Wind Energy Lecture slides

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WIND ENERGY

Keith Vaugh BEng (AERO) MEng

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Utilise the vocabulary associated with wind energy and its mechanics

Develop a comprehensive understanding of wind measurement and analysis, the workings of wind turbines, the various configurations and the components associated with the plant

Derive the governing equations for the power plant and the associated components

Determine the forces acting on the turbine blades and the supporting masts

OBJECTIVES

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The primary source of wind is Solar Radiation

Approximately 1-2% of the incident solar power (1.4 kW⋅m-2) is converted into wind

Radius of the Earth ∼ 6000 km therefore the CS area receiving solar radiation is about 1.13×1014 m2

Winds are variable in time and location

WIND SOURCE & CHARACTERISTICS

KV(Lutyens & Tarbuck 2000)

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Idealized winds generated by pressure gradient and Coriolis Force.

Actual wind patterns owing to land mass distribution..

(Lutyens & Tarbuck 2000)

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Geostrophic wind/Prevailing wind

Storms

TYPES OF WIND

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Geostrophic wind/Prevailing wind

Storms

TYPES OF WIND

Local winds/Sea breezes

Mountain wind/Valley wind

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Inter annual - Longer than 1 year variations

Annual - Seasonal or monthly variation

Diurnal - Daily variation

Short term - Turbulence and gusts

WIND VARIATION WITH TIME

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The Griggs - Putnam Index of deformity

Wind Atlas

Anemometers (measurement)

SITE ASSESSMENT

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TYPES OF WIND

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Anemometers

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Planning and Development Regulations 2008 (S.I. No. 235 of 2008), state that for; The erection of a mast for mapping meteorological conditions.

• No such mast shall be erected for a period exceeding 15 months in any 24 month period.

• The total mast height shall not exceed 80 metres.• The mast shall be a distance of not less than:

• the total structure height plus: • 5 metres from any party boundary, • 20 metres from any non-electrical overhead cables, • 20 metres from any 38kV electricity distribution lines, • 30 metres from the centreline of any electricity transmission line of 110kV or more.• 5 kilometres from the nearest airport or aerodrome, or any communication,

navigation and surveillance facilities designated by the Irish Aviation Authority, save with the consent in writing of the Authority and compliance with any condition relating to the provision of aviation obstacle warning lighting.

• Not more than one such mast shall be erected within the site.• All mast components shall have a matt, non-reflective finish and the blade shall be made

of material that does not deflect telecommunications signals.• No sign, advertisement or object, not required for the functioning or safety of the mast

shall be attached to or exhibited on the mast.

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It can be shown that the power, PD, a wind turbine delivers is proportional to the cube of the wind velocity

WIND MEASUREMENT

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PD= 162712ρu3Aη

WIND MEASUREMENT

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PD= 162712ρu3Aη

The mean power output over a period 0 toT isproportional to the cube of the mean cubicwind velocity,ū

u ≡ 1Tu3 dt0

T

∫⎛⎝⎜

⎞⎠⎟

13

WIND MEASUREMENT

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Wind speed varies with location and time

Weather has considerable influence on wind speed

Turbulence intensity, IT is a ratio of σT:u

IT depends on height and terrain,

σT increases as the steady wind speed increases

IT increases with surface roughness and varies approximately as;

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lnzz0

⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢⎢

⎤

⎦⎥⎥

−1

z - the height of turbine

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lnzz0

⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢⎢

⎤

⎦⎥⎥

−1 Terrain z0 (m)

Urban Area 3 - 0.4

Farmland 0.3 - 0.002

Open sea 0.001 - 0.0001z - the height of turbine

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If the average wind speed ū, is known for a given site, then it can be assumed that the wind obeys a Rayleigh distribution, i.e. the probability , p(u), of the wind having a velocity, u, is

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If the average wind speed ū, is known for a given site, then it can be assumed that the wind obeys a Rayleigh distribution, i.e. the probability , p(u), of the wind having a velocity, u, is

p u( ) = uσ 2e

−12uσ

⎛⎝⎜

⎞⎠⎟

2⎡

⎣⎢⎢

⎤

⎦⎥⎥

σ is the mode of the distribution, i.e. the value at which the probability distribution function (pdf) peaks. Although σ is not the mean value, there is a relationship between the average wind velocity and the mode of the Rayleigh pdf;

