Ch 9.7 Heat Fins

Overview

  • Introduction
  • Background
  • Model assumptions and approach
  • General compartmental model
  • Formulating the differential equation
  • Extensions and further problems

Introduction

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  • Recall Ch9.6 approach
  • Here we develop a model for heat fin (cooling device)
  • Our goal is to understand heat distribution along fin

Background

  • Heat fins enhance dissipation of heat from machinery susceptible to damage if heat is allowed to build up
  • Use in motorcycle engines, refrigerators, and computers
  • Fins increase surface area through which heat can be lost
  • Fans attached to force air across fins to reduce temperature more efficiently

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Examples of Fins

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A Typical Problem

  • Calculate the temperature distribution along a heat fin
  • Calculate the rate of heat dissipation
  • Determine the cooling efficiency for the heat fin
  • Our first step is to: formulate a differential equation for the changing temperature along the heat fin

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Model Assumptions

For our model we consider a rectangular heat fin of length \( l \), width \( w \), and thickness \( b \)

fin

Understanding our model assumptions

  • We are moving hot air to cool air
  • The fins increase surface area through which heat is lost
  • Fans force air across the fins

diagram

Assumptions

  • Thickness of fin \( b \) is small compared to length \( l \)
  • Heat lost is from surfaces of fins to surroundings according to Newton's law of cooling

fin diagram

Assumptions

  • Because most of heat loss occurs from faces of fin, both top and bottom, we neglect heat lost from sides.
  • Improving model by including heat lost from sides is not difficult and is explored in the exercises.

fin diagram

Assumptions

  • Neglect temperature variation over cross-section of fin.
  • This assumption is reasonable, provided fin is thin.
  • This will be the case if \( b << l \).

fin diagram

Assumptions

  • In doing this, we can effectively assume the equilibrium temperature to be a function of x alone.

fin diagram

Variables

  • Small section of the heat fin from \( x \) to \( x + \Delta x \).
  • Heat is conducted into section at \( x \).
  • Heat is conducted out of section at \( x + \Delta x \).
  • Heat lost from the top and bottom surfaces to surroundings.
  • Heat enters one way but exits two ways.

Figure

Compartment Diagram & Word Equation

Basic

\[ \small{ \begin{Bmatrix} \mathrm{rate \, of} \\ \mathrm{change \, of} \\ \mathrm{heat\, in \, section} \\ \end{Bmatrix} = \begin{Bmatrix} \mathrm{rate \, heat} \\ \mathrm{conducted} \\ \mathrm{in\, at\, x} \\ \end{Bmatrix} - \begin{Bmatrix} \mathrm{rate \, heat} \\ \mathrm{conducted} \\ \mathrm{out\, at\, x+\Delta x} \\ \end{Bmatrix} - \begin{Bmatrix} \mathrm{rate\, of\, heat} \\ \mathrm{lost\, to} \\ \mathrm{surroundings} \\ \end{Bmatrix}} \]

  • For equilibrium temperature, rate of change of heat is zero:

\[ \small{ \begin{Bmatrix} \mathrm{rate \, heat} \\ \mathrm{conducted} \\ \mathrm{in\, at\, x} \\ \end{Bmatrix} - \begin{Bmatrix} \mathrm{rate \, heat} \\ \mathrm{conducted} \\ \mathrm{out\, at\, x+\Delta x} \\ \end{Bmatrix} - \begin{Bmatrix} \mathrm{rate\, of\, heat} \\ \mathrm{lost\, to} \\ \mathrm{surroundings} \\ \end{Bmatrix} = 0} \]

Heat Conduction Terms

  • Recall from Ch9.5 (Heat Conduction through a Wall):

\[ \small{ \begin{align} \begin{Bmatrix} \mathrm{rate \, heat} \\ \mathrm{conducted} \\ \mathrm{in\, at\, x} \\ \end{Bmatrix} &= J(x)A(x) = J(x)bw \\ \begin{Bmatrix} \mathrm{rate \, heat} \\ \mathrm{conducted} \\ \mathrm{out\, at\, x+\Delta x} \\ \end{Bmatrix} &= J(x+\Delta x)A(x) = J(x+\Delta x)bw \end{align} } \]

  • Here, \( bw \) indicates area through which heat is conducted.

