Consider a 1.2-m-high and 2-m-wide double-pane window consisting of two 3 -mm- thick layers of glass \((k=\) \(0.78 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) separated by a \(12-\mathrm{mm}\)-wide stagnant air space \((k=0.026 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). 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 \(24^{\circ} \mathrm{C}\) 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.

Step by step solution

01

Calculate thermal resistances

First, we need to calculate the thermal resistances for convection and conduction across the window system. These thermal resistances are R1 for convection on inner glass surface, R2 for conduction in the inner glass layer, R3 for conduction in the air space, R4 for conduction in the outer glass layer, and R5 for convection on outer glass surface. We can calculate the convection and conduction resistances using the following equations: $$R_{conv} = \frac{1}{hA}$$ $$R_{cond} = \frac{L}{kA}$$ where h is the convection heat transfer coefficient, k is the thermal conductivity, A is the area, and L is the thickness.

02

Determine the total thermal resistance

Next, we will find the total thermal resistance by adding all the thermal resistances in the window system. $$R_{total} = R1 + R2 + R3 + R4 + R5$$

03

Calculate the steady rate of heat transfer

Now, we can calculate the steady rate of heat transfer (Q) by using the following equation: $$Q = \frac{T_{in} - T_{out}}{R_{total}}$$ where \(T_{in}\) is the temperature inside the room and \(T_{out}\) is the temperature outside.

04

Determine the temperature of the inner surface

Finally, we can find the temperature of the inner surface (T1) using the following equation: $$T1 = T_{in} - Q \cdot R1$$ Now, let's calculate the values.

05

(Calculations)

The area of the double-pane window is 1.2 m * 2 m = 2.4 m². All the resistances are in series. Thus, they will all have the same area, A = 2.4 m². $$R1 = \frac{1}{h_{1}A} = \frac{1}{(10 \frac{\text{W}}{\text{m}^{2} \text{K}})(2.4 \text{m}^{2})} = 0.04167\text{ K/W}$$ $$R2 = \frac{L}{kA} = \frac{0.003 \text{m}}{(0.78 \frac{\text{W}}{\text{m K}})(2.4 \text{m}^{2})} = 0.00161\text{ K/W}$$ $$R3 = \frac{L}{kA} = \frac{0.012 \text{m}}{(0.026 \frac{\text{W}}{\text{m K}})(2.4 \text{m}^{2})} = 0.01923\text{ K/W}$$ $$R4 = \frac{L}{kA} = \frac{0.003 \text{m}}{(0.78 \frac{\text{W}}{\text{m K}})(2.4 \text{m}^{2})} = 0.00161\text{ K/W}$$ $$R5 = \frac{1}{h_{2}A} = \frac{1}{(25 \frac{\text{W}}{\text{m}^{2} \text{K}})(2.4 \text{m}^{2})} = 0.01667\text{ K/W}$$

06

(Calculations)

Now, let's determine the total thermal resistance. $$R_{total} = R1 + R2 + R3 + R4 + R5 = 0.04167 + 0.00161 + 0.01923 + 0.00161 + 0.01667 = 0.08079\text{ K/W}$$

07

(Calculations)

We can now calculate the steady rate of heat transfer (Q). $$Q = \frac{T_{in} - T_{out}}{R_{total}} = \frac{24°C - (-5°C)}{0.08079\text{ K/W}} = 358.73\text{ W}$$

08

(Calculations)

Finally, let's find the temperature of the inner surface (T1). $$T1 = T_{in} - Q \cdot R1 = 24°C - (358.73\text{ W})(0.04167\text{ K/W}) = 9.05°C$$ ≈ \(9.0^{\circ}\mathrm{C}\). The steady rate of heat transfer through the double-pane window is approximately 358.73 W, and the temperature of its inner surface is approximately \(9.0^{\circ}\mathrm{C}\).

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