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Effects of real and reactive power mismatch

Short answer

Active power mismatch changes system frequency: a deficit slows rotors, lowering frequency; a surplus raises it. Reactive power mismatch alters local bus voltage magnitudes: a deficit causes voltage drop across line reactance, surplus causes rise. The two are decoupled due to high X/R ratio.

Explain how the mismatch in active power affects the system frequency and mismatch in reactive power affects the voltage magnitude in an interconnected power system.

In an interconnected system, a mismatch of active power (generation ≠ load + losses) shows up as a change in frequency, while a mismatch of reactive power shows up as a change in bus voltage magnitude. The two effects are nearly independent ("decoupled") because of the high X/R ratio of transmission lines.

Active power mismatch and frequency

All synchronous machines in an interconnected system run at the same electrical speed. The rotor of each machine obeys the swing equation:

2Hωsd2δdt2=Pm−Pe\frac{2H}{\omega_s}\frac{d^2\delta}{dt^2} = P_m - P_e
  • If load increases suddenly (Pe>PmP_e > P_m), the extra energy is first taken from the kinetic energy stored in the rotating masses (12Jω2\tfrac{1}{2}J\omega^2). The rotors slow down and system frequency falls.
  • If load drops (Pe<PmP_e < P_m), the surplus energy accelerates the rotors and frequency rises.
  • Since frequency is common to the whole network, the mismatch is shared by all units in proportion to their inertia and then their governor droop.

Frequency is restored in steps:

  1. Primary control (governor): each governor senses the speed change and opens/closes the steam or water valve. Steady-state change is Δf=−ΔPL∑(1/Ri)+D\Delta f = -\dfrac{\Delta P_L}{\sum (1/R_i) + D}, so a small offset remains.
  2. Secondary control (AGC / LFC): the speed changer is adjusted until Δf=0\Delta f = 0 and tie-line flows return to scheduled values (area control error ACE=ΔPtie+B Δf→0ACE = \Delta P_{tie} + B\,\Delta f \to 0).

Reactive power mismatch and voltage

For a line between buses with voltages V1V_1, V2V_2 and reactance XX (R neglected, small δ\delta):

Q12≈V1(V1−V2)X⇒ΔV≈QXVQ_{12} \approx \frac{V_1(V_1 - V_2)}{X} \quad\Rightarrow\quad \Delta V \approx \frac{Q X}{V}
  • Reactive power flows from the higher-voltage bus to the lower-voltage bus, and the voltage drop across a line depends mainly on the Q it carries.
  • If reactive demand exceeds local supply, Q must be imported over the reactance, so the voltage at that bus falls. Excess Q (e.g. a lightly loaded long line with its charging) makes the voltage rise (Ferranti effect).
  • Unlike frequency, voltage is a local quantity. Reactive power cannot be sent far efficiently, so the mismatch must be fixed near where it occurs.

Control: generator excitation (AVR), shunt capacitors/reactors, synchronous condensers, SVC/STATCOM and on-load tap-changing transformers.

Why P–f and Q–V are decoupled

Since X≫RX \gg R:

P≈V1V2Xsin⁡δ,Q≈V1(V1−V2cos⁡δ)XP \approx \frac{V_1V_2}{X}\sin\delta,\qquad Q \approx \frac{V_1(V_1 - V_2\cos\delta)}{X}

P depends mainly on the angle δ\delta (linked to rotor position and frequency); Q depends mainly on the voltage magnitudes. So ∂P/∂δ\partial P/\partial \delta and ∂Q/∂∣V∣\partial Q/\partial |V| are large, while ∂P/∂∣V∣\partial P/\partial |V| and ∂Q/∂δ\partial Q/\partial \delta are small.

MismatchAffectsNatureCorrected by
Active power PFrequency fSystem-wideGovernor, AGC
Reactive power QVoltage magnitudeLocalAVR, capacitors, tap changers

✓ Answer checked by Claude Sonnet 5.5

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