A lightly loaded full Journal bearing has Journal diameter

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

Given: d_j = 50 mm = 0.05 m
d_B = 50.05 = 0.05005 m
l_B = 20 mm = 0.02 m
N = 1200 rpm
μ = 0.03 Pa– sec
Frictional factor, f = Shear stress × Area

=	μU	× πd_jl_B
	h

Here, h = r_B – r_j
= 25.025 – 25 = 0.025 = 0.000025 m

U =	πd_jN	=	π × 0.05 × 1200	= 3.141
	60		60

∴ f	0.03 × 3.141	× π(0.05)(0.02)
	0.000025

= 11.841
Torque, T = f × r_j
= 11.841 × 0.025 = 0.2960
Power loss = T × ω

=	2πNT
	60

=	2π × 1200 × 0.2960
	60

= 37.196 W ≈ 37.2 W