Question

Steady heat conduction occurs through a 0.3 meter thick 9 m x 3 m composite wall at a rate of 1.2 kW. If the inner and outer surface temperatures of the wall are 15 degree C and 7 degree C, the effective thermal conductivity of the wall is:

A. 1.7 W/(m*K)
B. 2.2 W/(m*K)
C. 0.83 W/ (m*K)
D. 0.61 W/(m*K)
E. 5.1 W/(m*K)"

Answer 1

The effective thermal conductivity of the 0.3 meter thick composite wall, given the provided temperatures and heat transfer rate, is closest to option A, which is 1.7 W/(m*K).

Step-by-step explanation:

To calculate the effective thermal conductivity (k) of the composite wall, we can apply Fourier's Law of heat conduction, which is given by the formula Q = kA(T1 - T2) / d, where Q is the heat transfer rate, A is the area through which heat is transferred, T1 and T2 are the temperatures on the two sides of the wall, and d is the thickness of the wall.

In this case, the heat transfer rate (Q) is 1.2 kW or 1200 W, the surface area (A) of the wall is 9 m x 3 m = 27 m², the temperature difference (T1 - T2) is 15°C - 7°C = 8°C, and the thickness (d) is 0.3 m. Plugging these values into the formula, we get:

1200 W = k * 27 m² * 8°C / 0.3 m

To solve for k, we rearrange the formula:

k = (1200 W * 0.3 m) / (27 m² * 8°C) = 1.67 W/(m*K)

The effective thermal conductivity of the wall is therefore closest to option A, which is 1.7 W/(m*K).