The unit thermal resistances ( \(R\)-values) of both 40-mm and 90-mm vertical air spaces are given in Table 3-9 to be \(0.22 \mathrm{~m}^{2} \cdot \mathrm{C} / \mathrm{W}\), which implies that more than doubling the thickness of air space in a wall has no effect on heat transfer through the wall. Do you think this is a typing error? Explain.

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Explain your reasoning.

Step by step solution

01

Understanding Thermal Resistance (R-value)

Thermal resistance (R-value) represents the ability of a material to resist heat transfer in the presence of a temperature difference. The R-value is directly proportional to the thickness of a material, and its effectiveness is dependent on the material's thermal conductivity. The higher the R-value, the better the insulating performance.
02

Comparing R-values of air spaces with different thicknesses

Now that we have an understanding of R-values, let's examine the given values for 40-mm and 90-mm air spaces. Both thicknesses have the same R-value of 0.22 m²⋅C/W. This implies that increasing the air space thickness by more than double doesn't impact the heat transfer through the wall, which seems counterintuitive.
03

Analyzing the heat transfer in air spaces

For air spaces, the main heat transfer mechanism is conduction through air and radiation from the walls. When we increase the thickness of the air space, the conductive heat transfer through air may reduce as the temperature difference across the space will reduce. However, the radiative heat transfer may not be affected significantly with the increase in air space thickness. Thus, it's possible that increasing the air space thickness may not drastically affect the R-value, as the reduction in conductive heat transfer is compensated by the radiative heat transfer.
04

Concluding whether it's a typing error or not

Considering the analysis of heat transfer in air spaces, it could be concluded that it's not necessarily a typing error. Although it might seem counterintuitive, the R-value not changing significantly with the increase in air space thickness in a wall can be explained by the fact that conductive heat transfer decreases with increasing thickness, while radiative heat transfer remains relatively constant. This causes the overall heat transfer through the wall to not drastically change with the increase in air space thickness.

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Key Concepts

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

Heat Transfer
Heat transfer is a fundamental concept in thermodynamics and refers to the movement of thermal energy from one place to another as a result of temperature differences. This process occurs in three primary ways: conduction, convection, and radiation.

In the context of building insulation, understanding the principles of heat transfer is crucial for evaluating how effectively a material can maintain the desired temperature within a space. When we consider the thermal resistance or R-value of a particular setup, it's a measure of how well that setup can slow down the heat transfer.

For instance, in the provided exercise, two different thicknesses of air spaces both display the same R-value, suggesting that, unexpectedly, increasing the thickness does not improve the thermal resistance significantly. This indicates that there are other factors at play affecting the heat transfer through the air spaces beyond just their thickness.
Thermal Conductivity
Thermal conductivity is a physical property indicating a material's ability to conduct heat. It is usually denoted by the symbol \( k \) and measured in units of \( \frac{W}{m \cdot K} \). Materials with high thermal conductivity, such as metals, are efficient at transferring heat, while those with low thermal conductivity, such as insulating foams, are more effective as insulators.

In the exercise's scenario, the air spaces themselves have low thermal conductivity, meaning they don't transfer heat via conduction very effectively. This property plays a role in the observed R-value, which does not change appreciably with an increase in air space thickness. Essentially, the low thermal conductivity of the air signifies that once you reach a certain thickness, making the air space thicker doesn't significantly disrupt the heat flow, as air is not a good conductor.
Conductive Heat Transfer
Conductive heat transfer is the process by which thermal energy is transmitted through matter, from high temperature regions to cooler ones, via direct contact. Materials with high thermal conductivity facilitate this process more than those with low conductivity.

The exercise discusses the role of air space in conductive heat transfer. As thickness increases, one might expect a proportional decrease in heat flow because there's more material to traverse. However, air's thermal conductivity is relatively low, so once an air space reaches a certain thickness, further increasing it has minimal impact on conductive heat transfer. This explains why the 40-mm and 90-mm air spaces in the exercise have equivalent R-values. It's the limited conductive capabilities of the air that result in a surprising plateau in thermal resistance.
Radiative Heat Transfer
Radiative heat transfer refers to the emission of energy in the form of electromagnetic waves, primarily in the infrared spectrum. Unlike conduction and convection, it does not require a medium to occur and can even happen in a vacuum. Structures gain or lose heat radiatively based on their surface characteristics and temperature.

