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The rate of heat loss through a unit surface area of a window per unit temperature difference between the indoors and the outdoors is called the \(U\)-factor. The value of the \(U\)-factor ranges from about \(1.25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (or \(0.22 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\) ) for low-e coated, argon-filled, quadruple-pane windows to \(6.25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (or \(1.1 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\) ) for a single-pane window with aluminum frames. Determine the range for the rate of heat loss through a \(1.2-\mathrm{m} \times 1.8-\mathrm{m}\) window of a house that is maintained at \(20^{\circ} \mathrm{C}\) when the outdoor air temperature is \(-8^{\circ} \mathrm{C}\).

Short Answer

Expert verified
Based on the given data, calculate the range of heat loss through a 1.2m x 1.8m window for single-pane and quadruple-pane windows, considering indoor and outdoor temperatures are 20°C and -8°C, respectively.

Step by step solution

01

Identify the given data

We are given the following information: - U-factor of a quadruple-pane window = 1.25 W/m²·K - U-factor of a single-pane window = 6.25 W/m²·K - Window dimensions: 1.2 m × 1.8 m - Indoor temperature (T_indoor) = 20°C - Outdoor temperature (T_outdoor) = -8°C
02

Calculate the surface area of the window

To find the total surface area of the window, multiply its length and width: Surface_area = Length × Width Surface_area = 1.2 m × 1.8 m Surface_area = 2.16 m²
03

Calculate the temperature difference

We need to find the temperature difference between the indoor and outdoor temperatures: Temperature_difference = T_indoor - T_outdoor Temperature_difference = 20°C - (-8°C) Temperature_difference = 28°C
04

Calculate the range of heat loss for the single-pane and quadruple-pane windows

To find the range of heat loss, we'll use the U-factor formula: Heat_loss = U-factor × Surface_area × Temperature_difference For the single-pane window: Heat_loss_single_pane = 6.25 W/m²·K × 2.16 m² × 28°C Heat_loss_single_pane = 386.4 W For the quadruple-pane window: Heat_loss_quadruple_pane = 1.25 W/m²·K × 2.16 m² × 28°C Heat_loss_quadruple_pane = 77.28 W
05

Present the range of heat loss

The range of heat loss through the 1.2 m × 1.8 m window is between 77.28 W (quadruple-pane window) and 386.4 W (single-pane window).

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Key Concepts

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

U-factor
The U-factor is a critical measure in the field of thermal insulation. It quantifies the rate of heat transfer through a building element, such as a window or wall, given a specific temperature difference between the interior and exterior environments. Lower U-factor values indicate better insulating properties as they signify less heat escaping a building.

The U-factor varies based on the type of window. For instance, a quadruple-pane window might have a U-factor as low as 1.25 W/m²·K, while a single-pane window can reach up to 6.25 W/m²·K. This large difference highlights the importance of choosing the right window type for energy efficiency.

The calculation of heat loss using the U-factor is fairly straightforward: multiply the U-factor by the surface area of the window and the temperature difference between the inside and outside. Understanding the U-factor empowers homeowners to make more informed choices about their windows and other building materials, leading to significant energy savings over time.
Heat Transfer
Heat transfer refers to the movement of thermal energy from a warmer object or system to a cooler one. In buildings, this often occurs through windows, walls, and roofs. Effective thermal insulation slows down this process, helping to maintain a steady indoor temperature despite external climate conditions.

The three primary modes of heat transfer are conduction, convection, and radiation.
  • Conduction involves direct transfer of heat through materials, like window glazing.
  • Convection can occur when air or fluid moves, carrying heat with it.
  • Radiation involves heat transfer through electromagnetic waves.
The rate of heat transfer through windows can be significantly reduced by using materials with low U-factors. For instance, modern energy-efficient windows are designed to minimize conduction and convection, often using multiple panes with insulating gases like argon between them to slow down heat transfer.
Building Materials
Building materials play a vital role in determining the thermal performance and energy efficiency of structures. The choice of materials influences the U-factor and consequently the heat transfer capabilities of the building.

Materials such as low-emissivity glass and insulated frames are preferred in energy-efficient window designs. They create a barrier to heat transfer, enhancing a building's thermal insulation.

Quadruple-pane windows filled with inert gases are a prime example of advanced material use. They achieve a markedly lower U-factor than traditional single-pane windows, providing superior insulation. When selecting building materials, consider not only their initial cost but also their long-term impact on energy consumption and thermal comfort.
Energy Efficiency
Energy efficiency in buildings is the goal of reducing energy consumption without sacrificing comfort. Achieving energy efficiency often involves using materials and technologies that minimize heat loss or gain.

Improved energy efficiency has numerous benefits including:
  • Lower energy bills due to reduced heating and cooling needs.
  • A smaller carbon footprint, mitigating climate change effects.
  • Enhanced indoor comfort with stable temperatures year-round.
Investing in high-performance windows with low U-factors can significantly enhance a building's energy efficiency. It's also essential to consider the overall design and integration of materials throughout the building envelope to optimize insulation and minimize thermal bridges, further boosting energy efficiency.

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Most popular questions from this chapter

A transistor with a height of \(0.4 \mathrm{~cm}\) and a diameter of \(0.6 \mathrm{~cm}\) is mounted on a circuit board. The transistor is cooled by air flowing over it with an average heat transfer coefficient of \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the air temperature is \(55^{\circ} \mathrm{C}\) and the transistor case temperature is not to exceed \(70^{\circ} \mathrm{C}\), determine the amount of power this transistor can dissipate safely. Disregard any heat transfer from the transistor base.

What is metabolism? What is the range of metabolic rate for an average man? Why are we interested in the metabolic rate of the occupants of a building when we deal with heating and air conditioning?

We often turn the fan on in summer to help us cool. Explain how a fan makes us feel cooler in the summer. Also explain why some people use ceiling fans also in winter.

Consider steady heat transfer between two large parallel plates at constant temperatures of \(T_{1}=290 \mathrm{~K}\) and \(T_{2}=150 \mathrm{~K}\) that are \(L=2 \mathrm{~cm}\) apart. Assuming the surfaces to be black (emissivity \(\varepsilon=1\) ), determine the rate of heat transfer between the plates per unit surface area assuming the gap between the plates is (a) filled with atmospheric air, \((b)\) evacuated, \((c)\) filled with fiberglass insulation, and \((d)\) filled with superinsulation having an apparent thermal conductivity of \(0.00015 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

A 2.1-m-long, 0.2-cm-diameter electrical wire extends across a room that is maintained at \(20^{\circ} \mathrm{C}\). Heat is generated in the wire as a result of resistance heating, and the surface temperature of the wire is measured to be \(180^{\circ} \mathrm{C}\) in steady operation. Also, the voltage drop and electric current through the wire are measured to be \(110 \mathrm{~V}\) and \(3 \mathrm{~A}\), respectively. Disregarding any heat transfer by radiation, determine the convection heat transfer coefficient for heat transfer between the outer surface of the wire and the air in the room. Answer: \(156 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

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