/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 Heat is generated in a long wire... [FREE SOLUTION] | 91Ó°ÊÓ

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Heat is generated in a long wire of radius \(r_{o}\) at a constant rate of \(\dot{e}_{\text {gen }}\) per unit volume. The wire is covered with a plastic insulation layer. Express the heat flux boundary condition at the interface in terms of the heat generated.

Short Answer

Expert verified
Answer: The heat flux boundary condition at the interface between a wire and its insulation layer in terms of the heat generated in the wire is given by the equation: \(\frac{dT}{dr} = -\frac{\dot{e}_{\text {gen }} \cdot \pi {r_{o}}^2}{k}\)

Step by step solution

01

Determine the amount of heat generated per unit length

Given that heat is generated at a constant rate of \(\dot{e}_{\text {gen }}\) per unit volume in the wire, we can determine the total heat generated per unit length of the wire by multiplying the rate of heat generation by the volume per unit length of the wire: \(q_{gen} = \dot{e}_{\text {gen }} \cdot \pi {r_{o}}^2\) Where \(q_{gen}\) is the heat generated per unit length and \({r_{o}}^2\) is the cross-sectional area of the wire.
02

Apply Fourier's Law of Heat Conduction

Fourier's Law states that the heat flux (\(q''\)) is equal to the product of thermal conductivity (\(k\)) and the temperature gradient: \(q'' = -k \frac{dT}{dr}\) At the interface between the wire and the insulation layer, we are interested in the radial heat flux (\(q''_r\)). The temperature gradient will be defined as the change in temperature with respect to the radial distance.
03

Establish the boundary condition

In order to establish the boundary condition at the interface, we must equate the heat generated per unit length in the wire (\(q_{gen}\)) to the radial heat flux (\(q''_r\)) at the interface. This is because all heat generated in the wire has to be conducted through the interface to the insulation layer: \(q_{gen} = q''_r\) Substituting Step 1 and Step 2, we get: \(\dot{e}_{\text {gen }} \cdot \pi {r_{o}}^2 = -k \frac{dT}{dr}\)
04

Express the heat flux boundary condition

We will now express the heat flux boundary condition at the interface in terms of the heat generated in the wire. First, rearrange the equation from Step 3 to solve for \(\frac{dT}{dr}\): \(\frac{dT}{dr} = -\frac{\dot{e}_{\text {gen }} \cdot \pi {r_{o}}^2}{k}\) This equation represents the heat flux boundary condition at the interface between the wire and the insulation layer in terms of the heat generated in the wire.

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Key Concepts

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

Fourier's Law of Heat Conduction
Understanding the behavior of heat transfer is essential in many engineering applications, and Fourier's Law of Heat Conduction is the cornerstone of this understanding.

At its core, Fourier's Law states that the rate of heat transfer (heat flux, represented as \(q''\)) through a material is directly proportional to the negative of the temperature gradient across the material. In simpler terms, heat naturally flows from areas of high temperature to areas of lower temperature. This law can be mathematically represented as: \[q'' = -k \frac{dT}{dx}\]where \(k\) represents the material's thermal conductivity, and \(\frac{dT}{dx}\) is the temperature gradient in the direction of heat flow. In our exercise, we focused on the radial direction, which means the heat flux depends on changes in temperature with respect to the radial position, \(r\).
Thermal Conductivity
Thermal conductivity, symbolized by \(k\), is a fundamental property of materials that quantifies their ability to conduct heat.

Materials with high thermal conductivity, like metals, are able to transfer heat efficiently, making them ideal for cooking utensils and radiators. In contrast, materials with low thermal conductivity, such as plastics, are poor heat conductors and serve well as insulators in thermal applications. The units of thermal conductivity are watts per meter-kelvin (\(W/m\cdot K\)). It's important to note that \(k\) is not only intrinsic to the material but can also vary with temperature, pressure, and other factors.
Radial Heat Flux
In our problem, we deal with a cylindrical wire, indicating that heat moves radially outward from the center to the surface. This type of heat transfer is called radial heat flux, denoted as \(q''_r\). Unlike linear or planar heat flow, radial heat flow accounts for the curved geometry of the wire.

