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               102 Transmission system compensation




















                      Fig. 3.20 Series compensated transmission line.


                      very low, such a system would be hypersensitive to small changes. For this reason the
                      degree of series compensation is limited in practice to about 80%.
                        It is not practicable to distribute the capacitance in small units along the line, so in
                      practice lumped capacitors are installed at a small number of locations (typically one
                      or two) along the line. This makes for an uneven voltage profile.
                        The line in Figure 3.20 is assumed to be a lossless, symmetrical line with a
                      mid-point series capacitor with equal shunt reactors connected on either side. To
                      permit the line to be analysed in two halves, the capacitor is split into two equal series
                      parts.

                      3.6.1 Power-transfer characteristics and maximum transmissible
                             power

                      The general phasor diagram is shown in Figure 3.21. Note that V 2 leads V 1 in phase
                      as a result of the voltage V Cg inserted by the capacitor. Considering the sending-end
                      half, the conditions at its two ends are related by equation (3.2)
                                                         y          y
                                              E s ˆ V 1 cos  ‡ jZ 0 I 1 sin
                                                         2          2                   (3:46)
                                                   V 1   y        y
                                               I s ˆ j  sin  ˆ I 1 cos
                                                   Z 0   2        2
                      The receiving-end half behaves similarly. The capacitor reactance is X Cg ˆ 1/oC g
                      and the voltage across the capacitor is given by

                                              V Cg ˆ V 1   V 2 ˆ jI m X Cg              (3:47)
                      By symmetry, P ˆ V m I m , E ˆ E r , and
                                                      1           1
                                            V m ˆ V 1   = 2 V Cg ˆ V 2 ‡ = 2 V Cg       (3:48)
                      The currents I 1 and I 2 are given by
                                                       jV 1       V 2
                                               I m ˆ I 1 ‡  ˆ I 2   j                   (3:49)
                                                        X         X
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