--- title: "Fault current: two 50 MW generators, reactor inductance" subject: "Power Plant Equipment" chapter: 4 key: "semester_7/power_plant_equipment:fault-two-50mw-inductance" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/fault-two-50mw-inductance site: BCT Engineering --- # Fault current: two 50 MW generators, reactor inductance Power Plant Equipment · Chapter 4: Substation Equipments - Asked in: 2069 Chaitra, 8 marks - Video: [L10 switchgear and protection, Current Limiting Reactor](https://www.youtube.com/watch?v=Y6pcGSmGlec) — Base Technical Academy - 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-50mw-inductance - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-4 ## Question Figure below shows the single line diagram of parallel operated generators. Their ratings are: G1: capacity = 50 MW, 11kV, X1 = 0.16 pu based on its rating; G2: capacity = 50MW, 11kV, X2 = 0.16 pu based on its rating; Xl = 2 ohms. [Figure: G1 (through X1) and G2 (through X2), both star-grounded, connected to a common busbar; the feeder leaves the busbar through Xl] Calculate the fault current in the feeder during 3 phases to ground fault on the feeder. Calculate the value of inductance (in mH) of the reactor to be connected in series with Xl so that the fault current decreases by 40%. ## Short answer On a 50 MVA, 11 kV base, the feeder fault current is 2895 A. To reduce fault current by 40%, add a series reactor of 1.462 ohm, which is 4.65 mH at 50 Hz. ## Answer Ratings in MW are taken as MVA (unity power factor). Fault current $I_f = I_{base}/X_{eq}$; ohms to pu: $X_{pu} = X_\Omega\,\text{MVA}_b/\text{kV}^2$. **Base:** 50 MVA, 11 kV; $Z_{base} = 11^2/50 = 2.42\ \Omega$; $I_{base} = \dfrac{50\times10^6}{\sqrt3\times11\times10^3} = 2624.3\ \text{A}$. ```text G1 50MW,0.16 G2 50MW,0.16 | | ====+======+=======+==== 11 kV | Xl=2 ohm + new L | F ``` #### Fault current $$ \begin{aligned} X_1\parallel X_2 &= \frac{0.16}{2} = 0.08\ \text{pu} \\ X_l &= \frac{2}{2.42} = 0.82645\ \text{pu} \\ X_{eq} &= 0.08 + 0.82645 = 0.90645\ \text{pu} \\ I_f &= \frac{2624.3}{0.90645} = 2895.2\ \text{A} \end{aligned} $$ (Fault MVA $= 50/0.90645 = 55.16$ MVA.) #### Reactor for 40 % reduction in fault current New $I_f = 0.6\times2895.2 = 1737.1$ A, so the total reactance must become $X_{eq}/0.6$: $$ \begin{aligned} X_{eq,new} &= \frac{0.90645}{0.6} = 1.51074\ \text{pu} \\ X_{add} &= 1.51074 - 0.90645 = 0.60430\ \text{pu} \\ X_{add}(\Omega) &= 0.60430\times2.42 = 1.4624\ \Omega \\ L &= \frac{1.4624}{2\pi\times50} = 4.655\times10^{-3}\ \text{H} \end{aligned} $$ **Answer:** $I_f$ ≈ **2895 A**; series reactor ≈ **1.462 Ω, i.e. L ≈ 4.65 mH** (fault current falls to about 1737 A). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. On a 50 MVA, 11 kV base, what is the feeder Xl = 2 ohm in pu? A. 2 x 2.42 = 4.84 pu B. 2/2.42 = 0.826 pu C. 2/50 = 0.04 pu D. 2/11 = 0.18 pu Answer: B. Z_base = 11^2/50 = 2.42 ohm, so X_pu = 2/2.42 = 0.826 pu. (Key point: Per-unit conversion) 2. What are X_eq and the fault current in the feeder? A. 0.16 pu, 16.4 kA B. 0.906 pu, about 2895 A C. 0.90 pu, 55 A D. 1.65 pu, 1590 A Answer: B. X_eq = 0.16/2 + 0.826 = 0.906 pu and I_f = I_base/X_eq = 2624.3/0.906 = 2895 A. (Key point: Equivalent reactance and fault current) 3. To cut fault current by 40%, what must the total reactance become? A. X_eq x 0.6 B. X_eq / 0.6 C. X_eq x 0.4 D. X_eq + 0.4 Answer: B. Current is inversely proportional to reactance, so I = 0.6 I_old needs X_new = X_old/0.6 = 1.51 pu. (Key point: Condition for 40% reduction) 4. The added reactance is 1.462 ohm. How is L obtained at 50 Hz? A. L = X x 2 pi x 50 B. L = X/(2 pi x 50) = 4.65 mH C. L = X/50 = 29 mH D. L = X x 50 = 73 H Answer: B. X = 2 pi f L so L = 1.462/(314.16) = 4.65 mH. 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