Transcript Slide 1

Wind Power Variability in the Grid:
Regulation
J. McCalley
Outline
1. Power production: wind power equation
2. Variability: time and space
3. Governor response
2
Power production
Wind power equation
Swept area At of turbine blades:
v1
vt
Mass flow rate is the mass of
substance which passes through
a given surface per unit time.
v2
The disks have larger cross
sectional area from left to
right because
• v1 > vt > v2 and
• the mass flow rate must
be the same everywhere
within the streamtube
(conservation of mass):
Q1  Qt
 Q2
A1v1  At vt  A2v2
v
ρ=air density (kg/m3)
x Therefore, A < A < A
1
t
2
Reference: E. Hau, “Wind turbines: fundamentals, technologies, applications, economics,” 2 nd edition, Springer, 2006.
3
Power production
Wind power equation
x
1. Wind velocity: v 
t
2. Air mass flowing: m  Ax
m At x
Q


 At vt
3. Mass flow rate at swept area: t
t
t
4b. Force on turbine blades:
4a. Kinetic energy change:
v m
1
2
2
F


ma


m

v  Qt v1  v2 
KE  m v1  v2
t
t
2
5b. Power extracted:
5a. Power extracted:

P

KE 1 m 2 2 1

v1  v2  Qt v12  v22
t
2 t
2





P  Fvt  Qt vt v1  v2

6b. Substitute (3) into (5b):
6a. Substitute (3) into (5a):
P  (1/ 2)At vt (v12  v22 )
7. Equate
P  At vt2 (v  v )
1
2
 (1/ 2)vt (v12  v22 )  vt2 (v1  v2 )  (1/ 2)vt (v1  v2 )(v1  v2 )  vt2 (v1  v2 )  (1/ 2)(v1  v2 )  vt
8. Substitute (7) into (6b):P  At ((1 / 2)( v1  v2 )) 2 (v1  v2 )  At (v12  v22 )( v1  v2 )
3
4

A
v
v2 2
v2
t
1
3
9. Factor out v1 : P  4 (1  ( v ) )(1  v )
4
1
1
Power production
Wind power equation
10. Define wind
stream speed ratio, a:
a
11. Substitute a into
power expression of (9):
v2
v1
P
This ratio is fixed for a given
turbine & control condition.
At v13
12. Differentiate and find a
which maximizes function:
(we hold v1 constant here)
13. Find the maximum power
by substituting a=1/3 into (11):
4
(1  a 2 )(1  a)


P At v13

 2a (a  1)  (1  a 2 )  0
a
4
 2a 2  2a  1  a 2  3a 2  2a  1  0
(3a  1)(a  1)  0  a  1 / 3, a  1
P
At v13
4
At v13 8 4 8At v13
1 4
(1  )( ) 

9 3
4 93
27
5
Power production
Wind power equation
14. Define Cp, the power (or performance) coefficient, which
gives the ratio of the power extracted by the converter, P, to
the power of the air stream, Pin.
3
power extracted P  At v1 (1  a 2 )(1  a)
4
by the converter
KE 1 m 2
1
1
1
power of the
Pin 

v1  0  Q1v12  At v1v12  At v13
air stream
t
2 t
2
2
2
3
At v1
(1  a 2 )(1  a)
1
P
1
3
2
4
P

C
P

C

A
v
Cp 

 (1  a )(1  a)
p in
P
t 1
1
Pin
2
3
2
At v1
2


15. For a given v1, the maximum value of Cp occurs when its
numerator is maximum, i.e., when a=1/3:
Cp 
P 1 8 4 16
 ( )( ) 
 0.5926 The Betz Limit!
Pin 2 9 3 27
16. Cp depends on a=v2/v1. For a given v1, v2 depends on how
much energy is extracted by the converter, which depends
6
on pitch, θ, and rotational speed, ω.
Power production
Wind power equation
17. At the condition of maximum energy extraction, where
a=v2/v1=1/3, we have the reduce downstream speed is
1
v2  v1
3
From slide 4, we know that
vt  (1/ 2)(v1  v2 )
Substituting the downstream speed at maximum energy
extraction, the speed at the converter is:
1
2
vt  (1 / 2)( v1  v1 )  v1
3
3
At v13
P
(1  a 2 )(1  a)
4
Recall
Some points of interest:
a=1v2=v1=vt, and P=0;
a=1/3v2=v1/3, vt=2v1/3, and P=(ρAtv13/4)(32/27)
a=0v2=0, vt=v1/2, and P=ρAtv13/4
a=-1/3, v2=-v1/3, vt=v1/3, and P=(ρAtv13/4)(16/27)
a=-1v2=-v1, vt=0, and P=0
7
Power production
Cp vs. λ and θ
Tip-speed ratio:
u R
 
