Q. Two horizontal plates are placed 1.25 cm apart, the space between them being filled with oil of viscosity 14 poises. Calculate the shear stress in oil if upper plate is moved with a velocity of 2.5 m/s.
SOLUTION:
Given:
Distance between plates, dy 1.25 cm = 0.0125 m
Viscosity μ = 14 poise = 14/10 Ns/m²
Velocity of upper plate,u =2.5 m/sec.
Shear stress is given by equation as, τ = μ du/dy
Where du = change of velocity between plates
= น – 0 =2.5 m/sec.
dy= 0.0125 m
Thefore τ = 14/10 × 2.5/0.0125 = 280 N/m².
Q. The space between two square flat parallel plates is filled with oil. Each side of the plate is 60 cm. The thickness of the oil film is 12.5 mm. The upper plate, which moves at 2.5 metre per sec requires a force of 98.1 N to maintain the speed. Determine:
(i) The dynamic viscosity of the oil in poise, and
(ii) The kinematic viscosity of the oil in stokes if the specific gravity of the oil is 0.95.
SOLUTION:
Given:
Each side of a square plate
= 60cm = 0.6m
Area A = 0.6 * 0.6 = 0.36m²
Thickness of oil film dy = 12.5mm = 12.5 * 10 ^ – 3 m
Velocity of upper plate u= 2.5m /sec
Therefore Change of velocity between plates, du =2.5 m/sec
Force required on upper plate, F = 98.1N
Shear stress τ= Force/ Area = F/A = (98.1N)/0.36m²
(i) Let µ = Dynamic viscosity of oil
Using equation
τ = μ * du/dy (or) 98.1/0.36 = μ× 2.5/(12.5 * 10-³)
μ = 98.1/0.36 *( 12.5 * 10 -³)/2.5
= 1.3635Ns / m² [ 1Ns /m² = 10 poise]
(ii) Sp. gr. of oil, S = 0.95
Let v = kinematic viscosity of oil
Using equation,
Mass density of oil, ρ = S * 1000 = 0.95 * 1000 = 950kg /m³
Using the relation v = μ/ρ
We get v = (1.3635(Ns /m²)/950 =.001435 m²/sec
= 0.001435 * 10⁴ cm² / s
= 14.35 stokes.
