Question 2 of 7: Salient-Pole versus Cylindrical Rotor, and Two Operating Points

Question 2 (Problem 2): Salient-Pole versus Cylindrical Rotor, and Two Operating Points (5 + 10 + 10 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) — salient-pole versus cylindrical-rotor machines. A cylindrical-rotor (turbo) machine has a uniform air gap, so the reluctance seen by the armature magnetomotive force is the same wherever the rotor happens to be standing. One reactance therefore describes it completely: the direct- and quadrature-axis synchronous reactances are equal, $x_d = x_q = x_s$, and the steady-state power transfer to a bus of voltage $V$ is the familiar single sine term $P = E_{af}V\sin\delta / x_d$, whose maximum $E_{af}V/x_d$ occurs at exactly 90° of rotor angle.

A salient-pole machine, the low-speed hydro construction, presents a much larger air gap between the poles than under them. The quadrature axis is therefore the high-reluctance axis and $x_q$ is typically 0.55 to 0.70 of $x_d$. Blondel’s two-reaction theory resolves the armature current into direct- and quadrature-axis components and yields a second, excitation-independent term: $$\begin{aligned}&P \\ &=\dfrac{E_{af}V}{x_{d}}\sin\delta +\dfrac{V^{2}}{2}\left(\dfrac{1}{x_{q}}-\dfrac{1}{x_{d}}\right)\sin 2\delta\end{aligned}$$ The second term is the reluctance power: it exists even with the field unexcited, it peaks at 45° of rotor angle, and it makes the total maximum power larger than the cylindrical-rotor value while moving the peak to an angle below 90° (typically 70–80°). Practically, the salient-pole machine is therefore stiffer at small angles and reaches its stability limit earlier in angle but at a higher power.

Given. A 125 MVA, 11 kV, 50 Hz synchronous generator with x_d = 1.33 p.u. reaches rated open-circuit voltage at 325 A of field current, and feeds a network of 11 kV equivalent voltage through an equivalent reactance of 0.17 p.u. on the machine base while delivering 110 MW.

Find. For each of two operating conditions — unity power factor at the network equivalent bus, and rated terminal voltage — the excitation voltage, the field current, and the remaining terminal quantities.

Approach. Treat the machine, its synchronous reactance and the network reactance as one phasor chain, anchoring it at whichever bus the condition specifies and walking to the others.

1. Part (b) — put the loading on the machine base. Per unit on the machine rating, $$P=\dfrac{110}{125}=0.8800\ \text{p.u.}$$ Unity power factor at the network equivalent voltage means the current is in phase with $V_{EQ}=1.0\angle 0°$, so $$I_{a}=\dfrac{P}{V_{EQ}}=0.8800\angle 0°\ \text{p.u.}$$
2. Walk the current back through the network reactance to the terminals. The terminal voltage sits one network reactance upstream of the equivalent bus: $$\begin{aligned}V_{t}&=V_{EQ}+jx_{eq}I_{a}=1.0+j(0.17)(0.8800) \\ &=1.0000 + j0.1496 \\ &=\boxed{1.0111\angle 8.5084^\circ\ \text{p.u.}}\end{aligned}$$ which is 11.122 kV line to line.
3. Continue through the synchronous reactance to the excitation voltage. Both reactances are in cascade with the same current, so $$\begin{aligned}E_{af}&=V_{EQ}+j(x_{eq}+x_{d})I_{a} \\ &=1.0+j(1.50)(0.8800)=1.0000 + j1.3200\end{aligned}$$ $$\boxed{E_{af}=1.6560\angle 52.8533^\circ\ \text{p.u.}}$$
4. Convert the excitation voltage to a field current, then read the terminal power factor. On the air-gap line the open-circuit voltage is proportional to field current, and 325 A produces 1.0 p.u., so $$\begin{aligned}I_{f}&=|E_{af}|\,I_{f0}=(1.6560)(325) \\ &=\boxed{538.2\ \text{A}}\end{aligned}$$ The machine terminal power factor is not unity, because unity was imposed one reactance away. The current lags $V_t$ by the terminal-voltage angle: $$\begin{aligned}\text{p.f.}_{t}&=\cos(8.5084°) \\ &=0.9890\ \text{lagging}\end{aligned}$$
5. Part (c) — fix the terminal voltage instead and find the new angle. Now $|V_t| = 1.0$ p.u. and the same 110 MW crosses $x_{eq}$ to the equivalent bus. The power transfer across a pure reactance between two known magnitudes gives the angle directly: $$\begin{aligned}P&=\dfrac{|V_{t}||V_{EQ}|}{x_{eq}}\sin\delta_{t} \;\Rightarrow\;\sin\delta_{t} \\ &=\dfrac{(0.8800)(0.17)}{(1.0)(1.0)} \;\Rightarrow\;\delta_{t} \\ &=8.6037°\end{aligned}$$ so $V_{t}=0.988747 + j0.149600$ p.u.
6. Recover the armature current from the voltage drop it must produce. The current is whatever flows through $x_{eq}$ under that voltage difference: $$\begin{aligned}I_{a}&=\dfrac{V_{t}-V_{EQ}}{jx_{eq}} \\ &=\dfrac{-0.011253 + j0.149600}{j0.17} \\ &=0.880000 + j0.066196 \\ &=\boxed{0.8825\angle 4.3019^\circ\ \text{p.u.}}\end{aligned}$$ On the 125 MVA, 11 kV base the rated current is 6560.8 A, so this is 5789.8 A.
7. Excitation voltage, field current and power factor for this condition. Adding the drop across the synchronous reactance, $$\begin{aligned}E_{af}&=V_{t}+jx_{d}I_{a}=0.900705 + j1.320000\; \\ &\Rightarrow\;\boxed{E_{af}=1.5980\angle 55.6922^\circ\ \text{p.u.}}\end{aligned}$$ $$\begin{gathered}I_{f}=(1.5980)(325)=519.4\ \text{A}, \\ \text{p.f.}_{t}=\cos(8.6037°-4.3019°)=0.9972\ \text{lagging}\end{gathered}$$ Holding the terminal voltage down to 1.0 p.u. instead of letting it float to 1.0111 p.u. costs about 18.8 A of field current, which is the practical point of the comparison.