---
title: "EAC for a two machine system"
subject: "Power System Analysis II"
chapter: 6
key: "semester_5/power_system_analysis_ii:eac-two-machine-system"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/eac-two-machine-system
site: BCT Engineering
---

# EAC for a two machine system

Power System Analysis II · Chapter 6: Power System Stability

- Asked in: 2069 Chaitra, 2+8 marks
- Video: [Power System -ll || equal area criterion of stability part -1 ||](https://www.youtube.com/watch?v=fJpPnOJLIds) — Technical Subham Edu.
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/eac-two-machine-system
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6

## Question

What is equal area criterion for assessing the transient stability of a two machine system? Explain and justify with the help of a suitable example.

## Short answer

For a two-machine system, subtract individual swing equations to get an equivalent single machine with Meq = M1M2/(M1+M2). Stability holds if accelerating area equals decelerating area before delta_max = 180 deg - arcsin(Pm/Pmax). In the example, Pmax = 2 pu, delta0 = 23.58 deg, delta1 = 44.43 deg, A1 = 0.105, A2max = 0.629, so the system is stable with delta2 = 68.20 deg.

## Answer

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

```text
 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. According to the equal area criterion, the system is stable if:
   A. Accelerating area A1 equals decelerating area A2 before delta_max is passed
   B. A1 is always greater than A2
   C. Pm is greater than Pmax
   D. The fault is never cleared
   Answer: A. Kinetic energy gained in A1 must be returned as an equal decelerating area A2 before the maximum allowed angle. (Key point: EAC statement)

2. The equivalent inertia for a two-machine system is:
   A. M1 + M2
   B. M1 M2 / (M1 + M2)
   C. (M1 + M2)/2
   D. M1 - M2
   Answer: B. Subtracting the two scaled swing equations gives Meq = M1M2/(M1+M2). (Key point: Two-machine reduction)

3. For the equivalent single machine, Pmax equals:
   A. E1 + E2 over X
   B. X/(E1 E2)
   C. E1 E2 / X
   D. (E1 - E2)/X
   Answer: C. Pmax = E1E2/X, the maximum transfer between the two machines. (Key point: Equivalent Pmax)

4. When would the two machines fall out of step in the example?
   A. If H1 were increased
   B. If the angle stayed below 44.43 degrees
   C. If Pmax were larger
   D. If A1 exceeded the available decelerating area A2,max
   Answer: D. If A1 > A2,max the relative angle passes delta_max and synchronism is lost. (Key point: Condition for instability)

## Related in Power System Analysis II

- [Equal area criterion for transient stability](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aequal-area-criterion-concept) — Power System Analysis II, chapter 6, asked 4×
- [Stability types; EAC for parallel lines](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Astability-types-eac-parallel-lines) — Power System Analysis II, chapter 6

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