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               130 Power flows in compensation and control studies

                      4.4.5  Numerical example 3

                      The power transmission circuit shown in Figure 4.16 consists of two nodes, one
                      generator, one transmission line and one load. The load has a value of 1 ‡ j0:5p:u:
                      and the reactance of the transmission line is 0.1 p.u. The voltage magnitude and
                      phase angle at the generator point (Slack node) are kept at 1 p.u. and 0 radians,
                      respectively.
                        Using the Newton±Raphson power flow method, the nodal voltage magnitude and
                      phase angle at node two will be determined using 1 p.u. voltage magnitude and zero
                      phase angle at the load point as the initial condition.
                        Applying the theory presented in Section 4.3, the nodal admittance matrix is
                      formed for this system


                                                          j10   j10
                                           Y ˆ G ‡ jB ˆ             p:u:
                                                          j10   j10
                      The net active and reactive powers are calculated


                                                ?           net   gen   load     ?
                                  gen
                                        load
                            net
                           P   ˆ P     P   ˆ       p:u:   Q   ˆ Q      Q    ˆ       p:u:
                                                1                               0:5
                      First iteration
                      Using the initial voltage values V 1 j ˆ V 2 j ˆ 1, y 1 ˆ y 2 ˆ 0 and the values in the
                                                        j
                                                  j
                      nodal admittance matrix in equations (4.42) and (4.43), the following powers are
                      calculated
                                  P calc  ˆ V 2 kV 1 j G 21 cos (y 2   y 1 ) ‡ B 21 sin (y 2   y 1 )g
                                               f
                                        j
                                   2
                                      ‡ V 2 kV 2 j G 22 cos (y 2   y 2 ) ‡ B 22 sin (y 2   y 2 )g ˆ 0
                                        j
                                              f
                                                                                        (4:57)
                                 Q calc  ˆ V 2 kV 1 j G 21 sin (y 2   y 1 )   B 21 cos (y 2   y 1 )g
                                        j
                                               f
                                   2
                                      ‡ V 2 kV 2 j G 22 sin (y 2   y 2 )   B 22 cos (y 2   y 2 )g ˆ 0
                                              f
                                        j














                      Fig. 4.16 A two-node system.
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