Part 8 · Chapter 39

Power Quality and Renewable Integration

The grid no longer supplies a clean sinusoid to passive loads from spinning machines: power electronics distort the waveform at both ends of the wire, and converter-interfaced generation removes the rotating inertia that used to hold frequency steady — so quality and integration are two halves of one problem.

Electric Power Systems Prof. Mithun Mondal Reading time ≈ 55 min
i What you'll learn
  • Why voltage distortion is bought, not made: harmonic currents drawn by one customer become harmonic voltages for every customer through the shared source impedance.
  • How a fault three feeders away becomes a voltage sag at your terminals, and how the ITIC envelope decides whether the process survives it.
  • The negative-sequence unbalance factor \(|V_2|/|V_1|\) of Chapter 22 used as a quality index, and why a 2% unbalance is a serious matter for an induction motor.
  • The difference between THD and TDD, why only the second can be written into a contract, and how the IEEE 519 tables split responsibility between utility and customer.
  • Passive and active filters, the DVR, the DSTATCOM and the UPS — what each corrects and what it cannot.
  • The Betz limit \(C_p \le 16/27\) derived from momentum, the turbine power curve, and the DFIG and full-converter topologies.
  • The PV \(I\)–\(V\) curve, the maximum-power-point condition \(dI/dV = -I/V\), and what MPPT actually searches for.
  • Why displacing synchronous machines raises RoCoF, how \(df/dt = f_0\,\Delta P / 2E_{\text{kin}}\) follows from Chapter 27's swing equation, and what synthetic inertia can and cannot buy back.
Section 39-1

What Power Quality Means, and What It Costs

Every chapter so far has treated the supply as a balanced, undistorted, constant-frequency set of three sinusoids. Chapter 3 built the single-line diagram on that assumption; Chapter 22 relaxed the balance for the duration of a fault and then restored it; Chapter 33 allowed the frequency to wander by a fraction of a hertz and pulled it back. The assumption is an excellent one for a system of synchronous machines feeding motors, heaters and incandescent lamps. It has been quietly failing for forty years.

The load has changed. A modern factory draws its current through rectifiers, a modern office through switch-mode supplies, a modern railway through single-phase converters. None of these draws a sinusoid. At the same time the generation has changed: a wind farm or a photovoltaic plant reaches the network through an inverter, not through a shaft. Power quality is the name given to the whole family of deviations that result — in waveform, in magnitude, in balance and in frequency — together with the question of whether connected equipment can tolerate them.

It is worth being precise about the definition, because two different things are being measured. The voltage at a customer's terminals is the utility's product, and its quality is the utility's concern. The current the customer draws is the customer's product, and its distortion is the customer's concern. The two are not independent, and the link between them is the single most important equation in this half of the chapter. A harmonic current \(I_h\) of order \(h\) flowing out of a distorting load must flow back through the system impedance seen at the point of common coupling, and in doing so it produces a harmonic voltage there:

How a current problem becomes a voltage problem
\[ V_h = I_h\,Z_h \;\approx\; I_h\,\big(h\,X_s\big), \qquad X_s = \frac{V_{LL}^2}{S_{sc}} \]

The source reactance \(X_s\) is inversely proportional to the short-circuit level \(S_{sc}\) at the bus, which Chapter 25 taught us to compute. Two consequences follow immediately. A stiff bus — high fault level, low \(X_s\) — converts harmonic current into very little harmonic voltage, so a distorting load on a strong system disturbs nobody. A weak bus does the opposite. And because the impedance rises with \(h\), a high-order harmonic of modest amplitude can produce as much voltage distortion as a large fifth harmonic.

That equation also explains why power quality is a shared problem rather than a private one. The distorting customer suffers only mildly from the voltage he creates; his neighbours on the same bus, who may have installed nothing more offensive than a motor, suffer equally. Regulation therefore takes the form of limits on current injected by each customer and limits on voltage delivered by the utility, and Section 39-5 shows how the two are made to fit together.

🔑
Definition
Power quality is the degree to which the voltage waveform at a point of supply matches an ideal balanced sinusoid of constant magnitude and frequency, and the degree to which the current drawn there matches the same ideal.

A disturbance matters only if some load misbehaves because of it. A 3% voltage sag is invisible to a pump and fatal to a semiconductor stepper. Quality is therefore always assessed against equipment compatibility, never in the abstract.

The economics are what force the subject onto an engineer's desk. Chapter 30 costed energy in paise per kilowatt-hour. A power-quality event is not costed that way at all: the missing energy is negligible and the consequence is enormous. A 150 ms sag that trips a continuous-process plant destroys the material in the line, requires a cleaning and restart cycle of several hours, and delivers nothing for that time. The standard way to write this is

Annual cost of poor quality at one site
\[ C_{\text{annual}} = \sum_{k} N_k \Big( C_{\text{lost output}} + C_{\text{scrap}} + C_{\text{restart}} + C_{\text{equipment}} \Big)_k \]

where \(N_k\) is the expected number of events of type \(k\) per year, obtained from the site's fault statistics. For a plant of a few tens of megawatts a single deep sag routinely costs more than the entire annual electricity bill for one day of production, and a plant that sees twenty such events a year has an unanswerable case for mitigation. Surveys across industry consistently find that voltage sags and short interruptions account for the large majority of the total cost of poor quality — not harmonics, which are chronic rather than acute, and which mostly show up as extra losses and shortened equipment life.

The vocabulary is standardised. The categories below are those of IEEE 1159, and the rest of the first half of the chapter takes them in turn.

CategoryTypical durationTypical magnitudeUsual origin
Impulsive transient\(< 1\) msup to several puLightning, switching of inductive loads
Oscillatory transient0.3 – 50 ms1.2 – 2.0 puCapacitor bank energisation, cable switching
Sag (dip)0.5 cycle – 1 min0.1 – 0.9 puRemote faults, motor starting, transformer inrush
Swell0.5 cycle – 1 min1.1 – 1.8 puHealthy phases during an L-G fault, load rejection
Interruption0.5 cycle – 1 min (momentary/temporary)\(< 0.1\) puBreaker operation, auto-reclose sequence
Sustained interruption\(> 1\) min0 puPermanent fault, planned outage
Harmonic distortionsteady state0 – 20% THDConverters, saturated iron, arc devices
Voltage unbalancesteady state0.5 – 3%Single-phase loads, untransposed lines
Flickerintermittent, 0.5 – 25 Hz0.1 – 7% \(\Delta V/V\)Arc furnaces, welders, sawmills
Frequency deviationseconds to minutes\(\pm 0.5\) Hz and beyondGeneration–load imbalance (Chapter 33)
The same device is both culprit and victim. A variable-speed drive draws a badly distorted current, and its own DC-link undervoltage relay trips on a sag that a directly connected induction motor would ride through without noticing. The spread of power electronics has therefore increased both the amount of disturbance in the network and the sensitivity of the equipment connected to it — which is precisely why the subject grew from a curiosity into a discipline.
Section 39-2

Sags, Swells, Interruptions and the ITIC Curve

A voltage sag is a reduction of rms voltage to between 0.1 and 0.9 pu lasting from half a cycle to a minute. Almost every sag has the same cause: a short circuit somewhere on the network. Chapter 21 computed the current such a fault draws; the same calculation, read differently, gives the voltage that everyone else sees while it lasts.

Consider a load at a substation bus, and a fault on a feeder leaving the same bus. Let \(Z_S\) be the source impedance behind the bus and \(Z_F\) the impedance of the faulted feeder between the bus and the fault. During the fault the bus is simply the mid-point of a divider between the source e.m.f. and the short circuit:

The voltage-divider model of a sag
\[ V_{\text{sag}} = \frac{Z_F}{Z_S + Z_F}\,E \]

Everything that matters about sags is contained in this one expression. A fault close to the bus makes \(Z_F\) small and the sag deep; a distant fault leaves \(Z_F\) large and the sag shallow. A strong source — small \(Z_S\), high fault level — keeps the retained voltage high. And because \(Z_F\) grows with distance along the feeder, there is a definite vulnerability area: the set of fault positions on the network that would drive the voltage at this bus below the level at which this particular load trips. Multiply the length of that area by the fault rate per kilometre per year and the expected number of trips per year falls out. Example 4 carries the calculation through.

The duration of a sag is not set by the fault at all but by the protection of Chapter 36. A distribution feeder cleared in five cycles gives a 100 ms sag; a transmission fault cleared in three cycles gives a 60 ms sag; a fault cleared by backup after the main protection fails gives half a second. Faster protection is therefore power-quality mitigation, which is one reason numerical distance relays displaced electromechanical ones.

Sags are also not symmetric. A single line-to-ground fault — the commonest kind — produces a deep sag on the faulted phase and only a small change on the other two. What the load sees, though, depends on the transformers in between: Chapter 23 showed that a delta–star transformer blocks zero sequence, so a single-phase sag upstream appears downstream as a two-phase sag with a phase shift. A three-phase load therefore rarely sees the sag that a meter on the faulted circuit records, and sag classification (the "type A to type G" scheme) is nothing more than bookkeeping of which transformer connections lie between the fault and the load.

🔑
Sag depth, sag duration
\[ V_{\text{sag}} = \frac{Z_F}{Z_S+Z_F}E \quad\text{(set by the network)}, \qquad t_{\text{sag}} = t_{\text{relay}} + t_{\text{breaker}} \quad\text{(set by the protection)} \]

Depth and duration are decided by two entirely separate parts of the system, which is why a sag is always quoted as a pair of numbers and never as one.

A swell is the mirror image: an rms rise above 1.1 pu. The usual source is again a single line-to-ground fault, but seen on the healthy phases of an ungrounded or high-impedance-grounded system, where the neutral shifts and the sound phases rise toward line-to-line voltage — a factor of \(\sqrt3\) in the extreme. Load rejection produces the other common swell: a large block of load disconnects, the machines' excitation has not yet responded, and the voltage rises until the AVR of Chapter 34 pulls it back.

An interruption is a sag taken to its limit — below 0.1 pu. The important distinction is between the momentary interruption caused by an auto-reclose sequence (the breaker opens, waits half a second, and recloses onto a cleared arc) and the sustained interruption of a permanent fault. From the utility's reliability indices — SAIFI, SAIDI, CAIDI — momentary interruptions are usually excluded, since the supply is restored within a second. From the customer's process they are indistinguishable from an outage: a contactor whose coil is fed from the same supply drops out in about 20 ms and does not pick up again by itself.

What decides whether equipment survives a given depth-duration pair is not a formula but a measured envelope. The ITIC curve — the Information Technology Industry Council's revision of the older CBEMA curve — plots retained voltage against event duration and draws two boundaries. Between them, equipment is expected to operate normally. Above the upper boundary the overvoltage is expected to damage it; below the lower boundary the equipment may drop out, though it should not be damaged.

