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.

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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