--- title: "Reactance diagram: 20 MVA generator, 15 kV/25 MVA base" subject: "Power System Analysis I" chapter: 3 key: "semester_4/power_system_analysis_i:reactance-diagram-20mva-15kv-25mva" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/reactance-diagram-20mva-15kv-25mva site: BCT Engineering --- # Reactance diagram: 20 MVA generator, 15 kV/25 MVA base Power System Analysis I · Chapter 3: Computational Technique - Asked in: 2078 Chaitra, 10 marks - Video: [Impedance and Reactance diagram Problem-3 | Power System Analysis | PSA | per unit Reactance diagram](https://www.youtube.com/watch?v=TQ8W4-l9S2I) — Simple n Easy Learning - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/reactance-diagram-20mva-15kv-25mva - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3 ## Question A 20 MVA, 11 kV three-phase synchronous generator has a sub-transient reactance of 10%. It is connected through three identical single-phase Δ-Y connected transformers of 5000 kVA, 11/127.02 kV with a reactance of 15% to a high voltage transmission line having a total series reactance of j180 Ω. At the end of the HT transmission line, three identical single-phase star/star connected transformers of 5000 kVA, 127.02/12.702 kV with a reactance of 20%. The load is drawing 15 MVA at 20 kV at 0.9 pf lagging. Draw a single line diagram of the network and determine the reactance diagram. Choose a common base of 15 kV and 25 MVA. ## Short answer Using base 25 MVA, 15 kV at generator, line base 300 kV, load base 30 kV. Per-unit reactances: generator j0.0672, T1 j0.1344, line j0.05, T2 j0.1793. Load consumes 0.6 p.u. at 0.9 lag, represented as impedance 0.6667 + j0.3229 p.u. ## Answer #### Single line diagram ```text G T1 (3 x 1-ph) line T2 (3 x 1-ph) (~)--||--[ D | Y ]----------j180----[ Y | Y ]----> Load 20 MVA 5 MVA each 5 MVA each 15 MVA 11 kV 11/127.02 kV 127.02/ 20 kV X"=10% X=15% 12.702 kV 0.9 lag X=20% ``` #### Transformer bank ratings - T1 (Δ–Y): $3\times5 = 15$ MVA. Line voltages: 11 kV / $\sqrt3\times127.02 = 220$ kV. - T2 (Y–Y): 15 MVA, 220 kV / $\sqrt3\times12.702 = 22$ kV. #### Base values Assumption: the 15 kV base is taken in the generator circuit, with $S_b = 25$ MVA. | Section | Base kV | $Z_b$ (Ω) | |---|---|---| | Generator | 15 | 9 | | Line | $15\times220/11 = 300$ | $300^2/25 = 3600$ | | Load | $300\times22/220 = 30$ | 36 | #### Per-unit reactances $$ \begin{aligned} X_G &= 0.10\times\frac{25}{20}\left(\frac{11}{15}\right)^2 = 0.0672\\ X_{T1} &= 0.15\times\frac{25}{15}\left(\frac{11}{15}\right)^2 = 0.1344\\ X_{line} &= \frac{180}{3600} = 0.05\\ X_{T2} &= 0.20\times\frac{25}{15}\left(\frac{220}{300}\right)^2 = 0.1793 \end{aligned} $$ #### Load 15 MVA at 20 kV and 0.9 lagging: $S = 15/25 = 0.6$ p.u., $V = 20/30 = 0.6667$ p.u. $$ S = 0.6\angle25.84^\circ = 0.54 + j0.2615\ \text{p.u.} $$ As a series impedance: $$ Z_L = \frac{|V|^2}{S^*} = \frac{0.6667^2}{0.6}\angle25.84^\circ = 0.7407\angle25.84^\circ = 0.6667 + j0.3229\ \text{p.u.} $$ #### Reactance diagram ```text j0.0672 j0.1344 j0.05 j0.1793 +-/\/\------/\/\------/\/\------/\/\----+ | | (Eg) 0.6667 + j0.3229 | (load) +----------------- reference -----------+ ``` | Element | p.u. (25 MVA, 15 kV gen base) | |---|---| | Generator | j0.0672 | | T1 | j0.1344 | | Line | j0.0500 | | T2 | j0.1793 | | Load | 0.6667 + j0.3229 (0.54 + j0.2615 p.u. power) | ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Three single-phase 5 MVA, 11/127.02 kV units in a Δ-Y bank have what three-phase rating and line ratio? A. 5 MVA, 11/127.02 kV B. 15 MVA, 11/220 kV C. 15 MVA, 11/127.02 kV D. 5 MVA, 11/220 kV Answer: B. Rating = 3x5 = 15 MVA; the Y side line voltage is sqrt(3)x127.02 = 220 kV. (Key point: Transformer bank rating) 2. With 15 kV base at the generator, what is the base voltage of the line? A. 15 x 220/11 = 300 kV B. 15 x 127.02/11 = 173 kV C. 220 kV D. 15 kV Answer: A. Base voltage changes by the line-to-line transformer ratio 220/11, giving 300 kV. (Key point: Base voltage zones) 3. Why does the generator's 10% reactance get multiplied by (11/15)^2? A. Because the line is at 300 kV B. To convert from MVA to kVA C. To correct the rated voltage 11 kV to the base 15 kV D. Because the generator is star-connected Answer: C. Z_pu scales with (V_old/V_new)^2 as well as S_new/S_old; here the rated voltage 11 kV differs from the 15 kV base. (Key point: Per-unit change of base) 4. How is the load turned into a series impedance in p.u.? A. Z = S/|V|^2 B. Z = |V|^2 / S* with S = 0.6 p.u. and V = 0.6667 p.u. C. Z = V/S D. Z = 1/S Answer: B. Z_L = |V|^2/S*, giving 0.7407∠25.84° = 0.6667 + j0.3229 p.u. 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