Question 2 of 7: Reactive power in the system, and the two-reaction salient-pole machine

Problem 2 — Reactive power in the system, and the two-reaction salient-pole machine (20 points)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a) — why reactive power must be available everywhere

Reactive power is not a commodity that gets consumed and delivered the way real power is; it is an oscillation of energy between the magnetic and electric storage of the network, and its practical importance is that voltage magnitude is controlled by it. In a transmission network the series impedance is dominantly inductive, so the voltage drop along a branch is approximately $\Delta V \approx (RP + XQ)/V$ and, with $X \gg R$, the magnitude of the drop is governed almost entirely by the reactive flow. Real power transfer moves the phase angle; reactive power transfer moves the magnitude. It follows that a shortage of reactive supply shows up first as sagging voltages, and that it cannot be cured from a distance: reactive power travels badly, because pushing it down a reactive corridor consumes more of it ($I^2X$) the further it goes. This is the reason the requirement is expressed as “throughout” the system rather than “in total”.

The operational consequences of an inadequate local reserve are severe and progressive. Low voltage raises the current drawn by constant-power loads, which raises the reactive absorption of the lines feeding them, which lowers the voltage further — the positive feedback that ends in voltage collapse. Along the way, induction motors lose torque as the square of voltage and can stall; transformer tap-changers run out of range; generator field or armature current limiters bite and remove the very sources that were holding the area up; and losses rise everywhere because current rises for the same delivered power. Conversely, an excess of reactive supply at light load raises voltages above equipment ratings, so the requirement is really for a controllable reserve in both directions.

The major sources, in the order a system operator would call on them, are: synchronous generators, whose excitation is continuously controllable and which are the only fast source able to both supply and absorb (their capability is bounded by the rotor-heating, stator-heating and under-excitation limits of the capability chart); synchronous condensers, which are generators without prime movers, retained at key substations for the same continuous two-quadrant capability; shunt capacitor banks, fixed or switched, the cheapest bulk source but one whose output falls as the square of voltage exactly when it is most needed; shunt reactors, which absorb the surplus charging of long lightly-loaded lines; the transmission lines themselves, whose distributed shunt capacitance generates reactive power below the surge-impedance loading and absorbs it above; series capacitors, which reduce the effective corridor reactance and so reduce the reactive demand of the transfer; and power-electronic compensators — SVCs and STATCOMs — which give cycle-speed, continuously variable support and are increasingly the standard remedy at weak points in the network. Distribution capacitors and customer power-factor correction belong on the same list, since reactive power supplied at the load is reactive power that never has to be carried.

Part (b) — completing the two-reaction table

Given. A salient-pole machine on an infinite bus, $V = 1.00$ pu, $x_d = 0.6$ pu, $x_q = 0.3$ pu, armature resistance neglected and armature reaction (saturation) neglected. Three operating conditions, each with two of the four quantities $\{P, Q_2, E, \delta\}$ specified.

Find. The two missing entries in each row of the table, with $Q_2$ measured at the machine terminals (which, with the machine tied directly to the infinite bus, are the infinite-bus terminals).

Approach. Use the two-reaction (Blondel) power expressions for a salient-pole machine on an infinite bus: solve row A for $E$ linearly, evaluate row B directly, and solve row C as a quadratic in $\cos\delta$.

1. State the two-reaction relations. Neglecting armature resistance, the real and reactive power delivered by a salient-pole machine to an infinite bus of voltage $V$ at a rotor angle $\delta$ are $$\begin{aligned} P &= \frac{EV}{x_d}\sin\delta + \frac{V^{2}}{2}\left(\frac{1}{x_q}-\frac{1}{x_d}\right)\sin 2\delta \\[3pt] Q &= \frac{EV}{x_d}\cos\delta - V^{2}\left(\frac{\cos^{2}\delta}{x_d}+\frac{\sin^{2}\delta}{x_q}\right)\end{aligned}$$ With $V=1$, $x_d=0.6$ and $x_q=0.3$ the reluctance coefficient is $\frac{1}{2}\left(\frac{1}{0.3}-\frac{1}{0.6}\right)=0.833333$ pu.
2. Condition A — solve the $P$ equation for $E$. At $\delta = 45^{\circ}$ the reluctance term is at its maximum, $0.833333\sin 90^{\circ}=0.833333$ pu, so only $2.0-0.8333=1.1667$ pu is left for the excitation term: $$E=\frac{P-0.833333\sin 2\delta}{(V/x_d)\sin\delta}=\frac{1.166667}{1.666667\times 0.707107}=\boxed{0.9899\ \text{pu}}$$ Note that $E$ is below the bus voltage even though the machine is delivering 2.0 pu — the reluctance term is carrying 42 per cent of the load.
3. Condition A — evaluate $Q_2$ at that excitation. Substituting $E=0.989949$ and $\delta = 45^{\circ}$ into the reactive expression, $$Q_2=\frac{(0.9899)(1)}{0.6}\cos 45^{\circ}-\left(\frac{0.5}{0.6}+\frac{0.5}{0.3}\right)=1.1667-2.5=\boxed{-1.3333\ \text{pu}}$$ The machine is strongly under-excited: it delivers 2.0 pu of real power while absorbing 1.333 pu of reactive power from the bus.
4. Condition B — both unknowns are explicit. With $E = 1.25$ pu and $\delta = 40^{\circ}$ nothing has to be inverted; the two expressions are evaluated directly. The excitation term contributes $\frac{1.25}{0.6}\sin 40^{\circ}=1.3391$ pu and the reluctance term $0.833333\sin 80^{\circ}=0.8207$ pu, so $$\boxed{P=2.1598\ \text{pu}}$$
5. Condition B — the terminal reactive power. The same substitution in the reactive expression gives $$Q_2=\frac{1.25}{0.6}\cos 40^{\circ}-\left(\frac{\cos^{2}40^{\circ}}{0.6}+\frac{\sin^{2}40^{\circ}}{0.3}\right)=1.5959-2.3553=\boxed{-0.7594\ \text{pu}}$$ Still under-excited, but far less so than condition A — raising $E$ from 0.99 to 1.25 pu has bought back 0.574 pu of reactive support.
6. Condition C — turn $Q_2 = 0$ into a quadratic in $\cos\delta$. Writing $c=\cos\delta$ and using $\sin^{2}\delta = 1-c^{2}$, the reactive expression collapses to $$\left(\frac{1}{x_q}-\frac{1}{x_d}\right)c^{2}+\frac{EV}{x_d}c-\frac{V^{2}}{x_q}=0$$ which with these numbers is $1.666667c^{2}+1.916667c-3.333333=0$, or, after dividing through by 1.666667, the tidy form $c^{2}+1.15c-2=0$.
7. Take the admissible root. The two roots are $c=0.951638$ and $c=-2.101638$; the second lies outside $[-1,1]$ and is discarded as non-physical. Hence $$\boxed{\delta = 17.8918^{\circ}}$$ and back-substitution confirms $Q_2 = -0.000000$ pu, i.e. zero to within rounding.
8. Condition C — the real power at that angle. With $\delta=17.8918^{\circ}$ and $E=1.15$ pu, $$P=\frac{1.15}{0.6}\sin 17.89^{\circ}+0.833333\sin 35.78^{\circ}=0.5888+0.4873=\boxed{1.0761\ \text{pu}}$$ Condition C is therefore the unity-power-factor operating point of this machine at $E = 1.15$ pu.