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

Problem 2: Salient-Pole Machine on an Infinite Bus (5 + 15 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

The distinction is geometric and its consequences are entirely magnetic. A cylindrical-rotor (round-rotor) machine has a uniform air gap: the field winding is distributed in slots cut into a solid forged cylinder, so the reluctance seen by the armature magnetomotive force is the same whatever its position relative to the field axis. One reactance therefore describes the machine, the synchronous reactance $X_s$, and it is large — typically 1.0 to 2.0 per unit — because the small uniform gap is efficient at linking flux. These machines are the two- and four-pole turbo-generators driven by steam or gas turbines at 3600 or 1800 r/min, where the mechanical strength of a solid rotor is essential.

A salient-pole machine has projecting poles and hence a strongly non-uniform gap: the gap is small along the pole (the direct axis) and large in the interpolar space (the quadrature axis). Two reactances are needed, and because the direct axis presents the lower reluctance, $X_d > X_q$ always — the ratio $X_d/X_q$ is typically 1.5 to 2.5 for a hydro generator, and the machine in this problem has $12/9 = 1.33$. Salient-pole machines are the many-pole, low-speed units driven by hydraulic turbines and diesel engines, which is the dominant generation type in British Columbia, Quebec, Manitoba and Newfoundland.

The consequence for power transfer is the appearance of a second term. A cylindrical-rotor machine delivers $P=(EV/X_s)\sin\delta$, a pure sinusoid whose maximum $EV/X_s$ occurs at a power angle of exactly $90^{\circ}$ and which vanishes when the field is removed. The salient-pole machine adds a reluctance (saliency) term: $$P=\frac{EV}{X_d}\sin\delta+\frac{V^{2}}{2}\left(\frac{1}{X_q}-\frac{1}{X_d}\right)\sin 2\delta$$ This second term exists because the rotor prefers to align its low-reluctance axis with the stator field, so it produces torque even with the field current zero — the principle of the reluctance motor. Three practical consequences follow. The maximum power is larger than the cylindrical-rotor value for the same $E$ and $X_d$. The angle at which it occurs is less than $90^{\circ}$, because the double-frequency term peaks at $45^{\circ}$ and drags the composite peak back. And the synchronising torque $dP/d\delta$ near the origin is higher, so the salient-pole machine is inherently stiffer against small disturbances.

Part (b) — Maximum power and the angle at which it occurs

Given.

Find. The power angle $\delta$ at which the active power delivered to the infinite bus is a maximum, and the value of that maximum power.

Approach. Write the two-reaction power-angle expression on a per-phase basis, differentiate with respect to $\delta$, and solve the resulting quadratic in $\cos\delta$ for the physically admissible root.

1. Put the bus voltage on a per-phase basis and evaluate the two coefficients. The excitation is already line-to-neutral, so only the bus voltage needs converting: $$V=\frac{34\,500}{\sqrt{3}}=19{,}918.6\ \text{V (line-to-neutral)}$$ $$\begin{aligned} k_1&=\frac{EV}{X_d}=\frac{(30\,000)(19{,}918.6)}{12}=49.7965\ \text{MW/phase}\\ k_2&=\frac{V^{2}}{2}\left(\frac{1}{X_q}-\frac{1}{X_d}\right) =\frac{(19{,}918.6)^{2}}{2}\left(\frac{1}{9}-\frac{1}{12}\right)=5.5104\ \text{MW/phase} \end{aligned}$$ The reluctance coefficient is 11.1 per cent of the fundamental coefficient, so it is a real contributor rather than a rounding correction.
2. Differentiate the power-angle expression and set the derivative to zero. With $P=k_1\sin\delta+k_2\sin 2\delta$, $$\frac{dP}{d\delta}=k_1\cos\delta+2k_2\cos 2\delta=0$$ Using the identity $\cos 2\delta = 2\cos^{2}\delta - 1$ and writing $c=\cos\delta$, the stationarity condition becomes a quadratic: $$4k_2c^{2}+k_1c-2k_2=0$$
3. Solve the quadratic and discard the inadmissible root. Substituting the two coefficients in MW, $$c=\frac{-k_1\pm\sqrt{k_1^{2}+32k_2^{2}}}{8k_2} =\frac{-49.7965\pm 58.7483}{44.0833}$$ which gives $c=0.20307$ and $c=-2.4623$. The second root lies outside $[-1,1]$ and is rejected; it is the artefact of turning a trigonometric equation into an algebraic one. The admissible root gives $$\boxed{\delta_{\max}=\cos^{-1}(0.20307)=78.284^{\circ}}$$ As predicted in part (a), the peak has been pulled back from $90^{\circ}$ by the saliency.
4. Evaluate the maximum power. Substituting the angle into the two-term expression, per phase, $$\begin{aligned} P_{\max}&=k_1\sin78.284^{\circ}+k_2\sin(2\times78.284^{\circ})\\ &=48.7590+2.1913=50.9503\ \text{MW/phase} \end{aligned}$$ and for the three-phase machine $$\boxed{P_{\max}=3(50.9503)=152.851\ \text{MW}}$$
5. Check the result against the round-rotor value. Had the machine been cylindrical with the same $X_d$, the maximum would have been $3k_1=149.39$ MW at $\delta=90^{\circ}$. The saliency therefore buys 2.3 per cent more peak power and moves the peak 11.7° earlier — the qualitative statement of part (a) reproduced numerically, which is the check that the two halves of the answer agree.

Final results.