Set 40 — Power Quality and Renewable Integration
The last set of the book, and the one in which every earlier assumption is withdrawn. Problems 1 to 7 follow a single 11 kV plant: a measured spectrum becomes a distortion figure, the distortion figure becomes a compliance verdict, the current becomes a voltage through the source impedance of Set 25, and the power-factor bank installed under Set 30 turns out to resonate at exactly the harmonic the plant produces most of. Problems 8 to 12 treat the events rather than the steady state — sag depth from the fault-position divider of Set 21, the vulnerability radius that converts a fault rate into a trip rate, and the negative-sequence unbalance factor of Set 21 read as a quality index rather than as a fault quantity. Problems 13 to 19 replace the generation: the Betz limit, the power curve, photovoltaic string sizing at both temperature extremes, and the rate of change of frequency that follows from the swing equation of Set 27 when the rotating mass behind it is removed. Problem 20 puts the plant back together.
A harmonic current becomes a harmonic voltage through the source impedance. \(V_h = I_h\,hX_s\) with \(X_s = V_{LL}^2/S_{sc}\). The factor \(h\) is why a small high-order current matters as much as a large fifth, and \(S_{sc}\) is why the same load is harmless on a stiff bus and intolerable on a weak one.
THD normalises on the present fundamental; TDD normalises on the agreed demand. \(\text{TDD}/\text{THD}_I = I_1/I_L\). They agree only at full load. IEEE 519 limits TDD and never THD, because it is the injected ampere — not the ratio — that disturbs the neighbours.
Harmonic currents add in quadrature. \(I_{\text{rms}} = I_1\sqrt{1+\text{THD}_I^2}\), because harmonics of different order are orthogonal over a period. The same orthogonality splits power factor into a displacement factor and a distortion factor, and capacitors correct only the first.
A shunt capacitor resonates with the source at \(h_r = \sqrt{S_{sc}/Q_c}\). Setting \(h_rX_s = X_c/h_r\) with \(X_s = V^2/S_{sc}\) and \(X_c = V^2/Q_c\) gives it immediately. At that order the bus impedance is raised by the circuit's quality factor, and the injected current is unchanged — which is why the fault is invisible in the load's own spectrum.
A series reactor of \(p = X_L/X_C\) moves the branch's series resonance to \(1/\sqrt p\) and the branch's parallel resonance with the source down to \(h_p=\sqrt{X_C/(X_L+X_s)}\). A single-tuned filter is the same circuit with \(p = 1/h_n^2\) chosen so that \(X_C(1-1/h_n^2) = V^2/Q\) still delivers the reactive power wanted.
Sag depth is a voltage divider, sag duration is a protection setting. \(V_{\text{sag}} = Z_F E/(Z_S+Z_F)\) gives the depth and the phase-angle jump together; the relay and breaker of Set 36 give the duration. The pair is then read against the ITIC envelope, whose lower boundary is 0% below 20 ms, 70% to 0.5 s, 80% to 10 s and 90% thereafter.
Inverting the divider gives a length, and a length gives a trip rate. Solving \(|Z_F| = k|Z_S+Z_F|\) for the fault distance yields the vulnerability radius; multiplied by the number of exposed circuits and the fault rate per kilometre per year it becomes an expected number of process trips.
Unbalance is measured by \(\text{VUF} = |V_2|/|V_1|\), the same negative-sequence component Set 21 extracted for faults. Because an induction machine's negative-sequence impedance is its locked-rotor impedance, \(Z_1/Z_2 \approx 6\) and \(I_2/I_1 \approx 6\,\text{VUF}\) — the reason a 2% supply defect forces a 5% machine derating.
Wind power is cubic and capped by momentum. \(P = C_p\,\tfrac12\rho Av^3\) with \(C_p \le 16/27 = 0.5926\). Every wind answer is a chain: available power, Betz bound, rotor coefficient, drivetrain efficiency — and \(\rho\) is data, not a constant.
A photovoltaic string is bounded at both temperature extremes. The cold limit is an insulation limit on \(N V_{oc}\); the hot limit is a tracking limit on \(N V_{mp}\). Both use the same linear coefficients about 25 °C, and both must hold for the same integer \(N\).
RoCoF is the swing equation with no governor yet. \(df/dt = f_0\Delta P/2E_{\text{sys}}\) with \(E_{\text{sys}} = \sum H_iS_i\) in MW·s. Only synchronised rotating mass counts, and synthetic inertia buys back \(E_{\text{syn}} = f_0P_{\text{syn}}/2(df/dt)\) — a power requirement, not an energy one.
A six-pulse drive in an 11 kV plant draws a fundamental current of 420 A at full load. A power-quality analyser records the following harmonic currents, expressed as percentages of the drive's own fundamental:
| Order \(h\) | 5 | 7 | 11 | 13 | 17 | 19 | 23 | 25 |
|---|---|---|---|---|---|---|---|---|
| \(I_h/I_1\) (%) | 16.5 | 11.2 | 5.8 | 4.2 | 2.4 | 1.8 | 1.1 | 0.9 |
Find the current total harmonic distortion, the harmonic current in amperes, the true rms current and the distortion factor. If the displacement factor is 0.98 lagging, find the true power factor. Comment on the shape of the spectrum against the ideal \(I_h = I_1/h\).
The distortion is a ratio, so work in percent first. Harmonics of different orders are orthogonal over a period, so their rms values add in quadrature:
Note what is not in the list: no even orders, because a symmetric converter has half-wave symmetry, and no triplens, because a three-wire delta-fed bridge cannot produce a zero-sequence current.
The harmonic currents in amperes, at \(I_1 = 420\) A:
The true rms current follows from the same orthogonality:
A 21% distortion raises the rms current by only 2.3%. Squares are forgiving at these levels — which is exactly why heating alone is a poor detector of distortion, and why a plain moving-iron ammeter tells you nothing about it.
The distortion factor and the true power factor. With a sinusoidal supply voltage only the fundamental current transfers average power, so \(P = V_1I_1\cos\varphi_1\) while \(S = V_1I_{\text{rms}}\):
Set 30 would have installed capacitors to raise 0.98 to unity. Doing so here leaves a ceiling of 0.978, and the capacitors themselves make matters worse in a way Problem 4 makes explicit.
Against the ideal spectrum. The textbook six-pulse bridge gives \(I_h = I_1/h\), hence 20.0, 14.3, 9.1, 7.7, 5.9, 5.3, 4.3 and 4.0 per cent and \(\text{THD}_I = 29.0\%\) truncated at the 25th. The measured spectrum is uniformly below it:
The gap widens with order, which is the fingerprint of commutation overlap and a series line reactor: both round the corners of the quasi-square current, and rounding a waveform removes its high-order content first. A drive fitted with a 3% line reactor typically shows exactly this pattern.
The plant of Problem 1 is connected at an 11 kV point of common coupling with a three-phase fault level of 250 MVA. Its maximum demand under the connection agreement is 9 MVA. Assess IEEE 519 compliance at full load, where the drive draws 420 A of fundamental, and at 40% load, where it draws 168 A. State exactly which limits are breached in each case.
Fix the two reference currents, because everything in IEEE 519 is a percentage of one of them:
That band allows 7.0% for \(3\le h<11\), 3.5% for \(11\le h<17\), 2.5% for \(17\le h<23\), 1.0% for \(23\le h<35\) and 8.0% TDD — every figure a percentage of \(I_L\), never of the present fundamental.
Convert distortion to demand distortion. The scaling is a single ratio:
The THD is 21.45% at both operating points. The TDD differs by a factor of 2.5, and it is the TDD that tracks the amperes actually pushed into the network.
The individual orders, as percentages of \(I_L\). Each entry is the Problem 1 percentage multiplied by \(I_1/I_L\) — that is by 0.889 at full load and 0.356 at 40%:
The verdicts read straight off the table. At full load five limits are breached:
The 17th, 19th, 23rd and 25th are all inside their limits, because the measured spectrum falls faster than \(1/h\).
At 40% load the plant complies on every count — TDD 7.63% against 8.0%, the fifth 5.87% against 7.0%, and every other order comfortably inside. The compliance verdict has changed while nothing about the drive has:
This is the reason the standard specifies a fifteen- or thirty-minute demand interval and the maximum demand of the connection agreement. A measurement taken during a night shift would certify a plant that fails every day.
What is needed to comply at full load. Working backwards from the TDD limit:
The harmonic current must fall from 90.1 A to 37.8 A. A twelve-pulse arrangement removes the 5th, 7th, 17th and 19th and leaves \(\sqrt{5.8^2+4.2^2+1.1^2+0.9^2} = 7.32\%\), hence a TDD of 6.51% — compliant on the total, but the 11th alone is then \(5.8\times0.889 = 5.16\%\) against 3.5%. Pulse multiplication first, then a filter for the 11th and 13th. Problem 6 designs it.
For the same 11 kV bus of 250 MVA fault level, compute the harmonic voltage each order of the full-load drive current produces, the resulting voltage THD at the point of common coupling, and check both against the IEEE 519 voltage limits for a system above 1 kV and up to 69 kV. Assume no shunt capacitors are present.
The source reactance, from the fault level of Set 25:
The reactance rises linearly with order, so the driving impedance at the 25th is twenty-five times the value at the fundamental.
Each harmonic voltage is \(V_h = I_h\,hX_s\), expressed below as a percentage of the nominal phase voltage:
The voltage THD, again a quadrature sum:
Both are met — the largest individual is 2.64% against 3.0%, and the total is 4.96% against 5.0%. The margin on the total is 0.04 percentage points, which is no margin at all.
A word on the convention just used. Writing \(V_h = I_h\,hX_s\) assumes every harmonic ampere returns through the source. It does not: the plant's own resistive load is a shunt path, and at 6 MW it presents \(R_L = 121/6 = 20.17\ \Omega\) at every order:
The screening calculation is deliberately the pessimistic one, and the difference is small while the bus impedance is dominated by the source. It stops being small the moment anything makes the bus impedance large, which is Problem 4.
Why the low orders dominate here and not in the ideal case. The product \(I_h h\) is what sets each contribution:
For the ideal \(I_h = I_1/h\) spectrum this product is constant and every order contributes equally to \(\text{THD}_V\). This measured spectrum decays faster than \(1/h\), so the product falls and the fifth dominates — but only by a factor of 3.7 over the 25th, not by the factor of 18 that the current amplitudes suggest.
The utility's exposure, stated as a sensitivity. Since \(\text{THD}_V \propto X_s \propto 1/S_{sc}\):
A summer switching arrangement that takes one transformer out of service and drops the fault level to 180 MVA puts the bus outside the limit without any change in the plant's behaviour. This is why the compliance assessment is performed at minimum fault level, not at the normal one.
To correct the plant's displacement factor, a 10 MVAr capacitor bank is installed at the same 11 kV bus. The bus also carries 6 MW of resistive load, which is the only damping present; take the source \(X/R\) as 10 at the fundamental. Find the order of the parallel resonance, the bus impedance at that order, the fifth-harmonic voltage the drive now produces, the current in the capacitors, and state whether the bank survives. Then find the resonance order for bank sizes of 12.5, 7.5 and 5 MVAr.
The resonance order. The bank sits in parallel with the inductive source. Setting the two reactances equal at order \(h_r\), with \(X_s = V^2/S_{sc}\) rising as \(h\) and \(X_C = V^2/Q_c\) falling as \(1/h\):
Exactly the order at which this drive injects its largest current. The coincidence is not bad luck: 10 MVAr is what a 9 MVA plant at 0.85 power factor needs, and a fault level of 20 to 30 times the demand is normal — so \(\sqrt{S_{sc}/Q_c}\) lands between 4 and 6 on a very large number of industrial buses.
The element impedances at \(h = 5\):
The two reactances are equal and opposite, which is the resonance itself. What limits the impedance is the resistance in the loop — here almost entirely the resistive load.
The bus impedance seen by the injecting drive is the three branches in parallel:
The susceptances of the source and the capacitor cancel to four decimal places; the bank has converted a 2.42 Ω path into a 17.29 Ω one.
The fifth-harmonic voltage, with the injected current unchanged at 69.30 A:
Problem 3 gave 2.64% for the same current. The individual limit is 3.0% and the total limit 5.0%; both are now exceeded by a factor of six, and the load's own current spectrum is exactly what it was before the bank arrived.
The current at the point of common coupling is not unchanged, which is the part that surprises people. The drive still injects 69.3 A at the fifth, but a large current now circulates between the capacitor and the source:
The utility's meter at the point of common coupling records more fifth-harmonic current than the plant's entire agreed demand. The drive is injecting 69.3 A; the network is carrying 495 A; the difference is the resonant circulation, and it belongs to nobody's load.
The capacitor current, which is what actually fails. The harmonic voltage appears across a capacitive reactance that is five times smaller than at the fundamental:
Counting every other order as well — 94.3 A at the seventh, 30.6 A at the eleventh and so on — the figure is 1.39. The bank is overloaded continuously, at every hour the drive runs at full output. Fuses will operate on the weakest can within months, and each can lost raises \(X_C\), moves \(h_r\) upward, and changes the problem into a different one.
Other bank sizes, from the same square root:
Switching one 2.5 MVAr step out moves the peak between the fifth and the seventh, halving the distortion and bringing the capacitor duty to 1.16 pu — the classic diagnostic, and a genuine improvement without being a cure, since a peak at 5.77 still amplifies both its neighbours. Switching two steps out moves it onto the seventh and the symptom returns in full, which is how a maintenance team convinces itself the capacitors are not the problem.
The 10 MVAr bank of Problem 4 is to be kept, but detuned with a series reactor. Find the reactor percentage that places the branch's own series resonance at order 4.2 and at order 3.78, the reactive power each arrangement then delivers, and the order at which the detuned branch still resonates in parallel with the source. Recommend one of the two for this bus and say what would change the recommendation.
The reactor percentage. Writing \(p = X_L/X_C\), the branch is series-resonant where \(h X_L = X_C/h\):
These are the two standard catalogue values, and they exist for exactly this reason. Both place the series resonance in a gap between characteristic harmonics — 4.2 between the 3rd and 5th, 3.78 just below the 4th.
The reactive power delivered rises, because the net fundamental reactance is smaller than the capacitor's alone:
A gain of 6.0% and 6.7% respectively, following \(Q = Q_0/(1-p)\). This is a real effect and must be allowed for in the reactive plan of Set 30 — a bank detuned after installation will overcorrect at light load.
The parallel resonance does not disappear; it moves down. The branch is inductive above \(h_n\) and capacitive below it, so it can still cancel the source susceptance somewhere below \(h_n\). Setting the branch reactance equal and opposite to the source's:
The peak impedance at 3.22 is about 14 Ω — still a large impedance, but at an order where a six-pulse converter injects nothing.
The recommendation for this bus: 5.67%. The 7% reactor puts the parallel resonance at 3.02, which is the third harmonic to within 1%:
The plant's own load is a three-wire drive and produces no third harmonic, so 7% would work today. It would stop working the moment a four-wire supply of single-phase rectifier load is added on the same bus, or the moment the site transformer is driven a few per cent above its knee and its exciting current turns peaky. 5.67% costs nothing extra and removes that dependence.
What would change the recommendation. Two things, and both are switching states rather than equipment:
At minimum fault level even the 5.67% arrangement lands on the third. A detuning study must therefore be run at both the maximum and the minimum fault level, and if the two verdicts differ the bank is interlocked with the switching state or the reactor is sized for the worse case.
Design a single-tuned shunt filter for the same 11 kV bus that supplies 4 MVAr at the fundamental and is tuned to order 4.7. Find \(X_C\), \(X_L\), \(C\) and \(L\); the fundamental current and the capacitor voltage; the fraction of the drive's fifth-harmonic current the branch diverts, taking a branch quality factor of 50; and the fifth-harmonic voltage that remains at the bus. Then check how the same branch behaves at the seventh.
Two requirements fix the two elements. Tuning at \(h_n\) means \(X_L = X_C/h_n^2\); delivering \(Q\) at the fundamental means the net reactance is \(V^2/Q\):
The physical elements, at 50 Hz:
Per phase, star-connected. Check: \(1/2\pi\sqrt{LC} = 1/2\pi\sqrt{4.566\times10^{-3}\times100.5\times10^{-6}} = 235\) Hz, which is \(4.7\times50\).
The fundamental duty. The branch is capacitive at 50 Hz and carries
The capacitor sees 4.7% more than the system phase voltage, because the reactor's drop is subtracted from the supply and the capacitor must make up the difference. Its voltage rating must exceed the system value by at least \(1/(1-1/h_n^2)\), and this catches out designers who order capacitors at the nominal system voltage.
The branch resistance and the impedance at the fifth. A quality factor of 50 defined at the tuning frequency gives
Magnitude 0.8456 Ω. The branch is inductive at the fifth, because it is tuned below it — which is the deliberate choice, since a branch tuned above the fifth would be capacitive there and would form a parallel resonance with the source instead of a series path to earth.
Current division between the filter and the source:
Three quarters of the fifth is diverted, not all of it. Detuning to 4.7 buys stability at the cost of effectiveness, and a branch tuned at 4.9 would divert 88% — until its capacitance aged upward by 2% and it went inductive at the fifth.
The fifth-harmonic voltage that remains:
Against 2.64% with no bank at all (Problem 3) and 18.86% with the plain capacitor bank (Problem 4). The filter has done better than doing nothing, which is the whole point and is not automatic.
The branch rating, which must carry both currents:
At the seventh the same branch is much less useful:
Only 38%, because the branch's inductive reactance grows with \(h\) faster than the source's does once past the tuning point. A single branch is a single-order device; the standard arrangement is a 4.7 branch and a 6.7 branch working together, with the lower one always installed first so that the bus impedance seen by the upper one is already tamed.
A 415 V four-wire lighting and office circuit carries 100 A of fundamental in each phase, balanced, together with harmonics of 65% third, 20% fifth, 15% seventh, 10% ninth and 6% eleventh, all referred to the fundamental and identical in the three phases. Find the line rms current, the current THD, and the neutral current. State what conductor sizing follows and what the supply transformer must have.
The line current is a quadrature sum over every order present:
A distortion of 70% raises the rms by 22%. This is the spectrum of a population of single-phase rectifiers with capacitor input filters — every desk supply, LED driver and television in the building.
The neutral carries the sum of the three phase currents, and the sequence classification of Chapter 39 decides which orders survive that sum:
At order \(h\) the 120° displacement of the fundamental becomes \(120h\) degrees, which is a whole number of turns whenever \(h\) is a multiple of three. The three third-harmonic currents are therefore identical, not displaced.
The neutral current is three times the rms of the triplens alone:
The neutral carries 61% more than any line conductor, in a circuit that is perfectly balanced. Every rule of thumb that sizes a neutral at half the phase conductor, or omits it from the overload protection, is wrong here by a factor of three.
The conductor consequence. Sizing on rms current alone:
IEC 60364 requires the neutral of such a circuit to be sized as a current-carrying conductor and applies a derating factor to the cable when the third-harmonic content exceeds 33%. In practice the neutral is run at twice the phase cross-section, or two neutrals are run in parallel — and a four-pole device is used so that the neutral is protected and switched with the phases.
What the transformer must have: a delta winding. A zero-sequence current cannot enter a delta from the line, but it can circulate inside one:
That circulation is not free. The delta winding is heated by a current it does not deliver, and a transformer feeding heavy rectifier load must be derated by its K-factor — typically K-13 or K-20 for an office block, meaning the transformer is oversized by 30 to 50% for a load that draws its nameplate kVA.
A 33 kV substation has a source impedance of \(j2.5\ \Omega\) referred to that voltage. The feeders leaving the bus have an impedance of \(0.28+j0.42\ \Omega\) per kilometre. A three-phase fault occurs 4 km along one feeder and the protection clears it in 120 ms. Find the retained voltage and the phase-angle jump at the bus, plot the point against the ITIC envelope, and state the verdict for a load that follows ITIC and for one built to SEMI F47.
The bus is the mid-point of a divider between the source e.m.f. and the short circuit. Nothing else about the network matters while the fault is on:
The total impedance, which is where the source's high \(X/R\) shows up:
Read both numbers. The load retains 46.7% of nominal for 120 ms, and its voltage phasor jumps back by 18.7°:
The jump arises because the feeder's \(X/R\) of 1.5 differs from the source's \(X/R\) of infinity — the divider is complex, not real. A phase-controlled converter whose firing is referenced to a phase-locked loop can misfire on a jump of this size before its depth relay has even picked up.
The ITIC verdict. The lower boundary is 0% below 20 ms, 70% from 20 ms to 0.5 s, 80% from 0.5 s to 10 s and 90% beyond:
Equipment designed only to ITIC is expected to drop out. It is not expected to be damaged — the lower boundary is a functional limit, not a withstand limit, and the distinction matters when apportioning cost.
The SEMI F47 verdict is different, and that difference is the reason the specification exists:
Even the tighter specification is not met, by 3.3 percentage points. A fault one kilometre further out would have been survivable by an F47 machine and not by an ITIC one, which is precisely the band of events the specification was written to capture.
For the same 33 kV substation, invert the question. Find how far along a feeder a three-phase fault must lie before the retained voltage at the bus stays above 70%, and again above 80%. The substation has five feeders, each longer than the radii you find, and the three-phase fault rate is 0.09 per kilometre per year. Estimate the expected number of process trips per year for a load that follows ITIC and for one that trips at 80%, and comment on which of the two numbers a plant manager should be shown.
Set the retained voltage to \(k\) and solve for the distance. With \(Z_F = d(0.28+j0.42)\) and \(Z_S = j2.5\), work with squared magnitudes so no angles are needed:
The cross term \(2\times2.5\times0.42 = 2.10\) is what makes the relation non-linear: doubling the distance does not double the retained voltage.
The condition \(|Z_F|^2 = k^2|Z_S+Z_F|^2\) is a quadratic in \(d\):
Check the first: at 10.22 km, \(Z_F = 2.862+j4.292\), \(Z_S+Z_F = 2.862+j6.792\), ratio 5.158/7.370 = 0.700. Correct.
The vulnerability areas. Every three-phase fault inside these radii, on any of the five feeders, takes the bus below the stated level:
Multiply by the fault rate:
Ten percentage points of ride-through capability are worth 3.1 process trips a year at this substation. The relation between tolerance and trip rate is strongly non-linear, because the vulnerability radius grows faster than the tolerance does.
Which number to show, and why neither is the answer. Both are lower bounds, for four reasons that all push the same way:
A single line-to-ground fault gives a deep sag on one phase and a shallower one on the others, and what a three-phase load sees depends on the transformer connections in between — Set 23's zero-sequence blocking turns a one-phase sag upstream into a two-phase sag downstream. The honest figure to present is 7.74 as the three-phase contribution for the actual equipment tolerance, stated as a floor, with a measured sag survey commissioned to fill in the rest.
What the number is for. Multiplied by the cost of a trip it is the entire business case:
At ₹18 lakh per trip for a continuous process — lost output, scrapped material in the line and a four-hour restart — 7.74 trips a year is ₹1.4 crore. The DVR of Problem 10 costs a fraction of that, which is why the calculation is worth doing carefully rather than quoting a rule of thumb.
A 3 MVA load at 0.92 power factor on the bus of Problem 8 is to be protected against the 46.7%, 120 ms sag. Size a dynamic voltage restorer for it: the injected voltage, the converter rating, the real power drawn and the energy store required. Compare with a double-conversion UPS sized to ride through a one-minute interruption, and state which disturbances each device cannot handle.
The DVR inserts the missing phasor in series with the feeder. Its rating is set by what is missing, not by what the load needs:
The full load current passes through the injection transformer, so the current rating is 100%; only the voltage rating is fractional. That is the whole economy of the series device.
The real power and the energy, for in-phase injection:
Forty-nine watt-hours — less than a mobile telephone battery. The store is trivial because a sag is short; the converter is not, because a sag is deep.
The UPS for a one-minute interruption is a different machine entirely:
Nearly a thousand times the storage, twice the converter, and a permanent 4 to 8% conversion loss on the whole throughput all year. The DVR is idle and lossless between events.
The economic comparison, made explicit:
At 6% standing loss the UPS burns 166 kW continuously — 1.45 GWh a year, which at ₹7 per kWh is ₹1.0 crore of energy bought to protect against an event that lasts 120 ms.
What each cannot do. The limits are structural, not a matter of specification:
For this site the DVR is right, because Problem 9 counted sags and not interruptions. If the trip log showed momentary interruptions from an auto-reclose scheme, the DVR would ride through none of them and the answer would change completely — which is why the survey precedes the equipment.
A 415 V four-wire supply is measured at a distribution board as \(V_a = 240\angle0^\circ\), \(V_b = 228\angle{-}122^\circ\) and \(V_c = 244\angle120^\circ\) volts to neutral. Find all three sequence components, the voltage unbalance factor, and the NEMA figure. Explain why the two disagree, and state which of them belongs in a connection agreement.
Resolve into rectangular form first, with \(a = 1\angle120^\circ\):
The zero-sequence component, which exists here only because the supply is four-wire:
A 6 V neutral displacement. It drives current in the neutral conductor but it does not reach a delta-connected or three-wire load at all, which is why it plays no part in motor heating.
The positive-sequence component. The rotations bring the three phasors nearly onto the real axis:
Very nearly the arithmetic mean of the three magnitudes, 237.33 V, because the angle errors are small. That coincidence is useful as a check and disappears as soon as the angles depart further from 120°.
The negative-sequence component, which is the whole point of the calculation:
Check: \(V_0+V_1+V_2 = (-0.94+237.29+3.65)+j(5.99-2.65-3.33) = 240.00+j0.01 = V_a\). The three components reconstruct phase \(a\), as they must.
The unbalance factor of IEC 61000-4-30:
Just outside the 2% limit that almost every standard imposes on steady-state unbalance — a compliance failure, and only barely.
The NEMA figure uses magnitudes only:
Why they disagree, and by how much. The ratio is not a constant:
The last line is the fatal one. Three equal magnitudes at 0°, −110° and 120° are badly unbalanced and the NEMA index reports nothing at all, because it never looks at an angle. Here the deviations are unequal and there is a residual angle error, so the two indices part company by 89% rather than by the 73% of the symmetric case.
Which belongs in a contract: the sequence definition, and the reason is that it is the one that predicts damage:
The NEMA figure survives because it needs three voltmeter readings and no phasor measurement, which mattered in 1950 and does not now. Read as compliance numbers, 2.08% is a marginal failure and 3.93% is a gross one — so the choice of definition decides whether a supplier owes a remedy.
A 160 kW, 415 V induction motor is supplied from the board of Problem 11. Its negative-sequence impedance is one sixth of its positive-sequence impedance at rated slip. Find the negative-sequence current as a fraction of rated, the additional \(I^2R\) loss it produces, and the rise in total stator rms current. State the derating required and explain why the loss figure understates the damage.
The negative-sequence current follows from the two ratios. The machine presents its locked-rotor impedance to the backward field, because the rotor sees that field at a slip of very nearly 2:
A 2% defect in the supply produces a 12.5% current in the machine. The amplification is the impedance ratio, and it is the single most important number in the whole of unbalance analysis.
The extra loss goes as the square, so it looks small:
A thermal overload relay measuring rms current sees a rise of 0.78% and does nothing whatever. This is why unbalance protection is a separate function — a dedicated negative-sequence element — and not a matter of setting the overload a little tighter.
The derating is far larger than 1.56% suggests. NEMA MG-1 gives the factor as a curve, and at this level:
Eight kilowatts of a 160 kW machine given up to a 2% defect in somebody else's wiring. Above 5% the curve is not defined at all and NEMA simply forbids operation.
Why the loss figure understates the damage. Three effects compound, and none appears in \(I_2^2\):
So the rotor loss is two to three times the 1.56% figure and it is concentrated in a few square centimetres of bar and end ring. Insulation life halves for every 10 °C of hot-spot rise, and it is the hot spot rather than the mean that ends the machine.
The torque consequence, which is separate. The backward field produces a braking torque:
A double-frequency pulsating torque at 100 Hz appears on the shaft. On a large machine with a flexible coupling this excites torsional modes and is the reason unbalance limits are sometimes tighter for turbo-alternators than the thermal argument alone would require.
A three-bladed wind turbine has a rotor diameter of 126 m and operates in air of density 1.225 kg/m³. At a hub-height wind speed of 11 m/s its power coefficient is 0.46 and the combined gearbox and generator efficiency is 0.96. Find the power in the wind, the Betz maximum, the mechanical power at the shaft and the electrical output. Express the rotor's performance as a fraction of the Betz bound and the whole machine's as a fraction of the wind.
The swept area and the wind power. The mass flow through the disc is \(\rho Av\) and each kilogram carries \(v^2/2\) of kinetic energy:
The Betz bound, which is a property of momentum conservation and not of the blade:
The best possible actuator disc slows the far wake to one third of the upstream speed and the flow at the disc itself to two thirds. Extracting more would require the wake to stop, and stationary air cannot move aside to let the next parcel through.
The rotor and the drivetrain, in that order:
Note the order: \(C_p\) is an aerodynamic coefficient applied to the wind, and \(\eta\) is an electromechanical efficiency applied to the shaft. Multiplying the two together and calling the product an efficiency is the commonest error in this calculation, because it invites comparison with the Betz limit on the wrong quantity.
The two fractions, which say different things:
A modern rotor reaches about four fifths of a bound no design can pass. The remaining fifth goes to wake rotation, to the finite number of blades and to profile drag, and a century of aerodynamics has not recovered much of it — which is why the industry's growth has come from larger rotors rather than better ones.
The sensitivity that governs everything. Both the area and the cube law are strong levers:
A 9% taller tower that reaches 9% more wind buys more than a 9% larger rotor, and costs less. This single comparison is why hub heights have risen faster than rotor diameters at inland sites and why offshore machines, which have the wind already, grow the rotor instead.
The turbine of Problem 13 is rated at 5 MW. Find the rated wind speed at which that output is first reached, the rotor speed there and at 11 m/s for an optimum tip-speed ratio of 8, and the gearbox ratio needed to drive a 1500 rpm generator. At an offshore capacity factor of 0.45, find the annual energy and the equivalent full-load hours. Explain why the machine is not rated for the power available at 20 m/s.
Work backwards through the chain from the electrical rating to the wind:
Only 0.40 m/s above the 11 m/s of Problem 13, which produced 4.489 MW. The cube law makes the last 11% of output cost only 4% more wind.
The rotor speed follows from the tip-speed ratio. Holding \(\lambda = \omega R/v\) at its optimum is what variable speed is for:
Blade tip speed at rated is \(\omega R = 91.2\) m/s, which is close to the acoustic ceiling of about 80 m/s onshore and is one of the reasons offshore machines are allowed to be faster and therefore lighter.
The gearbox:
Three stages, and the single least reliable component in the machine. The alternative is a direct-drive permanent-magnet generator turning at 13.8 rpm, which for 5 MW needs of the order of 80 poles and a stator several metres across — heavier and dearer, but with nothing to wear out.
Annual energy at a capacity factor of 0.45:
The capacity factor is a statement about the wind at this site measured against a rating the manufacturer chose. It is not availability, which is separately about 97%, and it is not efficiency.
Why not rate the machine for 20 m/s? Compute what that would mean:
A generator, converter, transformer and cable of 27 MW instead of 5 MW — five and a half times the electrical plant — to capture the few dozen hours a year the wind blows at 20 m/s. Above \(v_{\text{rated}}\) the blades are pitched out of the wind to hold \(C_p\) down and the surplus is deliberately spilled, and at 25 m/s the machine shuts down and feathers because the structural loads, not the electrical ones, become unacceptable.
The economic statement of the same point:
A machine rated well below the peak available power has a high capacity factor and cheap plant; one rated near the peak has a low capacity factor and expensive plant. The optimum sits where the wind distribution says it does, which for most sites is near the mean wind speed multiplied by about 1.5.
The same 5 MW machine is proposed for an inland plateau site 1500 m above sea level with a mean air temperature of 30 °C. Take the standard atmosphere \(p = p_0(1-2.25577\times10^{-5}z)^{5.2559}\) with \(p_0 = 101\,325\) Pa, and the gas constant of dry air as 287.05 J/kg·K. Find the site air density, the output at 11 m/s, the new rated wind speed, and the power coefficient that would be needed to restore the sea-level output. Say what this does to the annual energy estimate and to the turbine's own control.
The site pressure, from the standard atmosphere:
The site density, from the ideal gas law at 30 °C:
Altitude accounts for 16.5 percentage points of the deficit and temperature for the remaining 4.2. Both matter; a cold high site loses much less than a hot one.
The output falls in exact proportion, because \(\rho\) is a simple multiplier on \(P_{\text{wind}}\):
A shortfall of 928 kW at the same wind speed, on the same machine, with nothing wrong with either. Twenty-one per cent of the plant's output is lost to the air it is standing in.
The rated wind speed rises as the cube root:
Nearly a metre per second higher. Since the wind speed distribution falls steeply in that region, the number of hours at or above rated collapses, and the capacity factor falls by considerably more than the 21% the density alone suggests.
Could a better rotor compensate? Solve for the coefficient that would restore 4.489 MW at 11 m/s:
No rotor has ever reached 85% of Betz, and momentum conservation forbids exceeding 100%. The answer is unambiguous: the air density cannot be engineered around at the rotor. The remedies are a larger rotor — \(D\propto\rho^{-1/2}\) gives \(126/\sqrt{0.7932} = 141\) m for the same output — or a different site.
What breaks in the paperwork. An energy estimate that uses 1.225 kg/m³ at this site is wrong by a factor of 1.26:
And the true figure is lower still, because the rated speed has moved. Four thousand megawatt-hours a year is the difference between a financeable project and an unfinanceable one, and it arises entirely from a constant copied out of a textbook.
What breaks in the turbine's own control. The controller holds \(\lambda\) at its optimum by matching generator torque to rotor speed with a fixed law:
The constant \(K\) contains \(\rho\). A controller commissioned with the sea-level value applies 26% too much torque at the plateau site, the rotor runs slow, \(\lambda\) falls below its optimum and \(C_p\) falls with it — so the machine loses a further one or two per cent on top of the density loss until \(K\) is re-entered. Site commissioning includes exactly this parameter.
A crystalline silicon module is rated at standard test conditions as \(V_{oc} = 49.5\) V, \(I_{sc} = 13.9\) A, \(V_{mp} = 41.2\) V and \(I_{mp} = 13.1\) A, with an area of 2.58 m². Find the maximum power, the fill factor and the efficiency. State what each of the four measured quantities depends on physically, and what the fill factor would tell an engineer about a degraded module.
The maximum power is the product at the knee:
The fill factor compares that rectangle with the one the open-circuit and short-circuit points would enclose:
Between 0.75 and 0.82 for a healthy crystalline module. It is a shape factor: it measures how square the corner of the \(I\)–\(V\) curve is, and nothing else.
The efficiency is referred to the aperture area at 1000 W/m²:
Equivalently 209 W per square metre. That figure, not the module wattage, is what sizes a roof — a 200 kW array needs 956 m² of module and rather more of roof.
What each quantity depends on:
Irradiance moves the current plateau up and down; temperature moves the voltage cliff left and right. The two act on different axes, which is why string design has to be checked at two quite separate operating points.
The fill factor as a diagnostic. Series and shunt resistance each degrade it in a distinguishable way:
A module whose \(V_{oc}\) and \(I_{sc}\) are both correct and whose fill factor has fallen has lost nothing at its two end points and 11% in between. This is the signature a flash test is looking for, and it is invisible to a multimeter reading open-circuit volts on a sunny day.
Temperature in the field, for completeness:
Sixteen per cent below the nameplate on a hot roof, and the nameplate was measured at a cell temperature of 25 °C that a module in service almost never sees. This is why plants are also quoted at NOCT or at PVUSA test conditions.
The module of Problem 16 is to be used on a 1500 V system whose inverter has an MPPT window of 850 to 1300 V. The record low ambient at the site is −8 °C and the highest expected cell temperature is 68 °C. Take \(V_{oc}\) to fall at 0.27% per °C and \(V_{mp}\) at 0.34% per °C, both referred to 25 °C. Find the largest and the smallest permissible number of modules in series, choose a string length, and verify it at both extremes.
The cold limit is an insulation limit, and it uses open-circuit voltage because that is the highest voltage the string can ever present — at dawn, before the inverter connects:
Twenty-eight would give \(28\times53.91 = 1509.5\) V, exceeding the system rating by 9.5 V on the coldest morning of the decade. There is no margin to negotiate here: the 1500 V figure is the cable and module insulation rating, and it is a safety limit rather than a performance one.
The hot limit is a tracking limit, and it uses the maximum-power voltage because that is where the inverter will try to operate:
A shorter string falls out of the bottom of the MPPT window on a hot afternoon. The inverter does not fail; it simply stops tracking and clamps at its lower limit, losing several per cent of the array's output at the very hours the irradiance is highest.
The permitted range and the choice:
Take the longest string the cold limit allows. A longer string means fewer strings for the same array power, hence fewer combiner ways, less DC cable and a lower current for the same power — all of which reduce cost and \(I^2R\) loss.
Verify at both extremes, which is the step most often skipped:
The middle line is the one that catches people out: on a cold bright morning the array is not at open circuit, it is at its maximum power point, and 1237 V must still be inside the tracking window. Here it is, by 63 V. Had the window's upper edge been 1200 V, 27 modules would have been legal and untrackable.
The string in operation:
String current is a module property and does not depend on the string length, which is why the cold-limit calculation is the only thing that sets how much power one DC circuit can carry.
What changes if the site is colder. At a record low of −20 °C:
Still 27, but only just — at −21 °C it becomes 26. A single degree of design temperature can remove a module from every string in a 200 MW plant, which is why the record low is taken from a long meteorological record and not from a typical year.
A 50 Hz system faces a largest credible loss of 400 MW. In a winter evening hour, demand is 4200 MW and 5200 MVA of synchronous plant is synchronised at an average inertia constant of 4.5 s. In a spring midday hour, demand is 2600 MW and only 1800 MVA of synchronous plant remains synchronised, the rest of the demand being met by converter-interfaced wind and solar; the same 400 MW infeed is at risk. Find the initial rate of change of frequency and the time to reach the first load-shedding stage at 49.2 Hz in each hour, if nothing responds.
Only synchronised rotating mass counts. The stored kinetic energy is the sum over machines actually connected, in megawatt-seconds:
Demand does not appear. A converter-interfaced plant has a rotor in the wind case and none in the solar case, and in neither does its energy reach the network, because the converter breaks the mechanical link between rotor speed and system frequency.
The swing equation of Set 27, evaluated at \(t = 0^+\):
No governor, droop or AGC term appears, because none of them has acted yet. The initial slope is pure physics and is fixed entirely by the size of the loss and the inertia synchronised at that instant.
The time to the first shedding stage, extrapolating the initial slope:
In the winter hour a governor delivering primary response in one to two seconds arrests the fall before shedding; in the spring hour nothing on the system is fast enough, and the first stage of under-frequency load shedding operates on an event that a full complement of plant would have absorbed unnoticed.
The comparison, made explicit:
RoCoF is inversely proportional to inertia and nothing else in this expression changed. The disturbance is identical — the same 400 MW — and the system's ability to absorb it has fallen by a factor of nearly three. The spring midday hour, with light demand and abundant renewables, is the dangerous one, which inverts every intuition built on a system of thermal plant.
For the spring hour of Problem 18, the operator must hold the initial RoCoF at or below 0.5 Hz/s. Find the deficit the remaining inertia can absorb at that slope, the synthetic-inertia power the converters must therefore supply, the virtual kinetic energy this represents and the energy actually delivered in the first second. Compare the cost of providing it from a battery with the cost of keeping synchronous plant on. Then examine what happens if a 600 MW embedded generation fleet is fitted with RoCoF relays set at 1.0 Hz/s.
How much loss the real inertia can take at 0.5 Hz/s. Rearrange the RoCoF expression for \(\Delta P\):
The 8100 MW·s of synchronised mass can absorb 162 MW at the permitted slope. The credible loss is 400 MW, so 238 MW of the disturbance has nothing to absorb it.
The synthetic-inertia power is the difference, delivered within the first cycle or two of the event:
Expressed as virtual kinetic energy, using the control law \(P_{\text{syn}} = (2E_{\text{syn}}/f_0)(df/dt)\):
The system must behave as though 11 900 MW·s of extra rotating mass were present — which at \(H = 4.5\) s is 2644 MVA of synchronous plant, a figure worth holding on to for the comparison below.
The energy that service actually delivers is trivial:
Sixty-six kilowatt-hours — the contents of a small electric car. A 238 MW battery of one-hour duration stores 238 MWh, some 3600 times what this service consumes. The binding constraint is the power rating and the response time, never the store.
The two ways of buying the same 0.5 Hz/s:
Keeping 2644 MVA of thermal plant synchronised at minimum stable generation displaces perhaps 800 MW of zero-fuel-cost renewable output every hour it is held — which is the minimum-generation curtailment that batteries and synchronous condensers were introduced to avoid. This is why fast frequency response is the product batteries win and bulk energy arbitrage is the one they struggle with.
The three limitations of synthetic inertia, none of which the arithmetic shows:
The second is the awkward one. A turbine that gives up rotor energy at \(t = 0\) must take it back at \(t = 5\) to 10 s, and that is often exactly when the frequency nadir arrives. Battery-sourced synthetic inertia has no such payback, which is why the two are not interchangeable even at the same MW.
The RoCoF relay fleet, which turns an event into a cascade. Without synthetic inertia the actual slope is 1.235 Hz/s, above the 1.0 Hz/s setting:
The slope trebles, and any generation still connected on a slower setting now trips too. Protection intended to detect islanding has converted a survivable single infeed loss into a system emergency — which is not a hypothetical, and is why RoCoF settings were raised from 0.125 Hz/s to about 1 Hz/s with added definite-time delay across whole national systems, one installation at a time.
With the synthetic inertia in place the relay problem disappears too:
One 238 MW service therefore buys two quite different things: it keeps the frequency nadir above the shedding stage, and it keeps the system-wide slope below the setting of a protection fleet nobody can afford to revisit quickly. Operators procure it for the second reason at least as often as the first.
Return to the 11 kV plant of Problems 1 to 6, with the 10 MVAr bank of Problem 4 in service. The utility has issued a non-compliance notice quoting a voltage THD of 19% and a fifth-harmonic current at the point of common coupling greater than the plant's whole agreed demand, and the plant's capacitor fuses have operated three times in six months. Evaluate four remedies against the IEEE 519 current and voltage limits, choose one, and state the residual risks. The fault level can fall to 180 MVA under a summer switching arrangement.
Establish the present state, evaluating the bus impedance of Problem 4 at every order and taking the current the utility actually measures — the source current, not the drive's:
The notice is confirmed on both counts, and both figures are dominated by one order. Above the seventh the bank has made matters better than the no-bank case of Problem 3, because it is a low impedance there — which is why a survey reporting only totals conceals the diagnosis.
Remedy 1 — switch one 2.5 MVAr step out. Free, reversible, and available this afternoon:
The capacitors are saved and the distortion is halved, but a peak at 5.77 amplifies both its neighbours and nothing complies. Its value is diagnostic: if the measured distortion halves when a step is switched out, the resonance hypothesis is proved at zero cost, and that proof is what justifies the capital for what follows.
Remedy 2 — detune the existing bank with a 5.67% series reactor. Three reactors, no other change:
Every individual limit and the total are met, and the voltage limits with room to spare. The reason is that a detuned bank is a very lossy filter: above order 4.2 the branch is inductive but still far below the source reactance, so it diverts a large share of every characteristic harmonic away from the utility. This is the property that makes detuned banks the commonest harmonic remedy in industry, and it is rarely stated in the catalogue.
Check remedy 2 at the minimum fault level, which is where remedy 1 fails outright:
Compare remedy 1 at 180 MVA: \(h_r = \sqrt{180/7.5} = 4.90\), back onto the fifth, with \(\text{THD}_V\) above 8%. The detuned arrangement holds because its resonance moves into an empty part of the spectrum as the fault level falls, while the plain bank's moves onto a populated one.
The capacitor duty under remedy 2:
The harmonic duty has become negligible, but the cans now stand at 6% above the system phase voltage because the reactor's drop must be made up. Detuned banks are therefore built with capacitors rated one voltage step higher — 12 kV cans on an 11 kV bank — and reusing the existing cans without checking this is a common and expensive error.
Remedy 3 — replace the bank with two tuned branches at 4.7 and 6.7, 4 MVAr each, following Problem 6:
Better than remedy 2 on every count, at the cost of six components instead of three and 8 MVAr of reactive support instead of 10.6 — which must be reconciled with the displacement-factor requirement of Set 30. Worth doing if the plant expands or if a neighbour's distortion appears on the bus; not worth doing merely to convert a pass into a better pass.
Remedy 4 — convert the drive to twelve-pulse and filter the residue. Two six-pulse bridges from a star and a delta secondary, 30° apart, cancel \(h = 5, 7, 17, 19\) and leave 11, 13, 23, 25:
The best result by an order of magnitude, and the only one that removes the harmonics at source rather than diverting them. It also holds at 180 MVA, giving 0.56% and 0.47%. Its cost is a new converter transformer and a drive rebuild, which is a capital project rather than a maintenance one.
The decision. Rank the four by what each achieves against what it costs:
Remedy 1 this week, to stop the fuse operations and prove the diagnosis; remedy 2 as the permanent answer, because it is the cheapest arrangement that meets every limit at both fault levels; remedies 3 and 4 held in reserve against an expansion. Doing 4 first would be technically excellent and would spend a capital budget on a problem three reactors can solve.
The residual risks, all of which are switching states or future load rather than equipment failures:
The second is the one to write into the site's connection rules: this bus may not carry single-phase rectifier load on a four-wire supply without the detuning being re-studied. The last is the one that produces disputes, since a low-impedance shunt is low-impedance to everybody's current, and a branch on a shared bus must be rated for the bus rather than for its owner.
Practice Problems
Work each on paper before opening the answer. The answer is given so you can check yourself; the method is deliberately not. Take \(f_1 = 50\) Hz and \(\rho = 1.225\) kg/m³ unless told otherwise.
P1. A load draws 180 A of fundamental together with 30 A at the 5th, 20 A at the 7th and 9 A at the 11th. Find \(\text{THD}_I\), the rms current and the distortion factor.
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\(I_{\text{harm}} = \sqrt{900+400+81} = 37.16\) A, so \(\text{THD}_I = \mathbf{20.6\%}\), \(I_{\text{rms}} = \mathbf{183.8}\) A, distortion factor \(\mathbf{0.979}\). Problem 1.P2. The same load is at a point of common coupling with \(I_L = 250\) A. Find the TDD.
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\(37.16/250 = \mathbf{14.9\%}\), or equivalently \(20.6\times180/250\). Problem 2.P3. A 33 kV bus has a fault level of 300 MVA and carries a 12 MVAr correction bank. At what order does it resonate with the source?
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\(h_r = \sqrt{300/12} = \mathbf{5.0}\) — the worst possible answer for a six-pulse load. Problem 4.P4. What series reactor, as a percentage of \(X_C\), places a bank's series resonance at order 4.5?
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\(p = 1/4.5^2 = \mathbf{4.94\%}\). The catalogue values 5.67% and 7% correspond to 4.2 and 3.78. Problem 5.P5. Design a single-tuned branch for a 6.6 kV bus delivering 3 MVAr at order 4.7. Find \(X_C\), \(X_L\), \(C\) and \(L\).
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\(X_C = 14.52/0.95473 = \mathbf{15.21\ \Omega}\), \(X_L = \mathbf{0.689\ \Omega}\), \(C = \mathbf{209.3\ \mu\text{F}}\), \(L = \mathbf{2.19}\) mH. Problem 6.P6. A 33 kV source of \(j2.0\ \Omega\) feeds feeders of \(0.3+j0.4\ \Omega\) per km. Find the retained voltage and phase jump for a fault 6 km out.
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\(Z_F = 3.0\angle53.13^\circ\), \(Z_S+Z_F = 4.754\angle67.75^\circ\), so \(V = \mathbf{0.631\angle{-}14.6^\circ}\) pu — inside the ITIC envelope for any duration under 0.5 s. Problem 8.P7. A supply gives \(|V_1| = 231\) V and \(|V_2| = 5.1\) V. Find the VUF and the negative-sequence current in a motor with \(Z_1/Z_2 = 6\).
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\(\text{VUF} = \mathbf{2.21\%}\); \(I_2/I_1 = 6\times2.21 = \mathbf{13.2\%}\), so derate by about 5%. Problems 11 and 12.P8. A 90 m rotor at \(C_p = 0.44\) with a drivetrain efficiency of 0.95 sits in a 9 m/s wind. Find the electrical output.
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\(A = 6361.7\ \text{m}^2\), \(P_{\text{wind}} = 2.841\) MW, \(P_{\text{elec}} = 0.44\times0.95\times2.841 = \mathbf{1.187}\) MW. Problem 13.P9. For the same machine and wind speed, what is the Betz maximum, and what fraction of it does the rotor achieve?
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\(0.5926\times2.841 = \mathbf{1.683}\) MW; the rotor reaches \(0.44/0.5926 = \mathbf{74.2\%}\) of it. Problem 13.P10. A module has \(V_{oc} = 45.0\) V at 25 °C falling at 0.30% per °C. How many may be placed in series on a 1500 V system if the record low is −5 °C?
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\(V_{oc} = 45(1+0.003\times30) = 49.05\) V, so \(\lfloor1500/49.05\rfloor = \mathbf{30}\) modules. Problem 17.P11. A 50 Hz system has 15 000 MW·s of synchronised kinetic energy and loses a 300 MW infeed. Find the initial RoCoF.
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\(50\times300/(2\times15\,000) = \mathbf{0.50}\) Hz/s. Problem 18.P12. How much synthetic-inertia power would halve that RoCoF, and what energy does it deliver in the first second?
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Halving needs \(E_{\text{syn}} = 15\,000\) MW·s more, so \(P_{\text{syn}} = (2\times15\,000/50)\times0.25 = \mathbf{150}\) MW — and \(150\ \text{MJ} = \mathbf{41.7}\) kWh. Problem 19.
Challenge Problems
Three problems that need an idea before they need a formula — each one a situation in which the obvious measurement gives the wrong answer.
C1 — The plant that passes every current test and keeps failing. A 33 kV works has a certified harmonic survey showing every IEEE 519 current limit met at the point of common coupling. Its 8 MVAr capacitor bank has nevertheless lost cans twice, and two variable-speed drives trip on DC-link overvoltage most afternoons. The fault level is 200 MVA. Explain what is happening, show why the survey is both correct and useless, and give the one measurement that settles it.
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The resonance order. \(h_r = \sqrt{200/8} = 5.0\) — the bank resonates with the source at exactly the fifth.
Why the survey passes. The certified survey was taken with a current clamp on the incomer during a night shift, when the drives ran at part load. Problem 2 showed that TDD scales with \(I_1/I_L\), so a plant at 40% output can be measured compliant and be non-compliant every afternoon. Worse, a survey of each load's current, rather than the incomer's, never sees the circulating current at all — that current flows between the capacitor and the source and passes through neither the drive nor the incoming meter's assumed path.
Why the survey is useless even if correct. A resonance changes the network's response to a current, not the current itself. Problem 4 gave the arithmetic: the injected 69 A became 495 A of circulation and an 18.9% harmonic voltage while the load's spectrum did not move by a milliampere. There is no current measurement at the load that can detect this, and that is a property of the physics rather than of the instrument.
Why the drives trip. A drive rectifier charges its DC link to the peak of the line voltage. A large fifth-harmonic voltage superimposed on the fundamental raises that peak by up to the harmonic's amplitude, so an 8–10% fifth-harmonic voltage can push the DC link 8–10% above nominal — into the overvoltage trip. The drives are victims of a distortion that other drives created and the capacitor amplified.
The one measurement. Record voltage harmonics at the point of common coupling, continuously, for a week, and switch one capacitor step out for one afternoon of it. If the fifth-harmonic voltage halves when the step is switched, the diagnosis is complete: \(h_r\) moved from 5.0 to \(\sqrt{200/5.33} = 6.1\) and left the fifth alone. Nothing else in a power system produces that signature.
The remedy and its order. Detune with a 5.67% reactor, which Problem 20 showed converts the bank from an amplifier into a partial filter. Check the result at minimum fault level — at 140 MVA the parallel resonance falls to \(\sqrt{X_C/(X_L+X_s)}\) with \(X_s = 33^2/140 = 7.78\) and \(X_C = 136.1\ \Omega\), giving \(h_p = \sqrt{136.1/15.5} = 2.96\), on the third. Since the works is three-wire and produces no third, that is acceptable, and it must be written into the site's connection rules so that it stays acceptable.
The general lesson: current limits protect the network from the customer, and voltage measurements protect the customer from the network. A survey that records only one of them can certify a plant that is destroying its own equipment.
C2 — Buying inertia that does not exist. A 50 Hz island system has 1400 MVA of synchronous plant at \(H = 4\) s and is adding 400 MW of wind. The largest infeed loss is 180 MW and the RoCoF limit is 0.5 Hz/s. The operator is offered three products: 60 MW of battery fast frequency response with a 250 ms delay; synthetic inertia from the wind farm's own converters at no extra capital cost; and a 300 MVA synchronous condenser at \(H = 3\) s. Evaluate all three and recommend a purchase.
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The starting position. \(E_{\text{sys}} = 4\times1400 = 5600\) MW·s, so
\[ \frac{df}{dt} = \frac{50\times180}{2\times5600} = 0.804\ \text{Hz/s} \]Already above the limit before the wind farm displaces anything. With wind running, synchronous plant falls to perhaps 900 MVA, \(E = 3600\) MW·s and RoCoF becomes 1.25 Hz/s.
What is needed. At \(E = 3600\) MW·s the permitted deficit is \(\Delta P = 2(3600)(0.5)/50 = 72\) MW, so 108 MW must come from somewhere — equivalently \(E_{\text{syn}} = 50(108)/(2\times0.5) = 5400\) MW·s.
Product 1, the battery. 60 MW is barely half of what is needed, and the 250 ms delay is the fatal objection rather than the shortfall. RoCoF is defined at \(t = 0^+\); a service that begins at \(t = 250\) ms cannot change the initial slope at all. By then the frequency has already fallen by \(1.25\times0.25 = 0.31\) Hz. It will help the nadir and will not help the RoCoF relays, and if the constraint is a relay fleet the product is worthless. Value: real, but for a different service.
Product 2, synthetic inertia from the wind farm. Free capital, and it comes with two structural problems. Its response is derived from a differentiated frequency measurement and is delayed by 100–200 ms for the same reason as the battery. And its energy is taken from the rotor, which must then be re-accelerated: the turbine's output falls below the available power for five to fifteen seconds afterwards, frequently coinciding with the nadir. On a small island, where the wind farm is a large fraction of generation, that payback can be larger than the original deficit. Value: genuine but bounded, and it must be modelled with the recovery included, never as a pure inertia constant.
Product 3, the synchronous condenser. \(E = 3\times300 = 900\) MW·s, raising the total to 4500 and the RoCoF to \(50(180)/(2\times4500) = 1.00\) Hz/s. Not sufficient alone either — but it is the only one of the three that acts at \(t = 0^+\) with no control, no delay and no payback, because it is a rotating mass. It also brings fault current and a voltage source for the converters' phase-locked loops to lock onto, neither of which appears in the RoCoF arithmetic and both of which a converter-dominated island needs.
The recommendation. None of the three alone; the condenser plus a constraint. Buy the condenser for the inherent inertia, fault level and voltage-source behaviour; contract the battery for fast frequency response to protect the nadir, which is what it is actually good at; take the wind farm's synthetic inertia for nothing but model its recovery honestly; and set a minimum-inertia constraint in the dispatch so that \(E_{\text{sys}}\) never falls below \(50(180)/(2\times0.5) = 9000\) MW·s, which on this system means keeping enough synchronous plant on to make up the difference.
The uncomfortable arithmetic. 9000 MW·s at \(H = 4\) s is 2250 MVA of synchronous plant on a system whose peak demand is under 1400 MW. The constraint cannot be met by dispatch alone, which is precisely why grid-forming converters — which supply inertial response as a property of their control rather than as a measured reaction — are a requirement rather than an option on small island systems.
C3 — The sag survey that disagrees with the trip log. A pharmaceutical plant records 14 process trips in a year. The vulnerability calculation of Problem 9, done for its 33 kV supply, predicts 4.6. The plant asks the utility to explain the difference. Identify the causes in order of likely contribution, quantify what you can, and state how each would be confirmed.
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Cause 1: only three-phase faults were counted. Single line-to-ground faults are five to ten times more numerous on a distribution system. They give a deep sag on one phase and shallower sags on the others, and what the plant sees depends on the transformer connections between: Set 23 showed a delta–star transformer blocks zero sequence, so a one-phase sag upstream arrives as a two-phase sag with a phase shift. A three-phase rectifier front end loses its DC link on a two-phase sag almost as readily as on a three-phase one. Confirm by comparing the utility's fault-type statistics with the site's disturbance recorder, event by event. Expected contribution: 5 to 7 trips a year, and this alone probably closes most of the gap.
Cause 2: the assumed trip threshold is wrong. Problem 9 showed that moving the threshold from 70% to 80% took the radius from 10.2 to 17.2 km and the count from 4.6 to 7.7. Most industrial contactors drop out at 80–85% and many drives trip on the phase jump rather than the depth. Confirm by a controlled ride-through test on the actual equipment, or by reading trip thresholds off the drive parameter lists. Expected contribution: a factor of 1.5 to 2 on whatever the fault count is.
Cause 3: transmission faults were excluded entirely. A fault anywhere on the 132 kV network above this substation sags every bus fed from it, and the vulnerability calculation stopped at the 33 kV feeders. Transmission faults are rarer per kilometre but the exposed length is enormous and the sags are shallow but very widespread. Confirm by correlating trip times against the transmission operator's event log; the signature is a trip with no local fault recorded. Expected contribution: 1 to 3 a year.
Cause 4: events with no fault at all. Large motor starting on the same bus, transformer energising inrush, and auto-reclose sequences that deliver the same sag two or three times a minute apart. A reclose onto a permanent fault produces two sags and is logged by the utility as one event. Confirm by the disturbance recorder's timestamps — a repeated sag at a 0.5 s or 5 s interval is a reclose signature and nothing else.
What cannot be the cause. The utility's protection being slow. Duration affects whether a point falls outside the ITIC envelope, and at 46.7% retained voltage the point of Problem 8 is outside it for any duration beyond one cycle. Halving the clearing time would not have saved a single one of these trips, so the common demand for faster protection is misdirected here.
The reconciliation. 4.6 three-phase, plus roughly 6 from L-G faults reaching the plant through the transformer connections, plus 2 transmission, plus 1 or 2 with no fault, gives 13 to 15 — which is the log. The prediction was not wrong; it answered a narrower question than the one the plant asked.
The action that follows. The remedy is unchanged by any of this. Faults are irreducible, the vulnerability area is fixed by impedances, and mitigation is on the load side: a DVR for the sensitive line, at the rating Problem 10 computed, sized on the deepest sag the survey records rather than on the average one.
Multiple-Choice Questions
MCQ 1. A drive's THD is 22% at full load. At one third of load its THD is:
(a) 7.3% (b) 22% (c) 66% (d) it depends on the supplyShow answer
(b). Every harmonic scales with the fundamental, so the ratio is unchanged. Option (a) is the TDD, which does fall by three — that confusion is precisely what IEEE 519 was written to prevent. Problems 1 and 2.MCQ 2. IEEE 519's current limits are graded by:
(a) system voltage (b) \(I_{sc}/I_L\) (c) the customer's THD (d) the harmonic order aloneShow answer
(b) — the stiffness of the bus. A customer small compared with his bus may distort more, because his current makes little voltage. Option (a) grades the voltage limits, which are the utility's side of the same treaty. Problem 2.MCQ 3. A 250 MVA bus carries a 10 MVAr bank. The parallel resonance is at order:
(a) 2.5 (b) 5 (c) 25 (d) 12.5Show answer
(b), from \(\sqrt{S_{sc}/Q_c} = \sqrt{25}\). Answer (c) is the ratio itself — forgetting the square root is the standard error, and it gives an order at which nothing is ever injected, so the mistake looks harmless and hides a real resonance. Problem 4.MCQ 4. At parallel resonance the current injected by the distorting load:
(a) rises sharply (b) is unchanged (c) falls (d) becomes zeroShow answer
(b). The load is a current source; the resonance changes the impedance it drives, not the current it delivers. This is why the fault is invisible in the load's own spectrum and obvious in the bus voltage. Problem 4.MCQ 5. A single-tuned filter for the fifth harmonic is deliberately tuned at 4.7 rather than 5.0 because:
(a) it is cheaper (b) capacitance drifts and a branch tuned above the harmonic becomes a parallel resonance (c) it improves the reactive output (d) the fifth is not the largest harmonicShow answer
(b). Detuning downward is a one-way safety margin: below the harmonic the branch is inductive there and remains a series path, while above it the branch is capacitive and forms a parallel resonance with the source. Option (c) is true — \(Q = Q_0/(1-1/h_n^2)\) — but it is a side effect, not the reason. Problem 6.MCQ 6. In a balanced four-wire circuit carrying 65% third-harmonic current, the neutral current is approximately:
(a) zero (b) 65% of a line current (c) 1.6 times a line current (d) 3 times a line currentShow answer
(c). The triplens add rather than cancel, giving \(3I_3\) in the neutral, but the line current also contains the fundamental — so the ratio is 1.61, not 3. Option (d) compares the neutral with the third harmonic alone. Problem 7.MCQ 7. A sag of 46.7% lasting 120 ms is judged against the ITIC envelope. The relevant boundary is:
(a) 0% (b) 70% (c) 80% (d) 90%Show answer
(b). The 70% boundary runs from 20 ms to 0.5 s. The 0% segment applies only below 20 ms — one cycle, which the smoothing capacitor of any electronic supply covers. Problem 8.MCQ 8. A load that drops out below 80% retained voltage, rather than below 70%, has an expected sag trip rate larger by roughly:
(a) nothing (b) 14% (c) 68% (d) 300%Show answer
(c). The vulnerability radius grows from 10.22 km to 17.19 km, so 4.60 trips become 7.74 — a 68% increase for ten percentage points of tolerance. Option (b) is the answer if one assumes the radius scales with the threshold, which it does not: the divider is non-linear in distance. Problem 9.MCQ 9. A DVR protecting a 3 MVA load against a 53% sag is rated at about:
(a) 0.16 MVA (b) 1.6 MVA (c) 3.0 MVA (d) 5.6 MVAShow answer
(b) — the injected voltage fraction times the load, since the full load current passes through the injection transformer but only 53% of the voltage. Option (c) is the UPS rating, which is what the DVR exists to avoid. Problem 10.MCQ 10. A supply has a VUF of 2%. The negative-sequence current in an induction motor is about:
(a) 0.33% (b) 2% (c) 12% (d) 2% of the rotor current onlyShow answer
(c). The machine presents its locked-rotor impedance to the backward field, so \(Z_1/Z_2 \approx 6\) and the voltage ratio is amplified sixfold. Option (a) inverts the ratio, which would make unbalance harmless and no standard would exist. Problems 11 and 12.MCQ 11. A photovoltaic string's maximum length on a 1500 V system is set by:
(a) \(V_{mp}\) at the highest cell temperature (b) \(V_{oc}\) at the lowest ambient temperature (c) \(V_{mp}\) at STC (d) the inverter's rated powerShow answer
(b) — the highest voltage the string can ever present, which occurs at open circuit on the coldest morning before the inverter connects. Option (a) sets the minimum length, from the bottom of the MPPT window. Both must hold for the same integer. Problem 17.MCQ 12. The initial RoCoF after a generation loss depends on:
(a) the loss and the synchronised kinetic energy (b) the loss and the total demand (c) the governor droop (d) the installed capacityShow answer
(a), and on nothing else at \(t = 0^+\). Droop, governors and AGC appear only in the seconds that follow, so (c) is a distractor that is correct about the nadir and wrong about the slope. Demand and installed capacity enter only through which machines happen to be synchronised. Problem 18.
Key Formulas
| Statement | Relation | Notes |
|---|---|---|
| Harmonic current into voltage | \(V_h = I_h\,hX_s\), \(X_s = V_{LL}^2/S_{sc}\) | The screening convention; ignores load damping |
| Total harmonic distortion | \(\text{THD}_I = \sqrt{\sum_{h\ge2}I_h^2}\big/I_1\) | Invariant with loading |
| Total demand distortion | \(\text{TDD} = \sqrt{\sum_{h\ge2}I_h^2}\big/I_L\) | \(\text{TDD}/\text{THD}_I = I_1/I_L\) |
| True rms current | \(I_{\text{rms}} = I_1\sqrt{1+\text{THD}_I^2}\) | Orthogonality of harmonics |
| Power factor | \(\text{PF} = \cos\varphi_1\big/\sqrt{1+\text{THD}_I^2}\) | Displacement × distortion |
| Parallel resonance order | \(h_r = \sqrt{X_C/X_s} = \sqrt{S_{sc}/Q_c}\) | Falls as \(\sqrt{S_{sc}}\) at minimum fault level |
| Detuning reactor | \(p = X_L/X_C = 1/h_n^2\) | 5.67% → 4.2; 7% → 3.78 |
| Detuned bank output | \(Q = Q_0/(1-p)\) | Capacitor voltage rises by the same factor |
| Residual parallel resonance | \(h_p = \sqrt{X_C/(X_L+X_s)}\) | Must be checked at minimum fault level |
| Single-tuned filter | \(X_C(1-1/h_n^2) = V^2/Q\), \(X_L = X_C/h_n^2\) | \(C = 1/\omega_1X_C\), \(L = X_L/\omega_1\) |
| Harmonic sequence | \(h = 3k{+}1\) positive, \(3k{+}2\) negative, \(3k\) zero | Triplens sum in the neutral as \(3I_h\) |
| Sag depth | \(V_{\text{sag}} = Z_FE/(Z_S+Z_F)\) | Complex — gives the phase jump too |
| Vulnerability radius | \(|Z_F|^2 = k^2|Z_S+Z_F|^2\) solved for \(d\) | Quadratic; \(N = \text{(circuits)}\times d\times\lambda\) |
| ITIC lower boundary | 0% below 20 ms, 70% to 0.5 s, 80% to 10 s, 90% after | SEMI F47 asks 50% for 200 ms |
| DVR rating | \(S = (1-V_{\text{sag}})S_{\text{load}}\), \(W = P_{\text{inj}}t\) | Rating from depth, store from duration |
| Voltage unbalance factor | \(\text{VUF} = |V_2|/|V_1|\times100\%\) | NEMA uses magnitudes only and is angle-blind |
| Unbalance in a motor | \(I_2/I_1 = \text{VUF}\times Z_1/Z_2 \approx 6\,\text{VUF}\) | Derate 0.95 at 2%, 0.75 at 5% |
| Power in the wind | \(P_{\text{wind}} = \tfrac12\rho Av^3\), \(A = \pi R^2\) | \(\rho\) is data, not a constant |
| Betz limit | \(C_p \le 16/27 = 0.5926\) | Wake retains \(v_1/3\); disc sees \(2v_1/3\) |
| Air density | \(\rho = p/RT\), \(p = p_0(1-2.25577{\times}10^{-5}z)^{5.2559}\) | \(R = 287.05\) J/kg·K |
| Tip-speed ratio | \(\lambda = \omega R/v\), optimum 7–9 | Fixes rotor speed once \(v\) is known |
| PV fill factor | \(\text{FF} = V_{mp}I_{mp}/V_{oc}I_{sc}\) | 0.75–0.82 healthy; falls with \(R_s\) or \(R_{sh}\) |
| String limits | \(N_{\max} = \lfloor V_{\text{sys}}/V_{oc}(T_{\min})\rfloor\), \(N_{\min} = \lceil V_{\text{mppt,lo}}/V_{mp}(T_{\max})\rceil\) | Both must hold for one integer |
| Rate of change of frequency | \(df/dt = f_0\Delta P/2E_{\text{sys}}\), \(E_{\text{sys}} = \sum H_iS_i\) | Only synchronised mass counts |
| Synthetic inertia | \(P_{\text{syn}} = (2E_{\text{syn}}/f_0)(df/dt)\) | A power and a response time, not a store |
Common Mistakes
Comparing a THD against an IEEE 519 limit. Every current limit in the standard is a TDD limit. The plant of Problem 2 has a THD of 21.45% at both operating points and is compliant at one and not at the other — Problems 1 and 2.
Forgetting the square root in \(h_r = \sqrt{S_{sc}/Q_c}\). The ratio itself is 25 for the bus of Problem 4, an order at which nothing is injected, so the error produces a comfortable and entirely wrong verdict — Problem 4.
Treating a resonance as a change in the load's current. The drive of Problem 4 injects the same 69.3 A before and after the bank is installed. What changes is the 495 A the meter records and the 18.9% harmonic voltage the bus carries — Problems 4 and 20.
Sizing power-factor capacitors on \(\tan\varphi\) alone. Set 30's calculation is incomplete on a bus with converter load, and the 10 MVAr that corrects the displacement factor is exactly the value that resonates at the fifth — Problems 4 and 20.
Tuning a filter exactly at the harmonic, or checking it at one fault level only. Capacitance drifts upward with age and downward when a fuse takes a can out, so a branch tuned at 5.0 can drift above the fifth and become a parallel resonance; and every resonance order scales as \(\sqrt{S_{sc}}\), so a bank that is safe at 250 MVA can land on a characteristic harmonic at 180 MVA — Problems 5, 6 and 20.
Sizing a neutral conductor at half the phase conductor. With 65% third-harmonic content the neutral carries 1.61 times the line current in a perfectly balanced circuit — Problem 7.
Quoting a sag as one number, or letting one number size the mitigation. Depth comes from the impedance divider and duration from the protection: the converter is rated on the first and the store on the second. The DVR of Problem 10 needs 1.60 MVA and 49 watt-hours, and swapping the two criteria gives a device that is useless and expensive at once — Problems 8 and 10.
Writing the NEMA unbalance figure into a contract, or reading 2% as a 2% problem. The NEMA index disagreed with the sequence definition by 89% in Problem 11 and reads zero for a supply unbalanced purely in angle; and the locked-rotor impedance amplifies whatever \(V_2\) survives sixfold into the rotor, where the cooling is worst and the resistance is raised by skin effect — Problems 11 and 12.
Comparing a turbine's overall efficiency with the Betz limit. Betz bounds \(C_p\), which is aerodynamic. The machine of Problem 13 converts 44.2% of the wind and its rotor reaches 77.6% of Betz — two different statements — Problem 13.
Using \(\rho = 1.225\) at a site that is not at sea level. The plateau site of Problem 15 has 79.3% of standard density, and no rotor can recover the difference because doing so would need 97.8% of the Betz limit — Problem 15.
Checking a photovoltaic string at one temperature. The cold limit uses \(V_{oc}\) and is a safety limit; the hot limit uses \(V_{mp}\) and is an economic one; and \(V_{mp}\) at the cold extreme must also sit inside the tracking window — Problem 17.
Counting converter-interfaced plant in \(E_{\text{sys}}\), then buying the shortfall in megawatt-hours. A full-converter turbine has a rotor with real inertia the network cannot reach and a photovoltaic plant has none at all; and the replacement service of Problem 19 is 238 MW for about a second — 66 kWh, which any battery has a thousand times over. Procure power and response time — Problems 18 and 19.
This is the last set of the book, and it is the one in which the assumptions of the first thirty-nine are withdrawn one at a time. Set 3's balanced undistorted sinusoid became a measured spectrum in Problem 1. Set 25's fault level became a source reactance in Problem 3 and a resonance in Problem 4. Set 30's power-factor capacitors became the cause of the problem rather than the cure. Set 21's symmetrical components, introduced for a fault lasting five cycles, became a steady-state quality index in Problem 11. Set 27's swing equation, written for a machine, became a constraint on the whole system in Problem 18 — and the quantity it depends on turned out to be the one nobody had ever been asked to provide.
What ties the four halves together is a single accounting question. Inertia, fault current, reactive capability, voltage-source behaviour and a clean waveform all arrived free with the synchronous machine, and every calculation in Parts 1 to 7 assumed they always would. Converter-interfaced generation and converter-interfaced load unbundle them: each must now be measured, specified, procured and paid for on its own. That is what a modern grid code is, what IEEE 519 is, and what the whole of this final set has been computing — the price of the things that used to come with the machine.