--- title: "Feeder reactor: 40 and 30 MVA, 50% reduction" subject: "Power Plant Equipment" chapter: 4 key: "semester_7/power_plant_equipment:fault-40-30mva-50-percent" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/fault-40-30mva-50-percent site: BCT Engineering --- # Feeder reactor: 40 and 30 MVA, 50% reduction Power Plant Equipment · Chapter 4: Substation Equipments - Asked in: 2078 Bhadra, 5 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_plant_equipment/fault-40-30mva-50-percent - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-4 ## Question Two generators of capacities 40MVA, 11kV and 30MVA, 11kV having their sub transient reactance of 5% and 4% respectively operating in parallel and supplying to a common load by single feeder of reactance 2Ω. Calculate the value of reactor to be connected in series with feeder so as to reduce the fault level at the end of the feeder by 50%. ## Short answer Existing fault level is 58.23 MVA. To reduce it by 50% to 29.11 MVA, double the total reactance from 0.687 pu to 1.374 pu by adding a series reactor of 0.687 pu (2.078 Ω). ## Answer Work in per unit on a common base. Fault MVA $=\text{MVA}_b/X_{eq}$; ohms to pu: $X_{pu} = X_\Omega\,\text{MVA}_b/\text{kV}^2$. **Base:** 40 MVA, 11 kV; $Z_{base} = 11^2/40 = 3.025\ \Omega$. ```text G1 40MVA,5% G2 30MVA,4% | | ====+=====+=======+==== 11 kV | 2 ohm feeder + X | F ``` #### Present fault level $$ \begin{aligned} X_{G1} &= 0.05\ \text{pu},\quad X_{G2} = 0.04\times\frac{40}{30} = 0.05333\ \text{pu} \\ X_{G1}\parallel X_{G2} &= \frac{0.05\times0.05333}{0.10333} = 0.025806\ \text{pu} \\ X_{feeder} &= \frac{2}{3.025} = 0.66116\ \text{pu} \\ X_{eq} &= 0.025806 + 0.66116 = 0.68696\ \text{pu} \\ \text{Fault MVA} &= \frac{40}{0.68696} = 58.23\ \text{MVA} \end{aligned} $$ #### Reactor for 50 % reduction New fault MVA $= 29.11$ MVA, so the total reactance must double: $$ \begin{aligned} X_{eq,new} &= 2\times0.68696 = 1.37393\ \text{pu} \\ X &= 1.37393 - 0.68696 = 0.68696\ \text{pu} \\ X(\Omega) &= 0.68696\times3.025 = 2.078\ \Omega \end{aligned} $$ **Answer:** Present fault level = 58.23 MVA; series reactor = **0.687 pu ≈ 2.078 Ω** (fault level becomes 29.11 MVA). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. On the 40 MVA base, what is G2 (30 MVA, 4%)? A. 0.0533 pu B. 0.03 pu C. 0.04 pu D. 0.0300 ohm Answer: A. 0.04 x 40/30 = 0.0533 pu. (Key point: G2 conversion) 2. What must happen to the total reactance for a 50% fault-level cut? A. It must halve B. It must double C. It must increase by 50% D. It stays the same Answer: B. Fault MVA is proportional to 1/X, so half the MVA needs twice the X. (Key point: Condition for 50% reduction) 3. Total X_eq = 0.687 pu. What reactor is needed? A. 0.343 pu = 1.04 ohm B. 1.374 pu = 4.16 ohm C. 0.687 pu = 2.078 ohm D. 0.687 ohm Answer: C. Added reactance equals existing X_eq = 0.687 pu; times 3.025 = 2.078 ohm. (Key point: Reactor value) ## Close variants set in other exams - [Feeder reactor: 40 and 30 MVA, 40% reduction](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-40-30mva-40-percent) — Power Plant Equipment, chapter 4, asked 4× - [Fault level: two 40 MVA generators, 2 Ω feeder](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-two-40mva-2-ohm-feeder) — Power Plant Equipment, chapter 4, asked 5× ## Related in Power Plant Equipment - [Fault current: two 50 MVA generators, feeder reactor](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-two-50mva-feeder-reactor) — Power Plant Equipment, chapter 4, asked 3× - [Fault level: two 50 MVA generators, 50% reduction](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-two-50mva-50-percent) — Power Plant Equipment, chapter 4 - [Fault current: two 50 MW generators, reactor inductance](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-two-50mw-inductance) — Power Plant Equipment, chapter 4 - [Bus-bar reactor scheme: 30 MVA and 50 MVA](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Abus-bar-reactor-30-50-mva) — Power Plant Equipment, chapter 4, asked 2× - [Fault level: three generators, feeder reactor](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-three-generators-feeder) — Power Plant Equipment, chapter 4 - [Fault level: two 25 MVA generators, 50% reduction](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-two-25mva-50-percent) — Power Plant Equipment, chapter 4 ## Related in other subjects - [Reactance diagram: 50 MVA, 13.8 kV base, three generators](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-50mva-13-8kv) — 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 - [Reactance diagram: three generators, 30 MVA 6.6 kV base](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-30mva-three-generators) — Power System Analysis I, chapter 3