Question

The thermal conductivity of a material is determined by fabricating the material into the shape of a hollow sphere, placing an electrical heater at the center, and measuring the surface temperature with thermocouples when steady state has been reached

Experimental data: r₁=1.12in,r₂=3.06in; for an electrical energy input at the rate of 11.1 W to the heater, T₁​=203∘F and T₂​=184∘F.

Determine
(a) the experimental value of the thermal conductivity,

Answer 1

The experimental value of the thermal conductivity is -0.0467 W/m·K.

The solution to determine the experimental value of the thermal conductivity:

Given:

Inner radius, r₁ = 1.12 inches (convert to meters: r₁ = 1.12 inches * 0.0254 m/inch = 0.02848 meters)

Outer radius, r₂ = 3.06 inches (convert to meters: r₂ = 3.06 inches * 0.0254 m/inch = 0.077664 meters)

Heat input, Q = 11.1 W

Inner temperature, T₁ = 203°F (convert to Kelvin: T₁ = 203°F + 459.67 = 675.15 K)

Outer temperature, T₂ = 184°F (convert to Kelvin: T₂ = 184°F + 459.67 = 642.15 K)

Formula:

The heat transfer through a hollow sphere can be calculated using the following formula:

Q = 4πk(T₂ - T₁)/(ln(r₂/r₁))

where:

Q is the heat transfer rate (W)

k is the thermal conductivity (W/m·K)

T₂ is the outer temperature (K)

T₁ is the inner temperature (K)

r₂ is the outer radius (m)

r₁ is the inner radius (m)

Solution:

Rearranging the formula to solve for k, we get:

k = (Q * ln(r₂/r₁))/(4π(T₂ - T₁))

Substituting the given values:

k = (11.1 W * ln(0.077664 m / 0.02848 m)) / (4π(642.15 K - 675.15 K))

k ≈ -0.0467 W/m·K.