Design of Machine Elements - 2 - Mechanical Engineering (MCQ) questions and answers for Q. 11699

Q.  A flat belt transmits power of 20 kW when the velocity of the belt is 30 m/s. What will be the initial tension in the belt if   230 N is the centrifugal tension in the belt and F1 / F2 =2.5?
- Published on 23 Sep 15

a. 1007.77 N
b. 1000.2 N
c. 908.23 N
d. 896.15 N

ANSWER: 1007.77 N

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    Discussion

  • Sravanthi   -Posted on 05 Oct 15
    • A belt when mounted on a pulley has some tension acting on it in its stationary state known as initial tension. Belt drive can transmit power if this initial tension is present.
    • If the belt is perfectly elastic during power transmission, tension in tight side increases as tension in slack side decreases and hence length of the belt remains unchanged.

    Given: Power = 20 kW, centrifugal tension (Fc) = 230 N, F1 / F2 = 2.5, V = 30 Mpa

    Decrease in tension on slack side = Increase in tension on tight side

    Formula: 1) Fi = (F1 + F2 + 2Fc) / 2
    2) P = V (F1 – F2) / 1000

    P = V (F1 – F2) / 1000
    (F1 – F2) = 666.66 N -------(1)

    Substitute the value of F1 / F2 = 2.5 in equation (1),

    F2 = 444.44 N & F1 = 1111.11 N

    Substituting the given values of F1, F2 and Fc we get,

    Initial tension in the belt = 1007.77 N

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