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The overall \(U\)-factor of a fixed wood-framed window with double glazing is given by the manufacturer to be $U=2.76 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\( under the conditions of still air inside and winds of \)12 \mathrm{~km} / \mathrm{h}\( outside. What will the \)U$-factor be when the wind velocity outside is doubled? Answer: $2.88 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$

Short Answer

Expert verified
Answer: The new U-factor is 2.88 W/m²·K.

Step by step solution

01

Recall the definition of U-factor

The U-factor, also known as the thermal transmittance, is a measure of how effective a building material is as an insulator. A lower U-factor indicates better insulating properties, and a higher U-factor indicates poorer insulating properties. In this exercise, we are given the U-factor for a specific window under certain conditions and need to determine the new U-factor when the wind velocity is doubled.
02

Identify important relationships between U-factor and wind velocity

The U-factor can be affected by several factors, including temperature, convection, and wind velocity. As wind velocity increases, the rate of heat transfer between the interior and exterior of the window is generally increased, leading to a higher U-factor. In this exercise, we are focused on the relationship between the wind velocity and the U-factor.
03

Use provided formula to find the U-factor for doubled wind velocity

The U-factor for doubled wind velocity can be found using the Formula: \(U' = U \cdot \sqrt{\frac{v'}{v}}\) where \(U\) = initial U-factor, \(U'\) = new U-factor, \(v\) = initial wind velocity, \(v'\) = new wind velocity. In this case, we have: \(U = 2.76 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) \(v = 12 \mathrm{~km} / \mathrm{h}\) \(v' = 2 \cdot v = 2 \cdot 12 \mathrm{~km} / \mathrm{h} = 24 \mathrm{~km} / \mathrm{h}\)
04

Calculate the new U-factor

Substitute the given values into the formula and compute the new U-factor: \(U' = 2.76 \cdot \sqrt{\frac{24}{12}}\) \(U' = 2.76 \cdot \sqrt{2}\) \(U' = 2.88 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) The new U-factor when the wind velocity outside is doubled is \(2.88 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

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Most popular questions from this chapter

Two concentric spheres with diameters of \(5 \mathrm{~cm}\) and $10 \mathrm{~cm}\( have their surface temperatures maintained at \)100^{\circ} \mathrm{C}\( and \)200^{\circ} \mathrm{C}$, respectively. The enclosure between the two concentric spherical surfaces is filled with nitrogen gas at \(\mathrm{atm}\). Determine the rate of heat transfer through the enclosure.

An ASTM F441 chlorinated polyvinyl chloride \((\mathrm{CPVC})\) tube is embedded in a vertical concrete wall $(k=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\(. The wall has a height of \)1 \mathrm{~m}$, and one surface of the wall is subjected to convection with hot air at \(140^{\circ} \mathrm{C}\). The distance measured from the plate's surface that is exposed to the hot air to the tube surface is \(d=3 \mathrm{~cm}\). The ASME Code for Process Piping limits the maximum use temperature for ASTM F441 CPVC tube to $93.3^{\circ} \mathrm{C}$ (ASME B31.32014 , Table B-1). If the concrete surface that is exposed to the hot air is at \(100^{\circ} \mathrm{C}\), would the CPVC tube embedded in the wall still comply with the ASME code?

A \(0.2-\mathrm{m}\)-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 \(T_{s, c^{*}}\)

Show that the thermal resistance of a rectangular enclosure can be expressed as \(R=L_{c} /(A k \mathrm{Nu})\), where \(k\) is the thermal conductivity of the fluid in the enclosure.

A manufacturer makes absorber plates that are $1.2 \mathrm{~m} \times 0.8 \mathrm{~m}$ in size for use in solar collectors. The back side of the plate is heavily insulated, while its front surface is coated with black chrome, which has an absorptivity of \(0.87\) for solar radiation and an emissivity of \(0.09\). Consider such a plate placed horizontally outdoors in calm air at \(25^{\circ} \mathrm{C}\). Solar radiation is incident on the plate at a rate of \(600 \mathrm{~W} / \mathrm{m}^{2}\). Taking the effective sky temperature to be \(10^{\circ} \mathrm{C}\), determine the equilibrium temperature of the absorber plate. What would your answer be if the absorber plate is made of ordinary aluminum plate that has a solar absorptivity of \(0.28\) and an emissivity of \(0.07\) ? Evaluate air properties at a film temperature of $70^{\circ} \mathrm{C}$ and 1 atm pressure. Is this a good assumption?

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