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Chapter 9: Problem 124

Determine the overall \(U\)-factor for a double-doortype wood-framed double-pane window with \(13-\mathrm{mm}\) air space and metal spacers, and compare your result with the value listed in Table 9-6. The overall dimensions of the window are \(2.00 \mathrm{~m} \times 2.40 \mathrm{~m}\), and the dimensions of each glazing are \(1.92 \mathrm{~m} \times 1.14 \mathrm{~m}\).

Short Answer

Expert verified
Question: Calculate the overall U-factor for a double-pane wooden window and compare it with the value given in Table 9-6. Answer: The overall U-factor for the double-pane wooden window is 6.78 W/m²·K. Compare this value to the one listed in Table 9-6 to check if it is within the expected range for this type of window.

Step by step solution

01

Calculate the glazing U-factor

We will use the following formula to calculate the glazing U-factor: \(U_{glazing} = \frac{1}{R_{out} + R_{gap} + R_{in}}\), where \(R_{out}\) and \(R_{in}\) are the resistance of the outer and inner surfaces of the glazing, and \(R_{gap}\) is the resistance of the air gap between the panes. We can find these values in Table 9-6 or similar reference material. For example, we can use \(R_{out} = 0.04 \mathrm{~m^2·K/W}\), \(R_{gap} = 0.18 \mathrm{~m^2·K/W}\), and \(R_{in} = 0.04 \mathrm{~m^2·K/W}\). Now, calculate the glazing U-factor: \(U_{glazing} = \frac{1}{0.04 + 0.18 + 0.04} = \frac{1}{0.26} = 3.85 \mathrm{~W/m^2·K}\)
02

Calculate the spacer U-factor

The spacer U-factor is given by: \(U_{spacer} = \frac{k_{spacer}}{t_{spacer}}\), where \(k_{spacer}\) is the thermal conductivity of the metal spacer and \(t_{spacer}\) is the thickness of the spacer. We can find these values in reference material or assume typical values, such as \(k_{spacer} = 20 \mathrm{~W/m·K}\) and \(t_{spacer} = 0.013 \mathrm{~m}\) (13 mm). Now, calculate the spacer U-factor: \(U_{spacer} = \frac{20}{0.013} = 1538.46 \mathrm{~W/m^2·K}\)
03

Calculate the overall window area and glazing area

The overall window area is given by: \(A_{window} = 2.00 \mathrm{~m} \times 2.40 \mathrm{~m} = 4.8 \mathrm{~m^2}\). The glazing area is given by: \(A_{glazing} = 1.92 \mathrm{~m} \times 1.14 \mathrm{~m} = 2.1888 \mathrm{~m^2}\).
04

Calculate the overall U-factor

Finally, we will use the following formula to calculate the overall U-factor: \(U_{overall} = \frac{A_{glazing}}{A_{window}} \cdot U_{glazing} + \frac{A_{spacer}}{A_{window}} \cdot U_{spacer}\), where \(A_{spacer} = A_{window} - A_{glazing}\). Now, calculate the overall U-factor: \(U_{overall} = \frac{2.1888}{4.8} \cdot 3.85 + \frac{(4.8-2.1888)}{4.8} \cdot 1538.46 = 1.7629 \cdot 3.85 + 0.5435 \cdot 1538.46 = 6.788 \mathrm{~W/m^2·K}\) The overall U-factor for the double-pane wooden window is 6.78 W/m²·K. Compare this value to the one listed in Table 9-6 to check if it is within the expected range for this type of window.

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Most popular questions from this chapter

A circular grill of diameter \(0.25 \mathrm{~m}\) has an emissivity of \(0.8\). If the surface temperature is maintained at \(150^{\circ} \mathrm{C}\), determine the required electrical power when the room air and surroundings are at \(30^{\circ} \mathrm{C}\).

A spherical block of dry ice at \(-79^{\circ} \mathrm{C}\) is exposed to atmospheric air at \(30^{\circ} \mathrm{C}\). The general direction in which the air moves in this situation is (a) horizontal (b) up \(\quad(c)\) down (d) recirculation around the sphere (e) no motion

A \(12-\mathrm{cm}\)-high and 20-cm-wide circuit board houses 100 closely spaced logic chips on its surface, each dissipating \(0.05 \mathrm{~W}\). The board is cooled by a fan that blows air over the hot surface of the board at \(35^{\circ} \mathrm{C}\) at a velocity of \(0.5 \mathrm{~m} / \mathrm{s}\). The heat transfer from the back surface of the board is negligible. Determine the average temperature on the surface of the circuit board assuming the air flows vertically upward along the 12 -cm-long side by (a) ignoring natural convection and ( \(b\) ) considering the contribution of natural convection. Disregard any heat transfer by radiation. Evaluate air properties at a film temperature of \(47.5^{\circ} \mathrm{C}\) and 1 atm pressure. Is this a good assumption?

A 0.2-m-long and \(25-\mathrm{mm}\)-thick vertical plate \((k=\) \(1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) separates the hot water from the cold air at \(2^{\circ} \mathrm{C}\). The plate surface exposed to the hot water has a temperature of \(100^{\circ} \mathrm{C}\), and the surface exposed to the cold air has an emissivity of \(0.73\). Determine the temperature of the plate surface exposed to the cold air \(\left(T_{s, c}\right)\). Hint: The \(T_{s, c}\) has to be found iteratively. Start the iteration process with an initial guess of \(51^{\circ} \mathrm{C}\) for the \(T_{s, c}\).

A vertical \(0.9\)-m-high and \(1.8\)-m-wide double-pane window consists of two sheets of glass separated by a \(2.2-\mathrm{cm}\) air gap at atmospheric pressure. If the glass surface temperatures across the air gap are measured to be \(20^{\circ} \mathrm{C}\) and \(30^{\circ} \mathrm{C}\), the rate of heat transfer through the window is (a) \(19.8 \mathrm{~W}\) (b) \(26.1 \mathrm{~W}\) (c) \(30.5 \mathrm{~W}\) (d) \(34.7 \mathrm{~W}\) (e) \(55.0 \mathrm{~W}\) (For air, use \(k=0.02551 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7296, v=\) \(1.562 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). Also, the applicable correlation is \(\mathrm{Nu}\) \(\left.=0.42 \mathrm{Ra}^{1 / 4} \mathrm{Pr}^{0.012}(H / L)^{-0.3}\right)\) (For air, use \(k=0.02588 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7282, v=1.608 \times\) \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}\) )
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