Gaussian Elimination-More Examples: Electrical Engineering
Chapter 04.06
Gaussian Elimination – More Examples
Electrical Engineering
Example 1
Three-phase loads are common in AC systems. When the system is balanced the analysis can be simplified to a single equivalent circuit model. However, when it is unbalanced the only practical solution involves the solution of simultaneous linear equations. In one model the following equations need to be solved.
[pic]
Find the values of [pic], [pic], [pic], [pic], [pic], and [pic] using naïve Gauss elimination.
Solution
Forward Elimination of Unknowns
Since there are six equations, there will be five steps of forward elimination of unknowns.
First step
Divide Row 1 by 0.7460 and multiply it by 0.4516, that is, multiply Row 1 by [pic].
[pic]
Subtract the result from Row 2 to get
[pic]
Divide Row 1 by 0.7460 and multiply it by 0.0100, that is, multiply Row 1 by [pic].
[pic]Subtract the result from Row 3 to get [pic]
Divide Row 1 by 0.7460 and multiply it by 0.0080, that is, multiply Row 1 by [pic].
[pic]Subtract the result from Row 4 to get
[pic]
Divide Row 1 by 0.7460 and multiply it by 0.0100, that is, multiply Row 1 by [pic].
[pic]Subtract the result from Row 5 to get
[pic]
Divide Row 1 by 0.7460 and multiply it by 0.0080, that is, multiply Row 1 by [pic].
[pic]Subtract the result from Row 6 to get
[pic]
Second step
Divide Row 2 by 1.0194 and multiply it by −0.0019464, that is, multiply Row 2 by [pic].
[pic]Subtract the result from Row 3 to get
[pic]
Divide Row 2 by 1.0194 and multiply it by 0.014843, that is, multiply Row 2 by [pic].
[pic]
Subtract the result from Row 4 to get
[pic]
Divide Row 2 by 1.0194 and multiply it by −0.0019464, that is, multiply Row 2 by [pic].
[pic]Subtract the result from Row 5 to get
[pic]
Divide Row 2 by 1.0194 and multiply it by 0.014843, that is, multiply Row 2 by [pic].
[pic]
Subtract the result from Row 6 to get
[pic]
Third step
Divide Row 3 by 0.77857 and multiply it by 0.52036, that is, multiply Row 3 by [pic].
[pic]
Subtract the result from Row 4 to get
[pic]
Divide Row 3 by 0.77857 and multiply it by 0.0098697, that is, multiply Row 3 by [pic].
[pic]
Subtract the result from Row 5 to get
[pic]
Divide Row 3 by 0.77857 and multiply it by 0.0078644, that is, multiply Row 3 by [pic].
[pic]
Subtract the result from Row 6 to get
[pic]
Fourth step
Divide Row 4 by 1.1264 and multiply it by −0.0012679, that is, multiply Row 4 by [pic].
[pic]
Subtract the result from Row 5 to get
[pic]
Divide Row 4 by 1.1264 and multiply it by 0.015126, that is, multiply Row 4 by [pic].
[pic]
Subtract the result from Row 6 to get
[pic]
Fifth step
Divide Row 5 by 0.80775 and multiply it by 0.60375, that is, multiply Row 5 by [pic].
[pic]
Subtract the result from Row 6 to get
[pic]
The six equations are
[pic] [pic]
[pic]
[pic]
[pic]
[pic]
Back Substitution
From the sixth equation
[pic]
[pic]
[pic]
Substituting the value of [pic] in the fifth equation,
[pic]
[pic]
[pic]
Substituting the value of [pic] and [pic] in the forth equation,
[pic]
[pic]
[pic]
Substituting the value of [pic], [pic] and [pic] in the third equation,
[pic]
[pic]
[pic]
Substituting the value of [pic], [pic], [pic] and [pic] in the second equation,
[pic]
[pic]
[pic]
Substituting the value of [pic], [pic], [pic], [pic], and [pic] in the first equation,
[pic]
[pic]
[pic]
Hence the solution vector is
[pic]
|SIMULTANEOUS LINEAR EQUATIONS | |
|Topic |Gaussian Elimination – More Examples |
|Summary |Examples of Gaussian elimination |
|Major |Electrical Engineering |
|Authors |Autar Kaw |
|Date |June 15, 2010 |
|Web Site | |
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