Voltage dips and transient stability
Rotor acceleration during a disturbance, grid recovery and loss of synchronism.
Before you begin: the power–angle characteristic and swing equation.
After this lesson you can: explain the first swing and investigate the effect of voltage-dip duration.
Work through the explanation, examples, practical experiment and independent exercises.
What changes during a dip?
At t = 1 s the bus voltage drops to kV, then recovers after the specified duration. Pmax falls by factor k during the dip. Mechanical power is unchanged unless a separate step is requested; the imbalance accelerates the rotor. This is an imposed grid-voltage experiment, not a detailed fault-network model. It does not calculate DC-offset or subtransient fault current.
Energy of the first swing
For D = 0 and constant Pm, the equal-area criterion compares accelerating energy during the disturbance with available decelerating energy after recovery. Integrate with respect to electrical angle, not time.
Adecelerating = ∫(Pe,after − Pm) dδ
δu = π − arcsin(Pm/Pmax,after)
The unstable post-disturbance equilibrium bounds the available braking region. Passing 90° alone does not prove loss of synchronism: the rotor may return. Estimate a critical duration through repeated simulations and refine the boundary.
Interpreting the laboratory output
If the angle drifts by one complete electrical revolution relative to its initial value, the simulation stops as a loss-of-synchronous-operation indicator. It also stops at |Δω| = 0.1 because the small-speed-deviation approximation is no longer appropriate. No stop within ten seconds means only that this event did not occur during that observation interval.
Repeat near-boundary cases with a smaller integration step. They are sensitive to both parameters and numerical accuracy. Voltage recovery alone does not guarantee return to equilibrium, and the final plotted sample is not automatically steady state.
Independent exercises
1. What is Pe at zero bus voltage in this model?
Show solution
Pe = 0 at every angle. Internal-source current can still be nonzero; this is not a detailed fault-current model.
2. Does Pm < Pmax after the disturbance guarantee synchronism?
Show solution
No. An equilibrium exists, but the rotor may have accumulated enough kinetic energy to leave its region of attraction.
Practical work
With P₀ = 0.8, ΔPm = 0 and residual voltage k = 0, compare dip durations 0.10, 0.20 and 0.40 s. Refine between returning and drifting trajectories. Repeat with D = 0 and explain the change.
Key takeaways
The rotor can accumulate too much kinetic energy before voltage recovers. Disturbance duration matters as much as depth.
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