mws_ele_sle_txt_seidel_examples

mws_ele_sle_txt_seidel_examples - 04.08 Gauss-Seidel Method...

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04.08.1 04.08 Gauss-Seidel Method – 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 the Gauss-Seidel method. Use = 20 20 20 20 20 20 ci cr bi br ai ar I I I I I I as the initial guess and conduct two iterations. Solution Rewriting the equations gives ( ) ( ) ( ) 7460 . 0 0080 . 0 0100 . 0 0080 . 0 0100 . 0 4516 . 0 120 ci cr bi br ai ar I I I I I I = 7460 0 0100 0 0080 0 0100 0 0080 0 4516 0 000 0 . I . I . I . I . I . . I ci cr bi br ar ai = ( ) ( ) ( ) 7787 0 0080 0 0100 0 5205 0 0080 0 0100 0 00 60 . I . I . I . I . I . . I ci cr bi ai ar br = 7787 0 0100 0 0080 0 5205 0 0100 0 0080 0 9 103 . I . I . I . I . I . . I ci cr br ai ar bi =
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04.08.2 Chapter 04.08 ( ) ( ) ( ) 8080 0 6040 0 0080 0 0100 0 0080 0 0100 0 00 60 . I . I . I . I . I . . I ci bi br ai ar cr = 8080 0 6040 0 0100 0 0080 0 0100 0 0080 0 9 103 . I
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This note was uploaded on 06/12/2011 for the course EML 3041 taught by Professor Kaw,a during the Spring '08 term at University of South Florida - Tampa.

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mws_ele_sle_txt_seidel_examples - 04.08 Gauss-Seidel Method...

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