Question 4 of 6: Cylindrical-rotor synchronous generator — excitation, limits and construction

Question 4: Cylindrical-rotor synchronous generator — excitation, limits and construction (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.

Given. A cylindrical-rotor (round-rotor) synchronous generator rated 100 MVA, 20 kV, with saturated synchronous reactance $X_s=1.5$ pu and armature resistance neglected. Taking its own rating as the base, the terminal voltage is held at $V=1.0\angle 0^\circ$ pu and the loading in parts (a)–(d) is half of rated apparent power, $S=0.5$ pu, at 0.8 power factor lagging.

Find. (a) the internal (excitation) voltage magnitude, in per unit and in volts; (b) the torque angle; (c) the steady-state power limit at that excitation; (d) which quantities change when the excitation is raised 10 % at constant real power, with a phasor diagram; (e) a sketch and description of the terminal short-circuit current; and (f) an annotated sketch of a four-pole salient-pole rotor.

Approach. Parts (a)–(d) all follow from one phasor equation, $E=V+jX_sI$, and its power form $P=(EV/X_s)\sin\delta$. Parts (e) and (f) are descriptive and are answered with an annotated figure and the physics behind each feature.

1. Part (a) — write the armature current for the stated loading. Half rated load at 0.8 power factor lagging, with the terminal voltage as reference, gives $$I=\frac{S}{V}\angle-\cos^{-1}(0.8)=0.5\angle-36.87^\circ=0.4-j0.3\ \text{pu}$$
2. Apply the round-rotor phasor equation. With armature resistance neglected the excitation voltage behind the synchronous reactance is $$E=V+jX_sI=1.0+j1.5\,(0.4-j0.3)=1.0+j0.6+0.45=1.45+j0.60\ \text{pu}$$ $$\boxed{E=1.5692\ \text{pu}\ \angle\,22.48^\circ}$$ Converting to volts on the 20 kV base, the excitation voltage corresponds to $1.5692\times 20=31.38$ kV line-to-line, or $31.38/\sqrt{3}=18.12$ kV per phase. Note the physical reading: to carry only half rated MVA at a lagging power factor the machine must be over-excited by $57\ \%$, which is what a synchronous reactance of 1.5 pu implies.
3. Part (b) — read the torque angle off the same phasor. The torque (power) angle is the angle by which $E$ leads the terminal voltage: $$\delta=\tan^{-1}\!\frac{0.60}{1.45}=\boxed{22.48^\circ}$$ As a check, $P=(EV/X_s)\sin\delta=(1.5692/1.5)\sin 22.48^\circ=0.400$ pu $=40$ MW, which is exactly $0.5\times 0.8$.
4. Part (c) — take the steady-state limit at fixed excitation. The power-angle characteristic $P=(EV/X_s)\sin\delta$ peaks at $\delta=90^\circ$, so $$P_{max}=\frac{EV}{X_s}=\frac{1.5692\times 1.0}{1.5}=1.0462\ \text{pu}$$ $$\boxed{P_{max}=1.0462\ \text{pu}=104.6\ \text{MW}=1.046\times 10^{8}\ \text{W}}$$ The machine is operating at $0.400/1.0462=38\ \%$ of its steady-state limit, a comfortable margin; in practice the excitation would be raised before approaching $\delta=90^\circ$, because operation beyond that point is statically unstable.
5. Part (d) — raise the excitation and hold the real power. The new excitation is $E'=1.1\times 1.5692=1.7262$ pu. Because $P=(E V/X_s)\sin\delta$ is unchanged at 0.400 pu and both $V$ and $X_s$ are unchanged, the product $E\sin\delta$ must be constant — the tip of the phasor $E$ slides along a horizontal line. Hence $$\sin\delta'=\frac{P X_s}{E'V}=\frac{0.400\times 1.5}{1.7262}=0.34759 \quad\Longrightarrow\quad \delta'=20.34^\circ$$ so the torque angle falls from $22.48^\circ$ to $20.34^\circ$.
6. Evaluate what else moves. The reactive output follows from $Q=(EV\cos\delta-V^{2})/X_s$: $$Q'=\frac{1.7262\cos 20.34^\circ-1}{1.5}=\frac{1.6185-1}{1.5}=0.4124\ \text{pu} \quad\text{(was }0.300\text{ pu)}$$ and the armature current is $$I'=\frac{E'\angle\delta'-V}{jX_s}=\frac{(1.6185+j0.600)-1}{j1.5}=0.400-j0.4124\ \text{pu}$$ $$\boxed{|I'|=0.5745\ \text{pu at }0.6963\ \text{pf lagging}\quad(\text{was }0.500\ \text{pu at }0.8)}$$ So: reactive output rises, armature current rises, the power factor becomes more lagging, the torque angle falls, and the steady-state limit rises to $E'V/X_s=1.1508$ pu. Unchanged are the real power (0.400 pu, fixed by the prime mover), the terminal voltage, the frequency and the shaft speed. In one sentence — over-exciting a machine on an infinite bus changes only its reactive output, and pays for it in armature current and stator heating.

