Stability classification and swing equation

Classification of power system stability

             Power system stability
       +----------------+-----------------+
  Rotor angle       Frequency          Voltage
  stability         stability          stability
   +------+                         +--------+
 Small-  Transient            Large-dist. Small-dist.
 signal  (large-dist.)
 (steady state / dynamic)

1. Rotor angle stability: ability of synchronous machines to remain in synchronism.
   • Steady state (small-signal) stability: small, gradual changes; limit $EV/X$.
   • Dynamic stability: small disturbances with controllers (AVR, governor, PSS) acting; concerns damping of oscillations over several seconds.
   • Transient stability: large sudden disturbances such as faults; first swing, about 1 s.
2. Voltage stability: ability to keep acceptable voltages at all buses; lost through lack of reactive power, leading to voltage collapse.
3. Frequency stability: ability to maintain frequency after a large imbalance between generation and load.

Derivation

Consider the rotor of a synchronous generator. Let

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

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

$$J\frac{d^2\theta_m}{dt^2} = T_a = T_m - T_e$$

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

$$\theta_m = \omega_{sm}t + \delta_m \;\Rightarrow\; \frac{d\theta_m}{dt} = \omega_{sm} + \frac{d\delta_m}{dt}, \quad \frac{d^2\theta_m}{dt^2} = \frac{d^2\delta_m}{dt^2}$$

so

$$J\frac{d^2\delta_m}{dt^2} = T_m - T_e$$

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

$$J\omega_m\frac{d^2\delta_m}{dt^2} = P_m - P_e$$

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

$$M\frac{d^2\delta_m}{dt^2} = P_m - P_e$$

Step 4: Introduce the inertia constant $H$. $H$ is the stored kinetic energy at synchronous speed per MVA rating $S$:

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

Substituting and dividing by $S$:

$$\frac{2H}{\omega_{sm}}\frac{d^2\delta_m}{dt^2} = \frac{P_m - P_e}{S} = P_m - P_e\ \ \text{(pu)}$$

Step 5: Use electrical angle. For a machine with $p$ poles, $\delta = \frac{p}{2}\delta_m$ and $\omega_s = \frac{p}{2}\omega_{sm}$, so $\delta/\omega_s = \delta_m/\omega_{sm}$:

$$\boxed{\frac{2H}{\omega_s}\frac{d^2\delta}{dt^2} = P_m - P_e} \qquad \text{or} \qquad \frac{H}{\pi f}\frac{d^2\delta}{dt^2} = P_m - P_e$$

with $\delta$ in electrical radians and $\omega_s = 2\pi f$. If $\delta$ is in electrical degrees, $\dfrac{H}{180f}\dfrac{d^2\delta}{dt^2} = P_m - P_e$.

For a machine connected to an infinite bus through reactance $X$, $P_e = \dfrac{EV}{X}\sin\delta = P_{max}\sin\delta$, so

$$\frac{H}{\pi f}\frac{d^2\delta}{dt^2} = P_m - P_{max}\sin\delta$$

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