Question

A concrete wall, which has a surface area of 20 m2 and is 0.30 m thick, separates conditioned room air from ambient air. The temperature of the inner surface of the wall is maintained at 25°C, and the thermal conductivity of the concrete is 1W/(m*K).

a. Determine the heat loss through the wall for outer surface temperatures ranging from -15°C to 38°C, which correspond to winter and summer extremes, respectively. Display your results graphically.
b. On your graph, also plot the heat loss as a function of the outer surface temperature for wall materials having thermal conductivities of 0.75 and 1.25 W/(m*K). Explain the family of curves you have obtained.

Answer 1

To determine the heat loss through a wall, you can use the formula Q = k * A * (T2 - T1) / d, where Q is the heat loss, k is the thermal conductivity of the wall material, A is the surface area of the wall, T2 is the temperature of the outer surface, T1 is the temperature of the inner surface, and d is the thickness of the wall. By substituting different values of T2 into the formula and plotting them on a graph, you can visualize the heat loss as a function of the outer surface temperature. Repeat this process for different thermal conductivities to obtain a family of curves.

Step-by-step explanation:

The heat loss through a wall can be calculated using the formula:

Q = k * A * (T2 - T1) / d

• Q is the heat loss
• k is the thermal conductivity of the wall material
• A is the surface area of the wall
• T2 is the temperature of the outer surface
• T1 is the temperature of the inner surface
• d is the thickness of the wall

To determine the heat loss for outer surface temperatures ranging from -15°C to 38°C, you can substitute different values of T2 into the formula. Plotting these values on a graph will give you a graphical representation of the heat loss as a function of the outer surface temperature. Repeat this process for wall materials with thermal conductivities of 0.75 W/(m*K) and 1.25 W/(m*K) to obtain a family of curves.