22-Elec-B7 Power Systems Engineering · December 2018

Question 2 of 7: Salient-Pole Machine on an Infinite Bus

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

Notes on this paper

Paper format. National Exams, December 2018 — 16-Elec-B7 Power Systems Engineering, 3 hours, open book, any non-communicating calculator permitted. The paper prints seven questions, headed “Problem 1” to “Problem 7”; the rubric states that any five questions constitute a complete paper and that all questions are of equal value. Every one of the seven is solved in full here, because the set is a study resource rather than a three-hour sitting. Sub-part point values are quoted as printed; note that the printed splits total 25 points on Problems 1, 2, 4, 5 and 6, 20 points on Problem 3, and Problem 7 carries no printed split at all — under the “equal value” rubric each question is worth one fifth of the paper regardless.

Reference texts. M. E. El-Hawary, Electrical Power Systems: Design and Analysis, rev. ed., IEEE Press / Wiley — the reference text named in the Engineers Canada syllabus for this examination code, and the source of the “cantilever” transformer model used in Problem 3. Supporting texts: J. D. Glover, M. S. Sarma and T. J. Overbye, Power System Analysis and Design, 6th ed.; J. J. Grainger and W. D. Stevenson, Power System Analysis; S. J. Chapman, Electric Machinery Fundamentals, 5th ed. Canadian practice context (protection philosophy, grounding) follows CSA and IEEE C37 relaying practice as applied by Canadian transmission utilities.

Check — two places where the printed data are not internally consistent, and what was done about each.

(1) Problem 1. The pair A = 0.95∠2°, B = 195∠87.5° forces a shunt parameter with a large real part (Re C = 3.00 × 10⁻⁴ S). No real transmission line has a shunt conductance of that size — on a physical line C lands within about a degree of +90°. The consequence is a transmission efficiency of only 80.9 %, of which about 156 MW of the 236 MW loss is charged to that fictitious conductance. The arithmetic below is exactly what the printed constants give and is what the marker will be looking for; the physical caveat is stated so the low efficiency is not read as a slip.

(2) Problem 6. Table (2) lists the line negative-sequence reactance as 0.320 pu against a positive-sequence 0.30 pu. A transmission line is a static, symmetric element, so physically X₋ must equal X₊. The printed 0.320 is used throughout (the paper governs). Taking 0.30 instead moves the answer to Problem 6(b) from 1.4169∠145.58° pu to 1.4212∠146.04° pu — a change of 0.3 %, so neither reading changes any conclusion.

Question 2: Salient-Pole Machine on an Infinite Bus (25 marks)

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. Infinite-bus voltage $V = 1.00$ pu, direct-axis reactance $X_d = 0.95$ pu, armature resistance and saturation neglected. Part (a): $P = 1.15$ pu at $E = 1.15$ pu and $\delta = 32^\circ$. Part (b): $X_q = 0.6$ pu, with the four operating conditions of Table (1).

Find. The quadrature-axis reactance implied by the part (a) operating point, then the two missing entries in each row of Table (1).

Approach. Use the two-reaction (Blondel) expressions for the real and reactive power delivered by a salient-pole machine to an infinite bus. Part (a) inverts the real-power expression for $X_q$; part (b) applies the same pair of expressions row by row, which in one case reduces to a quadratic in $\cos\delta$ and in another to a single transcendental equation.

The governing pair, with armature resistance neglected and all quantities in per unit, is

$$P=\frac{EV}{X_d}\sin\delta+\frac{V^{2}(X_d-X_q)}{2X_dX_q}\sin 2\delta$$ $$Q=\frac{EV}{X_d}\cos\delta-V^{2}\!\left(\frac{\cos^{2}\delta}{X_d}+\frac{\sin^{2}\delta}{X_q}\right)$$

The first term of $P$ is the ordinary cylindrical-rotor term; the second is the reluctance or saliency term, which exists only because the rotor offers a lower reluctance along the direct axis than along the quadrature axis. It peaks at $\delta = 45^\circ$ and pulls the whole characteristic to the left of the 90° cylindrical peak.

The salient-pole power-angle characteristic resolved into its two components, drawn for the part (a) operating point (E = 1.15 pu, Xd = 0.95 pu, Xq = 0.4578 pu). The reluctance term contributes 0.509 pu of the 1.15 pu output at 32 degrees.

