A cylindrical rod of stainless steel is insulated on its exterior surface except for the ends. The stead-state temperature distribution is T

Question

A cylindrical rod of stainless steel is insulated on its exterior surface except for the ends. The stead-state temperature distribution is T(x) = a – bx/L, where a = 305 K and b = 10 K. The diameter and length of the rod are D = 20mm and L = 100mm, respectively. Determine the heat flux along the rod, q”(sub x). Hint: The mass of the rod is M = 0.248 kg.

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Ngọc Hoa 2 months 2021-07-30T03:01:56+00:00 1 Answers 5 views 0

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    2021-07-30T03:02:58+00:00

    Answer:

    1490 W/m²

    Explanation:

    Given

    Length of the rod, L = 100 mm = 0.1 m

    Diameter of the rod, D = 20 mm = 0.02 m

    Steady-state temperature distribution is T(x) = a – bx/L, where a = 305 K and b = 10 K

    Mass of the rod, m = 0.248 kg

    The first step is to find out the density of the material.

    Volume = πD²L/4

    Volume = 3.142 * 0.02² * 0.1 / 4

    Volume = 0.00012568 / 4

    Volume = 0.00003142 m³

    Remember, density = mass / volume, so that

    Density = 0.248 / 0.00003142

    Density = 7893 kg/m³

    Knowing the density, we can find the thermal conductivity of the material from tables

    k for AISI 304 at 300 K is 14.9 W/mK

    Now, we use Fourier’s Law to find the heat flux

    q” = -k dT/dx,

    Remember T(x) = a – bx/L, on derivation, we have

    q” = -k * -b/L

    q” = -14.9 * -10/0.1

    q” = 149/0.1

    q” = 1490 W/m²

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