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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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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

A 1.4-m-diameter spherical steel tank filled with iced water at \(0^{\circ} \mathrm{C}\) is buried underground at a location where the thermal conductivity of the soil is \(k=0.55 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The distance between the tank center and the ground surface is \(2.4 \mathrm{~m}\). For ground surface temperature of \(18^{\circ} \mathrm{C}\), determine the rate of heat transfer to the iced water in the tank. What would your answer be if the soil temperature were \(18^{\circ} \mathrm{C}\) and the ground surface were insulated?

A 6-m-diameter spherical tank is filled with liquid oxygen \(\left(\rho=1141 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=1.71 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\) at \(-184^{\circ} \mathrm{C}\). It is observed that the temperature of oxygen increases to \(-183^{\circ} \mathrm{C}\) in a 144-hour period. The average rate of heat transfer to the tank is (a) \(249 \mathrm{~W}\) (b) \(426 \mathrm{~W}\) (c) \(570 \mathrm{~W}\) (d) \(1640 \mathrm{~W}\) (e) \(2207 \mathrm{~W}\)

We are interested in steady state heat transfer analysis from a human forearm subjected to certain environmental conditions. For this purpose consider the forearm to be made up of muscle with thickness \(r_{m}\) with a skin/fat layer of thickness \(t_{s f}\) over it, as shown in the Figure P3-138. For simplicity approximate the forearm as a one-dimensional cylinder and ignore the presence of bones. The metabolic heat generation rate \(\left(\dot{e}_{m}\right)\) and perfusion rate \((\dot{p})\) are both constant throughout the muscle. The blood density and specific heat are \(\rho_{b}\) and \(c_{b}\), respectively. The core body temperate \(\left(T_{c}\right)\) and the arterial blood temperature \(\left(T_{a}\right)\) are both assumed to be the same and constant. The muscle and the skin/fat layer thermal conductivities are \(k_{m}\) and \(k_{s f}\), respectively. The skin has an emissivity of \(\varepsilon\) and the forearm is subjected to an air environment with a temperature of \(T_{\infty}\), a convection heat transfer coefficient of \(h_{\text {conv }}\), and a radiation heat transfer coefficient of \(h_{\mathrm{rad}}\). Assuming blood properties and thermal conductivities are all constant, \((a)\) write the bioheat transfer equation in radial coordinates. The boundary conditions for the forearm are specified constant temperature at the outer surface of the muscle \(\left(T_{i}\right)\) and temperature symmetry at the centerline of the forearm. \((b)\) Solve the differential equation and apply the boundary conditions to develop an expression for the temperature distribution in the forearm. (c) Determine the temperature at the outer surface of the muscle \(\left(T_{i}\right)\) and the maximum temperature in the forearm \(\left(T_{\max }\right)\) for the following conditions: $$ \begin{aligned} &r_{m}=0.05 \mathrm{~m}, t_{s f}=0.003 \mathrm{~m}, \dot{e}_{m}=700 \mathrm{~W} / \mathrm{m}^{3}, \dot{p}=0.00051 / \mathrm{s} \\ &T_{a}=37^{\circ} \mathrm{C}, T_{\text {co }}=T_{\text {surr }}=24^{\circ} \mathrm{C}, \varepsilon=0.95 \\ &\rho_{b}=1000 \mathrm{~kg} / \mathrm{m}^{3}, c_{b}=3600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k_{m}=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\ &k_{s f}=0.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, h_{\text {conv }}=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}, h_{\mathrm{rad}}=5.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \end{aligned} $$

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: \(47.1 \mathrm{~kW}\)

A \(2.5 \mathrm{~m}\)-high, 4-m-wide, and 20 -cm-thick wall of a house has a thermal resistance of \(0.0125^{\circ} \mathrm{C} / \mathrm{W}\). The thermal conductivity of the wall is (a) \(0.72 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (b) \(1.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (c) \(1.6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (d) \(16 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (e) \(32 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\)

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