Calculating Radial Heat Flux

The calculation for radial heat flux requires understanding of how temperatures change over the radius of the cylinder. Because of the cylindrical geometry, the surface area through which heat is transferred increases with radius. Therefore, the radial heat flux diminishes as the distance from the axis increases, assuming heat generation is uniform.
Heat Generation Rate
In our exercise, the heat generation rate per unit volume is represented by \(\dot{e}_{\text {gen }}\). This tells us how much heat is produced by the wire's volume over time, a common scenario in electrical applications where resistance heating occurs.

For a cylindrical volume, we can relate this volumetric heat generation rate to the total heat produced per unit length of the wire, represented by \(q_{gen}\), with the following equation: \[q_{gen} = \dot{e}_{\text {gen }} \cdot \pi {r_{o}}^2\]Here, the term \(\pi {r_{o}}^2\) gives us the cross-sectional area through which the heat is generated. In thermal analysis, knowing the heat generation rate is crucial for predicting temperature distributions and ensuring the design's thermal performance meets the necessary requirements.

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Most popular questions from this chapter

A large steel plate having a thickness of \(L=4\) in, thermal conductivity of \(k=7.2 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\), and an emissivity of \(\varepsilon=0.7\) is lying on the ground. The exposed surface of the plate at \(x=L\) is known to exchange heat by convection with the ambient air at \(T_{\infty}=90^{\circ} \mathrm{F}\) with an average heat transfer coefficient of \(h=12 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\) as well as by radiation with the open sky with an equivalent sky temperature of \(T_{\text {sky }}=480 \mathrm{R}\). Also, the temperature of the upper surface of the plate is measured to be \(80^{\circ} \mathrm{F}\). Assuming steady onedimensional heat transfer, \((a)\) express the differential equation and the boundary conditions for heat conduction through the plate, \((b)\) obtain a relation for the variation of temperature in the plate by solving the differential equation, and \((c)\) determine the value of the lower surface temperature of the plate at \(x=0\).

Consider a large 5-cm-thick brass plate \((k=\) \(111 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) in which heat is generated uniformly at a rate of \(2 \times 10^{5} \mathrm{~W} / \mathrm{m}^{3}\). One side of the plate is insulated while the other side is exposed to an environment at \(25^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(44 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Explain where in the plate the highest and the lowest temperatures will occur, and determine their values.

A spherical container of inner radius \(r_{1}=2 \mathrm{~m}\), outer radius \(r_{2}=2.1 \mathrm{~m}\), and thermal conductivity \(k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is filled with iced water at \(0^{\circ} \mathrm{C}\). The container is gaining heat by convection from the surrounding air at \(T_{\infty}=25^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=18 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming the inner surface temperature of the container to be \(0^{\circ} \mathrm{C},(a)\) express the differential equation and the boundary conditions for steady one- dimensional heat conduction through the container, \((b)\) obtain a relation for the variation of temperature in the container by solving the differential equation, and \((c)\) evaluate the rate of heat gain to the iced water.

Consider a 20-cm-thick large concrete plane wall \((k=0.77 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) subjected to convection on both sides with \(T_{\infty 1}=22^{\circ} \mathrm{C}\) and \(h_{1}=8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on the inside, and \(T_{\infty 2}=8^{\circ} \mathrm{C}\) and \(h_{2}=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on the outside. Assuming constant thermal conductivity with no heat generation and negligible radiation, \((a)\) express the differential equations and the boundary conditions for steady one-dimensional heat conduction through the wall, \((b)\) obtain a relation for the variation of temperature in the wall by solving the differential equation, and \((c)\) evaluate the temperatures at the inner and outer surfaces of the wall.

A solar heat flux \(\dot{q}_{s}\) is incident on a sidewalk whose thermal conductivity is \(k\), solar absorptivity is \(\alpha_{s}\), and convective heat transfer coefficient is \(h\). Taking the positive \(x\) direction to be towards the sky and disregarding radiation exchange with the surroundings surfaces, the correct boundary condition for this sidewalk surface is (a) \(-k \frac{d T}{d x}=\alpha_{s} \dot{q}_{s}\) (b) \(-k \frac{d T}{d x}=h\left(T-T_{\infty}\right)\) (c) \(-k \frac{d T}{d x}=h\left(T-T_{\infty}\right)-\alpha_{s} \dot{q}_{s}\) (d) \(h\left(T-T_{\infty}\right)=\alpha_{s} \dot{q}_{s}\) (e) None of them

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