/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 111 Why are the windows considered i... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Why are the windows considered in three regions when analyzing heat transfer through them? Name those regions and explain how the overall \(U\)-value of the window is determined when the heat transfer coefficients for all three regions are known.

Short Answer

Expert verified
Answer: Three regions are considered when studying heat transfer through windows because they have different heat transfer characteristics. The three regions are the glazing or glass pane surfaces, the window spacer or edge seal, and the frame or supporting structure. The overall U-value of the window, which represents the rate of heat transfer through the entire window assembly, is determined using parallel and series resistance concepts for heat transfer calculations, and by considering the heat transfer coefficients for all three regions. Lower U-values indicate better insulation and energy efficiency for the window.

Step by step solution

01

Understanding the concept of heat transfer through windows

Heat transfer through windows occurs due to the temperature difference between the inside and outside environment. There are three primary modes of heat transfer: conduction, convection, and radiation. Considering windows in various regions helps to analyze the overall heat transfer between the inside and outside environment more effectively.
02

Identifying the three regions of windows

The three regions of windows where heat transfer is considered are: 1. The glazing or glass pane surfaces 2. The window spacer or edge seal 3. The frame or supporting structure These regions have different heat transfer characteristics, mainly due to their varying thermal conductivities and surface orientations.
03

Explaining heat transfer mechanisms in each region

1. In the glazing region, heat transfer occurs primarily through conduction and radiation. Single or multiple layers of glass panes can be used to reduce the heat transfer through this region. Additional insulating gases, such as argon or krypton, may be used between the glass panes to further minimize heat transfer. 2. The window spacer or edge seal, which maintains the gap between the glass panes, also acts as a heat transfer path. Heat transfer in this region occurs mainly due to conduction. Using low-conductivity materials such as silicon-based foam or rubber for the spacer helps to minimize heat transfer. 3. The frame or supporting structure of the window is another heat transfer region. It experiences heat transfer mainly through conduction and also convection (due to air contact). By using low-conductivity materials and proper insulation techniques, the frame's heat transfer can also be minimized.
04

Determining the overall U-value of the window

The overall U-value of the window, also known as the thermal transmittance, represents the rate of heat transfer through the entire window assembly, comprising all three regions. It is expressed in watts per square meter per kelvin (W/m²·K). When the heat transfer coefficients for all three regions are known, the overall U-value of the window can be determined using parallel and series resistance concepts for heat transfer calculations. 1. Find the thermal resistance for each region: \(R_{i} = \frac{1}{U_{i}}\), where \(R_{i}\) is the thermal resistance and \(U_{i}\) is the heat transfer coefficient for the respective region. 2. Combine the thermal resistances in parallel and series as required: For two resistances in series (like the glazing layers), add the resistances; for two resistances in parallel (like the spacer and frame in some cases), use the formula \(\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}\). 3. After obtaining the equivalent resistance for the entire window assembly, calculate the overall U-value as: \(U_{overall} = \frac{1}{R_{eq}}\). The overall U-value represents the ability of the window to transfer heat. Lower U-values indicate better insulation and energy efficiency for the window.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Thermal Transmittance
Thermal transmittance, commonly referred to as the U-value, is a crucial concept when it comes to understanding heat transfer through windows. It quantifies how well a window or any part of a building's envelope conducts heat. The U-value is measured in watts per square meter per kelvin (W/m²·K), and it represents the amount of heat that passes through one square meter of a structure when there's a one-degree temperature difference between the inside and outside environments.

Imagine the window as a gatekeeper for heat. A lower U-value, which is desirable, means less heat escapes in winter and less heat enters in summer, implying that the window is an effective barrier against unwanted heat transfer. This contributes to the overall energy efficiency of a building and can lead to significant savings in heating and cooling costs.

An efficient way to improve a window's U-value is by optimizing the three main regions—glazing, spacer, and frame—each holding a specific U-value. By considering the combination of these individual U-values through resistance calculations, one can compute the overall U-value of the entire window system, which is essential for architects and builders aiming to optimize energy consumption.
Thermal Resistance
If thermal transmittance tells us how easily heat conducts through the window, thermal resistance is its counterpart indicating how well the window resists heat flow. This resistance is given by the formula: \( R_{i} = \frac{1}{U_{i}} \), where \( R_{i} \) is the thermal resistance for a particular region of the window and \( U_{i} \) is the region's heat transfer coefficient.

Think of thermal resistance as a measure of the 'fight' against unwanted heat travel. Each region of the window has its own resistance, and the right materials can increase this resistance. For example, a glass pane might resist heat flow, but adding an insulating gas layer or a special coating can further enhance this resistance. Similarly, the material used for the window spacer can have a significant influence, as can the construction of the frame.

