Title: Chapter 3 Steady-State Conduction Multiple Dimensions
1Chapter 3 Steady-State Conduction Multiple
Dimensions
- CHAPER 3
- Steady-State Conduction Multiple Dimensions
23-1 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.
33-1 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
43-2 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.
53-2 Mathematical Analysis of Two-Dimensional
Heat Conduction
Consider a sine-wave distribution on the
upper edge, the boundary conditions are
63-2 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.
73-2 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.
83-2 Mathematical Analysis of Two-Dimensional
Heat Conduction
This function cannot fit the sine-function
boundary condition, so that the
solution may be excluded.
93-2 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.
103-2 Mathematical Analysis of Two-Dimensional
Heat Conduction
Let
The equation becomes
Apply the method of variable separation, let
113-2 Mathematical Analysis of Two-Dimensional
Heat Conduction
And the boundary conditions become
123-2 Mathematical Analysis of Two-Dimensional
Heat Conduction
Applying these conditions,we have
133-2 Mathematical Analysis of Two-Dimensional
Heat Conduction
accordingly,
and from (c),
This requires that
143-2 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.
153-2 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.
163-2 Mathematical Analysis of Two-Dimensional
Heat Conduction
Another set of boundary conditions
173-2 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
183-2 Mathematical Analysis of Two-Dimensional
Heat Conduction
This series is
then
The final solution is expressed as
193-2 Mathematical Analysis of Two-Dimensional
Heat Conduction
Transform the boundary condition
203-3 Graphical Analysis
neglect
213-4 The Conduction Shape Factor
- Consider a general one dimensional heat conduct-
- ion problem, from Fouriers Law
let
then
whereS is called shape factor.
223-4 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
233-4 The Conduction Shape Factor
243-4 The Conduction Shape Factor
253-4 The Conduction Shape Factor
263-4 The Conduction Shape Factor
273-4 The Conduction Shape Factor
283-4 The Conduction Shape Factor
293-4 The Conduction Shape Factor
30Example 3-1
31Example 3-2
32Example 3-3
33Example 3-4
343-5 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.
353-5 Numerical Method of Analysis
- 1?Discretization of the solving
363-5 Numerical Method of Analysis
2?Discrete equation
373-5 Numerical Method of Analysis
2?Discrete equation
Differential equation for two-dimensional
steady-state heat flow
383-5 Numerical Method of Analysis
2?Discrete equation
Discrete equation at nodal point (m,n)
no heat generation
?x ?y
393-5 Numerical Method of Analysis
2?Discrete equation
(1) Interior points steady-state no heat
generation
403-5 Numerical Method of Analysis
(1) Interior points
?x ?y
steady-state with heat generation
413-5 Numerical Method of Analysis
2?Discrete equation
(2) boundary points
423-5 Numerical Method of Analysis
(2) boundary points
?x ?y
?x ?y
433-5 Numerical Method of Analysis
(2) boundary points
?x ?y
443-5 Numerical Method of Analysis
3?Algebraic equation
453-5 Numerical Method of Analysis
- Iteration
- Simple Iteration Gauss-Seidel Iteration
46Example 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.
47Example 3-5
Solution The equations for nodes 1,2,4,5 are
given by
48Example 3-5
Solution Equations for nodes 3,6,7,8 are
The equation for node 9 is
49Example 3-5
The equation for node 9 is
50Example 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
51Example 3-5
The heat flow out the right face is
The heat flow out the bottom face is
The total heat flow out is
523-6 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
533-6 Numerical Formulation in Terms of Resistance
Elements
so
543-7 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.
553-7 Gauss-Seidel Iteration
56Example 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
57Example 3-6
Because each node has four resistance connected
to it and k is assumed constant, so
583-8 Accuracy Consideration
- Truncation Error Influenced by difference
scheme - Discrete Error Influenced by truncation error
?x - Round-off Error Influenced by ?x
59Summary
- Numerical Method
- Solving Zone
- Nodal equations
- thermal balance method Interior boundary
point - Algebraic equations
- Gauss-Seidel iteration
60Summary
(2)Resistance Forms
(3)Convergence
61Summary
(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
62Exercises
Exercises 3-16, 3-24, 3-48, 3-59