σ 2 = 2πu 2

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Rayleigh frequency distribution for a mean wind speed of 8 m⋅s-1

Rayleigh distribution

Prob

abili

ty o

f occ

uran

ce (

%)

Wind speed, u, m/s

Mean speed 8 m⋅s-1

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Hei

ght (

m)

Wind Speed (m⋅s-1)

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Hei

ght (

m)

Wind Speed (m⋅s-1)

A common describer of the dependence of u on height z is

u z( ) = us zzs

⎛

⎝⎜⎞

⎠⎟

α s

where;zs is height at which u is measured to be us, typically 10 mαs is the wind shear coefficient, which is dependent on the terrain

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Rotor diameter Slipstream

Tower

Nacelle

Blade

Hub

WIND TURBINE CONFIGURATION

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Determining the energy and power available in the wind requires an understanding of basic geometry & the physics of kinetic energy (KE).

“Kinetic Energy is the motion of waves, electrons, atoms, molecules, substances and objects”

Considering this statement and identifying air has mass, it will therefore move as a result of wind i.e. it has kinetic energy (KE).

The KE of an object (or a collection of objects, i.e. a car, train, etc...) with a total mass M and velocity v is given by;

ENERY AVAILABLE IN THE WIND

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Determining the energy and power available in the wind requires an understanding of basic geometry & the physics of kinetic energy (KE).

“Kinetic Energy is the motion of waves, electrons, atoms, molecules, substances and objects”

Considering this statement and identifying air has mass, it will therefore move as a result of wind i.e. it has kinetic energy (KE).

The KE of an object (or a collection of objects, i.e. a car, train, etc...) with a total mass M and velocity v is given by;

KE = 12mu2

where:m = Mass (kg), (1kg = 2.2 pounds)u = Velocity (Meters/second)

(1 meter = 3.281 feet = 39.39 inches)

ENERY AVAILABLE IN THE WIND

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In order to determine the KE of moving air molecules (i.e. wind), we can take a large air parcel in the shape of cylinder. This geometry will contain a collection of air molecules which will pass through the plane of the a wind turbines blades over a given time frame.

The volume of air contained within this parcel can be determined using established theory;

AREA

D

Air parcel

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Vol = A ×D

AREA

D

Air parcel

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Vol = A ×D

Vol = πd2

4×D or Vol = πr 2 ×D

AREA

D

Air parcel

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The air within this parcel also has a density ρ, and as density is mass per unit volume the density can therefore be expressed as;

ρ = mVol

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ρ = mVol

Transposing this formula to get it in terms of m, yields;

m = ρ × Vol

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ρ = mVol


m = ρ × Vol

Given that we have determined the Volume and the the density of the air parcel, we now must turn our attention to the velocity (u). If a time frame T is required for the parcel of air with thickness D to pass through the plane of the the wind turbine blades, then the parcel’s velocity can be expressed as;

u = DT

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ρ = mVol


m = ρ × Vol

Given that we have determined the Volume and the the density of the air parcel, we now must turn our attention to the velocity (u). If a time frame T is required for the parcel of air with thickness D to pass through the plane of the the wind turbine blades, then the parcel’s velocity can be expressed as;

u = DT

Transposing this formula

D = u × T

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Substituting expressions into original formula for kinetic energy

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KE = 12mu2

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Substituting for m;

KE = 12mu2

KE = 12× ρ × Vol( )× u2

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Substituting for m;

KE = 12mu2

KE = 12× ρ × Vol( )× u2

Vol can be replaced;

KE = 12× ρ × πd2

4

⎛

⎝⎜⎞

⎠⎟×D

⎛

⎝⎜⎞

⎠⎟× u2

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Substituting for m;

KE = 12mu2

KE = 12× ρ × Vol( )× u2


KE = 12× ρ × πd2

4

⎛

⎝⎜⎞

⎠⎟×D

⎛

⎝⎜⎞

⎠⎟× u2

D can be replaced;

KE = 12× ρ × πd2

4

⎛

⎝⎜⎞

⎠⎟× v × T

⎛

⎝⎜⎞

⎠⎟× u2

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Substituting for m;

KE = 12mu2

KE = 12× ρ × Vol( )× u2


KE = 12× ρ × πd2

4

⎛

⎝⎜⎞

⎠⎟×D

⎛

⎝⎜⎞

⎠⎟× u2

D can be replaced;

