Starting with an energy balance on a cylindrical shell volume element, derive the steady one-dimensional heat conduction equation for a long cylinder with constant thermal conductivity in which heat is generated at a rate of \(\dot{e}_{\text {gen }}\).

Short Answer

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Question: Derive the steady one-dimensional heat conduction equation for a long cylinder with constant thermal conductivity and a heat generation rate of \(\dot{e}_{\text {gen }}\). Answer: The steady one-dimensional heat conduction equation for a long cylinder with constant thermal conductivity and a heat generation rate of \(\dot{e}_{\text{gen}}\) is given by \(\frac{d}{dr}(\kappa r \frac{\partial T}{\partial r}) = \dot{e}_{\text{gen}} r\).

Step by step solution

01

Define a cylindrical shell volume element

Consider a cylindrical shell element of radius \(r\), thickness \(\delta r\), and length \(L\). The volume of the cylindrical shell (\(\delta V\)) is given by the formula \(\delta V = 2 \pi r L \delta r\).
02

Energy balance of the cylindrical shell volume element

Express the change in energy within the shell element due to the heat flow and heat generation. The rate of heat flow into the shell is \(-\kappa A_1 \frac{\partial T}{\partial r}\), and the rate of heat flow out of the shell is \(-\kappa A_2 \frac{\partial T}{\partial r}\), where \(\kappa\) is the thermal conductivity, \(A_1 = 2 \pi r L\), and \(A_2 = 2 \pi (r + \delta r) L\). The rate of energy generation within the shell is \(\dot{e}_{\text {gen}} \delta V= \dot{e}_{\text {gen}} (2 \pi r L \delta r)\). During steady-state conditions, the net energy flow into the shell element plus the energy generation within the shell must be equal to the energy flow out of the shell element. Mathematically, this can be expressed as: \(\big(-\kappa A_1 \frac{\partial T}{\partial r}\big) + \big(\dot{e}_{\text{gen}} (2 \pi r L \delta r)\big) = \big(-\kappa A_2 \frac{\partial T}{\partial r}\big)\)
03

Simplify the energy balance equation

Simplify the energy balance equation by canceling out the common terms, \(2 \pi L\), and combining the heat flow terms. \(-(\kappa r \frac{\partial T}{\partial r}) + (\dot{e}_{\text{gen}} r \delta r) = -(\kappa (r + \delta r) \frac{\partial T}{\partial r})\)
04

Take the limit of the equation as \(\delta r\) approaches zero

As the thickness of the cylindrical shell element approaches zero (\(\delta r \to 0\)), the energy balance equation becomes a differential equation. Taking the limit: \(\lim_{\delta r \to 0} \frac{-(\kappa (r + \delta r) \frac{\partial T}{\partial r}) - (\kappa r \frac{\partial T}{\partial r})}{\delta r} =\dot{e}_{\text{gen}} r \)
05

Apply the rules of differentiation

Apply the rules of differentiation to simplify the left side of the equation and obtain the one-dimensional heat conduction equation for a long cylinder: \(\frac{d}{dr}(\kappa r \frac{\partial T}{\partial r}) = \dot{e}_{\text{gen}} r\) This is the steady one-dimensional heat conduction equation for a long cylinder with constant thermal conductivity and a heat generation rate of \(\dot{e}_{\text {gen }}\).

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Key Concepts

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

Energy Balance Equation
The energy balance equation is fundamental when studying heat transfer, especially in systems like a cylindrical body, where heat is being generated or transferred. It is essentially an expression of the first law of thermodynamics, which states that energy cannot be created or destroyed, only transferred or converted from one form to another. For heat conduction in cylinders, we apply this principle to a small volume element of the cylinder.

In this context, the energy balance accounts for all the energy flowing into and out of the element, as well as any heat being generated within it. At steady state, which means there are no changes over time, every bit of energy that enters or is generated inside the volume element must also leave it, ensuring a constant temperature. The exercise's solution steps utilize this concept to express the energy changes through the surfaces of a cylindrical shell and equate them, ultimately leading to a differential equation describing the temperature distribution.
Thermal Conductivity
Thermal conductivity, denoted by the Greek letter kappa \(\kappa\), is a measure of a material's ability to conduct heat. It is a property that quantifies how easily heat is transferred through a material due to a temperature gradient. In the given exercise, \(\kappa\) is assumed to be constant, which simplifies the mathematical treatment of heat conduction.