The key to understanding thermal resistance in windows is to consider how these resistances add up, just as resistors do in an electric circuit. When analyzed individually and then combined, the resistances of the glazing, spacer, and frame can be used to calculate the window's overall thermal resistance. This concept is directly linked to the U-value, forming the backbone of the window's heat retention capability, and the overall energy efficiency of a building.
Heat Transfer Coefficients
Heat transfer coefficients are a detailed way to quantify how heat transfers through different materials or regions of a window. Defined for each specific area—the glazing, the spacer, and the frame—these coefficients have units of W/m²·K and directly influence the U-value calculation. Each of these coefficients \( (U_{i}) \) represents the effectiveness of heat transfer for the respective region.

The heat transfer coefficient for glazing involves conductivity and radiation aspects of the glass. Different types of glass and additional layers like coatings can alter this coefficient to improve thermal performance. Spacers and frames have their own unique coefficients that take into account their materials and geometry. The lower the heat transfer coefficient, the less heat can pass through that region, leading to better insulation and energy efficacy.

By knowing the heat transfer coefficients, we can compute the corresponding thermal resistances of each region of the window. This information contributes to the crucial step in our exercise solution—finding the cumulative thermal resistance and consequently the overall U-value. When evaluated together, these coefficients offer a comprehensive viewpoint on a window’s insulation characteristics and provide practical guidance for enhancing thermal performance in building design.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A \(50-\mathrm{cm} \times 50-\mathrm{cm}\) circuit board that contains 121 square chips on one side is to be cooled by combined natural convection and radiation by mounting it on a vertical surface in a room at \(25^{\circ} \mathrm{C}\). Each chip dissipates \(0.18 \mathrm{~W}\) of power, and the emissivity of the chip surfaces is 0.7. Assuming the heat transfer from the back side of the circuit board to be negligible, and the temperature of the surrounding surfaces to be the same as the air temperature of the room, determine the surface temperature of the chips. Evaluate air properties at a film temperature of \(30^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption?

A spherical tank \((k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with an inner diameter of \(3 \mathrm{~m}\) and a wall thickness of \(10 \mathrm{~mm}\) is used for storing hot liquid. The hot liquid inside the tank causes the inner surface temperature to be as high as \(100^{\circ} \mathrm{C}\). To prevent thermal burns on the skin of the people working near the vicinity of the tank, the tank is covered with a \(7-\mathrm{cm}\) thick layer of insulation \((k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and the outer surface is painted to give an emissivity of \(0.35\). The tank is located in a surrounding with air at \(16^{\circ} \mathrm{C}\). Determine whether or not the insulation layer is sufficient to keep the outer surface temperature below \(45^{\circ} \mathrm{C}\) to prevent thermal burn hazards. Discuss ways to further decrease the outer surface temperature. Evaluate the air properties at \(30^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption?

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 \(15^{\circ} \mathrm{C}\).

Consider an industrial furnace that resembles a 13 -ft-long horizontal cylindrical enclosure \(8 \mathrm{ft}\) in diameter whose end surfaces are well insulated. The furnace burns natural gas at a rate of 48 therms/h. The combustion efficiency of the furnace is 82 percent (i.e., 18 percent of the chemical energy of the fuel is lost through the flue gases as a result of incomplete combustion and the flue gases leaving the furnace at high temperature). If the heat loss from the outer surfaces of the furnace by natural convection and radiation is not to exceed 1 percent of the heat generated inside, determine the highest allowable surface temperature of the furnace. Assume the air and wall surface temperature of the room to be \(75^{\circ} \mathrm{F}\), and take the emissivity of the outer surface of the furnace to be \(0.85\). If the cost of natural gas is \(\$ 1.15 /\) therm and the furnace operates \(2800 \mathrm{~h}\) per year, determine the annual cost of this heat loss to the plant. Evaluate properties of air at a film temperature of \(107.5^{\circ} \mathrm{F}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption?

A \(1.5\)-m-diameter, 4-m-long cylindrical propane tank is initially filled with liquid propane, whose density is \(581 \mathrm{~kg} / \mathrm{m}^{3}\). The tank is exposed to the ambient air at \(25^{\circ} \mathrm{C}\) in calm weather. The outer surface of the tank is polished so that the radiation heat transfer is negligible. Now a crack develops at the top of the tank, and the pressure inside drops to \(1 \mathrm{~atm}\) while the temperature drops to \(-42^{\circ} \mathrm{C}\), which is the boiling temperature of propane at \(1 \mathrm{~atm}\). The heat of vaporization of propane at \(1 \mathrm{~atm}\) is \(425 \mathrm{~kJ} / \mathrm{kg}\). The propane is slowly vaporized as a result of the heat transfer from the ambient air into the tank, and the propane vapor escapes the tank at \(-42^{\circ} \mathrm{C}\) through the crack. Assuming the propane tank to be at about the same temperature as the propane inside at all times, determine how long it will take for the tank to empty if it is not insulated.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.