--- title: "SLG fault at mid-line through 4.033 Ω" subject: "Power System Analysis II" chapter: 5 key: "semester_5/power_system_analysis_ii:slg-fault-mid-line-motor" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/slg-fault-mid-line-motor site: BCT Engineering --- # SLG fault at mid-line through 4.033 Ω Power System Analysis II · Chapter 5: Unsymmetrical Faults on Power Systems - Asked in: 2075 Chaitra, 10 marks - Video: [Module 4 Lecture 13 - Problem 6 - LG,LL,LLG fault current calculation in a power system](https://www.youtube.com/watch?v=JnyP7a-5A6M) — Sumitha T L - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/slg-fault-mid-line-motor - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5 ## Question Two alternators are operating in parallel and supplying a synchronous motor which is receiving 60 MW power at 0.8 pf (lag) at 6 kV. Single line diagram for the system and its data are given below. Compute the fault current when a single line to ground fault occurs at the middle of the line through a fault resistance of 4.033 ohm. Data: G1 & G2: 11 kV, 100 MVA, xg1 = 0.20 pu, xg2 = xg0 = 0.10 pu; T1: 180 MVA, 11.5/115 kV, xT1 = 0.10 pu; T2: 170 MVA, 6.6/115 kV, xT2 = 0.10 pu; M: 6.3 kV, 160 MVA, xM1 = xM2 = 0.30 pu, xM0 = 0.10 pu; Line: xLINE1 = xLINE2 = 30.25 ohm, xLINE0 = 60.5 ohm. [Figure: G1 and G2 (star, grounded) in parallel on a bus - T1 - line - T2 - motor M (star, grounded)] ## Short answer On 100 MVA, 110 kV line base, pre-fault voltage 1.047 pu, Thevenin impedances Z1=j0.1624, Z2=j0.1449, Z0=j0.1562 pu and 3Rf=0.1 pu give sequence current 2.208 pu; fault current = 3 × I_a0 = 6.62 pu = 3.48 kA. ## Answer **Assumptions:** the figure does not mark transformer windings, so the usual arrangement is taken: T1 and T2 are Δ on the machine side and star-grounded on the 115 kV line side. Then the machine zero-sequence reactances are isolated by the deltas. Base: 100 MVA, 11 kV in the generator circuit. #### Base voltages and per-unit values $$ \begin{aligned} V_{b,line} &= 11 \times \frac{115}{11.5} = 110\ \text{kV}, \quad Z_{b,line} = \frac{110^2}{100} = 121\ \Omega \\ V_{b,motor} &= 110 \times \frac{6.6}{115} = 6.313\ \text{kV} \end{aligned} $$ | Element | Calculation | pu | |---|---|---| | G1 ∥ G2, $X_1$ | $0.2 \| 0.2$ | 0.10 | | G1 ∥ G2, $X_2$ | $0.1 \| 0.1$ | 0.05 | | T1 | $0.1 \times \frac{100}{180}\times(\frac{11.5}{11})^2$ | 0.0607 | | Line $X_1 = X_2$ | $30.25/121$ | 0.25 | | Line $X_0$ | $60.5/121$ | 0.50 | | T2 | $0.1 \times \frac{100}{170}\times(\frac{6.6}{6.313})^2$ | 0.0643 | | Motor $X_1 = X_2$ | $0.3 \times \frac{100}{160}\times(\frac{6.3}{6.313})^2$ | 0.1867 | | Fault resistance $R_f$ | $4.033/121$ | 0.0333 | #### Pre-fault voltage at the middle of the line Motor voltage $V_m = 6/6.313 = 0.9504\angle0^\circ$ pu (reference). Motor input $S = 60/0.8 = 75$ MVA $= 0.75$ pu. Half of the line is $X_{line}/2 = 0.125$ pu. $$ \begin{aligned} I_L &= \frac{0.75}{0.9504}\angle{-36.87^\circ} = 0.7891\angle{-36.87^\circ}\ \text{pu} \\ V_f &= V_m + I_L \cdot j(X_{T2} + X_{line}/2) = 0.9504 + 0.7891\angle{-36.87^\circ} \times j(0.0643 + 0.125) \\ &= 0.9504 + 0.7891\angle{-36.87^\circ} \times j0.1893 = 1.0469\angle{6.55^\circ}\ \text{pu} \end{aligned} $$ #### Thevenin impedances at the fault point (mid-line) $$ \begin{aligned} Z_1 &= j(0.10 + 0.0607 + 0.125)\ \|\ j(0.125 + 0.0643 + 0.1867) = j0.2857 \| j0.3760 = j0.1624\ \text{pu} \\ Z_2 &= j(0.05 + 0.0607 + 0.125)\ \|\ j0.3760 = j0.2357 \| j0.3760 = j0.1449\ \text{pu} \\ Z_0 &= j(0.0607 + 0.25)\ \|\ j(0.0643 + 0.25) = j0.3107 \| j0.3143 = j0.1562\ \text{pu} \end{aligned} $$ (Zero sequence: each half of the line, $X_0/2 = 0.25$, goes to ground through the grounded star of its transformer.) #### Fault current (sequence networks in series with $3R_f$) $$ \begin{aligned} I_{a0} = I_{a1} = I_{a2} &= \frac{V_f}{Z_1+Z_2+Z_0+3R_f} = \frac{1.0469\angle{6.55^\circ}}{0.1 + j0.4635} \\ &= 2.208\angle{-71.27^\circ}\ \text{pu} \\ I_f &= 3I_{a0} = 6.624\angle{-71.27^\circ}\ \text{pu} \end{aligned} $$ Base current on the line: $$ I_{base} = \frac{100\times10^6}{\sqrt3 \times 110\times10^3} = 524.9\ \text{A} $$ $$ |I_f| = 6.624 \times 524.9 = 3477\ \text{A} $$ **Answer: SLG fault current ≈ 6.62 pu ≈ 3.48 kA** (angle measured from the motor terminal voltage). Note: $3R_f = 0.1$ pu adds resistance in series, so it both reduces the current and makes it lag by less than 90°. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With T1 and T2 delta on the machine side and grounded star on the line side, what does the zero sequence network look like? A. Machines connected in series B. Each half-line connects to reference through its transformer; machines are isolated by the deltas C. Open circuit D. Same as positive sequence Answer: B. The deltas block zero sequence flow into the machines; Z0 = (0.0607 + 0.25) || (0.0643 + 0.25) = 0.1562 pu. (Key point: Zero sequence under assumed transformer connections) 2. How is the pre-fault voltage at the line midpoint found? A. Assumed 1 pu B. Motor voltage plus load-current drop through T2 and half the line, giving 1.047 pu C. Equal to generator voltage D. Zero Answer: B. The motor takes 60 MW at 0.8 pf, so Vf = Vm + IL j(XT2 + XL/2) = 1.0469 pu. (Key point: Pre-fault voltage) 3. What is the line-side base voltage when the generator base is 11 kV and T1 is 11.5/115 kV? A. 115 kV B. 110 kV C. 11.5 kV D. 100 kV Answer: B. Vb = 11 x 115/11.5 = 110 kV, so Zb = 110^2/100 = 121 ohm. (Key point: Base voltage change across transformers) 4. How is the 4.033 ohm fault resistance included? A. Rf = 0.0333 pu in series once B. Rf as 3Rf = 0.1 pu in the series connection C. Rf in parallel D. Ignored Answer: B. Rf = 4.033/121 = 0.0333 pu; the L-G loop contains 3Rf = 0.1 pu. (Key point: Fault resistance in the sequence loop) 5. What is the fault current? A. 2.208 pu B. 6.62 pu = 3.48 kA C. 10 pu D. 3.3 pu Answer: B. Ia0 = 1.0469/|0.1 + j0.4635| = 2.208 pu, If = 3Ia0 = 6.624 pu, times 524.9 A = 3477 A. 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