Wind Power

ENVT S106E

Wind Power Lecture Notes

Yehia Khalil, Ph.D., Sc.D. Professor of Chemical & Environmental Engineering [email protected] Yale College Office: M8, Mason Lab., 9 Hillhouse Ave Alternate emails: [email protected]

[email protected]

This lecture materials is a property of Professor Yehia Khalil. Please do not copy of distribute without a written permission from Prof. Khalil.

Outline 1) 2) 3) 4) 5)

Source of wind energy. Environmental impact and public acceptance. Failure mechanisms of wind turbines. Kinetic energy (KE) of wind. Types of wind turbines (WT): horizontal and vertical blade designs. 6) Dependence of the rotor’s power coefficient Cp on the tip-speed ratio (). 7) Power output of a wind turbine. 8) Greenhouse gases emissions and WT life cycle analysis (LCA) 9) Wind farms. 10) Economics of wind power.

An off-shore wind farm 2

Wind Energy • About 2% of the solar energy absorbed by the earth goes into the wind energy. • Solar radiation intensity that reaches the earth's is about 350 W/m2. • Given that only 2% is converted to wind energy, them ~ 7 W/m2 goes into wind energy.

• 35% of wind energy, that is 2.45 W/m2 (= 0.35 x 7 W/m2), is dissipated in the first kilometer (~3,281 ft) above Earth's surface.

Over a period of one year, the wind energy (E) is can be calculated as follows: E = wind energy intensity x Earth's surface area x seconds per year = (2.45 W/m2) (5.1 x 1014 m2) (3.2x107 s/yr) = 4.0 x 1022 J/yr Which is 200 times greater than the average energy consumption on Earth, estimated to be 2 x 1020 J/yr. 3

Are Wind Farms environmentally friendly as a source of energy? Explain.

4

Environmental Impact and Public Acceptance Benefits

Drawbacks

1. Renewable energy source.

1. Intermittent source of electricity.

2. No emissions of greenhouse gases.(1)

2. Unreliability due to failures of critical components (e.g., gearbox, generator, electronic components, etc) as demonstrated by field failure data.

3. No acid rain gases (SOx and NOx).(1) 4. No waste to dispose or radioactivity.

3. Visual impact. 5. No raw materials are

consumed.(1) 4. Noise level.

6. No mining, processing and transportation of fuel (that is wind).

5. Thread to birds.(2) (2) Examples:

(1)

At least during wind turbines operation’s phase. See lecture notes on life cycle impact assessment (LCIA). Some CO2 is produced during manufacturing the WT but it is much less than the emissions from burning an energy-equivalent amount of coal or natural gas.

• Over a 2-year period, 183 birds were killed in the Altamont wind farm in California. • An average of 0.13 birds killed per WT in 2003 in Spain (which has over 18 wind farms). 5

http://www.sfgate.com/business/article/Altamont-Pass-turbines-kill-fewer-birds-4230640.php#photo-4099594

6

Causes of Failures of Wind Turbines 1. Icing on the blades. 2. Lightning strikes.

3. High winds. 4. Failure of electronic and electric components. 5. High humidity. 6. Erosion caused by sand storms. 7. Corrosion specially for offshore wind farms.

8. Control algorithms failure. 9. Sensors failure.

10. Cyclic fatigue stresses on turbine blades.

7

Risk Mitigation Methods 1. Fault tolerant design. 2. Component redundancy (e.g., multiple sensors). 3. Components functional diversity. 4. Condition-based maintenance (CBM). 5. Reliability-centered maintenance (RCM). 6. Components testing after manufacturing to eliminate enfant mortality (i.e., early failures). 7. WT components manufacturing quality control and quality assurance (QC & QA). 8

http://www.google.com/#sclient=psy&hl=en&q=video+on+wind+turbin+failure&aq=f& aqi=&aql=&oq=&gs_rfai=&pbx=1&fp=652e0df67dbfff88 http://www.youtube.com/watch?v=sbCs7ZQDKoM

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How winds are formed?

