Sequence Circuits for Synchronous Generator

Section III: Sequence Circuits for Synchronous Generator

The three-phase equivalent circuit of a synchronous generator is shown in Fig. 1.16. This is redrawn in Fig. 7.6 with the neutral point grounded through a reactor with impedance Z_n . The neutral current is then given by

(7.32)

Fig. 7.6 Equivalent circuit of a synchronous generator with grounded neutral.

The derivation of Section 1.3 assumes balanced operation which implies I_a + I_b + I_c = 0. As per (7.32) this assumption is not valid any more. Therefore with respect to this figure we can write for phase-a voltage as

(7.33)

Similar expressions can also be written for the other two phases. We therefore have

(7.34)

Pre-multiplying both sides of (7.34) by the transformation matrix C we get

(7.35)

Since the synchronous generator is operated to supply only balanced voltages we can assume that E_an0 = E_an2 = 0 and E_an1 = E_an . We can therefore modify (7.35) as

(7.36)

We can separate the terms of (7.36) as

	(7.37)
	(7.38)
	(7.39)

Furthermore we have seen for a Y-connected load that V_a1 = V _an1 , V_a2 = V_an2 since the neutral current does not affect these voltages. However V_a0 = V_an0 + V_n . Also we know that V_n = - 3Z_nI_a0 . We can therefore rewrite (7.37) as

(7.40)

The sequence diagrams for a synchronous generator are shown in Fig. 7.7.

Fig. 7.7 Sequence circuits of synchronous generator: (a) positive, (b) negative and (c) zero sequence.

Section IV: Sequence Circuits for Symmetrical Transmission Line

The schematic diagram of a transmission line is shown in Fig. 7.8. In this diagram the self impedance of the three phases are denoted by Z_aa , Z_bb and Z_cc while that of the neutral wire is denoted by Z_nn . Let us assume that the self impedances of the conductors to be the same, i.e.,

Since the transmission line is assumed to be symmetric, we further assume that the mutual inductances between the conductors are the same and so are the mutual inductances between the conductors and the neutral, i.e.,

The directions of the currents flowing through the lines are indicated in Fig. 7.8 and the voltages between the different conductors are as indicated.

Fig. 7.8 Lumped parameter representation of a symmetrical transmission line.

Applying Kirchoff's voltage law we get

(7.41)

Again

(7.42)

(7.43)

Substituting (7.42) and (7.43) in (7.41) we get

(7.44)

Since the neutral provides a return path for the currents I_a , I_b and I_c , we can write

(7.45)

Therefore substituting (7.45) in (7.44) we get the following equation for phase-a of the circuit

(7.46)

Denoting

(7.46) can be rewritten as

(7.47)

Since (7.47) does not explicitly include the neutral conductor we can define the voltage drop across the phase-a conductor as

(7.48)

Combining (7.47) and (7.48) we get

(7.49)

Similar expression can also be written for the other two phases. We therefore get

(7.50)

Pre-multiplying both sides of (7.50) by the transformation matrix C we get

(7.51)

Now

Hence

Therefore from (7.51) we get

(7.52)

The positive, negative and zero sequence equivalent circuits of the transmission line are shown in Fig. 7.9 where the sequence impedances are

Fig. 7.9 Sequence circuits of symmetrical transmission line: (a) positive, (b) negative and (c) zero sequence.