Voltage and Reactive Power Control
Reactive power is what sets voltage, and it refuses to travel — so unlike frequency, which is one number for the whole grid, voltage is a separate problem at every busbar, solved by whatever source of vars happens to be nearby.
- Why frequency is a single system-wide number while voltage is a bus-by-bus problem, and what makes reactive power so expensive to transport.
- How the two-bus drop \(\Delta V \approx (RP+XQ)/V\) separates into a magnitude effect governed by \(Q\) and an angle effect governed by \(P\).
- The sensitivity \(dV/dQ\), the practical short-circuit-level rule \(\Delta V/V \approx \Delta Q/S_{sc}\), and where that sensitivity becomes infinite — the point of voltage collapse.
- The reactive balance of a generator, \(Q = (EV\cos\delta - V^2)/X_s\), and the AVR loop that regulates it — including why high gain and stability conflict.
- Why a shunt capacitor's output falls as \(V^2\) exactly when it is needed most, and how a STATCOM escapes that trap.
- How to compute the tap ratios \(t_s\) and \(t_r\) of a line's end transformers for a specified pair of busbar voltages.
- How the fast, slow and discrete controls are coordinated, and why an on-load tap changer can make a voltage collapse worse.
Why Voltage Is a Local Quantity
Chapter 33 was able to treat an entire interconnection as one machine with one speed. Every synchronous generator from one end of the system to the other runs at the same electrical frequency; a single measurement anywhere reports the health of the whole. Nothing of the kind is true of voltage. In a network of a thousand buses there are a thousand voltages, they differ from one another, and correcting one of them barely touches the rest. Understanding why is the key to the whole chapter.
The reason is that voltage is set by the local reactive balance, and reactive power cannot be moved any distance without destroying itself. Consider shipping power along a line of series impedance \(R+jX\) carrying current \(I\). The active power lost is \(I^2R\); the reactive power absorbed by the line's own reactance is \(I^2X\). On a transmission line \(X\) is typically five to fifteen times \(R\), so the reactive "loss" is an order of magnitude larger than the active loss for the same current. In per unit on a \(100\) MVA base, a line with \(R = 0.02\) and \(X = 0.20\) carrying \(1.0\) pu of current wastes \(2\%\) of the throughput as heat but consumes \(20\%\) of it as reactive power. Attempting to supply a distant load's vars from a central power station means paying that surcharge over and over.
Worse, 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; more current absorbs more vars. A system that tries to import reactive power over long distances is walking toward the cliff described in Section 34-3.
Two consequences follow immediately. First, reactive support must be installed where the reactive load is — a capacitor bank at the distribution substation is worth several times the same MVAr at the generating station. Second, there is no single controller of voltage; there is a hierarchy of local controllers, each responsible for its own bus, and the design problem is to keep them from fighting one another.
There is a further asymmetry. Frequency has memory: the kinetic energy stored in the rotors integrates the active power imbalance, so frequency drifts rather than jumping, and this gives control loops a few seconds of grace. Reactive power has no comparable storage in the network at large. Switch a large inductive load on and the voltage falls essentially instantaneously, limited only by the electromagnetic time constants of the machines. Voltage control must therefore be fast where it is automatic — the excitation system responds in tenths of a second — and the slow elements, tap changers and switched capacitors, act only on the settled value.
Finally, voltage matters for reasons of its own. Equipment insulation sets an upper limit; motor torque, which varies as \(V^2\), sets a lower one; transformer and generator losses rise at both extremes; and the load flow of Chapters 18 to 20 will simply not converge to an acceptable solution if the reactive resources are inadequate. Grid codes generally require transmission voltages to be held within about \(\pm5\%\) and distribution voltages within \(\pm6\%\).
The Voltage Drop and the P–Q Decoupling
The claim that \(Q\) controls voltage and \(P\) controls angle deserves a derivation rather than an assertion. Take the short-line model of Chapter 10: a sending bus at \(V_s\), a receiving bus at \(V_r\), a series impedance \(R+jX\), and a load \(P + jQ\) drawn at the receiving end. Choose \(V_r\) as the phase reference, so \(\mathbf{V}_r = V_r\angle 0^\circ\). Then
Separating the bracket into real and imaginary parts,
The in-phase component adds directly to the magnitude of \(\mathbf{V}_s\); the quadrature component, being at right angles, changes the magnitude only in second order but produces almost all of the angle between the two buses. Writing \(\Delta V\) for the in-phase drop and \(\delta V\) for the quadrature drop,
On a transmission line \(X \gg R\), so these reduce to \(\Delta V \approx XQ/V_r\) and \(\delta V \approx XP/V_r\). Reactive power flow produces the voltage magnitude difference; active power flow produces the angle difference. This is the same decoupling that justified the fast decoupled load flow of Chapter 20, seen here in its physical form.