10

• Original source of wind energy is the radiation from the sun where a major portion of the solar heat is absorbed by the sea and land. • In turn, sea and land heats up the surrounding air and set up motion of air current (lighter hotter air rises and colder air moves downwards).

• An enormous amount of power resides in the winds: ≈ 1-2% of the incident solar power (1.37 kW/m2) is converted into wind power. • The radius of the Earth (r) is ≈ 6000 km so the crosssectional area receiving solar radiation is ≈ 1014 m2 (= 3.142 x r2) and the power in the winds is ≈ 1015 W (= 1.37 kW/m2 x 1014 m2) ≈ 1,000 TW = 1015 J/sec • Land has a lower heat capacity than the sea and heats up quickly during the day.

A simplified Map of Global Wind Patterns

• During the day, the sea is, therefore, cooler than the land and this causes the cooler air to flow shorewards to replace the rising warm air on the land. • During the night, direction of air flow is reversed. TW = Tera Watt = 1012 Watt

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Specific Energy Output: It is the output power per unit of swept area and is useful when comparing turbines of different size or design.

Capacity Factor:

The Yaw control mechanism orients the turbine in the wind direction.

It is the ratio of the energy produced in a year to the energy that would be produced if the turbine operated at its rated power. The capacity factor is typically ≈ 1/3

WT Availability: Turbines operate typically for 65-80% of the time depending on demand and on whether the wind speed is below ucut-in or above ucut-out.

Horizontal-Axis Wind Turbine (HAWT) The ratio of the annual energy yield to that which would be produced at the rated power is called the “capacity factor” and it is typically ≈ 1/3 12

• Fixed-pitch constant speed WT. • With variable speed WT, the pitch can be altered.

http://en.wikipedia.org/wiki/File:EERE_illust_large_turbine.gif

13

Anemometer: Measures wind speed and transmits wind speed data to the controller. Blades: Most turbines have either two or three blades. Wind blowing over the blades causes the blades to "lift" and rotate. Brake: A disc brake, which can be applied mechanically, electrically, or hydraulically to stop the rotor in emergency situation. Controller: The controller starts up WT at wind speeds of about 8 to 16 miles per hour (mph) (3.6 to 7.2 m/s) and shuts off WT at about 55 mph (24.6 m/s). Turbines do not operate at wind speeds above about 55 mph because they might be damaged by the high winds. Gear box: Gears connect the low-speed shaft to the high-speed shaft and increase the rotational speeds from about 30 to 60 rotations per minute (rpm) to about 1000 to 1800 rpm, the rotational speed required by most generators to produce electricity. The gear box is a costly (and heavy) part of the wind turbine and engineers are exploring "direct-drive" generators that operate at lower rotational speeds and don't need gear boxes. Generator: Usually an off-the-shelf induction generator that produces 60-cycle AC electricity. High-speed shaft: Drives the generator. Low-speed shaft: The rotor turns the low-speed shaft at about 30 to 60 rotations per minute (rpm). Nacelle: The nacelle sits atop the tower and contains the gear box, low- and high-speed shafts, generator, 14 controller, and brakes. Some nacelles are large enough for a helicopter to land on.

Pitch: Blades are turned, or pitched, out of the wind to control the rotor speed (rpm) and keep the rotor from turning in winds that are too high or too low to produce electricity. Rotor: The blades and hub together are called the rotor. Tower: Towers are made from tubular steel, concrete, or steel lattice. Because wind speed increases with height, taller towers enable turbines to capture more energy and generate more electricity. Wind direction: "upwind" and "downwind” directions. Wind vane: Measures wind direction and communicates with the yaw drive to orient the turbine properly with respect to the wind. Yaw drive: Upwind turbines face into the wind; the yaw drive is used to keep the rotor facing into the wind as the wind direction changes. Downwind turbines don't require a yaw drive, the wind blows the rotor downwind. Yaw motor: Powers the yaw drive.