v1 v1
u: tangential velocity of blade tip
ω: rotational velocity of blade
R: rotor radius
v1: wind speed
Pitch: θ
GE SLE 1.5 MW
8
Power production
Wind Power Equation
1
3
P  C p Pin  C P ( ,  ) At v1
2
So power extracted depends on
1. Design factors:
• Swept area, At
2. Environmental factors:
• Air density, ρ (~1.225kg/m3 at sea level)
• Wind speed v13
3. Control factors affecting performance coefficient CP:
• Tip speed ratio through the rotor speed ω
9
• Pitch θ
Power production
Cp vs. λ and θ
One may plot P vs ω, for different wind
speeds v1, as shown below [*]. MPP is
the maximum power point.
1
3
P  C p Pin  C P ( ,  ) At v1
2
Important concept #1: The control strategy of all
US turbines today is to operate turbine at point of
maximum energy extraction, as indicated by the
locus of points in the figure.
Important concept #2: This strategy maximizes
energy produced. Any other strategy “spills” wind.
Important concepts #3:
• Cut-in speed>0 because for lower speeds, turbine
generates less power than it’s internal consumption.
• Generator should not exceed rated power
• Cut-out speed protects turbine in high winds
Important concepts #4:
• Pitch control used to limit power only if max pwr
exceeds rated; otherwise, blades fully pitched.
• Generator (rotor) control used to control speed.
For a given v1 and θ, one can control ω and thus λ
(using direct speed control or indirect speed control
[**]).
[*] B. Wu, et al., “Power conversion and control of wind energy systems,” Wiley, 2011.
[**] G. Abad, et al., “Doubly fed induction machine: modeling & control for wind energy generation,” Wiley, 2011.
10
Power production
Typical power curve
Cut-in speed (6.7 mph)
Cut-out speed (55 mph)
11
Wind Power Temporal & Spatial Variability
JULY2006
JANUARY2006
Blue~VERY LOW POWER;
Red~VERY HIGH POWER
Notice the temporal variability:
• lots of cycling between blue and red;
• January has a lot more high-wind power (red) than July;
Notice the spatial variability
• “waves” of wind power move through the entire Eastern Interconnection;
• red occurs more in the Midwest than in the East
12
12
Temporal variability of wind power
• For fixed speed machines, because the mechanical
power into a turbine depends on the wind speed, and
because electric power out of the wind generator depends
on the mechanical power in to the turbine, variations in
wind speed from t1 to t2 cause variations in electric power
out of the wind generator.
• Double-fed induction generators (DFIGs) also produce
power that varies with wind speed, although the torquespeed controller provides that this variability is less
volatile than fixed-speed machines.
• For a single turbine, this variability depends on three
features: (1) time interval; (2) location; (3) terrain.
13
Temporal variability of wind power –
time interval
Source: Task 25 of the International Energy Agency (IEA), “Design and operation of power systems with large amounts of wind
power: State-of-the-art report,” available at www.vtt.fi/inf/pdf/workingpapers/2007/W82.pdf.
14
Temporal variability of wind power –
time interval
Extreme weather events:
• Denmark: 2000 MW (83% of capacity) decrease in 6 hours or 12 MW (0.5%
of capacity) in a minute on 8th January, 2005.
• North Germany: over 4000 MW (58% of capacity) decrease within 10
hours, extreme negative ramp rate of 16 MW/min (0.2% of capacity) on
24th December, 2004
• Ireland: 63 MW in 15 mins (approx 12% of capacity at the time), 144 MW in
1 hour (approx 29% of capacity) and 338 MW in 12 hours (approx 68% of
capacity)
• Portugal: 700 MW (60% of capacity) decrease in 8 hours on 1st June, 2006
• Spain: Large ramp rates recorded for about 11 GW of wind power: 800 MW
(7%) increase in 45 minutes (ramp rate of 1067 MW/h, 9% of capacity), and
1000 MW (9%) decrease in 1 hour and 45 minutes (ramp rate -570 MW/h,
5% of capacity). Generated wind power between 25 MW and 8375 MW
have occurred (0.2%-72% of capacity).
• Texas, US: loss of 1550 MW of wind capacity at the rate of approximately
600 MW/hr over a 2½ hour period on February 24, 2007.
Source: Task 25 of the International Energy Agency (IEA), “Design and operation of power systems with large amounts of wind
power: State-of-the-art report,” available at www.vtt.fi/inf/pdf/workingpapers/2007/W82.pdf.
15
Temporal variability of wind power –
location and terrain
“In medium continental latitudes, the wind fluctuates greatly
as the low-pressure regions move through. In these
regions, the mean wind speed is higher in winter than in the
summer months.
The proximity of water and of land areas also has a
considerable influence. For example, higher wind speeds
can occur in summer in mountain passes or in river valleys
close to the coast because the cool sea air flows into the
warmer land regions due to thermal effects. A particularly
spectacular example are the regions of the passes in the
coastal mountains in California through to the lower lying
desert-like hot land areas in California and Arizona.”
Reference: E. Hau, “Wind turbines: fundamentals, technologies, applications, economics,” 2 nd edition, Springer, 2006.
16
Spatial variability of wind power –
geographical smooting from geo-diversity
1 turbine.
1 wind plant.
All turbines in region.
Variability of single turbine, as percentage of capacity, is significantly greater than the
variability of wind plant, which is significantly greater than variability of the region.
Source: Task 25 of the International Energy Agency (IEA), “Design and operation of power systems with large amounts of wind
power: State-of-the-art report,” available at www.vtt.fi/inf/pdf/workingpapers/2007/W82.pdf.
17
Spatial variability of wind power –
geographical smooting from geo-diversity
Single turbine reaches or
exceeds 100% of its capacity for
~100 hours per year, the area
