--- title: "Reactance diagram: 90 MVA generator (15%), two motors" subject: "Power System Analysis I" chapter: 3 key: "semester_4/power_system_analysis_i:reactance-diagram-90mva-generator-15" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/reactance-diagram-90mva-generator-15 site: BCT Engineering --- # Reactance diagram: 90 MVA generator (15%), two motors Power System Analysis I · Chapter 3: Computational Technique - Asked in: 2081 Chaitra (new course), 8 marks - Video: [Problem #2 | Impedance & Reactance diagram | PSA](https://www.youtube.com/watch?v=Mq3pAkxuIjc) — FOR STUDENTS - 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-90mva-generator-15 - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3 ## Question A 90 MVA, 11 kV 3-phase generator has a reactance of 15%. The generator supplies two motors through transformers and transmission line shown in figure below. The transformer T1 is a 3-phase transformer, 100 MVA, 10/132 kV, 6% reactance. The transformer T2 is composed of 3 single phase units each rated at 30 MVA, 66/10 kV with 5% reactance. The connection of T1 and T2 are as shown. The motors rated at 50 MVA and 40 MVA, both 10 kV and 20% reactance. Taking the generator rating as base, draw the reactance diagram and indicate the reactance in per unit. The reactance of the line is 100 Ω. [Figure: G (Y, grounded) → T1 (Δ–Y, Y grounded) → Line → T2 (Y grounded–Δ) → bus feeding motor M1 (Y, grounded) and motor M2 (Δ).] ## Short answer On a 90 MVA base, generator voltage 11 kV, line base 145.2 kV, motor base 12.70 kV. Per-unit reactances: G j0.15, T1 j0.0446, line j0.4269, T2 j0.0310, M1 j0.2231, M2 j0.2789. Values found by adjusting old p.u. with ratio of bases. ## Answer #### Choice of bases - $S_b = 90$ MVA and $V_b = 11$ kV in the generator circuit. - Line: through T1 (10/132 kV), $V_b = 11\times\dfrac{132}{10} = 145.2$ kV. - T2 is a Y–Δ bank of three single-phase 66/10 kV units. Its line-voltage ratio is $\sqrt3\times66 = 114.32$ kV / 10 kV, and its three-phase rating is $3\times30 = 90$ MVA. - Motor circuit: $V_b = 145.2\times\dfrac{10}{114.32} = 12.70$ kV. #### Per-unit reactances Use $X_{new} = X_{old}\,\dfrac{S_{b}}{S_{rated}}\left(\dfrac{V_{rated}}{V_b}\right)^2$. $$ \begin{aligned} G&: 0.15\ \text{(same base)}\\ T_1&: 0.06\times\frac{90}{100}\times\left(\frac{10}{11}\right)^2 = 0.0446\\ \text{Line}&: Z_b = \frac{145.2^2}{90} = 234.26\ \Omega,\quad X = \frac{100}{234.26} = 0.4269\\ T_2&: 0.05\times\frac{90}{90}\times\left(\frac{10}{12.70}\right)^2 = 0.0310\\ M_1&: 0.2\times\frac{90}{50}\times\left(\frac{10}{12.70}\right)^2 = 0.2231\\ M_2&: 0.2\times\frac{90}{40}\times\left(\frac{10}{12.70}\right)^2 = 0.2789 \end{aligned} $$ (T2 gives the same value on its HV side: $0.05\times(114.32/145.2)^2 = 0.0310$.) | Element | p.u. reactance (90 MVA base) | |---|---| | Generator | j0.150 | | T1 | j0.0446 | | Line | j0.4269 | | T2 | j0.0310 | | Motor M1 | j0.2231 | | Motor M2 | j0.2789 | #### Reactance diagram ```text j0.15 j0.0446 j0.4269 j0.031 +-/\/\-----/\/\------/\/\-----/\/\---+-----+ | | | (Eg) j0.2231 j0.2789 | | | | (Em1) (Em2) +------------- reference --------------+-----+ ``` The transformer winding connections (Y, Δ, earthing) do not change the positive-sequence p.u. reactances. They only introduce a phase shift, which is ignored in the reactance diagram. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The line base voltage is A. 11 kV B. 132 kV C. 145.2 kV D. 12.7 kV Answer: C. 11 x 132/10 = 145.2 kV. (Key point: Line base voltage) 2. T2 consists of three single-phase 66/10 kV units in Y-Delta. Its line ratio is A. 66/10 kV B. 114.32/10 kV C. 66/17.3 kV D. 198/10 kV Answer: B. Line voltage on the Y side is sqrt(3) x 66 = 114.32 kV; MVA rating is 3 x 30 = 90. (Key point: T2 ratio) 3. T1 on the 90 MVA, 11 kV base is A. 0.06 B. 0.0446 C. 0.0545 D. 0.1 Answer: B. 0.06 x 90/100 x (10/11)^2 = 0.0446. (Key point: T1 p.u.) 4. Transformer connections (Y, Delta, earthing) in the positive-sequence reactance diagram A. Change the p.u. values B. Only cause a phase shift that is ignored C. Add extra branches D. Double the reactance Answer: B. They do not change positive-sequence p.u. reactances. (Key point: Connections) ## Close variants set in other exams - [Reactance diagram: 90 MVA generator (25%), two motors](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-90mva-generator-25) — Power System Analysis I, chapter 3, asked 2× ## Related in Power System Analysis I - [Reactance diagram: 300 MVA, 20 kV generator, two motors](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-300mva-20kv) — Power System Analysis I, chapter 3 - [Reactance diagram: 20 MVA generator, 15 kV/25 MVA base](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-20mva-15kv-25mva) — Power System Analysis I, chapter 3 - [Reactance diagram: 100 MVA base, two parallel paths](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-90mva-two-paths) — Power System Analysis I, chapter 3 - [Impedance diagram: 25 MVA generator, two motors](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aimpedance-diagram-25mva-two-motors) — Power System Analysis I, chapter 3 - [Per unit values: 100 MVA generator, two motors](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aper-unit-100mva-6kv-two-motors) — Power System Analysis I, chapter 3 - [Reactance diagram: two generators, two motors, j50 Ω line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-two-gens-two-motors) — Power System Analysis I, chapter 3 ## Related in other subjects - [Design: 100 kVA 11/132 kV transformer with winding data](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-100kva-11-132kv) — Electric Machine Design, chapter 3, asked 3× - [L-L fault through Zf with two motors](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3All-fault-motors-zf-new) — Power System Analysis II, chapter 5 - [Differential protection: 11 kV, 100 MVA generator](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Agenerator-differential-11kv-6-ohm) — Switchgear and Protection, chapter 7 - [IDMT relays across 132/11 kV transformer (3500 A)](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Aidmt-132-11kv-3500a) — Switchgear and Protection, chapter 7