In the United States, building insulation is specified by the \(R\)-value (thermal resistance in \(\mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F} /\) Btu units). A homeowner decides to save on the cost of heating the home by adding additional insulation in the attic. If the total \(R\)-value is increased from 15 to 25 , the homeowner can expect the heat loss through the ceiling to be reduced by (a) \(25 \%\) (b) \(40 \%\) (c) \(50 \%\) (d) \(60 \%\) (e) \(75 \%\)

Short Answer

Expert verified
Answer: The percentage reduction in heat loss is 40%.

Step by step solution

01

Find the heat loss ratios

To find the heat loss ratio, we'll need to divide the initial R-value by the final R-value. Heat loss ratio = \(\frac{Initial R-value}{Final R-value}\).
02

Calculate the heat loss ratio with given R-values

Given the initial R-value of 15 and the final R-value of 25, we can calculate the heat loss ratio as: Heat loss ratio = \(\frac{15}{25}\).
03

Simplify the heat loss ratio

Now, we simplify the heat loss ratio: Heat loss ratio = \(\frac{3}{5}\).
04

Calculate the percentage reduction in heat loss

To find the percentage reduction in heat loss, multiply the heat loss ratio by 100 and subtract it from 100%: Percentage reduction in heat loss = \(100\% - (Heat loss ratio \times 100\%)\).
05

Plug in the heat loss ratio and find the answer

We have the heat loss ratio \(\frac{3}{5}\), so the percentage reduction in heat loss is: Percentage reduction in heat loss = \(100\% - (\frac{3}{5}\times 100\%) = 100\% - 60\% = 40\%\). The answer is (b) \(40 \%\).

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Key Concepts

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

Thermal Resistance
Thermal resistance, symbolized by an R-value, is a measurement of a material's ability to resist heat flow. The higher the R-value, the better the insulation's ability to prevent heat transfer.

Imagine a thermal barrier between two environments at different temperatures; the insulation slows down the heat trying to equalize the temperatures across this barrier. In the context of home insulation, materials with higher R-values will slow down the transfer of heat from inside the home to the outside during the winter, and vice versa during the summer.

Real-world Importance

The practical significance of thermal resistance comes into play when choosing materials for insulating homes. The chosen materials should have R-values that align with the climate and the specific heating and cooling needs of the house. In cold climates, a higher R-value is essential to keep heat indoors, while in milder climates, a lower R-value might be sufficient.

When selecting insulation, it's not just the R-value that matters, but also the insulation's thickness, density, and installation quality, as these factors will influence the overall effectiveness of the insulation in resisting heat flow.
Heat Loss Calculation
Calculating heat loss is critical for determining how well a house retains heat, and for making decisions about heating systems and insulation. Heat loss through a structure is influenced by several factors, including the thermal resistance of the materials, the temperature difference between inside and outside, and the surface area through which heat can escape.

Understanding the Calculation

To calculate the heat loss, one must understand the relationship between the R-value and the rate of heat loss. In essence, a low R-value means more heat loss, while a higher R-value suggests less heat escape. The heat loss calculation makes it possible to estimate the efficiency of insulation and the potential savings on energy costs by enhancing insulation.

For homeowners considering upgrades, a heat loss calculation can reveal how much improvement in insulation will reduce the overall heat loss. It solidifies the decision on whether an investment in insulation will yield substantial savings on energy expenses in the long run.
Home Insulation Efficiency
Home insulation efficiency refers to how effectively a building's insulation reduces energy consumption by minimizing heat transfer. The performance of insulation is a key factor in achieving energy efficiency in homes, which not only helps lower utility bills but also lessens environmental impact.

Maximizing Efficiency

To maximize insulation efficiency, one must consider the entire 'envelope' of the home, which includes the attic, walls, floors, windows, and doors. High-efficiency insulation maintains the desired temperature inside the home more consistently, requiring less energy for heating and cooling.