SOURSE: U.S. Department of Energy

Downwind WT

Upwind WT 15

Lightning conductor Wind direction indicator

Turbine Generator

Cup anemomet er for wind speed indication

16

Nacelle

Tower

http://en.wikipedia.org/wiki/Wind_turbine 17

http://www1.eere.energy.gov/windandhydro/wind_animation.html Highlight, right click and open hyperlink to go to the website and see animation 18

Power range: 0.4 kW up to 2 MW or 2.5 MW

Useful information for life cycle impact analysis (LCIA)

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Vertical-Axis Wind Turbine (VAWT)

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Kinetic Energy of Wind • For a wind speed u and air density  , the energy density E, i.e., energy per unit volume) is given by: E

1  .u 2 2

Kinetic Energy per unit volume

• The volume of air flowing per second through a cross-sectional area A normal to the direction of the wind = u.A • Hence, the K.E. of the volume of air flowing per second through this area = E.u.A

• This is called the wind power P (i.e., energy per second) and can be expressed as: P  E. A.u 

1 1 1  .u 2 . A.u   . A.u 3  .( .r 2 ).u 3 2 2 2

r = radius of blade

• Thus, the power in the wind “P” varies as the cube of the wind speed u. Key Insights: 3

P u

• Fluctuations in wind speed can cause WT power output to vary significantly. • Much more power is available at higher wind speeds.

21

Example:

Calculate the power (P) in a wind moving with a speed u = 5 m/s incident on a WT with blades of 100 m diameter (r = 50 m). How does the power change if the wind speed increases to u = 10 m/s? Assume the density of air  = 1.2 kg/m3 Solution: P

1 1  . A.u 3  x(1.2) x(3.142x502 ) x(53 )  0.6MW 2 2 3

 10 m / s    23 P    5 m/ s  Doubling the wind speed will increase the power P by a factor of 8 (=23), hence:

The wind power would increase to 8 x 0.6 MW = 4.8 MW 22

Can all the power in wind be extracted by the wind turbine?

23

Maximum Rotor Efficiency (CP) There are two extreme situations and none of them makes physical sense: 1) Downwind velocity is zero which means that the wind turbine extracts all of the wind’s kinetic energy. If this were to happen, no additional air can pass through the wind turbine as it will be stopped by the stagnant air. 2) Downwind velocity is the same as the upwind velocity and, hence, the wind turbine extracts none of the wind’s kinetic energy. If this were to happen, the turbine blades will not rotate.

In 1919, Albert Betz suggested that there must be some ideal slowing of the wind velocity so that the turbine can extract the maximum kinetic energy from the wind. 24

Maximum Rotor Efficiency (CP) • CP represents a constraint on the ability of a wind turbine to convert kinetic energy in the wind into mechanical power. • As wind passing though the turbine blades, it slows down and the pressure is reduced so it expands

25

Kinetic Energy Extracted by the Turbine Blades



1  KE b   m  v 2  vd2 2



ṁ = mass flow rate of air within stream tube v = upwind undisturbed wind speed vd = downwind wind speed Kinetic energy of the upstream wind is greater than the kinetic energy of the downstream wind.

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Determining Mass Flow Rate • The mass flow rate is easier to calculate at the blades of the rotor because the cross sectional area (swept area) A is known. • Then, the mass flow rate can be calculated as follows:

m   Avb

(6.19)

• Assume the velocity through the rotor vb is the average of upwind velocity v and downwind velocity vd, then:

v  vd vb = 2

 v  vd  m   A  2   27

Energy Extracted by the Turbine Blades 

Substituting

m in the kinetic energy equations gives: 1  v  vd  2 2 KE b     A     v  vd 2  2 





Let,

Downwind velocity vd   Wind Speed Ratio   Upwind velocity v 1  v  v  2 2 2 KE b     A    v   v 2  2 



 28

Energy Extracted by the Turbine Blades 1 v  v  2  2 2 KEPbb   A  v   v    2  2 

(6.22)