KE = 12× ρ × πd2

4

⎛

⎝⎜⎞

⎠⎟× v × T

⎛

⎝⎜⎞

⎠⎟× u2

Rewriting

KE = 12× ρ × πd2

4

⎛

⎝⎜⎞

⎠⎟× T

⎛

⎝⎜⎞

⎠⎟× u3

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Let us now consider the Power that can be achieved


Substituting for m;

KE = 12mu2

KE = 12× ρ × Vol( )× u2


KE = 12× ρ × πd2

4

⎛

⎝⎜⎞

⎠⎟×D

⎛

⎝⎜⎞

⎠⎟× u2

D can be replaced;

KE = 12× ρ × πd2

4

⎛

⎝⎜⎞

⎠⎟× v × T

⎛

⎝⎜⎞

⎠⎟× u2

Rewriting

KE = 12× ρ × πd2

4

⎛

⎝⎜⎞

⎠⎟× T

⎛

⎝⎜⎞

⎠⎟× u3

Power = KET

Power = EnergyTime

Power =

12× ρ × πd2

4

⎛⎝⎜

⎞⎠⎟× T

⎛

⎝⎜⎞

⎠⎟× u3

T

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Dividing by T

Power = 12× ρ × πd2

4

⎛

⎝⎜⎞

⎠⎟⎛

⎝⎜⎞

⎠⎟× u3

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Dividing by T

if we divide the Power by the cross-sectional area (A) of the parcel, then we are left with the expression;


4

⎛

⎝⎜⎞

⎠⎟⎛

⎝⎜⎞

⎠⎟× u3

Power

πd2

4

⎛⎝⎜

⎞⎠⎟

= 12× ρ( )× u3

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Dividing by T

if we divide the Power by the cross-sectional area (A) of the parcel, then we are left with the expression;

Examining this equation two important things can be identified • the Power is proportional to the cube of the wind speed• by dividing the Power by the area, an expression that is independent of the size of

the wind turbines rotor is achieved


4

⎛

⎝⎜⎞

⎠⎟⎛

⎝⎜⎞

⎠⎟× u3

Power

πd2

4

⎛⎝⎜

⎞⎠⎟

= 12× ρ( )× u3

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MOMENTUM BALANCE ACROSS

ROTOR BLADES

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Applied rotation


ROTOR BLADES

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u1

Applied rotation


ROTOR BLADES

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Undisturbed air flow at u1

u1

Applied rotation


ROTOR BLADES

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①


u1

Applied rotation


ROTOR BLADES

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① ②


u1

Applied rotation


ROTOR BLADES

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① ② ③


u1

Applied rotation


ROTOR BLADES

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④① ② ③


u1

Applied rotation


ROTOR BLADES

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④① ② ③


u1

u4

Applied rotation


ROTOR BLADES

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④① ② ③

Slipstream outer edge velocity

differential u4 > u1


u1

u4

Applied rotation


ROTOR BLADES

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④① ② ③

Slipstream area reduces in flow

direction




u1

u4

Applied rotation

(a) Air flow over a rotating aircraft propeller showing flow acceleration and slipstream boundary


ROTOR BLADES

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④① ② ③


direction




u1

u1u4

Applied rotation



ROTOR BLADES

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④① ② ③


direction




u1

u1u4

Applied rotation

Air flow driven rotation



ROTOR BLADES

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④① ② ③


direction




u1


u1u4

Applied rotation




ROTOR BLADES

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④① ② ③ ① ② ③


direction




u1


u1u4

Applied rotation




ROTOR BLADES

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④① ② ③ ④① ② ③


direction




u1


u1u4

Applied rotation




ROTOR BLADES

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④① ② ③ ④① ② ③


direction




u1


u1u4

u4

Applied rotation




ROTOR BLADES

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④① ② ③ ④① ② ③


direction




u1


u1


differential u4 < u1

u4

u4

Applied rotation




ROTOR BLADES

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④① ② ③ ④① ② ③


direction




u1


u1


differential u4 < u1

Slipstream area increases in flow direction

u4

u4

Applied rotation



(b) Air flow over a wind turbine showing flow deceleration and slipstream boundary


ROTOR BLADES

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Applying Bernoulli’s equation in the stream tube, upstream of the rotor between ① and ②, and downstream of the rotor between ③ and ④;

