Question

A 40 -cm-thick brick wall \((k=0.72 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=1.6 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ) is heated to an average temperature of \(18^{\circ} \mathrm{C}\) by the heating system and the solar radiation incident on it during the day. During the night, the outer surface of the wall is exposed to cold air at \(-3^{\circ} \mathrm{C}\) with an average heat transfer coefficient of $20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\(. Determine the wall temperatures at distances 15,30 , and \)40 \mathrm{~cm}\( from the outer surface for a period of \)2 \mathrm{~h}$.

Short answer

Answer: The wall temperatures after 2 hours are 13.25°C at 15 cm, 9.50°C at 30 cm, and 6.00°C at 40 cm from the outer surface.

Step-by-step solution

Variables and Constants

List the given variables and constants: - Thermal conductivity (k): 0.72 W/mK - Thermal diffusivity (α): 1.6 x 10^(-6) m^2/s - Wall thickness: 40 cm (0.4 m) - Initial temperature: 18°C - Night air temperature: -3°C - Heat transfer coefficient (h): 20 W/m²·K - Time: 2 hours (7200 s)

Fourier's Law of Heat Conduction and Heat Transfer Equation

Apply Fourier's Law of heat conduction, which states: q = -k × (dT/dx) and the heat transfer equation: q = h × (T_wall - T_ambient) Where "q" is the heat transfer rate, "dT/dx" is the temperature gradient in the wall, "T_wall" is the wall's temperature, and "T_ambient" is the ambient temperature.

Calculate Temperature Gradient

Since we are given the heat transfer coefficient (h) and the temperature of the cold air, we can rewrite the heat transfer equation as follows: dT/dx = (h × (T_wall - T_ambient)) / k Now, replace the known values into the equation and solve for the temperature gradient, dT/dx. dT/dx ≈ (20 × (18 - (-3))) / 0.72 ≈ 29.1667 K/m

Transient Conduction

Consider that the wall's temperature will change over time. We can use the following formula to account for the transient conduction: T(x,t) = T_initial + [(2αt/π)^(0.5)] × (dT/dx) × [exp(-x^2/(4αt))] × [1 - erf(x/(2 × (α × t)^(0.5)))] Plug in the known values and solve for the temperature at the given distances and time (T(0.15 m, 7200 s), T(0.3 m, 7200 s), and T(0.4 m, 7200 s)): T(0.15 m, 7200 s) ≈ 18 - 29.1667 × 0.15 × [1 - erf(0.15 / (2 × (α × t)^(0.5)))] ≈ 13.25 °C T(0.3 m, 7200 s) ≈ 18 - 29.1667 × 0.3 × [1 - erf(0.3 / (2 × (α × t)^(0.5)))] ≈ 9.5 °C T(0.4 m, 7200 s) ≈ 18 - 29.1667 × 0.4 × [1 - erf(0.4 / (2 × (α × t)^(0.5)))] ≈ 6.0 °C

Final Results

The wall temperatures at the given distances after 2 hours are: - At 15 cm from the outer surface: 13.25°C - At 30 cm from the outer surface: 9.50°C - At 40 cm from the outer surface: 6.00°C