--- title: "Reactance diagram: 90 MVA generator (25%), two motors" subject: "Power System Analysis I" chapter: 3 key: "semester_4/power_system_analysis_i:reactance-diagram-90mva-generator-25" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/reactance-diagram-90mva-generator-25 site: BCT Engineering --- # Reactance diagram: 90 MVA generator (25%), two motors Power System Analysis I · Chapter 3: Computational Technique - Asked in: 2075 Bhadra, 8 marks - Video: [How to find the perunit impedance of each element in electric electric power system using MATLAB??](https://www.youtube.com/watch?v=0VquSwY6zng) — Electrical Engineering simplified - 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-25 - 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 25%. The generator supplies two motors through transformers and transmission line shown in figure below. The transformer T1 is a 3-phase, 100 MVA, 10/132 kV, 6% reactance. The transformer T2 is composed of three single phase units each rated at 30 MVA, 66/10 kV with 5% reactance. The connection of T1 and T2 are shown in figure. The motors are rated at 50 MVA and 40 MVA, both 10 kV and 20% reactance. Taking the generator rating as base, draw the per unit reactance diagram. The reactance of the line is 100 Ω. [Figure: G1 → T1 (Δ on generator side, Y grounded on line side) → Line → T2 (Y grounded on line side, Δ on motor side) → bus feeding M1 (Y, grounded) and M2 (Δ).] ## Short answer On 90 MVA base: X_G=0.25, X_T1=0.0446, X_line=0.4269, X_T2=0.0310, X_M1=0.2231, X_M2=0.2789 pu. Base voltages: 11 kV, 145.2 kV, 12.70 kV. ## Answer Base: 90 MVA and 11 kV in the generator circuit (generator rating). Base voltages in other zones follow the line-to-line turns ratios. #### Base voltages - Generator zone: $V_{b1} = 11$ kV - Line zone (T1 is 10/132 kV): $V_{b2} = 11\times\frac{132}{10} = 145.2$ kV - T2: three single-phase 66/10 kV units connected Y (line side) – Δ (motor side), so line-to-line ratio is $66\sqrt{3}/10 = 114.32/10$ kV. Motor zone: $$ V_{b3} = 145.2\times\frac{10}{114.32} = 12.70\ \text{kV} $$ $$ Z_{b2} = \frac{145.2^2}{90} = 234.26\ \Omega $$ #### Per-unit reactances (on 90 MVA) $$ \begin{aligned} X_G &= 0.25\ \text{pu}\\ X_{T1} &= 0.06\times\frac{90}{100}\times\left(\frac{10}{11}\right)^2 = 0.0446\ \text{pu}\\ X_{line} &= \frac{100}{234.26} = 0.4269\ \text{pu}\\ X_{T2} &= 0.05\times\frac{90}{3\times30}\times\left(\frac{10}{12.70}\right)^2 = 0.0310\ \text{pu}\\ X_{M1} &= 0.2\times\frac{90}{50}\times\left(\frac{10}{12.70}\right)^2 = 0.2231\ \text{pu}\\ X_{M2} &= 0.2\times\frac{90}{40}\times\left(\frac{10}{12.70}\right)^2 = 0.2789\ \text{pu} \end{aligned} $$ | Element | X (pu, 90 MVA base) | |---|---| | Generator | 0.2500 | | T1 | 0.0446 | | Line (100 Ω) | 0.4269 | | T2 | 0.0310 | | Motor M1 | 0.2231 | | Motor M2 | 0.2789 | #### Reactance diagram ```text j0.0446 j0.4269 j0.0310 +---/\/\/-----/\/\/-----/\/\/---+--------+ | T1 line T2 | | j0.25 j0.2231 j0.2789 | | | (Eg) (Em1) (Em2) | | | +-------------------------------+--------+ reference bus ``` **Answer:** On 90 MVA base: $X_G = 0.25$, $X_{T1} = 0.0446$, $X_{line} = 0.4269$, $X_{T2} = 0.0310$, $X_{M1} = 0.2231$, $X_{M2} = 0.2789$ pu; base voltages 11 kV, 145.2 kV and 12.70 kV. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. T1 is 10/132 kV and T2 is Y-Δ 66/10 kV single-phase units. What is the motor-zone base? A. 10 kV B. 11 kV C. 12.70 kV D. 145.2 kV Answer: C. Line zone 145.2 kV, T2 ratio 114.32/10 gives 12.70 kV. (Key point: Motor zone base voltage) 2. What is T2's line-to-line ratio? A. 66/10 kV B. 66sqrt(3)/10 = 114.32/10 kV C. 66/(10 sqrt(3)) kV D. 198/10 kV Answer: B. The Y side is sqrt(3) x 66 = 114.32 kV. (Key point: T2 line ratio) 3. Why is X_T1 = 0.06 x 90/100 x (10/11)^2? A. Rated 10 kV of T1 LV winding differs from the 11 kV base B. T1 is a delta transformer C. Line is 145.2 kV D. The generator has 25% Answer: A. Both MVA and voltage corrections are required. (Key point: T1 correction) 4. What is the line p.u. reactance? A. 100/234.26 = 0.4269 B. 100/90 C. 100/145.2 D. 100/121 Answer: A. Zb = 145.2^2/90 = 234.26 Ω. (Key point: Line Zb) ## Close variants set in other exams - [Reactance diagram: 90 MVA generator (15%), two motors](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-90mva-generator-15) — 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 - [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 - [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 - [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