Two corollaries are worth stating. First, if the receiving-end load is at unity power factor, \(Q=0\) and the magnitude drop on a purely reactive line vanishes to first order — a line can carry a great deal of active power with very little voltage droop, provided its reactive requirement is met locally. Second, the term \(XQ/V_r\) changes sign with \(Q\). A capacitive load (leading, \(Q\) negative) makes the receiving-end voltage *higher* than the sending end. That is the Ferranti effect of Chapter 14 in one line of algebra: a lightly loaded long line is dominated by its own shunt charging current, which is capacitive.
The distribution engineer's version of the same equation is the reason power-factor correction (Chapter 30) improves voltage as well as billing. Cancelling \(Q\) at the load removes the \(XQ/V_r\) term from every upstream circuit simultaneously, which is worth far more than the copper losses it also saves.
The Sensitivity dV/dQ
Control engineering needs more than the direction of an effect; it needs its size. How many megavars must be injected at a bus to raise its voltage by one percent? The answer is the sensitivity \(dV/dQ\), and it can be obtained exactly for the two-bus system.
Neglect \(R\) for clarity — it changes the arithmetic but not the structure. From the phasor relation, with \(V_s\) held fixed by a strong source behind reactance \(X\) and a load \(P+jQ\) at the receiving bus,
This is a quadratic in \(u = V^2\), so for any given \(P\) and \(Q\) there are two positive solutions for the receiving voltage. The upper root is the normal operating point; the lower root is a high-current, low-voltage solution that is physically real but unstable under load-restoring dynamics. As the load increases the two roots approach one another and finally merge — the nose of the P–V curve, and the point of voltage collapse.
Differentiate the quartic implicitly with respect to \(Q\), holding \(P\) and \(V_s\) fixed. Writing \(F = V_s^2V^2 - (V^2+XQ)^2 - (XP)^2 = 0\),
Here \(Q\) is the reactive power drawn through the line, so the derivative is negative at any healthy operating point: more reactive demand, lower voltage. Injecting vars locally reduces \(Q\) and therefore raises \(V\). The denominator vanishes when \(V^2 + XQ = V_s^2/2\), which is exactly the condition that the two roots of the quartic coincide — at that point the sensitivity is infinite and no finite amount of reactive injection stabilises the bus.
The exact expression is useful for understanding, but system operators need something they can apply on the spot. Take the special case \(P = 0\), \(Q\) small and \(V \approx V_s\). Then \(V^2+XQ \approx V^2\) and the denominator becomes \(V(V_s^2 - 2V^2) \approx -V^3\), so \(dV/dQ \approx -X/V\). Expressing the source strength by its short-circuit level \(S_{sc} = V^2/X\) — the three-phase fault MVA at that bus, which is exactly the quantity computed in Chapter 25 — gives a rule of remarkable simplicity:
Injecting reactive power equal to \(1\%\) of the fault level at a bus raises its voltage by about \(1\%\). A strong bus (high fault level) is hard to move and hard to disturb; a weak bus (low fault level, remote, radially fed) swings violently for small changes in reactive load. This single line explains why voltage problems concentrate at the ends of long radial feeders and at buses fed through high-impedance transformers.
Where the Reactive Power Comes From and Goes
Before choosing a controller it is worth listing the reactive accounts of the network, because voltage control is nothing more than balancing them bus by bus.
The series reactance of every line and transformer absorbs \(I^2X\), and does so as the square of the loading. A transformer with \(10\%\) reactance at full load absorbs \(10\%\) of its rating in MVAr, which is a large number that is easily forgotten.
The shunt capacitance of every line generates \(V^2\omega C\), and does so as the square of the voltage — essentially constant, since voltage varies little. A \(400\) kV overhead line generates roughly \(0.5\) to \(0.7\) MVAr per kilometre; a high-voltage cable, with capacitance an order of magnitude larger (Chapter 9), generates ten to twenty times that, which is why long a.c. cables need shunt reactors at both ends.
The crossover between the two, where the line's own generation exactly meets its own absorption, is the surge impedance loading of Chapter 14. Below SIL a line is a net source of vars and its far end tends to rise; above SIL it is a net sink and its far end sags. Load flow calculations in Chapter 18 confirm this every time.
