J. McCalley Wind Power Variability in the Grid: Regulation

J. McCalley

Wind Power Variability in the Grid: Regulation

Outline1. Power production: wind power equation 2. Variability: time and space3. Governor response

2

Power productionWind power equation

v1 vt v2

v

x

Swept area At of turbine blades:

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):

ρ=air density (kg/m3)

Therefore, A 1 < At < A 2

2211

21

vAvAvA

Q Q Q

tt

t

Mass flow rate is the mass of substance which passes through a given surface per unit time.

3Reference: E. Hau, “Wind turbines: fundamentals, technologies, applications, economics,” 2nd edition, Springer, 2006.


ttt

t vAt

xA

t

mQ

3. Mass flow rate at swept area:

22

212

1vvmKE

1. Wind velocity:t

xv

xAm 2. Air mass flowing:

4a. Kinetic energy change:

5a. Power extracted: 2

221

22

21 2

1

2

1vvQvv

t

m

t

KEP t

6a. Substitute (3) into (5a):)()2/1( 2

221 vvvAP tt

4b. Force on turbine blades: 21 vvQv

t

m

t

vmmaF t

5b. Power extracted:

21 vvvQFvP ttt

6b. Substitute (3) into (5b):)( 21

2 vvvAP tt

ttttt vvvvvvvvvvvvvvvvv ))(2/1()())(()2/1()()()2/1( 21212

21212122

221

7. Equate

8. Substitute (7) into (6b): ))((4

)()))(2/1(( 2122

2121

221 vvvv

AvvvvAP t

t

9. Factor out v13: )1)()(1(

4 1

22

1

231

v

v

v

vvAP t

4


10. Define wind stream speed ratio, a:

1

2

v

va

)1)(1(4

231 aa

vAP t

11. Substitute a into power expression of (9):

12. Differentiate and find a which maximizes function:(we hold v1 constant here)

1,3/10)1)(13(

0123122

0)1()1(24

222

231

aaaa

aaaaa

aaavA

a

P t

This ratio is fixed for a given turbine & control condition.

13. Find the maximum power by substituting a=1/3 into (11):

27

8

3

4

9

8

4)

3

4)(

9

11(

4

31

31

31 vAvAvA

P ttt

5


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.

)1)(1(4

231 aa

vAP t

31

211

211

21 2

1

2

1

2

10

2

1vAvvAvQv

t

m

t

KEP ttin

power extracted by the converter

power of the air stream

)1)(1(2

1

21

)1)(1(4 2

31

231

aavA

aavA

P

PC

t

t

inp

15. For a given v1, the maximum value of Cp occurs when its numerator is maximum, i.e., when a=1/3:

5926.027

16)

3

4)(

9

8(

2

1

inp P

PC The Betz Limit!

312

1vACPCP tPinp

6

16. Cp depends on a=v2/v1. For a given v1, v2 depends on how much energy is extracted by the converter, which depends on pitch, θ, and rotational speed, ω.


17. At the condition of maximum energy extraction, where a=v2/v1=1/3, we have the reduce downstream speed is

7

))(2/1( 21 vvvt From slide 4, we know that

12 3

1vv

Substituting the downstream speed at maximum energy extraction, the speed at the converter is:

111 3

2)

3

1)(2/1( vvvvt

RecallSome points of interest:

a=1v2=v1=vt, and P=0;a=1/3v2=v1/3, vt=2v1/3, and P=(ρAtv1

3/4)(32/27)a=0v2=0, vt=v1/2, and P=ρAtv1

3/4a=-1/3, v2=-v1/3, vt=v1/3, and P=(ρAtv1

3/4)(16/27)a=-1v2=-v1, vt=0, and P=0

)1)(1(4

231 aa

vAP t

Power productionCp vs. λ and θ

Tip-speed ratio:11 v

R

v

u u: tangential velocity of blade tip

ω: rotational velocity of blade

R: rotor radiusv1: wind speed

Pitch: θ

GE SLE 1.5 MW 8

Power productionWind Power Equation

31),(

2

1vACPCP tPinp

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 v1

3

3. Control factors affecting performance coefficient CP: • Tip speed ratio through the rotor speed ω• Pitch θ

9

Power productionCp vs. λ and θ

10[*] 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.

31),(

2

1vACPCP tPinp

One may plot P vs ω, for different wind speeds v1, as shown below [*]. MPP is the maximum power point.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 windsImportant 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 [**]).