called “Denmark West” has a
maximum power production of
only about 90% throughout the
year, and the overall Nordic
system has maximum power
production of only about 80%.
 At other extreme, single
turbine output exceeds 0 for
about 7200 hours per year,
leaving 8760-7200=1560 hours it
is at 0. The area wind output
rarely goes to 0, and the system
Duration curve: provides number of hours on horizontal
wind output never does.
axis for which wind power production exceeds the
percent capacity on the vertical axis.
H. Holttinen, “The impact of large-scale power production on the Nordic electricity system,” VTT Publications 554, PhD Dissertation,
Helsinki University of Technology, 2004.
18
Temporal & spatial variability of wind power
Correlation coefficient between
time series of intervals T=5
min, 30 min, 1 hr, 2 hr, 4 hr, 12
hr, for multiple locations, as a
function of distance between
those locations.
The correlation coefficient
indicates how well 2 time
series follow each other. It will
be near 1.0 if the 2 time series
follow each other very well, it
will be 0 if they do not follow
each other at all.
For 5 min intervals, there is almost no correlation for locations separated by more than ~20
km, since wind gusts tend to occur for only relatively small regions. This suggests that that
even small regions experience geographical smoothing at 5min intervals. For 12hr intervals,
wind power production is correlated even for very large regions, since these averages are
closely linked to overall weather patterns that can be similar for very large regions.
H. Holttinen and Ritva Hirvonen, “Power System Requirements for Wind Power,” in “Wind Power in Power Systems,” editor, T.
Ackermann, Wiley, 2005.
19
Spatial variability of wind power –
geographical smooting from geo-diversity
The variability of the 1 farm, given as a
percentage of its capacity, is significantly
greater than that of the entire region of Western
Denmark.
International Energy Agency,”VARIABILITY OFWIND POWER AND OTHER RENEWABLES: Management options and strategies,”
June 2005, at http://www.iea.org/textbase/papers/2005/variability.pdf.
20
Temporal & spatial variability of wind power
If data used to develop the fig on the last slide is captured for a large number
of wind farms and regions, the standard deviation may be computed for each
farm or region. This standard deviation may then be plotted against the
approximate diameter of the farm’s or region’s geographical area. Above
indicates that hourly variation, as measured by standard deviation,
decreases with the wind farm’s or region’s diameter.
Task 25 of the International Energy Agency (IEA), “Design and operation of power systems with large amounts of wind power: Stateof-the-art report,” available at www.vtt.fi/inf/pdf/workingpapers/2007/W82.pdf.
21
Generation portfolio design to meet load
variability
A control area’s portfolio of conventional generation is designed to meet that
load variability. This is done by ensuring there are enough generators that
are on governor control, and that there are enough generators having ramp
rates sufficient to meet the largest likely load ramp. Typical ramp rates for
different kinds of units are listed below (given as a percentage of capacity):
 Diesel engines 40 %/min
 Industrial GT 20 %/min
 GT Combined Cycle 5 -10 %/min
 Steam turbine plants 1- 5 %/min
 Nuclear plants 1- 5 %/min
22
Variability of load
These plots show that the particular control
area responsible for balancing this load
must have capability to ramp 400 MW in
one hour (6.7 MW/min), 80 MW in 10
minutes (8 MW/min), and 10 MW in one
minute (10 MW/min) in order to meet all
MW variations seen in the system.
23
Net load
What happens to these requirements if wind is added
to the generation portfolio?
Because wind is not controllable the non-wind
generation must also meet its variability.
The composite variability that needs to be met by the
wind is called the net load.
PNL  PL  PW
If we plot the distribution (histogram) for net load
variation, we will find it is wider than the distribution
for load alone.
24
Variability of net load
Net load is red. Load is blue.
These plots show that if the
particular control area is
responsible for balancing only this
load, then it must have capability
to ramp 300 MW in one hour (5
MW/min), but…. If the control area
is responsible for balancing the
net load, it must have capability to
ramp 500 MW is one hour
(8.33MW/min).
25
Variability of net load
The net load relationship must treat PL and Pw (or
their deviations) as random variables.
PNL  PL  PW
So the real question is this: Given two random variables
x (load deviation) and y (wind power deviation) for which
we know the distributions fx(x) and fY(y), respectively,
how do we obtain the distribution of the net-load random
variable z=x-y, fz(z)?
Answer: If these random variables are independent,
then for the means, μz=μx-μy, and for the variances,
σz2= σx2+σy2.
26
Variability of net load
μz=μx-μy, σz2= σx2+σy2.
The impact on the means is of little interest since the
variability means, for both load and wind, will be ~0.
On the other hand, the impact on the variance is of great
interest, since it implies the distribution of the difference
will always be wider than either individual distribution.
Therefore we expect that when wind generation is
added to a system, the maximum MW variation seen in
the control area will increase, as seen on slide 25.
27
Correlation between ΔPL and ΔPw
It may be that load variation and wind variation are not
independent, but rather, that they may tend to change
in phase (both increase or decrease together) or in
antiphase (one increases when the other decreases).
We capture this with the correlation coefficient:
 xi   x yi   y 
N
i 1
rxy 
 xi   x   yi   y 
N
N
i 1
2
N
 xi   x yi   y 
N
2

