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Make a table listing the general solutions of all steady, unidimensional constant-properties heat conduction problemns in Cartesian, cylindrical and spherical coordinates, with and without uniform heat generation. This table should prove to be a very useful tool in future problem solving. It should include a total of 18 solutions. State any restrictions on your solutions. Do not include calculations.

Short Answer

Expert verified
The table lists solutions for Cartesian, cylindrical, and spherical systems both with and without heat generation.

Step by step solution

01

Introduction to Heat Conduction Equations

In heat conduction problems, we deal with the steady-state equation for temperature distribution. These equations are derived from the heat equation assuming no time dependence, with and without internal heat generation.
02

Cartesian Coordinates Solutions

In Cartesian coordinates, the steady-state heat conduction equation is: \[ \frac{d^2 T}{dx^2} + Q = 0 \] - Without heat generation (\(Q = 0\)): - Linear solution: \(T(x) = C_1 x + C_2\)- With heat generation (\(Q eq 0\)): - Parabolic solution: \(T(x) = -\frac{Q}{2k}x^2 + C_1 x + C_2\). Where \(C_1\) and \(C_2\) are constants.
03

Cylindrical Coordinates Solutions

In cylindrical coordinates, the radial steady-state equation is: \[ \frac{1}{r} \frac{d}{dr} \left( r \frac{dT}{dr} \right) + Q = 0 \]- Without heat generation: - Logarithmic solution: \(T(r) = C_1 \ln(r) + C_2\)- With heat generation: - Modified parabolic solution: \(T(r) = -\frac{Q}{4k}r^2 + C_1 \ln(r) + C_2\).
04

Spherical Coordinates Solutions

In spherical coordinates, the radial steady-state equation is: \[ \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{dT}{dr} \right) + Q = 0 \]- Without heat generation: - Inverse-linear solution: \(T(r) = \frac{C_1}{r} + C_2\)- With heat generation: - Quadratic modification: \(T(r) = -\frac{Q}{6k}r^2 + \frac{C_1}{r} + C_2\).
05

Table Compilation

The solutions can now be compiled into a table with the following headers:- Coordinate System- Without Heat Generation- With Heat Generation- Cartesian: - [Without: \( T(x) = C_1 x + C_2 \)] - [With: \( T(x) = -\frac{Q}{2k}x^2 + C_1 x + C_2 \)]- Cylindrical: - [Without: \( T(r) = C_1 \ln(r) + C_2 \)] - [With: \( T(r) = -\frac{Q}{4k}r^2 + C_1 \ln(r) + C_2 \)]- Spherical: - [Without: \( T(r) = \frac{C_1}{r} + C_2 \)] - [With: \( T(r) = -\frac{Q}{6k}r^2 + \frac{C_1}{r} + C_2 \)]These cover all steady-state unidimensional heat conduction situations in different coordinates.

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Key Concepts

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

Steady-State Equations
In heat conduction problems, steady-state equations are essential in understanding how temperature distributes within a material. These equations assume that time does not impact the temperature, meaning that the temperature remains constant over time. This simplification helps solve many practical problems such as insulation or heat dissipation in electronic components.

A steady-state equation takes into account the material's thermal properties like conductivity, and if applicable, includes internally generated heat, shown as a source term, generally denoted by \( Q \). A key feature of these equations is their reliance solely on spatial variables, eliminating time-dependent terms.
Cartesian Coordinates
When analyzing heat conduction in Cartesian coordinates, we describe the system using a straight-line path or along a single axis, usually denoted as \( x \). The steady-state heat conduction equation in one-dimensional Cartesian coordinates is represented by:
  • Without heat generation: \( \frac{d^2 T}{dx^2} = 0 \)
  • With heat generation: \( \frac{d^2 T}{dx^2} + Q = 0 \)
For a system without internal heat generation (\( Q = 0 \)), the temperature distribution will have a linear form: \( T(x) = C_1 x + C_2 \), where \( C_1 \) and \( C_2 \) are constants determined by boundary conditions.