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p1+ 1

2ρu

12 = p

2+ 1

2ρu

22 and p

3+ 1

2ρu

32 = p

4+ 1

2ρu

42

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p1+ 1

2ρu

12 = p

2+ 1

2ρu

22 and p

3+ 1

2ρu

32 = p

4+ 1

2ρu

42

The pressures at ① and ④ are the same and if the rotor is assumed to have minimal flow direction thickness, then the velocities at ② and ③ maybe considered to be identical by continuity. Therefore the equations can be combined so that;

p2− p

3= 12ρ u

12 − u

42( )

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p1+ 1

2ρu

12 = p

2+ 1

2ρu

22 and p

3+ 1

2ρu

32 = p

4+ 1

2ρu

42

The pressures at ① and ④ are the same and if the rotor is assumed to have minimal flow direction thickness, then the velocities at ② and ③ maybe considered to be identical by continuity. Therefore the equations can be combined so that;

p2− p

3= 12ρ u

12 − u

42( )

The thrust can be expressed as the sum of the pressures on either side of the rotor disc

T = A p2− p

3( )

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Substituting for p2 - p3 into the thrust formula

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T = A p2− p

3( )T = A 1

2ρ u

12 − u

42( )⎛

⎝⎜⎞⎠⎟

T = 12Aρ u

12 − u

42( )

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T = A p2− p

3( )T = A 1

2ρ u

12 − u

42( )⎛

⎝⎜⎞⎠⎟

T = 12Aρ u

12 − u

42( )

The thrust can also be expressed as

T = !m u1− u

4( )

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Equating the thrust formulae and combining with ṁ = ρu2A defines the velocity at the rotor as the average of the upstream and downstream velocities

u2= u

3=u1+ u

4( )2


T = A p2− p

3( )T = A 1

2ρ u

12 − u

42( )⎛

⎝⎜⎞⎠⎟

T = 12Aρ u

12 − u

42( )

The thrust can also be expressed as

T = !m u1− u

4( )

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The rotor power is the the product of the thrust, T and the velocity at the rotor, u2

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Pr= 12Aρ u

12 − u

42( )u2

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Pr= 12Aρ u

12 − u

42( )u2

The Power Coefficient is the ratio of the loss of kinetic energy in the airstream to the power of the airstream passing through the rotor

CP=

12Aρ u

12 − u

42( )u2

12Aρu

13

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which reduces to


Pr= 12Aρ u

12 − u

42( )u2

The Power Coefficient is the ratio of the loss of kinetic energy in the airstream to the power of the airstream passing through the rotor

CP=

12Aρ u

12 − u

42( )u2

12Aρu

13

CP=

12u1+ u

4( ) u12 − u42( )u13

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For a frictionless system, the maximum power coefficient can be expressed as

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where, ur = u4/u1CP= 121+ u

r− u

r2 − u

r3( )

CP=

12u1+ u

4( ) u12 − u42( )u13

=

12u1u12 − u

42( )+ u4 u12 − u42( )u13

=

12u13 − u

1u42 + u

4u12 − u

43

u13

= 12

u13

u13−u42

u12+u4

u1

−u43

u13

⎛

⎝⎜⎞

⎠⎟

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r− u

r2 − u

r3( )

CP=

12u1+ u

4( ) u12 − u42( )u13

=

12u1u12 − u

42( )+ u4 u12 − u42( )u13

=

12u13 − u

1u42 + u

4u12 − u

43

u13

= 12

u13

u13−u42

u12+u4

u1

−u43

u13

⎛

⎝⎜⎞

⎠⎟The maximum power coefficient is obtained for dCPdur

= 0

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r− u

r2 − u

r3( )

CP=

12u1+ u

4( ) u12 − u42( )u13

=

12u1u12 − u

42( )+ u4 u12 − u42( )u13

=

12u13 − u

1u42 + u

4u12 − u

43

u13

= 12

u13

u13−u42

u12+u4

u1

−u43

u13

⎛

⎝⎜⎞


= 0

dCP

dur

= 121− 2u

r− 3u

r2( ) = 0

ur= 13

12

1− 2ur− 3u

r2( ) = 0

0.5−1ur−1.5u

r2 = 0

1.5ur2 +1u

r− 0.5 = 0 rearranged

apply −b ± b2 − 4ac

2a to solve for u

r

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r− u

r2 − u

r3( )

