Chapter 3 Steady-State Conduction Multiple Dimensions

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Chapter 3 Steady-State Conduction Multiple
Dimensions

• CHAPER 3
• Steady-State Conduction Multiple Dimensions

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3-1 Introduction

• In Chapter 2 steady-state heat transfer
  was calculated in systems in which the
  temperature gradient and area could be expressed
  in terms of one space coordinate. We now wish to
  analyze the more general case of two-dimensional
  heat flow. For steady state with no heat
  generation, the Laplace equation applies.

(3-1)
The solution to this equation may be obtained by
analytical, numerical, or graphical techniques.
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3-1 Introduction

• The objective of any heat-transfer analysis
  is usually to predict heat flow or the
  temperature that results from a certain heat
  flow. The solution to Equation (3-1) will give
  the temperature in a two-dimensional body as a
  function of the two independent space coordinates
  x and y. Then the heat flow in the x and y
  directions may be calculated from the Fourier
  equations

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3-2 Mathematical Analysis of Two-Dimensional
Heat Conduction
Analytical solutions of temperature
distribution can be obtained for some simple
geometry and boundary conditions. The separation
method is an important one to apply.
Consider a rectangular plate. Three sides
are maintained at temperature T1, and the upper
side has some temperature distribution
impressed on it. The distribution can be a
constant temperature or something more complex,
such as a sine-wave.
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3-2 Mathematical Analysis of Two-Dimensional
Heat Conduction
Consider a sine-wave distribution on the
upper edge, the boundary conditions are
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3-2 Mathematical Analysis of Two-Dimensional
Heat Conduction
Substitute
We obtain two ordinary differential equations
in terms of this constant,
where ?2 is called the separation constant.
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3-2 Mathematical Analysis of Two-Dimensional
Heat Conduction
We write down all possible solutions and then
see which one fits the problem under
consideration.
This function cannot fit the sine-function
boundary condition, so that the
solution may be excluded.
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3-2 Mathematical Analysis of Two-Dimensional
Heat Conduction
This function cannot fit the sine-function
boundary condition, so that the
solution may be excluded.
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3-2 Mathematical Analysis of Two-Dimensional
Heat Conduction
It is possible to satisfy the sine-function
boundary condition so we shall attempt to
satisfy the other condition.
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3-2 Mathematical Analysis of Two-Dimensional
Heat Conduction
Let
The equation becomes
Apply the method of variable separation, let
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3-2 Mathematical Analysis of Two-Dimensional
Heat Conduction
And the boundary conditions become
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3-2 Mathematical Analysis of Two-Dimensional
Heat Conduction
Applying these conditions,we have
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3-2 Mathematical Analysis of Two-Dimensional
Heat Conduction
accordingly,
and from (c),
This requires that
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3-2 Mathematical Analysis of Two-Dimensional
Heat Conduction
then
We get
The final boundary condition may now be applied
which requires that Cn 0 for n gt1.
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3-2 Mathematical Analysis of Two-Dimensional
Heat Conduction
The final solution is therefore
The temperature field for this problem is shown.
Note that the heat-flow lines are perpendicular
to the isotherms.
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3-2 Mathematical Analysis of Two-Dimensional
Heat Conduction
Another set of boundary conditions
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3-2 Mathematical Analysis of Two-Dimensional
Heat Conduction
Using the first three boundary conditions, we
obtain the solution in the form of Equation
Applying the fourth boundary condition gives
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3-2 Mathematical Analysis of Two-Dimensional
Heat Conduction
This series is
then
The final solution is expressed as
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3-2 Mathematical Analysis of Two-Dimensional
Heat Conduction
Transform the boundary condition
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3-3 Graphical Analysis
neglect
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3-4 The Conduction Shape Factor

• Consider a general one dimensional heat conduct-
• ion problem, from Fouriers Law

let
then
whereS is called shape factor.
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3-4 The Conduction Shape Factor
Note that the inverse hyperbolic cosine can be
calculated from
For a three-dimensional wall, as in a
furnace, separate shape factors are used to
calculate the heat flow through the edge and
corner sections, with the dimensions shown in
Figure 3-4. when all the interior dimensions are
greater than one fifth of the thickness,
where A area of wall, L wall thickness, D
length of edge
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3-4 The Conduction Shape Factor
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3-4 The Conduction Shape Factor
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3-4 The Conduction Shape Factor
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3-4 The Conduction Shape Factor
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3-4 The Conduction Shape Factor
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3-4 The Conduction Shape Factor
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3-4 The Conduction Shape Factor
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Example 3-1
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Example 3-2
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Example 3-3
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Example 3-4
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3-5 Numerical Method of Analysis

• The most fruitful approach to the heat
  conduction is one based on ?nite-difference
  techniques, the basic principles of which we
  shall outline in this section.