Loads absorb. A composite distribution load runs at \(0.8\) to \(0.9\) power factor lagging, so it demands \(0.5\) to \(0.75\) MVAr per MW. Induction motors are the principal offenders, and their demand rises sharply when the voltage falls, because a motor is close to a constant-power device at its shaft and must draw more magnetising current at reduced voltage. This behaviour is the load-side mechanism of voltage collapse.
Generators both generate and absorb, according to their excitation, which is the subject of the next section. Their capability is bounded — by the field winding heating when overexcited and by end-region heating and steady-state stability when underexcited — so a generator is not an unlimited var source.
| Element | Role | Magnitude of \(Q\) | Varies as |
|---|---|---|---|
| Line series reactance | Absorbs | \(I^2X\) | Square of current |
| Line shunt capacitance | Generates | \(V^2\omega C\) | Square of voltage |
| Transformer reactance | Absorbs | \(I^2X\), up to \(0.1\) pu at full load | Square of current |
| Composite load | Absorbs | \(0.5\)–\(0.75\) MVAr per MW | Rises as voltage falls |
| Shunt capacitor bank | Generates | \(V^2/X_C\) | Square of voltage |
| Shunt reactor | Absorbs | \(V^2/X_L\) | Square of voltage |
| Generator / synchronous condenser | Either | \((EV\cos\delta-V^2)/X_s\) | Controlled by excitation |
| STATCOM | Either | \(V\,I_{\max}\) at the limit | Linearly with voltage |
Generator Excitation and the Automatic Voltage Regulator
The primary and most valuable source of controlled reactive power is the synchronous generator itself. From the machine model of Chapter 26, with terminal voltage \(V\), internal e.m.f. \(E\) behind synchronous reactance \(X_s\) and load angle \(\delta\), the complex power delivered is
The reactive expression is the whole of excitation control in one line. Raise the field current and \(E\) rises; if \(P\) is held constant by the governor, \(\delta\) adjusts so that \(E\sin\delta\) stays fixed, while \(E\cos\delta\) grows. The machine becomes overexcited and \(Q\) becomes positive — it delivers vars and behaves as a capacitor seen from the system. Reduce the field until \(E\cos\delta\) falls below \(V\) and \(Q\) turns negative: the machine is underexcited and absorbs vars like a reactor. The transition occurs at \(E\cos\delta = V\), which is unity power factor.
What the operator actually sets is not \(E\) but the terminal voltage, and it is regulated automatically. The automatic voltage regulator compares a rectified, filtered measurement of terminal voltage against a reference and drives the exciter accordingly.
Each element contributes a lag. The amplifier — once a rotating amplidyne, now a thyristor bridge — is fast, \(T_A\) of a few tens of milliseconds. The exciter, if it is a separate d.c. or a.c. machine, has \(T_E\) of a few tenths of a second; a static excitation system, fed from the machine terminals through a transformer and rectifier, removes this lag almost entirely. The generator field itself is the slowest element, with \(T_G\) equal to the open-circuit field time constant \(T'_{d0}\), several seconds unloaded but reduced to a fraction of a second by the closed loop.
The loop's difficulty is the classic one. With open-loop gain \(K = K_AK_EK_GK_R\), a step in the reference or a disturbance in the load leaves a steady-state error
so accuracy demands large \(K\). But three cascaded lags contribute up to \(270^\circ\) of phase lag, and the loop becomes oscillatory and then unstable well before \(K\) is large enough — Example 4 works out the exact limit for a representative set of time constants. The classical remedy is rate feedback: a stabilizing transformer wound on the exciter field produces a signal proportional to the derivative of the exciter output, \(sK_F/(1+sT_F)\), which supplies phase lead exactly where it is needed and permits a much higher forward gain. Modern regulators achieve the same with a proportional-integral-derivative structure, and integral action then removes the steady-state error altogether.
Two limiters complete the picture. The over-excitation limiter protects the field winding from prolonged overcurrent, and its operation during a system emergency removes reactive support at the worst possible moment — a recurring feature of voltage collapse incidents. The under-excitation limiter keeps the machine clear of the steady-state stability limit and of end-region heating. Together with the armature current limit they bound the generator capability curve, and every megavar of reactive planning must respect it.