Part (a) — separate the two power terms at the stated operating point. The cylindrical term is fixed by the given data: $$\begin{aligned}\frac{EV}{X_d}\sin\delta &= \frac{(1.15)(1.00)}{0.95}\sin 32^\circ \\ &= 1.210526\times 0.529919 = 0.641481\ \text{pu}\end{aligned}$$ so the saliency term must supply the balance, $$\begin{aligned}\frac{V^{2}(X_d-X_q)}{2X_dX_q}\sin 64^\circ &= 1.15-0.641481 \\ &= 0.508519\ \text{pu}\end{aligned}$$
Solve for the quadrature-axis reactance. Dividing by $\sin 64^\circ = 0.898794$ isolates the saliency coefficient, $$k=\frac{X_d-X_q}{2X_dX_q}=\frac{0.508519}{0.898794}=0.565779$$ Rearranging $X_d - X_q = 2kX_dX_q$ gives $X_d = X_q(1 + 2kX_d)$, and therefore $$\boxed{\begin{aligned}X_q &= \frac{X_d}{1+2kX_d} = \frac{0.95}{1+2(0.565779)(0.95)} \\ &= 0.4578\ \text{pu}\end{aligned}}$$ Substituting back reproduces $P = 1.1500$ pu exactly. The ratio $X_q/X_d = 0.48$ is on the low side for a hydro machine (0.6–0.7 is more usual), which is simply what this operating point implies — a strongly salient rotor is exactly what makes the reluctance term worth 0.509 pu here.
Part (b) — fix the saliency coefficient for the table. With $X_q = 0.6$ pu the two constants that recur in every row are $$\begin{aligned}\frac{V^{2}(X_d-X_q)}{2X_dX_q} &= \frac{0.95-0.6}{2(0.95)(0.6)} \\ &= 0.3070175 \\ \frac{1}{X_d} &=1.052632,\ \ \frac{1}{X_q}=1.666667\end{aligned}$$
Condition A ($Q_2 = 0$, $E = 1.10$) — the zero-reactive row is a quadratic in $\cos\delta$. Writing $c = \cos\delta$ and using $\sin^{2}\delta = 1 - c^{2}$, the reactive-power expression becomes $$\begin{aligned}\frac{EV}{X_d}c-\left[\frac{c^{2}}{X_d}+\frac{1-c^{2}}{X_q}\right] &=0 \\ \Longrightarrow\quad \left(\frac{1}{X_q}-\frac{1}{X_d}\right)c^{2}+\frac{EV}{X_d}\,c-\frac{1}{X_q} &=0\end{aligned}$$ Substituting the numbers, $$0.6140351\,c^{2}+1.1578947\,c-1.6666667=0$$ whose roots are $c = 0.955370$ and $c = -2.841084$. The second is outside $[-1,1]$ and is discarded, leaving $$\boxed{\delta_A=\arccos(0.955370)=17.18^\circ}$$ Substituting into the real-power expression, $$\boxed{P_A=1.157895\sin 17.18^\circ+0.307018\sin 34.36^\circ=0.5154\ \text{pu}}$$
Condition B ($E = 1.15$, $\delta = 38^\circ$) — both unknowns follow by direct substitution. For the real power, $$\begin{aligned}P_B &= \frac{(1.15)(1.00)}{0.95}\sin 38^\circ+0.307018\sin 76^\circ \\ &= 0.745274+0.297898\end{aligned}$$ $$\boxed{P_B=1.0432\ \text{pu}}$$ and for the reactive power, $$\begin{aligned}Q_B &= 1.210526\cos 38^\circ \\ &\quad -\left[1.052632\cos^{2}38^\circ+1.666667\sin^{2}38^\circ\right] \\ &= 0.953908-1.285375\end{aligned}$$ $$\boxed{Q_{2,B}=-0.3315\ \text{pu}}$$ The negative sign is the answer, not a slip: at this excitation the machine is under-excited and absorbs reactive power from the bus.
Condition C ($P = 1.9$, $\delta = 42^\circ$) — invert the real-power expression for $E$. The saliency term is independent of $E$, so it can be moved to the left-hand side and the remainder solved directly: $$\begin{aligned}E &= \frac{X_d\left[P-k_2\sin 2\delta\right]}{V\sin\delta} \\ &= \frac{0.95\left[1.9-0.307018\sin 84^\circ\right]}{\sin 42^\circ} \\ &= \frac{0.95(1.594664)}{0.669131}\end{aligned}$$ $$\boxed{E_C=2.2640\ \text{pu}}$$ With $E$ known the reactive power follows from the same expression as before, $$\begin{aligned}Q_C &= \frac{(2.2640)(1.00)}{0.95}\cos 42^\circ \\ &\quad -\left[1.052632\cos^{2}42^\circ+1.666667\sin^{2}42^\circ\right] \\ &= 1.771054-1.327557\end{aligned}$$ $$\boxed{Q_{2,C}=+0.4435\ \text{pu}}$$ An excitation of 2.26 pu is high but consistent: delivering 1.9 pu of real power at only 42° requires it, and the machine is now over-excited and exporting reactive power.
Condition D ($P = 1.2$, $E = 1.18$) — one transcendental equation in $\delta$. Here neither term can be isolated, so the real-power expression is solved numerically: $$1.242105\sin\delta+0.307018\sin 2\delta=1.2$$ Bisection between 1° and 80° converges to $$\boxed{\delta_D=45.98^\circ}$$ (the left-hand side evaluates to 1.200326 at 46.00° and 1.185319 at 45.00°, so the root is tightly bracketed). Substituting that angle into the reactive-power expression, $$\begin{aligned}Q_D &= 1.242105\cos 45.98^\circ \\ &\quad -\left[1.052632\cos^{2}45.98^\circ+1.666667\sin^{2}45.98^\circ\right] \\ &= 0.863185-1.370126\end{aligned}$$ $$\boxed{Q_{2,D}=-0.5069\ \text{pu}}$$

