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27 February, 23:31

An alloy fin with a thermal conductivity of 200 W/ሺm ∙ Kሻ has a length of 2.5 cm and a thickness of 3.5 mm. The base of the fin is held at 400o C and the ambient is at 20o C. Assuming that heat loss from the tip is negligible and that the convective heat transfer coefficient is 10 W/ሺmଶ ∙ Kሻ and the fin to be 1m wide, Find the heat transfer flux.

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  1. 28 February, 00:43
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    Answer: heat flux into the fun is 21.714 mW/m^2

    Explanation:

    Heat flux Q = q/A

    q = heat transfer rate W

    A = area m^2

    q = area * conductivity * temperature gradient

    Temperature gradient = difference in temperature of the metal faces divided by the thickness.

    Therefore Q = k * (temp. gradient)

    Q = 200 * ((400-20) / 3.5*10^-2)

    Q = 21714285.71 = 21.714 mW/m^2

    Answer 2: convective heat transfer flux between fin and air

    is 3800W/m^2

    Explanation:

    q = hA * (Ts-Ta)

    h = convective heat transfer coefficient

    Ts = temperature of fin

    Ta = temperature of air

    Q = q/A

    Q = h (Ts-Ta)

    Q = 10 (400 - 20)

    Q = 3800 W/m^2
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