i 1
N x y
i 1
where N is the number of points in the time series, and
μx, μy and σx, σy are the means and standard deviations,
respectively, of the two time series.
28
Correlation between ΔPL and ΔPw
The correlation coefficient measures strength and
direction of a linear relationship between two random
variables. It indicates how well two time series, x and y,
follow each other. It will be near 1.0 if the two time
series follow each other very well, it will be 0 if they do
not follow each other at all, and it will be near -1 if
increases in one occur with decreases in another.
29
Correlation between ΔPL and ΔPw
• If two variables are independent, they are uncorrelated and rxy=0.
• If rxy=0, then the two variables are uncorrelated but not necessarily
independent, because rxy detects only linear dependence between
variables.
• A special case exists if x and y are both normally distributed, then rxy=0
implies independence. This will be the case for our regulation data since
it is comprised of variations.
• If rxy=1 or if rxy=-1, then x and y are dependent.
• For values of rxy such that 0<|rxy|<1, the correlation coefficient reflects
the degree of dependence between the two variables, or conversely, the
departure of the two random variables from independence.
30
Using variance and correlation to measure regulation
coincidence
We will use four ideas in what follows:
• We assume the regulation component of load is normal.
• For normally distributed random variables, rxy can be used to
measure independence.
• Three standard deviations (3σ) of the net load is our measure
of “regulation burden.”
• We assign angles to the standard deviation of each component
comprising the net load, so we can treat them as vectors.
31
Using variance and correlation to measure regulation
coincidence
If two random variables are independent, then the variance of
their sum (or their difference) is  z2   x2   y2
σz
σy
Fig 1
σx
Consider if two time series
are perfectly correlated, i.e.,
rxy=1, implying x and y follow
each other perfectly:
σx
Fig 2
σz
σy
Consider if two time series
are perfectly anti-correlated,
i.e., rxy=-1, implying x and y
changes in anti-phase:
σz σ
y
Fig 3
σx
32
Using variance and correlation to measure regulation
coincidence
Define an angle θ:    (1  rxy )
2
where rxy ranges from -1 to 0 to 1, θ ranges from π to π/2 to 0, (as
in the three cases illustrated on slide 32), as summarized below.
Correlation
Perfectly correlated
Uncorrelated
Anti-correlated
rxy
1
0
-1
θ
0
π/2
π
Figure
7
6
8
The figure to the right illustrates the situation
for an angle θ, in this case corresponding to a
value rxy such that 0<rxy<1.
σz
Because wind is gen and netload is
load, we should use


