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A glass window of width \(W=1 \mathrm{~m}\) and height \(H=2 \mathrm{~m}\) is \(5 \mathrm{~mm}\) thick and has a thermal conductivity of \(k_{g}=\) \(1.4 \mathrm{~W} / \mathrm{m}=\mathrm{K}\). If the inner and outer surface temperatures of the glass are \(15^{\circ} \mathrm{C}\) and \(-20^{\circ} \mathrm{C}\), respectively, on a cold winter day, what is the rate of heat loss through the glass? To reduce heat loss through windows, it is customary to use a double pane construction in which adjoining panes are separated by an air space. If the spacing is \(10 \mathrm{~mm}\) and the glass surfaces in contact with the air have temperatures of \(10^{\circ} \mathrm{C}\) and \(-15^{\circ} \mathrm{C}\), what is the rate of heat loss from a \(1 \mathrm{~m} \times 2 \mathrm{~m}\) window? The themal conductivity of air is \(k_{a}=0.024 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Short Answer

Expert verified
The rate of heat loss through the single-pane glass window is 140 W, while the rate of heat loss through the double-pane glass window with air space is 22.4 W. Therefore, using the double-pane construction with an air space results in a significant reduction in heat loss through windows, which is a more efficient and energy-saving solution for homes and buildings.

Step by step solution

01

Calculate the Surface Area of the Glass Window

Since the width and height of the glass window are given, we can calculate the surface area (A) using the following formula: \(A = W \cdot H\) The window measures 1 m in width (W) and 2 m in height (H). So, \(A = 1\mathrm{~m} \cdot 2\mathrm{~m} = 2\mathrm{m^2}\).
02

Calculate the Rate of Heat Loss for Single-Pane Glass Window

Now, we can calculate the rate of heat loss (Q) through the single-pane glass window using the given formula: \(Q_{glass} = k_g \cdot A \cdot \frac{T_1 - T_2}{d}\) The values of the thermal conductivity of the glass (kg), inner (T1 = 15°C) and outer (T2 = -20°C) temperatures, and the glass thickness (d) are given. So, we can calculate the heat loss as follows: \(Q_{glass} = 1.4\frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}} \cdot 2\mathrm{m^2} \cdot \frac{15 - (-20)}{0.005\mathrm{~m}}\) \(Q_{glass} = 140\ \mathrm{W}\)
03

Calculate the Effective Thermal Conductivity for Double-Pane Glass Window with Air Space

For the double-pane glass window with air space, we have to calculate the effective thermal conductivity (ke) using the formula for conductances in series: \(\frac{1}{k_e} = \frac{1}{k_g} + \frac{1}{k_{air}}\) Given that the thermal conductivity of the air (k_air) is 0.024 W/m·K, we can calculate ke: \(\frac{1}{k_e} = \frac{1}{1.4} + \frac{1}{0.024}\) \(k_e = 0.0224~\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\)
04

Calculate the Rate of Heat Loss for Double-Pane Glass Window with Air Space

Now, we can apply the formula for the rate of heat loss through the double-pane glass window: \(Q_{double-pane} = k_e \cdot A \cdot \frac{T_1 - T_2}{d}\) We are given the surface temperatures in contact with the air (T1 = 10°C and T2 = -15°C) and the air spacing (d = 0.010 m). So, we can calculate the heat loss as follows: \(Q_{double-pane} = 0.0224\frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}} \cdot 2\mathrm{m^2} \cdot \frac{10 - (-15)}{0.01\mathrm{~m}}\) \(Q_{double-pane} = 22.4\ \mathrm{W}\)
05

