Question

A journal bearing has shaft diameter of 40 mm and a length of 40 mm. The shaft is rotating at 20 rad/s and the viscosity of the lubricant is 20 mPa.s. The clearance is 0.020 mm. The loss of torque due to the viscosity of the lubricant is approximately
1. 0.040 Nm
2. 0.252 Nm
3. 0.400 Nm
4. 0.652 Nm

Answer 1

Correct Answer - Option 1 : 0.040 Nm

Calculation:

Given: 

shaft diameter, d = 40 mm = 0.04 m⇒ r = 20 mm = 0.02 m,  length, l = 40 mm = 0.04 m, rotational speed of shaft, ω = 20 rad/s,

viscosity, μ = 20 mPa.s., clearnance, C = 0.020 mm 

velocity at suface of bearing, V = r × ω = 20 × 20 × 10^-3 m/s = 0.4 m/s

​\(\mathbf{\tau =μ \frac{{ V}}{C}}\)​

\(= \frac{{20\;×\; {{10}^{ - 3}} ×\:0.4}}{{0.02\; × \;{{10}^{ - 3}}}}\)

= 400 N/m²

Force, F = τ × A = τ × π d l

Torque = F × r

= 400 × 3.14 × 0.04 × 0.04 × 0.02

= 0.040 Nm