CP=

12u1+ u

4( ) u12 − u42( )u13

=

12u1u12 − u

42( )+ u4 u12 − u42( )u13

=

12u13 − u

1u42 + u

4u12 − u

43

u13

= 12

u13

u13−u42

u12+u4

u1

−u43

u13

⎛

⎝⎜⎞


= 0

dCP

dur

= 121− 2u

r− 3u

r2( ) = 0

ur= 13

12

1− 2ur− 3u

r2( ) = 0

0.5−1ur−1.5u

r2 = 0

1.5ur2 +1u

r− 0.5 = 0 rearranged

apply −b ± b2 − 4ac

2a to solve for u

r

Therefore substituting back into the efficiency formula

CP= 121+ 13− 19− 127

⎛⎝⎜

⎞⎠⎟= 1627

= 59.3%Betz or Lanchester-Betz limit

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Power delivered, PD, by a wind turbine delivers

where:CP = Power Coefficient 16⁄27

½ρu3A = Power available in the wind η = Efficiency of the aerodynamic, mech elec components

PD= C

P

12ρu3Aη

WIND TURBINE BLADE DESIGN

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Power delivered, PD, by a wind turbine delivers

where:CP = Power Coefficient 16⁄27

½ρu3A = Power available in the wind η = Efficiency of the aerodynamic, mech elec components

PD= C

P

12ρu3Aη

liftdrag

=

12ρC

Lu3A

12ρC

Du3A

WIND TURBINE BLADE DESIGN

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The rotational movement of the turbine blades induces an air velocity vector, rΩ

The air velocity vector, riΩ varies longitudinally from root to tip

ri represents the radius from 0 to 1, where 0 the turbine axis of rotation, and 1 is the radius at the outer periphery.

Rotation, Ω

rmax

r i

rmaxΩ

riΩ

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Plane ofRotation

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riΩ Plane ofRotation

riΩ -Air velocity component

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u1riΩ -Air velocity component

u1 - wind speed

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uT, i u1riΩ -Air velocity component

u1 - wind speed

uT, i - The total velocity vector

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uT, i u1

α


α - Angle of attack

u1 - wind speed


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uT, i u1

Φα



u1 - wind speed


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uT, i u1

FL

Φα

FL - Lift force



u1 - wind speed


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uT, i u1

FL

FD

Φα

FL - Lift force

FD - Drag force



u1 - wind speed


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uT, i u1

FL

FR

FD

Φα

FL - Lift force

FD - Drag force

FR - Resultant lift force



u1 - wind speed


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Plane ofRotation

FL

FR

FD

Φ

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FD

F L c

os Φ

FL sin Φ

F D s

in Φ

FD cos Φ

Resolved components:

FL

Axi

al

Tangential

The tangential and axial loads are established by resolving the vector diagrams for lift, drag and resultant lift

Plane ofRotation

FL

FR

FD

Φ

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FT= FLsinΦ − F

DcosΦ

FA= FLcosΦ − F

DsinΦ

FD

F L c

os Φ

FL sin Φ

F D s

in Φ

FD cos Φ

Resolved components:

FL

Axi

al

Tangential

The tangential and axial loads are established by resolving the vector diagrams for lift, drag and resultant lift

The tangential force provides thrust for power generation

The axial force imposes structural loading on the tower

Plane ofRotation

FL

FR

FD

Φ

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The air velocity component varies longitudinally as a result of rotational motion

As a consequence the angle of attack, α varies over the blade radius

Small angles of attack reduces CL, while large angles of attack results in a stall condition

Too large and too small an angle of attack reduces the power generated by the turbine

The extracted power is maximised when the CL:CD is maximised

CL:CD can be maximised by introducing a twist into the turbine blade thereby increasing the aerodynamic efficiency of the wind turbine.

