Question

A journal bearing has a 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

• A. $0.040Nm$
• B. $0.252Nm$
• C. $0.652Nm$
• D. $0.400Nm$

Answer 1

The correct option is A $0.040Nm$

$\eta =\frac{\mu v}{c}=\frac{20\times {10}^{-3}\times 20\times 20\times {10}^{-3}}{0.02\times {10}^{-3}}$
$=400N/m^2$
$\text{Force}=\eta A=400\times 3.14\times 0.04\times 0.04N$ $(\because F=\eta .\pi dl)$
$\text{Torque}=400\times 3.14\times 0.04\times 0.04\times 0.02(\because \text{Torque}=F\times r)$
$=0.040Nm$