--- title: "Fault level: four 25 MVA generators, bus reactor" subject: "Power Plant Equipment" chapter: 4 key: "semester_7/power_plant_equipment:fault-four-25mva-bus-reactor" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/fault-four-25mva-bus-reactor site: BCT Engineering --- # Fault level: four 25 MVA generators, bus reactor Power Plant Equipment · Chapter 4: Substation Equipments - Asked in: 2073 Shrawan, 5 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-four-25mva-bus-reactor - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-4 ## Question The figure below shows four identical generators, each rated 11 kV, 25 MVA and each having sub-transient reactance of 16% on its own rating. Find 3-ϕ fault level at one of the outgoing feeder. Also calculate the value of reactance to be connected in the bus bar between "B" and "C" so that fault level reduces by 40%. [Figure: four generators connected to a busbar at points A, B, C, D; outgoing feeders leave the busbar, with a 3-ϕ fault on the feeder between C and D] ## Short answer On 25 MVA base, four 0.16 pu generators in parallel give equivalent reactance 0.04 pu, so fault level is 625 MVA. To reduce to 375 MVA, a bus-bar reactor between B and C must be 0.32 pu, about 1.549 Ω. ## Answer All four generators are identical, so work on their own base. Fault MVA $=\text{MVA}_b/X_{eq}$. **Base:** 25 MVA, 11 kV; $X_g = 0.16$ pu; $Z_{base} = 11^2/25 = 4.84\ \Omega$. ```text G1 G2 G3 G4 | | | | A-------B---[X]----C-------D bus | F (feeder between C and D) ``` #### Fault level without reactor $$ \begin{aligned} X_{eq} &= \frac{0.16}{4} = 0.04\ \text{pu} \\ \text{Fault MVA} &= \frac{25}{0.04} = 625\ \text{MVA} \end{aligned} $$ ($I_f = 625/(\sqrt3\times11) = 32.80$ kA.) #### Reactor between B and C for 40 % reduction New fault MVA $= 0.6\times625 = 375$ MVA, so $X_{eq,new} = 25/375 = 0.06667$ pu. With the reactor, G3 and G4 feed the fault directly ($0.16\parallel0.16 = 0.08$ pu) and G1, G2 feed it through $X$ ($0.08 + X$): $$ \begin{aligned} \frac{1}{0.06667} &= \frac{1}{0.08} + \frac{1}{0.08+X} \\ 15 &= 12.5 + \frac{1}{0.08+X} \\ 0.08 + X &= 0.4 \\ X &= 0.32\ \text{pu} \end{aligned} $$ $$ X(\Omega) = 0.32\times4.84 = 1.549\ \Omega $$ **Answer:** Fault level = **625 MVA**; bus-bar reactor between B and C = **0.32 pu (32 %) on 25 MVA ≈ 1.549 Ω**, giving 375 MVA. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. On a 25 MVA base, how is the fault level at the bus obtained from the equivalent reactance? A. Fault MVA = base MVA / X_eq (pu) B. Fault MVA = base MVA x X_eq (pu) C. Fault MVA = X_eq / base MVA D. Fault MVA = sqrt(3) x 11 x X_eq Answer: A. Fault MVA = MVA_base / X_eq in per unit, so a smaller equivalent reactance gives a larger fault level. (Key point: Fault MVA formula) 2. Four identical generators of 0.16 pu feed the fault directly. What is X_eq and the fault level without the reactor? A. 0.64 pu, 39 MVA B. 0.16 pu, 156 MVA C. 0.04 pu, 625 MVA D. 0.08 pu, 312.5 MVA Answer: C. Four equal reactances in parallel give 0.16/4 = 0.04 pu, so fault MVA = 25/0.04 = 625 MVA. (Key point: Equivalent reactance of four generators) 3. What new fault MVA and X_eq are required for a 40% reduction in fault level? A. 250 MVA, 0.10 pu B. 375 MVA, 0.06667 pu C. 400 MVA, 0.0625 pu D. 1042 MVA, 0.024 pu Answer: B. Reducing by 40% leaves 0.6 x 625 = 375 MVA, giving X_eq = 25/375 = 0.06667 pu. (Key point: Target fault level) 4. With the reactor X between B and C, which description of the network seen by the fault is correct? A. All four generators are in series with X B. G3 and G4 feed directly (0.08 pu) and G1, G2 feed through (0.08 + X) C. Only G1 and G2 feed the fault, G3 and G4 are isolated D. G1 and G2 feed directly and G3, G4 feed through X Answer: B. The fault is beyond C, so G3 and G4 are on the fault side of the reactor, while G1 and G2 (0.16 || 0.16 = 0.08) must pass through X. 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