If the velocity distribution of a fluid over a plate is given by u=(3/4) y - y2, 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 dynamic viscosity of the fluid as 8.5 x 10-5 kg-sec/m2.
Find the shear stress developed in a lubricating oil, of viscosity 9.81 poise, filled between two parallel
plates 1 cm apart and moving with relative velocity of 2 m/s ?
Explanation: Shear stress is the stress that acts co-planar with cross section of the material.
Let the viscosity be n, distance between the plates be dz and relative velocity be du.
du= 2m/s
dz= 1 cm
= 0.01 m ( 1 m = 100 cm)
n= 9.81 poise
= 0.981 Pa s ( 1 poise= 0.1 Pa s)
The formula for the shear stress developed in the lubricating oil is given by the formula:
Shear stress
= n × du / dz
= 0.981 × 2 / 0.01
= 196.2 Pa
Therefore, the shear stress developed in the lubricating oil is 196.2 Pascal (Pa).
Explained in the notebook for detailed
The tank shown in the figure is 3m wide in the paper. Find the horizontal, vertical, and resultant
force on quarter-circle BC. Take the density of water as 1000kg/mÂ
1. Given the velocity field: V = (6 + 2xy + f²)i - ( xy² + 10t)j + 25k. What is the acceleration of a
particle at (3, 0, 2) at time t = 1?
The x component of velocity in a two dimensional incompressible flow is given by u=1.5x.
At the point (x, y) = (1, 0), the y component of velocity v = 0. The equation for the y
component of velocity is