Shunt Capacitors and Shunt Reactors
The cheapest reactive power in the system comes from a bank of capacitors connected in shunt. Its output is
Capacitors are cheap per MVAr, lossless to within a fraction of a percent, easy to install in modules and require no rotating plant. They have one serious defect, and it is written into the formula: their output falls as the square of the voltage. A bank rated \(30\) MVAr at nominal voltage produces only \(24.3\) MVAr at \(0.9\) pu — it deserts precisely when the voltage is low and the vars are needed. A capacitor supports a voltage that is already nearly correct; it cannot rescue one that is collapsing.
They also interact with the network in ways that must be checked. A bank of capacitance \(C\) behind a source of inductance \(L\) forms a parallel resonance at \(f_r = f_0\sqrt{S_{sc}/Q_C}\). If that frequency lands on a harmonic present in the load — the fifth and seventh from six-pulse converters are the usual suspects — the harmonic current is magnified and the bank may fail. Detuning reactors in series with each step move the resonance to a harmless frequency, and the same reactors limit the inrush when the step is switched. Chapter 39 returns to this.
Series capacitors are a different device with a different purpose: inserted in the line rather than across it, they cancel part of \(X\), which raises the transfer limit \(EV/X\sin\delta\) and reduces the drop \(XQ/V\). Their reactive output rises as \(I^2X_C\) — with load, not against it — which is exactly the desired behaviour, but they introduce sub-synchronous resonance risk with turbine-generator shafts and complicate protection. Chapter 38 treats them with the other series controllers.
Shunt reactors solve the opposite problem. On a long, lightly loaded EHV line, or on any cable of appreciable length, the line's own charging current raises the far-end voltage above the sending end (Chapter 14). A reactor connected at the line end absorbs \(V^2/X_L\) and cancels that rise. Reactors are switched in at light load — typically overnight — and out again as load returns, so the daily voltage-control routine of a transmission operator is largely a matter of switching reactors off in the morning and capacitors on, and reversing it at night.
Combining the short-circuit-level rule of Section 34-3 with the capacitor equation gives the size directly. To raise a \(132\) kV bus with a \(2000\) MVA fault level by \(2\%\), install about \(40\) MVAr. The estimate is good to about ten percent for small changes, and Example 2 checks it exactly.
Tap-Changing and Regulating Transformers
A transformer with a tapped winding does something no capacitor can: it changes the ratio between two voltage levels without generating or absorbing any reactive power itself. That distinction matters. Moving a tap redistributes voltage between the two sides; it does not add vars to the system. If the whole network is short of reactive power, tap changing shifts the problem rather than solving it — and Section 34-9 shows that it can make matters actively worse.
Two kinds exist. Off-circuit taps, usually \(\pm5\%\) in \(2.5\%\) steps, are set by hand with the transformer de-energised and compensate for the fixed position of a substation in the network. On-load tap changers (OLTC) switch under load using a diverter switch and transition impedance, and provide a wider range — commonly \(\pm10\%\) to \(\pm15\%\) in sixteen to thirty-two steps of \(1.25\%\) or so. The tap winding is placed on the high-voltage side, where the current is lower and the number of turns higher, giving finer resolution per turn.
Now the calculation the subject is known for. Consider a line of impedance \(R+jX\) with a transformer at each end, and let the busbar voltages outside the transformers be \(V_1\) at the sending end and \(V_2\) at the receiving end. Let \(t_s\) and \(t_r\) be the per-unit tap ratios, defined so that the line-side voltage at the sending end is \(t_sV_1\) and the line-side voltage at the receiving end is \(t_rV_2\). Both nominal ratios are already accounted for; \(t_s\) and \(t_r\) are the deviations from nominal.
The line drop, using the in-phase approximation of Section 34-2 with the receiving-end line voltage in the denominator, is
One equation cannot fix two unknowns, so a second condition is imposed: \(t_st_r = 1\). The reason is practical rather than mathematical. The two nominal ratios were chosen when the transformers were specified, and their product already performs the required transformation between the two voltage levels; requiring the deviations to be reciprocal means that whatever boost is applied at one end is matched by an equal buck at the other, so the overall transformation is unchanged and only the line voltage is raised. Substituting \(t_r = 1/t_s\) and multiplying through by \(t_s\),
\(P\) and \(Q\) are the three-phase load at the receiving end in MW and MVAr, \(R\) and \(X\) the per-phase line constants in ohms, and \(V_1\), \(V_2\) the line-to-line busbar voltages in kilovolts; the combination \((RP+XQ)/(V_1V_2)\) is then dimensionless. With no line drop the result collapses to \(t_s^2 = V_2/V_1\), as it must.
