Question

Consider a house that is maintained at a constant temperature of $22^{\circ} \mathrm{C}$. One of the walls of the house has three single-pane glass windows that are \(1.5 \mathrm{~m}\) high and \(1.8 \mathrm{~m}\) long. The glass $(k=0.78 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( is \)0.5 \mathrm{~cm}$ thick, and the heat transfer coefficient on the inner surface of the glass is $8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\(. Now winds at \)35 \mathrm{~km} / \mathrm{h}$ start to blow parallel to the surface of this wall. If the air temperature outside is \(-2^{\circ} \mathrm{C}\), determine the rate of heat loss through the windows of this wall. Assume radiation heat transfer to be negligible. Evaluate the air properties at a film temperature of $5^{\circ} \mathrm{C}\( and \)1 \mathrm{~atm}$.

Short answer

Answer: The rate of heat loss through the windows is 22,862.4 W.

Step-by-step solution

Calculate the area of one window

First, we need to calculate the area of one window. The dimensions of the window are given as (\(1.5 \mathrm{~m}\)) in height and (\(1.8 \mathrm{~m}\)) in length. The area of one window is calculated by multiplying the height by the length: \(A_w = 1.5 \times 1.8 = 2.7 \mathrm{~m}^2\)

Calculate the thermal resistance of the glass

Now that we have the area of one window, we can calculate the thermal resistance of the glass, given by the inverse of the product of the thermal conductivity \(k\) and the area divided by the thickness \(L\). Since there are three windows, we need to multiply the area by 3. The thermal conductivity of the glass is provided, k = 0.78 \(\mathrm{W/m K}\), and the thickness of the glass is given as \(L = 0.5 \mathrm{~cm}\) or \(0.005 \mathrm{~m}\): \(R_g = \frac{L}{kA_w} = \frac{0.005}{0.78 \times 2.7 \times 3} = 8.46 \times 10^{-4} \mathrm{~K/W}\)

Calculate the convective heat transfer coefficient

We know the wind speed is 35 km/h, which is 9.72 m/s. Next, we need to calculate the convective heat transfer coefficient (\(h_c\)) for this wind speed. Using the air properties evaluated at a film temperature of 5°C and 1 atm, the values for density (\(\rho\)), specific heat (\(c_p\)), thermal conductivity (\(k_a\)), and dynamic viscosity (\(\mu\)) can be found in a table or online sources, which are: $$ \rho = 1.217 \, \mathrm{kg/m^3}\\ c_p = 1006 \, \mathrm{J/kgK}\\ k_a = 0.0257 \, \mathrm{W/mK}\\ \mu = 1.81 \times 10^{-5} \, \mathrm{kg/ms}\\ $$ Now, we can calculate the Reynolds number Re and the Nusselt number Nu, which will help us find the convective heat transfer coefficient (\(h_c\)): $$ \mathrm{Re} = \frac{\rho v L}{\mu} = \frac{1.217 \times 9.72 \times 1.8}{1.81 \times 10^{-5}} = 1.14 \times 10^5\\ \mathrm{Nu} = 0.037 (\mathrm{Re})^{0.8} (\frac{c_p \mu}{k_a})^{1/3} = 0.037 (1.14 \times 10^5)^{0.8} (\frac{1006 \times 1.81 \times 10^{-5}}{0.0257})^{1/3} = 354.18\\ h_c = \frac{k_a \mathrm{Nu}}{L} = \frac{0.0257 \times 354.18}{1.8} = 5.09 \mathrm{W/m^2K}\\ $$

Calculate the convective resistance

Now that we have the convective heat transfer coefficient, we can calculate the convective resistance of the air (\(R_c\)), using the area of the windows multiplied by 3: \(R_c = \frac{1}{h_c A_w} = \frac{1}{5.09 \times 2.7 \times 3} = 2.07 \times 10^{-4} \mathrm{~K/W}\)

Calculate the total heat transfer

We can now add the thermal resistance of the glass and the convective resistance of the air to find the total thermal resistance (\(R_t\)), and the overall heat transfer coefficient (\(U\)): $$ R_t = R_g + R_c = 8.46 \times 10^{-4} + 2.07 \times 10^{-4} = 1.05 \times 10^{-3} \mathrm{K/W}\\ U = \frac{1}{R_t} = \frac{1}{1.05 \times 10^{-3}} = 952.4 \mathrm{W/K}\\ $$

Calculate the rate of heat loss

Finally, we can calculate the rate of heat loss through the windows by finding the temperature difference between the inside and outside, and multiplying by the overall heat transfer coefficient: \(\dot{Q} = U \times \Delta T = 952.4 \times (22 - (-2)) = 22862.4 \mathrm{~W}\) The rate of heat loss through the windows is 22,862.4 W.