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Ω

Φ at rmax < Φ at r1

rmaxΩ > r1Ω

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Blade r

adius, rmax

Axis of rotation

Ω


rmaxΩ > r1Ω

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Blade r

adius, rmax

Axis of rotation

Ω


rmaxΩ > r1Ω u1

u1

u1

u1

u1

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⑤rmax = 1.0

Blade r

adius, rmax

Axis of rotation

Ω


rmaxΩ > r1Ω u1

u1

u1

u1

u1

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⑤rmax = 1.0

Blade r

adius, rmax

Axis of rotation

Ω


rmaxΩ > r1Ω u1

u1

u1

u1

u1

r maxΩ

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⑤rmax = 1.0

Blade r

adius, rmax

Axis of rotation

Ω


rmaxΩ > r1Ω u1

u1

u1

u1

u1

r maxΩ

u T,m

ax

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⑤rmax = 1.0

Blade r

adius, rmax

Axis of rotation

Ω


rmaxΩ > r1Ω u1

u1

u1

u1

u1

r maxΩ

Φ

u T,m

ax

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①r1 = 0.2

⑤rmax = 1.0

Blade r

adius, rmax

Axis of rotation

Ω


rmaxΩ > r1Ω u1

u1

u1

u1

u1

r 1Ω

r maxΩ

Φ

u T,m

ax

uT, 1

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①r1 = 0.2

②r2 = 0.4

③r3 = 0.6

④r4 = 0.8

⑤rmax = 1.0

Blade r

adius, rmax

Axis of rotation

Ω


rmaxΩ > r1Ω u1

u1

u1

u1

u1

r 1Ω

r 2Ω

r 3Ω

r 4Ω

r maxΩ

Φ

u T,m

ax u T, 4 u T

, 3 u T, 2

uT, 1

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Plane ofRotation

uT, tip

u1

FL

FR

FD

Φα

FD

FR

FL

rtipΩ

rrootΩ Plane ofRotation

Φu1

uT, root

(a) Turbine tip

(b) Turbine root

Blade twist ensures that the angle of attack is optimised over the longitudinal length of the blade thereby maximising the lift:drag ratio

It should be noted that the angle of attack, α is dynamic

The blades thickness is also increased at the root provide structural support, reduce vibrations and to decrease rotational stresses

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Tip Speed Ratio, λ, a dimensionless term, describes the relationship between the rotational speed of the blades and the wind driving the turbine.

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Tip Speed Ratio, λ, a dimensionless term, describes the relationship between the rotational speed of the blades and the wind driving the turbine.

λ = rΩu1

When the Betz condition is satisfied, the angle Φ can be expressed in terms of r, R and λ

Betz condition, u1=

2uo

3

tanΦ =u

1

rΩ≡ 2R

3rΩλ

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A blade rotating at constant speed generates lift. As a consequence it has a reaction on the wind over a significant portion of the total annular area, dA;

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dA = 2πdr

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dA = 2πdr

In the time it takes for one blade to reach the position of the next blade, wind speed variations are modelled as the average wind speed over that time

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dA = 2πdr

In the time it takes for one blade to reach the position of the next blade, wind speed variations are modelled as the average wind speed over that time

= 2πnΩ

Time for one blade to reach position of next blade

n - number of blades

Ω -Annular speed of the turbine

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The total thrust developed from n blades on the annular area dA equals the rate of change of momentum;

dTn= ρu

122πrdr

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dTn= ρu

122πrdr When the Betz condition is satisfied

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dTn= ρu

122πrdr When the Betz condition is satisfied

Neglecting drag, the sum of the lift components dL from each section of blade between r and r + dr is equal also the the thrust

uT,i - Total velocity vector at blade section i

W - Width of blade or chord length at section i

dL = 12CLρuT , i2 Wdr

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Wind speed uT , i=u

1

sinΦ

Plane ofRotation

uT, i

u1

FL

FR

FD

ΦαriΩ

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Wind speed uT , i=u

1

sinΦ

Equating the thrust to the sum of the components of lift and substitute uT,i

Plane ofRotation

uT, i

u1

FL

FR

FD

ΦαriΩ

dTn= ρu

122πrdr = ndLcosΦ =

12nCLu12ρWdr cosΦ

sin2Φ

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Wind speed uT , i=u

1

sinΦ

Equating the thrust to the sum of the components of lift and substitute uT,i

Plane ofRotation

uT, i

u1

FL

FR

FD

ΦαriΩ

dTn= ρu

122πrdr = ndLcosΦ =

12nCLu12ρWdr cosΦ

sin2Φ

For the Betz condition to be satisfied the width at radius, r equals

W = 4πr tanΦ sinΦnCL

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The tip speed ratio, λ = rΩu1

Plane ofRotation

Φα

Plane ofRotation

Φ

(a) Turbine tip

(b) Turbine root

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The tip speed ratio, λ = rΩu1


Betz condition, u1=

2uo

3

tanΦ =u

1

rΩ= 2R

3rΩλ

Plane ofRotation

Φα

Plane ofRotation

Φ

(a) Turbine tip

(b) Turbine root

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The tip speed ratio,

W = 8πR sinΦ3λnC

L

λ = rΩu1


Betz condition, u1=

2uo

3

tanΦ =u

1

rΩ= 2R

3rΩλ

Substituting for tanΦ

As sinΦ increases and r decreases, the width of the blade increases from time to root.

