Engineering Ā· Simulator 084
Every power grid runs at one frequency, 50 or 60 Hz, as long as generation matches demand. Suddenly lose a power station or switch on a big load, and the frequency starts to fall. Change the grid's inertia, the generators' droop response and automatic control, and see how deep the dip goes and where the frequency settles.
All the spinning generators act like one big flywheel. When load exceeds generation, the missing power comes out of their kinetic energy and they slow down. In per-unit (p.u.) terms the swing equation is
2H dĪf/dt = ĪP_m ā ĪP_L ā D Īf
where H is the inertia constant (seconds of rated energy stored in rotation), ĪP_L the load step, ĪP_m the extra mechanical power from generators, and D the load damping (motors draw less power at lower frequency). Just after the step the rate of change of frequency is RoCoF = āĪP_L fā / 2H.
Primary control (droop): each governor raises output in proportion to the frequency error, ĪP_m ā āĪf / R, through a lag of time constant T_g. With a 5% droop, a 5% frequency drop would call for 100% more power. It stops the fall but leaves a steady error:
Īf_ss = āĪP_L / (1/R + D)
Secondary control (AGC): an integral controller slowly shifts the governors' set-points until the frequency is back at nominal.
A single-area, linearized model with one equivalent generator. No tie lines, no under-frequency load shedding, no inverter-based fast response, and no limits on generator ramp rates or reserves.
A 60 Hz grid with 4% droop and D = 1 suddenly gains 3% more load. Where does the frequency settle with primary control only? What RoCoF follows the step if H = 4 s?
Īf_ss = ā0.03 / (1/0.04 + 1) = ā0.03 / 26 ā ā0.00115 p.u., so the frequency settles at 60 Ć (1 ā 0.00115) ā 59.93 Hz. RoCoF = ā0.03 Ć 60 / (2 Ć 4) = ā0.225 Hz/s.