100% nominal 0100200 300400500 voltage (% of nominal) 1 ms10 ms100 ms 1 s10 s100 s event duration (log scale) 200%140%120% 70%80%90% 0% up to 20 ms prohibited — overvoltage damage acceptable — equipment rides through 58.5% for 150 ms — trips 80%, 100 ms — rides through
The ITIC compatibility envelope, with the sag of Example 4 plotted against it

The shape of the lower boundary repays attention. Below about 20 ms the boundary sits at zero: equipment is expected to survive a complete loss of voltage for one cycle, because the smoothing capacitor of any electronic supply holds the DC rail up for at least that long. From 20 ms to 0.5 s the boundary is 70%, from 0.5 s to 10 s it is 80%, and thereafter 90%. That is the whole curve, and a sag is judged simply by whether its point lies inside it.

Two cautions. First, the ITIC curve describes single-phase 120 V computer equipment; it is used far more widely than that, and industrial drives and contactors are frequently less tolerant than it suggests, tripping at 80–85% retained voltage. The SEMI F47 specification, which requires ride-through down to 50% for 200 ms, is the tighter standard used in semiconductor plants. Second, the curve says nothing about how often an excursion may occur — that is a reliability question, answered by counting events per year.

Why sags cannot be eliminated, only survived. Faults on a large network are irreducible: lightning, wind-borne debris, animals and excavators will always produce a few hundred per year on a distribution system of any size. The voltage divider guarantees that every one of them is seen as a sag by every customer inside its vulnerability area. Mitigation is therefore always about the load side — riding through, not preventing — and Section 39-6 is where the equipment for doing so appears.
Section 39-3

Transients and the Unbalance Factor

A transient is a disturbance measured in microseconds or milliseconds rather than in cycles, and it is classified by whether it rings. An impulsive transient is unidirectional: a lightning stroke to a line, or the collapse of the field of an inductive circuit when a contact opens. The standard test waveshape for the first is the 1.2/50 μs voltage impulse and the 8/20 μs current impulse of Chapter 37 — a front of 1.2 μs and a decay to half value at 50 μs. Its energy is small but its rate of rise is enormous, and it is the insulation, not the load, that is at risk. Surge arresters and the insulation coordination of Chapter 37 are the answer.

An oscillatory transient rings at a frequency set by the network's own inductance and capacitance. The archetype is capacitor-bank energisation. At the instant of closing, the capacitor voltage is zero while the system voltage is at some value \(V_0\); the capacitor must charge through the source inductance, and the circuit is a series \(LC\) loop:

Capacitor switching transient
\[ f_{\text{tr}} = \frac{1}{2\pi\sqrt{L_s C}} = f_1\sqrt{\frac{S_{sc}}{Q_c}}, \qquad v_C(t) = V_0\big[1-\cos(2\pi f_{\text{tr}} t)\big] \]

The cosine term is the whole story: an undamped \(LC\) charging transient overshoots to twice the driving voltage. Real circuits have resistance and the peak is 1.5–1.8 pu, at a frequency of a few hundred hertz to a few kilohertz. That is tolerable at the switched bus. The danger is voltage magnification: if a low-voltage power-factor-correction capacitor downstream, together with its transformer's leakage inductance, happens to be series-resonant near \(f_{\text{tr}}\), the transient at the customer's terminals can reach 3–4 pu and destroy the drive's DC-link capacitors. Pre-insertion resistors, synchronous closing at a voltage zero, and detuning the downstream bank are the three remedies.

🔑
Energisation overshoot
An uncharged capacitor switched onto a source through inductance rings to \(2V_0\) in the lossless limit, at \(f_{\text{tr}} = f_1\sqrt{S_{sc}/Q_c}\).

With \(S_{sc}=200\) MVA and \(Q_c=8\) MVAr the ring is at \(50\sqrt{25} = 250\) Hz. The same square root reappears in Section 39-5 as the harmonic resonance order, which is no coincidence — it is the same resonance, excited in one case by a step and in the other by a steady harmonic.

The remaining steady-state defect is unbalance. Its causes are all structural: single-phase loads that cannot be distributed evenly (traction, arc welders, rural distribution), untransposed lines whose phase inductances differ (Chapter 6), a blown fuse in one phase of a capacitor bank, or a single open conductor. Chapter 22 gave the exact instrument for measuring it. Resolve the three phase voltages into their symmetrical components and take the ratio of the negative-sequence magnitude to the positive-sequence magnitude:

The voltage unbalance factor
\[ V_2 = \tfrac13\big(V_a + a^2V_b + aV_c\big), \qquad V_1 = \tfrac13\big(V_a + aV_b + a^2V_c\big), \qquad a = 1\angle120^\circ \]
\[ \text{VUF} \;=\; \frac{|V_2|}{|V_1|}\times 100\% \]

This is the definition used by IEC 61000-4-30 and it is the one that predicts damage, for a reason that is worth stating carefully. In a three-phase induction machine the positive-sequence voltage drives the rotor forward; the negative-sequence voltage drives a field rotating backwards at synchronous speed, which the rotor sees at a slip of very nearly 2. The machine's negative-sequence impedance is therefore its locked-rotor impedance, typically one sixth of the positive-sequence impedance at rated slip. A negative-sequence voltage of 2% consequently drives a negative-sequence current of about 12% of rated — flowing in a rotor whose resistance is high at double frequency, and producing braking torque and heat rather than useful output.

🔑
Why 2% is the limit
\[ \frac{I_2}{I_1} \approx \frac{V_2/Z_2}{V_1/Z_1} = \text{VUF}\times\frac{Z_1}{Z_2} \approx 6\,\text{VUF} \]

The extra copper loss goes as \(I_2^2\), so at 2% unbalance the additional rotor and stator heating is enough to require a derating of roughly 5%, and at 5% unbalance the machine must be derated by about 25% or it will fail thermally. Most standards therefore cap steady-state unbalance at 2%.

A cruder index is often quoted and should be recognised: the NEMA voltage unbalance factor, defined as the maximum deviation of the three line voltages from their average, divided by that average. It requires no phasor measurement, only three voltmeter readings, which is its entire appeal. It agrees with the sequence definition only in special cases — Example 3 exhibits one, in which the two differ by exactly \(\sqrt3\) — and it is blind to unbalance that lies in the angles rather than the magnitudes. When a number is written into a contract, it should be the sequence one.

Chapter 22 was not only about faults. Symmetrical components were introduced to handle a momentary asymmetry lasting five cycles. Here the same transformation is applied to a permanent, small asymmetry and turned into a continuously monitored quality index. The mathematics does not care about the size or duration of the imbalance — which is why a single tool covers both a bolted line-to-ground fault and a 1.5% steady-state distortion of the supply.
Section 39-4

Harmonics: Sources, THD and TDD

Harmonics are the steady-state face of power quality. Any load whose current is a periodic but non-sinusoidal function of time at the supply frequency can be expanded in a Fourier series, and the terms above the fundamental are the harmonic currents. For a waveform with half-wave symmetry — \(i(t+T/2) = -i(t)\), which every symmetric converter produces — the even harmonics vanish identically, so in practice one deals with the odd orders only.

The three-phase structure imposes a second and more useful classification. Take a balanced set of harmonic currents of order \(h\), in which phase \(b\) lags phase \(a\) by \(120^\circ\) at the fundamental. At order \(h\) that displacement becomes \(120h\) degrees, and the sequence of the harmonic set follows from \(h \bmod 3\):

Harmonic orderOrdersSequenceConsequence in a three-phase system
\(h = 3k+1\)1, 4, 7, 10, 13, 16, 19 …PositiveRotates with the fundamental; adds to motor heating
\(h = 3k+2\)2, 5, 8, 11, 14, 17, 20 …NegativeRotates backwards; braking torque, extra rotor loss
\(h = 3k\)3, 6, 9, 12, 15 …ZeroIn phase in all three lines; sums in the neutral, trapped in a delta

The zero-sequence row is the one that surprises people. Third-harmonic currents in the three lines are identical, not displaced, so they do not cancel at the star point: they add. A four-wire office circuit feeding switch-mode supplies can carry a neutral current larger than any line current, which is why neutral conductors in such installations are oversized and why a delta winding is deliberately placed in the supply transformer — the triplens circulate inside it and never reach the network. The delta winding is not free: those circulating currents heat it, and a transformer supplying heavy non-linear load must be derated by its K-factor.

To quantify distortion with a single number, compare the rms of everything above the fundamental with the fundamental itself:

Total harmonic distortion
\[ \text{THD}_V = \frac{\sqrt{\sum_{h=2}^{H} V_h^2}}{V_1}, \qquad \text{THD}_I = \frac{\sqrt{\sum_{h=2}^{H} I_h^2}}{I_1}, \qquad I_{\text{rms}} = I_1\sqrt{1+\text{THD}_I^2} \]

The last identity follows from orthogonality: harmonics of different orders are orthogonal over a period, so their rms values add in quadrature. It has an immediate consequence for power factor. Suppose the supply voltage is sinusoidal — a good approximation on a stiff bus — so that only the fundamental current can transfer average power. Then \(P = V_1I_1\cos\varphi_1\) while \(S = V_1 I_{\text{rms}}\), and

Power factor with a distorted current
\[ \text{PF} = \frac{P}{S} = \underbrace{\frac{I_1}{I_{\text{rms}}}}_{\text{distortion factor}}\cdot\underbrace{\cos\varphi_1}_{\text{displacement factor}} = \frac{\cos\varphi_1}{\sqrt{1+\text{THD}_I^2}} \]

Chapter 30 improved power factor by cancelling \(\varphi_1\) with capacitors. That machinery is powerless against the distortion factor: a six-pulse drive with a displacement factor of unity and 31% current distortion still has a true power factor of only \(1/\sqrt{1.0967}=0.955\), and adding capacitors makes it worse, not better, because the capacitor is a low impedance to harmonics and will draw them in. This is the first place where the two halves of the subject collide.

Where does the distortion come from? Three families cover almost everything.

Line-commutated converters. A six-pulse bridge with a large smoothing inductance draws a quasi-square current of \(120^\circ\) conduction. Its Fourier series contains only the orders \(h = 6k\pm1\) — that is 5, 7, 11, 13, 17, 19, … — with amplitudes \(I_h = I_1/h\) in the idealised case. The rule generalises: a \(p\)-pulse converter produces \(h = pk\pm1\), so a twelve-pulse arrangement (two six-pulse bridges fed from a star and a delta secondary, \(30^\circ\) apart) cancels the 5th, 7th, 17th and 19th and leaves 11, 13, 23, 25. Pulse number is the cheapest harmonic control there is, and Chapter 38 built HVDC converters on exactly this principle.