When dealing with thermal processes, especially in conductive materials, we consider the rate at which heat \(Q\) flows through a substance. This rate depends on the thermal conductivity of the material, the area \(A\) perpendicular to the direction of heat flow, the temperature gradient \(-\frac{\partial T}{\partial r}\), and the thickness of the material. The negative sign indicates that heat flows from hotter to colder regions.
Steady-State Heat Transfer
Steady-state heat transfer refers to a situation where the temperature field in a material does not change over time. Mathematically, this implies that all partial derivatives of temperature with respect to time are zero. When analyzing heat conduction in cylinders for steady state, we assume the system has reached a point where temperatures remain static, and thus, the energy leaving any section of the cylinder equals the energy entering that section, adjusted for any generation or absorption of heat within the section itself.

This concept is crucial for simplifying the exercise as it allows for the assumption that the rate of heat flow into a volume element is equal to the rate of heat flow out of it plus any heat generated inside the element, all without considering time as a variable.
Differential Equation for Heat Conduction
A differential equation describes how a function's value changes in space and possibly in time. In the case of heat conduction, the differential equation arises from applying the limit as the thickness of a cylindrical shell --- represented by \(\delta r\) --- approaches zero.

The process of creating this differential equation involves analyzing how heat transfer rates alter in response to infinitesimally small changes in the radial coordinate \(r\). By doing so, we establish a relationship that expresses the steady-state heat conduction in terms of the change in temperatures across the cylinder's radius. This relationship takes the form of the equation presented in the last step of the exercise and provides a powerful tool for predicting temperature distribution in cylindrical systems.
Cylindrical Coordinates Heat Transfer
Heat transfer in cylindrical coordinates is characterized by variations in temperature across the radial \(r\), angular \(\theta\), and axial \(z\) dimensions. However, in the case of a long cylinder without end effects, such as in the presented exercise, we can simplify and focus solely on radial heat transfer since the angular and axial temperatures are considered constant.

The mathematical treatment takes into account the unique geometry of the cylinder, allowing the derivation of a heat conduction equation specific to this shape. The radius plays a critical role, affecting how heat dissipates. This specialized form of the heat conduction equation takes into account both the radial dimension and any internal heat generation, providing a comprehensive model for analyzing and predicting heat transfer in cylindrical systems.

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Most popular questions from this chapter

Consider a large 3 -cm-thick stainless steel plate \((k=\) \(15.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) in which heat is generated uniformly at a rate of \(5 \times 10^{5} \mathrm{~W} / \mathrm{m}^{3}\). Both sides of the plate are exposed to an environment at \(30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(60 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Explain where in the plate the highest and the lowest temperatures will occur, and determine their values.

When a long section of a compressed air line passes through the outdoors, it is observed that the moisture in the compressed air freezes in cold weather, disrupting and even completely blocking the air flow in the pipe. To avoid this problem, the outer surface of the pipe is wrapped with electric strip heaters and then insulated. Consider a compressed air pipe of length \(L=6 \mathrm{~m}\), inner radius \(r_{1}=3.7 \mathrm{~cm}\), outer radius \(r_{2}=4.0 \mathrm{~cm}\), and thermal conductivity \(k=14 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) equipped with a 300 -W strip heater. Air is flowing through the pipe at an average temperature of \(-10^{\circ} \mathrm{C}\), and the average convection heat transfer coefficient on the inner surface is \(h=30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming 15 percent of the heat generated in the strip heater is lost through the insulation, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the pipe, \((b)\) obtain a relation for the variation of temperature in the pipe material by solving the differential equation, and \((c)\) evaluate the inner and outer surface temperatures of the pipe.

Does heat generation in a solid violate the first law of thermodynamics, which states that energy cannot be created or destroyed? Explain.

Heat is generated uniformly at a rate of \(4.2 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\) in a spherical ball \((k=45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of diameter \(24 \mathrm{~cm}\). The ball is exposed to iced-water at \(0^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(1200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the temperatures at the center and the surface of the ball.

Consider steady one-dimensional heat conduction in a plane wall, long cylinder, and sphere with constant thermal conductivity and no heat generation. Will the temperature in any of these mediums vary linearly? Explain.

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