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3-5 Numerical Method of Analysis

• 1?Discretization of the solving

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3-5 Numerical Method of Analysis
2?Discrete equation

• Taylor series expansion

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3-5 Numerical Method of Analysis
2?Discrete equation
Differential equation for two-dimensional
steady-state heat flow
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3-5 Numerical Method of Analysis
2?Discrete equation
Discrete equation at nodal point (m,n)
no heat generation
?x ?y
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3-5 Numerical Method of Analysis
2?Discrete equation

• Thermal balance

(1) Interior points steady-state no heat
generation
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3-5 Numerical Method of Analysis

• Thermal balance

(1) Interior points
?x ?y
steady-state with heat generation
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3-5 Numerical Method of Analysis
2?Discrete equation

• Thermal balance

(2) boundary points
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3-5 Numerical Method of Analysis

• Thermal balance

(2) boundary points
?x ?y
?x ?y
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3-5 Numerical Method of Analysis

• Thermal balance

(2) boundary points
?x ?y
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3-5 Numerical Method of Analysis
3?Algebraic equation
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3-5 Numerical Method of Analysis

• Matrix notation

• Iteration
• Simple Iteration Gauss-Seidel Iteration

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Example 3-5

• Consider the square shown in the figure.
  The left face is maintained at 100? and the top
  face at 500?, while the other two faces are
  exposed to a environment at 100?. h10W/m2? and
  k10W/m?. The block is 1 m square. Compute the
  temperature of the various nodes as indicated in
  the figure and heat flows at the boundaries.

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Example 3-5
Solution The equations for nodes 1,2,4,5 are
given by
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Example 3-5
Solution Equations for nodes 3,6,7,8 are
The equation for node 9 is
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Example 3-5
The equation for node 9 is
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Example 3-5
We thus have nine equations and nine unknown
nodal temperatures. So the answer is
For the 500? face, the heat flow into the face is
The heat flow out of the 100? face is
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Example 3-5
The heat flow out the right face is

The heat flow out the bottom face is
The total heat flow out is
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3-6 Numerical Formulation in Terms of Resistance
Elements

• Thermal balance the net heat input to node i
  must be zero

qi heat generation, radiation, etc. i solving
node j adjoining node
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3-6 Numerical Formulation in Terms of Resistance
Elements
so
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3-7 Gauss-Seidel Iteration

• Steps
• Assumed initial set of values for Ti
• Calculated Ti according to the equation
• using the most recent values of
  the Ti
• Repeated the process until converged.

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3-7 Gauss-Seidel Iteration

• Convergence Criterion

• Biot number

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Example 3-6

• Apply the Gauss-Seidel technique to obtain the
  nodal temperature for the four nodes in the
  figure.

Solution All the connection resistance between
the nodes are equal, that is
Therefore, we have
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Example 3-6
Because each node has four resistance connected
to it and k is assumed constant, so
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3-8 Accuracy Consideration

• Truncation Error Influenced by difference
  scheme
• Discrete Error Influenced by truncation error
  ?x
• Round-off Error Influenced by ?x

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Summary

• Numerical Method
• Solving Zone
• Nodal equations
• thermal balance method Interior boundary
  point
• Algebraic equations
• Gauss-Seidel iteration

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Summary
(2)Resistance Forms
(3)Convergence

• Convergence Criterion

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Summary
(4)Accuracy

• Truncation Error
• Discrete Error
• Round-off Error

• Important conceptions
• Nodal equations thermal balance method
• Calculated temperature heat flow
• Convergence criterion
• How to improve accuracy

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Exercises
Exercises 3-16, 3-24, 3-48, 3-59