Single-phase rectifiers with capacitor filters. Every computer, LED driver and television draws current in a narrow spike near the peak of each half-cycle. The spectrum is dominated by the 3rd harmonic, at 60–80% of the fundamental in an uncorrected supply, and it is zero sequence. Individually trivial, these loads are numerous, and in aggregate they are the main source of triplen problems in commercial buildings.

Saturating iron and arc devices. A transformer driven above the knee of its magnetising curve draws a peaky exciting current rich in the 3rd and 5th; this is normally negligible, but a sustained overvoltage of a few per cent moves the operating point sharply up the curve and the exciting current can multiply several-fold. An arc furnace is worse than any of these. The arc is a non-linear, time-varying and randomly fluctuating resistance, so it produces not only odd harmonics but even harmonics and a continuous background of interharmonics at non-integer multiples of the supply frequency — which is exactly what causes the flicker of Section 39-5.

envelope Iₕ = I₁ / h 100% 20.014.39.1 7.75.95.3 4.34.0 157 111317 192325 harmonic order h (only h = 6k ± 1 present) % of fundamental THD_I = 29.0% (to h = 25), 31.1% (to ∞)
Spectrum of the ideal six-pulse converter current — the reference case for Examples 1 and 2

THD has one serious defect as a contractual quantity, and it is arithmetic rather than physical. The denominator is the present fundamental current. A drive running at one third of its rating draws one third of every harmonic, so its THD is unchanged — while the harmonic current it actually pushes into the network, and therefore the voltage distortion it creates, has fallen to one third. A load could be made to look compliant simply by being lightly loaded, or non-compliant by being efficient. The fix is to normalise against a fixed quantity, the maximum demand load current \(I_L\) at the point of common coupling, averaged over the fifteen or thirty minutes used for billing:

🔑
Total demand distortion
\[ \text{TDD} = \frac{\sqrt{\sum_{h\ge2} I_h^2}}{I_L}\times 100\%, \qquad \frac{\text{TDD}}{\text{THD}_I} = \frac{I_1}{I_L} \]

TDD and THD coincide at full load and diverge everywhere else. Because \(I_L\) is fixed by the connection agreement, TDD is proportional to the harmonic current actually injected — which is the quantity that produces voltage distortion for the neighbours. Every current limit in IEEE 519 is a TDD limit, never a THD limit.

Section 39-5

Limits, Resonance and Flicker

The whole regulatory structure rests on the equation of Section 39-1, \(V_h = I_h\,hX_s\). If every customer keeps his injected harmonic current below a share of what the bus can absorb, the resulting voltage distortion stays inside the level that equipment is designed for. IEEE 519 implements this as two tables applied at the point of common coupling — the nearest point on the network electrically common to this customer and at least one other. Distortion inside a plant, on a bus that only that plant uses, is nobody else's business.

The customer's current limits are graded by the stiffness of the bus, expressed as the ratio of the available short-circuit current to the maximum demand load current, \(I_{sc}/I_L\). A customer who is small compared with the bus he is connected to may distort more, because his current produces little voltage; a customer who dominates his bus must behave.

\(I_{sc}/I_L\)\(3\le h<11\)\(11\le h<17\)\(17\le h<23\)\(23\le h<35\)\(35\le h\le 50\)TDD
\(< 20\)4.02.01.50.60.35.0
20 – 507.03.52.51.00.58.0
50 – 10010.04.54.01.50.712.0
100 – 100012.05.55.02.01.015.0
\(> 1000\)15.07.06.02.51.420.0

All entries are percentages of \(I_L\); even harmonics are held to 25% of the odd-harmonic limit of the band they fall in. The high-order columns are much tighter than the low-order ones because the impedance factor \(h\) in \(V_h = I_h h X_s\) magnifies them, and because they excite resonances more readily.

The utility's side of the bargain is a voltage limit, and it depends only on the system voltage: at 1 kV and below, 5.0% for any individual harmonic and 8.0% THD; above 1 kV up to 69 kV, 3.0% and 5.0%; from 69 kV to 161 kV, 1.5% and 2.5%; above 161 kV, 1.0% and 1.5%. The tightening with voltage level reflects the fact that a transmission bus feeds a whole region, so distortion there propagates everywhere.

🔑
The division of responsibility
The customer limits the harmonic current he injects (TDD, graded by \(I_{sc}/I_L\)); the utility limits the harmonic voltage it delivers (THD, graded by system voltage).

Neither party can meet its obligation alone: the customer's current becomes voltage through the utility's impedance, and the utility's impedance is what makes the customer's current harmful. IEEE 519 is best read as a treaty rather than a specification.

There is one mechanism that can wreck compliance without any change in the injected current, and it is parallel resonance. Every power-factor-correction capacitor sits in parallel with the inductive source impedance behind it. Looking into the bus from the distorting load, the two form a parallel \(LC\) circuit whose impedance is enormous at its resonant frequency. Setting the source reactance \(X_s = V^2/S_{sc}\) equal to the capacitive reactance \(X_c = V^2/Q_c\) at order \(h_r\) gives \(h_r X_s = X_c/h_r\), so

Order of the parallel resonance
\[ h_r = \sqrt{\frac{X_c}{X_s}} = \sqrt{\frac{S_{sc}}{Q_c}} \]

A 200 MVA bus with an 8 MVAr capacitor bank resonates at \(h_r = \sqrt{25}=5\) — precisely the largest harmonic a six-pulse converter produces. A fifth-harmonic current of a few per cent then sees an impedance many times the short-circuit impedance, produces a large fifth-harmonic voltage, and drives a circulating current between the capacitor and the source that can be an order of magnitude greater than the injected current itself. Capacitors fail, fuses blow, and the cause is invisible from the load's spectrum, which has not changed at all. Switching one step of the bank out shifts \(h_r\) and the problem vanishes — the classic diagnostic test.

Chapter 30's remedy is Chapter 39's disease. Power-factor capacitors were sized in Chapter 30 purely on \(\tan\varphi\), with no reference to the spectrum of the load being corrected. On a modern plant that calculation is incomplete: the same bank that fixes the displacement factor may resonate with the supply at the fifth harmonic. The correction is small — add a series reactor of a few per cent, as Section 39-6 shows — but it must be made deliberately.

The last of the classical quality problems is flicker, and it is unique in being defined by human physiology rather than by equipment. A load whose current fluctuates — an arc furnace during melt-down, a rolling mill, a large welder, a wind turbine passing its tower — modulates the voltage at the bus. To first order the modulation is set by the reactive power swing against the short-circuit level:

Voltage fluctuation from a varying load
\[ \frac{\Delta V}{V} \;\approx\; \frac{\Delta P\,R + \Delta Q\,X}{V^2} \;\approx\; \frac{\Delta Q}{S_{sc}} \quad\text{on a predominantly reactive network} \]

An incandescent lamp turns that voltage modulation into a light modulation roughly as \(\Delta\Phi/\Phi \approx 3.4\,\Delta V/V\), and the human eye-brain system is most sensitive to light modulation at about 8.8 Hz, where a fluctuation of only 0.25% of the voltage is at the threshold of irritation. Sensitivity falls off sharply on both sides: at 1 Hz and at 25 Hz roughly ten times more modulation is needed for the same annoyance. The IEC flickermeter of IEC 61000-4-15 encodes this response as a weighted filter, and reports two numbers:

🔑
Short-term and long-term flicker severity
\[ P_{st} \;\text{ over 10 minutes},\qquad P_{lt} = \sqrt[3]{\frac{1}{12}\sum_{i=1}^{12}P_{st,i}^3} \;\text{ over 2 hours} \]

The scale is calibrated so that \(P_{st}=1\) is the threshold at which 50% of observers find the flicker of a 60 W incandescent lamp objectionable. Typical planning levels are \(P_{st}\le1.0\) and \(P_{lt}\le0.8\) at medium voltage. The cube in the \(P_{lt}\) definition weights the worst ten-minute periods heavily, so a single severe melt cycle is not averaged away.

The practical planning rule for a fluctuating load follows directly. An arc furnace of 60 MVA on a bus with a fault level of 2500 MVA swings perhaps 30 MVAr as the electrodes bore in, giving \(\Delta V/V = 30/2500 = 1.2\%\) at frequencies right in the sensitive band — some four to five times the irritation threshold. This is why furnace connections are specified by a short-circuit ratio, \(S_{sc}/S_{\text{furnace}} \ge 80\) being a common requirement, and why furnaces are connected at the highest available voltage and provided with a static VAr compensator whose response time is measured in milliseconds. The SVC of Chapter 38, sized for the reactive swing rather than for steady-state correction, is the standard flicker remedy and typically halves \(P_{st}\).

Section 39-6

Mitigation: Filters, DVR, DSTATCOM and UPS

Mitigation divides cleanly according to what is being corrected. Harmonic currents are diverted or cancelled by filters. Voltage events — sags, swells, unbalance, interruptions — are corrected by inserting a controlled voltage in series with the supply or a controlled current in shunt with it. The distinction matters because the two jobs need entirely different ratings.

The oldest device is the single-tuned passive filter: a series \(L\)–\(C\) branch connected in shunt at the bus, resonant at the harmonic to be removed. At resonance its impedance is just the small resistance \(R\), so it is nearly a short circuit to that harmonic and diverts the load's harmonic current into itself rather than into the source. At the fundamental it is capacitive, so it also supplies reactive power — usually the reason it is affordable at all. Its design follows from two requirements. Tuning at order \(h_n\) means \(h_n\omega_1 L = X_C/h_n\); supplying \(Q\) MVAr at the fundamental means the net fundamental reactance is \(X_C - X_L = V^2/Q\). Together:

Design of a shunt filter tuned to order \(h_n\) and rated \(Q\) MVAr
\[ X_L = \frac{X_C}{h_n^{\,2}} \;\Longrightarrow\; X_C\!\left(1-\frac{1}{h_n^{\,2}}\right) = \frac{V^2}{Q}, \qquad C = \frac{1}{\omega_1 X_C}, \quad L = \frac{X_L}{\omega_1} \]
\[ \text{For } V=11\text{ kV},\; Q=5\text{ MVAr},\; h_n=4.7:\quad X_C = \frac{24.2}{1-1/22.09} = 25.35\ \Omega,\; X_L = 1.148\ \Omega \]
\[ C = \frac{1}{2\pi(50)(25.35)} = 125.6\ \mu\text{F}, \qquad L = \frac{1.148}{2\pi(50)} = 3.65\ \text{mH} \]

Tuning slightly below the harmonic to be trapped — 4.7 rather than 5.0 — is deliberate. Capacitance drifts upward with age and temperature and downward when a fuse takes a can out of service; a filter tuned exactly at 5 could drift above the harmonic, at which point the branch becomes inductive at the fifth and forms a parallel resonance with the source instead of a series one. Detuning downward also guarantees that the unavoidable parallel resonance between the filter's capacitance and the source occurs at an order below 5, in the gap between the third and fifth harmonics where nothing is being injected. The same reasoning gives the detuned capacitor bank: an ordinary power-factor bank with a 5.67% or 7% series reactor, resonant at order 4.2 or 3.78, which cannot resonate with any characteristic harmonic and simply behaves as a capacitor above the fifth.

