--- title: "Fault level at end of two 200 km lines" subject: "Power System Analysis II" chapter: 3 key: "semester_5/power_system_analysis_ii:fault-two-stations-132kv-200km" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/fault-two-stations-132kv-200km site: BCT Engineering --- # Fault level at end of two 200 km lines Power System Analysis II · Chapter 3: Power System Fault Calculation - Asked in: 2082 Baishakh, 8 marks; 2068 Chaitra (asked 2 times) - Video: [Fault Analysis by Thevenin theorem | Solved Problem - 4 | Numericals | Power System Analysis | PSA](https://www.youtube.com/watch?v=dFepZEBPrr8) — Simple n Easy Learning - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/fault-two-stations-132kv-200km - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-3 ## Question Figure below shows generating stations feeding a 132-kV system. Determine the total fault current and fault MVA level for a 3-phase to ground bolted fault at the point marked by F. Consider the line is 200 km long. [Figure: G1 100 MVA, 11 kV, X = 15% feeding transformer T1 100 MVA, 11/132 kV, X = 10%; G2 50 MVA, 11 kV, X = 10% feeding transformer T2 50 MVA, 11/132 kV, X = 8%; both transformers feed the 132 kV bus, from which two parallel lines L1 and L2, each X = 0.2 Ω/phase/km and 200 km long, run to the fault point F] ## Short answer On 100 MVA base: G1 0.15, T1 0.10, G2 0.20, T2 0.16, parallel lines 0.1148 pu. Thevenin reactance at F = 0.2623 pu. Fault current = 1.667 kA, fault MVA = 381.2 MVA. ## Answer A bolted three-phase fault is balanced, so only the positive-sequence reactance diagram is needed. Base: 100 MVA, 11 kV on the generator side and 132 kV on the line side. Pre-fault voltage is taken as 1.0 pu and pre-fault load current is neglected (usual assumption for fault level calculation). As described, both transformers feed a common 132 kV bus, and two parallel 200 km lines run from that bus to F. #### Per-unit reactances $X_{new} = X_{old}\dfrac{S_{new}}{S_{old}}$ (voltage bases match the ratings): | Element | Rating | Given | On 100 MVA (pu) | |---|---|---|---| | G1 | 100 MVA | 15% | 0.15 | | T1 | 100 MVA | 10% | 0.10 | | G2 | 50 MVA | 10% | $0.10 \times 2 = 0.20$ | | T2 | 50 MVA | 8% | $0.08 \times 2 = 0.16$ | Line: $X_L = 0.2 \times 200 = 40\ \Omega$ per line. $$ \begin{aligned} Z_{base} &= \frac{132^2}{100} = 174.24\ \Omega \\ X_L &= \frac{40}{174.24} = 0.2296\ \text{pu} \end{aligned} $$ #### Reactance diagram ```text E=1 --[G1 j0.15]--[T1 j0.10]--+ | +--[L1]--+ 132 kV | |-- F | +--[L2]--+ E=1 --[G2 j0.20]--[T2 j0.16]--+ ``` #### Thevenin reactance at F $$ \begin{aligned} X_{G1T1} &= 0.15 + 0.10 = 0.25 \\ X_{G2T2} &= 0.20 + 0.16 = 0.36 \\ X_{gen} &= \frac{0.25 \times 0.36}{0.25 + 0.36} = 0.1475 \\ X_{lines} &= \frac{0.2296}{2} = 0.1148 \\ X_{th} &= 0.1475 + 0.1148 = 0.2623\ \text{pu} \end{aligned} $$ #### Fault current and fault MVA $$ \begin{aligned} I_f &= \frac{1}{X_{th}} = \frac{1}{0.2623} = 3.8121\ \text{pu} \\ I_{base} &= \frac{100 \times 10^6}{\sqrt{3} \times 132 \times 10^3} = 437.39\ \text{A} \\ I_f &= 3.8121 \times 437.39 = 1667.3\ \text{A} \\ \text{Fault MVA} &= \frac{S_{base}}{X_{th}} = \frac{100}{0.2623} = 381.21\ \text{MVA} \end{aligned} $$ **Answer:** total fault current $\approx 1.667$ kA (at 132 kV), fault level $\approx 381.2$ MVA. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is G2's 10% on 50 MVA expressed on a 100 MVA base? A. 0.20 pu B. 0.05 pu C. 0.10 pu D. 0.40 pu Answer: A. X_new = X_old x S_new/S_old = 0.10 x 100/50 = 0.20 pu. (Key point: Per-unit base change) 2. How is the Thevenin reactance at F built up? A. (G1+T1) parallel (G2+T2), in series with the two parallel lines B. All reactances added in series C. All reactances in parallel D. G1 and G2 in series with each line separately Answer: A. 0.25 || 0.36 = 0.1475 plus 0.2296/2 = 0.1148 gives X_th = 0.2623 pu. (Key point: Thevenin reactance) 3. A 40 ohm line on a 132 kV, 100 MVA base is how many pu? A. 0.2296 B. 0.4 C. 4.35 D. 0.0574 Answer: A. Z_base = 132^2/100 = 174.24 ohm, so 40/174.24 = 0.2296 pu. (Key point: Line reactance in pu) 4. What is the fault level obtained? A. About 381 MVA, i.e. 100/X_th B. About 262 MVA C. About 38 MVA D. About 1667 MVA Answer: A. Fault MVA = S_base/X_th = 100/0.2623 = 381.2 MVA, with I_f = 3.81 pu = 1.667 kA. 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