Compare Heat Loss Rates of Single-Pane and Double-Pane Windows

To determine the effectiveness of the double-pane windows in reducing heat loss, we will compare the rates of heat loss for both single-pane and double-pane windows: - Single-pane glass window: heat loss = 140 W - Double-pane glass window with air space: heat loss = 22.4 W So, using the double-pane construction with an air space results in a significant reduction in heat loss through windows, which is a more efficient and energy-saving solution for homes and buildings.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermal Conductivity
Thermal conductivity is a measure of how well a material conducts heat. It is represented by the symbol \( k \) and is expressed in units of watts per meter per degree Kelvin \( \frac{W}{m \, \cdot \, K} \). Thermal conductivity determines the rate at which heat is transferred through a material. In general, materials with high thermal conductivity will allow heat to flow through them quickly, while those with low conductivity will slow down the flow of heat.
This is important in building design and construction, especially in climates with extreme temperatures. By understanding and using materials with appropriate thermal conductivities, builders can improve energy efficiency and comfort in buildings.
  • High thermal conductivity: Metals like copper or aluminum.
  • Low thermal conductivity: Insulators like wood or fiberglass.
Single-Pane Window
A single-pane window is the most basic form of window design. As the term suggests, this type of window uses a single sheet of glass. Because glass has relatively high thermal conductivity compared to materials like air, heat easily passes through single-pane windows. This makes them less energy-efficient.
When temperatures drop outside, single-pane windows can allow significant heat loss, causing indoor heating systems to work harder to maintain a comfortable environment. The rate of heat loss through such a window can be calculated using the formula:
\[ Q = k_g \cdot A \cdot \frac{T_1 - T_2}{d} \]
Here, \( Q \) is the rate of heat transfer, \( k_g \) is the thermal conductivity of the glass, \( A \) is the area of the window, \( T_1 \) and \( T_2 \) are the temperatures on either side of the window, and \( d \) is the thickness of the glass.
Despite their inefficiency, single-pane windows were commonly used in the past because they are simple and inexpensive. However, they are increasingly being replaced with more efficient alternatives like double-pane windows, especially in areas where energy conservation is a priority.
Double-Pane Window
Double-pane windows provide a more energy-efficient solution compared to single-pane windows, primarily because they incorporate an insulating layer of air or gas between two layers of glass. This design reduces thermal conductivity, as air is a poor conductor of heat. The air space acts as a barrier, slowing down the rate at which heat is transferred between the inside and outside environments.
In the case of a double-pane window, the effective thermal conductivity is calculated by considering both the glass and the air spacer's thermal resistance:
\[ \frac{1}{k_e} = \frac{1}{k_g} + \frac{1}{k_{air}} \]
Where \( k_e \) is the effective thermal conductivity, \( k_g \) is the thermal conductivity of the glass, and \( k_{air} \) is that of air. This leads to significantly lower heat loss rates, making double-pane windows an effective option for reducing energy consumption in buildings.
Moreover, using double-pane glass with low-conductivity gases like argon or krypton instead of air further augments energy savings by reducing heat transfer.
Insulation
Insulation is the key to reducing heat transfer between the internal and external environments of a building. It acts by trapping air in tiny pockets, thus lowering the thermal conductivity of the composite material. Effective insulation balances energy usage by keeping warm air inside during winter and preventing outside heat from warming the interior during summer.
In window design, the shift from single-pane to double-pane windows is an example of how adding layers can enhance insulation. The air layer between double panes acts as an insulating barrier, drastically reducing the rate of heat loss. This not only saves energy but also reduces utility costs and enhances indoor comfort levels.
There are various common materials used for insulation:
  • Fiberglass: Widely used in walls and attics.
  • Foam board: Used for insulating areas that require a rigid material.
  • Reflective insulation: Often placed in attics to reflect heat.
Overall, insulation is a critical component in the design of energy-efficient homes and buildings.

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Most popular questions from this chapter

The diameter and surface emissivity of an electrically heated plate are \(D=300 \mathrm{~mm}\) and \(\varepsilon=0.80\), respectively. (a) Estimate the power needed to maintain a surface temperature of \(200^{\circ} \mathrm{C}\) in a room for which the air and the walls are at \(25^{\circ} \mathrm{C}\). The coefficient characterizing heat transfer by natural convection depends on the surface temperature and, in units of \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), may be approximated by an expression of the form \(h=0.80\left(T_{s}-T_{\infty}\right)^{1 / 3}\). (b) Assess the effect of surface temperature on the power requirement, as well as on the relative contributions of convection and radiation to heat transfer from the surface.

