22-Elec-B7 Power Systems Engineering · December 2016

Question 2 of 7: Excitation, Reactive Power and the Salient-Pole Loading Table

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, December 2016 — 07-Elec-B7 Power Systems Engineering. Open-book, 3 hours duration. Seven problems are printed; any five constitute a complete paper, only the first five appearing in the answer book are marked, and all questions are of equal value (20 points each). All seven are solved in full below, because this set is a study resource rather than a timed sitting.

Reference texts. This exam code follows M. E. El-Hawary, Electrical Power Systems: Design and Analysis (IEEE Press) — the notation is his throughout: the “cantilever” transformer equivalent of Problem 3, the two-reaction salient-pole construction of Problem 2, the ABCD two-port of Problem 1 and the sequence-network fault reduction of Problem 6. Corroborating references: J. D. Glover, M. S. Sarma and T. J. Overbye, Power System Analysis and Design, 6th ed. (ch. 3 transformers, ch. 4–5 transmission lines, ch. 6 power flow, ch. 7–9 faults and symmetrical components, ch. 11 transient stability); J. J. Grainger and W. D. Stevenson, Power System Analysis; and S. J. Chapman, Electric Machinery Fundamentals, 5th ed., for the machine and transformer chapters. Canadian practice references where grounding and protection are discussed: CSA C22.1 Canadian Electrical Code, Part I, CSA C22.2 and the IEEE C62 / IEEE 142 (Green Book) series on system grounding.

Check: four judgement calls carried through the solutions, all recorded here rather than hedged in the answers.

(i) Problem 1 does not state a frequency. Parts (c) and (d) need one, and the paper is a Canadian national exam, so 60 Hz is used throughout — consistent with the 60 Hz stated in Problems 3 and 7. At 50 Hz the capacitance and surge impedance are unchanged (they do not involve frequency) and only the propagation constant of part (d) rescales by 5/6.

(ii) Problem 2 prints “neglecting armature reaction”. Armature reaction is precisely what the direct- and quadrature-axis reactances represent, so the phrase is read as its standard counterpart, neglecting armature resistance — the only reading under which the two given reactances can be used at all.

(iii) Problem 6 prints a part label “c-” with no question text. The five parts that do carry text (a, b, d, e, f) are marked 4 points each and total the 20 points the paper allots to every problem, so nothing is missing from the assessment; the sequence networks that the empty label most plausibly asked for are drawn anyway.

(iv) Problem 5 gives no reactance between buses 2 and 3. Figure (3) confirms the topology: bus 3 hangs radially off bus 1 through the j0.12 pu line, and the only other tie is line 1–2.

Problem 2: Excitation, Reactive Power and the Salient-Pole Loading Table (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.

Given.

Quantity	Symbol	Value
Infinite-bus voltage	$V$	1.00 pu (constant)
Direct-axis reactance	$x_d$	0.95 pu
Quadrature-axis reactance	$x_q$	0.40 pu
Condition A	—	$Q_2=0.0$, $E=1.07$; $P$ and $\delta$ unknown
Condition B	—	$P=1.42$, $\delta=47^{\circ}$; $Q_2$ and $E$ unknown
Condition C	—	$E=1.27$, $\delta=40.00^{\circ}$; $P$ and $Q_2$ unknown
Armature resistance	$r_a$	neglected

Find. The meaning of over- and under-excitation and the mechanism by which a synchronous machine sources reactive power; then the six missing entries of Table (1), using the two-reaction (Blondel) power expressions for a salient-pole machine on an infinite bus.

The essay part — excitation and reactive power. For a synchronous machine tied to a fixed bus, the field current sets the internal emf $E$ while the terminal voltage is held by the system. The machine is over-excited when the field is strong enough that the internal emf, projected along the bus voltage, exceeds it — loosely, $E\cos\delta > V$ — and it is under-excited when the field is weak enough that $E\cos\delta < V$. The distinction is not about real power at all: $P$ is set by the prime mover (or, for a motor, by the shaft load) and the machine will hold that $P$ over a wide range of field current. What excitation controls is the reactive component.