In the exercise, it's highlighted that radiative heat transfer in air spaces remains largely unchanged regardless of the air space's thickness. This is because the efficiency of radiation depends more on the surfaces' emissivity and the temperature difference between those surfaces, rather than the air space itself. If the enclosing surfaces of the air space have low emissivity or are similar in temperature, adjustments to thickness will not significantly affect the R-value, as the conductive component is already minimal, and the radiative component remains constant.

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Most popular questions from this chapter

Consider a 25-m-long thick-walled concrete duct \((k=\) \(0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of square cross section. The outer dimensions of the duct are \(20 \mathrm{~cm} \times 20 \mathrm{~cm}\), and the thickness of the duct wall is \(2 \mathrm{~cm}\). If the inner and outer surfaces of the duct are at \(100^{\circ} \mathrm{C}\) and \(30^{\circ} \mathrm{C}\), respectively, determine the rate of heat transfer through the walls of the duct. Answer: \(47.1 \mathrm{~kW}\)

The plumbing system of a house involves a \(0.5-\mathrm{m}\) section of a plastic pipe \((k=0.16 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of inner diameter \(2 \mathrm{~cm}\) and outer diameter \(2.4 \mathrm{~cm}\) exposed to the ambient air. During a cold and windy night, the ambient air temperature remains at about \(-5^{\circ} \mathrm{C}\) for a period of \(14 \mathrm{~h}\). The combined convection and radiation heat transfer coefficient on the outer surface of the pipe is estimated to be \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the heat of fusion of water is \(333.7 \mathrm{~kJ} / \mathrm{kg}\). Assuming the pipe to contain stationary water initially at \(0^{\circ} \mathrm{C}\), determine if the water in that section of the pipe will completely freeze that night.

A row of 3 -ft-long and 1-in-diameter used uranium fuel rods that are still radioactive are buried in the ground parallel to each other with a center-to- center distance of 8 in at a depth of \(15 \mathrm{ft}\) from the ground surface at a location where the thermal conductivity of the soil is \(0.6 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\). If the surface temperature of the rods and the ground are \(350^{\circ} \mathrm{F}\) and \(60^{\circ} \mathrm{F}\), respectively, determine the rate of heat transfer from the fuel rods to the atmosphere through the soil.

The boiling temperature of nitrogen at atmospheric pressure at sea level ( 1 atm pressure) is \(-196^{\circ} \mathrm{C}\). Therefore, nitrogen is commonly used in low-temperature scientific studies since the temperature of liquid nitrogen in a tank open to the atmosphere will remain constant at \(-196^{\circ} \mathrm{C}\) until it is depleted. Any heat transfer to the tank will result in the evaporation of some liquid nitrogen, which has a heat of vaporization of \(198 \mathrm{~kJ} / \mathrm{kg}\) and a density of \(810 \mathrm{~kg} / \mathrm{m}^{3}\) at 1 atm. Consider a 3-m-diameter spherical tank that is initially filled with liquid nitrogen at 1 atm and \(-196^{\circ} \mathrm{C}\). The tank is exposed to ambient air at \(15^{\circ} \mathrm{C}\), with a combined convection and radiation heat transfer coefficient of \(35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The temperature of the thin-shelled spherical tank is observed to be almost the same as the temperature of the nitrogen inside. Determine the rate of evaporation of the liquid nitrogen in the tank as a result of the heat transfer from the ambient air if the tank is \((a)\) not insulated, \((b)\) insulated with 5 -cm-thick fiberglass insulation \((k=0.035 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), and (c) insulated with 2 -cm-thick superinsulation which has an effective thermal conductivity of \(0.00005 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

The heat transfer surface area of a fin is equal to the sum of all surfaces of the fin exposed to the surrounding medium, including the surface area of the fin tip. Under what conditions can we neglect heat transfer from the fin tip?

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