Two checks should always follow the arithmetic. First, is the required tap inside the physical range? A calculation that demands \(t_s = 1.25\) on a transformer with \(\pm10\%\) taps is telling you that the line needs reactive compensation, not a bigger tap changer. Second, taps are discrete: the nearest available step must be substituted and the achieved voltage recomputed, as Example 5 does.
A regulating transformer generalises the idea. A small booster transformer, its series winding in the line and its exciting winding fed from the same phase, injects a voltage in phase with the line voltage and so adjusts magnitude — the same effect as a tap, but obtainable without disturbing the main transformer and often used to trim one circuit of a parallel pair. Feed the exciting winding from the other two phases instead and the injected voltage is in quadrature; the magnitude is then almost unchanged but the angle shifts. Since Section 34-2 established that angle drives active power, such a phase-shifting transformer controls the megawatts flowing through a particular circuit — the only conventional device that does. It is the standard remedy where parallel paths of unequal impedance would otherwise share load in a ratio nobody wants.
Synchronous Condensers, SVC and STATCOM
A synchronous condenser is a synchronous machine running unloaded, its shaft connected to nothing, its excitation adjusted to make it draw or deliver reactive power exactly as the \(Q\) equation of Section 34-5 prescribes with \(P \approx 0\):
Overexcite it (\(E>V\)) and it supplies vars; underexcite it and it absorbs them. Its virtues are that the output is smoothly and continuously controllable across the full range from lagging to leading, that it contributes short-circuit level and inertia to a weak bus, and that under a voltage dip its output falls only linearly with \(V\) rather than as \(V^2\) — indeed a well-excited machine will briefly deliver far more than rated vars into a fault. Its vices are the losses of a spinning machine (a few percent of rating), the maintenance a rotating machine demands, and a response time of the order of a second, set by the field time constant.
Power electronics has largely displaced it. The static var compensator (SVC) is a shunt-connected combination of a thyristor-controlled reactor and fixed or thyristor-switched capacitor banks. Delaying the firing angle of the thyristors in series with the reactor chops the current, so the effective inductive susceptance is continuously variable; combined with fixed capacitance, the net susceptance can be varied smoothly from capacitive to inductive. Response takes one or two cycles. But the SVC remains an impedance: at its capacitive limit it is simply a capacitor, and its output still falls as \(V^2\).
The STATCOM escapes that limitation. It is a voltage-source converter connected to the bus through a coupling reactance, with a d.c. capacitor on the other side. Its control adjusts the magnitude of the converter's fundamental output voltage \(V_{\text{conv}}\); if that magnitude exceeds the bus voltage the device sources reactive power, and if it is less it sinks it, exactly as in the synchronous condenser but with no rotating part:
Its decisive advantage appears at the limit. The converter is limited by the current its semiconductors can carry, not by a fixed susceptance, so at low voltage it holds rated current and its output falls only as \(V\). At \(0.5\) pu voltage an SVC at its capacitive limit delivers \(25\%\) of rating; a STATCOM delivers \(50\%\). Response is a fraction of a cycle, and the footprint is a fraction of the SVC's because no large reactor is needed.
Both devices are given a deliberate droop slope of one to five percent in their control range, for exactly the reason the governors of Chapter 33 are given droop: several compensators on the same bus would otherwise fight for control of one voltage, and a small slope lets them share the duty in a defined proportion. The slope also keeps the device near the middle of its range in normal conditions, preserving headroom for the disturbance it was bought to handle.
| Device | Control | Speed | Output at \(0.5\) pu voltage | Adds fault level? |
|---|---|---|---|---|
| Fixed shunt capacitor | Switched, discrete | Seconds (mechanical) | \(25\%\) | No |
| Shunt reactor | Switched, discrete | Seconds | \(25\%\) | No |
| OLTC transformer | Discrete, ratio only | Tens of seconds | Adds no vars at all | No |
| Generator excitation | Continuous | \(0.1\)–\(1\) s | Limited by field | Yes |
| Synchronous condenser | Continuous | \(\approx 1\) s | \(\approx 50\%\), more transiently | Yes, plus inertia |
| SVC | Continuous | \(1\)–\(2\) cycles | \(25\%\) | No |
| STATCOM | Continuous | Sub-cycle | \(50\%\) | No |
Reactive-Power Planning and the Coordination of Controls
Given this catalogue, the planning question is what to install, how much, and where. Three principles organise the answer.
Compensate near the load. Every var supplied locally is a var that does not have to be pushed through the series reactance of the intervening network, and the saving compounds: the reduced current also reduces the \(I^2X\) absorbed upstream. This is why utilities install capacitors on distribution feeders and offer power-factor tariffs (Chapter 30) that push consumers to correct at the point of use.
