Numerical - Power rating of electric motor, given r.p.m & torque - Basic Mechanical Engineering

Q.  Determine power rating of an electric motor if it runs at 1440 r.p.m and line shaft transmits torque of 75 Nm.  Assume Reduction ratio = 1.6
- Published on 01 Sep 15

a. 10.36 kW
b. 11.3 kW
c. 7.068 kW
d. 9.12 kW

ANSWER: 7.068 kW

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    Discussion

  • Rohit Bagh   -Posted on 05 Jun 21
    Mechanical engineering
  • Uttamkumar Jagtap   -Posted on 19 Aug 17
    Very nice
    and Very help full sir
    ITI FITTER SEM-IV- FINAL EXAM- JULY-2016
    QUESTION NO. 19
    The tendency of force to rotate an object about an axis is called as torque or moment.
    Given: speed (n) : 1440 r.p.m, torque = 75 Nm, reduction ratio (I) = 1.6 
    Formula: P = 2 p N T / 60
    Solution: 
    i = n / N
    1.6 = 1440 / N
    N = 900 r.p.m
    P = (2 p N T)/ 60
    P = (2 x 3.14 x 900 x 75 ) /60
    P = 423900 /60
    = 7065 W 
    = 7.065 kW
    Power rating of an electric motor is 7.065 kW
  • jaypalsinh jadeja (asst. professor-amec dhari)   -Posted on 30 May 17
    nice explanation.....
  • Sravanthi   -Posted on 06 Nov 15
    The tendency of force to rotate an object about an axis is called as torque or moment.

    Given: speed (n) : 1440 r.p.m, torque = 75 Nm, reduction ratio (I) = 1.6

    Formula: P = 2 π N T / 60

    Solution:

    i = n / N

    1.6 = 1440 / N

    N = 900 r.p.m

    P = (2 π x 1440 x 75) / 60

    = 7068 W

    = 7.068 kW

    Power rating of an electric motor is 7.068 kW

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