3 2 3 3 3 3 v  v  v  v  v  v  2 2 2 +   v   v  = 2 2 2 2  2  v3 1    - 2 1     = 2 v3 1    1   2  =  2 1 1 3 KEPbb   Av  1    1   2  (6.22)   2 2

(KE)b= Energy in the wind

CP = Rotor efficiency

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Maximum Rotor Efficiency (CP) • Now, we need to determine the wind speed ratio λ which maximizes the rotor efficiency, CP • From the previous slide 2 3 1 1    CP  1    1   2   = - + 2 2 2 2 2

Set the derivative of rotor efficiency to zero and solve for λ:

CP =-2  1  3 2  0  CP =3 2  2  1  0  CP =  3  1   1  0 

Solve one equation in one unknown ()

This value will maximize the rotor efficiency

1  3 30

Maximum Rotor Efficiency (CP) • Plug the optimal value for λ = 1/3 back into CP to find the maximum rotor efficiency: 1  1  1   16 CP  1  1  2   =  59.3% 2  3  3   27

(6.26)

•

The maximum rotor efficiency of 59.3% occurs when the wind velocity is slowed to 1/3 of its upstream value.

•

This conclusions is called the “Betz efficiency” or “Betz’ law”

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Maximum Rotor Efficiency (CP) CP = 0.593

 = 1/3

Rotor efficiency CP vs. wind speed ratio λ

  Wind Speed Ratio 

Downwind velocity vd  Upwind velocity v

Wind Speed Ratio () 32

Summary of Key Insights • Wind turbines cannot capture more than 59.3% of wind’s energy (Betz, 1919).

• This maximum ratio of 59.3% is found at wind speed ration  = vd/v ≈ 1/3.

• Ideally, you want the turbine blades to slow the downstream wind velocity by 2/3 of its original speed v, that is, from v to 1/3v

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Tip-Speed Ratio (TSR) • WT Efficiency is a function of how fast the rotor turns (rpm). • Tip-Speed Ratio (TSR) is the speed of the outer tip of the blade divided by the upstream wind speed (v).

Rotor tip speed rpm   D Tip-Speed-Ratio (TSR)  = (6.27) Wind speed 60v

 rpm   .D 60   TSR  v

TSR is dimensionless

• D = rotor diameter (m) • v = upwind undisturbed wind speed (m/s) • rpm = rotor speed, (revolutions/min) 34

Tip-Speed Ratio (TSR) Cp = 59.3%

Rotors with two blades reach their maximum efficiency at higher tipspeed ratios compared to three blades type.

TSR for various rotor types. 35

Example Given Data: A 40-m diameter wind turbine with three-blades and 600 kW power output. The wind speed is 14 m/s and the air density () is 1.225 kg/m3 a. Find the rpm of the rotor if the wind turbine operates at a TSR of 4.0 b. Find the tip speed of the rotor c. What gear ratio is needed to match the rotor speed to the generator speed if the generator must turn at 1800 rpm? d. What is the efficiency of the wind turbine under these conditions? 36

Example (cont’d) a. Calculation of the rpm of the rotor if it operates at a TSR of 4.0 (rpm)(D) TSR 

60v

Tip-Speed-Ratio (TSR)  60v rpm  D 4.0  60sec/min 14m/s rpm  = 26.7 rev/min   40m/rev

We can also express this as seconds per revolution: 26.7 rev/min rpm  = 0.445 rev/sec or 2.24 sec/rev 60 sec/min

37

Example (cont’d) b) Rotor tip speed:

(rpm)(D) TSR  60v

Rotor tip speed  TSR.v

rpm   D Rotor tip speed= 60 sec/min Rotor tip speed = (rev/sec)   D

Rotor tip speed = 0.445 rev/sec   40 m/rev = 55.92 m/s c) Gear ratio:

