--- title: "Fault level: adding a third generator" subject: "Power Plant Equipment" chapter: 4 key: "semester_7/power_plant_equipment:fault-third-generator-added" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/fault-third-generator-added site: BCT Engineering --- # Fault level: adding a third generator Power Plant Equipment · Chapter 4: Substation Equipments - Asked in: 2080 Baisakh, 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-third-generator-added - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-4 ## Question Following figure shows two generators operating parallel. i) If a 3-phase to ground fault occurs on the outgoing feeder, calculate the fault current and fault level on the feeder. ii) If a 3rd generator (G3: 10MVA, 11kV, X3=0.2 pu) is added at point 'C', calculate the value of reactor to be added on the outgoing feeder so that the fault level remains same as before. [Figure: G1 20 MVA, 11 kV, X1 = 0.1 pu (path-1) and G2 20 MVA, 11 kV, X2 = 0.1 pu (path-2) connected to a common busbar at points A and B; busbar extends to point C; outgoing feeder from the busbar through a C.B., fault point 'F' on the outgoing feeder] ## Short answer Fault level is 400 MVA, fault current ≈ 20.99 kA. After adding G3, equivalent reactance drops to 0.04444 pu; to restore 0.05 pu total, add a feeder reactor of 0.00556 pu (0.0336 Ω). ## Answer Fault MVA $=\text{MVA}_b/X_{eq}$. The feeder has no reactance at first (none given). **Base:** 20 MVA, 11 kV; $Z_{base} = 11^2/20 = 6.05\ \Omega$; $I_{base} = \dfrac{20\times10^6}{\sqrt3\times11\times10^3} = 1049.7\ \text{A}$. ```text G1 20MVA G2 20MVA G3 10MVA (added) 0.1pu 0.1pu 0.2pu |A |B |C ==+==========+==========+=== 11 kV | C.B.--[X]-- F ``` #### (i) Fault current and fault level (two generators) $$ \begin{aligned} X_{eq} &= \frac{0.1}{2} = 0.05\ \text{pu} \\ \text{Fault MVA} &= \frac{20}{0.05} = 400\ \text{MVA} \\ I_f &= \frac{1}{0.05}\times1049.7 = 20{,}995\ \text{A} \approx 20.99\ \text{kA} \end{aligned} $$ #### (ii) Reactor after adding G3 G3 on 20 MVA base: $X_3 = 0.2\times\dfrac{20}{10} = 0.4\ \text{pu}$. $$ \begin{aligned} \frac{1}{X_{eq}'} &= \frac{1}{0.1}+\frac{1}{0.1}+\frac{1}{0.4} = 22.5 \\ X_{eq}' &= 0.04444\ \text{pu} \;\Rightarrow\; \text{Fault MVA} = 450\ \text{MVA (too high)} \end{aligned} $$ To keep 400 MVA the total must stay at 0.05 pu: $$ \begin{aligned} X_{reactor} &= 0.05 - 0.04444 = 0.005556\ \text{pu} \\ X_{reactor}(\Omega) &= 0.005556\times6.05 = 0.0336\ \Omega \end{aligned} $$ **Answer:** (i) Fault level = **400 MVA**, $I_f \approx$ **20.99 kA**. (ii) Feeder reactor ≈ **0.00556 pu = 0.0336 Ω** keeps the fault level at 400 MVA after G3 is added. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the fault level with G1 and G2 (20 MVA, 0.1 pu each)? A. 400 MVA B. 200 MVA C. 100 MVA D. 800 MVA Answer: A. X_eq = 0.05 pu; 20/0.05 = 400 MVA. (Key point: Two-generator fault level) 2. On the 20 MVA base, what is G3 (10 MVA, 0.2 pu)? A. 0.1 pu B. 0.4 pu C. 0.2 pu D. 0.8 pu Answer: B. 0.2 x 20/10 = 0.4 pu. (Key point: G3 conversion) 3. What does adding G3 do to the fault level without a reactor? A. Lowers it because more reactance is added B. Leaves it unchanged C. Raises it to 450 MVA because X_eq falls to 0.0444 pu D. Raises it to 800 MVA Answer: C. Another parallel source reduces X_eq, increasing the fault MVA. (Key point: Effect of G3) 4. What reactor restores the fault level to 400 MVA? A. 0.05 pu B. 0.0444 pu C. 0.4 pu D. 0.00556 pu (about 0.0336 ohm) Answer: D. Total must be 0.05 pu; 0.05 - 0.04444 = 0.00556 pu; times 6.05 = 0.0336 ohm. 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