mws_ele_sle_txt_gaussian_examples

# mws_ele_sle_txt_gaussian_examples - Chapter 04.06 Gaussian...

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04.06.1 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. = 9 . 103 00 . 60 9 . 103 00 . 60 000 . 0 120 8080 . 0 6040 . 0 0100 . 0 0080 . 0 0100 . 0 0080 . 0 6040 . 0 8080 . 0 0080 . 0 0100 . 0 0080 . 0 0100 . 0 0100 . 0 0080 . 0 7787 . 0 5205 . 0 0100 . 0 0080 . 0 0080 . 0 0100 . 0 5205 . 0 7787 . 0 0080 . 0 0100 . 0 0100 . 0 0080 . 0 0100 . 0 0080 . 0 7460 . 0 4516 . 0 0080 . 0 0100 . 0 0080 . 0 0100 . 0 4516 . 0 7460 . 0 ci cr bi br ai ar I I I I I I Find the values of ar I , ai I , br I , bi I , cr I , and ci I 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 60536 . 0 7460 . 0 4516 . 0 = . ( ) [ ] [ ] 643 . 72 0048429 . 0 0060536 . 0 0048429 . 0 0060536 . 0 27338 . 0 4516 . 0 60536 . 0 1 Row = × Subtract the result from Row 2 to get = 9 . 103 00 . 60 9 . 103 00 . 60 643 . 72 120 8080 . 0 6040 . 0 0100 . 0 0080 . 0 0100 . 0 0080 . 0 6040 . 0 8080 . 0 0080 . 0 0100 . 0 0080 . 0 0100 . 0 0100 . 0 0080 . 0 7787 . 0 5205 . 0 0100 . 0 0080 . 0 0080 . 0 0100 . 0 5205 . 0 7787 . 0 0080 . 0 0100 . 0 014843 . 0 0019464 . 0 014843 . 0 0019464 . 0 0194 . 1 0 0080 . 0 0100 . 0 0080 . 0 0100 . 0 4516 . 0 7460 . 0 ci cr bi br ai ar I I I I I I

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