Passive filters are cheap, lossless to a first approximation, and provide reactive power. Their weaknesses are structural: they are tuned to a fixed order, so a load that changes its spectrum defeats them; their performance depends on the source impedance, which changes with network switching; and — most awkwardly — they will absorb harmonic current arriving from other customers on the bus, so an over-effective filter can be overloaded by its neighbours' distortion.

The active power filter removes all three objections by abandoning tuning altogether. A voltage-source converter with a small DC capacitor measures the load current, extracts everything that is not fundamental positive-sequence, and injects the negative of it. The source then supplies only the fundamental. Because the reference is computed rather than tuned, the same hardware handles any spectrum, tracks changes within a cycle or two, and can be made to correct unbalance and displacement power factor at the same time. Its rating is set by the harmonic current it must supply — typically 20–30% of the load rating — and by the switching frequency required to reproduce the highest harmonic of interest, which is the practical limit on its bandwidth.

Turning to voltage events, three devices share one idea: a converter of Chapter 38's family, applied at distribution voltage.

The DVR (dynamic voltage restorer) is connected in series with the feeder through an injection transformer. When the supply voltage sags to \(V_{\text{sag}}\), the DVR synthesises and inserts the missing phasor \(V_{\text{inj}} = V_{\text{load}} - V_{\text{sag}}\) within a quarter-cycle, so the load never sees the event. Its rating is proportional to the depth of sag it must correct, not to the full load voltage, which is what makes it economic: correcting a 40% sag on a 2 MVA load needs an 0.8 MVA injection, not 2 MVA. Two injection strategies exist. In-phase injection is simplest but draws real power from the DC link for the duration; injecting at an angle that keeps the load current in quadrature with the injected voltage draws no real power at all, at the cost of a phase jump the load must tolerate. Since a sag lasts a few hundred milliseconds, the stored energy required is small — Example 4 works it out.

The DSTATCOM is the same converter connected in shunt. Injecting reactive current into the feeder raises the voltage at the load through the feeder impedance, which supports the bus during shallow sags, corrects steady-state unbalance by injecting negative-sequence current, cancels harmonics if the control includes them, and provides the millisecond-scale reactive swing needed against flicker. What it cannot do is correct a deep sag: raising the voltage by reactive injection alone requires a current inversely proportional to the feeder impedance, and on a stiff feeder that current is impossibly large. Deep sags need the series device; shallow, continuous problems need the shunt one.

The UPS is the only device on the list that addresses interruptions, because it is the only one with an energy store sized in minutes rather than cycles. In the double-conversion topology the load is permanently fed from an inverter running off a DC bus that a rectifier and a battery both support, so the load is completely decoupled from the supply and no transfer time exists. That isolation costs 4–8% of the throughput in conversion losses, all year, which is why UPS protection is reserved for loads where an interruption is intolerable and applied to those loads only rather than to a whole plant.

DeviceConnectionCorrectsTypical ratingCannot do
Detuned capacitor bankShuntDisplacement PF, avoids resonance= reactive demandReduce harmonic current
Single-tuned filterShuntOne harmonic order, plus reactive power= reactive demandAdapt to a changing spectrum
Shunt active filterShuntAll harmonics, unbalance, reactive power20–30% of loadSupport voltage during a deep sag
Series active filterSeriesVoltage harmonics, sag compensation10–20% of loadSink load harmonic current
DVRSeriesSags and swells, phase jumps30–50% of loadRide through an interruption for minutes
DSTATCOMShuntShallow sags, unbalance, flicker, harmonics10–30% of loadCorrect a deep sag on a stiff feeder
UPS (double conversion)Series (full flow)Everything, including interruptions100% of loadAvoid a permanent 4–8% loss
Match the device to the disturbance, not to the budget. The commonest and most expensive mistake in this field is protecting a whole plant with a large UPS when a single 30 kVA drive controller was the only sensitive item, or installing a harmonic filter to cure a problem that was in fact a resonance with an existing capacitor bank. A site survey — a week of recording at the point of common coupling, with the spectrum, the sag list and the load's own trip settings — costs a fraction of any of the equipment above and decides which one is needed.
Section 39-7

Wind: The Power Curve and the Betz Limit

The second half of this chapter changes the question. Sections 39-1 to 39-6 asked what happens to the waveform when the load is made of converters. The remaining sections ask what happens to the system when the generation is too.

Start with the resource. A mass of air of density \(\rho\) moving at speed \(v\) through an area \(A\) carries kinetic energy past that area at a rate

Power in the wind
\[ P_{\text{wind}} = \tfrac12\,\dot m\,v^2 = \tfrac12\big(\rho A v\big)v^2 = \tfrac12\rho A v^3 \]

Three features of this expression govern everything about wind engineering. The cube on \(v\) means a site with a 20% higher mean wind speed yields about 73% more energy, so siting dominates economics and hub heights keep rising. The area \(A = \pi R^2\) means output goes as the square of the rotor diameter, which is why machines have grown from 15 m rotors to 220 m. And \(\rho\) falls with altitude and rises in cold air, so the same machine produces some 10% less at 1000 m elevation than at sea level.

No turbine can capture all of \(P_{\text{wind}}\), and the reason is not a matter of engineering quality. If the rotor extracted all the kinetic energy, the air behind it would be at rest, and stationary air cannot move out of the way to let more air through. There must be a wake, and the wake must be moving. The optimum is a compromise, and Betz found it in 1919 with an argument that needs nothing but momentum conservation.

Model the rotor as an actuator disc: a permeable surface that extracts energy without changing the flow direction. Let the far-upstream speed be \(v_1\), the speed at the disc \(v_2\), and the far-downstream speed \(v_3\). The mass flow is the same everywhere, \(\dot m = \rho A v_2\). The thrust on the disc is the rate of change of momentum, \(F = \dot m(v_1-v_3)\), and the power extracted is also the rate at which the flow loses kinetic energy:

Actuator-disc balance
\[ P = F v_2 = \dot m (v_1-v_3)v_2, \qquad P = \tfrac12\dot m\big(v_1^2-v_3^2\big) \]
\[ \Longrightarrow\quad (v_1-v_3)v_2 = \tfrac12(v_1-v_3)(v_1+v_3) \quad\Longrightarrow\quad v_2 = \frac{v_1+v_3}{2} \]

The speed at the disc is the mean of the upstream and downstream speeds — a result obtained from nothing but the two ways of writing the same power. Substituting it back, and writing \(b = v_3/v_1\) for the fraction of speed retained in the wake,

Power coefficient of the actuator disc
\[ P = \tfrac12\rho A\!\left(\frac{v_1+v_3}{2}\right)\!\big(v_1^2-v_3^2\big) = \underbrace{\tfrac12\rho A v_1^3}_{P_{\text{wind}}}\cdot\underbrace{\tfrac12(1+b)(1-b^2)}_{C_p(b)} \]
\[ \frac{dC_p}{db} = \tfrac12\big(1-2b-3b^2\big) = 0 \;\Longrightarrow\; 3b^2+2b-1=0 \;\Longrightarrow\; b=\tfrac13 \]
\[ C_{p,\max} = \tfrac12\left(1+\tfrac13\right)\!\left(1-\tfrac19\right) = \tfrac12\cdot\tfrac43\cdot\tfrac89 = \frac{16}{27} = 0.5926 \]
🔑
The Betz limit
\[ P = C_p\,\tfrac12\rho A v^3, \qquad C_p \le \frac{16}{27} = 0.593 \]

The best possible rotor slows the wind to one third of its upstream speed in the far wake and to two thirds at the disc itself. Real three-bladed machines reach \(C_p\) of 0.45–0.50 at their best operating point, losing the remainder to wake rotation, finite blade number and drag — so they achieve about 80% of a bound that no design can pass.

\(C_p\) is not a constant. It depends on the blade pitch angle \(\beta\) and on the tip-speed ratio \(\lambda = \omega R/v\), the ratio of blade-tip speed to wind speed. There is a single peak: too slow and the air slips between the blades unused; too fast and each blade works in the turbulent wake of the one before. For a modern three-bladed rotor the peak sits near \(\lambda \approx 7\)–\(9\) at \(\beta = 0\). Since \(\lambda\) contains both \(\omega\) and \(v\), holding it at its optimum as the wind changes requires the rotor speed to be proportional to wind speed — which is the entire argument for variable-speed operation and hence for the converter interface.

Combining the cube law with the \(C_p\) peak and the machine's own ratings gives the power curve, the single characteristic by which a turbine is specified.

cut-in3 m/s rated 12 m/s cut-out25 m/s P = Cₚ · ½ρAv³ rated power 3 MW — pitch regulated spilled by pitching 0510 15202530 wind speed at hub height (m/s) 012 345 electrical output (MW)
The power curve: cubic below rated wind speed, flat above it, and zero outside the operating window

Four regions appear. Below the cut-in speed, typically 3–4 m/s, the available power does not cover the machine's own losses and the turbine idles. Between cut-in and rated wind speed, around 12 m/s, the controller holds \(\lambda\) at its optimum by varying rotor speed, so \(C_p\) is at its peak and the output follows the cube law exactly. At rated wind speed the generator and converter reach their limits; above it the blades are pitched out of the wind to reduce \(C_p\) deliberately, holding the output flat and spilling the rest. Beyond the cut-out speed of about 25 m/s the structural loads become unacceptable and the machine shuts down and feathers its blades — a genuine operating hazard, since a storm front can remove an entire wind farm's output in minutes.

Because the machine is rated at 12 m/s but the wind is usually well below that, the capacity factor — annual energy divided by rated power times 8760 h — is typically 0.25–0.35 onshore and 0.45–0.55 offshore. It is a statement about the wind, not about the machine's availability, which is separately about 97%.

How the rotor reaches the network defines four standard topologies. Type 1, the fixed-speed squirrel-cage generator connected directly through a soft-starter, is simple and obsolete: it runs at essentially one speed, so \(C_p\) is optimal at one wind speed only, it draws its magnetising reactive power from the grid and needs capacitors, and every tower-shadow pulse of torque appears at the terminals as a flicker-producing power pulse. Type 2 adds variable rotor resistance for a narrow speed range and does not solve the underlying problem.

