Question

A lightly loaded full journal bearing has journal diameter of 50 mm, bush bore of 50.05 mm and bush length of 20 mm. If rotational speed of journal is 1200 rpm and average viscosity of liquid lubricant is 0.03 Pa s, the power loss (in W) will be

• A. $118$
• B. $74$
• C. $237$
• D. $37$

Answer 1

The correct option is D $37$

Tangential velocity of shaft,
$u=\frac{\pi DN}{60}=\frac{\pi \times 50\times {10}^{-3}\times 1200}{60}$
$=3.14m/s$
Clearance, $y=\frac{50.05-50}{2}=0.025mm$
By $\tau =\mu \frac{du}{dy}$ Shear stress on shaft
$\tau =0.03\times \frac{3.14}{0.025\times {10}^{-3}}=3768N/m^2$
Shear force on shaft,
$F=\tau \times Area=3768\times \pi D\times L$
$=3768\times \pi \times 50\times {10}^{-3}\times 20\times {10}^{-3}$
$=11.83N$
Torque : $T=F\times \frac{D}{2}=11.83\times \frac{50\times {10}^{-3}}{2}$
$=0.2953Nm$
Power loss $=\frac{2\pi NT}{60}=\frac{2\pi \times 1200\times 0.2953}{60}$
$=37.1W$