Match the device to the duty. Reactive requirements divide into a large, slow, predictable component that follows the daily load curve, and a small, fast, unpredictable component that appears after a fault or a switching event. The first is met economically by switched capacitors and tap changers; the second demands continuous, fast plant — generator excitation, a STATCOM, an SVC or a synchronous condenser. Buying only the cheap kind leaves the system with no dynamic reserve, and it is the dynamic reserve that determines whether a disturbance is ridden through or becomes a collapse.
Preserve headroom. A compensator sitting at its capacitive limit in normal operation is a compensator that can do nothing when it is needed. Planning studies therefore constrain the steady-state output of dynamic devices to a fraction of rating, using static capacitors to carry the base requirement and freeing the fast plant to act.
Voltage stability studies quantify this with the \(Q\)–\(V\) curve at each critical bus: a fictitious var source is attached, its output varied, and the bus voltage traced. The minimum of that curve is the reactive margin — the megavars of additional demand the bus can tolerate before no solution exists. Planning criteria typically require a positive margin at every credible contingency.
Coordination between controls of different speeds is the remaining difficulty, and one instance of it deserves particular attention because it inverts the intuition of the whole chapter. During a slow voltage decline, the distribution-side voltage falls and each on-load tap changer dutifully acts to restore it. Restoring the customer voltage restores the customer load — a motor whose terminal voltage is brought back draws its full power again — and that restored load flows through the transmission network, deepening the transmission-side decline. The tap changer, doing exactly what it was designed to do, drives the system further toward collapse. Every major voltage-collapse event studied since the 1980s shows this signature, and the standard countermeasure is to block or reverse tap-changer action when the transmission voltage falls below a threshold.
Worked Examples
Problem. A load of \(P = 0.8\) pu, \(Q = 0.4\) pu is fed from an infinite bus of \(V_s = 1.0\) pu through a line of pure reactance \(X = 0.2\) pu. (a) Find both solutions for the receiving-end voltage. (b) Find \(dV/dQ\) at the operating point. (c) A shunt capacitor rated \(0.4\) pu MVAr at \(1.0\) pu voltage is connected at the load bus. Find the new voltage. (d) Find the maximum active power the line could deliver with \(Q\) held at \(0.4\) pu.
Solution (a). Use the quartic of Section 34-3 with \(u = V^2\):
The upper root is the operating point — already an unacceptable \(10.6\%\) below nominal. The lower root corresponds to the same power being delivered at five times the current; it satisfies the equations but no load-restoring control can hold the system there.
Solution (b). With \(V^2 + XQ = 0.80+0.08 = 0.88\):
Each \(0.01\) pu of extra reactive demand costs about \(0.0026\) pu of voltage — a weak bus by any standard.
Solution (c). If the capacitor supplied its full \(0.4\) pu the line would carry no reactive power at all, and
But the capacitor's output falls as \(V^2\), so at \(0.9868\) pu it delivers only \(0.4(0.9737)=0.3895\) pu and the line must still carry \(0.0105\) pu. Repeating the calculation with that residual gives \(V = 0.9846\); one further pass gives \(0.9842\), which is converged to three figures.
The voltage rises from \(0.894\) to \(0.984\) pu — a \(9\) percentage-point improvement bought entirely with local reactive support and no change anywhere else in the network. The linear estimate from part (b), \(0.894+0.259(0.4)=0.998\), overstates the gain, because the sensitivity weakens as the voltage recovers.
Solution (d). The two roots merge when the discriminant vanishes:
Had the reactive demand been \(1.0\) pu instead of \(0.4\), the same calculation gives \(P_{\max}=1.118\) pu — the transfer capability of the line is nearly halved by reactive burden alone, without a single ohm being added to \(X\).
Problem. A \(132\) kV substation busbar has a three-phase fault level of \(2000\) MVA. The voltage is to be raised by \(2\%\) with a shunt capacitor bank. Find the MVAr required, the capacitance per phase for a star-connected bank at \(50\) Hz, and verify the estimate from the source reactance.
Solution. By the short-circuit-level rule,
For a star-connected bank the three-phase output is \(Q_C = V_{LL}^2\,\omega C\), so
Check the rule directly. The Thévenin reactance behind the bus follows from the fault level:
The bank must be rated for the raised voltage, and its own action raises the voltage it sits on — a bank chosen for \(40\) MVAr at \(132\) kV delivers \(40(1.02)^2 = 41.6\) MVAr once connected, which is why the switching step is checked against the maximum permitted voltage rise, typically \(2\) to \(3\%\) per step.
