Determine the intensity of shear of an oil having viscosity = 1 poise. The oil is used for lubricating the clearance between a shaft of diameter 10 cm and its journal bearing. The clearance is 1.5 mm and the shaft rotates at 150 rpm

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 Q. Determine the intensity of shear of an oil having viscosity = 1 poise. The oil is used for lubricating the clearance between a shaft of diameter 10 cm and its journal bearing. The clearance is 1.5 mm and the shaft rotates at 150 rpm

SOLUTION:
Given:
μ= 1 poise- 10 Ns/m²
Dia of shaft D = 10cm = 0.1m
Distance between shaft and journal bearing.
dy = 1.5mm = 1.5 * 10-³ m
Speed of shaft N= 150r .p.m
Tangential speed of shaft is given by
μ = (π*DN)/60 =(π * 0.1 * 150)/60
= 0.785m / s
Using equation, τ = μ* du/dy
Where du =Change of velocity between shaft & bearing
= u + 0 = u
= 1/10 * 0.785/(1.5 * 10-³) = 52.33N / (m ^ 2)

 

Q. Calculate the dynamic viscosity of an oil, which is used for lubrication between a square plate of size of 0.8 * 0.8 m and an inclined plane with angle of inclination 30º as shown in figure The weight of the square plate is 300 N and it slides down the inclined plane with a uniform velocity of 0.3 m/s. The thickness of oil film is 1.5 mm.

SOLUTION:
Given:
Area of plate,A = 0.8 * 0.8 = 0.64m ^ 2
Area of plane,θ = 30°
Weight of plate,W = 300N
Velocity of plate,u = 0.3m / s
Thickness of oil film, t = dy = 1.5mm
= 1.5*10-³m
Let the viscosity of fluid between plate and inclined plane is µ.
Component of weight W, along the plane
= W * cos 60° = 300cos 60° = 150N
Thus the shear force, F, on the bottom surface of the plate = 150 N
and shear stress, τ=F/Area = 150 /0.64 N/m²
Now using equation, we have
τ = μ* du/dy
where du = Change of velocity= u – 0 = u = 0.3 m/s
dy = t = 1.5 * 10-³ * m
Therefore 150/0.64 = μ * 0.3/(1.5 * 10-³)
μ = (150 * 1.5 * 10-³)/(0.64 * 0.3)
= 1.17Ns / (m ^ 2) = 1.17 * 10
= 11.7 poise.

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