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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

Consider steady one-dimensional heat conduction through a plane wall, a cylindrical shell, and a spherical shell of uniform thickness with constant thermophysical properties and no thermal energy generation. The geometry in which the variation of temperature in the direction of heat transfer will be linear is (a) plane wall (b) cylindrical shell (c) spherical shell (d) all of them (e) none of them

Heat is generated in a long \(0.3-\mathrm{cm}\)-diameter cylindrical electric heater at a rate of \(180 \mathrm{~W} / \mathrm{cm}^{3}\). The heat flux at the surface of the heater in steady operation is (a) \(12.7 \mathrm{~W} / \mathrm{cm}^{2}\) (b) \(13.5 \mathrm{~W} / \mathrm{cm}^{2}\) (c) \(64.7 \mathrm{~W} / \mathrm{cm}^{2}\) (d) \(180 \mathrm{~W} / \mathrm{cm}^{2}\) (e) \(191 \mathrm{~W} / \mathrm{cm}^{2}\)

In a food processing facility, a spherical container of inner radius \(r_{1}=40 \mathrm{~cm}\), outer radius \(r_{2}=41 \mathrm{~cm}\), and thermal conductivity \(k=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is used to store hot water and to keep it at \(100^{\circ} \mathrm{C}\) at all times. To accomplish this, the outer surface of the container is wrapped with a 800 -W electric strip heater and then insulated. The temperature of the inner surface of the container is observed to be nearly \(120^{\circ} \mathrm{C}\) at all times. Assuming 10 percent of the heat generated in the heater is lost through the insulation, \((a)\) 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 material by solving the differential equation, and \((c)\) evaluate the outer surface temperature of the container. Also determine how much water at \(100^{\circ} \mathrm{C}\) this tank can supply steadily if the cold water enters at \(20^{\circ} \mathrm{C}\).

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