2
(1  rxy )
σy
θ
σx
33
Using variance and correlation to measure regulation
coincidence
Problem: Given the load has a standard deviation of σL and the
wind generation has a standard deviation of σw, and the
composite of the two (net load) has a standard deviation of σT,
then what component of σT can we attribute to the wind
generation? To solve this problem, we redraw our figure, where
the only change we have made is to re-label according to the
nomenclature of this newly-stated problem.
σT
σw
θ
σL
34
Using variance and correlation to measure regulation
coincidence
The contribution of σw to σT will be the projection of the σw vector
onto the σT vector, i.e., it will be the component of σw in the σT
direction. In the below figure, we have denoted this component as
X (we have also enhanced the figure to facilitate analytic
development).
σT
X
σT - X
Y
σw
θ
σL
35
Using variance and correlation to measure regulation
coincidence
The smaller right-triangle to the right provides
The larger right-triangle to the left provides:
Subtracting (**) from (*) results in:
Expanding the right-hand-side:
which simplifies to:
 L2  Y 2  ( T  X ) 2
**
 w2   L2  X 2  ( T2  2 T X  X 2 )
 w2   L2   T2
X 
2 T
X is the “regulation share” of the
“generator of interest.” It was
applied in [#] to different kinds of
resources, including wind and
solar. It can also be applied to
different kinds of loads.
*
 w2   L2  X 2  ( T  X ) 2
 w2   L2   T2  2 T X
Solving for X results in:
 w2  Y 2  X 2
Also, from law of cosines:
 T2   L2   w2  2 L w cos(180  )
  L2   w2  2 L w cos
σT
X
σT - X
[#] B. Kirby, M. Milligan, Y. Makarov, D. Hawkins, K. Jackson, H. Shiu “California Renewables Portfolio Standard Renewable
Generation Integration Cost Analysis, Phase I: One Year Analysis Of Existing Resources, Results And Recommendations, Final
Report,” Dec. 10, 2003, available at http://www.consultkirby.com/files/RPS_Int_Cost_PhaseI_Final.pdf.
Y
σw
θ
σL
36
What can you do with it?
Consider that a particular ISO projects 10 GW of additional
wind in the coming 5 years, and wind data together with
knowledge of where the wind farms will be located are
available. Then σw and rLw can be estimated, and since we
know σL, we may compute σT as:       2  cos



