Question

A bearing uses SAE 30 oil with a viscosity of 0.1 N·s/m2. The bearing is 30 mm in diameter, and the gap between the shaft and the casing is 2.0 mm. The bearing has a length of 3 cm. The shaft turns at ω = 350 rad/s. Assuming that the flow between the shaft and the casing is a Couette flow, find the torque required to turn the bearing.

Answer 1

Answer:

T = 2.78 x 10⁻³ N.m

Explanation:

given,

μ = 0.1 N.s/m²

diameter = 30 mm = 0.03 m

d y = 2 mm = 0.002 m

L = 0.03 m

ω = 350 rad/s

u = r ω

u = 0.015 x 350 = 5.25 N/s

$\tau =\mu\dfrac{du}{dy}$

$\tau =0.1\times \dfrac{5.25}{0.002}$

 τ = 262.5 N/m²

torque

T =  τ A r

T =  262.5 x π r² x r

T =  262.5 x π x 0.015³

T = 2.78 x 10⁻³ N.m