Read down the completed table and the machine tells a coherent story: it moves from mildly under-excited at condition A (where the reactive output is held at zero by construction), through progressively deeper reactive absorption at B and D, to strong reactive export at C where the excitation is nearly doubled. That is precisely the locus a synchronous condenser traces out as its field is varied at fixed real load, and it is why an over-excited machine is described as behaving like a capacitor seen from the bus.

Condition	$P$ (pu)	$Q_2$ (pu)	$E$ (pu)	$\delta$
Part (a) result	$X_q = 0.4578$ pu (with $X_d = 0.95$ pu, $X_q/X_d = 0.48$)
A	0.5154	0.0 (given)	1.10 (given)	17.18°
B	1.0432	−0.3315	1.15 (given)	38° (given)
C	1.9 (given)	+0.4435	2.2640	42° (given)
D	1.2 (given)	−0.5069	1.18 (given)	45.98°

Topic. Two-reaction (Blondel) theory of the salient-pole synchronous machine operating against an infinite bus.

Key relations. $P = (EV/X_d)\sin\delta + [V^{2}(X_d-X_q)/(2X_dX_q)]\sin 2\delta$; $Q = (EV/X_d)\cos\delta - V^{2}[\cos^{2}\delta/X_d + \sin^{2}\delta/X_q]$; the $Q = 0$ condition collapses to $(1/X_q - 1/X_d)\cos^{2}\delta + (EV/X_d)\cos\delta - 1/X_q = 0$.

Why this works. A salient rotor presents a different magnetic path to flux along the pole axis than to flux between the poles, so the armature reaction cannot be described by one synchronous reactance. Blondel's resolution splits the armature current into a direct-axis component seeing $X_d$ and a quadrature-axis component seeing $X_q$, and recombining the two produces the extra $\sin 2\delta$ term. That term carries no $E$, which is the key structural fact exploited twice above: it survives even at zero excitation (a reluctance machine develops torque with no field at all), and because it is independent of $E$ it can always be moved to the other side of the equation when $E$ is the unknown, as in condition C.

Common pitfall. Using the cylindrical-rotor reactive expression $Q = (EV\cos\delta - V^{2})/X_d$ on a salient machine is not a small refinement — on strongly salient rotors it gets the sign wrong. At condition D it would return roughly $+0.06$ pu against the true $-0.507$ pu, turning an absorbing machine into an exporting one. The second trap is discarding the wrong root of the condition-A quadratic; only the root inside $[-1,1]$ is a physical cosine, and the other is always large and negative.

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.

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