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Taking the acceleration due to gravity to be 10 m/s2, the separation factor of a cyclone 0.5 m in diameter and having a tangential velocity of 20 m/s near the wall is_____
 
Important : you should answer only the numeric value
    Correct answer is '160'. Can you explain this answer?
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    Taking the acceleration due to gravity to be 10 m/s2, the separation f...
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    Taking the acceleration due to gravity to be 10 m/s2, the separation f...
    Given:
    Diameter of cyclone = 0.5 m
    Acceleration due to gravity = 10 m/s^2
    Tangential velocity near the wall = 20 m/s

    To find:
    Separation factor

    Solution:

    The separation factor (K) for a cyclone is given by the expression:

    K = (r2 - r1) / (r1 x ω)

    where r2 = outer radius of the cyclone,
    r1 = inner radius of the cyclone,
    ω = angular velocity of the cyclone

    We know that the diameter of the cyclone is 0.5 m.
    So, the radius of the cyclone (r2) = 0.25 m (since diameter = 2 x radius)

    Let us assume that the inner radius of the cyclone (r1) is 0.1 times the outer radius (r2).

    So, r1 = 0.1 x 0.25 m = 0.025 m

    The tangential velocity near the wall is given as 20 m/s.
    We know that tangential velocity (Vt) = r x ω

    where r = radius of the cyclone,
    ω = angular velocity of the cyclone

    Substituting the values, we get:

    20 = 0.25 x ω

    ω = 80 rad/s

    Now, substituting the values of r2, r1 and ω in the expression for K, we get:

    K = (0.25 - 0.025) / (0.025 x 80)

    K = 160

    Therefore, the separation factor of the cyclone is 160.
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    Taking the acceleration due to gravity to be 10 m/s2, the separation factor of a cyclone 0.5 m in diameter and having a tangential velocity of 20 m/s near the wall is_____Important : you should answer only the numeric valueCorrect answer is '160'. Can you explain this answer?
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