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Chapter 3: Problem 224

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

Hot water at an average temperature of \(70^{\circ} \mathrm{C}\) is flowing through a \(15-\mathrm{m}\) section of a cast iron pipe \((k=52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) whose inner and outer diameters are \(4 \mathrm{~cm}\) and \(4.6 \mathrm{~cm}\), respectively. The outer surface of the pipe, whose emissivity is \(0.7\), is exposed to the cold air at \(10^{\circ} \mathrm{C}\) in the basement, with a heat transfer coefficient of \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The heat transfer coefficient at the inner surface of the pipe is \(120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Taking the walls of the basement to be at \(10^{\circ} \mathrm{C}\) also, determine the rate of heat loss from the hot water. Also, determine the average velocity of the water in the pipe if the temperature of the water drops by \(3^{\circ} \mathrm{C}\) as it passes through the basement.

Determine the winter \(R\)-value and the \(U\)-factor of a masonry cavity wall that is built around 4-in-thick concrete blocks made of lightweight aggregate. The outside is finished with 4 -in face brick with \(\frac{1}{2}\)-in cement mortar between the bricks and concrete blocks. The inside finish consists of \(\frac{1}{2}\)-in gypsum wallboard separated from the concrete block by \(\frac{3}{4}\)-in-thick (1-in by 3 -in nominal) vertical furring whose center- to-center distance is 16 in. Neither side of the \(\frac{3}{4}\)-in-thick air space between the concrete block and the gypsum board is coated with any reflective film. When determining the \(R\)-value of the air space, the temperature difference across it can be taken to be \(30^{\circ} \mathrm{F}\) with a mean air temperature of \(50^{\circ} \mathrm{F}\). The air space constitutes 80 percent of the heat transmission area, while the vertical furring and similar structures constitute 20 percent.

Determine the winter \(R\)-value and the \(U\)-factor of a masonry wall that consists of the following layers: \(100-\mathrm{mm}\) face bricks, 100 -mm common bricks, \(25-\mathrm{mm}\) urethane rigid foam insulation, and \(13-\mathrm{mm}\) gypsum wallboard.

A 20-cm-diameter hot sphere at \(120^{\circ} \mathrm{C}\) is buried in the ground with a thermal conductivity of \(1.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The distance between the center of the sphere and the ground surface is \(0.8 \mathrm{~m}\) and the ground surface temperature is \(15^{\circ} \mathrm{C}\). The rate of heat loss from the sphere is (a) \(169 \mathrm{~W}\) (b) \(20 \mathrm{~W}\) (c) \(217 \mathrm{~W}\) (d) \(312 \mathrm{~W}\) (e) \(1.8 \mathrm{~W}\)

A \(0.2\)-cm-thick, 10-cm-high, and 15 -cm-long circuit board houses electronic components on one side that dissipate a total of \(15 \mathrm{~W}\) of heat uniformly. The board is impregnated with conducting metal fillings and has an effective thermal conductivity of \(12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). All the heat generated in the components is conducted across the circuit board and is dissipated from the back side of the board to a medium at \(37^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(45 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the surface temperatures on the two sides of the circuit board. (b) Now a 0.1-cm-thick, 10-cm-high, and 15 -cm-long aluminum plate \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with \(200.2\)-cm-thick, 2-cm-long, and \(15-\mathrm{cm}\)-wide aluminum fins of rectangular profile are attached to the back side of the circuit board with a \(0.03-\mathrm{cm}-\) thick epoxy adhesive \((k=1.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). Determine the new temperatures on the two sides of the circuit board.
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