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

ANSWER: 0.126 kW

Related Content

 

    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

➨ Post your comment / Share knowledge


Enter the code shown above:

(Note: If you cannot read the numbers in the above image, reload the page to generate a new one.)
▲