--- title: "Equal area criterion for transient stability" subject: "Power System Analysis II" chapter: 6 key: "semester_5/power_system_analysis_ii:equal-area-criterion-concept" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/equal-area-criterion-concept site: BCT Engineering --- # Equal area criterion for transient stability Power System Analysis II · Chapter 6: Power System Stability - Asked in: 2073 Shrawan, 6 marks; 2073 Chaitra, 8 marks; 2072 Kartik, 8 marks; 2071 Shrawan, 6 marks (asked 4 times) - Video: [Equal Area Criterion - Power System Stability - Power System 3](https://www.youtube.com/watch?v=G6_VtQq0J2c) — Ekeeda - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/equal-area-criterion-concept - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6 ## Question Explain the Equal area criterion (concept) to study (evaluate) the transient stability of a single machine system connected to infinite bus with proper mathematical aids and diagram. ## Short answer Equal area criterion states a single machine-infinite bus system is transiently stable if the accelerating area (where Pm > Pe) equals the decelerating area (where Pe > Pm) on the power-angle curve. Derived by integrating the swing equation: ∫(Pm-Pe)dδ = 0 from δ0 to δmax. Stability requires fault clearing before critical clearing angle where areas just balance. ## Answer 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 ```text 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 $\delta_{cr} = \cos^{-1}[(\pi - 1.047)(0.5) - 0.866] = \cos^{-1}(0.181) = 79.6^\circ$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the equal area criterion require for stability? A. Accelerating area A1 is greater than A2 B. Pm = Pmax C. Pe is zero throughout D. Accelerating area A1 equals decelerating area A2 before delta reaches delta_max Answer: D. The integral of (Pm - Pe) must vanish at the angle where speed returns to synchronous. (Key point: Stability condition) 2. The criterion comes from integrating the swing equation, giving (d delta/dt)^2 proportional to: A. The integral of (Pm - Pe) d delta B. Pm x delta C. Pmax sin delta D. H only Answer: A. Multiplying by 2 d delta/dt and integrating gives speed deviation squared proportional to the area under Pa. (Key point: Origin of area rule) 3. The critical clearing angle is the clearing angle for which: A. A1 = 0 B. The fault lasts 1 s C. Pm = Pe during fault D. A2 just equals A1 with delta reaching delta_max Answer: D. Beyond it, available A2 is smaller than A1 and the machine loses synchronism. (Key point: Critical clearing angle) 4. For a terminal fault, what is true during the fault and for tcr? A. Pe = Pmax, tcr infinite B. Pe = 0, so the rotor accelerates under Pm and tcr = sqrt(4H(dcr - d0)/(ws Pm)) C. Pe = Pm D. Pm = 0 Answer: B. Pmax2 = 0 so Pe = 0, constant acceleration gives that tcr formula. 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