Power productionTypical power curve

Cut-in speed (6.7 mph) Cut-out speed (55 mph)

11

Wind Power Temporal & Spatial Variability

12

JULY2006JANUARY2006

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

Blue~VERY LOW POWER; Red~VERY HIGH POWER

12

Temporal variability of wind power

13

• 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 torque-speed 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.

Temporal variability of wind power – time interval

14Source: 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.

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.

Temporal variability of wind power – location and terrain

16

“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,” 2nd edition, Springer, 2006.

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.


18H. Holttinen, “The impact of large-scale power production on the Nordic electricity system,” VTT Publications 554, PhD Dissertation, Helsinki University of Technology, 2004.

Duration curve: provides number of hours on horizontal axis for which wind power production exceeds the percent capacity on the vertical axis.

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 wind output never does.

Temporal & spatial variability of wind power

19H. Holttinen and Ritva Hirvonen, “Power System Requirements for Wind Power,” in “Wind Power in Power Systems,” editor, T. Ackermann, Wiley, 2005.

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.

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.


20International Energy Agency,”VARIABILITY OFWIND POWER AND OTHER RENEWABLES: Management options and strategies,” June 2005, at http://www.iea.org/textbase/papers/2005/variability.pdf.

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.

Temporal & spatial variability of wind power

21Task 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.

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.

Generation portfolio design to meet load variability

22

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

Variability of load

23

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.

Net load

24

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.

WLNL PPP

If we plot the distribution (histogram) for net load variation, we will find it is wider than the distribution for load alone.

Variability of net load

25

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).


26

The net load relationship must treat PL and Pw (or their deviations) as random variables.

WLNL PPP

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, σz

2= σx2+σy

2.


27

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.

μz=μx-μy, σz2= σx

2+σy2.

Correlation between ΔPL and ΔPw

28

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:

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.

yx

N

iyixi

N

iyi

N

ixi

N

iyixi

xy N

yx

yxN

yx

r

1

1

2

1

2

1


29

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.


30

• 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.

Using variance and correlation to measure regulation coincidence

31

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.


32

If two random variables are independent, then the variance of their sum (or their difference) is 222

yxz

σx

σyσz

Consider if two time series are perfectly correlated, i.e., rxy=1, implying x and y follow each other perfectly:

σxσy

σz

Consider if two time series are perfectly anti-correlated, i.e., rxy=-1, implying x and y changes in anti-phase:

σy

σx

σz

Fig 1

Fig 2 Fig 3


33

Define an angle θ: )1(2 xyr

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 rxy θ FigurePerfectly correlated 1 0 7Uncorrelated 0 π/2 6Anti-correlated -1 π 8

σx

σy

σz

θ

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.

Because wind is gen and netload is load, we should use

)1(2 xyr


34

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.

σL

σw

σT

θ


35

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).

σL

σw

σT

θ

X

Y

σT - X


36

The smaller right-triangle to the right provides

σL

σw

σT

θ

X

Y

σT - X

222 XYw

The larger right-triangle to the left provides: 222 )( XY TL

Subtracting (**) from (*) results in:

*

**

2222 )( XX TLw

Expanding the right-hand-side: )2( 22222 XXX TTLw

which simplifies to: XTTLw 2222

Solving for X results in:T

TLwX

2

222

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.[#] 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.

Also, from law of cosines:

cos2

)180cos(222

222

wLwL

wLwLT

What can you do with it?

37

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:where Our new “regulation burden” will therefore be 3σT, and we will have to have a generation portfolio to handle this.

cos2222wLwLT

)1(2 Lwr

Example

38

)1(2 Lwr

T

TLwX

2

222 cos2222

wLwLT

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 10-minute 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.

• Determine the regulation share of the new wind farm.

4))5.0(1(

2)1(

2

Lwr

MWMW TT

wLwLT

3573119

1.1416)7071.0(5000625100004

cos)100)(25(2)25()100(cos2 22222

MWXT

TLw 11.20)119(2

1416110000625

2

222

What can you do with it?

39


How to determine the generation portfolio necessary to handle it?

T j

jj TRRSR 3,minregulating is

Then a guiding criteria is that for all operating conditions, we would like to satisfy the following relation:

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.

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.

SRj is up (or down) “headroom” for unit j, RR is ramp rate of unit j, ΔT is time interval over which σT is computed.

Obtaining regulation component

40Steve 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.

-100

-80

-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

RE

GU

LA

TIO

N I

N M

EG

AW

AT

TS

Regulation

=

+

Load Following Regulation

How to get the time series of the regulation component?


41B. 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...

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.,

12

......

