A 0.091-in-diameter electrical wire at 90°F is covered by 0.02-in-thick plastic insulation (k = 0.075 Btu/h·ft·°F). The wire is exposed to a

Question

A 0.091-in-diameter electrical wire at 90°F is covered by 0.02-in-thick plastic insulation (k = 0.075 Btu/h·ft·°F). The wire is exposed to a medium at 50°F, with a combined convection and radiation heat transfer coefficient of 2.5 Btu/h·ft2·°F. Calculate the critical radius (rcr) of the plastic insulation (in inches).

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Hải Đăng 2 months 2021-07-22T12:24:59+00:00 1 Answers 4 views 0

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    2021-07-22T12:26:57+00:00

    Answer:

    The critical radius of the plastic insulation is 0.72 inches.

    Explanation:

    Given that,

    Diameter = 0.091 in

    Thickness = 0.02 in

    Initial temperature = 90°F

    Final temperature = 50°F

    Heat transfer coefficient = 2.5 Btu/h.ft²°F

    Material conductivity = 0.075 Btu/h.ft °F

    We need to calculate the critical radius of the plastic insulation

    Using formula of critical radius

    r_{cr}=\dfrac{2K}{h}

    Where, k = Material conductivity

    h = Heat transfer coefficient

    Put the value into the formula

    r_{cr}=\dfrac{2\times0.075}{2.5}

    r_{cr}=0.06\ ft

    r_{cr}=0.72\ inches

    Hence, The critical radius of the plastic insulation is 0.72 inches.

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