Question 2 of 7: Armature Reaction, Reactive-Power Capability and the Round-Rotor Operating Table

Problem 2: Armature Reaction, Reactive-Power Capability and the Round-Rotor Operating Table (25 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.

(a) Armature reaction. When a synchronous machine is loaded, the balanced three-phase currents in the stator windings themselves produce a rotating magnetic field. That field turns in synchronism with the rotor field, so the two are stationary relative to one another and add vectorially to give the resultant air-gap flux. Armature reaction is precisely this effect of stator current on the air-gap flux: the machine no longer excites itself from the field winding alone. Its character depends on the phase of the current relative to the internal e.m.f. A lagging current produces a stator field that opposes the rotor field (demagnetising, so the terminal voltage sags and more field current is needed to hold it); a leading current produces a field that aids the rotor field (magnetising, so the terminal voltage rises); a current exactly in phase with the internal e.m.f. produces a cross-magnetising field at right angles to the rotor field, which distorts rather than weakens it and shifts the effective flux axis. Because the reaction flux is proportional to armature current and, in an unsaturated round-rotor machine, in quadrature with it, the whole effect can be represented by a single fictitious reactance in series with the internal e.m.f. Adding the true leakage reactance gives the synchronous reactance $X_s$ used in this problem — so the 0.3 pu reactance is not a stray inductance but mostly armature reaction expressed as a circuit element.

(b) When the machine is a source of reactive power. With armature resistance neglected, the reactive power delivered to the bus is$$Q=\frac{EV\cos\delta-V^{2}}{X_s},$$so the machine exports reactive power (acts as a capacitor seen from the network) precisely when $E\cos\delta > V$ — that is, when the internal e.m.f. projected onto the terminal voltage exceeds the terminal voltage. In operating language this is the over-excited condition: the field current is raised beyond the value that would make the machine deliver its real load at unity power factor. The armature current then lags the terminal voltage, armature reaction is demagnetising, and the extra field ampere-turns supply both the load flux and the exported vars. Under-excite the machine, so that $E\cos\delta$ falls below $V$, and the sign reverses: it absorbs vars, draws a leading current, and behaves as an inductor — the mode used at night on long lightly-loaded EHV lines to soak up line charging. At $E\cos\delta = V$ exactly the machine is at unity power factor, which is the locus conditions A and B sit on. The same statement holds at zero real load: an over-excited unloaded machine is a synchronous condenser, historically the standard var source before static compensators, and still used for inertia and short-circuit strength on networks with heavy inverter penetration.

(c) Given. Infinite bus $V = 1.00\ \text{pu}$; synchronous reactance $X_s = 0.3\ \text{pu}$; armature resistance neglected. The paper prints “neglecting armature reaction”; as part (a) explains, for a round-rotor machine armature reaction is already carried inside $X_s$, so removing it would leave no reactance to work with. The phrase is therefore read as the usual instruction to neglect armature resistance (and saturation), which is the only reading under which the table can be completed. Three operating conditions, each with two of the four quantities $(P,\ Q_2,\ E,\ \delta)$ known.

Find. The two missing entries in each row of the table, with $Q_2$ measured at the machine terminals, i.e. at the infinite bus to which the machine connects directly through $X_s$.

Approach. With resistance neglected the machine obeys the two standard round-rotor relations $P=(EV/X_s)\sin\delta$ and $Q=(EV\cos\delta-V^{2})/X_s$. Each row supplies two of the four unknowns, so each is a two-equation algebraic solve; the $Q_2 = 0$ rows collapse to the single condition $E\cos\delta = V$.

1. Condition A — unity power factor with a known excitation. Setting $Q_2 = 0$ in $Q=(EV\cos\delta-V^{2})/X_s$ gives $E\cos\delta=V$, so with $E = 1.12$:$$\cos\delta=\frac{V}{E}=\frac{1.00}{1.12}=0.892857\quad\Longrightarrow\quad\delta=26.766^\circ.$$The load angle follows from the excitation alone — no power balance needed yet.
2. Read the real power off the power-angle equation. Substituting into $P=(EV/X_s)\sin\delta$ with $\sin(26.766^\circ)=0.450340$:$$P=\frac{(1.12)(1.00)}{0.3}(0.450340)=3.73333\times0.450340,$$so that$$\boxed{\text{Condition A}:\quad P=1.6813\ \text{pu},\qquad \delta=26.77^\circ.}$$A useful cross-check: at unity power factor $|I_a| = P/V = 1.6813\ \text{pu}$, and $|V+jX_sI_a| = |1+j0.5044| = 1.1200$, recovering the given $E$.
3. Condition B — unity power factor with a known real power. The same two relations are now solved the other way. From $Q_2 = 0$, $E\cos\delta=V=1$; from $P = 2.7$, $E\sin\delta = PX_s/V = 2.7(0.3)=0.81$. These are the two rectangular components of the same phasor, so$$\begin{aligned}E&=\sqrt{(E\cos\delta)^2+(E\sin\delta)^2}=\sqrt{1^2+0.81^2}=\sqrt{1.6561},\\\tan\delta&=\frac{0.81}{1}=0.81.\end{aligned}$$
4. Evaluate condition B. The two components give$$\boxed{\text{Condition B}:\quad E=1.2869\ \text{pu},\qquad \delta=39.007^\circ.}$$More power at unity power factor demands both a larger angle and a larger excitation — on Figure 2 the operating point slides up the vertical unity-power-factor line, which is exactly the geometric meaning of the pair of equations.
5. Condition C — both electrical quantities known. Here $E = 1.1$ and $\delta = 37.5^\circ$ are given, so both outputs follow directly. With $\sin(37.5^\circ)=0.608761$ and $\cos(37.5^\circ)=0.793353$,$$P=\frac{(1.1)(1.00)}{0.3}(0.608761)=3.66667\times0.608761=2.2321\ \text{pu}.$$
6. Reactive power in condition C. Substituting the same numbers into the reactive relation,$$Q_2=\frac{EV\cos\delta-V^{2}}{X_s}=\frac{(1.1)(0.793353)-1}{0.3}=\frac{0.872688-1}{0.3}=\frac{-0.127312}{0.3},$$so that$$\boxed{\text{Condition C}:\quad P=2.2321\ \text{pu},\qquad Q_2=-0.4244\ \text{pu}.}$$The negative sign is physical, not an error: $E\cos\delta=0.8727$ is below $V=1$, so the machine is under-excited and absorbs 0.424 pu of reactive power from the bus. Its terminal power factor is $\cos[\arctan(-0.4244/2.2321)]=0.982$ leading.
7. Sanity-check the whole table against the stability limit. The steady-state pull-out power of each condition is $EV/X_s$, namely 3.733, 4.290 and 3.667 pu; every operating point sits comfortably below its own limit, and every load angle is well under $90^\circ$. Condition B is the most heavily loaded and also the most strongly excited, which is why it remains the stiffest of the three.

Final Results