Type 3, the doubly-fed induction generator, is the arrangement that made large wind economic. The stator is connected directly to the grid; the wound rotor is fed through slip rings from a back-to-back converter. In an induction machine the rotor circuit handles a fraction \(s\) of the air-gap power, so a converter designed for a slip range of \(\pm30\%\) about synchronous speed need only be rated at about 30% of the machine — a decisive cost saving over a full converter. Controlling the rotor current controls torque and stator reactive power independently, giving genuine variable speed and grid-code reactive capability from a partially rated converter. The price is the direct stator connection: a fault on the network appears immediately in the stator, induces a large current in the rotor, and threatens the converter. A crowbar that short-circuits the rotor through a resistance during the dip protects it, at the cost of turning the machine briefly into an uncontrolled induction generator drawing reactive power exactly when the system least wants it. Slip rings also mean brush maintenance.

Type 4, the full-converter machine, passes all the power through a back-to-back converter, so the generator — often a directly driven multi-pole permanent-magnet synchronous machine with no gearbox — is completely decoupled from the network frequency. The converter is rated at 100% and is correspondingly more expensive, but everything else improves: the generator can run at any speed, the grid-side converter presents a fully controllable current source that rides through faults without a crowbar and injects reactive current on demand, and harmonics are set by the converter's own switching rather than by the machine. Every large offshore machine built today is of this type, and Section 39-9 explains the one property it does not inherit from the rotor it replaced.

Section 39-8

Solar PV: The I–V Characteristic and MPPT

A photovoltaic cell is a large-area diode in which absorbed photons generate carriers. Its terminal behaviour is the superposition of a light-generated current source and the diode's own dark characteristic, with two parasitic resistances added for the metallisation and for leakage across the junction edge:

Single-diode model of a solar cell
\[ I = I_{ph} - I_0\left[\exp\!\left(\frac{q(V+IR_s)}{nkT}\right)-1\right] - \frac{V+IR_s}{R_{sh}} \]

The equation is implicit in \(I\) and must be solved iteratively, but its shape is easily read. At short circuit the exponential is negligible and \(I_{sc}\approx I_{ph}\), directly proportional to irradiance \(G\). At open circuit the net current is zero and the diode must absorb all of \(I_{ph}\), giving \(V_{oc} \approx (nkT/q)\ln(I_{ph}/I_0)\) — logarithmic in irradiance, hence nearly constant, and strongly dependent on temperature through \(I_0\). The curve is therefore a nearly horizontal current plateau that bends over into a nearly vertical voltage cliff, and the power \(P = VI\) is zero at both ends with a single maximum in the knee.

Figures of merit at standard test conditions (1000 W/m², 25 °C, AM1.5)
\[ \text{FF} = \frac{V_{mp}I_{mp}}{V_{oc}I_{sc}}, \qquad \eta = \frac{V_{mp}I_{mp}}{G\,A_{\text{module}}} \]
\[ \text{Typical module: } V_{oc}=45.6\ \text{V},\; I_{sc}=9.50\ \text{A},\; V_{mp}=37.2\ \text{V},\; I_{mp}=8.90\ \text{A} \]
\[ P_{mp}=331\ \text{W}, \qquad \text{FF} = \frac{331}{45.6\times9.50} = 0.764, \qquad \eta = \frac{331}{1000\times1.95} = 17.0\% \]

Temperature is the coefficient that matters in the field. Irradiance scales the current almost exactly, so on a bright day the current plateau simply rises. Temperature moves the voltage: \(V_{oc}\) falls at about \(-0.29\%\) per °C and \(P_{mp}\) at about \(-0.40\%\) per °C for crystalline silicon. A module at a cell temperature of 65 °C on a hot roof therefore delivers roughly 16% less than its nameplate. The same coefficient decides string sizing, and it does so at the cold extreme: at a record minimum of \(-5\) °C the open-circuit voltage rises to \(45.6[1+0.0029(30)] = 49.6\) V per module, so a 1000 V system permits at most \(1000/49.6 = 20\) modules in series. At the hot extreme the same string must still sit above the inverter's MPPT window: at 65 °C, \(V_{mp} = 37.2[1-0.0035(40)] = 32.0\) V per module, giving \(20\times32.0 = 640\) V, comfortably inside a 200–800 V window. Both limits must hold or the plant either breaks down its insulation in winter or stops tracking in summer.

Because the maximum-power point moves with both irradiance and temperature, it must be hunted continuously. The condition itself is elementary:

The maximum-power-point condition
\[ \frac{dP}{dV} = \frac{d(VI)}{dV} = I + V\frac{dI}{dV} = 0 \quad\Longrightarrow\quad \frac{dI}{dV} = -\frac{I}{V} \]
🔑
Incremental conductance
At the maximum power point the incremental conductance \(dI/dV\) equals the negative of the instantaneous conductance \(I/V\); to the left \(dI/dV > -I/V\), to the right \(dI/dV < -I/V\).

This gives an MPPT algorithm with a genuine stopping test: measure \(\Delta I/\Delta V\), compare with \(-I/V\), and move the operating voltage in the direction the inequality indicates. When the two are equal the search halts instead of oscillating.

The simpler perturb and observe method changes the converter's duty cycle by a small step, measures whether the power rose or fell, and keeps the direction that helped. It needs no differentiation and is what most inverters run, but it never settles — it dithers about the peak, losing a fraction of a per cent — and it can be fooled during a fast irradiance ramp, when power rises for reasons unconnected with the perturbation. The crudest method, holding \(V \approx 0.76\,V_{oc}\), exploits the fact that the ratio \(V_{mp}/V_{oc}\) is nearly constant, and survives in very small systems.

One failure mode defeats all of them. Under partial shading — a chimney, a leaf, a passing cloud edge — cells in one substring produce less current than the rest of the series string, and the bypass diodes across each substring conduct. The composite \(P\)–\(V\) curve then acquires several local maxima, and a hill-climbing tracker will happily settle on the wrong one and lose a third of the array's output. Global-search MPPT that periodically sweeps the whole voltage range, or module-level electronics that give each module its own tracker, are the two answers.

Both wind and solar share the property that distinguishes them from every generator in Chapters 26 to 33: their output is variable and only partly predictable, and it is not dispatchable upward. Three distinct effects must be separated.

Variability is the change in output over time, and it smooths dramatically with geography. Cloud passage over a single rooftop can take the output from full to 20% in seconds; over a 50 MW plant the same cloud takes minutes, because it must traverse the whole site. Across a region the fluctuations of \(N\) uncorrelated sites add in quadrature, so the relative variability of the aggregate falls roughly as \(1/\sqrt N\). Aggregation is thus the cheapest form of storage there is, and it is bought with transmission — one more reason the interconnection of Chapter 2 keeps growing.

Uncertainty is the error in the forecast, and it is what actually drives reserve procurement. A persistence forecast — "the next hour looks like this hour" — is surprisingly good at short horizons and useless beyond a few hours. Numerical weather prediction takes over from about six hours out, with day-ahead wind forecast errors of order 5–8% of installed capacity and solar errors dominated by cloud-cover timing. Because the error grows with horizon, an intraday market that lets positions be corrected an hour before delivery is worth far more to an operator than any amount of day-ahead accuracy.

Non-dispatchability is the structural fact that the resource cannot be increased on command. A wind farm can always be curtailed downward, and can hold a reserve margin by operating deliberately below its available power — but every megawatt of upward reserve so obtained is a megawatt of free fuel thrown away, which is why renewable plants are not the natural providers of the secondary reserve of Chapter 33.

The forecast is a system asset. Improving a day-ahead wind forecast by one percentage point of installed capacity reduces the reserve the operator must hold, and therefore the number of thermal units committed part-loaded overnight in Chapter 32's unit-commitment problem. On a system with 10 GW of wind that single percentage point is worth more each year than the metering it depends on costs to install — which is why forecasting is written into modern grid codes as an obligation on the generator, not an option for the operator.
Section 39-9

Inertia, RoCoF, Grid Codes and Storage

A synchronous generator does two things at once. It converts mechanical power into electrical power, and — because its rotor is a large mass turning in lock-step with the system frequency — it stores kinetic energy that the network can draw on without being asked. Chapter 27 wrote that store as the inertia constant \(H\), the seconds of rated output the machine could supply from its rotation alone:

Stored kinetic energy and the swing equation
\[ H = \frac{\tfrac12 J\omega_s^2}{S_{\text{rated}}}\ \text{[s]}, \qquad E_{\text{kin}} = H\,S \ \text{[MW·s]}, \qquad \frac{2H}{f_0}\frac{df}{dt} = P_m - P_e \ \text{(pu)} \]

Aggregate the machines that are synchronised at a given instant and the whole system behaves, for the first second after a disturbance, like one rotating mass of stored energy \(E_{\text{sys}} = \sum_i H_i S_i\). Rearranging the swing equation for a sudden loss of generation \(\Delta P\) megawatts gives the quantity that has become the defining constraint of modern system operation:

🔑
Rate of change of frequency
\[ \left.\frac{df}{dt}\right|_{t=0^+} = \frac{f_0\,\Delta P}{2\,E_{\text{sys}}} = \frac{f_0\,\Delta P}{2\sum_i H_iS_i} \]

RoCoF depends on the size of the loss and on the kinetic energy synchronised at that instant — and on nothing else. Governors, droop and AGC have not yet acted at \(t=0^+\); they appear only in the seconds that follow. The initial slope is pure physics.

Here is the difficulty. A converter-interfaced generator has no such store available. A full-converter wind turbine has a rotor with plenty of inertia, but the converter stands between that rotor and the network, so the rotor's speed is not tied to system frequency and its energy is not released by a frequency excursion. A photovoltaic plant has no rotating mass at all. Displacing synchronous plant with converter-interfaced plant therefore reduces \(E_{\text{sys}}\) directly — and since the largest credible loss \(\Delta P\) is set by the biggest single generator or interconnector and does not shrink, RoCoF rises in exact proportion to the inertia removed.

