Dinamica y Control De Sistema Electrico De Potencia

1- Matriz de Admitancia (Sistemas Acoplados).
El sistema de la Figura 1 posee dos líneas acopladas mutuamente, conforme es descripta en la Tabla 1. Calcular la matriz Ybus (matriz de admitancia debarras). Calcular;
a. La matriz incidencia de barras.
b. La matriz admitancia primitiva.
c. La matriz admitancia nodal del sistema.
Carga 1
Carga 2
(1)
(2)
1
3
4
5
2
6

Línea |Reactancia (pu) | AcopladaA la línea | Reactancia (pu) |
0 – 1 | 0.10 | | |
1 – 2 | 0.25 | | |
2 – 4 | 0.50 | | |
2 – 5 | 0.50 | 5 – 6 (2) | 0.10 |
3 – 6 | 0.20 | | |
0 – 3 | 0.10| | |
4 – 6 | 0.20 | | |
0 – 4 | 0.08 | | |
5 – 6 (1) | 0.40 | 5 – 6 (2) | 0.20 |
5 – 6 (2) | 0.40 | 5 – 6 (1) | 0.20 |
0 – 5 | 0.05 | | |

Grafo.

Matriz de IncidenciaElemento-Nodo.
| 0 | 1 | 2 | 3 | 4 | 5 | 6 |
a | 1 | -1 | 0 | 0 | 0 | 0 | 0 |
b | 0 | 1 | -1 | 0 | 0 | 0 | 0 |
c | 0 | 0 | 1 | 0 | -1 | 0 | 0 |
d | 0 | 0 | 1 | 0 | 0 | -1 | 0 |
e | 0 | 0 | 0 |1 | 0 | 0 | -1 |
f | 1 | 0 | 0 | -1 | 0 | 0 | 0 |
g | 0 | 0 | 0 | 0 | 1 | 0 | -1 |
h | -1 | 0 | 0 | 0 | 1 | 0 | 0 |
i | 0 | 0 | 0 | 0 | 0 | -1 | 1 |
j | 0 | 0 | 0 | 0 | 0 | -1 | 1 |
k | -1| 0 | 0 | 0 | 0 | 1 | 0 |

Matriz de Incidencia de Barras. (Ab).
| 1 | 2 | 3 | 4 | 5 | 6 |
a | -1 | 0 | 0 | 0 | 0 | 0 |
b | 1 | -1 | 0 | 0 | 0 | 0 |
c | 0 | 1 | 0 | -1 | 0 | 0 |
d | 0 |1 | 0 | 0 | -1 | 0 |
e | 0 | 0 | 1 | 0 | 0 | -1 |
f | 0 | 0 | -1 | 0 | 0 | 0 |
g | 0 | 0 | 0 | 1 | 0 | -1 |
h | 0 | 0 | 0 | 1 | 0 | 0 |
i | 0 | 0 | 0 | 0 | -1 | 1 |
j | 0 | 0 | 0 | 0 | -1| 1 |
k | 0 | 0 | 0 | 0 | 1 | 0 |

Matriz de Admitancia Primitiva (Yeq).

-10 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | -4 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | -2 | 0 | 0 | 0| 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | -2 | 0 | 0 | 0 | 0 | -10 | 0 | 0 |
0 | 0 | 0 | 0 | -5 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0 | -10 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0 | 0 | -5 |...