Question 2 of 7: Reactive Power Sources and a Salient-Pole Operating Table

Question 2: Reactive Power Sources and a Salient-Pole Operating Table (20 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.

Part (a) — why reactive power must be supplied locally, and from where. Reactive power is not consumed in the thermodynamic sense; it oscillates between the magnetic and electric energy stores of the network twice per cycle. What matters operationally is that it must nevertheless be supplied, because the series inductance of every line, cable and transformer absorbs $I^2X$ vars in proportion to the square of the current it carries, and because the load itself — induction motors above all — needs magnetising vars to establish its air-gap flux.

The first reason it must be supplied locally is voltage. On a predominantly inductive network the voltage drop along a branch is governed almost entirely by the reactive flow, $\Delta V \approx (RP + XQ)/V \approx XQ/V$ when $X \gg R$. Reactive power therefore cannot be transported far without collapsing the very voltage it is trying to support: shipping vars over a long line consumes more vars than it delivers. The second reason is loss. Reactive current occupies conductor cross-section and produces real $I^2R$ heating without delivering any useful energy, so a poor system power factor inflates both losses and the thermal loading of every element in the path. The third, and the one that ends in blackouts, is voltage stability: as loading rises the network approaches the nose of its P–V curve, where the incremental vars needed to hold voltage grow without bound. A system short of local reactive reserve slides down the lower branch of that curve and suffers voltage collapse, which is what took down the eastern interconnection in August 2003.

The major sources are, in rough order of controllability: synchronous generators, which can both generate and absorb vars by adjusting excitation within the limits of the capability curve, and which are the system's principal continuously-controlled reserve; shunt capacitor banks, switched in blocks at distribution and transmission buses — cheap, but their output falls with the square of voltage exactly when it is most needed; shunt reactors, which absorb vars on lightly loaded EHV lines and cables to control the Ferranti rise; synchronous condensers, unloaded synchronous machines run purely for var control, now returning to service for their inertia and short-circuit contribution; static var compensators and STATCOMs, thyristor- and IGBT-based devices giving continuous, sub-cycle control; series capacitors, which reduce the effective $X$ and so cut the var demand of a long line at source; on-load tap-changing transformers, which redistribute rather than create vars; and the lines and cables themselves, whose distributed shunt capacitance generates vars at light load.

Given. A salient-pole machine on an infinite bus.

Find. The six missing table entries, using the two-reaction (Blondel) model.

Approach. Write the two-reaction expressions for $P$ and $Q$ at the machine terminals, then treat each row as the algebra it happens to be: linear in $E$ for condition A, a direct evaluation for condition B, and a quadratic in $\cos\delta$ for condition C.

1. State the two-reaction power expressions. For a salient-pole machine delivering into a bus of voltage $V$ at internal angle $\delta$, $$P = \frac{EV}{x_d}\sin\delta + \frac{V^2(x_d - x_q)}{2x_dx_q}\sin 2\delta, \qquad Q = \frac{EV}{x_d}\cos\delta - V^2\!\left(\frac{\cos^2\delta}{x_d} + \frac{\sin^2\delta}{x_q}\right)$$ The second term of $P$ is the reluctance or saliency power — torque produced by the rotor's preference for aligning its low-reluctance direct axis with the stator field, present even at zero excitation. With the given data its coefficient is $$\frac{V^2(x_d - x_q)}{2x_dx_q} = \frac{1.0(0.6 - 0.3)}{2(0.6)(0.3)} = 0.8333\ \text{pu}$$ which is not a small correction — it is comparable with the fundamental term, so the saliency is load-bearing throughout this question.
2. Condition A — solve for E, which enters linearly. At $\delta = 45^\circ$, $\sin 2\delta = 1$ exactly, so $$2.05 = \frac{E(1.0)}{0.6}\sin 45^\circ + 0.8333 \quad\Longrightarrow\quad E = \frac{(2.05 - 0.8333)(0.6)}{\sin 45^\circ}$$ $$\boxed{E_A = 1.0324\ \text{pu}}$$ The reluctance term alone carries 0.833 of the 2.05 pu, i.e. 41 % of the output.
3. Condition A — then read off Q. Substituting into the reactive expression with $\cos^2 45^\circ = \sin^2 45^\circ = 0.5$: $$Q_A = \frac{1.0324}{0.6}(0.7071) - \left(\frac{0.5}{0.6} + \frac{0.5}{0.3}\right) = 1.2166 - 2.5000$$ $$\boxed{Q_A = -1.2833\ \text{pu}}$$ The machine is strongly under-excited: it is delivering 2.05 pu of real power while absorbing 1.28 pu of reactive power from the bus. That is consistent with an internal voltage of only 1.03 pu working through a 0.6 pu direct-axis reactance — the excitation is simply too low to cover the machine's own reactive demand.
4. Condition B — both unknowns are a direct evaluation. With $E = 1.27$ and $\delta = 47^\circ$ nothing is implicit: $$P_B = \frac{1.27}{0.6}\sin 47^\circ + 0.8333\sin 94^\circ = 1.5480 + 0.8313$$ $$\boxed{P_B = 2.3793\ \text{pu}}$$ $$Q_B = \frac{1.27}{0.6}\cos 47^\circ - \left(\frac{\cos^2 47^\circ}{0.6} + \frac{\sin^2 47^\circ}{0.3}\right) = 1.4436 - 2.5581$$ $$\boxed{Q_B = -1.1146\ \text{pu}}$$ Note that $\sin 2\delta = \sin 94^\circ$ is already past its peak: the reluctance contribution has begun to fall while the fundamental term is still rising.
5. Condition C — the zero-reactive condition is a quadratic in cosδ. Put $c = \cos\delta$ and use $\sin^2\delta = 1 - c^2$ in the reactive expression. Setting $Q = 0$ and collecting terms, $$c^2\!\left(\frac{1}{x_d} - \frac{1}{x_q}\right) - \frac{E}{x_d}c + \frac{1}{x_q} = 0$$ With $x_d = 0.6$, $x_q = 0.3$ and $E = 1.35$ the coefficient of $c^2$ is $1.6667 - 3.3333 = -1.6667$, so $$-1.6667c^2 - 2.25c + 3.3333 = 0 \quad\Longleftrightarrow\quad c^2 + 1.35c - 2 = 0$$ $$c = \frac{-1.35 \pm \sqrt{1.8225 + 8}}{2} = 0.89204 \quad\text{or}\quad -2.24204$$ The second root is outside $[-1, 1]$ and is discarded as unphysical — a cosine cannot exceed unity in magnitude. Hence $$\boxed{\delta_C = \arccos(0.89204) = 26.87^\circ}$$
6. Condition C — then the real power at that angle. $$P_C = \frac{1.35}{0.6}\sin 26.87^\circ + 0.8333\sin 53.74^\circ = 1.0166 + 0.6718$$ $$\boxed{P_C = 1.6888\ \text{pu}}$$ Substituting $E = 1.35$ and $\delta = 26.87^\circ$ back into the reactive expression returns $Q = 0$ to twelve figures, which closes the row.
7. Confirm all three points are on the stable side of the curve. The synchronising power coefficient is $$\frac{dP}{d\delta} = \frac{EV}{x_d}\cos\delta + \frac{V^2(x_d - x_q)}{x_dx_q}\cos 2\delta$$ which evaluates to $+1.217$, $+1.327$ and $+2.993$ pu/rad for conditions A, B and C. All are positive, so every row is a stable steady-state operating point — worth stating, because on a strongly salient machine the peak of the power-angle curve sits appreciably below $90^\circ$ and a large printed $\delta$ is not automatically admissible.