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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 \(r_{\text {inside. Discuss these limits. }}\)

Short Answer

Expert verified
The dimensionless temperature distribution simplifies to linear for \(r_o \gg r_i\) and radial for \(r_o \approx r_i\).

Step by step solution

01

Define the Problem

The problem is to derive a dimensionless temperature distribution for a spherical shell. The inner wall is at temperature \(T_i\) and the outer wall at temperature \(T_o\). We need to express this distribution in dimensionless form and analyze limiting cases.
02

Set Up the Differential Equation

Assume steady-state heat conduction in a spherical shell with no internal sources. The governing equation in spherical coordinates is given by\[ \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{dT}{dr} \right) = 0 \]where \(T\) is the temperature, and \(r\) is the radial coordinate.
03

Integrate the Differential Equation

Integrate the equation to obtain:\[ r^2 \frac{dT}{dr} = C_1 \]Integrate again:\[ T = -\frac{C_1}{r} + C_2 \]where \(C_1\) and \(C_2\) are integration constants.
04

Apply Boundary Conditions

Apply the boundary conditions:1. \( T = T_i \) at \( r = r_i \)2. \( T = T_o \) at \( r = r_o \)Solving these conditions, find the constants \(C_1\) and \(C_2\).
05

Solve for Temperature Constants

Substituting the boundary conditions:\[ T_i = -\frac{C_1}{r_i} + C_2 \]\[ T_o = -\frac{C_1}{r_o} + C_2 \]Subtract these equations to get:\[ C_1 = \frac{(T_o - T_i) r_i r_o}{r_o - r_i} \]\[ C_2 = T_i + \frac{C_1}{r_i} \]
06

Derive Dimensionless Temperature

Introduce dimensionless temperature \( \theta = \frac{T - T_i}{T_o - T_i} \):\[ \theta = \frac{-C_1/r + C_2 - T_i}{T_o - T_i} \]
07

Express Dimensionless Temperature

Substitute \(C_1\) and \(C_2\):\[ \theta = \frac{r_i/r_o}{1 - r_i/r_o} \left( \frac{r/r_o}{1 - r_i/r} - \frac{r_i/r}{1 - r_i/r} \right) \]
08

Simplify for Limit Cases

For \(r_o \gg r_i\), the shell is very thin, and the dimensionless temperature simplifies to a linear distribution:\[ \theta \approx \frac{r - r_i}{r_o - r_i} \]For \(r_o \approx r_i\),\[ \theta \approx \frac{r_i}{r} \]
09

Discuss Limit Cases

When \(r_o \gg r_i\), the spherical shell behaves like a planar wall, and the temperature distribution is linear. As \(r_o\) becomes close to \(r_i\), the geometry of the shell becomes negligible, and the distribution approaches the center of a solid sphere.

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Key Concepts

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

Dimensionless Temperature Distribution
When analyzing the heat distribution in a spherical shell, it's beneficial to express the temperature in a dimensionless form. This allows us to generalize the results and make comparisons regardless of size or conditions. The dimensionless temperature, often denoted by \( \theta \), is defined as:
  • \( \theta = \frac{T - T_i}{T_o - T_i} \)
Here, \(T\) represents the temperature at a given radius, \(T_i\) is the temperature of the inner wall, and \(T_o\) is that of the outer wall.

This transformation scales the problem, setting \(\theta = 0\) at the inner surface and \(\theta = 1\) at the outer surface. This creates a range from 0 to 1, making it easier to visualize changes. Once you have this dimensionless temperature, the heat distribution across the shell can be understood without needing specific temperature values, focusing instead on relative changes.
Steady-State Heat Conduction
Steady-state heat conduction refers to a situation where the temperature distribution doesn't change over time. In our spherical shell, this has significant implications: the heat entering any thin layer of the shell equals the heat leaving it. No heat is accumulated within the shell.

The differential equation used in the analysis arises from applying the law of conservation of energy, coupled with the spherical symmetry of the problem:
  • \( \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{dT}{dr} \right) = 0 \)
This results in a predictable pattern of heat flow, driven solely by the differences between the set temperatures of the inner and outer surfaces. The equation ensures a balance, creating a stable internal temperature field once the system reaches equilibrium.
Boundary Conditions
In the process of solving a differential equation for heat conduction, boundary conditions are crucial. They specify the temperatures at the inner and outer surfaces of the spherical shell:
  • \( T = T_i \) at \( r = r_i \)
  • \( T = T_o \) at \( r = r_o \)
These conditions are essential for integrating the differential equation and determining the integration constants \(C_1\) and \(C_2\).

Substituting these conditions into the integrated forms of the equation gives us the values needed to express our dimensionless temperature distribution. These anchor the solution to the physical situation presented in the problem.
Limiting Cases Analysis
Understanding the limiting cases helps reveal how the geometry of the system influences the temperature distribution.1. **Thin Shell Limit (\(r_o \gg r_i\))**: - As the outer radius becomes much larger than the inner radius, the shell approaches the form of a planar sheet. - The dimensionless temperature distribution simplifies to a linear equation: \( \theta \approx \frac{r - r_i}{r_o - r_i} \). - This mimics the heat conduction through a flat wall, as curvature effects become negligible.2. **Concentric Spheres (\(r_o \approx r_i\))**: - When the radii are nearly equal, the shell becomes very thin compared to its curvature. - The distribution now appears as \( \theta \approx \frac{r_i}{r} \), concentrating mostly around the inner radius. - This indicates a reduction in the geometrical complexity, with behavior aligning more with a solid sphere's center.

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Most popular questions from this chapter

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? \(T_{\infty}=\) \(20^{\circ} \mathrm{C}\).

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.

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.

Steam condenses in a \(2 \mathrm{~cm}\) I.D. thin-walled tube of \(99 \%\) aluminum at 10 atm pressure. There are circular fins of constant thickness, \(3.5 \mathrm{~cm}\) in diameter, every \(0.5 \mathrm{~cm}\) on the outside. The fins are \(0.8 \mathrm{~mm}\) thick and the heat transfer coefficient from them \(\bar{h}=6 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) (including both convection and radiation). What is the mass rate of condensation if the pipe is \(1.5 \mathrm{~m}\) in length, the ambient temperature is \(18^{\circ} \mathrm{C}\), and \(\bar{h}\) for condensation is very large? \(\left[\dot{m}_{\text {cond }}=0.802 \mathrm{~kg} / \mathrm{hr}\right.\). \(]\)

The heat tansfer coefficient, \(\bar{h}\), resulting from a forced flow over a flat plate depends on the fluid velocity, viscosity, density, specific heat, and thermal conductivity, as well as on the length of the plate. Develop the dimensionless functional equation for the heat transfer coefficient (cf. Section 6.5).

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