--- title: "Fault level: two 40 MVA generators, 2 Ω feeder" subject: "Power Plant Equipment" chapter: 4 key: "semester_7/power_plant_equipment:fault-two-40mva-2-ohm-feeder" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/fault-two-40mva-2-ohm-feeder site: BCT Engineering --- # Fault level: two 40 MVA generators, 2 Ω feeder Power Plant Equipment · Chapter 4: Substation Equipments - Asked in: 2081 Bhadra, 8 marks; 2075 Chaitra, 5+5 marks (asked 2 times) - Video: [SHORT CIRCUIT MVA / FAULT MVA WITH CURRENT LIMITING REACTANCE / KTU/ POWER SYSTEM ANALYSIS](https://www.youtube.com/watch?v=SB1tqp0fGYY) — praveen raj - 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-two-40mva-2-ohm-feeder - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-4 ## Question Two generators each of capacities 40 MVA, 11 kV 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 Ω. Calculate the fault level when 3-phase to ground fault occurs on the outgoing feeder. Also calculate the value of reactor to be connected in series with 2 Ω reactor to reduce the fault level by 40 %. ## Short answer On 40 MVA, 11 kV base, parallel generator reactance is 0.02222 pu, feeder 2 Ω = 0.66116 pu, total 0.68338 pu. Fault level = 40/0.68338 = 58.53 MVA, fault current ≈ 3072 A. To reduce by 40% (to 35.12 MVA), add series reactor of 1.378 Ω (0.4556 pu). ## Answer Work in per unit on a common base. Fault level $= \dfrac{\text{Base MVA}}{X_{pu}}$ and a reactor of $X$ ohms has $X_{pu} = X\cdot\dfrac{\text{MVA}_b}{\text{kV}_b^2}$. **Base:** 40 MVA, 11 kV. $$ Z_{base} = \frac{11^2}{40} = 3.025\ \Omega $$ ```text G1 40MVA,5% G2 40MVA,4% | | ====+=====+=======+==== 11 kV bus | X=2 ohm feeder | F (3-ph fault) ``` #### Fault level $$ \begin{aligned} X_{G1} &= 0.05\ \text{pu},\quad X_{G2} = 0.04\ \text{pu} \\ X_{G1}\parallel X_{G2} &= \frac{0.05\times0.04}{0.05+0.04} = 0.02222\ \text{pu} \\ X_{feeder} &= \frac{2}{3.025} = 0.66116\ \text{pu} \\ X_{eq} &= 0.02222 + 0.66116 = 0.68338\ \text{pu} \\ \text{Fault MVA} &= \frac{40}{0.68338} = 58.53\ \text{MVA} \end{aligned} $$ Fault current: $I_{base} = \dfrac{40\times10^6}{\sqrt3\times 11\times10^3} = 2099.5\ \text{A}$, so $I_f = 2099.5/0.68338 = 3072\ \text{A}$. #### Reactor to reduce fault level by 40 % New fault level $= 0.6\times58.53 = 35.12$ MVA. $$ \begin{aligned} X_{eq,new} &= \frac{40}{35.12} = \frac{0.68338}{0.6} = 1.13897\ \text{pu} \\ X_{added} &= 1.13897 - 0.68338 = 0.45559\ \text{pu} \\ X_{added}(\Omega) &= 0.45559\times3.025 = 1.378\ \Omega \end{aligned} $$ **Answer:** Fault level = 58.53 MVA (about 3.07 kA). A reactor of about **1.378 Ω** (0.4556 pu on 40 MVA) in series with the 2 Ω feeder reduces it by 40 % to 35.12 MVA. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which gives the equivalent reactance for the fault on the feeder? A. Generators in parallel, then feeder reactance in series with that B. All reactances in parallel C. Generators in series, then feeder in parallel D. Only the feeder reactance, ignoring generators Answer: A. G1 and G2 are in parallel at the bus, and the feeder is in series between bus and fault. (Key point: Equivalent reactance) 2. On a 40 MVA, 11 kV base, what is the 2 ohm feeder in pu? A. 0.0667 pu B. 0.661 pu C. 6.05 pu D. 2.0 pu Answer: B. Z_base = 11^2/40 = 3.025 ohm, so X = 2/3.025 = 0.661 pu. (Key point: Feeder in per unit) 3. What is the fault level (X_eq = 0.0222 + 0.661 = 0.683 pu)? A. 1800 MVA B. 27.4 MVA C. 58.5 MVA D. 40 MVA Answer: C. Fault MVA = base MVA/X_eq = 40/0.683 = 58.53 MVA. (Key point: Fault level) 4. How is the added reactance for a 40% reduction found? A. New X_eq = 0.6 X_eq; the reactor is the difference B. Reactor = 0.4 x 2 ohm = 0.8 ohm C. Reactor = 0.6 x 58.53 ohm D. New X_eq = X_eq/0.6; the reactor is the difference (about 0.456 pu = 1.378 ohm) Answer: D. Fault MVA becomes 0.6 times, so total reactance becomes 1/0.6 times; the extra is 0.4556 pu, which times 3.025 gives 1.378 ohm. (Key point: Reactor for 40% reduction) ## Close variants set in other exams - [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× - [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× - [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× ## Related in Power Plant Equipment - [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 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 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 - [Fault current: 50 MW and 40 MW generators](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-50mw-40mw-equal-current) — Power Plant Equipment, chapter 4 - [Fault level: adding a third generator](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-third-generator-added) — Power Plant Equipment, chapter 4 ## Related in other subjects - [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 - [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: 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