Question

consider a cylindrical roller bearing with a bore of 40 mm, subjected a radical load 15KN. the coefficient of friaction is 0.001 and the angular speed is 150​

Answer 1

Given :

d = 150 mm = 0.15 m W = 10 kN = 10000 N N = 1500 r.p.m.

= 1.5 d

c = 0.15 mm

Z = 0.011 kg/m-s We know that length of bearing,

= 1.5 d

=1.5 × 150

=225 mm

Bearing pressure,

= W / A

=W / ld

=10000 / (225 x150)

=0.296 N/mm

We know that coefficient of friction,

Answer 2

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A cylindrical roller bearing with a bore of 40 mm is subjected to a radial load of 15 kN. The coefficient of friction is 0.001 and the speed of rotation is 150 rad/s. What is the power lost in friction (in Watt)?

Concept:

• Both fluid friction and friction between moving elements in a bearing cause heat to be produced.
• Power is lost as a result of bearing friction.
• In contrast to ball bearings, which employ balls as their rolling parts, cylindrical roller bearings use cylinders.

Given:

• W = 15 kN = 15000 N
• D = 40 mm = 0.04 m
• coefficient of friction μ = 0.001
• Angular speed ω = 150 rad/s

Find:

• The power lost in friction

Solution:

Heat produced in a bearing or power loss in a bearing

P = μvW

v = rω

r = D/2 = 0.04/2 = 0.02

v = 0.02* 150 = 3 m/s

P = μvW

P = 0.001 * 3 * 15* 1000 = 0.045* 1000 = 45 W

The power lost is 45 W.

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