ECE 333 Renewable Energy Systems

Lecture 10: Wind Power Systems

Prof. Tom Overbye

Dept. of Electrical and Computer Engineering

University of Illinois at Urbana-Champaign

overbye@illinois.edu

Announcements

• Read Chapter 7• Quiz today on HW 4• HW 5 is posted on the website; there will be no quiz

on this material, but it certainly may be included in the exams

• First exam is March 5 (during class); closed book, closed notes; you may bring in standard calculators and one 8.5 by 11 inch handwritten note sheet – In ECEB 3017 and 3002

2

Variable-Slip Induction Generators

• Purposely add variable resistance to the rotor• External adjustable resistors - this can mean using a

wound rotor with slip rings and brushes which requires more maintenance

• Mount resistors and control electronics on the rotor and use an optical fiber link to send the rotor a signal for how much resistance to provide

3

Effect of Rotor Resistance on Induction Machine Power-Speed Curves

Real Power

Real Pow er

Slip10.950.90.850.80.750.70.650.60.550.50.450.40.350.30.250.20.150.10.050-0.05-0.1-0.15-0.2-0.25-0.3-0.35-0.4-0.45-0.5-0.55-0.6-0.65-0.7-0.75-0.8-0.85-0.9-0.95

Rea

l Pow

er

1.6

1.4

1.2

1

0.8

0.6

0.4

0.2

0

-0.2

-0.4

-0.6

-0.8

-1

-1.2

-1.4

-1.6

Real Power

Real Pow er

Slip10.950.90.850.80.750.70.650.60.550.50.450.40.350.30.250.20.150.10.050-0.05-0.1-0.15-0.2-0.25-0.3-0.35-0.4-0.45-0.5-0.55-0.6-0.65-0.7-0.75-0.8-0.85-0.9-0.95

Rea

l Pow

er

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0

-0.1

-0.2

-0.3

-0.4

-0.5

-0.6

-0.7

-0.8

-0.9

Left plot shows the torque-power curve from slip of -1 to 1 with external resistance = 0.05; right plot is with external resistance set to 0.99 pu.

4

Variable Slip Example: Vestas V80 1.8 MW

• The Vestas V80 1.8 MW turbine is an example in which an induction generator is operated with variable rotor resistance (opti-slip).

• Adjusting the rotor resistance changes the torque-speed curve

• Operates between 9 and 19 rpm

Source: Vestas V80 brochure

5

Induction Machine Circuit

• I, S into the machine (motor convention)• Rs = stator resistance (small)

• Xls = stator leakage flux

• Xm = magnetizing reactance, Xm >> Xls

• Xlr = inductance of rotor referred to stator

• Rr/s = represents energy transfer between electrical and mechanical side 6

Induction Motor Thevenin Equiv.

Find VTH and ZTH

looking into theleft

VTH = VOC

mOC a

s ls m

jXV =V

R jX jX

If Rs = 0, expression simplifies: mOC a

ls m

XV =V

X X7

Induction Motor Thevenin Equiv.

Short circuit Va to find ZTH

TH s ls mZ = R jX || jX

If Rs = 0, expression simplifies:TH ls mZ =jX ||jX

ls m ls m

THls m ls m

jX jX X XZ = j

j X +X X +X

Call this XTH

8

Simplified Circuit

• Assuming Rs = 0 simplifies the induction machine equivalent circuit and obtains this circuit which is easy to analyze

• We can rewrite Rr/s as in which the first termrepresents the rotor losses (heating) and the secondterm represents the mechanical power transfer

(1 )= Rr r

r

R s R

s s

9

Equivalent Circuit Example

• 2 pole, 60 Hz machine• Rs = 0 Ω

• Xls = 0.5 Ω

• Xm = 50 Ω

• Xlr = 0.5 Ω

• Rr = 0.1 Ω

• Slip = 0.05 • VLN = 5000° V

Find the input power.

Step 1: Calculate theequivalent circuit parameters

mTH a

ls m

XV =V

X X

TH

50V =500 495V

0.5 50

TH ls mX =X ||X

TH

0.5 50X =0.5||50= 0.495

0.5 50

10TH lrX +X = 0.495 0.5 0.995

Equivalent Circuit Example

equivalent circuitStep 3: Analyze the equivalent circuit

Step 2: Draw thecircuit

495 0I= 198 98.71

2 j0.995j A

S=VI*=98.2 48.9 kVAj11

Motor Starting

• Now let s=1 (standstill)

• Looks like a load to the system• A lot of reactive power is being transferred!• Ever notice that the lights dim when your air

conditioner comes on?

newI =50 497.5j A

newS=VI *=25+ 248.7 kVAj

2Q=V / X12

Calculating Torque-Speed Curve

• If you continue this analysis for different values of s, and plot the results, you’ll get the torque speed curve: torque * speed = power

• What if s = 0? (synchronous)• Like a jet flying at the same speed as another jet –

there is no relative motion• Rotor can’t see the stator field go by, so Rr looks

infinite and I is zero (open circuit)

13

Induction Generator Example

• Now let s = -0.05 (a generator)

• The negative resistance means that power is being transferred from the wind turbine to the grid

• A generator producing P but absorbing Q!

