Consider three similar double-pane windows with air gap widths of 5,10 , and \(20 \mathrm{~mm}\). For which case will the heat transfer through the window will be a minimum?

Short Answer

Expert verified
Based on the analysis of both conduction and convection processes, the heat transfer through the window will be a minimum for the case with an air gap width of 5 mm. This is due to the conduction process being inversely proportional to the air gap width and the convection process being less effective in smaller air gaps.

Step by step solution

01

Understand the heat transfer process

Heat transfer through a window occurs mainly due to conduction and convection. The conduction occurs through the glass and between the two panes of glass across the air gap, while the convection occurs within the air trapped in the gap. The total heat transfer rate through the window can be determined by analyzing these heat transfer processes.
02

Evaluate the conduction process

In the conduction process, the heat transfer rate is dependent on the thermal conductivity (k) of the material, the air gap width (d), and the temperature gradient (ΔT) across the gap. The rate of heat transfer by conduction (\(Q_{cond}\)) can be described by Fourier's Law: \(Q_{cond} = k \frac{A ΔT}{d}\), where \(A\) is the area of the window. For the same type of glass and the same temperature gradient, \(Q_{cond}\) will be inversely proportional to the air gap width d. \(Q_{cond} \propto \frac{1}{d}\).
03

Evaluate the convection process

In the convection process, the heat transfer rate is dependent on the convective heat transfer coefficient (h), the air gap width (d), and the temperature gradient (ΔT) across the gap. The rate of heat transfer by convection (\(Q_{conv}\)) can be described by Newton's Law of Cooling: \(Q_{conv} = h A ΔT\). For an air gap with a small width, the convective process will be less effective as it doesn't have enough space for air circulation. As a result, the convective heat transfer coefficient (\(h\)) will be smaller for smaller air gaps.
04

Combine conduction and convection processes

The total heat transfer rate (\(Q_{total}\)) through the window is the sum of the conduction and convection processes: \(Q_{total} = Q_{cond} + Q_{conv}\) Since \(Q_{cond}\) is inversely proportional to the air gap width d and \(h\) is smaller for smaller air gaps, the total heat transfer rate (\(Q_{total}\)) will be minimized when we have the minimum air gap width, which is 5 mm in this problem.
05

Conclusion

The heat transfer through the window will be a minimum for the case with an air gap width of 5 mm, since the conduction is inversely proportional to the air gap width and the convection process is less effective in smaller air gaps.

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Key Concepts

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

Understanding the Conduction Process
The conduction process is a fundamental concept in the study of heat transfer. It refers to the way heat moves through materials that are in direct contact. This process is driven by the temperature difference between two adjacent regions; heat flows from the hotter to the colder region. For instance, in a double-pane window, heat conduction occurs through the glass panes and the air trapped between them.

According to Fourier's Law of heat conduction, the heat transfer rate, denoted by the symbol \(Q_{cond}\), is directly proportional to the temperature difference across the material, \(ΔT\), and the cross-sectional area through which heat is being transferred, \(A\). This law mathematically expresses the relationship as \(Q_{cond} = k \frac{A ΔT}{d}\), where \(k\) is the thermal conductivity of the material and \(d\) is the thickness of the material.

When analyzing the efficiency of window insulation, the wider the air gap, the less conductive the window becomes, which means less heat is transferred through the window by conduction. Consequently, to minimize heat loss or gain, optimizing the air gap width is essential.
Analyzing the Convection Process
In the context of heat transfer through windows, the convection process plays a significant role in addition to the conduction process. Convection involves the transfer of heat by the physical movement of fluid, which in the case of windows is the air within the gap. This air can carry heat away from or towards the window panes, depending on the temperature difference between the indoor and outdoor environments.

The rate of heat transfer by convection, \(Q_{conv}\), is governed by Newton's Law of Cooling, stated as \(Q_{conv} = h A ΔT\), where \(h\) is the convective heat transfer coefficient. This coefficient \(h\) varies with the gap width; in narrower gaps, it diminishes as fluid movement is constrained, reducing the effectiveness of convection. Therefore, a smaller air gap can lead to less heat transfer due to the restricted air movement, which is why in the given example, a 5 mm gap will have the least convective heat transfer.
The Implications of Fourier's Law
Fourier's Law is indispensable when comprehending how different factors influence the rate of heat conduction. This applies not only to air gaps in windows but to all scenarios involving heat transfer through a solid. Fourier's Law quantifies that the rate of heat transfer by conduction \(Q_{cond}\) is proportional to the temperature gradient and the area perpendicularly through which heat is flowing, and inversely proportional to the distance over which the temperature change occurs.

