Set 35 — Voltage and Reactive Power Control
Set 34 treated an entire interconnection as one machine with one frequency. Nothing of the kind works here. Voltage is a different number at every busbar, it is set by the local reactive balance, and reactive power is so expensive to transport that importing it over any distance is self-defeating. These twenty problems work through the consequences: how far a bus voltage moves for a given injection, how large a capacitor must be and why its own success reduces its output, where the two-bus quadratic runs out of solutions altogether, what a tap changer can and cannot fix, and how a synchronous condenser, an SVC and a STATCOM differ at the one moment that matters — when the voltage has already fallen.
Every number below was computed from the stated data. Where an iteration is needed it is shown converging rather than quoted, and where the textbook rule of thumb is used it is checked against the exact expression it approximates.
The drop splits into two components, and they carry different cargo. \(\Delta V = (RP+XQ)/V_r\) lies in phase with \(\mathbf V_r\) and sets the magnitude difference; \(\delta V = (XP-RQ)/V_r\) lies in quadrature and sets the angle. With \(X \gg R\) on a transmission line, \(Q\) drives magnitude and \(P\) drives angle — the same decoupling that justified the fast decoupled load flow.
The exact two-bus relation is a quadratic in \(V^2\), not a linear drop. With \(R\) neglected, \(V^4 + (2XQ-V_s^2)V^2 + X^2(P^2+Q^2) = 0\). Two positive roots exist: the upper is the operating point, the lower is a high-current solution no load-restoring control can hold. Use the quadratic whenever the question is about collapse, maximum transfer, or a bus already well below nominal.
The sensitivity is exact and it changes sign at the nose. \(\dfrac{dV}{dQ} = \dfrac{X(V^2+XQ)}{V\left[V_s^2 - 2(V^2+XQ)\right]}\), negative on the upper branch and positive on the lower. The denominator vanishes when \(V^2+XQ = V_s^2/2\), which is precisely where the two roots merge.
The operator's version is one line. \(\Delta V/V \approx \Delta Q/S_{sc}\) with \(S_{sc} = V^2/X_{\text{source}}\). Injecting vars equal to one per cent of the fault level raises the bus by about one per cent — which is why weak, radially fed buses swing and strong ones do not.
A capacitor's output falls as \(V^2\), so sizing is implicit. A bank rated \(Q_{C0}\) at nominal delivers \(Q_{C0}V^2\) at the voltage it produces, and the voltage depends on what it delivers. Two or three passes settle it; ignoring the loop overstates the bank's effect.
A tap changer moves voltage between two sides and adds no vars whatever. With \(t_st_r=1\), \(t_s^2\left[1 - \dfrac{RP+XQ}{V_1V_2}\right] = \dfrac{V_2}{V_1}\). If the answer falls outside the physical range, the system needs reactive compensation, not a bigger tap changer.
A generator's reactive output is set entirely by its field. \(P = \dfrac{EV}{X_s}\sin\delta\) and \(Q = \dfrac{EV\cos\delta - V^2}{X_s}\). With \(P\) held by the governor, \(E\sin\delta\) is fixed, so raising \(E\) raises \(E\cos\delta\) and pushes vars out. Underexcitation absorbs vars but enlarges \(\delta\) and erodes stability margin.
The AVR's accuracy and its stability pull in opposite directions. \(e_{ss} = 1/(1+K)\) demands large gain; three cascaded lags forbid it. Rate feedback supplies phase lead where it is needed and an integral term removes the residue.
At depressed voltage the devices separate into two classes. A capacitor and an SVC at its limit are susceptances and fade as \(V^2\); a synchronous condenser and a STATCOM hold current and fade only as \(V\). At half voltage that is 25% of rating against 50% — and half voltage is when the outcome is decided.
A 150 km, 220 kV line has \(r = 0.08\ \Omega\)/km and \(x = 0.40\ \Omega\)/km. It delivers 150 MW at 0.9 power factor lagging with the receiving busbar at 220 kV.
- Find the active power lost and the reactive power absorbed by the line.
- Express each as a percentage of the delivered active power, and compare the reactive absorption with the load's own reactive demand.
- State what this implies about supplying the load's vars from the sending end.
The line constants and the current. Over 150 km,
The load's own reactive demand is \(Q_L = 150\tan(\arccos 0.9) = 72.65\) MVAr — the number the line is being asked to carry.
The two losses. Both are three-phase and both go as \(I^2\):
The ratio is exactly \(X/R = 5\), as it must be — the same current flows through both, so the ratio of the two "losses" is fixed by the line's geometry alone and is independent of loading.
As percentages of the delivered power:
Nearly a quarter of the throughput, in megavars, is swallowed by the line's own reactance. An engineer who tolerates 4.6% of copper loss without comment should be startled by the second figure.
The comparison that settles the argument. The load wants 72.65 MVAr; the line consumes another 34.44 MVAr in delivering it:
Supplying the load's vars from the sending end costs a 47% surcharge on this line alone. Two such lines in series and the surcharge approaches 100%: a megavar generated at the power station arrives as half a megavar.
And the surcharge is self-reinforcing. Reactive current is current, and it depresses the voltage along its path. A lower voltage requires more current for the same power, and more current absorbs more vars — the mechanism that Problem 6 follows to its conclusion. The active loss has no comparable feedback, because \(P\) and \(V\) are only weakly coupled.
The design consequence. Install the vars at the load. A 72.65 MVAr capacitor bank at the receiving busbar removes \(Q\) from the line entirely; the current falls to \(150\times10^6/(\sqrt3\times220\times10^3) = 393.6\) A, and both losses fall by the square of the current ratio:
A 19% cut in copper loss thrown in free with the voltage improvement — which is the whole content of the power-factor tariffs of Set 26.
A 33 kV feeder has \(R = 6.5\ \Omega\) and \(X = 9.0\ \Omega\) and supplies 8 MW at 0.86 power factor lagging, the receiving busbar being held at 33 kV.
- Find the in-phase and quadrature components of the drop, the sending-end voltage and the angle between the busbars.
- Compare the exact magnitude with the in-phase approximation.
- Repeat with the load corrected to unity power factor and comment on the size of the two effects.
The reactive demand. With \(\cos\phi = 0.86\), \(\tan\phi = 0.5934\):
Throughout, \(P\) and \(Q\) are three-phase in MW and MVAr, \(R\) and \(X\) per phase in ohms, and \(V\) line-to-line in kV; the combination \((RP+XQ)/V\) then comes out in kV directly.
The two components:
Note the arithmetic inside each bracket. In the in-phase term the two products add; in the quadrature term they subtract. That is the whole of the decoupling, visible in the signs.
The sending-end voltage and the angle:
The approximation, checked. Dropping the quadrature term altogether gives \(V_s \approx 33+2.870 = 35.870\) kV:
Six hundredths of one per cent. The quadrature component enters the magnitude only in second order, which is why every tap-changer and capacitor-sizing calculation in this set may use \(\Delta V\) alone and ignore \(\delta V\) entirely.
With the load corrected to unity power factor, a 4.747 MVAr capacitor at the load bus setting \(Q = 0\):
The comparison. The two components have swapped importance:
Removing 4.747 MVAr cut the magnitude drop by 45% and nearly halved the regulation, while the angle grew — the active power still has to get through, and it is the angle that carries it. The capacitor bought a voltage improvement, not a reduction in transfer.
A load of \(P = 0.7\) pu and \(Q = 0.35\) pu is fed from a source held at \(V_s = 1.05\) pu through a line of pure reactance \(X = 0.18\) pu. All quantities are on a 100 MVA base.
- Find both solutions for the receiving-end voltage.
- Find the current and the transmission angle at each.
- Explain why only one of them is an operating point.
Set up the quadratic. With \(R = 0\) the exact receiving-end relation is
A quadratic in \(u = V^2\). It is exact — no small-angle or small-drop assumption has been made — and it is the tool for every question about collapse.
Substitute the data:
Solve:
Both are positive and both satisfy the equations exactly. The quadratic does not distinguish between them; the physics does.
The currents and angles. The magnitude of the current is \(|S|/V\) with \(|S| = \sqrt{0.49+0.1225} = 0.7826\) pu, and the angle follows from the quadrature drop \(XP/V\) compared with \(V_s\):
The same power is delivered at 6.8 times the current and eight times the angle. The lower root is a genuine mathematical solution in which the line carries an enormous current at a very low voltage.
Why only the upper root is an operating point. Two independent reasons, and both matter:
The second is the deeper one. On the upper branch \(dV/dQ\) is negative, so a small increase in demand lowers the voltage a little and the system settles; on the lower branch it is positive, and Problem 4 evaluates both to show it.
A check on the upper root. Substituting back into the phasor relation:
For the system of Problem 3 (\(V_s = 1.05\), \(X = 0.18\), \(P = 0.7\), \(Q = 0.35\) pu):
- Evaluate \(dV/dQ\) at both roots and interpret the sign of each.
