Problem

  Question. The dynamic viscosity of an oil, used for lubrication between a shaft and sleeve is 6 poise. The shaft is diameter 0.4m and rotates at 190 r.p.m. Calculate the power lost in the bearing for a sleeve length of 90mm. The thickness of oil film is 1.5mm. 

Solution. Given :

Viscosity,    μ = 6 poise

                       = 6Ns/10m² = 0.6 Ns/m²

Dia. of shaft, D = 0.4 m

Speed of shaft, N = 190 r.p.m.

Sleeve length, L = 90mm = 90 × 10^-3 m

Thickness of oil film, t = 1.5 mm = 1.5 × 10^-3 m

Tangential velocity of shaft, u = π D N / 60

= π × 0.4 × 190 / 60

= 3.98 m/s

Using the relation,  τ = μ × (du/dy)

where, du = change in velocity = u – 0 = u= 3.98 m/s

           dy = change of distance = t = 1.5 × 10^-3 m

τ = 6 × 3.98 / 10 × 1.5×10^-3

= 1592 N/m²

This is shear stress on shaft

 Shear force on the shaft, F = shear stress × Area

= 1592 × πD × L

= 1592 × 0.4 × 90 × 10^-3 = 180.05 N

Torque on the shaft, T = Force × D/2

= 180.05 × 0.4/2 = 36.01 Nm

Power lost, = 2πNT/60

= 2π × 190 × 36.01 /60

= 716.48 W. Ans