In the example of double-pane windows with varying air gap widths—5 mm, 10 mm, and 20 mm—the implication of Fourier's Law is clear. As the air gap width increases, the heat transfer due to conduction decreases. This is critical to remember when selecting or designing windows for building insulation. By understanding Fourier's Law, engineers and architects can ensure that the windows they choose or design will effectively contribute to the energy efficiency of the building by minimizing undesired heat transfer.

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Most popular questions from this chapter

During a plant visit, it was observed that a \(1.5-\mathrm{m}\)-high and \(1-m\)-wide section of the vertical front section of a natural gas furnace wall was too hot to touch. The temperature measurements on the surface revealed that the average temperature of the exposed hot surface was \(110^{\circ} \mathrm{C}\), while the temperature of the surrounding air was \(25^{\circ} \mathrm{C}\). The surface appeared to be oxidized, and its emissivity can be taken to be \(0.7\). Taking the temperature of the surrounding surfaces to be \(25^{\circ} \mathrm{C}\) also, determine the rate of heat loss from this furnace. The furnace has an efficiency of 79 percent, and the plant pays \(\$ 1.20\) per therm of natural gas. If the plant operates \(10 \mathrm{~h}\) a day, 310 days a year, and thus \(3100 \mathrm{~h}\) a year, determine the annual cost of the heat loss from this vertical hot surface on the front section of the furnace wall.

Consider a thin 16-cm-long and 20-cm-wide horizontal plate suspended in air at \(20^{\circ} \mathrm{C}\). The plate is equipped with electric resistance heating elements with a rating of \(20 \mathrm{~W}\). Now the heater is turned on and the plate temperature rises. Determine the temperature of the plate when steady operating conditions are reached. The plate has an emissivity of \(0.90\) and the surrounding surfaces are at \(17^{\circ} \mathrm{C}\). As an initial guess, assume a surface temperature of \(50^{\circ} \mathrm{C}\). Is this a good assumption?

Write a computer program to evaluate the variation of temperature with time of thin square metal plates that are removed from an oven at a specified temperature and placed vertically in a large room. The thickness, the size, the initial temperature, the emissivity, and the thermophysical properties of the plate as well as the room temperature are to be specified by the user. The program should evaluate the temperature of the plate at specified intervals and tabulate the results against time. The computer should list the assumptions made during calculations before printing the results. For each step or time interval, assume the surface temperature to be constant and evaluate the heat loss during that time interval and the temperature drop of the plate as a result of this heat loss. This gives the temperature of the plate at the end of a time interval, which is to serve as the initial temperature of the plate for the beginning of the next time interval. Try your program for \(0.2\)-cm-thick vertical copper plates of \(40 \mathrm{~cm} \times 40 \mathrm{~cm}\) in size initially at \(300^{\circ} \mathrm{C}\) cooled in a room

Consider an \(L \times L\) horizontal plate that is placed in quiescent air with the hot surface facing up. If the film temperature is \(20^{\circ} \mathrm{C}\) and the average Nusselt number in natural convection is of the form \(\mathrm{Nu}=C \mathrm{Ra}_{L}^{n}\), show that the average heat transfer coefficient can be expressed as $$ \begin{aligned} &h=1.95(\Delta T / L)^{1 / 4} 10^{4}<\mathrm{Ra}_{L}<10^{7} \\ &h=1.79 \Delta T^{1 / 3} \quad 10^{7}<\mathrm{Ra}_{L}<10^{11} \end{aligned} $$

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 \(3-\mathrm{cm}\)-wide air space. Determine the steady rate of heat transfer through this window and the temperature of its inner surface for a day during which the room is maintained at \(20^{\circ} \mathrm{C}\) while the temperature of the outdoors is \(0^{\circ} \mathrm{C}\). Take the 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. Evaluate air properties at a film temperature of \(10^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption?

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