Two things then go wrong, and they are different in kind. The first is that the frequency nadir deepens: with a steeper initial slope, the frequency has fallen further by the time governor response arrives a few seconds later, and it may cross the first stage of under-frequency load shedding. The second is subtler and has caused real incidents. Small embedded generators are fitted with loss-of-mains protection that trips on RoCoF, historically set as low as 0.125 Hz/s, on the reasoning that no credible system event could produce such a slope. On a low-inertia system a large infeed loss now produces exactly that slope — and thousands of megawatts of embedded generation disconnect, making the original deficit worse. The remedy has been to raise RoCoF relay settings to around 1 Hz/s and lengthen their definite-time delays, a programme of retrofitting millions of installations.

first UFLS stage, 49.2 Hz loss of 250 MW infeed −0.21 Hz/s −0.63 Hz/s nadir 49.55 Hz nadir 49.06 Hz — load shed E = 30 000 MW·s E = 10 000 MW·s 50.049.849.6 49.449.249.0 0510 1520 time after the event (s) system frequency (Hz)
The same 250 MW loss on a high-inertia and a low-inertia system — Example 6

The response has been to write into grid codes the services that used to be free by-products of synchronous machines. A modern connection code for a wind or solar plant specifies, at minimum:

RequirementWhat is demandedWhy it was once automatic
Fault ride-throughStay connected for a voltage dip following a specified depth–duration profile, typically to 0 pu for 150 ms then a linear recovery to 0.85 pu by 1.5 sA synchronous machine cannot disconnect itself; it swings and resynchronises
Dynamic reactive currentDuring the dip inject \(I_q \ge K(1-V)\,I_{\text{rated}}\) with \(K\approx2\), up to full rated currentThe machine's internal e.m.f. forces reactive current into a depressed terminal voltage automatically
Frequency responseReduce output on a droop of 3–5% above a dead band; provide fast frequency response within 1 s if contractedGovernor droop of Chapter 33
Inertial responseDeliver power proportional to \(df/dt\) for the first seconds (synthetic inertia)Kinetic energy of the rotor, released by physics
Ramp-rate limitationRestrict the rate of increase of output, typically 10% of rated per minuteThermal plant cannot ramp fast even when asked
Reactive capability at the point of connectionOperate at any power factor within \(\pm0.95\) at full outputThe excitation system of Chapter 34
Forecasting and telemetrySubmit availability forecasts and real-time output to the control centreDispatchable plant is instructed, not observed

Fault ride-through deserves a word about why it exists. In the earliest wind connections it was regarded as safe for a generator to disconnect on a voltage dip, and codes required it. That is sound when the plant is 2% of the system and disastrous when it is 40%: a single transmission fault, cleared normally in 100 ms, would then trip several gigawatts of generation across a whole region and convert a routine event into a system emergency. The requirement was therefore inverted. The plant must now remain connected through a fault that its own protection is not clearing, and must actively help by injecting reactive current in proportion to the depth of the dip — precisely the behaviour the synchronous machine of Chapter 26 exhibits for free, now written as a control law.

Synthetic inertia is the corresponding replacement for the missing rotor. A converter that measures \(df/dt\) and injects power proportional to it emulates a rotating mass of chosen size: setting \(P_{\text{syn}} = 2E_{\text{syn}}\,(df/dt)/f_0\) inserts \(E_{\text{syn}}\) MW·s of virtual kinetic energy into the RoCoF equation. On a wind turbine the energy comes from the rotor, briefly slowed below its optimum \(\lambda\); on a battery it comes from the cells. Three limitations must be understood. Differentiating a noisy frequency measurement is delicate and introduces a delay of a hundred milliseconds or more, so the response is not truly instantaneous the way a rotor's is. On a wind turbine the extracted energy must be paid back — the rotor has to be re-accelerated, which means a period of reduced output a few seconds later, sometimes right at the frequency nadir. And a converter has no overload capability worth the name: a synchronous machine will deliver several times rated current for a second, whereas a converter's semiconductors will not.

The deeper answer is the grid-forming converter, which abandons the current-source control philosophy entirely and instead regulates its output as a voltage of controlled magnitude and phase behind an impedance — that is, it behaves as a synchronous machine does, supplying inertial response and fault current as an inherent property of its control rather than as a measured reaction. Grid-forming control also solves a problem that grid-following converters cannot: a phase-locked loop needs a stable voltage to lock onto, so a system consisting entirely of grid-following converters has nothing to define its own frequency. Some minimum of grid-forming capability, synchronous or converter-based, is a hard requirement, and system operators now specify it explicitly as a minimum inertia or minimum short-circuit level to be maintained at all times.

🔑
Synthetic inertia is a power problem, not an energy problem
\[ P_{\text{syn}} = \frac{2E_{\text{syn}}}{f_0}\frac{df}{dt}, \qquad E_{\text{delivered}} \approx P_{\text{syn}}\,t \ \text{for } t \sim 1\ \text{s} \]

Replacing 2500 MW·s of lost inertia at a RoCoF of 0.5 Hz/s calls for 50 MW — held for about a second, which is 50 MJ, or 14 kWh. A battery of a few megawatt-hours has the energy a thousand times over; what it must have is the power and the response time. This is why fast frequency response is the service batteries win and bulk energy arbitrage is the one they struggle with.

When the network cannot absorb what the renewables can produce, the output is curtailed. Three causes must be distinguished, because the remedies differ. Network curtailment occurs when a line or transformer would be overloaded — a wind farm in a windy region connected by inadequate transmission — and the remedy is a line or a dynamic line rating. Minimum-generation curtailment occurs when the thermal units that must stay on for inertia, voltage support or reserve are already at their minimum stable output, so the renewable energy has nowhere to go; the remedy is grid-forming converters, synchronous condensers, storage, or demand that can be moved. Economic curtailment occurs when the market price goes negative because subsidised generation bids below zero, and it is a design feature of the market rather than a fault.

Storage attacks all three. A battery system is characterised by its power rating in MW, its energy rating in MWh, the ratio of the two (its duration, commonly 1–4 hours), its round-trip efficiency (85–90% for lithium-ion at the AC terminals) and its cycle life. It shifts the solar peak into the evening ramp, holds reserve without committing a thermal unit part-loaded, and — because a converter's response is limited only by control bandwidth — delivers fast frequency response an order of magnitude faster than any governor. Pumped hydro remains the only mature technology for the multi-day storage that a system dominated by weather-driven generation eventually needs; its round-trip efficiency of 75–80% is lower, but its energy is cheap once the reservoirs exist.

Every service the grid needs was once a free by-product. Inertia, fault current, reactive capability, voltage-source behaviour and predictable output all came bundled with the synchronous machine, and the whole of Parts 6 and 7 of this book was written on the assumption that they always would. Converter-interfaced generation unbundles them: each must now be specified, procured and paid for separately. That is the single organising idea behind every modern grid code, and it is the thread that ties this final chapter back to the swing equation of Chapter 27 and the load-frequency control of Chapter 33.
Section 39-10

Worked Examples

1 THD of a six-pulse converter current

Problem. A six-pulse rectifier with a large smoothing inductance draws a fundamental current of 100 A. Take the idealised spectrum \(I_h = I_1/h\) for \(h = 6k\pm1\). Compute the total harmonic distortion up to the 25th, the true rms current, and the distortion factor. Compare with the value obtained when the series is summed to infinity.

Solution. The characteristic orders below 25 are 5, 7, 11, 13, 17, 19, 23 and 25. Their amplitudes are \(100/h\) amperes:

\(h\)57111317192325
\(I_h\) (A)20.0014.299.097.695.885.264.354.00
\(I_h^2\) (A²)400.0204.182.659.234.627.718.916.0
Distortion and rms
\[ \sum_{h\ge5} I_h^2 = 843.1\ \text{A}^2 \;\Longrightarrow\; I_{\text{harm}} = \sqrt{843.1} = 29.04\ \text{A} \]
\[ \text{THD}_I = \frac{29.04}{100} = 29.04\%, \qquad I_{\text{rms}} = \sqrt{100^2 + 843.1} = \sqrt{10\,843} = 104.13\ \text{A} \]
\[ \text{distortion factor} = \frac{I_1}{I_{\text{rms}}} = \frac{100}{104.13} = 0.960 \]

Summing to infinity is a short exercise in series. The characteristic orders are the odd integers that are not multiples of three, so

The exact limit
\[ \sum_{h\;\text{odd},\,3\nmid h}\frac{1}{h^2} = \underbrace{\frac{\pi^2}{8}}_{\text{all odd}} - \underbrace{\frac{1}{9}\cdot\frac{\pi^2}{8}}_{\text{odd multiples of }3} = \frac{\pi^2}{9} = 1.09662 \]
\[ \text{THD}_I = \sqrt{1.09662 - 1} = \sqrt{0.09662} = 0.3108 = 31.08\% \]

Truncating at the 25th therefore captures 29.0 of the 31.1 percentage points: the tail beyond the 25th contributes little to THD, because the amplitudes fall as \(1/h\) and their squares as \(1/h^2\). It does not follow that the tail is harmless — the voltage distortion each harmonic causes is \(I_h\,hX_s\), which is the same for every order when \(I_h = I_1/h\). Every characteristic harmonic of an ideal six-pulse converter produces an equal share of the voltage distortion, and the high orders are limited far more tightly for exactly that reason.

2 THD, TDD and an IEEE 519 assessment

Problem. A plant is supplied at 11 kV from a bus with a three-phase fault level of 115 MVA. Its maximum demand under the connection agreement is 7.5 MVA. A six-pulse drive within the plant draws 300 A of fundamental current at full load, with the measured spectrum (as a percentage of its own fundamental) 5th 18.0, 7th 12.0, 11th 6.0, 13th 4.5, 17th 2.0, 19th 1.5, 23rd 0.7, 25th 0.6. Assess compliance at full load and at one-third load. Then evaluate the effect of converting the drive to a twelve-pulse arrangement, and of adding tuned filters that remove 90% of the 11th and 13th.

Solution. First fix the two reference currents.

Short-circuit ratio at the point of common coupling
\[ I_{sc} = \frac{115\times10^6}{\sqrt3\,(11\,000)} = 6035\ \text{A}, \qquad I_L = \frac{7.5\times10^6}{\sqrt3\,(11\,000)} = 393.6\ \text{A} \]
\[ \frac{I_{sc}}{I_L} = \frac{6035}{393.6} = 15.3 \;\;(<20) \;\Longrightarrow\; \text{TDD limit } 5.0\%,\; h<11 \text{ limit } 4.0\%,\; 11\le h<17 \text{ limit } 2.0\% \]

The distortion of the drive current itself is independent of loading:

THD of the drive current
\[ \text{THD}_I = \sqrt{18.0^2+12.0^2+6.0^2+4.5^2+2.0^2+1.5^2+0.7^2+0.6^2}\,\% = \sqrt{531.4}\,\% = 23.05\% \]

Now convert to demand distortion at the two operating points, using \(\text{TDD} = \text{THD}_I\times I_1/I_L\):

Full load and one-third load
\[ I_1 = 300\ \text{A}: \quad \text{TDD} = 23.05\times\frac{300}{393.6} = 17.57\% \]
\[ I_1 = 100\ \text{A}: \quad \text{TDD} = 23.05\times\frac{100}{393.6} = 5.86\% \]

The THD is 23.05% in both cases; the TDD differs by a factor of three. This is the whole argument for TDD. The limit is 5.0%, so the drive fails at full load by a factor of 3.5 and fails marginally even at one third of its rating. The individual limits fail too: the fifth harmonic is \(18.0\times300/393.6 = 13.7\%\) of \(I_L\) against a limit of 4.0%.