Generator rpm 1800 Gear Ratio = = = 67.4 Rotor rpm 26.7 38

Example (cont’d) d. Efficiency of the complete wind turbine (blades, gear box, generator) under these conditions: Overall wind turbine efficiency:

1 1   3 PW   Av = 1.225    402 143  2112 kW 2 2 4  1 Ideal WT Power  Av 3 2

1 Actual WT Power  Av 3 2

The actual WT power is given in the problem statement as 600 kW

Actual WT Power  Ideal WT Power

600 kW   28.4% 2112 kW

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• The speed V of the blade at a radius r is given by: V

r.Vtip

where Vtip is the speed of the blade tip and R is max radius R of the blade

• The tip-speed ratio () is the ratio of the speed of the blade at the tip, Vtip, to the speed of the incident wind uo



Wind Incident on a Rotating Turbine

Vtip u0

 is the same as TSR which was defined before. 40

Example: A wind turbine (WT) with three blades is operating in a mean wind speed of 8 m/s. The turbine rotate at 15 rpm. Each blade is 40 m long. Estimate Vtip and tip-speed ratio () Solution: The time  (in seconds) for one revolution of the tip of a bade of length R is:



2R Distance  Vtip Velocity

The number of revolutions per minutes, nrpm can be calculated as follows: 60 nrpm   The tip speed and tip-speed ratio (TSR) can be calculated as follows: Vtip 

2R



 2R.

nrpm 60

TSR   

 2 (40)(15) / 60  62.8 m / s Vtip

Vmean _ wind _ speed



62.8  7.85 8

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Output Power Curve

Ucut-in = minimum required for the WT to operate.

Ucut-out = maximum safe operating wind speed. WT overspeed is a failure mode.

Output Power vs. Wind Speed

The blade pitch can be altered to shed power by reducing the angle of attack (feathering) or the blades stall by design. The turbine can be stopped by application of the shaft brake.

• The output power curve for a typical fixe- pitch, constant speed turbine is shown in this figure. • The rated wind speed is such that the wind is strong enough to produce the maximum output power (Prated) of the turbine generator. • The tip-speed ratio () and hence the rotor power coefficient Cp change with wind speed for constant-speed turbines. 42

Example: Calculate the average power output (P) of a WT with blades of 85 m diameter (D) operating in wind with a mean speed of 7 m/s. At this speed, the rotor power coefficient Cp is 0.45. The rated rotor output power is 1.5 MW when the wind speed is greater than 13 m/s. What is the power coefficient Cp at a wind speed of 13 m/s? Assume the density of air is 1.2 kg/m3

Solution:

Cp 

P

1  .u03 . A 2 Hence,

P





 0.45

(1 / 2).(1.2)( .852 ).(7 3 ) 4

P  526 kW  0.526 MW



C p  1.5 x106 /{(1 / 20.(1.2)( .852 ).(133 )}  0.2 4

In this example, the power coefficient (Cp) at the rated wind speed (i.e., 13 m/s) is just below half its value (from 0.45  0.2) at the mean wind speed (i.e., 7 m/s). 43

Wind Turbine / Electrolyzer Integrated System

Hydrogen Gas

44

GLOBAL WARMING POTENTIAL (GWP) & GREENHOUSE GASES (GHG) • The global warming potential (GWP) of the WTs/Electrolyzer system is a combination of CO2, CH4, and N2O emissions expressed as CO2equivalence. • For a 100-year time horizone, the capacity of CH4 and N2O to contribute to the warming of the atmosphere is 21 and 310 times higher than CO2, respectively.

• The GWP and the contribution from each compound is given in the Table below.

45

Life Cycle GWP (CO2 -equivalent)

757 + 43 + 170 = 970 g CO2

• The above Figure shows how the CO2-equivalent emissions are divided among the different process blocks for the wind farm/electrolysis system. • Because of the steel and concrete requirements, the construction and operation of the wind turbines account for 78% of the total GWP. • Hydrogen storage and compression accounts for 18% of the GWP. This is due primarily to the production of the steel used in the storage tanks. 46

Wind Farms When the turbines are placed on a square grid (d x d), the power per unit land area ca be calculated as follows: 1 3 2 v d power 8 2 land area nd 

where n.d is the number of blade diameters between turbines.

