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:

$T = 1.06 \times 10^{-3} N. mm$

Explanation:

Given data:

$\mu = 0.1 N-s /m^2$

d = 30 mm = 0.03 m

dy = 2.0 mm

L = 3 cm

$\omega = 350 rad/s$

we know

$u = r\omega$

$u = 0.15 \times 350 = 52.5 N/s$

$\tau = \mu \frac{du}{dy} = 0.1 \times \frac{1}{.002} = 100 N/m^2$

$T = \tau A r$

 $= 100 \times \frac{\pi}{4} 0.03^2 \times 0.015$

$T = 1.06 \times 10^{-3} N. mm$