- Use the sensitivity to predict the voltage after 10 MVAr of local injection, and compare with the exact quadratic.
- State the condition under which the sensitivity becomes infinite.
The expression. Differentiating \(F = V_s^2V^2 - (V^2+XQ)^2 - (XP)^2 = 0\) implicitly with respect to \(Q\) gives
Here \(Q\) is the reactive power drawn through the line. Injecting vars locally reduces it, so the voltage moves the other way.
At the upper root, \(V = 0.9776\), \(V^2 = 0.955736\):
Negative: more reactive demand, lower voltage. Each 0.01 pu (1 MVAr on this base) of extra demand costs 0.2% of a per unit, so about 0.2% of voltage.
At the lower root, \(V = 0.1441\), \(V^2 = 0.020765\):
Positive — on the lower branch, increasing the reactive demand raises the voltage. That perverse sign is what makes the branch unstable: any control that injects vars when the voltage is low drives the system further from equilibrium rather than back to it. Here is the dynamic argument promised in Problem 3.
The linear prediction. A 10 MVAr capacitor on the 100 MVA base is \(\Delta Q = -0.10\) pu through the line:
The exact value. Re-solving the quadratic with \(Q = 0.25\):
Excellent, because the sensitivity barely changes over this range. It would not be excellent for a 40 MVAr injection, as Problem 7 will show — the sensitivity weakens as the voltage recovers, so a linear extrapolation always overstates the gain.
Where the sensitivity blows up. The denominator vanishes when
which is exactly the condition that the two roots of the quadratic coincide — the discriminant is zero there. At that point no finite reactive injection stabilises the bus: the derivative is infinite because the two branches have become tangent. Problem 6 finds the loading at which it happens.
A 220 kV substation busbar has a three-phase fault level of 3500 MVA. The voltage is to be raised by 2.5% with a shunt capacitor bank.
- Find the MVAr required from the short-circuit-level rule.
- Find the capacitance per phase for a star-connected bank at 50 Hz, and its reactance.
- Verify the rule against the exact Thévenin reactance, and state what the bank actually delivers once connected.
The rule. From the special case of the sensitivity with \(P\) small and \(V \approx V_s\), \(dV/dQ \approx -X/V\), and with \(S_{sc} = V^2/X\),
The fault level is the quantity computed in Set 33; here it is being reused as a measure of how hard the bus is to move.
The capacitance. For a star-connected bank the three-phase output is \(Q_C = V_{LL}^2\,\omega C\) with \(C\) the per-phase capacitance:
The verification. The Thévenin reactance behind the busbar follows from the fault level:
Exactly the target, because the two calculations are algebraically identical: \(X\Delta Q/V\) divided by \(V\) is \(\Delta Q/(V^2/X) = \Delta Q/S_{sc}\). The "rule of thumb" is the sensitivity in disguise.
What the bank actually delivers. Its own action raises the voltage it sits on, and its output goes as the square of that voltage:
4.9% more than rated. That works in the engineer's favour here, but it is the reason a capacitor bank's own rating must be checked against the raised voltage, and why utilities limit the step size so that the rise per switching operation stays within 2–3%.
Where the rule breaks down. It assumed \(V \approx V_s\) and small \(\Delta Q\). Both fail at a depressed bus:
Here \(\Delta Q/S_{sc} = 0.025\), so the rule is comfortably inside its range. Problem 7 works a case where it is not.
Again for the system of Problem 3 (\(V_s = 1.05\) pu, \(X = 0.18\) pu, 100 MVA base):
- Find the maximum active power the line can deliver with the reactive demand held at 0.35 pu, and the voltage at that point.
- Find the same two quantities if the reactive demand is fully compensated at the load bus.
- Express the gain as a percentage and comment on what it costs.
The condition. The two roots merge when the discriminant of the quadratic in \(u\) vanishes:
Beyond that loading the quadratic has no real root at all: no steady state exists, at any voltage. That is what voltage collapse means algebraically.
With \(Q = 0.35\) pu:
The voltage at the nose is the repeated root, \(u = (V_s^2-2XQ)/2\):
Thirty per cent below nominal, and reached long before any thermal limit. On this line the limit is voltage, not heat.
With the reactive demand compensated locally, so that the line carries \(Q = 0\):
The compensated case collapses to a familiar result: maximum transfer at \(V = V_s/\sqrt2\), which is the matched-impedance condition. The uncompensated case is a perturbation of it.
The gain:
And it costs 35 MVAr of capacitors, not one ohm of new line. Compare the alternative: to raise \(P_{\max}\) by 13.9% by reducing \(X\) would need the reactance cut from 0.18 to 0.161 pu — a second conductor bundle, or a series capacitor, over the whole route.
The general result behind both numbers. Writing the nose condition out and cancelling the \(Q^2\) terms gives a striking simplification for the reactive limit at fixed \(P\):
Linear in nothing, but exact and free of square roots. At \(P = 0.7\) pu it gives \(Q_{\max} = 1.1025/0.72 - 0.18(0.49)/1.1025 = 1.53125 - 0.08 = 1.4513\) pu — so the bus can absorb 145 MVAr before collapsing, against the 35 MVAr it actually draws. Challenge C2 turns that difference into a planning criterion.
A load of \(P = 0.8\) pu and \(Q = 0.5\) pu is fed from a 1.0 pu source through a line of reactance \(X = 0.20\) pu on a 100 MVA base.
- Find the uncompensated load-bus voltage.
- Find the capacitor rating, quoted at nominal voltage, that would raise the bus to exactly 0.98 pu.
- A standard 50 MVAr bank is installed instead. Find the voltage it actually produces, showing the iteration.
The uncompensated voltage. From the quadratic,
13.2% below nominal — unacceptable, and the lower root sits at 0.2175 pu. This is a bus in real trouble, and the linear rule of Problem 5 is outside its range here.
Invert the question for part 2. Rather than guess a rating, fix \(V = 0.98\) and solve the phasor relation for the reactive power the line must then carry:
So the capacitor must supply \(0.5 - 0.032252 = 0.467748\) pu at 0.98 pu voltage.
Convert to a nameplate rating, which is quoted at nominal voltage:
Sizing the bank at 46.8 MVAr — the value actually needed at the operating voltage — would fall 4% short, because the nameplate is defined at a voltage the bus never reaches.
Part 3: a real bank of 50 MVAr. Its output is \(0.5V^2\) pu, which depends on the voltage it produces. Start from \(V = 1.0\) and iterate, solving the quadratic afresh each pass:
Converged to four figures by the fourth pass. The iteration is contracting because \(dQ_C/dV\) is small compared with the network's own stiffness — which is why two or three passes always suffice in practice.
The converged answer:
A 50 MVAr bank delivering 48.34 MVAr and lifting the bus from 0.8676 to 0.9833 pu — an improvement of 11.6 percentage points, bought entirely locally with no change anywhere else in the network.
What the first pass alone would have claimed. Ignoring the \(V^2\) dependence gives 0.9868 pu, an overstatement of 0.35 percentage points. And the linear sensitivity is worse still: at the uncompensated point \(V^2+XQ = 0.752704+0.1 = 0.852704\), so
Because the sensitivity weakens sharply as the voltage recovers, from \(-0.279\) at 0.868 pu to about \(-0.20\) at 0.983 pu. A linear extrapolation over a large correction always promises more than it delivers.
A 25 MVAr capacitor bank is to be connected to a 33 kV busbar whose three-phase fault level is 900 MVA. The feeder supplies a six-pulse converter installation.
- Find the bank's output at 0.95 and 0.90 pu voltage.
- Find the parallel resonant frequency of the bank against the source, and say why the answer is unacceptable.
- A 6% detuning reactor is added in series with the bank. Find the new resonant frequency, the bank's series tuning order and its fundamental output.
The fade. A capacitor is a fixed susceptance, so \(Q_C = V^2/X_C\):
A 10% voltage dip costs 19% of the bank's output — it deserts precisely when the vars are wanted. Everything in Problems 17 and 19 follows from this one line.
The two reactances. Both referred to 33 kV:
\(X_{s1}\) is the source reactance at fundamental frequency; at harmonic order \(h\) it is \(hX_{s1}\) while the capacitor's is \(X_C/h\).
The parallel resonance. Source inductance and bank capacitance resonate where the two magnitudes are equal:
Sixth harmonic — and that is the trouble. A six-pulse converter produces characteristic harmonics of order \(6k\pm1\), so the fifth and seventh sit either side of the resonance and both are strongly magnified. A resonance between two large harmonics is nearly as dangerous as one on top of either.
The detuning reactor. With \(X_L = 0.06X_C = 2.6136\ \Omega\) in series with the bank, the branch is series-resonant at
Below \(h_s\) the branch is capacitive, above it inductive. At the fifth harmonic and beyond the bank now presents an inductance, and cannot resonate with the source there at all.
