📘 Multi-Source Symmetrical Fault – Detailed Advanced Problem
Now we solve a system with two generators feeding a fault through a transformer and line. This is advanced GATE level.
🔹 Problem Statement
Two generators supply a common bus. Generator 1: 50 MVA, 11 kV, X1 = 0.2 pu Generator 2: 25 MVA, 11 kV, X2 = 0.25 pu Both connected to a 50 MVA base system. Transformer: 50 MVA, 11/132 kV, Xt = 0.1 pu Transmission line reactance = 30 Ω A 3-phase fault occurs at the receiving end of the line. Find: 1) Thevenin reactance 2) Fault current (pu) 3) Fault MVA🔹 Step 1 – Choose Base
Select: S_base = 50 MVA Voltage base: Generator side = 11 kV Line side = 132 kV🔹 Step 2 – Convert Generator 2 to New Base
Generator 2 rating = 25 MVA But system base = 50 MVA Change of base formula: X_new = X_old × (S_new / S_old) = 0.25 × (50 / 25) = 0.25 × 2X2_new = 0.5 pu
Generator 1 already on 50 MVA base: X1 = 0.2 pu🔹 Step 3 – Parallel Combination of Generators
Both connected in parallel at same bus. Equivalent reactance: 1 / Xeq = 1/X1 + 1/X2 = 1/0.2 + 1/0.5 = 5 + 2 = 7 Xeq = 1/7X_gen_eq ≈ 0.143 pu
🔹 Step 4 – Convert Line Reactance to PU
Line side base impedance: Z_base = (132²) / 50 = 17424 / 50 = 348.48 Ω Line per unit: X_line = 30 / 348.48 ≈ 0.086 puLine reactance ≈ 0.086 pu
🔹 Step 5 – Total Thevenin Reactance
Now series combination: Generator equivalent = 0.143 Transformer = 0.1 Line = 0.086 Total: Z_th = 0.143 + 0.1 + 0.086 = 0.329 puZ_th ≈ 0.329 pu
🔹 Step 6 – Fault Current in PU
I_fault = 1 / Z_th = 1 / 0.329 ≈ 3.04 puFault current ≈ 3.04 pu
🔹 Step 7 – Fault MVA
Fault MVA = Base MVA / Z_th = 50 / 0.329 ≈ 152 MVAFault MVA ≈ 152 MVA
🔹 Step 8 – Conceptual Understanding
✔ Multiple sources increase fault level ✔ Parallel reactances reduce Thevenin impedance ✔ Lower Z_th → Higher fault current ✔ Fault MVA increases when more generators connected🎯 Final Answers
- Z_th ≈ 0.329 pu
- I_fault ≈ 3.04 pu
- Fault MVA ≈ 152 MVA
Multi-Source Fault = Parallel Reduction + Series Addition + 1/Z
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