Problem. A round-rotor generator with \(X_s = 1.2\) pu is connected to a \(1.0\) pu bus and delivers \(P = 0.8\) pu. Find the load angle and reactive output for excitation \(E = 2.0\), \(E = 1.1\) pu, and find the excitation at which the machine operates at unity power factor.
Solution. Since \(P\) is fixed by the governor, \(E\sin\delta = PX_s/V = 0.8(1.2)/1.0 = 0.96\) pu for every excitation level. That single relation does most of the work.
The first case delivers \(0.629\) pu of vars at a power factor of \(0.8/\sqrt{0.8^2+0.629^2}=0.786\) lagging; the second absorbs \(0.386\) pu at \(0.901\) leading. Nothing but the field current changed.
For unity power factor, \(Q=0\) requires \(E\cos\delta = V = 1.0\), and \(E\sin\delta = 0.96\) always, so
Note also what happened to the load angle: the underexcited machine sits at \(60.8^\circ\), far closer to the \(90^\circ\) steady-state stability limit of Chapter 28 than the overexcited one at \(28.7^\circ\). Reducing excitation to absorb vars therefore erodes stability margin, which is precisely what the under-excitation limiter exists to prevent.
Problem. An excitation system has \(T_A = 0.1\) s, \(T_E = 0.4\) s, \(T_G = 1.0\) s, with \(K_E = K_G = K_R = 1\) and \(T_R\) negligible. Find the largest amplifier gain \(K_A\) for which the loop is stable, the frequency of the resulting sustained oscillation, and the best steady-state regulation obtainable.
Solution. The characteristic equation is \(1 + K_AG(s) = 0\) with the three lags in cascade:
Build the Routh array. The first-column entry of the \(s^1\) row decides stability, because a sign change there means a root has crossed into the right half-plane:
At \(K_A = 19.25\) the \(s^1\) entry vanishes and the auxiliary equation formed from the \(s^2\) row gives the crossing frequency:
Even at the very edge of instability the loop cannot hold the terminal voltage closer than about \(5\%\), and a usable gain — say \(K_A = 8\) for a reasonable damping — leaves \(11\%\). This is the conflict that forces the rate-feedback loop of Section 34-5 into the design: the stabilizing transformer supplies phase lead near \(6\) rad/s, moves the stability boundary to several hundred, and an integral term then removes the residual error entirely.
Problem. A three-phase \(132\) kV line has \(R = 25\ \Omega\) and \(X = 66\ \Omega\) per phase. Tap-changing transformers are installed at both ends with \(t_st_r = 1\). The load at the receiving end is \(50\) MW at \(0.9\) power factor lagging, and both busbars are to be held at \(132\) kV. (a) Find \(t_s\) and \(t_r\). (b) If the taps are limited to \(\pm10\%\) in steps of \(1.25\%\), find the receiving busbar voltage actually achieved.
Solution (a). The reactive demand of the load:
The sending transformer must boost by \(9.34\%\) and the receiving transformer buck by \(8.54\%\). Check the arithmetic against the line itself: the line-side voltages are \(t_sV_1 = 144.33\) kV and \(t_rV_2 = 120.73\) kV, and the drop is \(2848.3/120.73 = 23.59\) kV. Since \(120.73+23.59=144.32\) kV, the solution is consistent.
Solution (b). The nearest available step below \(1.0934\) is \(1.0875\) (seven steps of \(1.25\%\)). With \(t_s\) fixed at that value, \(V_2\) becomes the unknown in the same relation:
The receiving busbar settles at \(130.25\) kV, or \(98.7\%\) of nominal — comfortably within a \(\pm5\%\) band, and as close as a discrete tap changer can come. Had the load been doubled, \((RP+XQ)/(V_1V_2)\) would rise to \(0.327\) and the requirement to \(t_s = 1.219\), which is outside the tap range entirely. The correct response then is not a larger tap changer but reactive compensation at the receiving end: injecting \(24.22\) MVAr locally removes the \(XQ\) term and drops the requirement to \(t_s^2 = 1/(1-1250/17424)\), that is \(t_s = 1.0379\) — well inside the range.
Problem. A \(33\) kV bus is supported by a \(30\) MVAr device. A contingency depresses the bus to \(31\) kV, and a severe fault momentarily depresses it to \(16.5\) kV. Compare the reactive output of (a) a fixed capacitor bank, (b) an SVC at its capacitive limit, and (c) a STATCOM rated for the same \(30\) MVAr, in each condition.
