EAC for a two machine system

The equal area criterion (EAC) is a direct method for checking transient stability without solving the swing equation. A system is stable after a disturbance if the area representing the kinetic energy gained by the rotor (accelerating area $A_1$) can be returned as an equal decelerating area $A_2$ before the rotor angle passes the maximum allowed angle. The criterion applies to a one-machine–infinite-bus system, or to a two-machine system after it has been reduced to an equivalent single machine.

Reduction of a two-machine system

Consider two finite machines connected by a reactance $X$, with internal voltages $E_1\angle\delta_1$ and $E_2\angle\delta_2$ (lossless network). Their swing equations are

$$M_1\frac{d^2\delta_1}{dt^2} = P_{m1} - P_{e1}, \qquad M_2\frac{d^2\delta_2}{dt^2} = P_{m2} - P_{e2}$$

For a lossless link, $P_{e2} = -P_{e1}$. Let $\delta = \delta_1 - \delta_2$. Subtract the second equation divided by $M_2$ from the first divided by $M_1$:

$$\begin{aligned} \frac{d^2\delta}{dt^2} &= \frac{P_{m1}-P_{e1}}{M_1} - \frac{P_{m2}+P_{e1}}{M_2} \\ \frac{M_1M_2}{M_1+M_2}\frac{d^2\delta}{dt^2} &= \frac{M_2P_{m1}-M_1P_{m2}}{M_1+M_2} - P_{e1} \end{aligned}$$

So the system behaves like one machine on an infinite bus:

$$M_{eq}\frac{d^2\delta}{dt^2} = P_{m,eq} - P_{max}\sin\delta$$

with $M_{eq} = \dfrac{M_1M_2}{M_1+M_2}$, $P_{m,eq} = \dfrac{M_2P_{m1}-M_1P_{m2}}{M_1+M_2}$ and $P_{max} = \dfrac{E_1E_2}{X}$.

Equal area criterion

Multiply the equivalent swing equation by $2\,\dfrac{d\delta}{dt}$ and integrate:

$$\left(\frac{d\delta}{dt}\right)^2 = \frac{2}{M_{eq}}\int_{\delta_0}^{\delta}(P_{m,eq}-P_e)\,d\delta$$

The relative angle stops changing ($d\delta/dt = 0$) and the system is stable if

$$\int_{\delta_0}^{\delta_{2}}(P_{m,eq}-P_e)\,d\delta = 0 \;\;\Longrightarrow\;\; A_1 = A_2$$

This must happen at some $\delta_2 \le \delta_{max} = 180^\circ - \sin^{-1}(P_{m,eq}/P_{max})$.

 Pe
  |          _______
  |        /    A2   \     Pmax sin(delta)
 P1 ------/----------- \------------
  |   A1 /|             \
 P0 ----/-|-             \
  |    /  |               \
  +---d0--d1-----d2--------dmax-- delta

Example (justification)

Assume two machines linked by a line, with $E_1 = 1.2$ pu, $E_2 = 1.0$ pu, $X = 0.6$ pu, and $H_1 = 4$, $H_2 = 6$ MJ/MVA on a common base. The machines are initially transferring $P_0 = 0.8$ pu, and the equivalent input then rises suddenly to $P_1 = 1.4$ pu.

• $P_{max} = \dfrac{1.2\times 1.0}{0.6} = 2.0$ pu and $H_{eq} = \dfrac{4\times 6}{4+6} = 2.4$ MJ/MVA. ($H_{eq}$ affects the speed of the swing but not the angles.)
• Initial angle: $\delta_0 = \sin^{-1}(0.8/2) = 23.58^\circ$.
• New equilibrium: $\delta_1 = \sin^{-1}(1.4/2) = 44.43^\circ$. Limit: $\delta_{max} = 180^\circ - 44.43^\circ = 135.57^\circ$.

The accelerating area is

$$A_1 = P_1(\delta_1-\delta_0) - P_{max}(\cos\delta_0-\cos\delta_1) = 0.105\ \text{pu·rad}$$

The decelerating area available up to $\delta_{max}$ is

$$A_{2,max} = P_{max}(\cos\delta_1-\cos\delta_{max}) - P_1(\delta_{max}-\delta_1) = 0.629\ \text{pu·rad}$$

Since $A_{2,max} > A_1$, the system is stable. Setting $A_1 = A_2$, that is $P_1(\delta_2-\delta_0) = P_{max}(\cos\delta_0-\cos\delta_2)$, and solving numerically gives $\delta_2 = 68.20^\circ$. The relative angle swings from $23.58^\circ$ to $68.20^\circ$ and then oscillates about $44.43^\circ$.

If the step were large enough that $A_1 > A_{2,max}$, the relative angle would pass $\delta_{max}$ and the machines would fall out of step. This shows how the EAC decides stability from areas alone.