/*! 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

How do differential equations with constant coefficients differ from those with variable coefficients? Give an example for each type.

Consider a large plane wall of thickness \(L=0.4 \mathrm{~m}\), thermal conductivity \(k=1.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and surface area \(A=\) \(30 \mathrm{~m}^{2}\). The left side of the wall is maintained at a constant temperature of \(T_{1}=90^{\circ} \mathrm{C}\) while the right side loses heat by convection to the surrounding air at \(T_{\infty}=25^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=24 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming constant thermal conductivity and no heat generation in the wall, \((a)\) express the differential equation 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 rate of heat transfer through the wall. Answer: (c) \(7389 \mathrm{~W}\)

Exhaust gases from a manufacturing plant are being discharged through a 10 - \(\mathrm{m}\) tall exhaust stack with outer diameter of \(1 \mathrm{~m}\), wall thickness of \(10 \mathrm{~cm}\), and thermal conductivity of \(40 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The exhaust gases are discharged at a rate of \(1.2 \mathrm{~kg} / \mathrm{s}\), while temperature drop between inlet and exit of the exhaust stack is \(30^{\circ} \mathrm{C}\), and the constant pressure specific heat of the exhaust gasses is \(1600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). On a particular day, the outer surface of the exhaust stack experiences radiation with the surrounding at \(27^{\circ} \mathrm{C}\), and convection with the ambient air at \(27^{\circ} \mathrm{C}\) also, with an average convection heat transfer coefficient of \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Solar radiation is incident on the exhaust stack outer surface at a rate of \(150 \mathrm{~W} / \mathrm{m}^{2}\), and both the emissivity and solar absorptivity of the outer surface are 0.9. Assuming steady one-dimensional heat transfer, (a) obtain the variation of temperature in the exhaust stack wall and (b) determine the inner surface temperature of the exhaust stack.

Consider a steam pipe of length \(L\), inner radius \(r_{1}\), outer radius \(r_{2}\), and constant thermal conductivity \(k\). Steam flows inside the pipe at an average temperature of \(T_{i}\) with a convection heat transfer coefficient of \(h_{i}\). The outer surface of the pipe is exposed to convection to the surrounding air at a temperature of \(T_{0}\) with a heat transfer coefficient of \(h_{o^{*}}\) Assuming steady one-dimensional heat conduction through the pipe, \((a)\) express the differential equation and the boundary conditions for heat conduction through the pipe material, \((b)\) obtain a relation for the variation of temperature in the pipe material by solving the differential equation, and (c) obtain a relation for the temperature of the outer surface of the pipe.

Consider a short cylinder of radius \(r_{o}\) and height \(H\) in which heat is generated at a constant rate of \(\dot{e}_{\text {gen. }}\). Heat is lost from the cylindrical surface at \(r=r_{o}\) by convection to the surrounding medium at temperature \(T_{\infty}\) with a heat transfer coefficient of \(h\). The bottom surface of the cylinder at \(z=0\) is insulated, while the top surface at \(z=H\) is subjected to uniform heat flux \(\dot{q}_{H}\). Assuming constant thermal conductivity and steady two-dimensional heat transfer, express the mathematical formulation (the differential equation and the boundary conditions) of this heat conduction problem. Do not solve.

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