An aluminum plate \(4 \mathrm{~mm}\) thick is mounted in a horizontal position, and its bottom surface is well insulated. A special, thin coating is applied to the top surface such that it absorbs \(80 \%\) of any incident solar radiation, while having an emissivity of \(0.25\). The density \(\rho\) and specific heat \(c\) of aluminum are known to be \(2700 \mathrm{~kg} / \mathrm{m}^{3}\) and \(900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively. (a) Consider conditions for which the plate is at a temperature of \(25^{\circ} \mathrm{C}\) and its top surface is suddenly exposed to ambient air at \(T_{\infty}=20^{\circ} \mathrm{C}\) and to solar radiation that provides an incident flux of \(900 \mathrm{~W} / \mathrm{m}^{2}\). The convection heat transfer coefficient between the surface and the air is \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the initial rate of change of the plate temperature? (b) What will be the equilibrium temperature of the plate when steady-state conditions are reached? (c) The surface radiative properties depend on the specific nature of the applied coating. Compute and plot the steady-state temperature as a function of the emissivity for \(0.05 \leq \varepsilon \leq 1\), with all other conditions remaining as prescribed. Repeat your calculations for values of \(\alpha_{S}=0.5\) and \(1.0\), and plot the results with those obtained for \(\alpha_{S}=0.8\). If the intent is to maximize the plate temperature, what is the most desirable combination of the plate emissivity and its absorptivity to solar radiation?

A freezer compartment is covered with a 2 -mm-thick layer of frost at the time it malfunctions. If the compartment is in ambient air at \(20^{\circ} \mathrm{C}\) and a coefficient of \(h=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) characterizes heat transfer by natural convection from the exposed surface of the layer, estimate the time required to completely melt the frost. The frost may be assumed to have a mass density of \(700 \mathrm{~kg} / \mathrm{m}^{3}\) and a latent heat of fusion of \(334 \mathrm{~kJ} / \mathrm{kg}\).

If \(T_{s} \approx T_{\text {sur }}\) in Equation \(1.9\), the radiation heat transfer coefficient may be approximated as $$ h_{r, a}=4 \varepsilon \sigma \bar{T}^{3} $$ where \(\bar{T} \equiv\left(T_{s}+T_{\text {sur }}\right) / 2\). We wish to assess the validity of this approximation by comparing values of \(h_{r}\) and \(h_{r, a}\) for the following conditions. In each case, represent your results graphically and comment on the validity of the approximation. (a) Consider a surface of either polished aluminum ( \(\varepsilon=\) \(0.05)\) or black paint \((\varepsilon=0.9)\), whose temperature may exceed that of the surroundings \(\left(T_{\text {sur }}=25^{\circ} \mathrm{C}\right)\) by 10 to \(100^{\circ} \mathrm{C}\). Also compare your results with values of the coefficient associated with free convection in air \(\left(T_{\infty}=T_{\text {sur }}\right)\), where \(h\left(\mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)=0.98 \Delta T^{1 / 3}\). (b) Consider initial conditions associated with placing a workpiece at \(T_{s}=25^{\circ} \mathrm{C}\) in a large furnace whose wall temperature may be varied over the range \(100 \leq\) \(T_{\text {sur }} \leq 1000^{\circ} \mathrm{C}\). According to the surface finish or coating, its emissivity may assume values of \(0.05\), \(0.2\), and \(0.9\). For each emissivity, plot the relative error, \(\left(h_{r}-h_{r, a}\right) / h_{r}\), as a function of the furnace temperature.

Liquid oxygen, which has a boiling point of \(90 \mathrm{~K}\) and a latent heat of vaporization of \(214 \mathrm{~kJ} / \mathrm{kg}\), is stored in a spherical container whose outer surface is of \(500-\mathrm{mm}\) diameter and at a temperature of \(-10^{\circ} \mathrm{C}\). The container is housed in a laboratory whose air and walls are at \(25^{\circ} \mathrm{C}\). (a) If the surface emissivity is \(0.20\) and the heat transfer coefficient associated with free convection at the outer surface of the container is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the rate, in \(\mathrm{kg} / \mathrm{s}\), at which oxygen vapor must be vented from the system? (b) Moisture in the ambient air will result in frost formation on the container, causing the surface emissivity to increase. Assuming the surface temperature and convection coefficient to remain at \(-10^{\circ} \mathrm{C}\) and \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively, compute the oxygen evaporation rate \((\mathrm{kg} / \mathrm{s})\) as a function of surface emissivity over the range \(0.2 \leq \varepsilon \leq 0.94\).

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