📘 Equal Area Criterion – Numerical Problem (Step by Step)
We analyze stability of a single machine infinite bus system after a temporary fault.
🔹 Problem Statement
Given: E = 1.2 pu V = 1 pu X_pre = 0.6 pu X_fault = 1.8 pu Mechanical Power Pm = 0.8 pu Find: ✔ Initial angle δ₀ ✔ Maximum power ✔ Critical clearing angle (conceptually)🔹 Step 1 – Pre-Fault Maximum Power
Power angle equation:Pmax = EV / X
Pre-fault: Pmax = (1.2 × 1) / 0.6 Pmax = 2 pu🔹 Step 2 – Initial Rotor Angle δ₀
At steady state: Pm = Pe So: 0.8 = 2 sin δ₀ sin δ₀ = 0.4δ₀ = sin⁻¹(0.4) ≈ 23.6°
🔹 Step 3 – During Fault
New maximum power: Pmax_fault = (1.2 × 1) / 1.8 = 0.67 pu Since: Pm = 0.8 pu > 0.67 pu Rotor accelerates. Accelerating power: Pa = Pm − Pe_fault🔹 Step 4 – Equal Area Condition
Area A1 (Acceleration): A1 = ∫(Pm − Pe_fault) dδ Area A2 (Deceleration): A2 = ∫(Pe_post − Pm) dδ For stability:A1 = A2
We use graphical solution in exam.🔹 Step 5 – Maximum Post Fault Power
Assume post-fault reactance same as pre-fault: Pmax_post = 2 pu Now deceleration occurs when: Pe_post > Pm🔹 Step 6 – Critical Clearing Angle Concept
At critical clearing angle δc: Area A1 = Area A2 If fault cleared before δc → Stable If after δc → Unstable In GATE, usually asked conceptually or numerically simplified.🎯 Important Observations
- Higher reactance during fault → lower Pe
- Higher Pm → less stability margin
- Clearing time critical
- δ must remain below 180°
Equal Area = Energy Balance of Rotor
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