12

1 11

T

LLLLLL

TLF TkTkkTkTk

Tk

Tkiik

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

15

...... 7667 kkkkk

kLLLLL

LF

When k=20 (the 20th time point), then

15

...... 272620141320

LLLLLLF


42R. Hudson, B. Kirby, and Y. Wan, “Regulation Requirements for Wind Generation Facilities,” available at http://www.consultkirby.com/files/AWEA_Wind_Regulation.pdf.

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 kkk LFLLR 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.

Governor response

43

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:

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:

RsTsTsTsT

PP

GTGT

CM

11

1

11

ˆˆ

s

ˆ

s

PP C

C

ˆ

RsTsTssTsTs

PP

GTGT

CM

1

1111ˆ

Governor response

44

Consider ΔPM(t) for very large values of t, i.e., for the steady-state, using the final value theorem: )(ˆlim)(lim

0sfstf

st

RPP

RsTsTsTsT

P

RsTsTs

s

sTsTs

Ps

PstPP

CM

GTGT

C

s

GTGT

C

s

Ms

Mt

M

1

1111lim

1

1111lim

ˆlim)(lim

0

0

0

Applying FVT

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.

Governor response

45

Assuming that the local behavior as characterized by previous equation can be extrapolated to a larger domain, we can draw the below figure:

PM

ω

PC1

ω0

PC2

CM RPRP

Δω ΔPM

MPR Slope=-R

“Droop” characteristic: Primary control system acts in such a way so that steady-state frequency “droops” with increasing mechanical power.

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.

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%).

Governor response

46

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:

The total change in generation will equal ΔP, so

Solving for Δf results in

60/

60/ f

R

SP

SP

fR

pui

RiMi

RiMipui

60...

1

1 f

R

S

R

SP

Kpu

RK

pu

R

Kpu

RK

pu

R

R

S

R

S

Pf

...60

1

1

Substitute previous expression into top expression for ΔPMi results in

Kpu

RK

pu

Rpui

Ri

pui

RiMi

R

S

R

S

P

R

Sf

R

SP

...60

1

1

If all units have the same per-unit droop constant, i.e., Rpui=R1pu=…=RKpu, then

RKR

Ri

pui

RiMi SS

PSf

R

SP

...60 1

Governor response

47

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.

RKR

Ri

pui

RiMi SS

PSf

R

SP

...60 1

Governor response

48

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.

Generator control view of a standard DFIG

49

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

Generator control view of a standard DFIGfor maximum power extraction from the wind

50

A torque set point, TSP, is det-ermined by the rotor charac-teristic 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.

View of a standard DFIG with inertial emulation

51

Introduce a signal proportional to rate of change of frequency

View of a standard DFIG with inertial emulation & primary control

52

Introduce a signal proportional to rate of change of frequency

Δω

Introduce a signal proportional frequency deviation

Illustration of primary speed control: Reg-down

53

“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.”

R. Zavadil, N. Miller, A. Ellis, E. Muljadi, E. Camm, and B. Kirby, “Queuing up,” IEEE power & energy magazine, Nov/Dec, 2007.

2.4 MW/sec!

Illustration of primary speed control – Reg down

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R. Walling (GE-Energy), “Making the wind work on the plant side,” presentation slides, Distributech Conf & Exhibition., http://www.uwig.org/DTECH2007/Walling.pdf.

Droop: R=Δω/ΔP=(1.2/60)/(11.5/23)=0.04 pu

Same picture but cleaned up a little….

Illustration of primary speed control – Reg down

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R. Walling (GE-Energy), “Making the wind work on the plant side,” presentation slides, Distributech Conf & Exhibition., http://www.uwig.org/DTECH2007/Walling.pdf.

Consider a 200 MW gas turbine at 20%/min:40MW/min=0.67MW/sec.

Looks look at 2.4 MW/sec….


Consider a 50 MW diesel engine 40%/min:20MW/min=0.33MW/sec.

Wind farms are ABLE to move fast!

Illustration of primary speed control – Reg up & down

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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?

Illustration of primary speed control – Reg up

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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.


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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/indus-act/reliability.asp#anchor.


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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/indus-act/reliability.asp#anchor.

Illustration of primary speed control – ramp rate control

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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,

S. Hartge (GE-Energy), “Integrating wind energy into the grid,” presentation slides, 2009.

It is also useful to limit ramps when tracking the maximum power point.

Illustration of primary speed control – start up control

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Turbines sequenced onat 20 sec intervals.

Desired ramp rate limit is 3 MW/min


Illustration of primary speed control: shut-down control

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Shutdown interval set to 5 minutes


Documents

J. McCalley Wind Power Variability in the Grid: Regulation