Question

Consider a 1.2-m-high and 2 -m-wide glass window with a thickness of $6 \mathrm{~mm}\(, thermal conductivity \)k=0.78 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(, and emissivity \)\varepsilon=0.9$. The room and the walls that face the window are maintained at \(25^{\circ} \mathrm{C}\), and the average temperature of the inner surface of the window is measured to be $5^{\circ} \mathrm{C}\(. If the temperature of the outdoors is \)-5^{\circ} \mathrm{C}$, determine \((a)\) the convection heat transfer coefficient on the inner surface of the window, \((b)\) the rate of total heat transfer through the window, and (c) the combined natural convection and radiation heat transfer coefficient on the outer surface of the window. Is it reasonable to neglect the thermal resistance of the glass in this case?

Short answer

Additionally, can the thermal resistance of the glass be reasonably neglected in this case? Answer: The convection heat transfer coefficient on the inner surface of the window is 6.5 W/(m²·K), the rate of total heat transfer through the window is 153.2 W, and the combined natural convection and radiation heat transfer coefficient on the outer surface of the window is 130 W/(m²·K). Since the thermal resistance of the glass accounts for only 16.8% of the total resistance, it is reasonable to neglect it in this case.

Step-by-step solution

Calculate the temperature difference across the window

To calculate the overall heat transfer, we first need to find the temperature difference across the window. The inner surface is at 5°C, and the outdoor temperature is -5°C. The temperature difference is given by: \(\Delta T = T_\text{indoors} - T_\text{outdoors} = 5 - (-5) = 10 \thinspace{°C}\)

Calculate the resistance of the glass due to conduction

Next, we'll find the thermal resistance due to conduction through the glass. The resistance is given by the following formula: \(R_\text{cond} = \frac{L}{kA}\) Where: \(L\) is the thickness of the glass (0.006 m) \(k\) is the thermal conductivity (0.78 W/(m·K)) \(A\) is the area of the window (1.2 * 2 = 2.4 m²) \(R_\text{cond} = \frac{0.006}{0.78 \times 2.4} = 0.0032 \thinspace \mathrm{m^2 \cdot K / W}\)

Calculate the convection heat transfer coefficient on the inner surface of the window

The heat transfer through the glass window is given by the formula: \(q = \frac{\Delta T}{R_\text{cond} + R_\text{conv, inner}}\) We need to determine the convection resistance on the inner surface, which is given by: \(R_\text{conv, inner} = \frac{1}{h_\text{inner} A}\) where \(h_\text{inner}\) is the convection heat transfer coefficient. We know from the problem statement that the inner surface of the window is at 5°C, and the room temperature is 25°C, which means that the temperature difference between the two is 20°C. Now, we can set up a balance between the heat transfers due to conduction in the glass and convection at the inner surface: \(q = kA\frac{\Delta T_\text{cond}}{L} = h_\text{inner} A\Delta T_\text{conv, inner}\) Solving for \(h_\text{inner}\): \(h_\text{inner} = \frac{k\Delta T_\text{cond}}{L\Delta T_\text{conv, inner}} = \frac{0.78 \times 10}{0.006 \times 20} = 6.5 \thinspace \mathrm{W/(m^2 \cdot K)}\)

Calculate the rate of total heat transfer through the window

We now have all the necessary resistances to find the total heat transfer rate through the window: \(q = \frac{\Delta T}{R_\text{cond} + R_\text{conv, inner}} = \frac{10}{0.0032 + (1 / (6.5 \times 2.4))} = 153.2 \thinspace \mathrm{W}\)

Find the combined natural convection and radiation heat transfer coefficient on the outer surface of the window

The heat transfer on the outer surface of the window has contributions from both natural convection and radiation. Therefore, we can write the combined resistance of these two processes as: \(R_\text{conv, outer} = \frac{1}{h_\text{outer,total} A}\) where \(h_\text{outer,total} = h_\text{conv, outer} + h_\text{rad, outer}\) We know the heat transfer through the window, so we can find the total outer heat transfer coefficient using a similar balance between conduction and the combined convection and radiation on the outer surface: \(q = kA\frac{\Delta T_\text{cond}}{L} = h_\text{outer,total} A\Delta T_\text{conv, outer}\) Solving for \(h_\text{outer,total}\): \(h_\text{outer,total} = \frac{k\Delta T_\text{cond}}{L\Delta T_\text{conv, outer}} = \frac{0.78 \times 10}{0.006 \times 10} = 130 \thinspace \mathrm{W/(m^2 \cdot K)}\)

Check if the thermal resistance of the glass can be neglected

To determine whether it is reasonable to neglect the thermal resistance of the glass, we can compare it to the other resistances: \(\frac{R_\text{cond}}{R_\text{conv, inner} + R_\text{cond}} = \frac{0.0032}{0.0032 + (1 / (6.5 \times 2.4))} = 0.168\) Since the ratio is just 16.8%, we can conclude that it is reasonable to neglect the thermal resistance of the glass in this case. In conclusion, the convection heat transfer coefficient on the inner surface of the window is 6.5 W/(m²·K), the rate of total heat transfer through the window is 153.2 W, and the combined natural convection and radiation heat transfer coefficient on the outer surface of the window is 130 W/(m²·K). Since the thermal resistance of the glass accounts for only 16.8% of the total resistance, it is reasonable to neglect it in this case.