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

$u = r\omega$

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

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

$T = \tau A r$

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

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

Answer 2

The torque required to turn the bearing assuming that the flow between the shaft and the casing is a Couette flow is; T = 2.78 × 10⁻³ N/m

Calculation of Torque

We are given;

Viscosity; μ = 0.1 N.s/m²

diameter; d = 30 mm = 0.03 m

radius; r = d/2 = 0.015 m

Gap between shaft and casing; dy = 2 mm = 0.002 m

Length of bearing; L = 0.03 m

Angular velocity; ω = 350 rad/s

du = rω

du = 0.015 × 350 = 5.25 N/s

Formula for stree is;

τ = μ(du/dy)

τ = 0.1 × (5.25/0.002)

τ = 262.5 N/m²

To find torque, we will use the formula;

T = τ*A*r

T = 262.5 * π * 0.015² * 0.015

T = 2.78 × 10⁻³ N/m

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