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Heat Exchange Terms

  • Recall from Ch9.3 (Hot Water Heater):

\[ \small{ \begin{Bmatrix} \mathrm{rate\, heat} \\ \mathrm{exchanged\, with} \\ \mathrm{surroundings} \\ \end{Bmatrix}} = \pm hS \Delta U \]

  • Parameters:
    • \( h \) = Newton cooling coefficient
    • \( S \) = surface area from which heat is exchanged
    • \( \Delta U \) = temperature difference between fin and surroundings

Formulating the Differential Equation

\[ \small{ \begin{Bmatrix} \mathrm{rate\, heat} \\ \mathrm{exchanged\, with} \\ \mathrm{surroundings} \\ \end{Bmatrix}} = \pm hS \Delta U \]

  • Parameters:
    • \( h \) depends on material
    • \( S = 2w \Delta x \) (neglect side faces)
    • \( \Delta U = U(x^*)-u_s \)
    • \( x < x^* < x + \Delta x \)

Formulating the Differential Equation

  • Thus

\[ \small{ \begin{Bmatrix} \mathrm{rate\, heat} \\ \mathrm{exchanged\, with} \\ \mathrm{surroundings} \\ \end{Bmatrix} } = hS \Delta U = 2hw\Delta x[U(x^*)-u_s] \]

  • The differential equation is now

\[ J(x)bw - J(x+\Delta x)bw - 2hw\Delta x[U(x^*)-u_s] = 0. \]

  • Alternatively,

\[ J(x)bw - J(x+\Delta x)bw - 2hw\Delta x[U(x + \lambda \Delta x)-u_s] = 0. \]

  • \( x^* = x + \lambda \Delta x \) and \( 0 \leq \lambda \leq 1 \).

Units (Optional)

  • Determine units of \( h \):

\[ \begin{aligned} \big[ hS \Delta U \big] &= \big[J(x)bw \big], \\ \big[ h \big] \times m^2 \times C &= \frac{ W }{m^2} \times m^2, \\ \big[ h \big] &= \frac{ W }{ C \times m^2 }, \\ \big[ h \big] &= \frac{ J }{ C \times m^2 \times sec } \end{aligned} \]

Example 9.11

  • Recall

\[ J(x)bw - J(x+\Delta x)bw - 2hw\Delta x[U(x + \lambda \Delta x)-u_s] = 0 \]

  • Divide this equation by \( - bw\Delta x \):

\[ \frac{J(x+\Delta x)-J(x)}{\Delta x}+ \frac{2h}{b}\left[U(x+ \lambda \Delta x)-u_{s}\right] = 0 \]

  • Let \( \Delta x \rightarrow 0 \) to obtain

\[ \frac{dJ}{dx} = - \frac{2h}{b}\left[U(x)-u_{s}\right] \]

Example 9.11

  • From the previous slide,

\[ \frac{dJ}{dx} = - \frac{2h}{b}\left[U(x)-u_{s}\right] \]

  • Recall Fourier's Law:

\[ \frac{dJ}{dx} = - k \frac{dU}{dx} \]

  • We thus obtain

\[ k\frac{d^2U}{dx^2} = \frac{2h}{b}[U(x)-u_{s}] \]

Example 9.11

  • From the previous slide,

\[ k\frac{dJ}{dx} = \frac{2h}{b}\left[U(x)-u_{s}\right] \]

  • Let

\[ \beta = \frac{2h}{kb} \]

  • Then our equation becomes

\[ \frac{d^2 U}{dx^2} = \beta [U - u_s] \]

Extensions and further problems

  • Heat fins come in many shapes and sizes
  • If we consider a heat fin with different cross-sections (ie: circular), the derivation of differential equation is identical to Equation 9.35, but value for \( \beta \) would differ.
  • Though complex, we are equipped now to develop a differential equation for when cross-section varies with distance from origin
  • A heat production term could be considered for problems where temperature of surroundings may be higher than temperature of fin (where fin gains heat rather than loses it)