Can the generator stay in step?
Classical modelA fault interrupts power transfer. Clear it early enough and the rotor keeps swinging around its operating point. Leave it too long and it loses synchronism.
60 Hz
Try an experiment
Rotor angle over time
Fault starts at 0 s · No dampingWhere does the accelerating energy go?
Predict, change, explain
- Increase clearing time until the first swing escapes. Where is the boundary?
- Double H. How much more clearing time does the model allow?
- Reduce the post-fault transfer limit. Does a faster breaker compensate for a weaker network?
Model, assumptions, and sources
A classical synchronous generator connected to an ideal infinite bus. All powers share the machine MVA base. Electrical frequency is 60 Hz; pre-fault Pmax = 2 pu. The fault starts at t = 0 and forces electrical power to zero. Clearing switches instantaneously to the selected post-fault network.
dδ/dt = Δω
dΔω/dt = (2πf / 2H) [Pₘ − Pmax sin(δ)]
δ is electrical radians; Δω is the electrical angular-speed deviation in rad/s. Pmax = 0 during the fault. Initial conditions are δ₀ = asin(Pₘ / 2), Δω₀ = 0. The equal-area boundary gives the critical clearing angle; constant fault-on acceleration then gives the critical clearing time. Post-fault trajectories use fourth-order Runge–Kutta with a 1 ms step and an 8 s horizon. The chart ends early if the angle exceeds one full revolution.
Stable means a bounded first swing in this ideal undamped model; oscillations do not decay. This tool does not model excitation, governors, inverter controls, storage dynamics, or a real multi-machine network. A weaker post-fault network can be unstable even with immediate clearing.
Learning references: NPTEL: swing equation · NPTEL: equal-area criterion.