22-Elec-B7 Power Systems Engineering · December 2015
Question 2 of 7: Excitation State and the Two-Reaction Salient-Pole Generator
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2015 — 07-Elec-B7 Power Systems Engineering. Open-book, 3 hours. Seven problems are printed; any five constitute a complete paper and all questions are of equal value (25 points each). All seven are solved in full below, because the set is a study resource rather than a timed sitting.
Reference texts. The syllabus for this code follows M. E. El-Hawary, Electrical Power Systems: Design and Analysis (IEEE Press) — the notation used throughout this paper (the cantilever transformer model, the ABCD two-port, the two-reaction salient-pole construction) is his. Corroborating references: J. D. Glover, M. S. Sarma and T. J. Overbye, Power System Analysis and Design, 6th ed.; J. Grainger and W. Stevenson, Power System Analysis; and, for the machine chapters, S. J. Chapman, Electric Machinery Fundamentals, 5th ed. Canadian practice references where grounding and protection are discussed: CSA C22.1 Canadian Electrical Code, Part I and CSA C22.3 No. 1 for overhead systems.
Check: two points read off the printed figures. (i) Figure (1) of Problem 4 prints the line admittance as y = −j10 pu, i.e. a series reactance of j0.10 pu. The negative sign is the physically correct one for an inductive line and is used here. (ii) In Figure (3-a) of Problem 6, transformer T2 carries its delta on the generator side and T1 carries its delta on the motor side; both line-side windings are wye with a solidly earthed neutral. That reading governs the zero-sequence network built in Problem 6(a).
Problem 2: Excitation State and the Two-Reaction Salient-Pole Generator (25 points)
Part (a) — over-excitation, under-excitation and var control
The excitation state of a synchronous machine is a statement about the
magnitude of its internal generated voltage $E$ relative to the terminal voltage
$V_t$ it is tied to. A machine is over-excited when its field current is
raised far enough that $E$ exceeds $V_t$, and under-excited when the
field is weakened so that $E$ falls below $V_t$. Because the machine is
connected to the system through a mostly reactive impedance, that comparison
decides which way reactive power flows.
The mechanism is easiest to see on a round-rotor machine, where the
reactive power delivered to the terminals is
$$Q=\frac{EV_t\cos\delta - V_t^{2}}{X_d}$$
The numerator changes sign exactly when $E\cos\delta$ passes through $V_t$.
Over-excited, $E\cos\delta > V_t$, the numerator is positive and the machine
exports reactive power: it looks like a capacitor to the network and
its armature current lags the terminal voltage. Under-excited, $E\cos\delta <
V_t$, $Q$ is negative and the machine absorbs reactive power, looking
inductive, with a leading armature current. At the boundary the machine runs at
unity power factor and exchanges no vars at all.
This is why a synchronous machine is the most flexible reactive-power
resource on a power system: the field current is a continuously adjustable knob
that moves the operating point smoothly from var absorption to var production
without switching anything. To make the machine appear as a source of
reactive power, the operator simply raises the field current until $E$ is
comfortably above $V_t$ — on an automatic voltage regulator this happens by
itself whenever the terminal-voltage set point is raised above the system
voltage. A machine run with no mechanical power at all, purely for this purpose,
is a synchronous condenser, and several have been recommissioned in
Canada in recent years to provide both vars and inertia on networks with high
inverter-based generation. The limits on how far the field may be pushed are the
rotor field-winding heating limit at the over-excited end and the stator
end-region heating limit, together with the steady-state stability limit, at the
under-excited end; both appear as boundaries of the machine capability
diagram.
Parts (b) and (c) — two-reaction calculation of E
Given. A salient-pole generator with $x_d = 1.0$ pu and $x_q = 0.6$ pu, armature resistance neglected, running at rated terminal voltage $V_t = 1.0$ pu and rated kVA so that $I_a = 1.0$ pu, first at 0.8 power factor lagging and then at 0.8 leading.
Find. The generated (excitation) voltage $E$ in per unit for each of the two power factors.
Figure 2.1 — two-reaction phasor construction for part (b), 0.8 pf lagging. Adding j x(q) I(a) to the terminal voltage locates the q-axis; the armature current is then resolved along the two axes and E is built up on the q-axis alone.