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Wind Farms • Good Wind farm sites require an average wind speed > 6 m/s.

• Onshore suitable locations include high altitude plains, exposed ridges, open lands and coastal areas. • Offshore wind farms need undersea electric cables to land and installation & maintenance costs which are higher than onshore farms. • Spacing of WTs in a wind farm: 5 to 10 rotor diameter (D) downwind by 5D crosswind.

• On a wind farm, WTs must be spaced out enough so that they do not interfere with each other.

• One common way of spacing WTs out is ensuring there is at least 5 rotor diameters between the turbines as shown in the figure. 48

Economics of Wind Turbines 1. Capital cost: ~ $2 to $3 million per WT of 3 MW capacity (roughly $1M per 1 MW WT). 2. O&M cost is about 2% of the capital cost. Unfortunately, WT field experience showed that is an underestimation of O&M costs. 3. Lifetime of a WT is about 20 to 30 years. Typically 20 years 4. Payback period is ~ 1 year 5. Electricity generation cost: WT is ~ 3 to 10 cents/kWh vs. 4 to 9 cents/kWh for coal-fired plants.

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Useful References 1. Siemens Energy and Automation, Inc. Wind Turbine (online). 2.

3. 4.

5.

http://www2.sea.siemens.com/NR/rdonlyres/1F91AFE0-BB27-4D13-91F7153AEA0D6C98/0/WindTurbine.jpg [9 June 2009]. Cullum A, Kwan C, Macdonald K. British Columbia Wind Energy Feasibility Study (online). http://www.geog.ubc.ca/courses/geog376/students/class05/cskwan/intro.h tml [4 May 2009]. Dodge, Darrel. Part 1 - Early History Through 1875: Wind Power's Beginnings (online). Illustrated History of Wind Power Development. http://www.telosnet.com/wind/early.html [10 June 2009]. Aubrecht GJ. Solar Energy: Wind, Photovoltaics, and Large-Scale Installatons. In: Energy – Physical, Environmental, and Social Impact (3), edited by Erik Fahlgren. Upper Saddle River, NJ: Pearson Education Inc., 2006, chapt. 21, 461-465. Kump, L.R., Kasting, J.F., and Crane, R.G. The Atmospheric Circulation System. In: The Earth System (2), edited by Patrick Lynch. Upper Saddle River, New Jersey, USA: 2004, chapter 4, pp. 55-82. 50

Useful References 6. Environment Canada. Canadian Atlas Level 0 (online). http://collaboration.cmc.ec.gc.ca/science/rpn/modcom/eole/CanadianAtlas 0.html [20 May 2009]. 7. Gustavson MR. Limits to Wind Power Utilization. Science 204: 13 – 17, 1979. 8. MacKay DJC. Sustainable Energy – Without the Hot Air (Online). UIT Cambridge. http://www.inference.phy.cam.ac.uk/sustainable/book/tex/ps/253.326.pdf [4 May 2009]. 9. Danish Wind Industry Association. Guided Tour on Wind Energy (online). http://www.windpower.org/en/tour.htm [4 May 2009]. 10. Learning (online). Solacity Inc. http://www.solacity.com/SiteSelection.htm [20 May 2009]. 11. D’Emil B, Jacobsen M, Jensen MS, Krohn S, Petersen KC, and Sandstørm, K. Wind with Miler (online). Danish Wind Industry Association. http://www.windpower.org/en/kids/index.htm [4 May 2009].