The new parallel resonance moves below \(h_s\), because the reactor adds to the source's inductance:
Between the third and fifth harmonics, where a balanced three-phase converter load produces very little: the third is a zero-sequence set that a delta winding traps, and the converter's own spectrum starts at the fifth. The reactor did not remove the resonance; it moved it into a quiet part of the spectrum. That distinction is worth insisting on.
The fundamental output rises, because the reactor subtracts from the capacitor's reactance:
6.4% more than the nameplate — \(1/(1-0.06)\) exactly. The capacitor units must also withstand a raised voltage across themselves, so a detuned bank is specified with capacitor cans rated above the system voltage.
And the reactor earns its cost twice. Besides detuning, it limits the inrush when the step is switched, which for a bank switched against an existing energised bank can otherwise reach a hundred times rated current at several kilohertz.
A 300 km, 400 kV overhead line has \(L = 1.0\) mH/km and \(C = 11.5\) nF/km per phase, at 50 Hz.
- Find the reactive power the line generates and its surge impedance loading.
- Find the net reactive balance when it carries 800 MW at unity power factor.
- Verify that the balance is zero at surge impedance loading, and account for any discrepancy.
The two line constants:
The charging generation, which goes as the square of the voltage and is therefore almost constant:
Comfortably inside the 0.5–0.7 MVAr/km that 400 kV construction gives. An unloaded line of this length is a 173 MVAr capacitor that cannot be switched off.
The surge impedance loading:
At 800 MW, which is 1.47 times SIL:
Above SIL the line is a net sink and its far end sags. Two hundred megavars must come from somewhere, and by Problem 1 they had better come from the receiving end.
At SIL itself:
Exactly zero, and the identity is easy to see: \(3I^2X = V^2b\ell\) requires \(3(P/\sqrt3V)^2\,\omega L\ell = V^2\omega C\ell\), that is \(P^2 L = V^4C\), that is \(P = V^2\sqrt{C/L} = V^2/Z_c\).
The discrepancy that ought to appear, and why it does not here. This calculation lumps the whole shunt capacitance at the ends and the whole series reactance in the middle. A distributed treatment (Set 12) puts the charging current at every point, so the current is not uniform along the line and the true \(I^2X\) integral differs:
An electrical length of 18°, at which the nominal-\(\pi\) and exact models differ by under 1%. The lumped answer is safe here; at 600 km it would not be, and the ABCD constants would be needed.
The operating consequence. The line's reactive character reverses at 542.6 MW:
Which is the whole daily routine of a transmission operator: reactors out and capacitors in through the morning, and the reverse at night.
A 25 km, 132 kV XLPE cable circuit has a capacitance of 0.22 µF/km per phase. An overhead line of the same length and voltage would have 9.5 nF/km.
- Find the charging MVAr of each and the ratio between them.
- Size a shunt reactor to absorb 80% of the cable's charging, and find its inductance per phase.
- Find the cable's charging current and state why it limits the useful length of an a.c. cable.
The cable's charging. Total capacitance \(C = 0.22(25) = 5.5\ \mu\text{F}\) per phase:
The overhead line, for comparison. \(C = 9.5(25) = 0.2375\ \mu\text{F}\):
Twenty-three times, and the reason is geometric: the cable's conductor and its earthed screen are millimetres apart with a dielectric of relative permittivity 2.3 between them, while the line's conductors are metres apart in air. Set 30 derived the capacitance itself; this is what it costs in operation.
The reactor. To absorb 80% of the cable's generation:
Eighty per cent rather than a hundred is deliberate: full compensation would leave the circuit reactively neutral at no load but net absorbing under load, and the residual 6 MVAr of generation is useful. Reactor compensation of 60–80% is the usual choice.
The charging current. Per phase, with the phase voltage \(132/\sqrt3 = 76.21\) kV:
Equivalently \(30.11\times10^6/(\sqrt3\times132\times10^3) = 131.7\) A. It flows whether or not any load is connected.
Why it limits the length. The charging current is proportional to length and it occupies the conductor's thermal capacity. If the cable's continuous rating is 600 A, then
At 114 km the cable is full of its own charging current and can carry no load at all. Real designs stop far short of that, and reactors placed at intervals along the route push the limit out — but beyond a few tens of kilometres at transmission voltage, the honest answer is d.c. (Chapter 38), which has no charging current whatever.
A round-rotor generator with \(X_s = 1.4\) pu is connected to a 1.0 pu busbar and delivers \(P = 0.85\) pu, the governor holding that output constant.
- Find the load angle, reactive output and power factor for excitations of \(E = 2.2\) pu and \(E = 1.3\) pu.
- Find the excitation for unity power factor, and the excitation to deliver 0.6 pu of reactive power.
- Comment on the stability margin in each case.
The relation that does most of the work. Since the governor fixes \(P\),
Raising the field lengthens the phasor \(E\) but its vertical component is pinned. Only \(E\cos\delta\) can change, and that is exactly what the reactive equation depends on.
Overexcited, \(E = 2.2\) pu:
Underexcited, \(E = 1.3\) pu:
The machine now absorbs vars and behaves as a reactor seen from the system. Nothing but the field current changed; the turbine is doing precisely what it was doing before.
Unity power factor requires \(Q = 0\), that is \(E\cos\delta = V = 1.0\), while \(E\sin\delta = 1.19\) always:
For \(Q = +0.6\) pu, rearranging the reactive equation for \(E\cos\delta\):
Consistent with part 1: at \(E = 2.2\) the output was 0.6074 pu, so 0.6 pu needs a shade less field. The two calculations are inverses of one another and each checks the other.
The stability margin. Collect the load angles:
Absorbing vars costs stability margin, and steeply. At \(E = 1.3\) the machine sits at two-thirds of the way to the steady-state limit of Set 24, and a modest further reduction in field would take it over. This is precisely what the under-excitation limiter exists to prevent — that, and the end-region heating that underexcited operation causes in the stator core.
And the practical reading. A generator is the cheapest continuously variable var source on the system, because the plant is already there and only the field current changes. But its capability is bounded on three sides — field heating when overexcited, end-region heating and stability when underexcited, and armature current at both — so it is never an unlimited source, and reactive planning must respect the capability curve rather than the rating plate.
An excitation system has \(T_A = 0.08\) s, \(T_E = 0.6\) s and \(T_G = 1.2\) s, with \(K_E = K_G = K_R = 1\) and the sensing lag negligible.
- Find the largest amplifier gain \(K_A\) for which the loop is stable.
- Find the frequency of the sustained oscillation at that gain.
- Find the best steady-state regulation the loop could ever achieve, and the regulation at a usable gain of \(K_A = 10\).
The characteristic equation. Three lags in cascade with unity feedback:
Expand the product. First two factors:
The Routh array. The first-column entry of the \(s^1\) row decides stability:
The crossing frequency. At \(K_A = 27.20\) the \(s^1\) entry vanishes and the auxiliary equation from the \(s^2\) row gives the pair of imaginary roots:
Just under one hertz — and uncomfortably close to the 0.2 to 2 Hz band in which the machine's own electromechanical mode lives. That coincidence is why exciter tuning and power system stabiliser tuning cannot be done independently.
The regulation. With unity feedback and a step disturbance,
And \(K_A = 27.2\) is not a usable gain — it is the boundary of instability, at which the loop oscillates for ever. A gain giving reasonable damping is a third of it, and the regulation is then 9%, five times worse than any grid code allows.
The conflict stated plainly:
The resolution, and it is the reason the block diagram has a feedback path in it. A stabilising transformer wound on the exciter field returns \(sK_F/(1+sT_F)\), a derivative term that supplies phase lead near 6 rad/s — exactly where it is needed. The stability boundary moves to several hundred, and a proportional-integral term then removes the steady-state error entirely rather than merely shrinking it. Problem 13 puts numbers on what that is worth.
With its AVR out of service, a generator's terminal voltage falls by 12% when full reactive load is applied.
- Find the fall with the AVR in service at an open-loop gain of \(K = 40\).
- Find the gain needed to hold the fall within 0.2%.
- Using the loop of Problem 12, say whether that gain is attainable, and what must be done if it is not.
What closed-loop gain does to a disturbance. The open-loop change \(\Delta V_{ol}\) is divided by \(1+K\), because the regulator responds to the error it produces:
The same expression as \(e_{ss} = 1/(1+K)\), read as a disturbance-rejection ratio rather than as a tracking error. Both are the sensitivity function of the loop evaluated at zero frequency.
At \(K = 40\):
A factor of 41 improvement for no change in the machine — the AVR is doing all of it.
The gain for 0.2%:
Is 59 attainable? Problem 12 found the boundary at \(K_A = 27.2\) for the uncompensated three-lag loop, and a usable gain at about a third of that:
Not attainable — and not even at the boundary of instability, where the loop would oscillate for ever at 0.909 Hz. The uncompensated loop cannot meet the specification by any choice of gain whatever. That is a structural conclusion, not a tuning problem.