(1  rLw )
where
2
T
2
L
2
w
L
w
2
Our new “regulation burden” will therefore be 3σT, and we
will have to have a generation portfolio to handle this.
37
Example
 w2   L2   T2
X
2 T
 T2   L2   w2  2 L w cos


2
(1  rLw )
Several very large wind farms in close proximity to one another and within the
service area of a certain balancing authority (BA) have 10-minute variability
characterized by 3 standard deviations equal to 75 MW. The BA net load previous
to interconnection of these wind farms has 10-minute variability characterized by 3
standard deviations equal to 300 MW. The correlation coefficient between the 10minute variation of the wind farms and that of the BA net load previous to
interconnection is -0.5.
• Determine the regulation burden of the BA net load after interconnection of the
new wind farms.


2
(1  rLw ) 

2
(1  (0.5)) 

4
 T2   L2   w2  2 L w cos  (100) 2  (25) 2  2(25)(100) cos

4
 10000 625 5000(0.7071)  1416.1
  T  119MW  3 T  357MW
• Determine the regulation share of the new wind farm.
X
 w2   L2   T2 625 10000 14161

 20.11MW
2 T
2(119)
38
What can you do with it?
How to determine the generation portfolio necessary to handle it?
First, you should ramp rates (RR) of gens in your existing portfolio.
You should also know RR of gens that can be added to your portfolio.
 Diesel engines 40 %/min
 Industrial GT 20 %/min
 GT Combined Cycle 5 -10 %/min
 Steam turbine plants 1- 5 %/min
 Nuclear plants 1- 5 %/min
You should also know the typical operating conditions you expect, and the
committed units at throughout those operating conditions (from historical data
or a production cost simulation) spinning reserve criteria. This information will
be highly influenced by the spinning reserve requirement.
Then a guiding criteria is that for all operating conditions, we would like to
satisfy the following relation:
SR is up (or down)
 minSR , RR  T  3
j
j is regulating
j
j
T
“headroom” for unit j, RR
is ramp rate of unit j, ΔT is
time interval over which σT
is computed.
39
Obtaining regulation component
=
100
80
+
REGULATION IN MEGAWATTS
60
40
20
0
-20
-40
-60
-80
-100
07:00
07:20
07:40
08:00
08:20
08:40
09:00
09:20
09:40
10:00
Regulation
Load Following
Regulation
How to get the time series of the regulation component?
Steve Enyeart, “Large Wind Integration Challenges for Operations / System Reliability,” presentation by Bonneville Power
Administration, Feb 12, 2008, available at http://cialab.ee.washington.edu/nwess/2008/presentations/stephen.ppt.
40
Obtaining regulation component
Define Lk, LFk, and LRk as the load, load following
component, and regulation component, respectively, at
time k∆t. Assume that the load, Lk, is given for k=1,…,N.
The load-following component is given by a moving
average of length 2T time intervals, i.e.,
Lk T  Lk T 1  ...  Lk  ...  Lk T 1  Lk T
1 k T
LFk 
L

 i
2T  1 i k T
2T  1
Example: Load data taken at ∆t=2 minute intervals, and compute LFk
based on a 28 minute rolling average. So T was chosen as 7, resulting in
LFk 
Lk 7  Lk 6  ...  Lk  ...  Lk 6  Lk 7
15
When k=20 (the 20th time point), then
LF20 
L13  L14  ...  L20  ...  L26  L27
15
B. Kirby and E. Hirst, “Customer-Specific Metrics for The Regulation and Load-Following Ancillary Services,” Report
ORNL/CON-474, Oak Ridge National Laboratories, Energy Division, January 2000...
41
Obtaining regulation component
A result of this computation, where it is clear that the moving average
tends to smooth the function.
Once the load following component is obtained, then the regulation
component can be computed from LRk  Lk  LFk
Then you can compute the deviations for a desired time interval, e.g., 2,
5, 10 minutes, and then obtain the corresponding standard deviation of
the resulting distribution.
R. Hudson, B. Kirby, and Y. Wan, “Regulation Requirements for Wind Generation Facilities,” available at
http://www.consultkirby.com/files/AWEA_Wind_Regulation.pdf.
42
Governor response
A conventional synchronous generator, for both steam-turbines and hydro
turbines, can control the mechanical power seen by the generator in response
to either a change in set-point, ∆PC, or in response to change in frequency,
∆ω. The dynamics of this feedback control system are derived in EE 457 and
EE 553, the conclusion of which is:
ˆ

P
1
ˆ
C
ˆ
PM 

1  TT s 1  TG s  1  TT s 1  TG s  R
where TT is the time constant of the turbine, and TG is the time constant of the
speed-governor, and the circumflex indicates Laplace domain.
Now consider a step-change in power of ΔPC and in frequency of Δω, which
in the LaPlace domain is:
PC

