--- title: "Fault level: three generators, feeder reactor" subject: "Power Plant Equipment" chapter: 4 key: "semester_7/power_plant_equipment:fault-three-generators-feeder" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/fault-three-generators-feeder site: BCT Engineering --- # Fault level: three generators, feeder reactor Power Plant Equipment · Chapter 4: Substation Equipments - Asked in: 2081 Baisakh, 6 marks - Video: [Short Circuit Fault Level Calculation](https://www.youtube.com/watch?v=ahLb3xONOqY) — Ratss AESQUARE ALL ABOUT ELECTRICAL ENGINEERING - 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-three-generators-feeder - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-4 ## Question The given figure shows three generators operating in parallel. Find the fault current and fault level if three phase to ground fault occurs in the outgoing feeder. What should be the value of feeder reactor to be added to reduce the fault level by 40%? [Figure: G1 20 MVA, 11 kV, X1 = 0.1 pu; G2 30 MVA, 11 kV, X2 = 0.2 pu; G3 40 MVA, 11 kV, X3 = 0.3 pu; all three connected to a common busbar (points A, B, C); an outgoing feeder from the busbar through a circuit breaker, with the fault point on the feeder beyond the C.B.] ## Short answer On 100 MVA base, G1: 0.5 pu, G2: 0.6667 pu, G3: 0.75 pu, parallel Xeq = 0.2069 pu. Fault level = 100/0.2069 = 483.33 MVA, fault current ≈ 25.37 kA. To reduce by 40% (to 290 MVA), series reactor needed = 0.1379 pu = 0.167 Ω. ## Answer Convert all reactances to one base, combine the three parallel generators, and use Fault MVA $=\text{MVA}_b/X_{eq}$. The feeder reactance is not given, so it is taken as zero (fault just beyond the C.B.). **Base:** 100 MVA, 11 kV. $X_{new} = X_{old}\dfrac{\text{MVA}_{new}}{\text{MVA}_{old}}$; $Z_{base} = 11^2/100 = 1.21\ \Omega$. ```text G1 20MVA G2 30MVA G3 40MVA 0.1pu 0.2pu 0.3pu |A |B |C ==+==========+==========+=== 11 kV | C.B. |---- F ``` #### Reactances on 100 MVA base | Generator | Calculation | $X$ (pu) | |---|---|---| | G1 | $0.1\times100/20$ | 0.5 | | G2 | $0.2\times100/30$ | 0.6667 | | G3 | $0.3\times100/40$ | 0.75 | $$ \begin{aligned} \frac{1}{X_{eq}} &= \frac{1}{0.5}+\frac{1}{0.6667}+\frac{1}{0.75} = 2+1.5+1.3333 = 4.8333 \\ X_{eq} &= 0.20690\ \text{pu} \end{aligned} $$ #### Fault level and fault current $$ \begin{aligned} \text{Fault MVA} &= \frac{100}{0.2069} = 483.33\ \text{MVA} \\ I_f &= \frac{483.33\times10^6}{\sqrt3\times11\times10^3} = 25{,}368\ \text{A} \approx 25.37\ \text{kA} \end{aligned} $$ #### Feeder reactor for 40 % reduction New fault MVA $= 0.6\times483.33 = 290$ MVA. $$ \begin{aligned} X_{eq,new} &= \frac{100}{290} = 0.34483\ \text{pu} \\ X_{reactor} &= 0.34483 - 0.20690 = 0.13793\ \text{pu} \\ X_{reactor}(\Omega) &= 0.13793\times1.21 = 0.1669\ \Omega \end{aligned} $$ **Answer:** Fault level = 483.33 MVA, fault current ≈ 25.37 kA; feeder reactor ≈ **0.167 Ω** (0.1379 pu on 100 MVA) brings the level down to 290 MVA. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why must reactances be converted to a common base? A. The generators have different MVA ratings, so pu values on their own bases cannot be combined B. To make the voltages different C. To remove the feeder reactance D. To convert them to ohms Answer: A. X_new = X_old x MVA_new/MVA_old puts all on one base. (Key point: Common base) 2. On a 100 MVA base, what is G2 (30 MVA, 0.2 pu)? A. 0.06 pu B. 0.667 pu C. 0.2 pu D. 1.5 pu Answer: B. 0.2 x 100/30 = 0.667 pu. (Key point: Converted value) 3. With 1/X_eq = 2 + 1.5 + 1.333, what is the fault level? A. 20.7 MVA B. 100 MVA C. 483.3 MVA D. 2000 MVA Answer: C. X_eq = 0.2069 pu, so 100/0.2069 = 483.3 MVA (25.37 kA). (Key point: Fault level) 4. What feeder reactor gives a 40% reduction? A. 0.0828 pu B. 0.345 pu C. 0.138 ohm D. 0.138 pu (about 0.167 ohm) Answer: D. New X_eq = 100/290 = 0.3448 pu; minus 0.2069 = 0.138 pu; times 1.21 = 0.167 ohm. 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