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Swing equation and swing curve

Short answer

Swing equation: (2H/ωs)d2δ/dt2 = Pm - Pe in per unit, derived from Newton's law; for an infinite bus, Pe = (EV/X) sinδ. A swing curve plots δ versus t and shows stability (δ peaks then decreases) or loss of synchronism (δ keeps increasing), plus maximum rotor angle and stability margin.

Derive swing equation of a synchronous machine to be applicable in the study of power system stability. What is meant by swing curve? What information is supplied by the swing curve?

The swing equation relates the rotor angle δ\delta of a synchronous machine to the difference between mechanical input and electrical output power; it is the basic equation of power system stability studies.

Derivation

Consider the rotor of a synchronous generator. Let

  • JJ = moment of inertia of rotor and turbine (kg·m²)
  • θm\theta_m = angular position of the rotor with respect to a stationary axis (mech. rad)
  • TmT_m = mechanical (shaft) torque, TeT_e = electromagnetic torque (N·m)

Step 1: Newton's law of rotation. Accelerating torque = moment of inertia × angular acceleration:

Jd2θmdt2=Ta=Tm−TeJ\frac{d^2\theta_m}{dt^2} = T_a = T_m - T_e

Step 2: Measure angle from a synchronously rotating frame. The rotor angle δm\delta_m is the angle of the rotor relative to a reference axis rotating at synchronous speed ωsm\omega_{sm}:

θm=ωsmt+δm  ⇒  dθmdt=ωsm+dδmdt,d2θmdt2=d2δmdt2\theta_m = \omega_{sm}t + \delta_m \;\Rightarrow\; \frac{d\theta_m}{dt} = \omega_{sm} + \frac{d\delta_m}{dt}, \quad \frac{d^2\theta_m}{dt^2} = \frac{d^2\delta_m}{dt^2}

so

Jd2δmdt2=Tm−TeJ\frac{d^2\delta_m}{dt^2} = T_m - T_e

Step 3: Convert torque to power. Multiply by the rotor speed ωm\omega_m (power = torque × speed):

Jωmd2δmdt2=Pm−PeJ\omega_m\frac{d^2\delta_m}{dt^2} = P_m - P_e

M=JωmM = J\omega_m is the angular momentum (inertia constant). Since the speed changes very little during a swing, ωm≈ωsm\omega_m \approx \omega_{sm} and MM is taken as constant:

Md2δmdt2=Pm−PeM\frac{d^2\delta_m}{dt^2} = P_m - P_e

Step 4: Introduce the inertia constant HH. HH is the stored kinetic energy at synchronous speed per MVA rating SS:

H=12Jωsm2S  (MJ/MVA or s)  ⇒  M=Jωsm=2HSωsmH = \frac{\tfrac12 J\omega_{sm}^2}{S}\ \ \text{(MJ/MVA or s)} \;\Rightarrow\; M = J\omega_{sm} = \frac{2HS}{\omega_{sm}}

Substituting and dividing by SS:

2Hωsmd2δmdt2=Pm−PeS=Pm−Pe  (pu)\frac{2H}{\omega_{sm}}\frac{d^2\delta_m}{dt^2} = \frac{P_m - P_e}{S} = P_m - P_e\ \ \text{(pu)}

Step 5: Use electrical angle. For a machine with pp poles, δ=p2δm\delta = \frac{p}{2}\delta_m and ωs=p2ωsm\omega_s = \frac{p}{2}\omega_{sm}, so δ/ωs=δm/ωsm\delta/\omega_s = \delta_m/\omega_{sm}:

2Hωsd2δdt2=Pm−PeorHπfd2δdt2=Pm−Pe\boxed{\frac{2H}{\omega_s}\frac{d^2\delta}{dt^2} = P_m - P_e} \qquad \text{or} \qquad \frac{H}{\pi f}\frac{d^2\delta}{dt^2} = P_m - P_e

with δ\delta in electrical radians and ωs=2πf\omega_s = 2\pi f. If δ\delta is in electrical degrees, H180fd2δdt2=Pm−Pe\dfrac{H}{180f}\dfrac{d^2\delta}{dt^2} = P_m - P_e.

For a machine connected to an infinite bus through reactance XX, Pe=EVXsin⁡δ=Pmaxsin⁡δP_e = \dfrac{EV}{X}\sin\delta = P_{max}\sin\delta, so

Hπfd2δdt2=Pm−Pmaxsin⁡δ\frac{H}{\pi f}\frac{d^2\delta}{dt^2} = P_m - P_{max}\sin\delta

This is the swing equation. It is a second-order non-linear differential equation; its solution δ(t)\delta(t) describes how the rotor swings after a disturbance. If damping is included, a term DdδdtD\frac{d\delta}{dt} is added on the left.

Swing curve

A swing curve is the plot of rotor angle δ\delta against time tt, obtained by solving the swing equation (usually numerically, e.g. point-by-point or Runge-Kutta method) for a given disturbance and clearing time.

 delta
  ^        unstable (fault cleared late)
  |                 /
  |              _/
  |   .--.     _/       stable: delta peaks
  |  /    \  _/         and swings back
  | /  .-. \/  .-.
  |/  /   \___/   \__
  +--+-------------------> t
  d0 tc

Information supplied by the swing curve

  1. Stable or unstable: if δ\delta reaches a maximum and then decreases (oscillates about a new value), the system is stable; if δ\delta keeps increasing, the machine loses synchronism.
  2. Maximum rotor swing δmax\delta_{max} and the margin from the critical angle.
  3. Critical clearing time: by plotting swing curves for different clearing times, the largest time that still gives a stable curve is found; this sets the required speed of relays and breakers.
  4. Frequency and damping of oscillations of the rotor after the disturbance.
  5. In multi-machine systems, relative angles between machines show which machines swing together and which separate.

✓ Answer checked by Claude Sonnet 5.5

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