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Useful References 12. Clarke S. Electricity Generation Using Small Wind Turbines At Your Home Or Farm (Online). Ontario Ministry of Agriculture, Foods and Rural Affairs. http://www.omafra.gov.on.ca/english/engineer/facts/03-047.htm#noise [25 May 2009]. 13. Marris E and Fairless D. Wind Farms' Deadly Reputation Hard to Shift. Nature 447: 126, 2007. 14. Keith D. Wind Power and Climate Change (online). University of Calgary. http://www.ucalgary.ca/~keith/WindAndClimateNote.html [20 May 2009]. 15. Accio Energy. About Accio Energy (online). http://www.hydrowindpower.com/ [12 June 2009].

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Backup Slides

53

Principles of a Horizontal-Axis Wind Turbine • Not all the power in wind can be extracted by a WT. • Some of K.E is carried down stream of the turbine in order to maintain flow of air.

• This effect places a theoretical maximum efficiency of 59% for extracting power from the wind, known as the Betz limit. u2  u1  u0 u0 . A0  u1. A1  u2 . A2

• Upstream, the speed of wind is u0 and it passes through an area A0. • By the time the wind reaches the turbine it has slowed to u1 and the area of the stream-tube has increased to A1, which is the area swept out by the blades of the turbine.

• Downstream of the turbine, the wind’s cross-sectional area is A2 and its speed is u2. 54

Principles of a Horizontal-Axis Wind Turbine (HAWT) • The drop in speed of the wind before and after the turbine gives rise to the pressure drop across the turbine, through Bernoulli’s theorem, so there is a thrust on the turbine blades. • The maximum power is generated when downstream of the turbine the wind speed is 1/3 of the upstream speed u0 and at the turbine the wind speed is 2/3 of u0, that is: 1 2 u2  u0 and u1  u0 3 3

Hence:

From u1 

u0  u 2 , then : 2

1 1 1 1 4 2 u1  (u0  u2 )  (u0  u0 )  ( u0 )  u0 2 2 3 2 3 3 2 4 1 u2  2u1  u0  2( u0 )  u0  u0  u0  u0 3 3 3

• The fraction of power extracted by the turbine which is called the “Power Coefficient” Cp is give by: P Cp 

1  . A.u03 2

55

Principles of a Horizontal-Axis Wind Turbine (HAWT) • The limit of Cp is 59.3% (= 16/27) of the incident wind power is called the Betz or Lanchester-Betz limit and was first derived by Lanchester in 1915 and independently by Betz in 1921. 1 16 Pmax   . A.( ).u03 2 27

• The thrust T exerted on the turbine by the wind is equal to the rate of change of momentum: T

dm .(u0  u2 ) dt

• Where dm/dt is the mass of wind flowing through the stream tube per second. • Power extracted from wind = (thrust) x (air speed at the turbine) P  T .u1 

dm .(u0  u2 ).u1 dt

• The power P can also be expressed as the rate of loss of K.E of the wind: 1 dm 2 P . .(u0  u22 ) 2 dt

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Principles of a Horizontal-Axis Wind Turbine (HAWT) • From the continuity equation, we have: dm   . A.u1 dt • Hence: • Let:

u1  (1  a).u0

P  2  .u12 . A.(u0  u1 )

where a is called the " induction factor"

Then : P 

1 .u03 . A.{4a.(1  a) 2 } 2

• The power coefficient Cp which represents the fraction of the power in the wind that is extracted by the turbine is given by: Cp 

P 1  .u03 . A 2

 4a.(1  a) 2

• Maximizing P by setting dCp/dt to zero and solving for the factor “a” gives a = 1/3 Then, the maximum power extracted Pmax when a = 1/3 1 Pmax   .u03 . A.{16 / 27} 2

• The limit for the rotor’s power coefficient Cp of 16/27 (= 59.3%) of the incident wind power is called the Betz or Lanchester-Betz limit. 57

The Rayleigh frequency distribution for a mean wind speed of 8 m/s

Typical Variation of Wind Speed with Height The commonly used form to describe the dependence of u on height z is given by:

u ( z )  us .(

z s ) zs

s is called the wind shear coefficient and zs is the height at which u is measured to be us and is typically 10 m. s also shows a large variation over a 24 hr period and can change from < 0.15 during the day to > 0.5 at night. This is because the at night the surface temperature drops as the ground loses heat by radiation. After sunrise, the ground is heated by the Sun and warms the air in contact which then rises causing mixing and reducing wind shear.

