ece 333 renewable energy systems lecture 10: wind power systems prof. tom overbye dept. of...
TRANSCRIPT
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
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