Question 2 of 7: The Reactive Capability Curve and a Round-Rotor Operating Table

Question 2: The Reactive Capability Curve and a Round-Rotor Operating Table (20 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) — the reactive capability curve and what sets each of its boundaries.

The capability curve is drawn in the plane of active power against reactive power at the machine terminals, with lagging (supplied) reactive power conventionally to the right. Every boundary on it is a thermal or a stability limit, and each has a different physical origin.

The armature-current limit is the arc of a circle centred on the origin. With the terminal voltage held at its rated value the apparent power is $S=V_tI_a$, so a fixed maximum stator current is a fixed maximum $\sqrt{P^{2}+Q^{2}}$. What it protects is stator copper loss and therefore the temperature of the armature winding insulation. It is the binding constraint over the middle of the diagram, near unity power factor.

The field-current limit is the arc of a second circle, centred on the negative reactive axis at $-V_t^{2}/X_s$ and of radius $E_{max}V_t/X_s$. It comes from the same two power expressions used in part (b): eliminating $\delta$ between $P=(E V_t/X_s)\sin\delta$ and $Q=(EV_t\cos\delta-V_t^{2})/X_s$ gives $$P^{2}+\left(Q+\frac{V_t^{2}}{X_s}\right)^{2}=\left(\frac{EV_t}{X_s}\right)^{2}$$ so a ceiling on the excitation is a ceiling on $E$ and hence a circle. It protects the rotor winding from over-heating and binds in the over-excited region, where the machine is exporting reactive power.

In the under-excited region two further limits appear. The end-region heating limit arises because a weakly excited machine has a low rotor field, so the armature end-winding leakage flux is no longer opposed and penetrates the stator core end laminations in the axial direction, inducing eddy currents in the end packets and the clamping structure and producing local hot spots. Close beside it lies the steady-state stability limit: as $\delta$ approaches ninety degrees the synchronising torque coefficient $dP/d\delta$ falls to zero and the machine can no longer hold synchronism against a disturbance, so a practical margin of the order of ten per cent in power is left. Finally the horizontal prime-mover limit caps the active power at whatever the turbine can deliver, which is a mechanical rather than an electrical constraint, and a lower boundary at minimum stable turbine load often applies as well. Reducing the terminal voltage shrinks the armature-limit circle in proportion to $V_t$ and moves the field-limit circle inward and to the left, so the reactive capability of a generator is always poorest exactly when the system most needs it, during a low-voltage event.

Given. A cylindrical-rotor machine on an infinite bus, with the three partly specified operating conditions of the printed table.

Find. The two missing entries in each row of the table, so that all three operating points are fully specified.

Approach. Both unknowns in every row follow from the two round-rotor power transfer equations written at the infinite bus; each condition supplies two of the four quantities $P$, $Q_2$, $E$, $\delta$, and the pair of equations then closes algebraically with no iteration.

1. Part (b) — write the two governing equations at the bus. With armature resistance neglected the machine sees the infinite bus through the synchronous reactance alone, so the complex power delivered to the bus is $S_2=V\,I_a^{*}$ with $I_a=(E\,/\,\delta-V)/(jX_s)$. Separating real and imaginary parts, $$P=\frac{EV}{X_s}\sin\delta \qquad Q_2=\frac{EV\cos\delta-V^{2}}{X_s}$$ With $V=1.00$ and $X_s=0.375$ these become $P=2.66667\,E\sin\delta$ and $Q_2=(E\cos\delta-1)/0.375$. It is worth restating the two of them in the form used below: $E\sin\delta=0.375P$ and $E\cos\delta=1+0.375Q_2$, which makes the pair a simple rectangular decomposition of the phasor $E$.
2. Condition A: the reactive equation fixes the angle first. Here $E=1.25$ and $Q_2=-0.1$ are known, so $$E\cos\delta=1+0.375(-0.1)=0.9625 \quad\Longrightarrow\quad \cos\delta=\frac{0.9625}{1.25}=0.7700$$ which gives $$\boxed{\delta_A=39.6461^{\circ}}$$ Only the positive root is admissible because the machine is generating.
3. Condition A: the active power then follows directly. With $\sin(39.6461^{\circ})=0.638044$, $$P_A=\frac{(1.25)(1.00)}{0.375}\sin\delta_A=3.33333\times0.638044=\boxed{P_A=2.1268\ \text{pu}}$$ The pull-out power at this excitation is $EV/X_s=3.3333$ pu, so the machine is loaded to sixty-four per cent of its steady-state limit — comfortable, but the slightly negative $Q_2$ shows it is drawing a little reactive power from the system, that is, running under-excited.
4. Condition B: zero reactive power collapses the second equation. Setting $Q_2=0$ makes $$E\cos\delta=V=1.000$$ which says that the tip of the phasor $E$ lies on the vertical line through $V$ in the chart below. The active power then supplies the other component, $$E\sin\delta=0.375P=0.375(1.35)=0.50625$$
5. Condition B: recover the excitation and the angle from those two components. Dividing and then combining, $$\tan\delta_B=\frac{0.50625}{1.000} \quad\Longrightarrow\quad \boxed{\delta_B=26.8508^{\circ}}$$ $$E_B=\sqrt{(1.000)^{2}+(0.50625)^{2}}=\sqrt{1.256289}=\boxed{E_B=1.12084\ \text{pu}}$$ Delivering 1.35 pu at exactly unity power factor therefore needs only about twelve per cent more excitation than the bus voltage itself.
6. Condition C: both equations are explicit. With $E=1.3$ and $\delta=40^{\circ}$ nothing has to be inverted: $$P_C=\frac{(1.3)(1.00)}{0.375}\sin40^{\circ}=3.46667\times0.642788=\boxed{P_C=2.2283\ \text{pu}}$$ $$Q_{2C}=\frac{(1.3)\cos40^{\circ}-1}{0.375}=\frac{0.995858-1.000000}{0.375}=\boxed{Q_{2C}=-0.01105\ \text{pu}}$$ The reactive output is very slightly negative, and the reason is visible in the numerator: $E\cos\delta=0.9959$ falls just short of $V=1.000$, so the machine sits a hair on the under-excited side of the unity-reactive-power line. This is not a rounding artefact — the sign is the answer, and it means the machine is absorbing about eleven kVAr per MVA of rating.
7. Read the three points off the operating chart. Plotting $E\sin\delta$ against $E\cos\delta$ turns the two governing equations into a pair of rulings: horizontal lines are constant $P$ and vertical lines are constant $Q_2$, with the line $E\cos\delta=V$ separating over- from under-excited operation. Conditions A and C both sit marginally to the left of that line and condition B sits exactly on it, which is the geometric statement of the three answers.
8. Closure check on the armature current. Recomputing $I_a=(E\,/\,\delta-V)/(jX_s)$ from each completed row and forming $S_2=VI_a^{*}$ reproduces the tabulated $P$ and $Q_2$ to six decimal places, with $|I_a|$ equal to 2.1292, 1.3500 and 2.2284 pu for A, B and C respectively. Condition B has the smallest current for a substantial power output, which is the expected reward for operating at unity power factor.