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29 October, 04:37

A Micro - Hydro turbine generator rotor is accelerating uniformly from an angular velocity of 610 rpm to its operating angular velocity of 837 rpm. The radius of the rotor is 0.62 m and its rotational acceleration is 5.9 rad/s2. What is the rotor's angular velocity (in rad/s) after 3.2 s?

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  1. 29 October, 06:44
    0
    Given:

    wo = 610 rpm

    To rad/s,

    = 610 rpm * 2pi rad/60 s

    = 63.88 rad/s

    wf = 837 rpm

    = 87.65 rad/s

    ao = 5.9 rad/s^2

    Using equations of circular motion,

    wf = wo + aot

    87.65 = 63.88 + 5.9*t

    t = 4.025 s

    Using the same equation,

    At t = 3.2s (first 3.2 s of the motion),

    wf = 63.88 + 5.9 * 3.2

    = 82.76 rad/s

    = 82.8 rad/s
  2. 29 October, 07:11
    0
    The rotor's angular velocity is 82.73rad/s

    Explanation:

    It is a curvilinear movement of a constant radius. If there is uniform angular acceleration, then it is a circular motion with constant acceleration, whose equations are analogous to that of the translational motion.

    Calculating the initial velocity of the rotor, V1 in rad/s

    V1 = 610rev/minute * 6.28 * 1miute/60secs

    V1 = 63.85rad/s

    Using kinematic equation to calculate the final velocity of the rotor

    Given:

    Angular acceleration = 5.9rad/s^2

    Time, t = 3.2seconds

    V2 = V1 + a * t

    V2 = 63.85 + (5.9) * (3.2)

    V2 = 63.85 + 18.88

    V2 = 82.73rads/s
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