How does the generator neutral reactance Xn appear in the zero sequence network?

3Xn in series with the machine X0 to the reference

Xn in series with X0 of the machine

Why are the grounded-star/grounded-star transformers kept as series branches in the zero sequence network?

Both grounded neutrals give a path for zero sequence current through them

They block zero sequence current

They are replaced by open circuits

2079 Bhadra · 8 marks

Sequence networks with parallel transformer banks

Short answer

On 50 MVA, 220 kV base: G1,G2 X1=X2=0.40 pu, X0=0.16 pu; neutral reactances 0.10 pu give 3Xn=0.30 pu in zero-sequence. Transformers X=0.375 pu, lines X1=X2=0.0517 pu, X0=0.0129 pu. Positive/negative network: parallel paths give 0.4008 pu between buses, machines 0.40 pu. Zero-sequence: bus tie 0.3815 pu, each machine branch 0.46 pu (0.16 + 0.30).

Draw the positive, negative and zero sequence impedance networks for the power system shown in figure below. Choose a base of 50 MVA, 220 kV in 50 Ω transmission lines and mark all reactances in pu. The ratings of the generators and transformers are: Generator 1: 25 MVA, 11 kV, X'' = 20%; Generator 2: 25 MVA, 11 kV, X'' = 20%; Three-phase transformer (each): 20 MVA, 11 Y/220 kV, x = 15%. The negative sequence reactance of each synchronous machine is equal to its subtransient reactance. The zero sequence reactance of each machine is 8%. Assume that the zero sequence reactances of lines are 25% of their positive sequence reactances. [Figure: machine 1 (star, neutral grounded through X = 5% on machine 1 rating) at bus 1; from bus 1 two parallel paths, each consisting of a Y-grounded/Y-grounded transformer, a j50 Ω line and another Y-grounded/Y-grounded transformer, to bus 2; machine 2 (star, neutral grounded through X = 5% on machine 2 rating) at bus 2]

Sequence networks are drawn by replacing each element with its positive, negative or zero sequence reactance on a common base. Only the generators have EMFs (positive network); neutral reactances appear as $3X_n$ in the zero sequence network only.

Base values

$S_b = 50$ MVA; $V_b = 220$ kV on the lines and 11 kV at the machines. $Z_{b,line} = 220^2/50 = 968\ \Omega$ .

Per unit reactances

Element	Positive	Negative	Zero
G1, G2: $0.20 \times 50/25$	0.40	0.40	$0.08 \times 2 = 0.16$
Neutral reactance: $0.05 \times 50/25 = 0.10$	–	–	$3X_n = 0.30$
Each transformer: $0.15 \times 50/20$	0.375	0.375	0.375
Each line: $50/968$	0.0517	0.0517	$0.25 \times 0.0517 = 0.0129$

Each parallel path (transformer + line + transformer):

Positive/negative: $0.375 + 0.0517 + 0.375 = 0.8017$ pu; two paths in parallel $= 0.4008$ pu
Zero: $0.375 + 0.0129 + 0.375 = 0.7629$ pu; two paths in parallel $= 0.3815$ pu

All transformers are star-grounded/star-grounded, so zero sequence current passes through them (series path, as in the positive network).

Positive sequence network

  1                                     2
  o---j0.375--j0.0517--j0.375---o
  |                             |
  +---j0.375--j0.0517--j0.375---+
  |                             |
 j0.40                        j0.40
  |                             |
 (E1)                          (E2)
  |                             |
  +======= reference bus =======+

Negative sequence network

Same as positive but without EMFs: G1 and G2 are j0.40 each, connected straight to the reference; the two paths of 0.8017 pu each between buses 1 and 2.

Zero sequence network

  1                                     2
  o---j0.375--j0.0129--j0.375---o
  |                             |
  +---j0.375--j0.0129--j0.375---+
  |                             |
 j0.16                        j0.16
  |                             |
 j0.30 (3Xn)                  j0.30 (3Xn)
  |                             |
  +===== reference (ground) ====+

Each machine branch in the zero sequence network: $0.16 + 0.30 = 0.46$ pu.

Summary: Positive: buses 1-2 joined by $j0.4008$ (equivalent), machines $j0.40$ with EMFs. Negative: same, no EMFs. Zero: buses joined by $j0.3815$ (equivalent), machines $j0.46$ to ground.

✓ Answer checked by Claude Sonnet 5.5

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