Question

The roof of a house consists of a \(15-\mathrm{cm}\)-thick concrete slab \(\left(k=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\) and \(\left.\alpha=0.69 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) that is \(18 \mathrm{~m}\) wide and \(32 \mathrm{~m}\) long. One evening at $6 \mathrm{p} . \mathrm{m}$., the slab is observed to be at a uniform temperature of \(18^{\circ} \mathrm{C}\). The average ambient air and the night sky temperatures for the entire night are predicted to be \(6^{\circ} \mathrm{C}\) and \(260 \mathrm{~K}\), respectively. The convection heat transfer coefficients at the inner and outer surfaces of the roof can be taken to be $h_{i}=5 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\( and \)h_{o}=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, respectively. The house and the interior surfaces of the walls and the floor are maintained at a constant temperature of \(20^{\circ} \mathrm{C}\) during the night, and the emissivity of both surfaces of the concrete roof is \(0.9\). Considering both radiation and convection heat transfers and using the explicit finite difference method with a time step of \(\Delta t=5 \mathrm{~min}\) and a mesh size of $\Delta x=3 \mathrm{~cm}$, determine the temperatures of the inner and outer surfaces of the roof at 6 a.m. Also, determine the average rate of heat transfer through the roof during that night.

Short answer

Answer: To find the temperatures of the inner and outer surfaces of the roof at 6 a.m., you will need to apply the explicit finite difference method using the provided equations and initial temperature conditions. After performing 144 time steps, you will find the temperatures of the inner and outer surfaces as \(T_{inner}= T^{144}_5\) and \(T_{outer}= T^{144}_0\). To calculate the average rate of heat transfer through the roof during the night, use the Fourier law for heat conduction at the inner surface and divide the result by the total time of 12 hours to find \(Q\).

Step-by-step solution

List given information and variables

Concrete slab thickness: \(d=15 cm=0.15 m\) Thermal conductivity of concrete: \(k=1.4 W/m \cdot K\) Thermal diffusivity of concrete: \(\alpha=0.69\times10^{-6} m^2/s\) Width: \(W=18m\) Length: \(L=32m\) Initial temperature of slab: \(T_i=18^{\circ}C\) Temperature inside the house: \(T_{house}=20^{\circ}C\) Air temperature: \(T_{air}=6^{\circ}C\) Night sky temperature: \(T_{sky}=260K\) Convection heat transfer coefficient: \(h_{i}=5 W/m^2 \cdot K\) \(h_{o}=12 W/m^2 \cdot K\) Emissivity: \(\epsilon = 0.9\) Time step: \(\Delta t=5 \, \text{min}= 300 \, s\) Mesh size: \(\Delta x=3 cm=0.03 m\)

Calculate the time steps and number of meshes

To calculate the temperatures at 6 a.m., we need the total amount of time steps: Number of hours: 12 hours Total time steps: \((12*3600)/300=144\) time steps We also need to calculate the number of meshes inside the concrete slab: Number of meshes: \(d/\Delta x = 0.15/0.03 = 5\)

Set up the explicit finite difference method equations

We will need to set up the equations for the explicit finite difference method to calculate the temperature distribution inside the concrete slab. For the internal nodes (\(1\leq i\leq 4\)), the equation is: \(T^{n+1}_{i} = T^{n}_{i} + \frac{\alpha \Delta t}{(\Delta x)^2}\left(T^{n}_{i+1} - 2T^{n}_{i} + T^{n}_{i-1}\right)\) For \(i = 0\) (outer surface node): \(T^{n+1}_{0} = T^n_{0} + \frac{\alpha\Delta t}{(\Delta x)^2}(T^n_{1} - T^n_{0}) + \frac{2 \alpha\Delta t}{\Delta x}\left(\frac{h_o(T_{air} - T^n_{0}) + \epsilon \sigma (T_{sky}^4 - (T^n_{0}+273.15)^4)}{k}\right)\) For \(i = 5\) (inner surface node): \(T^{n+1}_{5} = T^n_{5} + \frac{\alpha\Delta t}{(\Delta x)^2}(T^n_{4} - T^n_{5}) + \frac{2 \alpha\Delta t}{\Delta x}\left(\frac{h_i(T_{house} - T^n_{5})}{k}\right)\)

Calculate the temperature distribution

Using the equations from Step 3, perform calculations for 144 time steps given the initial \(T_i\) for each mesh. Apply the method to update the node temperatures at each time step.

Temperatures of inner and outer surfaces at 6 a.m.

After performing calculations for the 144 time steps, determine the temperatures of inner and outer surfaces of the concrete roof: \(T_{inner}= T^{144}_5\) \(T_{outer}= T^{144}_0\)

Calculate the average rate of heat transfer

To determine the average rate of heat transfer through the roof during the entire night (12 hours), we will use the Fourier law for heat conduction at the inner surface: $$q = k A\frac{\Delta T}{d}$$ where \(A = WL\) is the area of the roof, and \(\Delta T = T_{inner} - T_{outer}\). Now we can find the average rate of heat transfer through the roof: \(Q = \frac{q}{(12 \,\text{hours} \cdot 3600 \,\text{s/hour})}\). The final results are the temperatures of the inner and outer surfaces of the roof at 6 a.m. and the average rate of heat transfer through the roof during the night.