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Reactive power injection and generation limits

Short answer

Reactive power injection raises bus voltage; consumption lowers it. For a generator on an infinite bus, changing excitation varies Q at constant terminal voltage. Maximum Q is limited by field heating and armature current; minimum Q by steady-state stability (δ ≤ 90°) and end-region heating.

What is the role of reactive power injection or consumption in a specified bus? For a synchronous generator connected to infinite bus discuss the concept of maximum and minimum reactive power generation limit.

Role of reactive power injection or consumption at a bus

Bus voltage depends mainly on the net reactive power at that bus, since ΔV≈QX/V\Delta V \approx QX/V for a line with X≫RX \gg R.

  • Injecting Q at a bus (generator over-excited, capacitor, condenser) reduces the reactive current drawn over the lines, so the voltage rises.
  • Consuming Q at a bus (inductive load, reactor, under-excited generator) increases the drop across line reactance, so the voltage falls.

So Q injection/absorption is the main tool to hold the bus voltage at its specified value; this is why a PV bus has its Q left free (within limits) to keep ∣V∣|V| fixed.

Generator on an infinite bus

For a cylindrical-rotor generator with internal emf EfE_f, terminal voltage VV (fixed by the infinite bus), synchronous reactance XsX_s and load angle δ\delta:

P=EfVXssin⁡δ,Q=EfVcos⁡δ−V2XsP = \frac{E_fV}{X_s}\sin\delta,\qquad Q = \frac{E_fV\cos\delta - V^2}{X_s}

With P fixed by the turbine, Q is set by the excitation EfE_f:

  • Efcos⁡δ>VE_f\cos\delta > V (over-excited): Q > 0, generator supplies Q (lagging pf).
  • Efcos⁡δ<VE_f\cos\delta < V (under-excited): Q < 0, generator absorbs Q (leading pf).

Maximum reactive power limit

The upper limit is set by field (rotor) heating. The maximum field current gives Ef,maxE_{f,max}, so

Qmax=VXsEf,max2−(PXsV)2−V2XsQ_{max} = \frac{V}{X_s}\sqrt{E_{f,max}^2 - \left(\frac{PX_s}{V}\right)^2} - \frac{V^2}{X_s}

On the P–Q capability chart this is a circle of radius VEf,max/XsVE_{f,max}/X_s centred at (0,−V2/Xs)(0, -V^2/X_s). At high P, the armature current limit (circle of radius VIa,maxVI_{a,max} about the origin) may set the bound instead.

Minimum reactive power limit

When the generator is under-excited:

  1. Steady-state stability limit: as EfE_f is reduced, δ\delta increases for the same P; at δ=90∘\delta = 90^\circ the machine loses synchronism. A practical margin (e.g. δ≤70∘\delta \le 70^\circ) defines the limit.
  2. Stator end-region heating in under-excited operation, which further restricts absorption.
  3. A minimum excitation limit set by the exciter.
        Q (supplied)
        ^    field-heating limit
        |  .---.
        | /     \  armature-current limit
  ------+--------+-----> P
        | \     /  stability / end-heating limit
        |  '---'
        v Q (absorbed)

So in load-flow studies a PV bus has Qmin≤QG≤QmaxQ_{min} \le Q_G \le Q_{max}; if a limit is hit, Q is fixed at that limit and the bus is treated as a PQ bus.

✓ Answer checked by Claude Sonnet 5.5

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