Approach. A salient-pole machine has two different
reactances, so the single-reactance phasor $E = V_t + jx_dI_a$ is not available.
Blondel’s two-reaction method fixes this: add $jx_qI_a$ to the terminal
voltage first, which locates the q-axis and hence the rotor angle $\delta$;
resolve $I_a$ into its direct- and quadrature-axis components; then build $E$
along the q-axis using $x_d$ for the direct-axis component alone.
Part (b) — locate the rotor angle at 0.8 pf lagging.
With the terminal voltage as reference and the current lagging by
$\phi=\cos^{-1}0.8=36.87^{\circ}$, the phasor $V_t+jx_qI_a$ lies along the
q-axis, and its angle from $V_t$ is the rotor angle:
$$\tan\delta=\frac{x_qI_a\cos\phi}{V_t+x_qI_a\sin\phi}
=\frac{(0.6)(1.0)(0.8)}{1.0+(0.6)(1.0)(0.6)}=\frac{0.48}{1.36}=0.35294$$
$$\boxed{\delta=19.44^{\circ}}$$
The lagging current adds to the denominator, which is why a lagging load pulls
the rotor angle down relative to what a leading load of the same magnitude would
give.
Resolve the armature current onto the two axes. The
angle between the armature current and the q-axis is $\delta+\phi$, so
$$\begin{aligned}
I_d &= I_a\sin(\delta+\phi)=1.0\sin(19.44^{\circ}+36.87^{\circ})=\sin 56.31^{\circ}=0.8321\ \text{pu}\\
I_q &= I_a\cos(\delta+\phi)=\cos 56.31^{\circ}=0.5547\ \text{pu}
\end{aligned}$$
A useful self-check falls out here: the q-axis was defined so that the component
of terminal voltage perpendicular to it is $x_qI_q$, and indeed
$V_t\sin\delta = \sin 19.44^{\circ} = 0.3328$ equals
$x_qI_q = 0.6(0.5547) = 0.3328$.
Build the excitation voltage along the q-axis. Because
the demagnetising direct-axis current is the only component that sees $x_d$,
$$E=V_t\cos\delta+x_dI_d=(1.0)\cos 19.44^{\circ}+(1.0)(0.8321)=0.9430+0.8321$$
$$\boxed{E=1.775\ \text{pu at }19.44^{\circ}\ \text{ahead of }V_t}$$
The machine needs roughly 78 per cent more excitation voltage than terminal
voltage to carry rated current at this lagging power factor; it is strongly
over-excited, which is exactly the state part (a) describes for a machine
exporting vars.
Part (c) — repeat at 0.8 power factor leading.
Only the sign of $\phi$ changes, and the effect on the denominator is
dramatic:
$$\tan\delta=\frac{x_qI_a\cos\phi}{V_t-x_qI_a\sin\phi}
=\frac{(0.6)(0.8)}{1.0-(0.6)(0.6)}=\frac{0.48}{0.64}=0.75
\ \Longrightarrow\ \delta=36.87^{\circ}$$
The rotor angle has come out numerically equal to the power-factor angle. That
is not a coincidence to be glossed over, it is the whole answer: it means
$\delta+\phi = 36.87^{\circ}-36.87^{\circ}=0$, so the armature current lies
exactly along the q-axis.
Evaluate E for the leading case. With the current
purely quadrature-axis,
$$I_d=I_a\sin(\delta+\phi)=\sin 0^{\circ}=0,\qquad I_q=I_a\cos 0^{\circ}=1.0\ \text{pu}$$
and the direct-axis reactance drops out of the calculation altogether:
$$E=V_t\cos\delta+x_dI_d=(1.0)\cos 36.87^{\circ}+0=0.8+0$$
$$\boxed{E=0.800\ \text{pu at }36.87^{\circ}\ \text{ahead of }V_t}$$
Since $E = 0.8 < V_t = 1.0$, the machine is under-excited and is
absorbing reactive power — consistent with the leading armature current
stated in the question and with the definitions given in part (a). The
saliency has no influence at all on this particular operating point, because
there is no direct-axis armature current for $x_d$ to act on.
