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5 April, 21:34

For two 0.2 m long rotating concentric cylinders, the velocity distribution is given by u (r) = 0.4/r - 1000r m/s. If the diameters are 2 cm and 4 cm, respectively, calculate the fluid viscosity if the torque on the inner cylinder is measured to be 0.0026 N*m.

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  1. 6 April, 00:13
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    5.9*10^-3 Pa*s

    Explanation:

    The fluid will create a tangential force on the surface of the cylinder depending on the first derivative of the speed respect of the radius.

    τ = μ * du/dr

    u (r) = 0.4/r - 1000*r

    The derivative is:

    du/dr = - 1/r^2 - 1000

    On the radius of the inner cylinder this would be

    u' (0.02) = - 1/0.02^2 - 1000 = - 3500

    So:

    τ = - 3500 * μ

    We don't care about the sign

    τ = 3500 * μ

    That is a tangential force per unit of area.

    The area of the inner cylinder is:

    A = h * π * D

    And the torque is

    T = F * r

    T = τ * A * D/2

    T = τ * h * π/2 * D^2

    T = 3500 * μ * h * π/2 * D^2

    Then:

    μ = T / (3500 * h * π/2 * D^2)

    μ = 0.0026 / (3500 * 0.2 * π/2 * 0.02^2) = 5.9*10^-3 Pa*s
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