Part (e) — terminal three-phase short circuit. When a solid three-phase fault is applied at the terminals of an unloaded machine, the armature current is not a constant sinusoid: its envelope decays through three distinct regimes, and each phase additionally carries a unidirectional offset. The waveform is sketched below.

The significant features, in the order in which they appear, are these. The subtransient period occupies roughly the first one to three cycles. Currents induced in the amortisseur (damper) windings and in the rotor iron oppose the change of flux and hold the effective reactance down to $X_d''$, typically 0.15–0.25 pu, so this is the largest symmetrical current the machine ever produces; its envelope decays with the subtransient time constant $T_d''$, of the order of 0.03 s. As the damper currents die away the transient period takes over, lasting perhaps half a second to two seconds: the flux is now held only by the field winding, the effective reactance rises to $X_d'$ (typically 0.2–0.4 pu) and the envelope decays with $T_d'$. Finally the field current returns to its pre-fault value and the sustained (steady-state) period is reached, in which the current is limited by the full synchronous reactance $X_d$ — here 1.5 pu, so the sustained fault current is less than rated current and would not by itself operate an overcurrent relay.

Superimposed on all three is the d.c. offset. Because the armature circuit is almost purely inductive the current cannot change instantaneously, so each phase acquires a unidirectional component sized to force $i(0^{+})=0$; its magnitude depends on the point on the voltage wave at which the fault strikes, is zero in one phase and maximum in another, and it decays with the armature time constant $T_a$ (of the order of 0.1 s). The offset is what makes the first peak asymmetrical, and it is the reason a circuit breaker's making capacity is specified in terms of an asymmetrical peak while its breaking capacity is quoted as a symmetrical r.m.s. value. Envelope extrapolation back to $t=0$ — drawing the transient envelope back through the subtransient region — is the standard graphical test used to measure $X_d''$, $X_d'$ and $X_d$ from an oscillogram.

Part (f) — four-pole salient-pole rotor. The sketch below shows the required construction. The four poles are spaced $90^\circ$ apart mechanically (which is $180^\circ$ electrically, since electrical angle equals mechanical angle multiplied by the number of pole pairs).

The direct axis (d-axis) passes through the centre of a pole, along the path of the main field flux; it is the low-reluctance direction, which is why $X_d > X_q$. The quadrature axis (q-axis) lies midway between adjacent poles, $90^\circ$ electrical — and therefore $45^\circ$ mechanical on a four-pole machine — from the d-axis; the flux path there crosses the wide interpolar air gap, so the reluctance is high. The field winding is a concentrated d.c. coil wound around each pole shank, all four coils in series and fed from the exciter through slip rings or a brushless rotating rectifier; the polarities alternate N, S, N, S around the periphery. The amortisseur (damper) winding consists of copper or brass bars driven axially through the pole faces and short-circuited at both ends by end rings, forming a partial squirrel cage; it damps rotor oscillations by acting as an induction machine whenever the rotor slips relative to the synchronous field, provides a starting torque for a synchronous motor, and carries the induced current that produces the subtransient reactance $X_d''$ of part (e). Salient-pole construction is used on low-speed hydraulic units — the great majority of British Columbia's installed capacity — because a large pole number is needed to make 60 Hz at turbine speed.