What must be done. Two remedies, in the order they were historically adopted:
With integral action the question "what gain do I need for 0.2%?" stops being asked at all: the steady-state error is zero for a step, and what remains is a settling time rather than an offset. Modern digital regulators are all PID, and the 0.2% specification is met with margin to spare.
One reason not to drive the error to zero, however. Several machines on one busbar with perfectly integrating AVRs would each insist on its own reference and would fight, exchanging enormous circulating vars — exactly the problem that droop solves for governors in Set 34. So a deliberate reactive droop or cross-current compensation of 2–5% is introduced, and each machine accepts a small voltage error in proportion to the vars it is producing. The var sharing then divides in a defined ratio instead of being decided by whichever regulator has the highest gain.
A three-phase 220 kV line has \(R = 18\ \Omega\) and \(X = 52\ \Omega\) per phase. Tap-changing transformers at both ends satisfy \(t_st_r = 1\). The receiving-end load is 90 MW at 0.88 power factor lagging and both busbars are to be held at 220 kV.
- Find \(t_s\) and \(t_r\).
- Verify the answer against the line drop itself.
- State what would be needed if the load were doubled.
The reactive demand:
The tap relation. With \(t_sV_1\) the sending line-side voltage and \(t_rV_2\) the receiving line-side voltage, the in-phase drop gives \(t_sV_1 = t_rV_2 + (RP+XQ)/(t_rV_2)\). Imposing \(t_r = 1/t_s\) and clearing:
The reciprocal condition is a practical convention, not a mathematical necessity: the two nominal ratios already perform the transformation between voltage levels, so requiring the deviations to be reciprocal means a boost at one end is matched by a buck at the other, and only the line voltage changes.
Substitute:
The sending transformer boosts by 4.58% and the receiving transformer bucks by 4.38%. Both are comfortably inside a \(\pm10\%\) range.
The verification. Compute the line-side voltages and the drop between them:
Consistent to the last figure. Notice what the tap changers have done: the busbars are both at 220 kV while the line itself runs at 230 kV at one end and 210 kV at the other. The 20 kV of drop has been pushed entirely onto the line, where nobody's equipment is connected.
If the load doubled, to 180 MW at the same power factor:
A 9.85% boost — right at the edge of the range, with nothing left for a contingency. The correct response is not a wider tap changer but compensation at the receiving end.
What compensation would buy. Injecting the doubled load's 97.15 MVAr locally removes the \(XQ\) term entirely:
A 3.53% boost instead of 9.85%, from 97 MVAr of capacitors. The tap changer was never short of range; the system was short of vars, and the tap changer was being asked to substitute for something it cannot supply.
The transformers of Problem 14 have a range of \(\pm10\%\) in steps of 1.25%.
- Choose the nearest available step for \(t_s\).
- With that step fixed and \(t_r = 1/t_s\), find the receiving busbar voltage actually achieved.
- Comment on the second root of the resulting quadratic, and on whether the reciprocal condition survives contact with a real tap changer.
The available steps are \(1 + 0.0125n\) for \(n = -8 \ldots +8\). Around the required 1.0458:
Rearrange the tap relation with \(V_2\) as the unknown. From \(t_s^2\left[V_1 - (RP+XQ)/V_2\right] = V_2\),
The load \(P\) and \(Q\) are held at their Problem 14 values, so \(RP+XQ = 4146.0\) throughout. Strictly a constant-power load; a voltage-dependent one would need a second iteration.
Substitute \(t_s^2 = 1.1025\):
The two roots:
Well inside a \(\pm5\%\) band, and as close as a discrete tap changer can come. The excess over 220 kV is the price of rounding upward: the exact ratio was 1.0458 and the nearest step is 1.05, a 0.4% over-boost that appears at the busbar as +0.89%.
The second root is the same phenomenon as Problem 3. At 20.59 kV the same 90 MW would be drawn at
A factor of 10.8. Every voltage relation in this chapter that is quadratic carries a low-voltage companion solution, and the tap equation is no exception. Discard it on the same grounds as before.
Does \(t_st_r=1\) survive? Not exactly. The required \(t_r = 0.952381\) is not a step:
Choosing \(t_r = 0.95\) gives \(t_st_r = 1.05(0.95) = 0.9975\), not unity. The reciprocal condition is a convenience for hand calculation; a load-flow program simply carries both ratios independently and does not impose it at all.
What the discrepancy is worth. A product of 0.9975 rather than 1 means the whole transformation is 0.25% low, so the receiving busbar would settle about 0.25% below the 221.96 kV computed — near 221.4 kV, still inside band. The convention costs a quarter of a per cent and saves an unknown; that is a good trade for an examination and an unnecessary one for a computer.
A transformer of leakage reactance 0.10 pu connects bus \(i\) to bus \(j\). Its tap is on the bus-\(i\) winding at an off-nominal ratio \(a = 1.05\).
- Derive the bus admittance entries and hence the equivalent \(\pi\) circuit.
- Evaluate every branch numerically.
- Interpret the two shunt branches, and reconcile them with the claim that a tap changer supplies no reactive power.
Set up the model. Represent the transformer as an ideal unit of ratio \(a:1\) at the bus-\(i\) end followed by the leakage admittance \(y\) to bus \(j\). Call the internal node \(t\), so \(V_t = V_i/a\). The ideal transformer conserves complex power, so \(I_i = I_t/a\).
The two terminal equations:
Symmetric off-diagonals, as any bilateral element must give, but unequal diagonals — the tap has broken the symmetry between the two ends.
Read off the \(\pi\) equivalent. For a \(\pi\) with series \(y_{se}\) and shunts \(y_{sh,i}\), \(y_{sh,j}\):
This is the standard model used by every load-flow program from Set 19 onward, and it is why a tap-changing transformer can be entered into \(\mathbf Y_{\text{bus}}\) without any special-case code.
Numerically, with \(y = 1/(j0.10) = -j10\) pu and \(a = 1.05\):
Check against the diagonals:
Interpretation. The signs are worth dwelling on:
A capacitor on one side and a reactor on the other. At 1.0 pu voltage the first injects 0.4535 pu of vars and the second absorbs 0.4762 pu — a net absorption of 0.0227 pu, which is precisely the extra \(I^2X\) that the redistributed current produces. No vars have been created.
Reconciling with the chapter's claim. The shunt branches are an artefact of forcing a two-port with unequal diagonals into a \(\pi\) shape. They have no physical existence:
Reverse the tap to \(a = 0.95\) and the two shunts swap character — the reactor moves to bus \(i\) and the capacitor to bus \(j\). A device that genuinely produced vars could not have its output reverse merely by relabelling which side the tap is on.
A 132 kV busbar has a fault level of 1800 MVA. A credible contingency depresses it by 4.5%, and a synchronous condenser is to be specified to restore it. The machine offered has \(X_s = 1.6\) pu on its own rating.
- Find the MVAr required and choose a standard rating.
- Find the excitation needed at full leading output, and the maximum reactive power the machine can absorb.
- Compare its output at 0.5 pu voltage with that of a capacitor bank of the same rating.
The requirement, from the short-circuit-level rule:
Headroom is not padding. A compensator sitting at its limit in normal operation can do nothing when it is needed, and reactive planning constrains the steady-state output of dynamic plant precisely to preserve it.
The condenser's characteristic. A synchronous machine on no mechanical load has \(\delta \approx 0\), so the reactive equation of Problem 11 collapses:
Overexcite (\(E > V\)) and it supplies vars; underexcite and it absorbs them. Smoothly, continuously, and across the whole range — which is the machine's distinguishing virtue.
The excitation at rated leading output. On the machine's own 90 MVA base, rated output is \(Q = 1.0\) pu at \(V = 1.0\) pu:
Two and a half times the terminal voltage of internal e.m.f. — which is why a synchronous condenser is built with an unusually generous field winding and a correspondingly large exciter. A generator of the same MVA would never need 2.6 pu.
The absorbing limit. The field cannot be reversed, so the extreme case is \(E = 0\):
Only 62.5% of rating, and the range is inherently asymmetric: a condenser is a better capacitor than it is a reactor. In practice the underexcitation limiter stops it well short of zero field, so 40–50% of rating is the usable absorbing capability.
At 0.5 pu voltage, the field held at its 2.6 pu ceiling:
Because the machine holds an internal e.m.f. that the bus voltage cannot touch, while the capacitor holds only a susceptance. And this is the condition that matters — half voltage, motors stalled, magnetising current heavy, the seconds in which the disturbance is decided one way or the other.
The three things the machine gives that a capacitor cannot. Beyond the output at depressed voltage:
The last two have returned to prominence: converter-dominated grids are short of both, and old condensers retired in the 1980s have in several countries been recommissioned, or new ones built, for exactly these reasons.
Against which stand its costs — a few per cent of rating in losses, the maintenance a rotating machine demands, and a response time of about a second set by the field time constant. Problem 19 puts a power-electronic alternative beside it.
