Question

3. A 100 mm diameter shaft is supported by a bearing 50 mm long with a minimum oil-film thickness of 0.02 mm and radial clearance of 0.06 mm. It is lubricated by SAE 30 oil. The bearing carries a load of 4 kN of at 900 rpm. Employing the design charts, determine The temperature of the oil film • • The coefficient of friction • The friction power

Answer 1

Given:

D = 100 mm, L = 50 mm, h₀ = 0.02 mm, c = 0.06 mm, SAE 30, F = 4000 N, N_rpm = 900 rpm

1. As the vicosity of the oil varies logarithmically with temperature and as neither of them have been given. Then, assuming vicosity of the oil to be 30mPa.s.

Then, according to the viscosity-temperature chart temperature is found to be 65 ⁰C.

2. Coefficient of friction -

According to Petroff's equation -

247 = 1

where, f = Coefficient of friction

r = radius = D/2 = 50 mm

c = radial clearance = 0.06 mm

$\mu$ = viscosity of oil (Ns/mm²) = 30mPa.s = 30 * 10-9 N.s/mm²

n = journal speed (revolution/sec) = 900 rpm = 900/60 rps = 15 revol./sec

p = unit bearing pressure (N/mm²) = F/A = 4000/ 100 * 50 = 0.8 N/mm²    A = projected Area(DL)

f = 2 * (pi)² * (50/0.06) * (30 * 10-9 * 15 / 0.8) = 0.009

Coefficient of friction = 0.009

3. Frictional Power -

F.P = $\omega T$ = Angular speed * Torque

where, $\omega$ = 2 * (pi) * n = 2 * pi * 15 = 94.24 rad/s

T = f * F * r = 0.009 * 4000 * 50 = 1800 N.mm = 1.8 N.m

F.P. = 94.24 * 1.8 = 169.632

Frictional Power = 169.63 W

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