EEE 431 Computational Methods in Electrodynamics
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Transcript of EEE 431 Computational Methods in Electrodynamics
2
FINITE DIFFERENCE METHODS (cont).
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Finite Difference Method
Solve the diffusion Equation (Parabolic D.E.)
Subject to the boundary conditions:
And initial condition
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2, 0 1xtx
(0, ) 0 (1, ) 0, 0t t t
( ,0) 100x
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Finite Difference Method
Mathematical model of a temperature distribution of 1m long rod, with its ends in contacts with ice blocks which is initially at .0100 C
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Finite Difference Method
Using explicit method:
Solve the problem for since it is symmetric.
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0.1, 1/ 2, 1
( ) 0.005
x r k
t k r x
0 0.5x
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Finite Difference Method
1( , 1) ( 1, ) ( 1, )
21 1
(1,2) (2,1) (0,1) (1,3) (2,2) (0,2)2 21 1
(2,2) (3,1) (1,1) (2,3) (3,2) (1,2)2 21 1
(3,2) (4,1) (2,1) (3,3) (4,2) (2,2)2 2
. .
. .
. .
i j i j i j
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Xt
0 0.1 0.2 0.3 0.4 0.5 0.6 … 1.0
0 50 100 100 100 100 100 100 50
0.005 0 75.0 100 100 100 100 100 0
0.01 0 50 87.9 100 100 100 100 0
0.015 0 43.75 75 93.75 100 100 100 0
0.02 0 37.5 68.75 87.5 96.87 100 96.87 0
0.025 0 34.37 62.5 82.81 93.75 96.87 93.75 0
0.03 0 31.25 58.59 78.21 89.84 93.75 89.84 0
.
.
.
1.0 0 14.16 27.92 38.39 45.18 47.44 45.18 0
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Finite Difference Method
Implicit Method Choose The values at the fixed nodes are
calculated as it is calculated in the implicit formulation.
20.2, 1, 1 , ( ) 0.04x r k So t k r x
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Finite Difference Method
For the free nodes we use the formula obtained :
( 1, 1) 4 ( , 1) ( 1, 1) ( 1, ) ( 1, )i j i j i j i j i j
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Finite Difference Method
The value of for the first time step:
1 2
1 2 3
2 3 4
3 4
0 4 50 100
4 100 100
4 100 100
4 0 100 50
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Finite Difference Method
Solution of 4 simultaneous equations gives the values of at t=0.04.
Using these values of and applying the same equation, a set of simultaneous equations can be obtained for t=0.08.