An SVC on a 132 kV busbar consists of a fixed capacitor bank of 80 MVAr and a thyristor-controlled reactor rated 140 MVAr at full conduction, both at nominal voltage.
- Find the range of net output.
- Find the firing angle that gives a net output of +30 MVAr at nominal voltage, taking \(\alpha\) from the voltage peak.
- Find the conduction angle, and the net output at that same firing angle if the busbar falls to 0.95 pu.
The TCR susceptance. Delaying the thyristors chops the reactor current, and Fourier analysis of the chopped waveform gives the fundamental susceptance
At \(\alpha = 90^\circ\) the bracket is \(\pi\) and \(B_L = 1/X_L\) — full conduction, the reactor fully in. At \(\alpha = 180^\circ\) the bracket vanishes and the reactor is out. Between them the susceptance varies continuously, which is the whole point of the device.
The range of net output. The capacitor is fixed and the reactor subtracts from it:
A 140 MVAr control range from a 220 MVAr installation. The asymmetry — 80 capacitive against 60 inductive — is a design choice, and a busbar whose main problem is Ferranti rise at night would be given the reverse split.
The reactor duty for +30 MVAr:
Solve the transcendental equation by bracketing. Working in degrees for readability but radians in the first term:
Note how insensitive the function is near 122°: a whole degree of firing angle moves the output by about 2.2 MVAr, so the control resolution is generous. Near 90° and near 180° the function is far steeper.
The conduction angle follows directly:
Each thyristor conducts for 115.85° of every cycle instead of the full 180°, and the two antiparallel devices share the duty.
At 0.95 pu voltage, with the firing angle unchanged. Both branches are susceptances, so the whole SVC scales as \(V^2\):
The controller of course would not leave \(\alpha\) alone — it would advance the firing towards 180° to restore the output. But the ceiling it is working towards has fallen too: at 0.95 pu the maximum available is \(80(0.9025) = 72.2\) MVAr, not 80. Once the SVC has run out of firing-angle range it is simply a capacitor, and it fades like one.
The harmonic price. A chopped reactor current is rich in odd harmonics — 13.8% third, 5.0% fifth, 2.6% seventh at the worst firing angle. Delta-connecting the TCR traps the triplens, and the capacitor branch is usually split into tuned filters for the fifth and seventh. That filtering is a substantial part of an SVC's cost and footprint, and it is one of the reasons the STATCOM of Problem 19 displaced the SVC in new installations.
A 132 kV busbar is to be supported by a 100 MVAr device. Compare a fixed capacitor bank, an SVC at its capacitive limit, and a STATCOM of the same rating at busbar voltages of 0.95, 0.85 and 0.50 pu.
- Tabulate the output of each.
- Account for the difference in the underlying models.
- The busbar feeds a large induction-motor load. State which device you would specify and why.
The two models. Everything follows from what is held constant at the limit:
The STATCOM is a voltage-source converter behind a coupling reactance, and its limit is the current its semiconductors can carry — a device rating, not a network quantity. Nothing about the bus voltage changes what a transistor can conduct.
The table:
At a mild depression the three are within 5% of one another and the capacitor, at a small fraction of the cost, is plainly the right purchase. At half voltage the STATCOM delivers twice as much.
The STATCOM's governing relation, for completeness:
Identical in form to the synchronous condenser's \(V(E-V)/X_s\) of Problem 17, with the converter's fundamental output voltage playing the part of the internal e.m.f. and the coupling reactor the part of \(X_s\). The similarity is not a coincidence: a STATCOM is a synchronous condenser without a rotor.
And its short-term overload. The current limit is thermal, so a converter will hold 1.2 pu current for a second or two:
A capacitor has no equivalent capability whatever: its output at a given voltage is fixed by physics and no controller can raise it.
Why the motor load settles the choice. An induction motor is close to a constant-power device at its shaft, so when the terminal voltage falls its current rises, and its magnetising demand rises with it. If the voltage falls far enough the motor stalls, and a stalled motor draws locked-rotor current at a power factor near 0.2 — five or six times normal current, almost all of it reactive:
The demand spikes exactly when the voltage is lowest, and the reactive source that fades as \(V^2\) is fading at the same moment. That is the mechanism of a fault-induced delayed voltage recovery, and it is why the 0.50 pu column of the table is the one that decides the specification.
The recommendation, and it is a split one. Not one device but two:
Buying 100 MVAr of capacitors would satisfy every steady-state study and fail the motor-reacceleration case. Buying 100 MVAr of STATCOM would satisfy both and cost several times as much for capability that sits idle 99% of the time. Problem 20 costs the split arrangement out.
A 220 kV substation busbar with a fault level of 2500 MVA has a reactive demand that swings by 120 MVAr between night and peak. Post-contingency studies additionally require 45 MVAr to be supplied within two seconds, with the busbar depressed to 0.85 pu at the moment the requirement arises. Budgetary rates are ₹8 lakh per MVAr for switched capacitors, ₹35 lakh for an SVC and ₹60 lakh for a STATCOM, all installed.
- Choose a capacitor step size that keeps the switching voltage step within 2%.
- Cost an all-capacitor scheme, an all-STATCOM scheme, and a split scheme, and say which meets the requirements.
- State the headroom rule that the chosen scheme must operate under.
The step size. By the short-circuit-level rule, a step of \(Q\) MVAr moves the busbar by \(Q/S_{sc}\):
Four steps of 30 covers the swing with a 1.2% voltage step, comfortably inside the limit and fine enough that customers do not notice the switching. Two steps of 60 would have given 2.4% and been rejected.
Scheme A: all capacitors. The dynamic requirement is 45 MVAr at 0.85 pu, so a capacitor sized at nominal must be larger:
Cheap — and it fails. Mechanically switched capacitors take several hundred milliseconds of circuit-breaker time plus a controller delay, and a bank once switched cannot be adjusted; the requirement is for 45 MVAr within two seconds and continuously controllable. Scheme A also has no answer at all if the voltage falls further than 0.85 pu, since its output would keep falling with it.
Scheme B: all STATCOM. A single device covering both duties:
It meets every requirement and it costs seven times Scheme A. Most of the 120 MVAr of capability is being bought at STATCOM prices to perform a duty a capacitor does perfectly well — a slow, predictable, daily swing.
Scheme C: the split. Capacitors for the base duty, dynamic plant for the emergency:
The STATCOM is sized at 55 MVAr because at 0.85 pu it delivers \(55(0.85) = 46.8\) MVAr, clearing the 45 MVAr requirement with 4% to spare. Note how much smaller it can be than the capacitor equivalent of Scheme A: 55 against 65, because it fades linearly rather than quadratically.
An SVC instead? At ₹35 lakh per MVAr it is cheaper per unit, but it fades as \(V^2\):
Cheaper than Scheme C by ₹10 crore, and a defensible answer — provided the study's 0.85 pu is a firm floor. If the credible contingency set includes a fault that takes the busbar to 0.6 pu, the SVC yields \(65(0.36) = 23.4\) MVAr against the STATCOM's \(55(0.6) = 33\) MVAr, and the ranking reverses. The decision turns entirely on the depth of the worst credible dip, which is a study result, not a preference.
The headroom rule. Whatever dynamic device is chosen must not be used for steady-state duty:
The capacitor steps carry the daily swing and the STATCOM trims the residue between steps — that residue being at most half a step, 15 MVAr, comfortably within its slope range. If the STATCOM is allowed to drift to its capacitive limit in normal operation, the substation has paid ₹33 crore for a device that will be at its stop when the contingency arrives.
And the coordination that must accompany it. The three controls act at different speeds and the ordering must be preserved:
Each layer slower than the one beneath it. Where the separation is lost, the controls hunt and the equipment wears out chasing each other — and the tap-changer blocking is not optional, for the reason Challenge C1 works out in detail.
Practice Problems
Work each on paper before opening the answer. The answer is given so you can check yourself; the method is deliberately not.
P1. A 66 kV busbar has a fault level of 1200 MVA. A 20 MVAr bank is switched in. Estimate the voltage rise, and find the capacitance per phase for a star-connected bank at 50 Hz.
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Rise \(= 20/1200 = \mathbf{1.67\%}\); \(C = 20\times10^6/[2\pi(50)(66\times10^3)^2] = \mathbf{14.61\ \mu\text{F}}\).P2. A bank rated 40 MVAr at 132 kV finds its busbar at 125 kV. What does it deliver?
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\(40(125/132)^2 = \mathbf{35.87}\) MVAr — 10% short for a 5.3% dip. Problem 8.P3. A load of \(P = 1.0\), \(Q = 0.6\) pu is fed through \(X = 0.25\) pu from a 1.0 pu source. Find both solutions for the load-bus voltage.
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\(u^2-0.7u+0.085=0\) gives \(u = 0.5436\) or \(0.1564\), so \(V = \mathbf{0.7373}\) or \(\mathbf{0.3954}\) pu. Problem 3.P4. For the system of P3, find the maximum active power the line can deliver at that reactive demand, and the voltage there.
