Transcript ECE 310 - University of Illinois at Urbana–Champaign

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
[email protected]
Announcements
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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
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In ECEB 3017 and 3002
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Variable-Slip Induction Generators
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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
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Effect of Rotor Resistance on Induction
Machine Power-Speed Curves
Real Pow er
Real Pow er
0.9
1.6
0.8
1.4
0.7
1.2
0.6
1
0.5
0.8
0.4
0.6
0.3
0.2
Real Power
Real Power
0.4
0.2
0
0.1
0
-0.1
-0.2
-0.2
-0.4
-0.3
-0.6
-0.4
-0.8
-0.5
-1
-0.6
-0.7
-1.2
-0.8
-1.4
-0.9
-1.6
-0.95
-0.9
-0.85
-0.8
-0.75
-0.7
-0.65
-0.6
-0.55
-0.5
-0.45
-0.4
-0.35
-0.3
-0.25
-0.2
-0.15
-0.1
-0.0500.050.10.150.20.250.30.350.40.450.50.550.60.650.70.750.80.850.90.951
Slip
-0.95
-0.9
-0.85
-0.8
-0.75
-0.7
-0.65
-0.6
-0.55
-0.5
-0.45
-0.4
-0.35
-0.3
-0.25
-0.2
-0.15
-0.1
-0.0500.050.10.150.20.250.30.350.40.450.50.550.60.650.70.750.80.850.90.951
Slip
Real Pow er
Real Pow er
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.
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Variable Slip Example: Vestas V80
1.8 MW
• The Vestas V80 1.8 MW turbine is an
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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
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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
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Induction Motor Thevenin Equiv.
Find VTH and ZTH
looking into the
left
VTH = VOC
jX m
VOC =Va
R s  jXls  jXm
If Rs = 0, expression simplifies: V =V
OC
a
Xm
X ls  X m
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Induction Motor Thevenin Equiv.
Short circuit Va to
find ZTH
ZTH =  Rs  jXls  ||  jXm 
If Rs = 0, expression simplifies:
jXls  jXm
Xls X m
ZTH =
j
j  Xls +X m 
 Xls +Xm 
ZTH =jXls ||jXm
Call this XTH
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Simplified Circuit
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Assuming Rs = 0 simplifies the induction machine
equivalent circuit and obtains this circuit which is
easy to analyze
Rr
(1  s ) Rr
We can rewrite Rr/s as
= Rr 
s
s
in which the first term
represents the rotor losses (heating) and the second
term represents the mechanical power transfer
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Equivalent Circuit Example
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2 pole, 60 Hz machine Find the input power.
Rs = 0 Ω
Step 1: Calculate the
equivalent circuit parameters
Xls = 0.5 Ω
Xm
Xm = 50 Ω
VTH =Va
X ls  X m
Xlr = 0.5 Ω
50
VTH =500
 495V
Rr = 0.1 Ω
0.5  50
Slip = 0.05
XTH =Xls ||Xm
VLN = 5000° V
0.5  50
X TH =0.5||50=
 0.495
0.5  50
X TH +X lr = 0.495  0.5  0.995
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Equivalent Circuit Example
Step 2: Draw the
circuit
Step 3: Analyze the
equivalent circuit
equivalent circuit
4950
I=
 198  j98.71A
2  j0.995
S=VI*=98.2  j 48.9 kVA
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Motor Starting
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Now let s=1 (standstill)
Inew =50  j 497.5 A
S=VInew *=25+j 248.7 kVA
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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?
Q=V2 / X
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Calculating Torque-Speed Curve
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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)
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Induction Generator Example
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Now let s = -0.05 (a generator)
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The negative resistance means that power is being
transferred from the wind turbine to the grid
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495
I=
 200.4  j99.7 A
-2  j 0.995
S=VI*=  100.2+j 49.8 kVA
A generator producing P but absorbing Q!
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Reactive Power Support
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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
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Induction Generator Rotor Losses
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What about rotor losses?
Rr = 0.1
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2
P= I R r =5 kW
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!
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Doubly-Fed Induction Generators
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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
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GE 1.5 MW DFIG Example
GE 1.5 MW turbines were the
best selling wind turbines
in the US in 2011
Source: GE Brochure/manual
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Indirect Grid Connection Systems
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Wind turbine is allowed to spin at any speed
Variable frequency AC from the generator goes
through a rectifier (AC-DC) and an inverter (DCAC) to 60 Hz for grid-connection
Good for handling rapidly changing windspeeds
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Wind Turbine Gearboxes
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A significant portion of the weight in the nacelle is due
to the gearbox
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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)
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Average Power in the Wind
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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:
Pavg
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1
3
   Av    A  v 3 
avg
2
avg 2
This is why we can’t use average wind speed vavg to
find the average power in the wind
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Average Windspeed
v   hours @ v 

miles of wind


total hours
  hours @ v 
i
vavg
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vavg   vi   fraction of total hours @ vi 
i
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vi = wind speed (mph)
The fraction of total hours at vi is also the probability
that v = vi
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Average Windspeed
vavg   vi   fraction of total hours @ vi 
i
vavg   vi   probability that v = vi 
i
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This is the average wind speed in probabilistic terms
Average value of v3 is found the same way:
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v
 
avg
  vi 3   probability that v = vi 
i
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Example Windspeed Site Data
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Wind Probability Density Functions
Windspeed probability density function (pdf):
between 0 and 1, area under the curve is equal to 1
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Windspeed p.d.f.
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f(v) = wind speed pdf
Probability that wind is between two wind speeds:
v2
p  v1  v  v2    f (v)dv
v1

p  0  v      f (v)dv = 1
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# of hours/year that the wind is between two wind
speeds:
v2
hrs / yr  v1  v  v2   8760   f (v)dv
v1
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Average Windspeed using p.d.f.
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This is similar to earlier summation, but now we have a
continuous function instead of discrete function
vavg   vi  p  v = vi 
i
discrete

vavg   v  f (v)dv
0
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continuous
Same for the average of (v3)
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v
 
avg
  vi 3  p  v = vi 
i
discrete
3
v
 

avg
  v 3  f (v)dv
0
continuous
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Weibull p.d.f.
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Starting point for characterizing statistics of wind
speeds
k-1
k v
f (v )     e
cc
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v
- 
c
k
Weibull pdf
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
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Weibull p.d.f.
k=2 looks
reasonable
for wind
Weibull p.d.f. for c = 8
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Where did the Weibull PDF Come From
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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
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But this approximation is not going to work well for every
data set!!
Reference: http://www.barringer1.com/pdf/Chpt1-5th-edition.pdf
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