--- title: "Feeder reactor: 40 and 30 MVA, 40% reduction" subject: "Power Plant Equipment" chapter: 4 key: "semester_7/power_plant_equipment:fault-40-30mva-40-percent" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/fault-40-30mva-40-percent site: BCT Engineering --- # Feeder reactor: 40 and 30 MVA, 40% reduction Power Plant Equipment · Chapter 4: Substation Equipments - Asked in: 2074 Chaitra, 6 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-40-percent - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-4 ## Question Two generators of capacities 40MVA, 11kV and 30 MVA, 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 2 ohm. 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 40%. ## Short answer On 40 MVA base, generator reactances are 0.05 pu and 0.05333 pu in parallel, plus feeder 0.66116 pu gives 0.68696 pu. Present fault level is 58.23 MVA. To reduce by 40%, required series reactor is 0.458 pu, about 1.385 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 $$ \begin{aligned} X_{G1} &= 0.05\ \text{pu},\quad X_{G2} = 0.04\times\frac{40}{30} = 0.05333\ \text{pu} \\ X_{G1}\parallel X_{G2} &= \frac{0.05\times0.05333}{0.10333} = 0.025806\ \text{pu} \\ X_{feeder} &= \frac{2}{3.025} = 0.66116\ \text{pu} \\ X_{eq} &= 0.68696\ \text{pu} \\ \text{Fault MVA} &= \frac{40}{0.68696} = 58.23\ \text{MVA} \end{aligned} $$ #### Reactor for 40 % reduction New fault MVA $= 0.6\times58.23 = 34.94$ MVA. $$ \begin{aligned} X_{eq,new} &= \frac{0.68696}{0.6} = 1.14494\ \text{pu} \\ X &= 1.14494 - 0.68696 = 0.45798\ \text{pu} \\ X(\Omega) &= 0.45798\times3.025 = 1.385\ \Omega \end{aligned} $$ **Answer:** Present fault level = 58.23 MVA; series reactor ≈ **0.458 pu = 1.385 Ω** (fault level becomes 34.94 MVA). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which is the correct present equivalent reactance? A. Generators in parallel (0.0258 pu) plus feeder (0.661 pu) = 0.687 pu B. Generators in series plus feeder C. Only the feeder 0.661 pu D. Generators in parallel with feeder Answer: A. The generators are parallel at the bus and the feeder is in series. (Key point: Equivalent reactance) 2. What total reactance gives a 40% reduction? A. 0.6 x X_eq = 0.412 pu B. X_eq/0.6 = 1.145 pu C. X_eq x 1.4 = 0.96 pu D. 2 x X_eq = 1.374 pu Answer: B. Fault MVA falls to 0.6 times, so reactance rises by 1/0.6. (Key point: New reactance for 40% reduction) 3. What reactor is added? A. 0.687 pu = 2.078 ohm B. 1.145 pu = 3.46 ohm C. 0.458 pu = 1.385 ohm D. 0.458 ohm Answer: C. 1.145 - 0.687 = 0.458 pu; times 3.025 = 1.385 ohm. (Key point: Reactor) ## Close variants set in other exams - [Feeder reactor: 40 and 30 MVA, 50% reduction](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-40-30mva-50-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: 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: four 25 MVA generators, bus reactor](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-four-25mva-bus-reactor) — Power Plant Equipment, chapter 4 - [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× ## Related in other subjects - [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 - [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: 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