A twelve-pulse arrangement retains only \(h = 12k\pm1\), cancelling the 5th, 7th, 17th and 19th:

Twelve-pulse residual
\[ \text{THD}_I = \sqrt{6.0^2+4.5^2+0.7^2+0.6^2}\,\% = \sqrt{57.10}\,\% = 7.56\% \]
\[ \text{TDD} = 7.56\times\frac{300}{393.6} = 5.76\% \quad (\text{limit } 5.0\%) \]

A large improvement — the harmonic current has fallen to a third — but still not compliant, and the 11th alone stands at \(6.0\times0.762 = 4.57\%\) against its 2.0% limit. Pulse multiplication has removed the biggest harmonics and left the ones the standard treats most severely.

Adding shunt branches tuned near the 11th and 13th that divert 90% of each leaves 0.60% and 0.45% of the fundamental at those orders:

With twelve-pulse plus tuned filters
\[ \text{THD}_I = \sqrt{0.60^2+0.45^2+0.7^2+0.6^2}\,\% = \sqrt{1.4125}\,\% = 1.19\% \]
\[ \text{TDD} = 1.19\times0.762 = 0.91\%, \qquad \text{largest individual} = 0.7\times0.762 = 0.53\%\ (23\text{rd, limit }0.6\%) \]

Every limit is now met with margin. The lesson is the order of the remedies: pulse multiplication first, because it costs a transformer winding and no tuning; filters second, sized only for what pulse multiplication could not remove. Reversing the order gives a filter three times larger doing a job the transformer would have done for less.

3 The negative-sequence unbalance factor

Problem. A three-phase supply is measured as \(V_a = 400\angle0^\circ\), \(V_b = 380\angle{-}120^\circ\), \(V_c = 390\angle120^\circ\) volts. Find the sequence components and the unbalance factor. Compare with the NEMA definition, and estimate the negative-sequence current in an induction motor whose negative-sequence impedance is one sixth of its positive-sequence impedance.

Solution. With \(a = 1\angle120^\circ\), the transformation of Chapter 22 gives

Positive sequence
\[ V_1 = \tfrac13\big(V_a + aV_b + a^2V_c\big) = \tfrac13\big(400\angle0^\circ + 380\angle0^\circ + 390\angle0^\circ\big) = \tfrac{1170}{3} = 390\ \text{V} \]

The three rotations bring all three phasors onto the real axis, so the positive-sequence voltage is simply the arithmetic mean — which is exactly what one expects when the angles are already correct and only the magnitudes differ.

Negative sequence
\[ V_2 = \tfrac13\big(V_a + a^2V_b + aV_c\big) = \tfrac13\big(400\angle0^\circ + 380\angle120^\circ + 390\angle240^\circ\big) \]
\[ = \tfrac13\Big[(400 - 190 - 195) + j(0 + 329.09 - 337.75)\Big] = \tfrac13(15 - j8.66) = 5.00 - j2.89\ \text{V} \]
\[ |V_2| = \sqrt{5.00^2 + 2.89^2} = 5.77\ \text{V} \]
Unbalance factor
\[ \text{VUF} = \frac{|V_2|}{|V_1|}\times100 = \frac{5.77}{390}\times100 = 1.48\% \]

The NEMA estimate uses only magnitudes. The average is \((400+380+390)/3 = 390\) V and the largest deviation from it is 10 V, so

NEMA comparison
\[ \text{NEMA} = \frac{10}{390}\times100 = 2.56\%, \qquad \frac{\text{NEMA}}{\text{VUF}} = \frac{2.56}{1.48} = 1.73 = \sqrt3 \]

The factor \(\sqrt3\) is not general. It arises here because the deviations \((+10,-10,0)\) sum to zero and are purely real, so \(V_2 = \tfrac13(10 - 10a^2) = \tfrac{10}{3}(1-a^2)\) and \(|1-a^2| = \sqrt3\). Change the pattern of deviations, or introduce an angle error, and the two definitions part company — the NEMA figure being blind to angle unbalance altogether. Read as a compliance number, 1.48% is inside the usual 2% limit while 2.56% is outside it, so the choice of definition is not academic.

For the motor, the negative-sequence voltage of 5.77 V is applied to an impedance one sixth of the positive-sequence value:

Consequence for the machine
\[ \frac{I_2}{I_1} = \frac{|V_2|/Z_2}{|V_1|/Z_1} = 0.0148\times6 = 8.9\% \]
\[ \text{extra loss} \;\propto\; I_2^2 \;\Longrightarrow\; \Delta P_{\text{loss}} \approx (0.089)^2 = 0.79\%\ \text{of the rated } I^2R \text{ loss, concentrated in the rotor} \]

Expressed as a fraction of total losses that seems small, but the negative-sequence current flows in the rotor at a slip near 2, where the rotor resistance is raised by skin effect and the cooling is worst. It is the local hot spot, not the total loss, that determines insulation life — which is why NEMA MG-1 asks for a derating of about 5% at 2% unbalance and about 25% at 5%.

4 Sag depth, the vulnerability area, and a DVR

Problem. A 33 kV substation has a source impedance of \(j2.0\ \Omega\) referred to that voltage. Feeders leaving the bus have an impedance of \(0.3+j0.4\ \Omega\) per kilometre. A three-phase fault occurs 5 km out on one feeder and is cleared in 150 ms. (a) Find the retained voltage and the phase-angle jump at the bus. (b) Determine how far along a feeder a fault must be before the retained voltage stays above 70%. (c) If the substation has six such feeders and the fault rate is 0.1 faults per kilometre per year, estimate the number of ITIC violations per year. (d) Size a DVR to protect a 2 MVA, 0.9 power-factor load against this sag.

Solution (a). The bus sits at the mid-point of the divider between the source and the fault:

Voltage divider
\[ Z_F = 5(0.3+j0.4) = 1.5+j2.0 = 2.50\angle53.13^\circ\ \Omega, \qquad Z_S+Z_F = 1.5+j4.0 = 4.272\angle69.44^\circ\ \Omega \]
\[ V_{\text{sag}} = \frac{Z_F}{Z_S+Z_F} = \frac{2.50\angle53.13^\circ}{4.272\angle69.44^\circ} = 0.585\angle{-}16.3^\circ\ \text{pu} \]

The load retains 58.5% of nominal for 150 ms, and its voltage phasor also jumps back by \(16.3^\circ\). That phase jump is not a curiosity: a thyristor converter that fires from a phase-locked loop can misfire when the reference moves this far this quickly, and some equipment trips on the jump rather than on the depth. Plotted on the ITIC envelope of Section 39-2, the point \((150\ \text{ms},\,58.5\%)\) lies well below the 70% boundary — the load is expected to drop out.

Solution (b). Set the retained voltage to 0.70 with the fault at distance \(d\):

Boundary of the vulnerability area
\[ |Z_F|^2 = 0.25d^2, \qquad |Z_S+Z_F|^2 = (0.3d)^2 + (2+0.4d)^2 = 0.25d^2 + 1.6d + 4 \]
\[ 0.25d^2 = 0.49\big(0.25d^2+1.6d+4\big) \;\Longrightarrow\; 0.1275d^2 - 0.784d - 1.96 = 0 \]
\[ d = \frac{0.784 + \sqrt{0.6147+0.9996}}{0.255} = \frac{0.784+1.2705}{0.255} = 8.06\ \text{km} \]

Every three-phase fault within 8.06 km of the bus, on any feeder, drives this load below the ITIC boundary. That length of line is the vulnerability area.

Solution (c). Six feeders, each exposed over the first 8.06 km:

Expected event count
\[ N = 6\times8.06\ \text{km}\times0.1\ \text{faults/km/year} = 4.8\ \text{events per year} \]

Roughly five process trips a year from three-phase faults alone — and since single line-to-ground faults are several times more numerous and produce shallower but still troublesome sags, the true figure is higher. This number, multiplied by the cost of a trip from Section 39-1, is the entire business case for what follows.

Solution (d). The DVR must make up the difference between the retained voltage and the voltage the load needs:

DVR rating
\[ V_{\text{inj}} = 1.000 - 0.585 = 0.415\ \text{pu}, \qquad S_{\text{DVR}} = 0.415\times2.0 = 0.83\ \text{MVA} \]
\[ P_{\text{inj}} = 0.415\times2.0\times0.9 = 0.747\ \text{MW}, \qquad W = 0.747\times10^6\times0.150 = 112\ \text{kJ} = 0.031\ \text{kWh} \]

A converter of 0.83 MVA — 42% of the load rating — and an energy store of 31 watt-hours. The rating is set by the injected voltage, the store by the sag duration, and the second is trivially small. A UPS protecting the same load against a one-minute interruption would need 2 MVA and 33 kWh, a thousand times the storage. That contrast is the reason the DVR exists.

5 Wind turbine output, Betz and capacity factor

Problem. A three-bladed turbine has a rotor diameter of 90 m and operates in air of density 1.225 kg/m³. At a hub-height wind speed of 12 m/s its power coefficient is 0.45 and the combined gearbox and generator efficiency is 0.95. Find (a) the power in the wind and the Betz maximum, (b) the electrical output, (c) the output at 8 m/s, (d) the rotor speed if the optimum tip-speed ratio is 8, (e) the wind speed at which a 3 MW rating is reached, and (f) the annual energy at a capacity factor of 0.32.

Solution (a).

Swept area and available power
\[ A = \pi R^2 = \pi(45)^2 = 6361.7\ \text{m}^2 \]
\[ P_{\text{wind}} = \tfrac12(1.225)(6361.7)(12)^3 = 3896.6\times1728 = 6.733\ \text{MW} \]
\[ P_{\text{Betz}} = \tfrac{16}{27}\times6.733 = 0.5926\times6.733 = 3.990\ \text{MW} \]

Solution (b). The rotor takes \(C_p = 0.45\) of the wind power, and the drivetrain passes 95% of that:

Electrical output at 12 m/s
\[ P_{\text{mech}} = 0.45\times6.733 = 3.030\ \text{MW}, \qquad P_{\text{elec}} = 0.95\times3.030 = 2.878\ \text{MW} \]

The rotor achieves \(0.45/0.5926 = 76\%\) of the Betz maximum, which is typical of a good three-bladed machine at its design point.

Solution (c). Below rated wind speed the controller holds \(\lambda\), and hence \(C_p\), constant, so the output follows the cube law exactly:

Scaling to 8 m/s
\[ P(8) = 2.878\left(\frac{8}{12}\right)^3 = 2.878\times0.2963 = 0.853\ \text{MW} \]

Two thirds of the wind speed gives a little over a quarter of the power. This single number explains why annual energy is so sensitive to siting and hub height, and why the capacity factor is far below unity even on a machine that is available all year.

Solution (d).

