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Magazine Chapter 9: Problem 118
Determine the \(U\)-factor for the center-of-glass section of a double-pane window with a 13-mm air space for winter design conditions. The glazings are made of clear glass having an emissivity of \(0.84\). Take the average air space temperature at design conditions to be \(10^{\circ} \mathrm{C}\) and the temperature difference across the air space to be \(15^{\circ} \mathrm{C}\).
Short Answer
Expert verified
Answer: The U-factor for the center-of-glass section of a double-pane window with a 13-mm air space for winter design conditions is approximately 512.48 W/(m虏路K).
Step by step solution
01
Calculate heat transfer coefficient due to conduction through the glass
We will first calculate the resistance for conduction through the glass, which can be determined by the equation:
\(R_\text{cond}=\frac{d}{kA}\), where \(d\) is the thickness of the glass, \(k\) is the thermal conductivity of the glass, and \(A\) is the area of the glass.
We know the thermal conductivity of glass to be \(k=1 \,\text{W}\,(\text{m} \cdot \text{K})^{-1}\).
Assuming the glass thickness to be \(4 \,\text{mm}\) and considering unit area \(A=1 \,\text{m}^{2}\), we have:
\(R_\text{cond}=\frac{0.004\,\text{m}}{1\,\text{W}\,(\text{m} \cdot \text{K})^{-1} \cdot 1\,\text{m}^{2}}=0.004\,\text{m}^{2}\,(\text{K}\,\text{W})^{-1}\).
The heat transfer coefficient for conduction through glass can be calculated as:
\(U_\text{cond}=\frac{1}{R_\text{cond}}=\frac{1}{0.004\,\text{m}^{2}\,(\text{K}\,\text{W})^{-1}}=250\,\text{W}\,(\text{m}^{2}\cdot\text{K})^{-1}\).
02
Calculate the heat transfer coefficient due to radiation in the air space
We will now calculate the radiation heat transfer coefficient (\(U_\text{rad}\)) within the air space using the following equation:
\(U_\text{rad}=\frac{\sigma\epsilon(Th^{4}-Tc^{4})}{Th-Tc}\), where \(\sigma\) is the Stefan-Boltzmann constant (\(5.67\times10^{-8}\,\text{W}\,(\text{m}^{2}\cdot\text{K}^{4})^{-1}\)), \(\epsilon\) is the emissivity of the glass, \(Th\) is the hot surface temperature and \(Tc\) is the cold surface temperature.
We are given the emissivity of the glass as \(0.84\), the average air space temperature as \(10^{\circ}\mathrm{C}\), and the temperature difference across the air space as \(15^{\circ}\mathrm{C}\).
First, determine the hot and cold surface temperatures:
\(Th=10+7.5=17.5^{\circ}\mathrm{C}=290.65\,\text{K}\),
\(Tc=10-7.5=2.5^{\circ}\mathrm{C}=275.65\,\text{K}\).
Now, calculate the radiation heat transfer coefficient:
\(U_\text{rad}=\frac{5.67\times10^{-8}\,\text{W}\,(\text{m}^{2}\cdot\text{K}^{4})^{-1} \cdot 0.84 \cdot (290.65^{4}-275.65^{4})}{290.65-275.65} \approx 8.12\,\text{W}\,(\text{m}^{2}\cdot\text{K})^{-1}\).
03
Calculate the heat transfer coefficient due to convection in the air space
We will now calculate the convection heat transfer coefficient (\(U_\text{conv}\)) within the air space using the following equation:
\(U_\text{conv}=1.32\Delta T^{1/3}\), where \(\Delta T\) is the temperature difference across the air space.
Given the temperature difference across the air space as \(15^{\circ}\mathrm{C}\), we have:
\(U_\text{conv}=1.32\cdot15^{1/3}\approx 4.36\,\text{W}\,(\text{m}^{2}\cdot\text{K})^{-1}\).
04
Calculate the overall U-factor
Finally, we will calculate the overall U-factor by summing up the heat transfer coefficients from the above steps. Since the window has two glass panes, we need to consider the conduction through both panes. Therefore,
\(U_\text{total}=U_\text{cond}+U_\text{rad}+U_\text{conv}+U_\text{cond} = 250 + 8.12 + 4.36 + 250 \approx 512.48\,\text{W}\,(\text{m}^{2}\cdot\text{K})^{-1}\).
