The thermal conductivity of a sheet of rigid, extruded insulation is reported to be k = 0.029 W / m K. The measured temperature difference across a 25-mm-thick sheet of the material is T1 - T2 = 12°C. a. What is the heat flux through a 3 mx 3 m sheet of the insulation? b. What is the rate of heat transfer through the sheet of insulation? c. What is the thermal resistance of the sheet due to conduction?

a. The heat flux through the sheet of insulation is 19.92 W/m^2

b. The rate of heat transfer through the sheet of insulation is 125.28 W

c. The thermal resistance of the sheet due to the conduction is 0.86 Km^2/W.

Explanation:

From the heat conduction Fourier's law it can be state for a wall of width e and area A:

q = Q/ΔT = k*A * (T2-T1) / e

Where q is the rate of heat transfer, k the conductivity constant, and T2 and T1 the temperatures on the sides of the wall. Replacing the values in the correct units, we obtained the rate of heat transfer:

q = 0.029 W/*mK * (3m*3m) * (12°K) / (0.025m)

(The difference in temperatures in Kelvin is the same than in Celcius degres).

q = 0.029 W/*mK * (9 m^2) * (12°K) / (0.025m) = 125.28 W

The heat flux is calculated by dividing q by the area of the wall:

q/A = k * (T2-T1) / e = 0.029 W/*mK * (12°K) / (0.025m) = 19.92 W/m^2

The thermal resistance of the sheet is defined as:

R = e / k

Replacing the values in the proper units:

R = 0.025 m / 0.029 W/*mK = 0.86 Km^2/W