The mechanism is visible in the armature current. The current is driven by the phasor difference $\mathbf{E}-\mathbf{V}$ across the synchronous reactance, so when $E$ is large the current leads the bus voltage and the machine delivers reactive power to the system, behaving exactly like a capacitor; when $E$ is small the current lags and the machine absorbs reactive power, behaving like an inductor. An over-excited generator is therefore a source of reactive power, and an under-excited one is a sink. Because the field current is continuously adjustable and responds in a fraction of a second through the automatic voltage regulator, the synchronous machine is the fastest and most flexible reactive-power resource on a power system — far more so than switched capacitor banks, which are discrete and slow. Run at essentially zero real power with an over-excited field, the machine becomes a synchronous condenser, a pure reactive source used to hold voltage at a weak bus or at an HVDC terminal; the modern equivalent duty is performed by static compensators, but retired generators are still converted to condensers for exactly this reason.

Two limits bound the useful range and both appear on the machine capability curve: over-excitation is limited by rotor (field-winding) heating, and under-excitation is limited first by stator-end-iron heating and ultimately by steady-state stability, since weakening the field shrinks the maximum power the machine can transmit.

Figure 2 — Two-reaction power-angle characteristics of the machine at the three excitations found below. Each operating point lies on its own curve; note that all three sit on the rising limb, so every condition is steady-state stable. The reluctance term skews the peak to the left of 90°, which is the visible signature of saliency.

Approach. Write the Blondel two-reaction expressions for $P$ and $Q$ at the machine terminals against a 1.0 pu bus, then solve each condition for whichever pair of the four quantities is missing — a quadratic in $\cos\delta$ for Condition A, a linear solve for $E$ in Condition B, and a direct evaluation for Condition C.

Write the two-reaction power expressions and put the numbers in once. For a salient-pole machine on an infinite bus with negligible armature resistance, $$\begin{aligned}P&=\frac{EV}{x_d}\sin\delta+\frac{V^{2}}{2}\left(\frac{1}{x_q}-\frac{1}{x_d}\right)\sin 2\delta\\Q&=\frac{EV}{x_d}\cos\delta-V^{2}\left(\frac{\cos^{2}\delta}{x_d}+\frac{\sin^{2}\delta}{x_q}\right)\end{aligned}$$ With $V=1.00$, $x_d=0.95$ and $x_q=0.40$ these become $P=1.052632\,E\sin\delta+0.723684\sin 2\delta$ and $Q=1.052632\,E\cos\delta-\left(1.052632\cos^{2}\delta+2.5\sin^{2}\delta\right)$. The second term of $P$ is the reluctance power: it exists even with the field removed and it is what makes the machine salient.
Condition A — the $Q=0$ requirement is a quadratic in $\cos\delta$. Putting $c=\cos\delta$ and $\sin^{2}\delta=1-c^{2}$ into the $Q$ expression and setting it to zero gives $$\left(\frac{1}{x_q}-\frac{1}{x_d}\right)c^{2}+\frac{E}{x_d}\,c-\frac{1}{x_q}=0\;\Longrightarrow\;1.447368\,c^{2}+1.126316\,c-2.5=0$$ whose roots are $c=0.981553$ and $c=-1.759735$. The second lies outside $[-1,1]$ and is discarded, leaving $\delta_A=\arccos(0.981553)=11.02^{\circ}$.
Evaluate the real power at Condition A. Substituting $\delta_A=11.0223^{\circ}$ and $E=1.07$, $$P_A=1.052632(1.07)\sin(11.0223^{\circ})+0.723684\sin(22.0446^{\circ})=0.21522+0.27147\;\Longrightarrow\;\boxed{\ P_A=0.4870\ \text{pu},\ \delta_A=11.02^{\circ}\ }$$ Substituting back into the $Q$ expression returns $-2\times10^{-16}$, confirming the root. Note that more than half of this power comes from the reluctance term — at small $\delta$ the $\sin 2\delta$ term is the larger of the two.
Condition B — the excitation is the only unknown in the $P$ equation. With $\delta=47^{\circ}$ the reluctance term is fixed at $0.723684\sin(94^{\circ})=0.721921$, so $$1.052632\,E\sin(47^{\circ})=1.42-0.721921=0.698079\;\Longrightarrow\;E_B=\frac{0.698079}{0.769846}=0.9068\ \text{pu}$$ The excitation is below the bus voltage, so Condition B is under-excited before any reactive power is computed — a useful advance check on the sign of the answer.
Complete Condition B with the reactive power. Substituting $E_B=0.906777$ and $\delta=47^{\circ}$ into the $Q$ expression, $$Q_B=1.052632(0.906777)\cos(47^{\circ})-\left(1.052632\cos^{2}47^{\circ}+2.5\sin^{2}47^{\circ}\right)=0.650968-1.826799$$ $$\boxed{\ E_B=0.9068\ \text{pu},\qquad Q_{2,B}=-1.1758\ \text{pu}\ }$$ The machine is drawing 1.18 pu of reactive power from the system while delivering 1.42 pu of real power — deeply under-excited operation, which in practice would be blocked by the under-excitation limiter.
Condition C — both unknowns follow by direct substitution. With $E=1.27$ and $\delta=40.00^{\circ}$, $$P_C=1.052632(1.27)\sin 40^{\circ}+0.723684\sin 80^{\circ}=0.859306+0.712690$$ $$Q_C=1.052632(1.27)\cos 40^{\circ}-\left(1.052632\cos^{2}40^{\circ}+2.5\sin^{2}40^{\circ}\right)=1.024032-1.650649$$ $$\boxed{\ P_C=1.5720\ \text{pu},\qquad Q_{2,C}=-0.6266\ \text{pu}\ }$$ Even at $E=1.27$, well above the bus voltage, the machine is still absorbing reactive power: at $40^{\circ}$ the projection $E\cos\delta=0.973$ is below $V=1.0$, which is the exact criterion stated in the essay part.
Check that every condition is steady-state stable, and that saliency really matters. The synchronising coefficient is $\mathrm{d}P/\mathrm{d}\delta=(EV/x_d)\cos\delta+V^{2}(1/x_q-1/x_d)\cos 2\delta$, which evaluates to $+2.447$, $+0.550$ and $+1.275$ pu rad$^{-1}$ at A, B and C — all positive, so all three points sit on the rising limb of Figure 2. Finally, note what happens if the machine is wrongly treated as round-rotor: $Q=(EV\cos\delta-V^{2})/x_d$ would give $-0.0286$ pu at Condition C against the true $-0.6266$ pu. The saliency term is not a refinement here, it is the answer.