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\(P_{\max} = \sqrt{0.49/0.25-0.36} = \mathbf{1.2649}\) pu at \(V = \sqrt{0.35} = \mathbf{0.5916}\) pu. Problem 6.P5. Evaluate \(dV/dQ\) at the upper root of P3, and say what a 10 MVAr injection would buy on a 100 MVA base.
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\(V^2+XQ = 0.6936\), so \(dV/dQ = \mathbf{-0.6073}\) pu/pu — about \(\mathbf{+0.061}\) pu of voltage, though the exact quadratic gives rather less. Problem 4.P6. A generator with \(X_s = 1.0\) pu on a 1.0 pu busbar delivers 0.9 pu of active power. Find the excitation and load angle for 0.5 pu of reactive output.
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\(E\cos\delta = 1.5\), \(E\sin\delta = 0.9\), so \(E = \mathbf{1.749}\) pu at \(\delta = \mathbf{30.96^\circ}\). Problem 11.P7. An AVR has an open-loop gain of 60. What steady-state regulation does it give?
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\(1/(1+60) = \mathbf{1.64\%}\). Problem 13.P8. A 132 kV line with \(R = 20\ \Omega\), \(X = 50\ \Omega\) supplies 40 MW at 0.9 lagging, both busbars at 132 kV and \(t_st_r=1\). Find \(t_s\) and \(t_r\).
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\(Q = 19.373\), \(RP+XQ = 1768.6\), ratio 0.10151; \(t_s = \mathbf{1.0550}\), \(t_r = \mathbf{0.9479}\). Problem 14.P9. A 40 MVAr device sits on a busbar depressed to 0.6 pu. Compare an SVC at its capacitive limit with a STATCOM.
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SVC \(40(0.36) = \mathbf{14.4}\) MVAr; STATCOM \(40(0.6) = \mathbf{24.0}\) MVAr. Problem 19.P10. A 30 MVAr bank is proposed for a busbar with a 750 MVA fault level. Is there a harmonic objection?
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\(h = \sqrt{750/30} = \mathbf{5}\) — resonance exactly on the fifth harmonic. Yes; detune it. Problem 8.P11. A synchronous condenser with \(X_s = 1.8\) pu is excited to \(E = 2.5\) pu on a 1.0 pu busbar. Find its output.
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\(Q = V(E-V)/X_s = 1.5/1.8 = \mathbf{0.833}\) pu. Problem 17.P12. A transformer of admittance \(-j20\) pu has its tap on the bus-\(i\) side at \(a = 0.95\). Give the \(\pi\) equivalent.
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Series \(\mathbf{-j21.053}\); shunt at \(i\) \(\mathbf{-j1.108}\) (inductive); shunt at \(j\) \(\mathbf{+j1.053}\) (capacitive) — the mirror image of Problem 16.
Challenge Problems
Three problems that need an idea rather than a formula — the awkward corners where the chapter's rules turn against the engineer applying them.
C1 — When raising the tap lowers the voltage. A load behaving as a constant impedance at its low-voltage busbar is fed from a 1.0 pu source through a reactance \(X\) and an on-load tap changer of ratio \(n\), so that \(V_L = nV_t\). At \(V_L = 1.0\) pu the load is \(0.9 + j0.5\) pu. Obtain \(V_L(n)\) in closed form, find the tap ratio at which raising the tap begins to lower the customer voltage, and evaluate it for a strong and a weak feed.
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The closed form. A constant-impedance load draws \(P = P_0V_L^2\) and \(Q = Q_0V_L^2\). Substituting into the two-bus quadratic with \(u = V_t^2\) and \(V_L = nV_t\):
\[ u^2 + \Big(2XQ_0n^2u - V_s^2\Big)u + X^2\big(P_0^2+Q_0^2\big)n^4u^2 = 0 \]Every term carries a factor \(u\), so one power divides out and the quartic becomes linear in \(u\):
\[ V_t^{\,2} = \frac{V_s^2}{1 + 2XQ_0n^2 + X^2K\,n^4}, \qquad V_L^{\,2} = \frac{V_s^2\,n^2}{1 + 2XQ_0n^2 + X^2K\,n^4}, \qquad K = P_0^2+Q_0^2 \]Two consequences at once. First, a purely constant-impedance load always has a solution — there is no nose and no collapse, because the load's own demand falls as fast as the voltage. Second, the collapse mechanism must therefore come from something that restores the load, and the tap changer is exactly such a device.
The turning point. Write \(s = n^2\) and differentiate \(V_L^2 = V_s^2 s/(1+2XQ_0s+X^2Ks^2)\):
\[ \frac{d}{ds}\left(V_L^2\right) = 0 \;\Longrightarrow\; 1 - X^2Ks^2 = 0 \;\Longrightarrow\; n_{\text{crit}} = \frac{1}{\sqrt{X\sqrt{K}}} \]Remarkably, \(Q_0\) drops out of the turning-point condition altogether — only \(|S_0| = \sqrt{K}\) and \(X\) survive. With \(P_0 = 0.9\), \(Q_0 = 0.5\): \(K = 1.06\), \(\sqrt K = 1.0296\).
\[ \begin{array}{lccc} X\ (\text{pu}) & \text{SCR} = 1/(X|S_0|) & n_{\text{crit}} & \text{Inside a } \pm10\%\ \text{range?} \\ \hline 0.10 & 9.71 & 3.117 & \text{no, far outside} \\ 0.30 & 3.24 & 1.799 & \text{no} \\ 0.60 & 1.62 & 1.272 & \text{no, but close} \\ 0.90 & 1.08 & 1.039 & \textbf{yes — reached at the third step} \\ 1.20 & 0.81 & 0.900 & \text{yes — reversed at every tap} \end{array} \]The strong feed, \(X = 0.30\). The tap changer works exactly as intended:
\[ \begin{array}{ccc} n & V_L & V_t \\ \hline 1.000 & 0.8465 & 0.8465 \\ 1.050 & 0.8730 & 0.8314 \\ 1.100 & 0.8973 & 0.8158 \end{array} \]Eight steps raise the customer voltage by 5.1 percentage points — at the cost of 3.1 points at the transmission busbar, and of restoring the load from 0.645 to 0.724 pu of active power. The tap changer succeeded, and it made the upstream problem worse while doing so. That trade is present at every tap operation, everywhere, always.
The weak feed, \(X = 0.90\), a short-circuit ratio of 1.08 — a genuinely weak radial feed:
\[ \begin{array}{ccc} n & V_L & V_t \\ \hline 1.0000 & 0.60208 & 0.60208 \\ 1.0375 & 0.60267 & 0.58089 \\ 1.0500 & 0.60262 & 0.57393 \\ 1.1000 & 0.60135 & 0.54668 \end{array} \]The customer voltage peaks at \(n = 1.0389\) and falls thereafter. Beyond the third tap step the controller is doing harm: it sees a low voltage, raises the tap, and the voltage goes down. It then raises the tap again. The loop runs to the end of its range, dragging the transmission busbar down 5.5 percentage points for a customer-side gain of nothing at all.
Why the real events are worse than this model. Three effects the algebra above deliberately excludes:
\[ \begin{array}{ll} \text{Motor loads} & \text{restore toward constant power, so } K\ \text{itself rises as } V\ \text{falls} \\ \text{Field-current limiters} & \text{remove generator var support after 10–20 s of overload} \\ \text{Cascading} & \text{each tripped circuit raises } X\ \text{and lowers } n_{\text{crit}} \end{array} \]Each shifts \(n_{\text{crit}}\) downward into the tap range on systems that the static calculation would call safe. The standard countermeasure is unchanged since the 1980s: block or reverse tap-changer action when the transmission-side voltage falls below a threshold, typically 0.90 pu, and accept the customer-side undervoltage as the lesser harm.
The lesson in one line. A control loop whose plant gain reverses sign is not a badly tuned loop; it is a loop that must be taken out of service, and it must be taken out on a signal from the other side of the transformer, because on its own side everything still looks as though it needs more tap.
C2 — The Q–V curve and the reactive margin. A busbar draws \(P = 1.0\) pu and \(Q = 0.5\) pu through \(X = 0.22\) pu from a 1.0 pu source. Derive a closed form for the maximum reactive power the busbar can absorb, evaluate the reactive margin, and find what a contingency raising \(X\) to 0.30 pu does to it. State the planning conclusion.
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The construction. A Q–V study attaches a fictitious variable var source at the busbar, sweeps its output \(Q_{\text{inj}}\), and records the busbar voltage. The net reactive power drawn through the line is \(Q_{\text{net}} = 0.5 - Q_{\text{inj}}\), and each value is solved from the quadratic of Problem 3.
\[ \begin{array}{cc} Q_{\text{inj}}\ (\text{pu}) & V\ (\text{pu}) \\ \hline +0.2 & 0.8957 \\ +0.1 & 0.8655 \\ \ \ \ 0.0 & 0.8323 \\ -0.1 & 0.7949 \\ -0.2 & 0.7511 \\ -0.3 & 0.6957 \\ -0.4 & 0.6017 \\ -0.4164 & 0.5463\ \text{(the minimum)} \end{array} \]The curve turns back on itself at the bottom, and the minimum is the reactive margin: the additional megavars of demand the busbar can tolerate before no solution exists.
The closed form. The bottom of the curve is where the two roots merge, so set the discriminant to zero and solve for \(Q\) rather than for \(P\):
\[ \big(V_s^2-2XQ\big)^2 = 4X^2\big(P^2+Q^2\big) \;\Longrightarrow\; V_s^4 - 4XQV_s^2 + 4X^2Q^2 = 4X^2P^2 + 4X^2Q^2 \]The \(Q^2\) terms cancel exactly, leaving a linear equation:
\[ \boxed{\ Q_{\max} = \frac{V_s^{2}}{4X} - \frac{XP^{2}}{V_s^{2}}\ } \]Two clean limits: with no active power the busbar absorbs \(V_s^2/4X\), a quarter of the fault level; every megawatt of active transfer erodes that allowance quadratically.
Evaluate:
\[ X = 0.22:\quad Q_{\max} = \frac{1}{0.88} - 0.22 = 1.1364-0.22 = 0.9164\ \text{pu} \]\[ \text{margin} = 0.9164 - 0.5 = \mathbf{0.4164\ \text{pu}} = 41.6\ \text{MVAr on a 100 MVA base} \]And the voltage at the bottom, \(V = \sqrt{(V_s^2-2XQ_{\max})/2} = \sqrt{0.2984} = 0.5463\) pu — far below anything operable, which is the point: the margin is measured to the cliff edge, not to a comfortable operating point.
After the contingency, \(X = 0.30\) pu:
\[ Q_{\max} = \frac{1}{1.2} - 0.30 = 0.8333-0.30 = 0.5333\ \text{pu} \]\[ \text{margin} = 0.5333-0.5 = \mathbf{0.0333\ \text{pu}} = 3.3\ \text{MVAr} \]A 36% increase in reactance destroyed 92% of the margin. That non-linearity is the single most important fact about voltage stability planning, and it is why margins must be evaluated for the post-contingency network rather than scaled from the intact one. A busbar with 41.6 MVAr of margin looks entirely healthy; the same busbar one circuit down has 3.3 MVAr, which is less than the reactive demand of a single large motor starting.
The planning conclusion. Typical criteria require a post-contingency reactive margin of at least 5% of the busbar's own reactive demand — here \(0.05(0.5) = 0.025\) pu, which 0.0333 pu just satisfies — and many utilities apply a stiffer absolute floor as well. This busbar would be flagged: it passes numerically but with no allowance for load growth, for a second contingency, or for the error in the load model itself. The remedy is either a second circuit, which raises \(Q_{\max}\) by lowering \(X\), or local dynamic support, which lowers the \(Q\) the line must carry. The Q–V curve prices both on the same axis, which is why it is the standard tool.
C3 — Diagnosing a busbar that will not hold voltage. A 132 kV distribution grid substation runs 4% low at peak despite a 30 MVAr capacitor bank in service and its incoming transformer taps at maximum boost. Adding a second 30 MVAr bank raises the voltage by only 0.8%, where 1.5% was expected. Identify the four candidate explanations in order of likelihood, state the measurement that distinguishes each, and say what the disappointing result reveals about the busbar.
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Start from what the observation itself measures. The rule \(\Delta V/V = \Delta Q/S_{sc}\) can be inverted. The second bank delivered roughly \(30(0.96)^2 = 27.6\) MVAr and moved the busbar 0.8%:
\[ S_{sc}^{\text{implied}} = \frac{\Delta Q}{\Delta V/V} = \frac{27.6}{0.008} = 3450\ \text{MVA} \]If the expectation of 1.5% came from an assumed fault level of about 1840 MVA, then either the assumption was wrong or something else is absorbing the vars. That one division is the whole diagnostic, and it costs nothing.
Cause 1: the bank is not delivering its nameplate. Commonest and cheapest to check. At 0.96 pu it delivers 92% of rating before anything else is wrong; add a blown fuse on one or two cans in a large bank and the shortfall reaches 15–20% with no external symptom. Distinguish by measuring the bank's own current per phase and comparing with \(Q_{\text{rated}}V/(\sqrt3 V_{\text{rated}}^2)\); an unbalance protection relay reading is the same evidence.
Cause 2: the fault level is genuinely higher than assumed. If the network has been reinforced since the planning study, the busbar is stiffer than the model says — and a stiff busbar is hard to raise as well as hard to depress. This is the benign explanation: the voltage is low for a different reason and the capacitors were never the right remedy. Distinguish by a short-circuit study on the current network model, or by the measured voltage step when a known load is switched.
Cause 3: voltage-dependent load is eating the improvement. Raising the busbar raises the load — a constant-impedance component draws \(V^2\), so a 1.5% rise increases the reactive demand by 3%, which on a 200 MVAr load is 6 MVAr, a fifth of the bank. The improvement is real but partly self-cancelling. Distinguish by recording the feeder MW and MVAr before and after switching; a genuine load increase will be visible in both.
Cause 4: the busbar is close to its nose. The least likely and the most serious. On the flat part of the P–V curve the sensitivity is small, and the observed \(\Delta V/\Delta Q\) would be smaller than expected — but this cause predicts the opposite, a sensitivity larger than expected, because near the nose \(dV/dQ\) grows without bound. Distinguish by the sign of the discrepancy: an under-responsive busbar is a stiff busbar, and a stiff busbar is not near collapse.
And that last point is the answer to the final question. The disappointing result is reassuring. A busbar that moves less than expected for a given injection has a higher effective fault level than the model assumed, and by Problem 4 the sensitivity is large only where the operating point is close to the merger of the two roots. The observation is therefore evidence against a voltage-stability problem and in favour of causes 1 to 3 — a defective bank, a stale network model, or a load that grows when you feed it.
What to do about the 4%. If causes 1 and 2 are cleared, the busbar is low not because it lacks vars but because the whole upstream network is low, and no amount of local capacitance will fix a deficit that exists two voltage levels up. The tap changers at maximum boost are the giveaway: they have already moved all the voltage they can, and by Problem 16 they added no vars while doing it. The remedy lies upstream.
The general rule. When a reactive injection produces less voltage than predicted, invert the measurement to find the fault level it implies, and compare that with the model before touching anything else. A discrepancy in \(S_{sc}\) is a network-data problem; agreement with the model points at the plant. The two are diagnosed by one division and are almost never confused after it.
