Question

A hollow aluminum sphere, with an electrical heater in the center, is used in tests to determine the thermal conductivity of insulating materials. The inner and outer radii of the sphere are 0.18 and 0.21 m, respectively, and testing is done under steady-state conditions with the inner surface of the aluminum maintained at 250°C. In a particular test, a spherical shell of insulation is cast on the outer surface of the sphere to a thickness of 0.15 m. The system is in a room for which the air temperature is 20°C and the convection coefficient at the outer surface of the insulation is 30 W/m2 ⋅ K. If 80 W are dissipated by the heater under steady-state conditions, what is the thermal conductivity of the insulation?

Answer 1

To calculate the thermal conductivity of the insulation, we must find the temperature difference across the insulation, T_insulation_outer, and apply Fourier's Law of heat conduction in spherical coordinates considering steady-state conditions and using given parameters.

Step-by-step explanation:

To find the thermal conductivity of the insulation in the described scenario, we can use Fourier's Law of heat conduction, which in steady-state for a spherical shell is given as:

Q = − k ∙ (4πr² ∙ ΔT) / L

Where Q is the heat transfer rate, k is the thermal conductivity, r is the radius at the position where temperature gradient is measured, ΔT is the temperature difference, and L is the thickness of the material.

Given that the rate of heat dissipation (Q) is 80 W, the outer radius of the aluminum sphere (r) is 0.21 m, the thickness of the insulation (L) is 0.15 m, the inner surface temperature (T_inner) is 250°C, and the room temperature (T_outer) is 20°C, we can use the convection at the outer surface of the insulation to find the temperature at the outer surface of the insulation (T_insulation_outer).

For steady-state conditions and assuming the only heat transfer from the outer surface of the insulation is by convection, we have:

Q = h ∙ A ∙ (T_insulation_outer − T_outer)

Solving for T_insulation_outer when Q = 80 W, h = 30 W/m²∙K, and using A for the surface area of the outer sphere of radius (0.21 m + 0.15 m) gives us T_insulation_outer.

Using this T_insulation_outer and T_inner, we can find the temperature difference (ΔT) across the insulation. Subsequently, we can rearrange Fourier's Law to solve for the thermal conductivity (k) of the insulation material:

k = Q ∙ L / (4πr² ∙ ΔT)

Using all known values in the above formula will yield the thermal conductivity of the insulation.