With uniform heat generation, we derive a parabolic solution: \( T(x) = -\frac{Q}{2k}x^2 + C_1 x + C_2 \). This indicates that temperature changes become nonlinear, significantly influenced by the internal heat source.
Cylindrical Coordinates
In cylindrical coordinates, we consider heat flow in a radial direction, typically seen in systems like pipes or cylindrical containers. The radial steady-state conduction equation is given by:
  • Without heat generation: \( \frac{1}{r} \frac{d}{dr} \left( r \frac{dT}{dr} \right) = 0 \)
  • With heat generation: \( \frac{1}{r} \frac{d}{dr} \left( r \frac{dT}{dr} \right) + Q = 0 \)
In absence of heat generation, the solution is logarithmic: \( T(r) = C_1 \ln(r) + C_2 \). The logarithmic nature represents how heat dissipates radially.When heat generation is present, a modified parabolic solution emerges: \( T(r) = -\frac{Q}{4k}r^2 + C_1 \ln(r) + C_2 \), introducing a quadratic term. This accounts for the heat contributing to increasing temperatures as it spreads outward from the center.
Spherical Coordinates
Spherical coordinates apply to scenarios where heat conduction occurs outward from a spherical source, like a planet or a spherical tank. The radial steady-state equation is structured as follows:
  • Without heat generation: \( \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{dT}{dr} \right) = 0 \)
  • With heat generation: \( \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{dT}{dr} \right) + Q = 0 \)
For systems free from internal heat sources, the temperature distribution is inverse-linear: \( T(r) = \frac{C_1}{r} + C_2 \). This inverse relationship captures how temperature changes diminish with distance from the center.

Adding uniform heat generation introduces a quadratic component: \( T(r) = -\frac{Q}{6k}r^2 + \frac{C_1}{r} + C_2 \). This indicates that the influence of internal heat becomes substantial, especially closer to the center of the sphere.

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Most popular questions from this chapter

The highest heat flux that can be achieved in nucleate boiling (called \(q_{\max }-\) see the qualitative discussion in Section 9.1) depends upon \(\rho_{g}\), the saturated vapor density; \(h_{f g}\), the latent heat vaporization; \(\sigma\), the surface tension; a characteristic length, \(l\); and the gravity force per unit volume, \(g\left(\rho_{f}-\rho_{g}\right)\), where \(\rho_{f}\) is the saturated liquid density. Develop the dimensionless functional equation for \(q_{\max }\) in terms of dimensionless length.

Thin fins with a \(0.002 \mathrm{~m}\) by \(0.02 \mathrm{~m}\) rectangular cross section and a thermal conductivity of \(50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) protrude from a wall and have \(\bar{h} \simeq 600 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) and \(T_{0}=170^{\circ} \mathrm{C}\). What is the heat flow rate into each fin and what is the effectiveness? \(T_{\infty}=\) \(20^{\circ} \mathrm{C}\).

A thermocouple epoxied onto a high conductivity surface is intended to measure the surface temperature. The thermocouple consists of two each bare, \(0.51 \mathrm{~mm}\) diameter wires. One wire is made of Chromel (Ni-10\% Cr with \(\left.k_{\mathrm{cr}}=17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) and the other of constantan (Ni-45\% Cu with \(\left.k_{\mathrm{cn}}=23 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\). The ends of the wires are welded together to create a measuring junction having has dimensions of \(D_{w}\) by \(2 D_{w}\). The wires extend perpendicularly away from the surface and do not touch one another. A layer of epoxy \(\left(k_{\mathrm{ep}}=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\) separates the thermocouple junction from the surface by \(0.2 \mathrm{~mm}\). Air at \(20^{\circ} \mathrm{C}\) surrounds the wires. The heat transfer coefficient between each wire and the surroundings is \(\bar{h}=28 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), including both convection and radiation. If the thermocouple reads \(T_{\text {tc }}=40^{\circ} \mathrm{C}\), estimate the actual temperature \(T_{s}\) of the surface and suggest a better arrangement of the wires.

Develop the dimensionless temperature distribution in a spherical shell with the inside wall kept at one temperature and the outside wall at a second temperature. Reduce your solution to the limiting cases in which \(r_{\text {outside }} \gg r_{\text {inside }}\) and in which \(r_{\text {outside }}\) is very close to \(r_{\text {inside. Discuss these limits. }}\)

Steam condenses on the inside of a small pipe, keeping it at a specified temperature, \(T_{i}\). The pipe is heated by electrical resistance at a rate \(\dot{q} \mathrm{~W} / \mathrm{m}^{3}\). The outside temperature is \(T_{\infty}\) and there is a natural convection heat transfer coefficient, \(\bar{h}\) around the outside. (a) Derive an expression for the dimensionless expression temperature distribution, \(\Theta=\left(T-T_{\infty}\right) /\left(T_{i}-T_{\infty}\right)\), as a function of the radius ratios, \(\rho=r / r_{o}\) and \(\rho_{i}=r_{i} / r_{o}\); a heat generation number, \(\Gamma=\dot{q} r_{o}^{2} / k\left(T_{i}-T_{\infty}\right)\); and the Biot number. (b) Plot this result for the case \(\rho_{i}=2 / 3, \mathrm{Bi}=1\), and for several values of \(\Gamma\). (c) Discuss any interesting aspects of your result.

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