Sequence networks of power systems are very useful for computing unsymmetrical faults at different points of a power system network. The knowledge of positive- sequence network is necessary for load-studies on power systems. If the stability studies involve unsymmetrical faults, then negative- and zero-sequence networks are required also.
A power system network consists of synchronous machines, transformers and lines. Using these, complete sequence networks of a power system can be easily drawn. The positive-sequence network is drawn by examining one line diagram of the power system. In fact, the single line reactance diagram, as employed for calculation of symmetrical fault current, is the positive-sequence diagram of the power system.
The negative-sequence network is quite similar to positive-sequence network—only generators or rotating machines may have different sequence impedances and the negative-sequence network does not contain any voltage source. The negative-sequence impedances for transmission lines and transformers are the same as the positive-sequence impedances.
In many cases only one sequence network is drawn for positive- and negative-sequence representation. The reference bus for positive- and negative-sequence networks is the system neutral. Any impedance connected between a neutral and ground is not included in the positive- and negative-sequence networks as neither of these sequence currents can flow through such an impedance.
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Zero-sequence sub-networks for different elements of a power system can be easily combined to form complete zero-sequence network. The zero-sequence network does not contain any voltage source. Any impedance included in generator or transformer neutral becomes three times its value in a zero-sequence network. Special care needs to be taken in connecting the zero-sequence impedance of transformer.
All the sequence impedances are expressed in per-unit values and referred to the same base MVA and base kV.
The procedure for drawing sequence networks is illustrated through the following example:
Example:
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Draw the sequence networks for the power system shown in Fig. 3.18:
Solution:
Let base MVA for complete system be 50 MVA, base kV for generator side 11 kV and transmission side 220 kV.
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Per unit reactances of generator G1 –
X1 = X2 = j 0.2 pu
and X0 = j 0.05 pu …(Given)
Per unit reactances of generator G2 –
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X1 = X2 = [j 0.15 x (50/30)] = j 0.25 pu
and X0 = [j 0.03 x (50/30)] = j 0.05 pu
Per unit reactances of transformer T1 –
X1 = X2 = X0 = j 0.10 pu
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Per unit reactances of transformer T2 –
X1 = X2 = X0 = [j 0.075 x (50/30)] = j 0.125 pu
Per unit reactances of transmission line 1 –
X1 = X2 = 0.1 pu …(Given)
Assuming zero-sequence reactance of transmission line 3.5 times of its positive-sequence reactance –
X0 = 0.35 pu
Per unit reactances of transmission line 2 –
X1 = X2 = j 0.12 pu (Given)
X0 = j 0.12 x 3.5 = j 0.42 pu
The neural reactance of j 0.024 and j 0.03 comes out to be [j 0.024 x 950/30)] i.e., 0.4 pu and [j 0.03 x (50/30)] i.e., j 0.05 pu.
These neutral reactances appear as 3 x j 0.04 i.e., j 0.12 pu and 3 x j 0.05 i.e., j 0.15 pu in the zero sequence networks.
The positive-, negative- and- zero sequence networks are shown in Figs. 3.19 (a), 3.19 (b) and 3.19 (c) respectively:
