--- title: "Sequence networks with parallel transformer banks" subject: "Power System Analysis II" chapter: 4 key: "semester_5/power_system_analysis_ii:sequence-networks-parallel-transformers-2079" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/sequence-networks-parallel-transformers-2079 site: BCT Engineering --- # Sequence networks with parallel transformer banks Power System Analysis II · Chapter 4: Unbalanced System Analysis - Asked in: 2079 Bhadra, 8 marks - Video: [POSITIVE, NEGATIVE, ZERO SEQUENCE REACTANCE DIAGRAM / KTU/ POWER SYSTEM ANALYSIS](https://www.youtube.com/watch?v=zWNFHgJ7b5s) — praveen raj - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/sequence-networks-parallel-transformers-2079 - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-4 ## Question 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] ## 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). ## Answer 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 ```text 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 ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does the generator neutral reactance Xn appear in the zero sequence network? A. It does not appear B. Xn in series with X0 of the machine C. Xn in parallel with X0 D. 3Xn in series with the machine X0 to the reference Answer: D. Zero sequence current of all three phases flows through Xn, so it appears as 3Xn. (Key point: Neutral reactance in zero sequence) 2. How is the generator X'' = 20% on 25 MVA converted to the 50 MVA base? A. 0.20 x 25/50 = 0.10 B. 0.20 x 50/25 = 0.40 C. 0.20 x 220/11 D. 0.20 unchanged Answer: B. X_new = X_old x (S_base_new/S_base_old) = 0.20 x 50/25 = 0.40 pu. (Key point: Per unit conversion) 3. Which networks contain generator EMF sources? A. All three B. Negative only C. Positive sequence network only D. Zero only Answer: C. Machines generate only balanced positive sequence voltages, so negative and zero networks are passive. (Key point: Where EMFs appear) 4. Why are the grounded-star/grounded-star transformers kept as series branches in the zero sequence network? A. They have zero reactance B. Both grounded neutrals give a path for zero sequence current through them C. They block zero sequence current D. They are replaced by open circuits Answer: B. A Yg-Yg transformer passes zero sequence current, so its leakage reactance appears in series as in the positive network. 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