--- title: "Generator-reactor system fault MVA" subject: "Power System Analysis II" chapter: 3 key: "semester_5/power_system_analysis_ii:generator-reactor-fault-mva" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/generator-reactor-fault-mva site: BCT Engineering --- # Generator-reactor system fault MVA Power System Analysis II · Chapter 3: Power System Fault Calculation - Asked in: 2072 Chaitra, 10 marks - Video: [102. Short Circuit MVA of a Power System](https://www.youtube.com/watch?v=wxYQ8QoCHsk) — Bright Engineers - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/generator-reactor-fault-mva - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-3 ## Question An interconnected generator reactor system has been shown in figure below. The base values for the given % reactance are the rating of individual pieces of equipments. Determine the fault current and fault MVA for a 3-phase short circuit fault at F. Assume bus bar voltage as 11 kV. [Figure: G1 10 MVA, 10% connected at the faulted bus F; reactor 10 MVA, 5% between bus F and the bus of G2 (20 MVA, 15%); reactor 8 MVA, 4% between the G2 bus and G3 (20 MVA, 15%)] ## Short answer On 20 MVA base, Thevenin reactance at F is 0.0984 pu. Fault current = 10.16 pu = 10.67 kA at 11 kV. Fault MVA = 20 / 0.0984 = 203 MVA. ## Answer The fault MVA is found by converting every reactance to one common MVA base, reducing the network to a single Thevenin reactance seen from F, and using $\text{Fault MVA} = \text{Base MVA}/X_{th}$ with prefault voltage 1 pu. #### Step 1: Choose base and convert reactances Base: $S_b = 20$ MVA, $V_b = 11$ kV. Use $X_{new} = X_{old}\,\dfrac{S_b}{S_{rated}}$ (same voltage, so no voltage correction). | Element | Rating | Given X | X on 20 MVA (pu) | |---|---|---|---| | G1 | 10 MVA | 10% | $0.10 \times 20/10 = 0.20$ | | Reactor R1 (F to G2 bus) | 10 MVA | 5% | $0.05 \times 20/10 = 0.10$ | | G2 | 20 MVA | 15% | 0.15 | | Reactor R2 (G2 bus to G3) | 8 MVA | 4% | $0.04 \times 20/8 = 0.10$ | | G3 | 20 MVA | 15% | 0.15 | #### Step 2: Reactance diagram All generator EMFs are equal (1 pu) and their neutral is the reference, so for a 3-phase fault every source branch ends on the reference bus. ```text F (G1 bus) R1=0.1 B (G2 bus) R2=0.1 ---o----------------/\/\/\---------o-------/\/\/\---+ | | | G1 0.2 G2 0.15 G3 0.15 | | | ===+=========== reference =======+================+=== ``` #### Step 3: Reduce the network as seen from F $$ \begin{aligned} X_{R2+G3} &= 0.10 + 0.15 = 0.25 \\ X_B &= 0.15 \parallel 0.25 = \frac{0.15 \times 0.25}{0.40} = 0.09375 \\ X_{R1+B} &= 0.10 + 0.09375 = 0.19375 \\ X_{th} &= 0.20 \parallel 0.19375 = \frac{0.20 \times 0.19375}{0.39375} = 0.09841\ \text{pu} \end{aligned} $$ #### Step 4: Fault current and fault MVA $$ \begin{aligned} I_f &= \frac{1}{X_{th}} = \frac{1}{0.09841} = 10.16\ \text{pu} \\ I_b &= \frac{S_b}{\sqrt{3}\,V_b} = \frac{20\times 10^6}{\sqrt{3}\times 11\times 10^3} = 1049.7\ \text{A} \\ I_f &= 10.16 \times 1049.7 = 10\,667\ \text{A} \approx 10.67\ \text{kA} \\ \text{Fault MVA} &= \frac{S_b}{X_{th}} = \frac{20}{0.09841} = 203.2\ \text{MVA} \end{aligned} $$ Check: $\sqrt{3} \times 11 \times 10.667 = 203.2$ MVA. The fault MVA does not depend on the base chosen. With a 10 MVA base every pu reactance is halved, so $X_{th} = 0.04920$ pu and $10/0.04920 = 203.2$ MVA, the same result. **Answer: Fault current ≈ 10.67 kA (10.16 pu on 20 MVA base); fault MVA ≈ 203.2 MVA.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Reactor R2 (8 MVA, 4%) on a 20 MVA base is A. 0.04 x 8/20 = 0.016 pu B. 0.04 pu C. 0.04 x 20/8 = 0.10 pu D. 0.04 x 20 = 0.8 pu Answer: C. X_new = X_old x S_base/S_rated = 0.04 x 20/8 = 0.10 pu. (Key point: Common base conversion) 2. Seen from F, how is the network reduced? A. R2 + G3 = 0.25, parallel with G2 = 0.09375, plus R1 = 0.19375, parallel with G1 gives X_th B. All five reactances in parallel C. All five reactances in series D. G1 in series with G2 then parallel with G3 Answer: A. Work from the far end: (R2 + G3) || G2, then add R1, then put in parallel with G1 at F. (Key point: Network reduction order) 3. With X_th = 0.09841 pu on 20 MVA, fault MVA is A. 20 x 0.0984 = 1.97 MVA B. 20/0.09841 = 203.2 MVA C. 10/0.09841 MVA D. 1/0.0984 = 10.2 MVA Answer: B. Fault MVA = S_base/X_th = 203.2 MVA, and I_f = 10.67 kA at 11 kV. (Key point: Fault MVA) 4. If a 10 MVA base were used instead, the fault MVA would be A. Half as large B. Twice as large C. The same 203.2 MVA D. Undefined Answer: C. All pu values double but S_base halves, so S_base/X_th is unchanged; the physical fault level does not depend on the base. 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