Electrical Machines 101 · The Swing Equation

A step in mechanical power: the equal-area criterion, and the swing it predicts

Pmax = 2.00 pu · H = 3.5 pu-s · 50 Hz
Verdict Stable rotor turns back
New equilibrium δ₁ 48.6° from δ₀ = 30.0°
Peak swing δ₂ 69.9° overshoot 21.3°
A₁ accelerating 0.078 pu·rad
A₂ available 0.478 pu·rad · 6.2× A₁
Margin left 84% of A₂ unused

The step

Mechanical power steps from Pm0 = 1.00 pu to Pm0 + ΔP. The network is untouched, so there is only one power-angle curve. The rotor cannot move instantly, so δ stays at 30° and the accelerating power at that first instant is exactly ΔP. Positive — the governor opens the valves and the rotor swings up; negative — load rejection or valve closure, and it swings down.

+0.50 pu
−2.50 (load rejection)+1.00 pu (pickup)
Pm: 1.00 → 1.50 pu
0
0 — classical12
ζ = 0.000 · rings for ever

Equal-area criterion

A₁ is swept out accelerating from δ₀ to δ₁; A₂ is swept out decelerating beyond δ₁. The rotor turns back where the two are equal.

Power-angle curve with accelerating and decelerating areas shaded
A₁ accelerating A₂ decelerating margin unused

Time-domain response

The same step, integrated step by step from the swing equation below. δ₂ is where the rotor turns around, not where it settles.

Rotor angle and machine frequency deviation against time after the step
rotor angle δ(t) frequency deviation Δf(t)

The swing equation

Newton's second law for the rotor, and both plots above are solutions of it. The rotor is a mass on a shaft: whatever torque is left once the electromagnetic torque has been supplied goes into accelerating it. δm is measured from a synchronously rotating reference, so it holds still in steady state and moves only when the torques stop balancing.

J d2δmdt2 = Tm − Te − Ddδmdt

inertia × angular acceleration mechanical torque in electromagnetic torque out damping

Numbers behind the plots

Angles in electrical degrees; areas in per-unit power × radians, which is energy per unit of the machine's rating.

Quantity Symbol Value Note