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Power Output of a Wind Turbine The power in the wind Pw at a given site is given by: Pw 

1 1  . A.  u 3   . A. {u ( z )}3 . f (u ).du 2 2

Where u(z) is the wind speed at the height of turbine hub and f(u) is the wind frequency distribution. The average output power P0 is given by: 1 P0   . A C p ( ).{u ( z )}3 . f (u ).du 2

Where the tip-speed ratio  = Vtip / u. • For a variable pitch or speed turbine, Cp can be kept close to its maximum value over a range of wind speeds. • For a fixed pitch turbine, Cp changes with the wind speed as shown in the figure.

If is the mean wind speed at the height of the turbine hub , the average output (P0) for a fixed speed and pitch turbine is:

P0  0.2 D 2 .  u ( z )  3

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Example: Estimate the power output of a wind farm consisting of 25 fixed pitch and constant speed 1 MW turbines. The hub is z = 85 m and diameter is D = 55 m. The wind has an average speed of u = 7 m/s at a height of 10 m. The land is characterized by a surface roughness parameter z0 = 0.001 m

Solution:

P0  0.2 D 2 .  u ( z )  3 s 

1 0.001 0.2 ( )  0.08 2 10

s is called the wind shear coefficient

Uhub = 7(85/10)0.08 = 8.3 m/s P0 = (0.2)(55)2.(8.3)3 = 346 kW

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Fatigue in Wind Turbines • Fatigue in the bade material is an important concern due to the very large number of stress cycles, typically on the order of 100 million, O(108) over a 30-year design life of the turbine.

• The rotation of the blades causes the loads experienced by the turbine to change repeatedly, i.e., cyclical stress. These loads weaken the structure through fatigue. Cyclic stress versus log10 (# of cycles to failure)

• For a 30-year lifetime, an 80 m diameter turbine operating for 80% of the time at  = 8 in a wind speed of 10 m/s will make about 2.4x108 stress cycles (N). • This means that the maximum stresses (force per unit area) must be lower than in other structures to avoid failure through fatigue. 62

Fatigue Failure in Wind Turbines • Fatigue failure is the fracture of material after it has been subjected to repeated cycles of stress changes at levels considerably below its initial static strength. • The number of cycles to failure (N) decreases as the alternating stress level increases.

• The stress level can be characterized by the mean stress and its range:  imax   imin mean i  2

Range   imax   imin • Fatigue involves the initiation and growth of cracks in a material under repeated stress cycles. • Discontinuities such as sharp corners or flaws in the material are prime sites. • Fatigue can be quantified by using the Palmer-Milner linear damage rule (called Milner’s rule)

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Fatigue Failure in Wind Turbines • This method breaks down the cyclic stresses that a structure undergoes into the number of cycles, ni, at each stress level, i , that occur. • The total damage DM sustained by a structure is given by: is

DM   i 1

ni n n  1  2  .... N i N1 N 2

• Where Ni is the number of cycles to failure at the stress level , i • Milner’s rule states that failure will occur when DM = 1

• The fatigue strength of a material is the value of the stress level, , required to cause failure after a specified number of cycles N. The results can be expressed as an S vs. N plot (S-N plot) where S is the ratio /0 • The stress 0 is the static strength of the material. The data can be represented by the equation:

S   /  0  1  b. log10 ( N )

where b is a positive cons tan t 64

• The good quality material has a value of b = 0.1, i.e., the fatigue strength decreases by 10% for each decade increase in the number of cycles. • The poor quality material has b = 0.14. • As can be seen, at stress ratio S = 0.2, the good quality material has over two orders of magnitude longer lifetime.

S-N curves for two fiberglass composites

The equation in the figure shows that the stress (S) that can be tolerated decreases with the number of cycles.

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