Rotor dynamics for incremental angle change

For a small change in rotor angle, the swing equation can be linearised about the operating point. The result is a second-order linear differential equation that shows the rotor oscillates about $\delta_0$ with a natural frequency set by the synchronising power coefficient and the inertia.

Swing equation

For a machine connected to an infinite bus through a total reactance $X$:

$$M\frac{d^2\delta}{dt^2} = P_m - P_e = P_m - P_{max}\sin\delta, \qquad M = \frac{H}{\pi f}\ \text{(pu)}$$

Linearising for a small change

Let the rotor angle change by a small amount $\Delta\delta$ from the steady-state angle $\delta_0$, with $P_m$ constant:

$$\begin{aligned} \delta &= \delta_0 + \Delta\delta \\ M\frac{d^2(\delta_0+\Delta\delta)}{dt^2} &= P_m - P_{max}\sin(\delta_0+\Delta\delta) \end{aligned}$$

Expand the sine. For small $\Delta\delta$: $\cos\Delta\delta \approx 1$, $\sin\Delta\delta \approx \Delta\delta$, so

$$\sin(\delta_0+\Delta\delta) \approx \sin\delta_0 + \Delta\delta\cos\delta_0$$

At steady state $P_m = P_{max}\sin\delta_0$ and $d^2\delta_0/dt^2 = 0$. These terms cancel, which leaves

$$M\frac{d^2\Delta\delta}{dt^2} + \left(P_{max}\cos\delta_0\right)\Delta\delta = 0$$

The term $P_s = \left.\dfrac{dP_e}{d\delta}\right|_{\delta_0} = P_{max}\cos\delta_0$ is the synchronising power coefficient. So:

$$M\frac{d^2\Delta\delta}{dt^2} + P_s\,\Delta\delta = 0$$

Including damping

If there is damping power $D\,\dfrac{d\Delta\delta}{dt}$ (from damper windings), the equation becomes

$$M\frac{d^2\Delta\delta}{dt^2} + D\frac{d\Delta\delta}{dt} + P_s\,\Delta\delta = 0$$

Its characteristic equation is $Ms^2 + Ds + P_s = 0$, which gives

$$\omega_n = \sqrt{\frac{P_s}{M}} = \sqrt{\frac{\pi f P_s}{H}}\ \text{rad/s}, \qquad \zeta = \frac{D}{2\sqrt{M P_s}}$$

The damped frequency of oscillation is $\omega_d = \omega_n\sqrt{1-\zeta^2}$.

Interpretation

• If $P_s > 0$ (that is, $0 < \delta_0 < 90^\circ$), the roots have negative real parts or lie on the imaginary axis. $\Delta\delta$ then oscillates about $\delta_0$, and the system is steady-state stable.
• If $P_s < 0$ ($\delta_0 > 90^\circ$), one root is real and positive. $\Delta\delta$ then grows exponentially and the machine loses synchronism.
• The machine is most stable at small $\delta_0$, where $P_s$ is large. The limit is reached at $\delta_0 = 90^\circ$, where $P_s = 0$.
• A larger $H$ lowers the frequency of the rotor oscillation; it does not change the stability limit.