Dynamic Stability & the Swing Equation

A generator stays in step with the grid only as long as the magnetic "spring" coupling its rotor to the system can pull it back. Power transfer follows P = (E·V/X)·sin δ, where δ is the rotor angle. Push δ too far — with too much load or a fault that lasts too long — and the rotor slips over the top and loses synchronism. The rotor behaves exactly like a ball rolling in an energy well; trip a fault below and watch it swing back, or roll away.

Network
Generator EMF E
Grid voltage V
Reactance X (gen+line)
Mechanical power Pm
Machine Dynamics
Damping D
Disturbance — Three-Phase Fault
Clearing time

A nearby bolted fault drops power transfer to ~0 while the breaker is open, so the rotor accelerates. Clear it before the critical time or it slips.

One-Line Diagram
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Power-Angle Curve & Equal-Area Criterion
P = (EV/X)·sin δ Mechanical power Pm ▦ Accelerating area A₁ ▦ Decelerating area A₂
Energy Well — the Rolling Ball
Potential energy Total energy (KE + PE) ● Rotor "ball"
Rotor Angle Swing δ(t)
Readout

The Physics

Power flows from the machine to the grid in proportion to the sine of the rotor angle δ — the angle between the generator's internal EMF and the grid voltage:

Pe = (E·V / X)·sin δ  ·  Pmax = E·V / X (at δ = 90°)

In steady state the electrical power out equals the mechanical power in, so δ settles where Pmax·sin δ = Pm. There are two such angles, but only the one below 90° is stable: there the synchronizing power dP/dδ = Pmax·cos δ is positive, so a nudge to a higher angle produces more opposing power that pushes the rotor back. Past 90° that slope flips and the machine pulls out of step.

The swing equation is Newton's law for the rotor. Its inertia constant H resists angle changes; the net of mechanical drive minus electrical load accelerates it:

(2H / ωs)·d²δ/dt² = Pm − Pe(δ) − D·dδ/dt

This is identical to a ball of mass ∝ H rolling in a potential well whose shape is the integral of (Pe − Pm). The bottom of the well is the stable angle; the rim is the unstable equilibrium at 180°−δ₀. A fault injects kinetic energy (the rotor speeds up while it can't deliver power). If that energy is less than the height of the rim, the ball rolls back and oscillation damps out — stable. If it's more, the ball goes over the top — loss of synchronism.

📐 Equal-area criterion. During the fault the rotor gains an accelerating area A₁ on the power-angle curve. After the breaker clears, the curve above Pm provides a decelerating area A₂. The machine survives only if A₂ can equal A₁ before δ reaches the unstable point. The largest fault you can clear in time defines the critical clearing time — push the clearing-time slider past it and the swing runs away.
⚙ What helps: more inertia H (slower swing, more time to clear), lower reactance X (taller curve, bigger A₂), faster protection (smaller A₁), and damping (faster settle). Try them and watch the critical clearing time move.