# Numerical - Power lost in friction, given radial load, diameter & speed

Q.  A journal of 120 mm diameter rotates in a bearing at a speed of 1000 rpm. What is the power lost during friction if 8 kN radial load acts on the journal and coefficient of friction is 2.525 x 10-3?
- Published on 19 Oct 15

a. 0.126 kW
b. 0.253 KW
c. 2.365 kW
d. 7.615 kW

#### Discussion

• KETAN    -Posted on 19 Nov 20
Ans is wrong 0.08636 ans is comming.
• SYED MUHAMMAD SHAH JAMI   -Posted on 27 Nov 16
The solution provided above doesn't end up with the answer mentioned kindly recheck the calculations please. thanks
• Sravanthi   -Posted on 05 Oct 15
Given: Journal diameter = 120 mm, Rotating speed of journal = 1000 rpm, radial load acting on journal = 8 kN, coefficient of friction (f) = 2.525 x 10-3

Rate of heat generated is equal to power lost in friction.
Power lost in friction is the product of frictional torque and angular velocity.

Therefore, power lost in friction = [(2 π ns) / (f W r)] / 106 kJ /s

where,
ns = speed of journal in rps
F = coefficient of friction
W = radial load acting on journal
r = radius of journal

convert speed into rps
Therefore, 1000 rpm = 1000 / 60 = 16.66 rps

Substituting the given values in formula,
Power lost in friction = [(2 π ns) / (f W r)] / 106
= [ (2 x π x 16.66) / (2.525 x 10-3 x 8000 x 60)]
= 0.126 kW

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