Solution. A capacitor and an SVC at its limit are both fixed susceptances, so their output varies as \(V^2\). A STATCOM at its limit is a constant-current source, so its output varies as \(V\).
At a mild depression the three are within \(6\%\) of one another and the capacitor, costing a small fraction of the alternatives, is plainly the right purchase. At half voltage the STATCOM delivers twice what the impedance devices manage, and it is in exactly that region — the seconds after a fault, when motors are stalled and drawing heavy magnetising current — that the outcome of the disturbance is decided. The comparison is the whole economic argument of reactive planning in miniature: buy static plant for the daily duty, dynamic plant for the emergency, and do not confuse the two.
Chapter Summary
Moving vars costs \(I^2X\), an order of magnitude more than the \(I^2R\) of active power. Compensate at the load.
\(\Delta V=(RP+XQ)/V\) sets the magnitude; \(\delta V=(XP-RQ)/V\) sets the angle. With \(X\gg R\), \(Q\) drives \(V\) and \(P\) drives \(\delta\).
\(\dfrac{dV}{dQ}=\dfrac{X(V^2+XQ)}{V[V_s^2-2(V^2+XQ)]}\); practically, \(\Delta V/V \approx \Delta Q/S_{sc}\).
The quartic has two roots; they merge at \(V^2+XQ = V_s^2/2\), where \(dV/dQ\) is infinite.
\(Q=(EV\cos\delta - V^2)/X_s\). Overexcited machines supply vars; underexcited absorb them, at the cost of stability margin.
\(e_{ss}=1/(1+K)\) demands high gain, three lags forbid it; rate feedback resolves the conflict.
\(t_s^2\left[1-\dfrac{RP+XQ}{V_1V_2}\right]=\dfrac{V_2}{V_1}\) with \(t_st_r=1\) — and taps add no vars at all.
Capacitors and SVCs fade as \(V^2\); condensers and STATCOMs only as \(V\). Dynamic reserve is what survives a disturbance.
Problems
Take \(f = 50\) Hz throughout. Where a per-unit system is used, state the base explicitly before starting — most of the difficulty in reactive-power problems is bookkeeping.
- A \(33\) kV feeder has \(R = 8\ \Omega\) and \(X = 12\ \Omega\). It supplies \(6\) MW at \(0.85\) power factor lagging. Find the sending-end voltage using the in-phase and quadrature drop components, and then repeat with the load corrected to unity power factor. Comment on the size of the two effects.
- A busbar has a fault level of \(750\) MVA at \(66\) kV. A \(15\) MVAr capacitor bank is switched in. Estimate the voltage rise, and find the capacitance per phase for a star-connected bank.
- A load of \(P = 1.0\) pu, \(Q = 0.6\) pu is fed through a line of reactance \(X = 0.25\) pu from a \(1.0\) pu source. Find both solutions for the load-bus voltage, the sensitivity \(dV/dQ\) at the operating point, and the reactive injection required to raise the voltage to \(0.98\) pu.
- For the system of Problem 3, find the maximum active power the line can deliver at the given reactive demand, and the voltage at that point. How much does the maximum rise if the load's reactive demand is fully compensated locally?
- A generator with \(X_s = 1.0\) pu on a \(1.0\) pu bus delivers \(0.9\) pu of active power. Find the excitation required to deliver \(0.5\) pu of reactive power, and the excitation at which the machine begins to absorb reactive power. State the load angle in each case.
- An excitation system has \(T_A = 0.05\) s, \(T_E = 0.5\) s and \(T_G = 1.0\) s with unity gains elsewhere. Find the maximum stable amplifier gain and the corresponding oscillation frequency, and state the best steady-state regulation available.
- A \(220\) kV line with \(R = 20\ \Omega\), \(X = 60\ \Omega\) supplies \(80\) MW at \(0.85\) power factor lagging. Tap changers at both ends satisfy \(t_st_r=1\) and both busbars are to be held at \(220\) kV. Find \(t_s\) and \(t_r\). If the receiving-end transformer's range is \(\pm10\%\), is the requirement met?
- A \(25\) MVAr device supports an \(11\) kV bus. Tabulate its output at \(11\), \(10\) and \(6\) kV for a fixed capacitor, an SVC at its capacitive limit and a STATCOM of the same rating. Which device would you specify if the bus feeds a large induction-motor load, and why?