495I= 200.4 99.7 A

-2 0.995j

j

S=VI*= 100.2+ 49.8 kVAj

14

Reactive Power Support

• Wind turbine generators can produce real power but consume reactive power

• This is especially a problem with Types 1 and 2 wind turbines which are induction machines, like this model

• Capacitors or other power factor correction devices are needed

• Types 3 and 4 can provide reactive support, details beyond the scope of this class

15

Induction Generator Rotor Losses

• What about rotor losses?

• This means before getting out to the stator and producing the 100 kW, there are 5 kW being lost in the rotor.

• That means what was actually captured from the wind was 105 kW, but 5 was lost!

2

rP= I R =5 kWRr = 0.1

16

Doubly-Fed Induction Generators

• Another common approach is to use what is called a doubly-fed induction generator in which there is an electrical connection between the rotor and supply electrical system using an ac-ac converter

• This allows operation over a wide-range of speed, for example 30% with the GE 1.5 MW and 3.6 MW machines

• Called Type 3 wind turbines

17

GE 1.5 MW DFIG Example

Source: GE Brochure/manual

GE 1.5 MW turbines were the best selling wind turbines in the US in 2011

18

Indirect Grid Connection Systems

• Wind turbine is allowed to spin at any speed• Variable frequency AC from the generator goes

through a rectifier (AC-DC) and an inverter (DC-AC) to 60 Hz for grid-connection

• Good for handling rapidly changing windspeeds

19

Wind Turbine Gearboxes

• A significant portion of the weight in the nacelle is due to the gearbox– Needed to change the slow blade shaft speed into the higher

speed needed for the electric machine

• Gearboxes require periodic maintenance (e.g., change the oil), and have also be a common source of wind turbine failure

• Some wind turbine designs are now getting rid of the gearbox by using electric generators with many pole pairs (direct-drive systems)

20

Average Power in the Wind

• How much energy can we expect from a wind turbine?

• To figure out average power in the wind, we need to know the average value of the cube of velocity:

• This is why we can’t use average wind speed vavg to find the average power in the wind

3 31 1

2 2avg avgavg

P Av A v

21

Average Windspeed

hours@miles of wind

total hours hours@

i ii

avgi

i

v vv

v

• vi = wind speed (mph)

• The fraction of total hours at vi is also the probability that v = vi

fraction of total hours@ avg i ii

v v v

22

Average Windspeed

• This is the average wind speed in probabilistic terms• Average value of v3 is found the same way:

fraction of total hours@ avg i ii

v v v

probability that = avg i ii

v v v v

3 3 probability that = i iavgi

v v v v

23

Example Windspeed Site Data

24

Wind Probability Density Functions

Windspeed probability density function (pdf): between 0 and 1, area under the curve is equal to 1

25

Windspeed p.d.f.

• f(v) = wind speed pdf• Probability that wind is between two wind speeds:

• # of hours/year that the wind is between two wind speeds:

2

1

1 2 ( ) v

v

p v v v f v dv

0

0 ( ) = 1 p v f v dv

2

1

1 2/ 8760 ( ) v

v

hrs yr v v v f v dv 26

Average Windspeed using p.d.f.

• This is similar to earlier summation, but now we have a continuous function instead of discrete function

• Same for the average of (v3) 0

( ) avgv v f v dv

= avg i i

i

v v p v v

3 3 = i iavgi

v v p v v

3 3

0

( ) avg

v v f v dv

discrete

continuous

continuous

discrete

27

Weibull p.d.f.

• Starting point for characterizing statistics of wind speeds

k-1-

( ) e Weibull pdf

kv

ck vf v

c c

• k = shape parameter • c = scale parameter• Keep in mind actual data is key. The idea of

introducing the Weibull pdf is to see if we can get a an equation that approximates the characteristics of actual wind site data

28

Weibull p.d.f.

k=2 looks reasonable

for wind

Weibull p.d.f. for c = 829

Where did the Weibull PDF Come From

• Invented by Waloddi Weibull in 1937, and presented in hallmark American paper in 1951

• Weibull's claim was that it fit data for a wide range of problems, ranging from strength of steel to the height of adult males

• Initially greeted with skepticism – it seemed too good to be true, but further testing has shown its value

• Widely used since it allows a complete pdf response to be approximated from a small set of samples– But this approximation is not going to work well for every

data set!!

30Reference: http://www.barringer1.com/pdf/Chpt1-5th-edition.pdf