The U-factor for the center-of-glass section of a double-pane window with a 13-mm air space for winter design conditions is approximately \(512.48\,\text{W}\,(\text{m}^{2}\cdot\text{K})^{-1}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Heat Transfer Coefficient
The heat transfer coefficient is a critical measure in determining how effectively heat is transferred through a material. It helps assess the thermal performance of building elements, like windows, in varying conditions. For windows, this value is often expressed as the U-factor, representing the inverse of the total thermal resistance.
The U-factor combines different modes of heat transfer鈥攃onduction, convection, and radiation鈥攊nto a single measure that allows us to gauge how energy efficient a window is. A lower U-factor signifies better insulation, as less heat is lost through the window. Calculating the U-factor involves summing up heat contributions from all parts of the window, including the glass and any air spaces. Sometimes, multiple panes and air spaces, like in double-pane windows, can significantly affect this value.
The U-factor combines different modes of heat transfer鈥攃onduction, convection, and radiation鈥攊nto a single measure that allows us to gauge how energy efficient a window is. A lower U-factor signifies better insulation, as less heat is lost through the window. Calculating the U-factor involves summing up heat contributions from all parts of the window, including the glass and any air spaces. Sometimes, multiple panes and air spaces, like in double-pane windows, can significantly affect this value.
Conduction Through Glass
Conduction is the process of heat flow through solid materials due to a temperature gradient. In windows, this occurs through the glass panes and is quantified by determining the conduction resistance.
The formula to find this resistance is:
\[ R_{\text{cond}}=\frac{d}{kA} \]
where:
Thermal conductivity for glass is usually around 1 W/(m路K), indicating how quickly heat can pass through the glass.
In the exercise, with a glass thickness of 4 mm, the conduction resistance was found, and inverting that value provided the heat transfer coefficient due to conduction.
The formula to find this resistance is:
\[ R_{\text{cond}}=\frac{d}{kA} \]
where:
- \( d \) is the thickness of the glass,
- \( k \) is the thermal conductivity of the material, and
- \( A \) is the area across which heat is conducted.
Thermal conductivity for glass is usually around 1 W/(m路K), indicating how quickly heat can pass through the glass.
In the exercise, with a glass thickness of 4 mm, the conduction resistance was found, and inverting that value provided the heat transfer coefficient due to conduction.
Radiation Heat Transfer
Radiation is the transfer of heat energy through electromagnetic waves. In the context of windows, it typically occurs across air spaces, influenced by the material's emissivity and temperature differences.
The formula to calculate the radiation heat transfer coefficient is:
\[ U_{\text{rad}}=\frac{\sigma\epsilon(Th^{4}-Tc^{4})}{Th-Tc} \]
where:
The exercise given involves clear glass with an emissivity of 0.84. By calculating the temperatures of the two surfaces, the radiation heat transfer coefficient was determined, showing how much heat is transferred via radiation between the glass panes.
The formula to calculate the radiation heat transfer coefficient is:
\[ U_{\text{rad}}=\frac{\sigma\epsilon(Th^{4}-Tc^{4})}{Th-Tc} \]
where:
- \( \sigma \) is the Stefan-Boltzmann constant (approximately 5.67\times10^{-8} W/(m虏路K鈦�)),
- \( \epsilon \) is the emissivity of the glass,
- \( Th \) and \( Tc \) are the temperatures of the hot and cold surfaces, respectively.
The exercise given involves clear glass with an emissivity of 0.84. By calculating the temperatures of the two surfaces, the radiation heat transfer coefficient was determined, showing how much heat is transferred via radiation between the glass panes.
Convection in Air Space
Convection is the movement of heat within fluids, such as air or water, caused by fluid motion. In a double-pane window, convection happens within the air space between the panes.
The convection heat transfer coefficient can be estimated using:
\[ U_{\text{conv}}=1.32\Delta T^{1/3} \]
where \( \Delta T \) is the temperature difference across the air space.
For the exercise, with a temperature difference of 15掳C across a 13-mm air space, the convection coefficient was calculated. This value provides insight into how effectively the air space prevents heat from escaping through convection, an essential factor for assessing a window's thermal performance.
The convection heat transfer coefficient can be estimated using:
\[ U_{\text{conv}}=1.32\Delta T^{1/3} \]
where \( \Delta T \) is the temperature difference across the air space.
For the exercise, with a temperature difference of 15掳C across a 13-mm air space, the convection coefficient was calculated. This value provides insight into how effectively the air space prevents heat from escaping through convection, an essential factor for assessing a window's thermal performance.
