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Why do we characterize the heat conduction ability of insulators in terms of their apparent thermal conductivity instead of their ordinary thermal conductivity?

Short Answer

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Answer: Insulators are characterized by their apparent thermal conductivity because it takes into account the combined effects of conduction, radiation, and convection, while ordinary thermal conductivity only measures the conduction of heat. As insulating materials often have low ordinary thermal conductivity values, other heat transfer mechanisms, such as radiation and convection, become more significant. Apparent thermal conductivity gives a more accurate representation of the insulators' overall heat conduction ability in practical applications.

Step by step solution

01

Define ordinary thermal conductivity

Ordinary thermal conductivity is the property of a material that quantifies its ability to conduct heat. It is typically denoted by the symbol 'k' and is expressed in watts per meter-kelvin (W/mK). This value is constant for a given material and can be used to determine the rate of heat transfer through the material under steady-state conditions.
02

Define apparent thermal conductivity

Apparent thermal conductivity, on the other hand, is a property that takes into account both the conduction of heat through the material as well as other heat transfer mechanisms such as radiation and convection. It is also expressed in watts per meter-kelvin (W/mK) but represents an effective value that is not solely dependent on the material's intrinsic conduction ability.
03

Differences between ordinary and apparent thermal conductivity

Ordinary thermal conductivity measures only the conduction of heat through a material and assumes that other heat transfer mechanisms are negligible. However, this may not always be true, especially for insulating materials, since they often have low ordinary thermal conductivity values and other heat transfer mechanisms become more significant. Apparent thermal conductivity takes these additional factors into account and gives a more accurate representation of the overall heat conduction ability of the material.
04

Suitability of apparent thermal conductivity for insulators

Insulators are materials that typically have low ordinary thermal conductivity values. As in insulating materials, the effects of radiation and convection could be more significant than conduction itself. Thus, using ordinary thermal conductivity to describe an insulator's ability to conduct heat may not give an accurate representation of its performance in practical applications. On the other hand, apparent thermal conductivity takes into account the combined effects of conduction, radiation, and convection, resulting in a more accurate value for insulators. This is why we characterize insulators' heat conduction ability using apparent thermal conductivity instead of their ordinary thermal conductivity.

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Most popular questions from this chapter

The critical heat flux (CHF) is a thermal limit at which a boiling crisis occurs whereby an abrupt rise in temperature causes overheating on a fuel rod surface that leads to damage. A cylindrical fuel rod \(2 \mathrm{~cm}\) in diameter is encased in a concentric tube and cooled by water. The fuel generates heat uniformly at a rate of \(150 \mathrm{MW} / \mathrm{m}^{3}\). The average temperature of the cooling water, sufficiently far from the fuel rod, is \(80^{\circ} \mathrm{C}\). The operating pressure of the cooling water is such that the surface temperature of the fuel rod must be kept below \(300^{\circ} \mathrm{C}\) to prevent the cooling water from reaching the critical heat flux. Determine the necessary convection heat transfer coefficient to prevent the critical heat flux from occurring.

Using the conversion factors between W and Btu/h, \(\mathrm{m}\) and \(\mathrm{ft}\), and \(\mathrm{K}\) and \(\mathrm{R}\), express the Stefan-Boltzmann constant $\sigma=5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^{4}\( in the English unit \)\mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot \mathrm{R}^{4}$.

A 1000-W iron is left on an ironing board with its base exposed to the air at \(20^{\circ} \mathrm{C}\). The convection heat transfer coefficient between the base surface and the surrounding air is $35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\(. If the base has an emissivity of \)0.6$ and a surface area of \(0.02 \mathrm{~m}^{2}\), determine the temperature of the base of the iron. Answer: \(674^{\circ} \mathrm{C}\)

Water enters a pipe at \(20^{\circ} \mathrm{C}\) at a rate of $0.25 \mathrm{~kg} / \mathrm{s}\( and is heated to \)60^{\circ} \mathrm{C}$. The rate of heat transfer to the water is (a) \(10 \mathrm{~kW}\) (b) \(20.9 \mathrm{~kW}\) (c) \(41.8 \mathrm{~kW}\) (d) \(62.7 \mathrm{~kW}\) (e) \(167.2 \mathrm{~kW}\)

A solid plate, with a thickness of \(15 \mathrm{~cm}\) and a thermal conductivity of \(80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), is being cooled at the upper surface by air. The air temperature is $10^{\circ} \mathrm{C}$, while the temperatures at the upper and lower surfaces of the plate are 50 and \(60^{\circ} \mathrm{C}\), respectively. Determine the convection heat transfer coefficient of air at the upper surface, and discuss whether the value is reasonable or not for forced convection of air.

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