ˆ
ˆ
 
PC 
s
s
PC

1
PˆM 

s1  TT s 1  TG s  s1  TT s 1  TG s  R
43
Governor response
Consider ΔPM(t) for very large values of t, i.e., for the steady-state, using the
final value theorem: lim f (t )  lim sfˆ ( s)
t 
Applying
FVT
s 0
PM  lim PM (t )  lim sPˆM
t 
s 0

sPC
s
1
 lim


s 0 s 1  T s 1  T s 



s
1

T
s
1

T
s
R
T
G
T
G



PC

1
 lim


s 0 1  T s 1  T s 



1

T
s
1

T
s
R
T
G
T
G



PM  PC 
R
Some important cautions:
 Δ PM, ΔPC, and Δω in the above are time-domain variables (not Laplace)
 Δ PM, ΔPC, and Δω are steady-state values of the time-domain variables
(the values after you wait a long time)
 Δ PM, ΔPC, and Δω are not the cause of the condition but rather the result
of a ΔPL somewhere in the network, and the above shows how they are
related in the steady-state.
44
Governor response
Assuming that the local behavior as characterized by previous equation
can be extrapolated to a larger domain, we can draw the below figure:
“Droop” characteristic: Primary
control system acts in such a way so
that steady-state frequency “droops”
with increasing mechanical power.
   RPM
ω
The R constant, previously called the
regulation constant, is also referred to as the
droop setting. When power is specified in
units of MW and frequency in units of rad/sec,
then R has units of rad/sec/MW.
Slope=-R
ΔPM
ω0
Δω
  RPM  RPC
PC1
PC2
PM
When power and frequency are specified in pu, then R is dimensionless and
relates fractional changes in ω to fractional changes in PM. In North America,
governors are set with Rpu=0.05, i.e., if a disturbance occurs which causes a
5% increase in steady-state frequency (from 60 to 57 Hz), the corresponding
change in unit output will be 1 pu (100%).
45
Governor response
Now consider a general multimachine system having K generators. From eq.
(6), for a load change of ΔP MW, the ith generator will respond according to:
 S Ri f
f / 60
R pui  
 PMi 
PMi / S Ri
R pui 60
 S R1
S RK
 ... 
The total change in generation will equal ΔP, so P   
R Kpu
 R1 pu

f
 P

60  S R1
S RK 
Solving for Δf results in
 ... 


RKpu 
 R1 pu
 f

 60
Substitute previous expression into top expression for ΔPMi results in
 S Ri f
S Ri
P
PMi 

R pui 60 R pui  S R1
S RK 
 ... 


RKpu 
 R1 pu
If all units have the same per-unit droop constant, i.e., Rpui=R1pu=…=RKpu, then
 S Ri f
S Ri P
PMi 

