Question

Consider a \(1.5\)-m-high and 2.4-m-wide doublepane window consisting of two \(3-\mathrm{mm}\)-thick layers of glass $(k=0.78 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( separated by a \)12-\mathrm{mm}\(-wide stagnant airspace \)(k=0.026 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$. Determine the steady rate of heat transfer through this double-pane window and the temperature of its inner surface for a day during which the room is maintained at $21^{\circ} \mathrm{C}\( while the temperature of the outdoors is \)-5^{\circ} \mathrm{C}$. Take the convection heat transfer coefficients on the inner and outer surfaces of the window to be \(h_{1}=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(h_{2}=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and disregard any heat transfer by radiation.

Short answer

Answer: The steady rate of heat transfer through the double-pane window is 374.64 W, and the temperature of the inner surface is 10.59 °C.

Step-by-step solution

Calculate the area of the window

First, we need to calculate the area of the window. It is given that the window has a height of 1.5 m and a width of 2.4 m. Area = Height × Width = 1.5 m × 2.4 m = 3.6 m^2

Calculate the thermal resistances

Next, we can find the thermal resistances for each layer of the window. For the glass panes, it is given that they have a thickness of 3 mm and a thermal conductivity of 0.78 W/m·K. For the air space, the thickness is 12 mm, and the thermal conductivity is 0.026 W/m·K. We also have the convection heat transfer coefficients for the inner and outer surfaces, h1 = 10 W/m^2·K, and h2 = 25 W/m^2·K. Using these parameters, we can find the thermal resistances as follows: R1 (convection) = 1 / (h1 × Area) = 1 / (10 W/m^2·K × 3.6 m^2) = 0.0278 m^2·K/W R2 (glass) = Thickness / (k × Area) = 0.003 m / (0.78 W/m·K × 3.6 m^2) = 0.00137 m^2·K/W R3 (air space) = 0.012 m / (0.026 W/m·K × 3.6 m^2) = 0.0377 m^2·K/W R4 (glass) = 0.003 m / (0.78 W/m·K × 3.6 m^2) = 0.00137 m^2·K/W R5 (convection) = 1 / (h2 × Area) = 1 / (25 W/m^2·K × 3.6 m^2) = 0.00111 m^2·K/W

Calculate the total thermal resistance

Now we can calculate the total thermal resistance of the system by summing up the individual resistances: R_total = R1 + R2 + R3 + R4 + R5 = 0.0278 + 0.00137 + 0.0377 + 0.00137 + 0.00111 = 0.0694 m^2·K/W

Calculate the steady rate of heat transfer

We can calculate the steady rate of heat transfer through the double-pane window using the overall resistance and the temperature difference between the indoor and outdoor environments: Q = ΔT / R_total = (21 - (-5)) °C / 0.0694 m^2·K/W = 26 K / 0.0694 m^2·K/W = 374.64 W

Calculate the temperature of the inner surface

Now we can find the temperature of the inner surface using the heat transfer rate and the convection resistance at the inner surface: ΔT1 = Q × R1 = 374.64 W × 0.0278 m^2·K/W = 10.41 K The temperature of the inner surface is then: T_inner = T_room - ΔT1 = 21 °C - 10.41 K = 10.59 °C

Final Results

The steady rate of heat transfer through the double-pane window is 374.64 W, and the temperature of the inner surface is 10.59 °C.