Multiple-Choice Questions
MCQ 1. On a transmission line with \(X \gg R\), the voltage magnitude difference between two busbars is set mainly by:
(a) \(P\) (b) \(Q\) (c) the line length (d) the angleShow answer
(b). \(\Delta V = (RP+XQ)/V\) and with \(X\gg R\) the \(XQ\) term dominates. Option (d) inverts the relation: the angle is the consequence of \(P\), not a cause of the magnitude. Problem 2.MCQ 2. The two-bus relation \(V^4+(2XQ-V_s^2)V^2+X^2(P^2+Q^2)=0\) has two positive roots because:
(a) the algebra is approximate (b) two physically distinct states satisfy the same power demand (c) one root is spurious (d) \(Q\) can be either signShow answer
(b). Both roots satisfy the equations exactly — the same power delivered at low voltage and high current, or high voltage and low current. Neither is spurious; the lower one is merely unstable. Problem 3.MCQ 3. Injecting reactive power equal to 2% of a busbar's fault level raises its voltage by about:
(a) 0.2% (b) 2% (c) 4% (d) 20%Show answer
(b) — \(\Delta V/V \approx \Delta Q/S_{sc}\), one for one. Problem 5.MCQ 4. The sensitivity \(dV/dQ\) becomes infinite when:
(a) \(Q = 0\) (b) \(V = V_s\) (c) \(V^2+XQ = V_s^2/2\) (d) \(X = 0\)Show answer
(c), which is exactly the condition that the two roots merge. Note (d) gives the opposite: with \(X = 0\) the sensitivity is zero, since an infinitely stiff busbar cannot be moved at all. Problem 4.MCQ 5. A shunt capacitor bank rated 40 MVAr delivers, at 0.9 pu voltage:
(a) 40 MVAr (b) 36 MVAr (c) 32.4 MVAr (d) 44.4 MVArShow answer
(c) — \(40(0.81)\). Option (b) is the trap: the output goes as \(V^2\), not as \(V\). That distinction is the whole of Problems 8 and 19. Problem 8.MCQ 6. The parallel resonant order of a shunt bank against the source is:
(a) \(S_{sc}/Q_C\) (b) \(\sqrt{S_{sc}/Q_C}\) (c) \(\sqrt{Q_C/S_{sc}}\) (d) \(Q_C/S_{sc}\)Show answer
(b). A larger bank on a weaker busbar resonates lower, which is where the strong harmonics live — so (c), which reverses the ratio, gets the danger exactly backwards. Problem 8.MCQ 7. A line carries exactly its surge impedance loading. Its net reactive balance is:
(a) generating (b) absorbing (c) zero (d) depends on the power factorShow answer
(c) — \(I^2X\) absorbed exactly equals \(V^2\omega C\) generated, which is the definition of SIL. Problem 9.MCQ 8. A generator's governor holds \(P\) constant while the field is increased. Then:
(a) \(\delta\) rises and \(Q\) falls (b) \(\delta\) falls and \(Q\) rises (c) both rise (d) both fallShow answer
(b). \(E\sin\delta\) is pinned by the governor, so a larger \(E\) forces a smaller \(\sin\delta\) and a larger \(E\cos\delta\) — hence more vars and a smaller angle. Reactive support and stability margin move together. Problem 11.MCQ 9. An AVR loop has three cascaded lags. Raising the forward gain:
(a) improves accuracy and stability (b) improves accuracy, worsens stability (c) worsens both (d) affects neitherShow answer
(b) — \(e_{ss} = 1/(1+K)\) falls while the closed-loop poles march toward the imaginary axis. Rate feedback is what resolves the conflict. Problems 12 and 13.MCQ 10. A tap-change calculation returns \(t_s = 1.24\) on a transformer with \(\pm10\%\) taps. The correct response is:
(a) specify a wider tap range (b) install reactive compensation (c) accept the nearest step (d) reverse the buck and boost endsShow answer
(b). A tap changer redistributes voltage and adds no vars; a demand outside its range is a statement about the reactive balance. Problem 14 showed local compensation cutting a 9.85% requirement to 3.53%.MCQ 11. At 0.5 pu busbar voltage, an SVC at its capacitive limit and a STATCOM of the same rating deliver, respectively:
(a) 50% and 50% (b) 25% and 50% (c) 50% and 25% (d) 25% and 25%Show answer
(b). At its limit the SVC is a susceptance and fades as \(V^2\); the STATCOM holds rated current and fades as \(V\). Problem 19.MCQ 12. During a slow voltage decline, on-load tap changers acting to restore distribution voltage:
(a) always help (b) have no effect on the transmission side (c) restore the load and deepen the transmission decline (d) inject reactive powerShow answer
(c). Restoring the customer voltage restores the customer load, and that restored load flows through the transmission network. Option (d) is the standing misconception the \(\pi\)-model shunts of Problem 16 encourage. Challenge C1.
Key Formulas
| Statement | Relation | Notes |
|---|---|---|
| In-phase drop | \(\Delta V = (RP+XQ)/V_r\) | Sets the magnitude difference |
| Quadrature drop | \(\delta V = (XP-RQ)/V_r\) | Sets the angle; second order in magnitude |
| Exact two-bus relation | \(V^4+(2XQ-V_s^2)V^2+X^2(P^2+Q^2)=0\) | Two positive roots; \(R\) neglected |
| Sensitivity | \(\dfrac{dV}{dQ} = \dfrac{X(V^2+XQ)}{V\left[V_s^2-2(V^2+XQ)\right]}\) | Negative upper branch, positive lower |
| Short-circuit-level rule | \(\Delta V/V \approx \Delta Q/S_{sc}\), \(S_{sc}=V^2/X\) | Good while \(V \approx 1\) and \(\Delta Q \ll S_{sc}\) |
| Nose of the P–V curve | \((V_s^2-2XQ)^2 = 4X^2(P^2+Q^2)\) | \(V_{\text{crit}}^2 = (V_s^2-2XQ)/2\) |
| Reactive limit at fixed \(P\) | \(Q_{\max} = \dfrac{V_s^2}{4X} - \dfrac{XP^2}{V_s^2}\) | Bottom of the Q–V curve; the margin |
| Shunt capacitor | \(Q_C = V^2\omega C = Q_{\text{rated}}V^2\) | Sizing needs iteration |
| Bank resonance | \(h = \sqrt{S_{sc}/Q_C}\) | Detune with 5–7% series reactor |
| Line reactive balance | \(Q_{\text{net}} = I^2X - V^2\omega C\), zero at SIL | \(\text{SIL} = V^2/Z_c\) |
| Generator output | \(P = \dfrac{EV}{X_s}\sin\delta\), \(Q = \dfrac{EV\cos\delta-V^2}{X_s}\) | \(E\sin\delta\) pinned by the governor |
| Synchronous condenser | \(Q = V(E-V)/X_s\) | Absorbing limit \(-V^2/X_s\) |
| AVR regulation | \(e_{ss} = 1/(1+K)\) | Rate feedback lifts the gain limit |
| Tap ratios | \(t_s^2\left[1-\dfrac{RP+XQ}{V_1V_2}\right] = \dfrac{V_2}{V_1}\), \(t_r = 1/t_s\) | Adds no vars; round to a real step |
| Achieved voltage at a fixed tap | \(V_2^2 - t_s^2V_1V_2 + t_s^2(RP+XQ)=0\) | Take the upper root |
| Off-nominal tap \(\pi\) | \(\dfrac{y}{a}\); shunts \(\dfrac{y(1-a)}{a^2}\) and \(\dfrac{y(a-1)}{a}\) | Shunts are modelling artefacts |
| TCR susceptance | \(B_L(\alpha) = \dfrac{2(\pi-\alpha)+\sin2\alpha}{\pi X_L}\) | \(\alpha\) from the voltage peak, \(\sigma = 2(\pi-\alpha)\) |
| STATCOM output | \(Q = \dfrac{V(V_{\text{conv}}-V)}{X_{\text{coupling}}}\), \(Q = VI_{\max}\) at the limit | Falls as \(V\), not \(V^2\) |
| OLTC control reversal | \(n_{\text{crit}} = 1/\sqrt{X\sqrt{P_0^2+Q_0^2}}\) | Constant-impedance load; Challenge C1 |
Common Mistakes
Answering a collapse question with the linear drop formula. \(\Delta V = (RP+XQ)/V\) has one solution; the phenomenon has two, and the nose exists only in the quadratic. Use the approximation for tap and capacitor sizing, the quadratic for anything about limits — Problems 3 and 6.
Taking a capacitor's output as proportional to \(V\). It goes as \(V^2\). A 10% dip costs 19% of the output, and at half voltage three-quarters of it is gone — Problems 8 and 19.
Sizing a capacitor bank without the iteration. Its output depends on the voltage it produces. The first pass always overstates the improvement; here by 0.35 percentage points, and a linear sensitivity extrapolation overstated it by 2.4 — Problem 7.
Expecting a tap changer to fix a reactive shortage. It moves voltage between two sides and generates nothing. A required ratio outside the physical range is a message about vars, not about the tap changer — Problems 14 and 16.
Quoting a tap ratio to four decimal places and stopping. Taps are discrete. The nearest step must be substituted and the achieved voltage recomputed, and the reciprocal condition \(t_st_r=1\) generally does not survive the rounding — Problem 15.
Reading the shunt branches of the off-nominal tap model as real reactive plant. They are an artefact of forcing an unsymmetrical two-port into a \(\pi\) shape, and they swap character when the tap is reversed — Problem 16.
Specifying an AVR by its steady-state regulation alone. The gain that meets a 0.2% specification may exceed the gain the loop's own lags permit, in which case no setting works and the loop itself must change — Problems 12 and 13.
Comparing reactive plant only at nominal voltage. Every device meets the specification there. The honest comparison is at 0.5 pu, where a capacitor and a STATCOM of equal rating differ by a factor of two — Problems 17 and 19.
Counting total installed MVAr as the reserve. A system with ample static capacitors and no dynamic plant has no reserve at all in the seconds after a fault, which is when reserve is defined — Problem 20.
Assuming an on-load tap changer always helps. Restoring the customer voltage restores the customer load, and on a weak feed raising the tap can lower the very voltage it is chasing — Challenge C1.
Scaling a reactive margin from the intact network. A 36% increase in reactance destroyed 92% of the margin in Challenge C2. Margins must be computed for the post-contingency case, never extrapolated.
Part 7 now has both halves of the control problem. Frequency was one number and one integral; voltage has been twenty numbers and a hierarchy of controllers that must be kept from fighting one another. What both share is that they are the operator's problem — decided in seconds and minutes, on a network whose plant is already installed.
Part 8 turns to the plant itself. Set 36 takes up the switchgear that has appeared throughout this set as a constraint — the circuit breakers whose rupturing capacity limits how much fault level a busbar may usefully have, and whose closing time decides whether a mechanically switched capacitor can be called dynamic reserve. The arc that a breaker must interrupt, the transient recovery voltage that decides whether it succeeds, and the ratings that follow from both, are the subject there.