Rpui 60 S R1  ...  S RK 
46
Governor response
PMi 
 S Ri f
S Ri P

Rpui 60 S R1  ...  S RK 
Conclusion of previous slide is that, for a MW imbalance of ΔP (caused by, for
example, a generation trip), following the action of primary control (but before
action of AGC), units “pick up” in proportion to their MVA ratings.
So full primary control action generally occurs within about 30 seconds
following an outage. AGC will generally not have too much influence until
about 1-2 minutes following an outage. So the above relation may be thought
of as characterizing the state of the power system in the 30 second to 2
minute time frame.
47
Governor response
I performed a brief review of the websites from TSOs (in Europe), reliability
councils (i.e., NERC and regional organizations) and ISOs (in North America)
in 2009 and concluded that there are no requirements regarding use of
primary frequency control in wind turbines.
I did a little more searching this past week and conclude that the situation in
the US has not yet changed much, but it is possible I may have missed
something. Please let me know if you come across any North American
requirements for wind turbines to have primary control capability.
The technology for implementing primary control is available from most utility-scale
turbine manufacturers, however.
48
Generator control view of a standard DFIG
G. Ramtharan, J.B. Ekanayake and N. Jenkins, “Frequency support from double fed induction wind turbines,”
IET Renew. Power Gener., 2007, 1, (1), pp. 3–9
49
Generator control view of a standard DFIG
for maximum power extraction from the wind
A torque set point, TSP, is determined by the rotor characteristic curve for max power
extraction. A reference rotor
current iqrref is derived from
the electromagnetic torque
equation & then compared
with the measured current iqr.
A proportional integral (PI)
controller is used to determine
the required rotor voltage, vqr.
50
View of a standard DFIG with inertial emulation
Introduce a signal
proportional to rate of
change of frequency
51
View of a standard DFIG with inertial
emulation & primary control Δω
Introduce a signal
proportional
frequency
deviation
Introduce a signal
proportional to rate of
change of frequency
52
Illustration of primary speed control: Reg-down
“The figure illustrates the power response of a 60-MW wind plant with GE turbines to a 2% increase in system
frequency. During this test, the site was initially producing slightly less than 23 MW. The system overfrequency
condition was created using test software that injected a 2% controlled ramp offset into the measured frequency
signal. The resulting simulated frequency increased at a 0.25 Hz/s rate from 60 Hz to 61.2 Hz. While the
frequency is increasing, the farm power drops at a rate of 2.4 MW/s. After 4.8 s the frequency reaches 61.2 Hz &
the power of the farm is reduced by approximately 50%. The overfrequency condition is removed with a controlled
ramp back to 60 Hz at the same 0.25 Hz/s rate. The plant power then increases back to an unconstrained power
level. This level is slightly higher than the unconstrained level prior to the test due to an increase in the wind
speed. These rates of frequency change are representative of relatively severe system disruptions. The plant
response is adjustable with control settings. The ramp rate power limiter becomes disabled whenever the system
is responding to frequency-related grid conditions and automatically becomes active again once the system
frequency is within the droop deadband.”
2.4 MW/sec!
R. Zavadil, N. Miller, A. Ellis, E.
Muljadi, E. Camm, and B.
Kirby, “Queuing up,” IEEE
power & energy magazine,
Nov/Dec, 2007.
53
Illustration of primary speed control – Reg down
Same picture but cleaned up a little….
R. Walling (GE-Energy),
“Making the wind work on the
plant side,” presentation slides,
Distributech Conf & Exhibition.,
http://www.uwig.org/DTECH20
07/Walling.pdf.
Droop:
R=Δω/ΔP
=(1.2/60)/(11.5/23)
=0.04 pu
54
Illustration of primary speed control – Reg down
Looks look at 2.4 MW/sec….
 Diesel engines 40 %/min
 Industrial GT 20 %/min
 GT Combined Cycle 5 -10 %/min
 Steam turbine plants 1- 5 %/min
 Nuclear plants 1- 5 %/min
Consider a 50 MW diesel engine 40%/min:
20MW/min=0.33MW/sec.
Consider a 200 MW gas turbine at 20%/min:
40MW/min=0.67MW/sec.
Wind farms are ABLE to move fast!
R. Walling (GE-Energy),
“Making the wind work on the
plant side,” presentation slides,
Distributech Conf & Exhibition.,
http://www.uwig.org/DTECH20
07/Walling.pdf.
55
Illustration of primary speed control – Reg up & down
R. Nelson
(Siemens), “Active
power control in
Siemens wind
turbines,”
presentation slides,
2011,
But how can we achieve reg-up if we are operating a maximum power extraction?
56
Illustration of primary speed control – Reg up
Answer: We cannot….. unless
the windfarms operate in “delta control.”
requires spilling wind,
using pitch control, but it
may be the least expensive
way to provide capability.
57
Illustration of primary speed control – Reg up & down
However, the below reference provides some cautions, as follows (pg. 65).
J. Undrill, “Power and Frequency Control as it Relates to Wind-Powered Generation,” publication LBNL-4143E of the
Lawrence-Berkley National Lab, December 2010, available at http://www.ferc.gov/industries/electric/indusact/reliability.asp#anchor.
58
Illustration of primary speed control – Reg up & down
And more…. (pg. 70).
J. Undrill, “Power and Frequency Control as it Relates to Wind-Powered Generation,” publication LBNL-4143E of the
Lawrence-Berkley National Lab, December 2010, available at http://www.ferc.gov/industries/electric/indusact/reliability.asp#anchor.
59
Illustration of primary speed control – ramp rate control
Some ISOs (e.g., ERCOT, some
Canadian provinces) require
ramp rate control to smoothly
transition from one output level
to another during and after
curtailments.
R. Nelson (Siemens), “Active power control in Siemens wind
turbines,” presentation slides, 2011,
It is also useful to limit
ramps when tracking the
maximum power point.
S. Hartge (GE-Energy), “Integrating wind energy
into the grid,” presentation slides, 2009.
60
Illustration of primary speed control – start up control
Turbines
sequenced on
at 20 sec intervals.
Desired ramp rate
limit is 3 MW/min
S. Hartge (GE-Energy), “Integrating wind energy into the grid,”
presentation slides, 2009.
61
Illustration of primary speed control: shut-down control
Shutdown interval
set to 5 minutes
S. Hartge (GE-Energy), “Integrating wind energy into the grid,”
presentation slides, 2009.
62