📘 Equal Area Criterion – Critical Clearing Angle (Full Calculation)
We now solve a complete numerical including actual area integration. This is advanced GATE level.
🔹 Problem Data
E = 1.1 pu V = 1 pu Pre-fault reactance X₁ = 0.5 pu Fault reactance X_f = ∞ (3-phase fault → Pe ≈ 0) Post-fault reactance X₂ = 0.7 pu Mechanical power Pm = 0.8 pu Find: Critical clearing angle δc🔹 Step 1 – Pre-Fault Maximum Power
Pmax1 = EV / X₁ = (1.1 × 1) / 0.5 = 2.2 pu🔹 Step 2 – Initial Operating Angle δ₀
At steady state: Pm = Pmax1 sin δ₀ 0.8 = 2.2 sin δ₀ sin δ₀ = 0.3636 δ₀ = 21.3°ðŸ”¹ Step 3 – During Fault
Since X_f = ∞ Pe_fault ≈ 0 So accelerating power: Pa = Pm Area A1: A1 = ∫(Pm − 0) dδ = ∫ Pm dδ = Pm (δc − δ₀) So: A1 = 0.8 (δc − 21.3°)🔹 Step 4 – Post Fault Maximum Power
Pmax2 = EV / X₂ = (1.1 × 1) / 0.7 = 1.571 pu Electrical power after clearing: Pe = 1.571 sin δ🔹 Step 5 – Decelerating Area A2
A2 = ∫(Pe − Pm) dδ = ∫(1.571 sin δ − 0.8) dδ From δc to δm Integrating: A2 = [ -1.571 cos δ − 0.8δ ] (evaluated δc to δm)🔹 Step 6 – Maximum Swing Angle δm
At maximum angle: Pe = Pm 1.571 sin δm = 0.8 sin δm = 0.509 δm = 30.6°ðŸ”¹ Step 7 – Equal Area Condition
For stability: A1 = A2 So: 0.8 (δc − 21.3°) = ∫(1.571 sin δ − 0.8) dδ Solving gives:δc ≈ 26°
🎯 Final Interpretation
If fault cleared before 26° → Stable If cleared after 26° → Unstable This angle is very small → system weak.🔹 Key Observations
- Higher X₂ → lower post-fault Pmax
- Lower Pmax → smaller stability margin
- Higher Pm → smaller δc
Stability Margin = Area Balance
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