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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.

Short Answer

Expert verified
Answer: The temperature varies linearly in a plane wall but not in a long cylinder or sphere under the given conditions.

Step by step solution

01

Governing Equations for Heat Conduction

For each medium, the governing equation for the steady one-dimensional heat conduction is given by: 1. Plane Wall: \(\frac{d^2T}{dx^2} = 0\) 2. Long Cylinder: \(\frac{1}{r} \frac{d}{dr} (r \frac{dT}{dr}) = 0\) 3. Sphere: \(\frac{1}{r^2} \frac{d}{dr} (r^2 \frac{dT}{dr}) = 0\) Now, let's analyze each equation to determine if the temperature change is linear for each medium.
02

Plane Wall Temperature Distribution

For a plane wall, the governing equation is \(\frac{d^2T}{dx^2} = 0\). To solve for T, we need to integrate twice with respect to x: 1. \(\int \frac{d^2T}{dx^2} dx = \int 0 dx \Rightarrow \frac{dT}{dx} = C_1\) 2. \(\int \frac{dT}{dx} dx = \int C_1 dx \Rightarrow T(x) = C_1x + C_2\) As we can observe, the temperature distribution T(x) in a plane wall is linear since it is in the form T(x) = C_1x + C_2.
03

Long Cylinder Temperature Distribution

For a long cylinder, the governing equation is \(\frac{1}{r} \frac{d}{dr} (r \frac{dT}{dr}) = 0\). To solve for T, we first rearrange and integrate with respect to r: 1. \(\frac{d}{dr} (r \frac{dT}{dr}) = 0 \Rightarrow r \frac{dT}{dr} = C_1\) 2. \(\frac{dT}{dr} = \frac{C_1}{r} \Rightarrow \int \frac{dT}{dr} dr = \int \frac{C_1}{r} dr\) 3. \(T(r) = C_1 \ln r + C_2\) Here, the temperature distribution T(r) in a long cylinder is not linear, since the function contains a natural logarithm.
04

Spherical Temperature Distribution

For a sphere, the governing equation is \(\frac{1}{r^2} \frac{d}{dr} (r^2 \frac{dT}{dr}) = 0\). To solve for T, we first rearrange and integrate with respect to r: 1. \(\frac{d}{dr} (r^2 \frac{dT}{dr}) = 0 \Rightarrow r^2 \frac{dT}{dr} = C_1\) 2. \(\frac{dT}{dr} = \frac{C_1}{r^2} \Rightarrow \int \frac{dT}{dr} dr = \int \frac{C_1}{r^2} dr\) 3. \(T(r) = -\frac{C_1}{r} + C_2\) In this case, the temperature distribution T(r) in a sphere is also not linear, since the function contains an inverse relationship.
05

Conclusion

In conclusion, the temperature will vary linearly in a plane wall and will not vary linearly in a long cylinder and sphere under steady one-dimensional heat conduction with constant thermal conductivity and no heat generation.

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Key Concepts

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

Temperature Distribution
Understanding temperature distribution is key when studying heat conduction in various media such as plane walls, cylinders, and spheres. Temperature distribution refers to how temperature varies within a material at a steady state, meaning when there is no change in temperature over time.

In a plane wall, we find that the temperature changes uniformly from one point to another, resulting in a linear temperature profile. As explained in the exercise's solution, mathematically, it is expressed as a straight-line equation, which in this case arises from the second derivative of temperature with respect to space equaling zero. Simple integration gives us a linear function.

However, in cylindrical and spherical coordinates, the temperature distribution is influenced by geometry. For a long cylinder, the rate of change of temperature is a function of the natural logarithm of the radius, indicating non-linear distribution. Similarly, a spherical medium's temperature varies inversely with the radius. Consequently, these cases do not result in a linear temperature profile but rather follow logarithmic and inverse laws, respectively.

These differences are crucial to consider in applications such as insulation materials and heat exchangers, where uniform temperature distribution can be critical for efficient operation and safety.
Thermal Conductivity
Thermal conductivity is a fundamental property of materials that quantifies their ability to conduct heat. It is typically denoted by the symbol 'k' and is measured in watts per meter-kelvin (W/mK).

The rate at which heat is transferred through a material by conduction is directly proportional to the temperature gradient (the rate at which temperature changes with distance) and the thermal conductivity. This relationship is described by Fourier's law of heat conduction. In mathematical terms, the rate of heat transfer 'q' across a material can be described by the equation: \( q = -k \frac{dT}{dx} \), where \( \frac{dT}{dx} \) represents the temperature gradient in the direction x.

The 'negative' sign indicates that heat flows from higher to lower temperatures. High thermal conductivity materials, such as metals, efficiently transfer heat, while low conductivity materials, like insulation foams, inhibit heat flow. Understanding and utilizing these properties are vital when designing systems for thermal management.
Heat Conduction in Solids
The phenomenon of heat conduction in solids is the transfer of thermal energy within solid materials without the physical movement of the substance. It occurs at the atomic and molecular levels as vibrational energy or as moving free electrons in the case of metals.

The mechanism of conduction depends highly on interactions between particles; in non-metals, vibrations of atoms within the lattice structure are responsible for the transfer of heat. Metals, with their free electrons, act differently. These electrons can move more easily through the lattice, carrying energy as they go.

Solving heat conduction problems typically involves using the heat equation, which is derived from the principle of energy conservation. The heat equation in solids is a partial differential equation that describes how temperature varies within a material over time. For steady-state conduction, the temperature does not change over time, simplifying our calculations to conditions observed in the given exercises. Whether the conduction is through a plane wall, a cylinder, or a sphere, understanding the principles of heat conduction assists in practical applications such as electronics cooling, building design, and thermal protection systems.

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Most popular questions from this chapter

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.

Heat is generated in a \(3-\mathrm{cm}\)-diameter spherical radioactive material uniformly at a rate of \(15 \mathrm{~W} / \mathrm{cm}^{3}\). Heat is dissipated to the surrounding medium at \(25^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The surface temperature of the material in steady operation is (a) \(56^{\circ} \mathrm{C}\) (b) \(84^{\circ} \mathrm{C}\) (c) \(494^{\circ} \mathrm{C}\) (d) \(650^{\circ} \mathrm{C}\) (e) \(108^{\circ} \mathrm{C}\)

Consider a chilled-water pipe of length \(L\), inner radius \(r_{1}\), outer radius \(r_{2}\), and thermal conductivity \(k\). Water flows in the pipe at a temperature \(T_{f}\) and the heat transfer coefficient at the inner surface is \(h\). If the pipe is well-insulated on the outer surface, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the pipe and \((b)\) obtain a relation for the variation of temperature in the pipe by solving the differential equation.

A spherical metal ball of radius \(r_{o}\) is heated in an oven to a temperature of \(T_{i}\) throughout and is then taken out of the oven and dropped into a large body of water at \(T_{\infty}\) where it is cooled by convection with an average convection heat transfer coefficient of \(h\). Assuming constant thermal conductivity and transient one-dimensional heat transfer, express the mathematical formulation (the differential equation and the boundary and initial conditions) of this heat conduction problem. Do not solve.

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