📘 Two Machine System – Fault Stability Numerical
We analyze transient stability when a 3-phase fault occurs between two generators.
🔹 Problem Data
Generator 1: H1 = 6 MJ/MVA Generator 2: H2 = 4 MJ/MVA Pm1 = 1 pu Pm2 = 0.7 pu Pre-fault maximum power transfer: Pmax = 1.5 pu During fault: Pmax_fault = 0.2 pu Post-fault: Pmax_post = 1.2 pu Initial rotor angle δ₀ = 25° Find: ✔ Relative accelerating power ✔ Stability condition conceptually🔹 Step 1 – Equivalent Inertia
ωs = 314 rad/sec M1 = 2H1 / ωs = 12/314 = 0.0382 M2 = 2H2 / ωs = 8/314 = 0.0255 1/M_eq = 1/M1 + 1/M2 = 26.18 + 39.21 = 65.39 M_eq = 0.0153🔹 Step 2 – Equivalent Mechanical Power
Relative mechanical input: Pm = Pm1 − Pm2 = 1 − 0.7Pm = 0.3 pu
🔹 Step 3 – During Fault
Electrical power during fault: Pe_fault = 0.2 sin δ At δ₀ = 25°: Pe_fault = 0.2 × sin25° = 0.2 × 0.422 = 0.084 pu Accelerating power: Pa = Pm − Pe_fault = 0.3 − 0.084Pa = 0.216 pu
Rotor accelerates strongly.🔹 Step 4 – Post Fault Power
Pe_post = 1.2 sin δ System decelerates when: Pe_post > Pm 1.2 sin δ > 0.3 sin δ > 0.25 δ > 14.5° Since δ₀ already 25°, Deceleration possible after clearing.🔹 Step 5 – Stability Condition
Use Equal Area Criterion: Area A1 (Acceleration during fault) Area A2 (Deceleration after fault) If: A1 ≤ A2 → Stable A1 > A2 → Unstable Because fault power extremely low (0.2 pu), Acceleration area large → Stability margin small.🎯 Final Interpretation
✔ Severe fault reduces Pe drastically ✔ Higher mechanical mismatch increases instability ✔ Fast clearing required ✔ Two-machine behaves like equivalent SMIBFault Strength Controls Stability Margin
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