Ch9.6 Radial Heat Flow

Introduction

  • Radial heat flow occurs in cylinders and spheres.
  • We formulate an ODE with respect to spatial variable \( r \) using heat flux \( \small{J(r)} \) for equilibrium temperature \( \small{U(r)} \) in wall.
  • Assumes angular and lateral symmetry, with \( \small{U(r,\theta) = U(r)} \).
  • Approach is similar to previous sections, except area \( \small{A(r)} \) through which heat conducts will depend on \( r \).

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Background

  • Examples of radial heat conduction include planetary cooling (spherical) and hot water piping (cylindrical).
  • Insulation on pipes is sometimes used in an effort to retain heat and prevent pipes from freezing and rupturing.
    • Modeling heat conduction in pipes can help determine whether insulation is effective and how much to use.

Problem Statement

Formulate differential equation for equilibrium temperature inside a cylinder as function of distance from center line.

Assumptions

  • Cylinder is comprised of homogeneous material.
  • Interior and exterior held uniformly constant temperatures.
  • Interior of cylinder is warmer than exterior.
  • Heat flux through wall is by conduction only.
  • Heat is conducted according to Fourier's law.
  • Conductivity of material is constant.
  • Temperature has angular and lateral symmetry and depends only on radial distance from center line.
  • Temperature inside cylinder is in equilibrium state.
  • Model temperature at a point inside cylinder by considering thin section and then letting thickness of section tend to zero.

Compartment Diagram

  • For assumptions stated above, compartment diagram will have the following form.
  • Since temperature of section is at equilibrium, heat conducted into and out of section will be same.

Word Equation

Using the balance law, the word equation for our assumptions and compartment diagram is shown below.

\[ \small{ \begin{aligned} \begin{Bmatrix} \mathrm{change \, in \, heat } \\ \mathrm{content \, of \, shell } \end{Bmatrix} &= \begin{Bmatrix} \mathrm{heat \, conducted } \\ \mathrm{into \, shell } \end{Bmatrix} - \begin{Bmatrix} \mathrm{heat \, conducted } \\ \mathrm{out \, of \, shell } \end{Bmatrix} \\ & = 0 \end{aligned} } \]

Variables and Parameters

  • Distance \( r \) meters from center line, measured from warmer side \( r=a \) to cooler side \( r=b \).
  • Thin shell ranges from \( r \) to \( \small{r+ \Delta r} \).
  • Conductivity of material is \( \small{k>0} \).
  • Surface area of shell at \( r \) is \( \small{A(r) \, m^2} \).
  • Temperature at \( r \) is \( \small{U(r) \, \,^\circ C} \).
  • Radial form for flux, in Watts/\( \small{m^2} \), is

\[ \small{ J(r) = -k\frac{dU(r)}{dr}} \]

Formulation of Differential Equation

  • Heat conduction is expressed in terms of Joules/sec.
  • Units of \( \small{J(r)A(r)} \) are \( \small{\left(\frac{W}{m^2}\right)(m^2) = W = \frac{Joules}{sec}} \).
  • Starting with the word equation, we proceed as follows:

\[ \small{ \begin{aligned} \begin{Bmatrix} \mathrm{change \, in \, heat } \\ \mathrm{content \, of \, shell } \end{Bmatrix} &= \begin{Bmatrix} \mathrm{heat \, conducted } \\ \mathrm{into \, shell } \end{Bmatrix} - \begin{Bmatrix} \mathrm{heat \, conducted } \\ \mathrm{out \, of \, shell } \end{Bmatrix} \\ \\ 0 & = J(r)A(r) - J(r+ \Delta r)A(r+ \Delta r) \end{aligned} } \]

  • Let \( \small{h = \Delta r} \), divide both sides by \( \small{-h} \), and let \( \small{h \rightarrow 0} \):

\[ \small{ 0 = \lim_{h \rightarrow 0} \frac{J(r+ h)A(r+h) - J(r)A(r)}{h} = \frac{d}{dr}\left[J(r)A(r)\right] } \]

Formulation of Differential Equation

From previous slide:

\[ \small{ \frac{d}{dr}\left[J(r)A(r)\right] = 0 } \]

Since \( \small{A(r) = 2\pi rl} \), we have

\[ \small{ 2\pi l\frac{d}{dr}\left[rJ(r)\right] = 0 } \]

Using Fourier's law \( \small{J(r) = -kU'(r)} \):

\[ \small{ \frac{d}{dr}\left(r\frac{dU}{dr}\right) = 0 } \]

2nd Order ODE for Radial Conduction

From previous slide,

\[ \small{ \frac{d}{dr}\left(r\frac{dU}{dr}\right) = 0 } \]

Using the product rule, we obtain

\[ \small{r \frac{d^2 U}{dr^2} + \frac{d U}{dr} = 0 } \]

Discussion of Results

  • Assuming thermal equilibrium with angular and lateral symmetry gives model \( \small{U(r)} \) as a function of \( r \) only.
    • This leads to our second order ODE for \( \small{U(r)} \).
    • Otherwise we have \( \small{U(r,\theta,z,t)} \) in cylindrical coordinates and \( \small{U(\rho,\theta,\phi,t)} \) in spherical coordinates, with PDE model.
  • Analytical solution of our ODE will be explored in Ch11.4.
  • Numerical solution of 2nd order ODE can entail boundary conditions, finite differences, and matrix-vector formulations (similar to Ch21.4 for PDEs in Math 366).

\[ \small{r \frac{d^2 U}{dr^2} + \frac{d U}{dr} = 0 } \]