Question

A concrete wall with a surface area of \(20 \mathrm{~m}^{2}\) and a thickness of \(0.30 \mathrm{~m}\) separates conditioned room air from ambient air. The temperature of the inner surface of the wall \(\left(T_{1}\right)\) is maintained at \(25^{\circ} \mathrm{C}\). (a) Determine the heat loss \(\dot{Q}(\mathrm{~W})\) through the concrete wall for three thermal conductivity values of \(0.75,1\), and $1.25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$ and outer wall surface temperatures of \(T_{2}=-15,-10,-5,0,5,10,15,20,25,30\), and \(38^{\circ} \mathrm{C}\) (a total of 11 data points for each thermal conductivity value). Tabulate the results for all three cases in one table. Also provide a computer-generated graph [Heat loss, \(\dot{Q}(\mathrm{~W})\) vs. Outside wall temperature, $\left.T_{2}\left({ }^{\circ} \mathrm{C}\right)\right]$ for the display of your results. The results for all three cases should be plotted on the same graph. (b) Discuss your results for the three cases.

Short answer

Based on the given information, explain how the heat loss through a concrete wall is affected by its thermal conductivity and the temperature difference across the wall.

Step-by-step solution

(Step 1: Prepare the given values)

(First, let's write down the given values: - Surface area of the wall, \(A = 20 \mathrm{~m}^2\) - Thickness of the wall, \(D = 0.30 \mathrm{~m}\) - Inner surface temperature, \(T_1 = 25^{\circ} \mathrm{C}\) (convert to Kelvin by adding 273.15, so \(T_1 = 298.15 \mathrm{K}\)) - Thermal conductivities: \(k_1 = 0.75 \mathrm{W/m \cdot K}\), \(k_2= 1.0 \mathrm{W/m \cdot K}\), and \(k_3= 1.25 \mathrm{W/m \cdot K}\) - Outer surface temperatures: \(T_{2i} = \{-15, -10, ..., 38\}^{\circ} \mathrm{C}\) (Convert each to Kelvin by adding 273.15) Now, we will calculate the heat loss for each given \(k\) and \(T_2\) using the heat transfer formula. )

(Step 2: Calculate the heat loss for each case)

(For each thermal conductivity value and outer surface temperature, calculate the heat loss (\(\dot{Q}\)) using the heat transfer formula: \(\dot{Q} = k*A*\dfrac{T_{1}-T_{2}}{D}\) For each value of \(k_i\), plug it into the formula along with the given values of \(A\), \(D\), and \(T_1\) and iterate through all outer surface temperatures \(T_{2i}\). Tabulate the results for all cases: | \(k\) (W/m·K) | \(T_2\) (°C) | \(\dot{Q}\) (W) | |-------|------|-------| | 0.75 | -15 | ... | | ... | ... | ... | | 1.25 | 38 | ... | Create a graph showing the heat loss, \(\dot{Q}(\mathrm{~W})\), vs. outside wall temperature, \(T_{2}\left({}^{\circ} \mathrm{C}\right)\), for each case. Plot all three cases on the same graph. )

(Step 3: Discuss the results)

(Examine the results of each case and discuss the relationship between heat loss, thermal conductivity, and outer surface temperature. Expect to see an increase in heat loss with higher thermal conductivity values as the material allows more heat to pass through it. Additionally, expect heat loss to increase as the outer surface temperature decreases, since the temperature difference between the inner and outer surfaces of the wall will become larger, allowing more heat to transfer from inside the room to the outside environment.