  • Upgrade Insulation: Increasing the R-value of insulation in key areas, like the attic and walls, can significantly reduce energy costs.
  • Seal Gaps: Ensuring that there are no gaps or leaks where air can pass through is crucial for maintaining the insulation's effectiveness.
  • Insulation Maintenance: Regular checks and proper maintenance of insulation can prevent degradation over time.
By focusing on these aspects, homeowners can enhance their home's insulation efficiency, leading to more sustainable living and cost savings.

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Most popular questions from this chapter

Cold conditioned air at \(12^{\circ} \mathrm{C}\) is flowing inside a \(1.5\)-cm- thick square aluminum \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) duct of inner cross section \(22 \mathrm{~cm} \times 22 \mathrm{~cm}\) at a mass flow rate of \(0.8 \mathrm{~kg} / \mathrm{s}\). The duct is exposed to air at \(33^{\circ} \mathrm{C}\) with a combined convection-radiation heat transfer coefficient of \(13 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The convection heat transfer coefficient at the inner surface is \(75 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the air temperature in the duct should not increase by more than \(1^{\circ} \mathrm{C}\) determine the maximum length of the duct.

A hot plane surface at \(100^{\circ} \mathrm{C}\) is exposed to air at \(25^{\circ} \mathrm{C}\) with a combined heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The heat loss from the surface is to be reduced by half by covering it with sufficient insulation with a thermal conductivity of \(0.10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assuming the heat transfer coefficient to remain constant, the required thickness of insulation is (a) \(0.1 \mathrm{~cm}\) (b) \(0.5 \mathrm{~cm}\) (c) \(1.0 \mathrm{~cm}\) (d) \(2.0 \mathrm{~cm}\) (e) \(5 \mathrm{~cm}\)

Consider a flat ceiling that is built around \(38-\mathrm{mm} \times\) \(90-\mathrm{mm}\) wood studs with a center-to-center distance of \(400 \mathrm{~mm}\). The lower part of the ceiling is finished with 13-mm gypsum wallboard, while the upper part consists of a wood subfloor \(\left(R=0.166 \mathrm{~m}^{2} \cdot{ }^{\circ} \mathrm{C} / \mathrm{W}\right)\), a \(13-\mathrm{mm}\) plywood, a layer of felt \(\left(R=0.011 \mathrm{~m}^{2} \cdot{ }^{\circ} \mathrm{C} / \mathrm{W}\right)\), and linoleum \(\left(R=0.009 \mathrm{~m}^{2} \cdot{ }^{\circ} \mathrm{C} / \mathrm{W}\right)\). Both sides of the ceiling are exposed to still air. The air space constitutes 82 percent of the heat transmission area, while the studs and headers constitute 18 percent. Determine the winter \(R\)-value and the \(U\)-factor of the ceiling assuming the 90 -mm-wide air space between the studs ( \(a\) ) does not have any reflective surface, (b) has a reflective surface with \(\varepsilon=0.05\) on one side, and ( ) has reflective surfaces with \(\varepsilon=0.05\) on both sides. Assume a mean temperature of \(10^{\circ} \mathrm{C}\) and a temperature difference of \(5.6^{\circ} \mathrm{C}\) for the air space.

The fins attached to a surface are determined to have an effectiveness of \(0.9\). Do you think the rate of heat transfer from the surface has increased or decreased as a result of the addition of these fins?

A total of 10 rectangular aluminum fins \((k=\) \(203 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) are placed on the outside flat surface of an electronic device. Each fin is \(100 \mathrm{~mm}\) wide, \(20 \mathrm{~mm}\) high and \(4 \mathrm{~mm}\) thick. The fins are located parallel to each other at a center- tocenter distance of \(8 \mathrm{~mm}\). The temperature at the outside surface of the electronic device is \(60^{\circ} \mathrm{C}\). The air is at \(20^{\circ} \mathrm{C}\), and the heat transfer coefficient is \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine \((a)\) the rate of heat loss from the electronic device to the surrounding air and \((b)\) the fin effectiveness.

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