Table (1), completed.

Row	Real power	Reactive power / angle
Condition A	$P=0.4870$ pu	$Q_2=0.0$ (given)
Condition A (cont.)	$E=1.07$ (given)	$\delta=11.02^{\circ}$
Condition B	$P=1.42$ pu (given)	$Q_2=-1.1758$ pu
Condition B (cont.)	$E=0.9068$ pu	$\delta=47^{\circ}$ (given)
Condition C	$P=1.5720$ pu	$Q_2=-0.6266$ pu
Condition C (cont.)	$E=1.27$ (given)	$\delta=40.00^{\circ}$ (given)

Topic. Two-reaction (Blondel) theory of the salient-pole machine, and the excitation–reactive-power link.

Key relations. $P=\dfrac{EV}{x_d}\sin\delta+\dfrac{V^{2}}{2}\left(\dfrac{1}{x_q}-\dfrac{1}{x_d}\right)\sin 2\delta$; $Q=\dfrac{EV}{x_d}\cos\delta-V^{2}\left(\dfrac{\cos^{2}\delta}{x_d}+\dfrac{\sin^{2}\delta}{x_q}\right)$; synchronising coefficient $\mathrm{d}P/\mathrm{d}\delta$; round-rotor limit $x_q\to x_d$ recovers $P=(EV/x_s)\sin\delta$ and $Q=(EV\cos\delta-V^{2})/x_s$.

Why this works. A salient-pole rotor presents a low-reluctance path along the pole axis and a high-reluctance path between the poles, so the armature reaction flux — and hence the reactance it sees — depends on where the rotor is. Resolving the armature current into direct- and quadrature-axis components with separate reactances is the whole of two-reaction theory, and the extra $\sin 2\delta$ term that falls out is real torque: a salient machine will hold synchronism with no field at all, which is how reluctance motors work. Because the reluctance term peaks at $45^{\circ}$ while the excitation term peaks at $90^{\circ}$, the composite curve peaks somewhere between, and always below $90^{\circ}$. The reactive expression has the same structure with sine and cosine interchanged, and setting it to zero is what turns Condition A into a quadratic rather than a transcendental equation.

Common pitfall. Two recurring errors. First, taking both roots of the Condition-A quadratic seriously: one always lies outside $[-1,1]$ and must be discarded, and a student who does not check this reports a nonsensical angle. Second, confusing $Q_2$, the reactive power at the machine terminals, with the reactive loss in the reactance; the expression above already gives the terminal quantity, and no further subtraction is needed.

Source. El-Hawary, Electrical Power Systems: Design and Analysis, ch. 6 (synchronous machine, two-reaction model, power-angle characteristics); Chapman, Electric Machinery Fundamentals, 5th ed., ch. 4–5 (synchronous generator operation, V-curves and capability); Glover, Sarma & Overbye, §11.2.

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