Análise matricial sistemas de energia

CURSO: ANÁLISE MATRICIAL DE SISTEMAS DE ENERGIA ELÉTRICA PROF.: DR. CARLOS A. CASTRO

Capítulo 1 Exercício 1 Item 1.a) Usando as regras de formação da matriz Y: 1. Elemento da diagonal Ykk = admitâncias dos ramos conectados ao nó k. 2. Elemento fora da diagonal Ykm = - admitância do ramo que conecta o nó k ao nó m →Senão há ramo conectando os nós k e m → Ykm = 0. A matriz para o sistema proposto será:

1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
Y1,1 -Y2,1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

2
-Y1,2 Y2,2 -Y3,2 0 0 0 0 0 -Y9,2 0 0 0 0 0 0 0 0 0 0 0 0 0

3
0 -Y2,3 Y3,3 -Y4,3 0 0 0 0 0 0 0 -Y12,3 0 0 0 0 0 0 0 0 0 0

4
0 0 -Y3,4 Y4,4 -Y5,4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

5
0 0 0 -Y4,5 Y5,5 0 0 0 0 0 0 0 0 -Y14,5 0 0 0 0 0 0 0 0

6
0 0 0 0 -Y5,6 Y6,6 -Y7,6 0 0 0 0 0 0 0 0 0 -Y17,6 0 0 0 0 0

7
0 0 0 0 0 -Y6,7 Y7,7 -Y8,7 0 0 0 0 0 0 0 0 0 0 0 -Y20,7 0 0

8
0 0 0 0 0 0 -Y7,8 Y8,8 0 0 0 0 0 0 0 0 0 0 0 0 0 0

9
0 -Y2,9 0 0 0 0 0 0 Y9,9 -Y10,9 -Y11,9 0 0 0 0 0 0 0 0 0 0 0

10
0 0 0 0 0 0 0 0 -Y9,10 Y10,10 0 0 0 0 0 0 0 0 0 0 0 0

11
0 0 0 0 0 0 0 0 -Y9,11 0 Y11,11 0 0 0 0 0 0 0 0 0 0 0

12
0 0 -Y3,12 0 0 0 0 0 0 0 0 Y12,12 -Y13,12 0 0 0 0 0 0 0 0 0

13
0 0 0 0 0 0 0 0 0 0 0 -Y12,13 Y13,13 0 0 0 0 0 0 0 0 0

14
0 0 0 0 -Y5,14 0 0 0 0 0 0 0 0 Y14,14 -Y15,14 0 0 0 0 0 0 0

15
0 0 0 0 0 0 0 0 0 0 0 0 0 -Y14,15 Y15,15 -Y16,15 0 0 0 0 0 0

16
0 0 0 0 0 0 0 0 0 0 0 0 0 0 -Y15,16 Y16,16 0 0 0 0 0 0

17
0 0 0 0 0 -Y6,17 0 0 0 0 0 0 0 0 0 0 Y17,17 -Y18,17 -Y19,17 0 0 0

18
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -Y17,18 Y18,18 0 0 0 0

19
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -Y17,19 0 Y19,19 0 0 0

20
0 0 0 0 0 0 -Y7,20 0 0 0 0 0 0 0 0 0 0 0 0 Y20,20 -Y21,20 -Y22,20

21
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -Y20,21 Y21,21 0

22
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -Y20,22 0 Y22,22

Item 1.b)

Grau de Esparsidade = ((NB2 – ( NB + 2 . NR ))/NB2).100 NB = 22 NR = 21 GE = ((222 – ( 22 + 2 . 21 ))/222).100 GE = 86,78%

Item 1.c) Com a injeção I5 no nó cinco temos (1) I5 =