Rotor speed at the optimum tip-speed ratio
\[ \lambda = \frac{\omega R}{v} \;\Longrightarrow\; \omega = \frac{\lambda v}{R} = \frac{8\times12}{45} = 2.133\ \text{rad/s}, \qquad N = \frac{60\,\omega}{2\pi} = 20.4\ \text{rpm} \]

Twenty revolutions a minute against a generator that wants 1500 — hence either a gearbox of ratio 74:1 or, in a direct-drive machine, a generator with a very large number of poles and a correspondingly large diameter.

Solution (e). Rated output requires \(3.0/0.95 = 3.158\) MW at the shaft, hence \(3.158/0.45 = 7.018\) MW in the wind:

Rated wind speed
\[ v^3 = \frac{7.018\times10^6}{3896.6} = 1801 \;\Longrightarrow\; v = 12.16\ \text{m/s} \]

Above 12.16 m/s the blades must be pitched to hold \(C_p\) below 0.45 and spill the surplus — the flat portion of the power curve in Section 39-7.

Solution (f).

Annual energy
\[ E = 3.0\ \text{MW}\times8760\ \text{h}\times0.32 = 8410\ \text{MWh per year} \]

Equivalently, 2803 full-load hours. Note what the capacity factor is not: it is not availability, which is separately about 97%, and it is not efficiency. It is a statement about the wind resource at this site measured against a rating chosen by the manufacturer.

6 RoCoF before and after the renewables arrive

Problem. A 50 Hz system has a demand of 5000 MW met by synchronous plant totalling 6000 MVA at an average inertia constant of 5 s. The largest credible loss is a 250 MW infeed. (a) Find the initial RoCoF and the time to reach the first under-frequency load-shedding stage at 49.2 Hz if nothing responds. (b) Repeat for a high-renewable hour in which demand is 3000 MW and only 2000 MVA of synchronous plant remains synchronised, the same 250 MW infeed being at risk. (c) A RoCoF-based loss-of-mains relay fleet is set at 0.5 Hz/s. Comment. (d) Find the synthetic-inertia power that would restore the 0.5 Hz/s limit, and the energy it represents.

Solution (a). The stored kinetic energy synchronised is

High-inertia case
\[ E_{\text{sys}} = \sum H_iS_i = 5\times6000 = 30\,000\ \text{MW·s} \]
\[ \left.\frac{df}{dt}\right|_{0^+} = \frac{f_0\,\Delta P}{2E_{\text{sys}}} = \frac{50\times250}{2\times30\,000} = 0.208\ \text{Hz/s} \]
\[ t_{49.2} = \frac{50.0-49.2}{0.208} = \frac{0.8}{0.208} = 3.84\ \text{s} \]

Almost four seconds of margin before load shedding — comfortably longer than the two to three seconds a governor needs to deliver primary response, so in practice the frequency is arrested near 49.55 Hz and never reaches the shedding stage.

Solution (b). The same loss on one third of the inertia:

Low-inertia case
\[ E_{\text{sys}} = 5\times2000 = 10\,000\ \text{MW·s}, \qquad \frac{df}{dt} = \frac{50\times250}{2\times10\,000} = 0.625\ \text{Hz/s} \]
\[ t_{49.2} = \frac{0.8}{0.625} = 1.28\ \text{s} \]

RoCoF trebles and the margin collapses to 1.28 s — shorter than the response time of any governor on the system. The frequency now passes the first load-shedding stage before help arrives, which is the trace drawn in red in Section 39-9. It is worth being explicit that the disturbance has not changed: the same 250 MW was lost. Only the system's ability to absorb it did.

Solution (c). At 0.625 Hz/s the system-wide RoCoF exceeds the 0.5 Hz/s relay setting, so every embedded generator equipped with such a relay disconnects within its definite-time delay. If that fleet totals, say, 400 MW, the deficit grows from 250 MW to 650 MW while the inertia stays at 10 000 MW·s, and the RoCoF jumps to \(50\times650/20\,000 = 1.63\) Hz/s. The protection intended to detect islanding has converted a survivable event into a cascade — the reason RoCoF settings have been raised to about 1 Hz/s with added time delay across whole systems.

Solution (d). To hold the initial RoCoF at 0.5 Hz/s, the converters must supply the difference between the actual deficit and the deficit the remaining inertia can absorb at that slope:

Synthetic inertia requirement
\[ \Delta P_{\text{allowed}} = \frac{2E_{\text{sys}}}{f_0}\left(\frac{df}{dt}\right)_{\!\lim} = \frac{2\times10\,000}{50}\times0.5 = 200\ \text{MW} \]
\[ P_{\text{syn}} = 250 - 200 = 50\ \text{MW}, \qquad E_{\text{syn}} = \frac{f_0\,P_{\text{syn}}}{2\,(df/dt)} = \frac{50\times50}{2\times0.5} = 2500\ \text{MW·s} \]
\[ W \approx 50\ \text{MW}\times1\ \text{s} = 50\ \text{MJ} = 13.9\ \text{kWh} \]

Fifty megawatts, held for roughly a second: fourteen kilowatt-hours of energy. A 50 MW battery with a one-hour duration stores 50 MWh, some 3600 times what this service consumes. The binding constraints are the power rating and the speed of the response, never the store — which is why fast frequency response and bulk energy shifting are priced as entirely different products.

Review

Chapter Summary

Current becomes voltage

\(V_h = I_h\,hX_s\) with \(X_s = V^2/S_{sc}\): a weak bus turns one customer's harmonics into everyone's distortion.

Sags

\(V_{\text{sag}} = Z_F E/(Z_S+Z_F)\) sets the depth, the protection sets the duration, the ITIC envelope decides survival.

Unbalance

\(\text{VUF}=|V_2|/|V_1|\); since \(Z_2\approx Z_1/6\), a 2% unbalance drives 12% negative-sequence current into the rotor.

THD versus TDD

THD normalises on the present fundamental and hides light loading; TDD normalises on \(I_L\) and is what IEEE 519 limits.

Resonance

A capacitor bank resonates with the source at \(h_r=\sqrt{S_{sc}/Q_c}\) — detune it with a 5.67% or 7% series reactor.

Series or shunt

Deep sags need a series device (DVR); harmonics, unbalance and flicker need a shunt one (filter, DSTATCOM).

Betz

\(P = C_p\,\tfrac12\rho A v^3\) with \(C_p\le16/27\); the wake must keep one third of the upstream speed.

MPPT

The maximum-power point is where \(dI/dV=-I/V\); partial shading creates local maxima that fool hill-climbing.

RoCoF

\(df/dt = f_0\Delta P/2E_{\text{sys}}\): displacing synchronous plant raises the slope in exact proportion.

Unbundled services

Inertia, fault current, reactive capability and voltage-source behaviour must now be specified and paid for one by one.

Practice

Practice Problems

Take \(f_1 = 50\) Hz and \(\rho = 1.225\) kg/m³ unless stated otherwise. Where a standard is invoked, use the IEEE 519 tables of Section 39-5 and the ITIC envelope of Section 39-2.

  1. A load draws a fundamental current of 250 A together with harmonics of 40 A at the 5th, 28 A at the 7th, 15 A at the 11th and 11 A at the 13th. Find \(\text{THD}_I\), the true rms current, the distortion factor, and the true power factor if the displacement factor is 0.96 lagging.
  2. The load of Problem 1 is connected at a point of common coupling where the fault level gives \(I_{sc} = 9000\) A and the agreed maximum demand current is \(I_L = 500\) A. Compute \(I_{sc}/I_L\), the TDD, and state which individual limits are met and which are not.
  3. A 6.6 kV bus has a short-circuit level of 150 MVA and carries a 6 MVAr power-factor-correction bank. Find the order of the parallel resonance. If a six-pulse drive on the same bus injects 4% fifth-harmonic current, explain qualitatively what happens and compute the size of series reactor, as a percentage of \(X_C\), that moves the resonance to order 4.2.
  4. Measured phase voltages are \(V_a = 235\angle0^\circ\), \(V_b = 228\angle{-}122^\circ\) and \(V_c = 240\angle119^\circ\) volts. Compute \(V_0\), \(V_1\) and \(V_2\), the unbalance factor, and the NEMA figure. Explain why the two disagree by more than they did in Example 3.
  5. A 132 kV source of impedance \(j8\ \Omega\) feeds a substation through a line of \(0.12+j0.40\ \Omega\) per kilometre. A three-phase fault occurs 12 km down an adjacent circuit and is cleared in 90 ms. Find the retained voltage and phase-angle jump, plot the point against the ITIC envelope, and state whether a DVR is needed for a load that trips below 75%.
  6. Design a single-tuned shunt filter for a 6.6 kV bus that must supply 3 MVAr at the fundamental and be tuned to order 4.7. Find \(X_C\), \(X_L\), \(C\) and \(L\), and the fundamental current the branch carries.
  7. A turbine of 110 m rotor diameter operates at \(C_p = 0.47\) with a drivetrain efficiency of 0.96. Find the electrical output at 10 m/s, the rated wind speed for a 4.5 MW machine, and the rotor speed at a tip-speed ratio of 7.5 in a 10 m/s wind. Comment on why the machine is not rated at the power available at 20 m/s.
  8. A photovoltaic module has \(V_{oc}=49.2\) V, \(I_{sc}=11.4\) A, \(V_{mp}=41.0\) V, \(I_{mp}=10.7\) A and an area of 2.1 m². Find \(P_{mp}\), the fill factor and the efficiency at 1000 W/m². If \(V_{oc}\) falls at \(-0.28\%\) per °C, find the largest number of modules that may be placed in series on a 1500 V system when the record low temperature is \(-10\) °C.
  9. A 50 Hz island has 1200 MVA of synchronous plant at \(H = 4\) s and 600 MW of converter-interfaced generation with no inertial response. Find the RoCoF following the loss of a 120 MW unit. Then find the synthetic-inertia power needed to keep the RoCoF below 0.4 Hz/s, and the energy that service delivers in the first second.
  10. An arc furnace of 80 MVA is to be connected at a bus with a fault level of 3200 MVA, and swings 45 MVAr during boring-in. Estimate \(\Delta V/V\) and the short-circuit ratio, compare with the usual requirement \(S_{sc}/S_{\text{furnace}}\ge80\), and state what must change if the requirement is not met.
Tip: in every problem in this chapter, identify the reference quantity before computing anything. Harmonic problems turn on whether the denominator is \(I_1\) or \(I_L\); sag problems on whether the impedance divider is written from the source or from the fault; RoCoF problems on whether the inertia is quoted as \(H\) in seconds on a machine base or as \(E\) in MW·s on the system. Almost every wrong answer in this material is a right calculation performed against the wrong base — which is the same discipline Chapter 4 imposed with the per-unit system, applied here to quantities the per-unit system never covered.