Ch 9.6 - Introduction to Radial Heat Flow

Radial Flow

Cross Sectional Areas (3D \( \rightarrow \) 2D)

Plane Wall (x) \( \rightarrow \) Rectangle

Cylinder (r) \( \rightarrow \) Area of the outside of a cylinder (Rectangle)

Sphere (r) \( \rightarrow \) Area of a sphere

How do these areas change as we increase x or r?

Fourier's Law for Radial Heat Conduction

Main Ideas:

1) Heat spreading out as it moves

2) \( J(r) \) is heat flux at some radius \( r \) from center

3) Rate of flow of heat:

\[ \begin{Bmatrix} \mathrm{rate\, of} \\ \mathrm{flow\, of} \\ \mathrm{heat} \\ \end{Bmatrix} = J(r)A(r) \]

where \( A(r) \) is cross-sectional area at \( r \).

Fourier's Law for Radial Heat Conduction (Continued)

Fourier's Law of Heat Conduction

\[ J(x) = -k\frac{dU(x)}{dx} \]

Fourier's Law for Radial Heat Conduction

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

Fourier's Law for Radial Heat Conduction (Continued again)

Fourier's Law for Radial Heat Conduction

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

Terms

\( J(r) \) is the heat flux at \( r \) (Watts)

\( U(r) \) is the temperature at \( r \) (\( ° \) C, \( ° \) F, \( ° \) K)

\( k \) is the conductivity

Heat flux is directly proportional to temperature gradient (heat flows from hot to cold)

Model Approach

  • Consider a cylindrical object: hottest inside & coolest out
  • \( l= \) length of cylinder

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Defining Variables

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  • \( b= \) outer radius of the cylinder
  • \( a= \) inner radius
  • Cylinder becomes solid as \( a \) approaches \( 0 \)
  • Cylinder is a hollow pipe for \( a \) close to \( b \)
  • \( r= \) radial distance from its centerline
  • \( t= \) time

Recall Parameters

  • \( J(r)= \) heat flux at \( r \)
  • \( A(r) = \) area through which heat flows
  • \( U(r)= \) temperature at \( r \)
  • \( k= \) conductivity

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A Visualization

  • Radial heat flow in a cylinder
  • Heat flows from the center to the outside of the cylinder in all directions

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  • The cross section of the cylinder is on left
  • An arbitrary cylindrical shell inside the cylindrical region shown on right; from \( r \) to \( r + \Delta r \)

Stating Assumptions

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  • Heat flows in the radial direction only; the temperature inside depends on \( r \) and \( t \) only
  • Thermal equilibrium; thus equilibrium temperature is function of just \( r \)

General Compartmental Model

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General Compartmental Model

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General Compartmental Model

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General Compartmental Model

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General Compartmental Model

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General Compartmental Model

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Formulating the Differential Equation

  • Recall that \( U(r) \) is the equilibrium temperature from the distance of \( r \) and \( J(r) \) is the heat flux at a distance \( r \).
  • We want to formulate the Differential Equation needed for the figure 9.9.

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Annular shell

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  • Heat is entering cylindrical shell with area

\[ A(r) = 2\pi rl \]

  • Heat is leaving through cylindrical shell with area

\[ A(r+\Delta r) = 2\pi (r+\Delta r)l \]

Solution Equation

  • Thus

\[ \begin{aligned} \begin{Bmatrix} \mathrm{rate\, heat} \\ \mathrm{conducted } \\ \mathrm{in\, at} ~ r \\ \end{Bmatrix} & = J(r)A(r) \\ \\ \begin{Bmatrix} \mathrm{rate\, heat} \\ \mathrm{conducted } \\ \mathrm{out\, at} ~ r + \Delta r \\ \end{Bmatrix} & = J(r + \Delta r)A(r + \Delta r) \end{aligned} \]

Substituting Equations

  • For equilibrium temperatures, LHS of word equation is zero.
  • Thus

\[ J(r)A(r) - J(r + \Delta r)A(r + \Delta r) = 0 \]

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Rewriting The Equation (Eva)

  • Rewrite the previous equation:

\[ -[J(r+\Delta r)A(t+\Delta r) - J(r)A(r)] = 0 \]

  • Dividing by \( \Delta r \), we obtain

\[ -\left[\frac{J(t+\Delta r)A(r+\Delta r)-J(r)A(r)}{\Delta r}\right] = 0 \]

  • Letting \( \Delta r \) approach 0, we obtain

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

Constant Rate of Heat Flow

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  • From the previous slide:

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

  • Recall rate of heat flow is

\[ J(r)A(r) \]

  • Thus rate of heat flow is constant inside cylinder.

Differential Equation

  • Since \( A(r) = 2\pi rl \) and \( J = -k\frac{dU}{dr} \), we have

\[ -\frac{d}{dr}(-2k\pi lr \frac{dU}{dr}) = 0 \]

  • Note that \( 2k\pi l \) is a constant, and hence

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

  • This is our differential equation for radial heat flow via conduction.