Question 2 of 7: Excitation state and the salient-pole generated voltage

Question 2: Excitation state and the salient-pole generated voltage

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) — over-excitation and under-excitation

Excitation state is defined by comparing the internal generated voltage $E$ with the terminal voltage $V_t$. A machine is over-excited when the field current is large enough that $E\cos\delta$ exceeds $V_t$; it is under-excited when the field current is reduced until $E\cos\delta$ falls below $V_t$. The dividing case, $E\cos\delta = V_t$, is unity power factor, where the machine exchanges only real power with the system.

The consequence for reactive power follows from the round-rotor relation

which is positive when $E\cos\delta > V_t$ and negative when it is less. An over-excited generator therefore delivers reactive power to the system and operates at a lagging power factor; an under-excited generator absorbs reactive power and operates at a leading power factor. The same machine is thus a continuously adjustable source or sink of vars, controlled entirely by the field rheostat or the automatic voltage regulator, with no switching and no discrete steps.

To make a synchronous machine appear as a pure source of reactive power, it is run at or near zero real load and heavily over-excited — the classical synchronous condenser. Uncoupled from any prime mover, the machine draws only enough real power to cover its own losses, so $\delta \approx 0$, $P\approx 0$ and $Q \approx (E-V_t)V_t/X_d$, which is large and positive when $E$ is pushed well above $V_t$. Reversing the argument, reducing the field until $E$ falls below $V_t$ makes the same machine absorb vars, which is how synchronous condensers and under-excited generators hold down voltage on lightly loaded transmission at night. Because the response is continuous and inherently fast, and because a rotating machine also contributes short-circuit current and inertia, synchronous condensers remain in service at converter terminals and at the ends of long lines even where static compensators are cheaper.

Two limits bound this freedom, and both appear on the machine's capability curve: the field winding's thermal rating caps over-excitation, and under-excitation is limited by armature-end heating and, ultimately, by the steady-state stability limit as $E$ becomes too small to hold synchronism.

Parts (b) and (c) — generated voltage by two-reaction theory

Given. A salient-pole synchronous generator operating at rated terminal voltage and rated apparent power, with the two axis reactances listed below.

Find. The per-unit generated (excitation) voltage E and the power angle δ for 0.8 lagging power factor, and again for 0.8 leading power factor.

Approach. A salient-pole machine has no single synchronous reactance, so $E \ne V_t + jX_sI_a$. Instead, construct the auxiliary phasor $E_q = V_t + jX_qI_a$, which lies exactly along the q-axis and therefore fixes $\delta$; resolve $I_a$ into its d- and q-axis components using $\psi = \delta + \varphi$; then add the extra direct-axis drop $(X_d - X_q)I_d$ to reach $E$.

1. Part (b), fix the terminal phasors for 0.8 lagging. Rated kVA at rated voltage means $|I_a| = 1.0$ pu, and $\varphi = \cos^{-1}0.8 = 36.87^\circ$ lagging, so $V_t = 1.0\angle 0^\circ$ and $I_a = 1.0\angle-36.87^\circ$ pu.
2. Locate the quadrature axis. The whole method rests on the fact that $V_t + jX_qI_a$ is collinear with $E$, because the q-axis component of the armature reaction is already fully accounted for by $X_q$: $$E_q=V_t+jX_qI_a = 1.0 + j(0.6)(1.0\angle-36.87^\circ) = 1.0+(0.36+j0.48)=1.36+j0.48$$ $$\begin{aligned} |E_q| &= 1.44222\ \text{pu} \\ \delta &= \tan^{-1}\!\frac{0.48}{1.36}=19.440^\circ \end{aligned}$$ so $\boxed{\delta = 19.44^\circ}$ for the lagging case.
3. Resolve the armature current onto the axes. The angle between $I_a$ and the q-axis is $\psi = \delta + \varphi = 19.440^\circ + 36.870^\circ = 56.310^\circ$, hence $$\begin{aligned} I_d &= |I_a|\sin\psi = 0.83205\ \text{pu} \\ I_q &= |I_a|\cos\psi = 0.55470\ \text{pu} \end{aligned}$$
4. Add the saliency term. Along the q-axis the remaining drop is $(X_d - X_q)I_d$, so $$E = |E_q| + (X_d-X_q)I_d = 1.44222 + (1.0-0.6)(0.83205) = 1.44222+0.33282$$ $$\boxed{E = 1.775\ \text{pu at }\delta = 19.44^\circ}$$ An independent check is available directly from the axis components: $E = V_t\cos\delta + X_dI_d = 0.94296 + (1.0)(0.83205) = 1.77504$ pu, and the orthogonality condition $V_t\sin\delta = X_qI_q$ gives $0.33282 = (0.6)(0.55470)$, which closes exactly.
5. Part (c), repeat for 0.8 leading. Only the sign of the phase angle changes: $I_a = 1.0\angle+36.87^\circ$ pu. Rebuilding the auxiliary phasor, $$E_q = 1.0 + j(0.6)(1.0\angle 36.87^\circ) = 1.0 + (-0.36+j0.48) = 0.64+j0.48$$ $$\begin{aligned} |E_q| &= 0.800\ \text{pu} \\ \delta &= \tan^{-1}\!\frac{0.48}{0.64} = 36.870^\circ \end{aligned}$$
6. Evaluate the saliency term for the leading case. Now $\psi = \delta + \varphi = 36.870^\circ - 36.870^\circ = 0^\circ$, so $I_d = \sin 0^\circ = 0$ and the entire armature current lies on the quadrature axis. The saliency correction vanishes and $$\boxed{E = 0.800\ \text{pu at }\delta = 36.87^\circ}$$ This clean result is a genuine coincidence of the chosen numbers, not a general rule: with $\varphi = -\delta$ the current sits exactly on the q-axis, the machine produces no direct-axis armature reaction at all, and $E$ collapses to $|E_q|$.
7. Interpret the two operating points. At 0.8 lagging the machine needs $E = 1.775$ pu, well above terminal voltage, so it is strongly over-excited and exporting vars. At 0.8 leading it needs only $E = 0.800$ pu, below terminal voltage, so it is under-excited and absorbing them — the same conclusion Part (a) reaches from the sign of $E\cos\delta - V_t$, which is $+0.673$ pu in the first case and $-0.360$ pu in the second. The field current must therefore be cut to roughly 45 per cent of its lagging value to move between the two points at constant real load.