Question 2 of 7: Excitation Control and the Salient-Pole Power-Angle Equations

Problem 2: Excitation Control and the Salient-Pole Power-Angle Equations (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) — over-excitation, under-excitation, and the machine as a reactive source. For a synchronous machine tied to a constant-voltage bus, the excitation (field) current sets the magnitude of the internal emf $E$ while the prime mover sets the real power $P$. The machine is called over-excited when $E$ is raised above the value that makes the terminal reactive power zero: the internal emf then exceeds the bus voltage sufficiently that the armature current lags the terminal voltage, and the machine delivers reactive power to the system, behaving as a capacitor seen from the bus. It is under-excited when $E$ is reduced below that value: the armature current leads the terminal voltage and the machine absorbs reactive power, behaving as an inductor.

Physically, the reactive flow is set by the difference between the internal emf projected on the bus voltage and the bus voltage itself. In the simplest round-rotor form, $Q=(EV\cos\delta - V^2)/x_d$, so $Q$ changes sign exactly when $E\cos\delta$ crosses $V$. Raising the field current raises $E$, raises $E\cos\delta$, and pushes $Q$ positive — which is the whole of reactive control on a generator. Because the real power is held by the turbine, raising the excitation at constant $P$ simply slides the operating point along a constant-power line on the capability chart towards the lagging (over-excited) side; the armature current rises and its angle swings from leading to lagging.

Operating a machine purely as a source of reactive power is therefore just the limiting case $P\to0$ with $E$ held well above $V$: the machine is run unloaded (or motored on a small amount of power to cover its own losses) and heavily over-excited, so it draws a nearly pure leading current and injects vars. A machine used this way is a synchronous condenser, and it has the practical advantages over a shunt capacitor bank that its output is continuously controllable by the exciter, that the vars do not collapse with the square of the voltage when the system is in trouble, and that it contributes short-circuit strength and inertia. Canadian transmission utilities have returned to synchronous condensers for exactly this reason at points where retiring synchronous plant has left the grid short of both vars and fault level. The same machine under-excited is equally useful for absorbing the charging vars of a lightly loaded EHV line.

Given. A salient-pole machine on an infinite bus, armature resistance neglected, with three partially specified operating points.

Find. The three missing entries of Table (1): $P$ and $\delta$ for condition A, $Q_2$ and $E$ for condition B, $P$ and $Q_2$ for condition C.

Approach. Use the two-reaction (Blondel) power-angle expressions for a salient-pole machine on an infinite bus, then solve each condition for whichever pair of quantities is missing — condition A needs a quadratic in $\cos\delta$, conditions B and C are direct substitutions.

1. Write the governing pair of equations. With armature resistance neglected and $V$ as reference, the two-reaction theory gives $$P=\frac{EV}{x_d}\sin\delta+\frac{V^{2}(x_d-x_q)}{2x_dx_q}\sin2\delta,$$ $$Q=\frac{EV}{x_d}\cos\delta-V^{2}\left(\frac{\cos^{2}\delta}{x_d}+\frac{\sin^{2}\delta}{x_q}\right).$$ The second term of $P$ is the reluctance power, which exists even with the field unexcited because the rotor is magnetically not round.
2. Evaluate the constants once. With $V=1.00$, $x_d=0.95$ and $x_q=0.40$, $$\frac{V}{x_d}=1.052632,\ \frac{V}{x_q}=2.500000,\ \frac{V^{2}(x_d-x_q)}{2x_dx_q}=\frac{0.55}{0.76}=0.723684.$$ Every number below is a substitution into the two equations with these three constants.
3. Condition A — impose $Q_2=0$ and solve for the angle. Setting $Q=0$ with $E=1.08$ and writing $c=\cos\delta$ turns the reactive equation into a quadratic (using $\sin^2\delta = 1-c^2$): $$\frac{E}{x_d}c=\frac{c^{2}}{x_d}+\frac{1-c^{2}}{x_q}\ \Rightarrow\ \left(\frac{1}{x_q}-\frac{1}{x_d}\right)c^{2}-\frac{E}{x_d}c+\frac{1}{x_q}=0,$$ $$1.447368\,c^{2}+1.136842\,c-2.500000=0.$$ Its roots are $c=0.978953$ and $c=-1.764408$; only the first is a physical cosine, so $$\boxed{\delta_A=\cos^{-1}0.978953=11.776^\circ}$$
4. Complete condition A by substituting that angle into the real-power equation. Both terms of $P$ are now known: $$P_A=1.136842\sin11.776^\circ+0.723684\sin23.552^\circ=0.232035+0.289147,$$ $$\boxed{P_A=0.5212\ \text{pu at }\ Q_2=0}$$ The reluctance term contributes more than half of it — at this small angle the field term is still climbing while $\sin2\delta$ is already well developed.
5. Condition B — solve the real-power equation for the internal emf. Here $\delta=45^\circ$ makes $\sin2\delta=1$ exactly, so the reluctance term is at its maximum and the equation is linear in $E$: $$1.45=\frac{E}{0.95}\sin45^\circ+0.723684\ \Rightarrow\ \frac{E}{0.95}(0.707107)=0.726316,$$ $$\boxed{E_B=0.9758\ \text{pu}}$$ Note that the machine is delivering 1.45 pu of real power on an internal emf below the bus voltage — possible only because the reluctance term is carrying half the load.
6. Complete condition B from the reactive equation. Substituting $E_B$ and $\delta=45^\circ$, where $\cos^2\delta=\sin^2\delta=0.5$: $$Q_B=\frac{0.975807}{0.95}(0.707107)-\left(\frac{0.5}{0.95}+\frac{0.5}{0.40}\right)=0.726316-1.776316,$$ $$\boxed{Q_{2,B}=-1.0500\ \text{pu (under-excited, absorbing)}}$$
7. Condition C — both quantities are direct substitutions. With $E=1.3$ and $\delta=37.5^\circ$ ($\sin\delta=0.608761$, $\cos\delta=0.793353$, $\sin2\delta=0.965926$): $$P_C=1.368421(0.608761)+0.723684(0.965926)=0.833042+0.699025,$$ $$Q_C=1.368421(0.793353)-\left(\frac{0.629410}{0.95}+\frac{0.370590}{0.40}\right)=1.085534-1.588905,$$ $$\boxed{P_C=1.5321\ \text{pu},\ Q_{2,C}=-0.5034\ \text{pu}}$$
8. Read the completed table back as machine behaviour. Only condition A sits on the unity-power-factor line. Conditions B and C are both under-excited — the machine is absorbing 1.05 pu and 0.50 pu of reactive power respectively — which is characteristic of a strongly salient machine with a small $x_q$: the $\sin^{2}\delta/x_q$ term in the reactive equation grows quickly with load angle and it is subtractive. A candidate who used the round-rotor formula $Q=(EV\cos\delta-V^{2})/x_d$ would have reported $Q_C=+0.033$ pu, i.e. the wrong sign, so the saliency matters to the conclusion here and not merely to the third decimal.

The completed table, with the given entries shown in place for comparison: