Solved Problems · Set 38

Wind Energy Conversion Systems

Part 7 · Renewables and Grid Interface — a source with a cube law, a hard physical ceiling and a great deal of inertia. Six problems on wind power conversion.

Prof. Mithun Mondal 6 solved problems GATE · ESE · University

Set 38 — Wind Energy Conversion Systems

The photovoltaic array of Set 37 had one operating point worth holding and no inertia at all. A wind turbine has the same structure with every variable made harder.

Its optimum is a tip speed ratio rather than a voltage, so tracking it means controlling the speed of a 90-metre rotor. Its power scales with the cube of wind speed rather than linearly with irradiance, so a modest gust is a large power excursion. And where a panel responds instantly, a multi-megawatt rotor stores tens of megajoules of kinetic energy — which turns out to be both the control problem and, during a grid fault, the answer.

Part 7 · Chapter 28 · 6 solved problems

i Method Recap
  • The rotor extracts a fraction of the wind's kinetic power:

    \[ P = \tfrac12\rho Av^3C_p(\lambda,\beta), \qquad \lambda = \frac{\omega R}{v} \]
  • Betz caps that fraction:

    \[ C_{p,max} = \frac{16}{27} = 0.593 \]
  • Holding \(\lambda\) optimal makes the power a cubic in rotor speed:

    \[ P_{opt} = k\omega^3, \qquad k = \frac{\rho AR^3C_{p,max}}{2\lambda_{opt}^3} \]
  • A DFIG's converter handles only the slip power:

    \[ P_{rotor} = -sP_{airgap}, \qquad \frac{S_{conv}}{P_{rated}} \approx \frac{|s|_{max}}{1-s} \]
  • Direct drive means very high torque and very many poles:

    \[ T = \frac{P}{\omega}, \qquad P_{poles} = \frac{120f_e}{N_{rpm}} \]
  • Fault ride-through needs somewhere to put the power:

    \[ E = Pt_{fault}, \qquad C = \frac{2E}{V_{max}^2-V_{nom}^2}, \qquad R_{chopper} = \frac{V_{dc}^2}{P} \]
Problem 1CoreWhat the Wind Will Give

A 90 m diameter turbine operates at 12 m/s with a power coefficient of 0.45 in air of density 1.225 kg/m³. Find the swept area, the extracted power, the fraction of the Betz limit achieved, and the rotor speed at an optimal tip speed ratio of 7.5.

Solution

The swept area:

\[ A = \pi R^2 = \pi(45)^2 = 6362\ \text{m}^2 \]

Two-thirds of a hectare of moving air, intercepted by three blades. Power scales with \(D^2\), which is why turbines have grown relentlessly — doubling the diameter quadruples the output from the same wind and the same tower footprint.

The power:

\[ P = \tfrac12\rho Av^3C_p = \tfrac12(1.225)(6362)(12)^3(0.45) \]
\[ = 3.03\ \text{MW} \]

The \(v^3\) is the dominant term and the source of every control difficulty in this set. A 10% increase in wind speed is a 33% increase in power, and a gust from 12 to 15 m/s nearly doubles it.

Against the Betz limit:

\[ C_{p,max} = \frac{16}{27} = 0.593 \]
\[ \frac{0.45}{0.593} = 75.9\%\ \text{of the theoretical maximum} \]

The Betz limit is not an engineering constraint but a conservation argument: extracting all the kinetic energy would leave the air stationary behind the rotor, and stationary air cannot get out of the way of the air behind it. The optimum slows the wind to a third of its upstream speed, and 16/27 is what falls out.

The rotor speed. The tip speed ratio is the dimensionless variable that determines \(C_p\):

\[ \lambda = \frac{\omega R}{v} = 7.5 \;\Longrightarrow\; \omega = \frac{(7.5)(12)}{45} = 2.0\ \text{rad/s} = 19.1\ \text{rpm} \]

Nineteen revolutions per minute — and the blade tips are moving at \((2.0)(45) = 90\) m/s, seven and a half times the wind speed. That tip speed is what caps rotor size: beyond about 90 m/s the blades erode from rain and the aerodynamic noise becomes unacceptable, so larger rotors must turn proportionally slower.

Why \(\lambda\) has an optimum rather than simply increasing:

\(\lambda\)Behaviour\(C_p\)
Too lowblades stall; wind slips through between themlow
7–8optimal for a 3-blade rotor0.45–0.50
Too higheach blade works in the wake of the lastfalls

The optimum depends on blade count — a two-blade rotor peaks nearer \(\lambda = 10\) and a multi-blade water pumper near 1. Three blades at \(\lambda \approx 7.5\) is the compromise between \(C_p\), tip speed, noise and structural cost that the industry converged on.

The immediate consequence for the drive train:

\[ T = \frac{P}{\omega} = \frac{3.03\times10^6}{2.0} = 1.52\ \text{MN}\!\cdot\!\text{m} \]

One and a half meganewton-metres — six hundred times the mill drive of Set 31. Either a gearbox steps the speed up by a factor of about a hundred so a conventional generator can be used, or the generator is built to take that torque directly. Problems 3 and 4 take those two paths.

The cube law and the Betz limit set everything else. Power goes as \(v^3\), so the machine must be rated for a wind speed it rarely sees; and no rotor can exceed 59.3%, so the only route to more energy is more swept area — which drives rotor speed down and torque up until the drive train becomes the hard problem.
Answera\(A = 6362\ \text{m}^2,\ P = 3.03\ \text{MW}\)   b 75.9% of the Betz limit   c\(\omega = 2.0\ \text{rad/s} = 19.1\ \text{rpm}\)
Problem 2CoreFour Regions and a Cube

Describe the four operating regions of a variable-speed turbine. Derive the optimal torque control law, evaluate its constant for this machine, and find the power at 8 and 6 m/s.

3.03 MW 3 m/s 12 m/s 25 m/s I II: P ∝ v³ III: pitch IV Cp tracking
The cube law region is where the converter earns its keep
Solution

The four regions:

RegionWind speedControl objectiveActuator
I< 3 m/s (cut-in)none — parked
II3–12 m/smaximise \(C_p\) — hold \(\lambda_{opt}\)generator torque
III12–25 m/shold rated powerblade pitch
IV> 25 m/s (cut-out)shut down and featherpitch to 90°

Two different actuators for two different regions. Below rated the converter controls the torque and the pitch stays fixed; above rated the pitch spills the excess and the converter simply holds rated power. The handover between them is one of the trickier parts of turbine control.

The optimal tracking law. In region II, hold \(\lambda = \lambda_{opt}\) so that \(C_p = C_{p,max}\). Substituting \(v = \omega R/\lambda_{opt}\):

\[ P = \tfrac12\rho A\left(\frac{\omega R}{\lambda_{opt}}\right)^3C_{p,max} = k\omega^3 \]
\[ k = \frac{\rho AR^3C_{p,max}}{2\lambda_{opt}^3} \]

Every wind-speed term has vanished. The controller does not need to measure the wind at all — it commands a torque based purely on the measured rotor speed, and the aerodynamics do the rest.

Evaluate the constant:

\[ k = \frac{(1.225)(6362)(45)^3(0.45)}{2(7.5)^3} = 3.79\times10^5 \]
\[ \text{check: } P = k\omega^3 = \left(3.79\times10^5\right)(2.0)^3 = 3.03\ \text{MW}\ \checkmark \]

Equivalently, as a torque command:

\[ T_{opt} = k\omega^2 \]

Which is all the region II controller needs: measure the speed, square it, multiply by a constant, command that torque. No wind measurement, no search, no hill climbing — a striking contrast with the PV tracker of Set 37, and it works because the turbine's characteristic is known in advance where a panel's is not.

The power at lower wind speeds:

\[ P(8) = (3.03)\left(\frac{8}{12}\right)^3 = 0.898\ \text{MW} \]
\[ P(6) = (3.03)\left(\frac{6}{12}\right)^3 = 0.379\ \text{MW} \]
WindPower% of rated\(\omega\)
6 m/s0.379 MW12.5%9.5 rpm
8 m/s0.898 MW30%12.7 rpm
10 m/s1.75 MW58%15.9 rpm
12 m/s3.03 MW100%19.1 rpm

Halving the wind speed leaves an eighth of the power. That is why turbine siting is dominated by mean wind speed rather than by anything else — a site with 20% more wind yields 73% more energy, which dwarfs any conceivable improvement in the machine.

Why variable speed is worth the converter. A fixed-speed turbine holds \(\omega\) constant, so \(\lambda\) varies inversely with wind speed:

\[ \text{fixed speed} \;\Longrightarrow\; C_p = C_{p,max}\ \text{at only one wind speed} \]
Benefit of variable speedValue
Energy capture+5 to +10% annually
Drive train loadsgusts absorbed as speed, not torque
Acoustic noiseslower rotation at low wind
Grid interfacedecoupled — reactive control possible
Power qualityno tower-shadow flicker at \(3P\)

The second row is worth as much as the first. A gust hitting a fixed-speed turbine becomes a torque spike through the gearbox; hitting a variable-speed one, it accelerates the rotor instead and the energy is recovered later. The rotor's inertia becomes a buffer, and gearbox life improves markedly.

Optimal wind tracking needs no wind measurement and no search. Holding the tip speed ratio makes the optimum a cubic in rotor speed alone, so the controller commands \(T = k\omega^2\) and the aerodynamics settle at the right point — where the PV tracker of Set 37 had to hunt because a panel's characteristic is not known in advance.
Answera four regions: cut-in, \(C_p\) tracking, pitch-regulated, cut-out   b\(P_{opt} = k\omega^3\) with \(k = 3.79\times10^5\)   c 0.898 MW at 8 m/s, 0.379 MW at 6
Problem 3Exam levelA Converter for a Third of the Machine

A doubly-fed induction generator operates over a slip range of ±30%. Show that its converter needs to handle only the slip power, compute the rating for this 3 MW machine, and identify the drawbacks that have made it less popular.

Solution

The arrangement. The stator connects directly to the grid; the rotor windings are brought out through slip rings to a back-to-back converter:

\[ \text{stator} \to \text{grid directly}; \qquad \text{rotor} \to \text{converter} \to \text{grid} \]

Feeding the rotor at slip frequency lets the machine run at any speed while the stator stays synchronised at 50 Hz — hence "doubly fed". It is a wound-rotor induction machine used as Set 34's equations describe, with the slip power recovered rather than dissipated.

The power split. From the induction machine relations of Set 34:

\[ P_{stator} = P_{airgap}, \qquad P_{rotor} = -sP_{airgap}, \qquad P_{mech} = (1-s)P_{airgap} \]
\[ P_{total} = P_{stator}+P_{rotor} = P_{airgap}(1-s) = P_{mech}\ \checkmark \]

The converter handles only \(sP_{airgap}\) — and above synchronous speed the rotor delivers that power to the grid rather than absorbing it, which is why the converter must be bidirectional.

The rating at the worst case:

\[ s = -0.3\ \text{(30\% above synchronous)},\quad P_{mech} = 3.03\ \text{MW} \]
\[ P_{airgap} = \frac{3.03}{1.3} = 2.33\ \text{MW} \]
\[ P_{rotor} = (0.3)(2.33) = 0.70\ \text{MW} \]
\[ \frac{0.70}{3.03} = \mathbf{23\%\ of\ the\ machine\ rating} \]

Under a quarter. That is the DFIG's whole reason for existing: a 3 MW turbine with a 0.7 MW converter, saving three quarters of the semiconductors, the filters, the cooling and the losses.

The complete picture:

AdvantageDetail
Converter at ~25% ratingmajor cost and loss saving
Independent P and Q controlvia the rotor-side converter
Standard generatorwound-rotor induction — mature technology
Lower converter lossesonly a quarter of the power is processed
DrawbackDetail
Slip rings and brusheswear items, 90 m up a tower
Speed range limited to ±30%beyond it the converter rating grows
Poor fault ride-throughthe stator is directly connected — sees every grid disturbance
Gearbox still requiredthe highest-maintenance component
Crowbar needed for faultsprotects the rotor converter

The third row is what turned the industry away. Because the stator is bolted to the grid, a voltage dip appears directly across the machine, inducing large rotor currents that would destroy the converter — so a crowbar must short the rotor, at which point all control is lost precisely when the grid code demands controlled reactive support.

Why maintenance dominates the economics. The turbine is at the top of a tower, often offshore:

\[ \text{brush replacement} \approx \text{every 6–12 months} \]

A routine job at ground level becomes a vessel, a weather window and a crew offshore. The same argument that removed brushes from industrial drives in Set 34 applies here with the costs multiplied — which is why offshore turbines have moved almost entirely to the full-converter arrangement of Problem 4.

Where it still makes sense:

SituationDFIG?
Onshore, accessible, cost-drivenyes — still widely installed
Offshoreno — maintenance access
Weak grid, strict codesno — ride-through
Existing product linesyes — amortised development
Feeding the rotor instead of the stator shrinks the converter by four. Only the slip power passes through it, so ±30% of speed range needs 23% of rating — and the price is slip rings at the top of a tower and a stator hard-wired to a grid whose faults it must then survive.
Answera\(P_{rotor} = -sP_{airgap}\)   b 0.70 MW — 23% of the 3.03 MW machine   c slip rings, gearbox and poor fault ride-through
Problem 4Exam levelPaying for the Full Converter

The alternative is a direct-drive permanent-magnet generator with a full-rated back-to-back converter. Find the generator torque and the pole count for a 15 Hz electrical frequency, and explain what the full converter buys for its cost.

Solution

The torque, with no gearbox:

\[ T = \frac{P}{\omega} = \frac{3.03\times10^6}{2.0} = 1.52\ \text{MN}\!\cdot\!\text{m} \]

Machine size scales with torque, not power — so a direct-drive generator for 3 MW is enormous, typically 5–7 m in diameter and weighing 50–80 tonnes. That mass sits at the top of the tower and drives the whole structural design.

The pole count:

\[ P_{poles} = \frac{120f_e}{N} = \frac{(120)(15)}{19.1} = 94 \quad\to\quad 96\ \text{poles} \]

Ninety-six poles — forty-eight pole pairs, against the four of the industrial machines in Part 6. High pole count is how a slow rotor produces a usable electrical frequency, and it is also why the machine must be large in diameter: the poles have to fit around the circumference.

What the full converter buys:

BenefitWhy
Complete grid decouplingthe generator never sees the grid
Full speed range0 to rated, not ±30%
Excellent fault ride-throughthe grid-side converter handles the fault alone
Reactive support at any outputeven at zero wind
No slip ringsno brushes to replace
No gearbox (direct drive)removes the highest-failure component
Any generator typePMSG, wound-field or induction

Rows one and three are the decisive pair. With a full converter the machine is completely isolated from grid events — a voltage dip is handled entirely by the grid-side converter, which keeps controlling current throughout and can inject the reactive support that grid codes demand. The DFIG's crowbar has no counterpart here.

What it costs:

CostMagnitude
Converter rating100% instead of 23% — 4×
Converter lossesall power processed twice
Generator size and mass50–80 t nacelle mass
Magnet costhundreds of kg of rare earth
Tower and foundationsized for the nacelle mass
\[ \text{loss: } \approx 2\text{–}3\%\ \text{through the converter, against} \approx 0.7\%\ \text{for a DFIG} \]

Two extra points of loss on every megawatt-hour, forever. Against that, the gearbox removed was itself worth 1–2% and was the single largest source of unplanned downtime — so the lifetime energy comparison is much closer than the instantaneous efficiency suggests.

The three architectures compared:

Fixed speedDFIGFull converter
Converter rating0%23%100%
Speed range~1%±30%full
Energy capturebaseline+5%+8%
Gearboxyesyesoptional
Slip ringsnoyesno
Fault ride-throughpoormoderateexcellent
Reactive supportnonepartialfull
Typical uselegacyonshoreoffshore, new large

The last two rows show what decided it. As grid codes tightened to demand ride-through and reactive support — the requirements of Problem 5 — the full converter's advantages stopped being optional refinements and became conditions of connection.

Grid codes, not efficiency, chose the full converter. Its extra two points of loss and four-times converter rating were tolerable once network operators began requiring turbines to ride through faults and supply reactive current — things a directly-connected stator fundamentally cannot do.
Answera\(T = 1.52\ \text{MN}\!\cdot\!\text{m}\)   b 96 poles for 15 Hz at 19.1 rpm   c buys full decoupling, ride-through and reactive support for 4× the converter
Problem 5ChallengeStaying Connected Through a Fault

Grid codes require the turbine to remain connected through a dip to 15% of nominal voltage for 150 ms. Compute the energy that cannot be exported, show that no practical capacitor can absorb it, size the braking chopper, and describe the alternative that uses the rotor.

Solution

The problem. During a deep voltage dip the grid-side converter's export capability collapses:

\[ P_{export} \propto V_{grid} \;\Longrightarrow\; \text{at }0.15\ \text{pu, only }15\%\ \text{can leave} \]

Meanwhile the wind keeps blowing and the generator keeps producing. Power arrives at the DC link at nearly the full rate and cannot depart, so the link capacitor charges — and its voltage rises until something gives.

The energy involved:

\[ E = Pt = \left(3.03\times10^6\right)(0.15) = 455\ \text{kJ} \]

Nearly half a megajoule in a sixth of a second. For comparison, the entire link capacitor of the drive in Set 26 stored a few hundred joules.

Could a capacitor absorb it? Allowing the link to rise from 1100 to 1200 V:

\[ C = \frac{2E}{V_{max}^2-V_{nom}^2} = \frac{2\left(455\times10^3\right)}{1200^2-1100^2} = \frac{9.1\times10^5}{2.3\times10^5} \]
\[ C = \mathbf{3.95\ farads} \]

Four farads at 1200 V. That is not a capacitor bank; it is a room full of them, costing more than the converter and occupying more space than the nacelle allows. The approach is not marginal — it is off by orders of magnitude.

So the energy must be dissipated. A braking chopper — the same circuit that Set 33 used for a non-regenerative DC drive:

\[ R = \frac{V_{dc}^2}{P} = \frac{1100^2}{3.03\times10^6} = 0.40\ \Omega \]
ParameterValue
Resistance0.40 Ω
Peak power3.03 MW
Energy per event455 kJ
Dutyintermittent — rated thermally, not continuously
Controlchopper modulated to hold \(V_{dc}\)

A three-megawatt resistor sounds extravagant until the duty is considered: it operates for a fraction of a second, a handful of times a year. Like the soft starter of Set 30, it is rated for a transient rather than a continuous load, so it is far smaller than its peak power suggests.

The alternative uses the rotor. A 3 MW rotor stores an enormous amount of kinetic energy:

\[ \text{reduce the generator torque} \;\Longrightarrow\; \text{the rotor accelerates and stores the surplus} \]

Instead of dissipating the power, stop taking it — the wind then accelerates the rotor, banking the energy as speed, to be recovered once the grid returns. The pitch system simultaneously begins to feather. This works because 150 ms is short compared with the rotor's mechanical time constant, so the overspeed is only a few per cent.

What the grid code actually demands goes beyond merely staying connected:

RequirementTypical specification
Low-voltage ride-throughstay connected at 15% for 150 ms
Reactive current injectionup to 100% of rated during the dip
Active power recoveryback to 90% within 1 s of clearance
Frequency responsereduce output above 50.2 Hz
Synthetic inertiarelease rotor energy on a frequency drop
Voltage controlhold a setpoint at the connection point

The second row is the one that requires a full converter. Injecting reactive current during a dip — helping to support the very voltage that has collapsed — needs a converter in full control of its output, which a directly-connected stator with a crowbar across its rotor cannot provide.

Synthetic inertia is the same rotor energy used for a different purpose:

\[ E_{kinetic} = \tfrac12J\omega^2 \quad\text{— tens of MJ for a 3 MW rotor} \]

A converter-connected turbine is invisible to grid frequency — its rotor speed has nothing to do with the grid's — so as conventional synchronous plant retires, system inertia falls and frequency excursions become sharper. Synthetic inertia deliberately couples the two: on a frequency drop, the converter briefly takes more power than the wind is providing, slowing the rotor and releasing its stored energy. The rotor's inertia was always there; the control makes it available.

Nothing electrical can store half a megajoule in a nacelle, but the rotor already does. Absorbing a 150 ms fault in capacitance would take four farads; absorbing it as rotor speed takes a few per cent of overspeed — which is why modern turbines ride through faults by not taking the power rather than by dissipating it.
Answera\(E = 455\ \text{kJ}\)   b a capacitor would need 3.95 F — impossible   c chopper \(R = 0.40\ \Omega\), or store it as rotor speed
Problem 6ChallengeA Wind Farm

Specify the electrical system for a 30-turbine, 90 MW wind farm: the collection network, the converter architecture per turbine, and the farm-level control. Explain what changes for an offshore installation at 80 km.

Solution

The turbine terminals. Each machine produces low voltage and is stepped up immediately:

\[ \text{generator } 690\ \text{V} \to \text{converter} \to \text{transformer} \to 33\ \text{kV} \]
\[ I_{33\ \text{kV}} = \frac{3.03\times10^6}{\sqrt3(33{,}000)} = 53\ \text{A per turbine} \]

The transformer sits in the tower base or the nacelle. Stepping up at the turbine is essential: 3 MW at 690 V is 2500 A, which cannot be carried more than a few tens of metres economically.

The collection network:

ElementSpecification
Collection voltage33 kV (or 66 kV for larger farms)
Topologyradial strings of 6–10 turbines
String currentup to ~530 A for 10 turbines
Cabletapered along the string — smaller at the far end
Substation33/132 kV, 90 MVA
Protectiondirectional — fault current flows both ways

The last row matters. In a conventional distribution network fault current flows one way from the source; in a wind farm every turbine is a source, so protection must be directional and coordinated with converters whose fault contribution is limited to about 1.1 pu rather than the several per unit a synchronous machine delivers.

Farm-level control is what the network operator actually interacts with:

FunctionPurpose
Active power curtailmentrespond to a network limit or negative prices
Reactive power / voltage setpointhold voltage at the connection point
Frequency responsecurtail above 50.2 Hz, release inertia below
Wake steeringyaw upwind turbines slightly to help those behind
Ramp rate limitingtypically 10% of rated per minute

The fourth row is a genuinely modern idea: deliberately misaligning an upwind turbine costs it a little power but moves its wake off the machine behind, and the farm total improves by a few per cent. It is only possible because every turbine's converter allows independent control.

Wake losses are the farm-scale version of Set 37's shading:

\[ \text{a turbine in a wake sees reduced and more turbulent wind} \]
\[ \text{typical array losses: } 5\text{–}15\%\ \text{of the sum of individual outputs} \]

And because power goes as \(v^3\), a 10% velocity deficit costs 27% of that turbine's output. Spacing is therefore a direct trade against land or seabed area — typically 5–7 rotor diameters crosswind and 8–10 downwind.

Offshore at 80 km changes the transmission question entirely. AC cable has capacitance:

\[ I_{charging} = \omega CV\ell \quad\text{— grows with length, carries no power} \]
DistanceTechnologyReason
< 50 kmHVACsimple, cheap, no converters
50–80 kmHVAC with compensationreactors at each end
> 80 kmHVDC (VSC)charging current consumes the cable's capacity

Beyond about 80 km the charging current fills the cable and there is no room left for real power — so the connection becomes a VSC-HVDC link with an offshore converter platform. That is Set 40, and it is the largest single cost item in a far-offshore project.

The complete offshore specification:

ElementChoiceReason
Turbinedirect-drive PMSG, full converterno gearbox or brushes to service at sea
Collection66 kVhalves the current, fewer strings
Offshore substation66/220 kV or an HVDC platform
TransmissionVSC-HVDC at 80 kmcable charging current
Redundancydesigned ina vessel and a weather window per repair
Condition monitoringextensivepredict failures before they strand a crew

Every choice traces to the same fact: access costs a boat. The technology that wins offshore is whichever needs the fewest visits, which is why the full-converter direct-drive machine — more expensive and slightly less efficient than a DFIG — dominates there completely.

Offshore, maintenance access decides the electrical architecture. A gearbox and a set of slip rings are minor items onshore and major liabilities 90 metres above the North Sea — so the more expensive, slightly less efficient full-converter direct-drive machine wins, and beyond 80 km the connection itself becomes an HVDC project.
Answera 33 kV radial strings, 53 A per turbine, 33/132 kV substation   b curtailment, voltage setpoint, wake steering   c beyond 80 km, VSC-HVDC
Formulas

Key Formulas

QuantityRelationNotes
Rotor power\(P = \tfrac12\rho Av^3C_p\)3.03 MW here
Swept area\(A = \pi R^2\)6362 m²
Betz limit\(C_{p,max} = 16/27 = 0.593\)Momentum theory
Tip speed ratio\(\lambda = \omega R/v\)7–8 optimal, 3 blades
Optimal power law\(P_{opt} = k\omega^3\)No wind measurement needed
Its constant\(k = \dfrac{\rho AR^3C_{p,max}}{2\lambda_{opt}^3}\)\(3.79\times10^5\)
Optimal torque\(T_{opt} = k\omega^2\)The region II command
Cube scaling\(P \propto v^3\)Half the wind, an eighth the power
DFIG rotor power\(P_{rotor} = -sP_{airgap}\)Bidirectional
DFIG converter rating\(\dfrac{|s|}{1-s}\) of machine rating23% at \(s = -0.3\)
Direct-drive torque\(T = P/\omega\)1.52 MN·m
Pole count\(P_{poles} = 120f_e/N\)96 poles
Fault energy\(E = Pt_{fault}\)455 kJ
Capacitor needed\(C = \dfrac{2E}{V_{max}^2-V_{nom}^2}\)3.95 F — impractical
Chopper resistance\(R = V_{dc}^2/P\)0.40 Ω
Rotor kinetic energy\(\tfrac12J\omega^2\)Ride-through and synthetic inertia
Cable charging\(I = \omega CV\ell\)HVDC beyond ~80 km
Pitfalls

Common Mistakes

  1. Using the rotor radius where the diameter is given. A factor of four in area — Problem 1.

  2. Treating the Betz limit as an engineering target. It is a conservation bound; 0.45 is a good real machine — Problem 1.

  3. Assuming higher tip speed ratio is always better. Beyond the optimum each blade works in the last one's wake — Problem 1.

  4. Thinking the MPPT controller needs an anemometer. The wind terms cancel in the cubic law — Problem 2.

  5. Scaling power linearly with wind speed. Half the wind gives an eighth of the power — Problem 2.

  6. Computing the DFIG converter rating as \(sP_{mech}\). It is \(sP_{airgap} = sP_{mech}/(1-s)\) — Problem 3.

  7. Overlooking that a DFIG's stator is directly grid-connected. Every disturbance reaches the machine — Problem 3.

  8. Sizing a direct-drive generator by power. Machine size follows torque, and the torque is 1.5 MN·m — Problem 4.

  9. Proposing a capacitor bank for ride-through. Four farads — Problem 5.

  10. Assuming AC transmission works at any offshore distance. Charging current fills the cable beyond ~80 km — Problem 6.

Looking Ahead

Six problems on a source that gives everything or nothing depending on the cube of a number nobody controls. The Betz limit capped extraction at 59.3% and a real rotor reached 76% of that; holding the tip speed ratio made the tracking law a cubic in rotor speed with no wind measurement needed at all; the DFIG got away with 23% of a converter and paid in slip rings and fault behaviour; and riding through a 150 ms fault turned out to need four farads of capacitance or, far more sensibly, a few per cent of rotor overspeed.

That last result is the theme of what follows. Both renewable sources produce power when the resource is available rather than when it is wanted, and both were rescued in this set by storage — the rotor's kinetic energy, used for milliseconds. Extending that idea from milliseconds to hours, and from kinetic to electrochemical, gives the technology that both sources ultimately depend on.

Next: Set 39 — Energy Storage and EV Charging, where a 75 kWh battery pack is charged at 2C, a bidirectional converter moves 150 kW in both directions with four interleaved phases, a dual active bridge is sized for an 11 kW onboard charger, and the move to 800 V architectures is shown to quarter the conduction loss at 350 kW.