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               124 Power flows in compensation and control studies











                      Fig. 4.13 Representation of series capacitor.


















                      Fig. 4.14 A three-node test system.

                      It should be mentioned that in power flow studies the contribution of generators and
                      loads is made through the vector of nodal currents as opposed to the nodal admit-
                      tance matrix. In this example, the admittance elements in equation (4.29) were placed
                      in locations (1, 1), (1, 2), (2, 1) and (2, 2) of the matrix in equation (4.31) since the
                      transformer is connected between nodes one and two of the network. Similarly, the
                      admittances in equation (4.28) were placed in locations (2, 2), (2, 3), (3, 2) and (3, 3)
                      since the transmission line is connected between nodes 2 and 3. The contribution of
                      the shunt capacitor is only in location (3, 3). It should be noted that zero entries exist
                      in locations (1, 3) and (3, 1) since there is no transmission element directly linking
                      nodes one and three in this network.

                      4.4.3  Nodal impedance based power flow method

                      Power flow solutions may be achieved quite simply by using equation (4.18), which is
                      the nodal admittance matrix equation of the network. The simplest case corresponds
                      to a power network where only one generator exists in the system. By definition, this
                      would be the Slack generator and the voltage magnitude and phase angle become
                      known at its point of connection, say node one,
                                             2                         3
                                     2   3                              2   3
                                              Y 11   Y 12  Y 13      Y 1n
                                       I 1                                V 1
                                             6                         7
                                       I 2   6  Y 21  Y 22  Y 23      Y 2n 7 V 2
                                     6   7                              6   7
                                     6   7                              6   7
                                             6                         7
                                     6   7                              6   7
                                       I 3 7 ˆ  6  Y 31  Y 32  Y 33                     (4:32)
                                                .      .   .   .     .
                                     6                             Y 3n 7 V 3 7
                                                                        6
                                             6                         7
                                     6  . 7     .      .   .    .       6  . 7
                                       . . 5  6  .     .   .     .   . 7   . . 5
                                     4                                  4
                                             4                       . 5
                                       I n    Y n1   Y n2  Y n3      Y nn  V n
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