meter per second at a distance y meter above the plate, determine the shear stress at y = 0 and y = 0.15 m. Take dynamic viscosity of fluid as 8.63 poises.

Solution

m = 8.63 poise = 0.863 Pa·s
Assuming the fluid to be Newtonian,

Shear stress, T = m (du/dy)
u = (2/3)y - y²
du/dy = 2/3 - 2y
Therefore,
t = m [2/3 - 2y]
(i) At y = 0
T = m(2/3) = (2 × 0.863)/3 = 0.5756 N/m²
(ii) At y = 0.15 m
T = 0.863[(2/3) - 2(0.15)] = 0.3167 N/m²