Equal area criterion for transient stability

The equal area criterion (EAC) is a graphical method to check the transient stability of a single machine connected to an infinite bus (or of two machines) without solving the swing equation step by step. It states that the system is stable if, after a disturbance, the area of accelerating energy under the power-angle curve can be balanced by an equal area of decelerating energy.

Basis

For a machine connected to an infinite bus, the swing equation (in pu, resistance and damping neglected) is

$$\frac{2H}{\omega_s}\frac{d^2\delta}{dt^2} = P_m - P_e = P_a, \qquad P_e = P_{max}\sin\delta = \frac{EV}{X}\sin\delta$$

Multiplying both sides by $2\dfrac{d\delta}{dt}$:

$$\frac{2H}{\omega_s}\cdot 2\frac{d\delta}{dt}\frac{d^2\delta}{dt^2} = 2(P_m - P_e)\frac{d\delta}{dt} \;\Rightarrow\; \frac{d}{dt}\left(\frac{d\delta}{dt}\right)^2 = \frac{\omega_s}{H}(P_m - P_e)\frac{d\delta}{dt}$$

Integrating from the initial angle $\delta_0$ (where $d\delta/dt = 0$, the rotor runs at synchronous speed):

$$\left(\frac{d\delta}{dt}\right)^2 = \frac{\omega_s}{H}\int_{\delta_0}^{\delta}(P_m - P_e)\,d\delta$$

For the machine to be stable, the rotor angle must stop increasing at some maximum angle, i.e. $d\delta/dt = 0$ at $\delta_{max}$:

$$\int_{\delta_0}^{\delta_{max}}(P_m - P_e)\,d\delta = 0$$

This means the area under the $P_a$–$\delta$ curve must be zero: the accelerating area $A_1$ (where $P_m > P_e$) must be equal to the decelerating area $A_2$ (where $P_e > P_m$):

$$A_1 = \int_{\delta_0}^{\delta_c}(P_m - P_e)\,d\delta = A_2 = \int_{\delta_c}^{\delta_{max}}(P_e - P_m)\,d\delta$$

Power-angle diagram

 Pe
  ^          Pmax1 sin d (pre-fault)
  |           .-''-.
  |         .' .--. '.    Pmax3 sin d
  |        / .' A2 '. \   (post-fault)
 Pm-------+-+-------+--\------
  |      /|A1|        \ \
  |     / |..|.....     \ \  Pmax2 sin d
  +----+--+--+--------+--+--> d
       0  d0 dc     dmax 180
 A1: accelerating area, A2: decelerating area

Applying it to a fault (sudden disturbance)

1. Before the fault: the machine runs at $\delta_0$, where $P_m = P_{max1}\sin\delta_0$.
2. During the fault: the transfer reactance rises, the power curve drops to $P_{max2}\sin\delta$ ($P_{max2} = 0$ for a fault at the generator bus). Since $P_m > P_e$, the rotor accelerates and $\delta$ increases. The energy gained is area $A_1$.
3. Fault cleared at $\delta_c$: the post-fault curve $P_{max3}\sin\delta$ applies. Now $P_e > P_m$, so the rotor decelerates, but $\delta$ keeps increasing until the kinetic energy is given back. This is area $A_2$.
4. Stability test: the system is stable if $A_2$ can become equal to $A_1$ before $\delta$ reaches $\delta_{max} = 180^\circ - \sin^{-1}(P_m/P_{max3})$. The rotor then swings back and oscillates about the new operating point. If the available $A_2 < A_1$, the machine loses synchronism.

Critical clearing angle

The largest clearing angle for which $A_1 = A_2$ just holds (with $\delta$ reaching exactly $\delta_{max}$) is the critical clearing angle:

$$\cos\delta_{cr} = \frac{P_m(\delta_{max} - \delta_0) + P_{max3}\cos\delta_{max} - P_{max2}\cos\delta_0}{P_{max3} - P_{max2}}$$

For a fault at the generator terminals with no change in network ($P_{max2} = 0$, $P_{max3} = P_{max}$, $\delta_{max} = \pi - \delta_0$):

$$\delta_{cr} = \cos^{-1}\left[(\pi - 2\delta_0)\sin\delta_0 - \cos\delta_0\right]$$

and, since $P_e = 0$ during the fault, the critical clearing time is

$$t_{cr} = \sqrt{\frac{4H(\delta_{cr} - \delta_0)}{\omega_s P_m}}$$

Assumptions and limitation

• Mechanical input $P_m$ constant during the transient; damping and resistance neglected.
• Machine represented by a constant voltage behind transient reactance.
• Valid only for one machine against an infinite bus (or two machines); for multi-machine systems, numerical solution of swing equations is needed.

Example: with $P_m = 1$ pu and $P_{max} = 2$ pu, $\delta_0 = 30^\circ$; for a terminal fault $\delta_{cr} = \cos^{-1}[(\pi - 1.047)(0.5) - 0.866] = \cos^{-1}(0.181) = 79.6^\circ$.