Numerical - Power loss, given frictional torque & speed

Q.  A cylindrical roller bearing is subjected to frictional torque of 500 N-mm. What is the approximate power loss if the shaft rotates at 1500 r.p.m?
- Published on 23 Sep 15

a. 0.0785 kW
b. 0.0392 kW
c. 0.0526 W
d. 0.0236 W

ANSWER: 0.0785 kW
 

    Discussion

  • Sravanthi   -Posted on 01 Oct 15
    Given: Frictional torque = 500 N-mm, Speed = 1500 r.p.m

    Formula: Frictional power loss = (2 π nT) / (60 x 1000)

    Convert frictional torque into N-m
    Therefore, T = 0.5 N-m

    Substituting the given values we get,

    Frictional power loss = ( 2 π nT) / (60 x 1000)
    = ( 2 x π x 1500 x 0.5) / (60 x 1000)
    Frictional power loss = 0.0785 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.)