Sunday, 22 February 2026

📘 Symmetrical Fault – Detailed Step-by-Step Problem

We solve a complete generator–transformer–line fault problem clearly. Follow every step carefully.

🔹 Problem Statement

A power system consists of: Generator: 50 MVA, 11 kV, Xg = 0.2 pu Transformer: 50 MVA, 11/132 kV, Xt = 0.1 pu Transmission line reactance = 40 Ω A 3-phase fault occurs at the receiving end of the line. Find: 1) Fault current in pu 2) Fault MVA 3) Actual fault current in kA

🔹 Step 1 – Choose Base Values

Since generator rating is 50 MVA, Choose: S_base = 50 MVA Voltage base: Generator side = 11 kV Line side = 132 kV Base values automatically adjust across transformer.

🔹 Step 2 – Convert Line Impedance to PU

Line side base impedance: Z_base = (V_base)^2 / S_base = (132^2) / 50 = 17424 / 50 = 348.48 Ω Line per unit reactance: X_line_pu = 40 / 348.48 ≈ 0.115 pu

Line reactance ≈ 0.115 pu

🔹 Step 3 – Total Thevenin Reactance

All in per unit: Generator = 0.2 pu Transformer = 0.1 pu Line = 0.115 pu Total: Z_th = 0.2 + 0.1 + 0.115 = 0.415 pu

Thevenin reactance = 0.415 pu

🔹 Step 4 – Fault Current in PU

For 3-phase fault: I_fault = 1 / Z_th = 1 / 0.415 ≈ 2.41 pu

Fault current = 2.41 pu

🔹 Step 5 – Fault MVA

Fault MVA = Base MVA / Z_th = 50 / 0.415 ≈ 120.5 MVA

Fault MVA ≈ 120.5 MVA

🔹 Step 6 – Convert Fault Current to Actual

We calculate base current on line side. I_base = S_base / (√3 × V_base) = 50 × 10^6 / (1.732 × 132000) ≈ 218.7 A Now actual fault current: I_actual = 2.41 × 218.7 ≈ 527 A

Actual fault current ≈ 0.527 kA

🔹 Important Clarifications

✔ Always convert everything into per unit first ✔ Add reactances directly in pu ✔ Assume prefault voltage = 1 pu ✔ For 3-phase fault → only positive sequence used

🎯 Final Answers

Fault current = 2.41 pu
Fault MVA ≈ 120.5 MVA
Actual fault current ≈ 0.527 kA

Symmetrical Fault Analysis = Per Unit + Series Reactance Addition

Shaktimatha 369 Technologies

Labels