Question 2 of 7: Armature reaction, reactive-power capability and the round-rotor operating table

Question 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.

Part (a) — armature reaction. When a synchronous machine is loaded, the three stator windings carry balanced currents, and those currents themselves set up a rotating magnetic field in the air gap. That stator field rotates at exactly the same speed as the rotor field, so the two are stationary relative to one another and combine into a single resultant air-gap field. Armature reaction is the name given to this effect of the stator (armature) current on the air-gap flux: the field the machine actually works with is not the field the excitation winding alone would produce, but that field modified by the load.

The direction of the modification depends on the phase of the stator current relative to the excitation voltage. A lagging (inductive) armature current produces a stator field that is largely in opposition to the rotor field, so the resultant flux is reduced — the machine is demagnetised, and the terminal voltage falls unless the field current is raised. A leading (capacitive) armature current does the reverse and magnetises the machine, raising the terminal voltage. A current in quadrature with the excitation voltage produces a cross-magnetising field that distorts rather than strengthens or weakens the flux, and is what transfers real power across the air gap.

The engineering value of the idea is that, for a round-rotor machine in the linear (unsaturated) region, the whole effect is proportional to the armature current and can therefore be represented by a single reactance in series with the internal voltage. That reactance, lumped together with the true leakage reactance of the stator winding, is the synchronous reactance $X_s$ — here 0.3 pu. The instruction in part (c) to "neglect armature reaction" is therefore best read as an instruction to neglect saturation and armature resistance and to use the constant-$X_s$ linear model; the reaction itself is already inside $X_s$ and cannot be removed from it.

Part (b) — when the machine supplies reactive power. Connected to an infinite bus of voltage $V$ through the synchronous reactance, the machine's terminal reactive power is $$Q_2=\frac{EV\cos\delta-V^{2}}{X_s}$$ This expression changes sign at $E\cos\delta=V$. When the excitation is raised so far that $E\cos\delta>V$ the numerator is positive and the machine delivers reactive power to the network; it is then said to be over-excited, and it behaves towards the system exactly like a capacitor. When $E\cos\delta<V$ the machine is under-excited and absorbs reactive power, behaving like a reactor. At $E\cos\delta=V$ it exchanges none.

Two practical consequences follow. First, because $\delta$ is fixed by the mechanical power and not by the field current, reactive output is controlled almost independently of real output by moving the field rheostat — which is why generators are the primary voltage-control resource on a transmission system, and why a synchronous machine run at no load purely for this purpose is called a synchronous condenser. Second, the ability to absorb reactive power is limited: driving $E$ too low brings the rotor angle towards the steady-state stability limit and also risks overheating the stator end-iron, which is why the under-excited region of a generator's capability chart is much smaller than the over-excited region.

Given. A round-rotor machine on an infinite bus, with $V=1.00$ pu held constant, $X_s=0.30$ pu, negligible armature resistance and no saliency; three operating conditions each specifying two of the four quantities $P$, $Q_2$, $E$, $\delta$.

Find. The two missing entries of each row of the table.

Approach. Both table entries in every row follow from the same pair of power-angle relations; the only trick is to notice that $Q_2=0$ is the single condition $E\cos\delta=V$, which decouples the two equations and makes rows A and B closed-form.

1. State the two governing relations. For a cylindrical-rotor machine with negligible resistance, connected to a bus of voltage $V$ through $X_s$, $$P=\frac{EV}{X_s}\sin\delta,\qquad Q_2=\frac{EV\cos\delta-V^{2}}{X_s}$$ With $V=1.00$ and $X_s=0.30$ these become $P=(E/0.3)\sin\delta$ and $Q_2=(E\cos\delta-1)/0.3$.
2. Condition A — excitation and zero reactive output given. Setting $Q_2=0$ gives $E\cos\delta=V$, so with $E=1.15$, $$\cos\delta=\frac{V}{E}=\frac{1.00}{1.15}=0.86957 \quad\Longrightarrow\quad \boxed{\delta_A=29.592^\circ}$$ Then $\sin\delta=0.49382$ and $$P_A=\frac{EV}{X_s}\sin\delta=3.8333\times0.49382=\boxed{P_A=1.8930\ \text{pu}}$$ The machine is at the boundary between the over- and under-excited regions: it delivers 1.8930 pu of real power and no reactive power at all.
3. Condition B — real power and zero reactive output given. The same $Q_2=0$ condition again fixes $E\cos\delta=1.00$, while the real-power equation fixes the quadrature component: $$E\sin\delta=\frac{P X_s}{V}=2.75\times0.30=0.8250$$ Dividing one by the other removes $E$ entirely, so $$\tan\delta=\frac{0.8250}{1.00}\quad\Longrightarrow\quad \boxed{\delta_B=39.523^\circ}$$ and taking the square root of the sum of squares removes $\delta$, $$E_B=\sqrt{1.00^{2}+0.8250^{2}}=\boxed{E_B=1.2964\ \text{pu}}$$ Geometrically this is just Pythagoras on Figure 2.1: the tip of $E$ lies on the vertical $Q_2=0$ line at a height equal to $PX_s/V$.
4. Condition C — excitation and angle given. Both unknowns now follow by direct substitution. With $E=1.05$ and $\delta=37.5^\circ$, so that $\sin\delta=0.60876$ and $\cos\delta=0.79335$, $$P_C=3.5000\times0.60876=\boxed{P_C=2.1307\ \text{pu}}$$ $$Q_C=\frac{(1.05)(0.79335)-1.00}{0.30}=\frac{0.83302-1.00}{0.30} =\boxed{Q_{2,C}=-0.5566\ \text{pu}}$$ The negative sign is the answer, not a slip: $E\cos\delta=0.83302$ is less than the bus voltage of 1.00, so by the criterion established in part (b) the machine is under-excited and is absorbing -0.5566 pu of reactive power from the network even while it exports 2.1307 pu of real power. Figure 2.1 shows point C lying to the left of the dashed locus, which is exactly this statement drawn rather than asserted.