Plane ofRotation

Φα

Plane ofRotation

Φ

(a) Turbine tip

(b) Turbine root

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(a) A three bladed wind turbine operates in a mean wind speed of 8 m·s-1. The turbine rotates at 15 RPM, each blade is 40 m long and has an angle of attack, α of 5.4º. Determine;

(i) the speed of the tip, (ii) the tip speed ratio, comment on this result (iii) the width and the and angle the blade makes with

the plane of rotation for r = 0, r = R/2 and r = R. Assume CL ≈ 1

(b) What is the significance of introducing a twist into wind turbine blade design?

EXAMPLE : BLADE DESIGN

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The time, t for one revolution of the blade tip for a turbine of length R

t = 2πRrΩ

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t = 2πRrΩ

The number of revolutions, nRPM per minute RPM

nRPM

= 60t

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t = 2πRrΩ

The number of revolutions, nRPM per minute RPM

nRPM

= 60t

The tip speed rΩ and the tip speed ratio, λ are

rΩ = 2πRt

=2πRn

RPM

60=2π 40m( ) 15RPM( )

60= 62.8m

s

λ = rΩu1

=62.8m

s8ms

= 7.85

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Given;

tanΦ =u1

rΩ= 2R3rΩλ

→Φtip= tan−1 2

3λ= 4.85°

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Given;

tanΦ =u1

rΩ= 2R3rΩλ

→Φtip= tan−1 2

3λ= 4.85°

Substituting for Φtip into width equation yields the width at the tip

W = 8πR sinΦ3λnC

L

=8π 40( ) sin4.85°( )3 7.85( ) 3( ) 1( ) =1.2m

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Given;

tanΦ =u1

rΩ= 2R3rΩλ

→Φtip= tan−1 2

3λ= 4.85°


W = 8πR sinΦ3λnC

L

=8π 40( ) sin4.85°( )3 7.85( ) 3( ) 1( ) =1.2m

Angle (º)

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Given;

tanΦ =u1

rΩ= 2R3rΩλ

→Φtip= tan−1 2

3λ= 4.85°


W = 8πR sinΦ3λnC

L

=8π 40( ) sin4.85°( )3 7.85( ) 3( ) 1( ) =1.2m

Angle (º)

Radius (m) Width (m) Twist (Φ - α) Wind Φ

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Given;

tanΦ =u1

rΩ= 2R3rΩλ

→Φtip= tan−1 2

3λ= 4.85°


W = 8πR sinΦ3λnC

L

=8π 40( ) sin4.85°( )3 7.85( ) 3( ) 1( ) =1.2m

Angle (º)

Radius (m) Width (m) Twist (Φ - α) Wind Φ40 1.2046 -0.5454 4.855

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Given;

tanΦ =u1

rΩ= 2R3rΩλ

→Φtip= tan−1 2

3λ= 4.85°


W = 8πR sinΦ3λnC

L

=8π 40( ) sin4.85°( )3 7.85( ) 3( ) 1( ) =1.2m

Angle (º)


20 2.3837 4.2405 9.640

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Given;

tanΦ =u1

rΩ= 2R3rΩλ

→Φtip= tan−1 2

3λ= 4.85°


W = 8πR sinΦ3λnC

L

=8π 40( ) sin4.85°( )3 7.85( ) 3( ) 1( ) =1.2m

Angle (º)


20 2.3837 4.2405 9.640

10 4.5787 13.3641 18.764

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CP - λ curve for a high tip speed ration wind turbine (Andrews &

Jelley, 2007)

DEPENDENCE OF CP ON λ

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Output power versus wind speed (Andrews & Jelley, 20007)

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TYPICAL WIND TURBINE

CONFIGURATION

Image source: http://www.popsci.com/content/next-gen-wind-turbine-examined

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POWER OUTPUT OF A WIND

TURBINE

The power in the wind, Pw at a given site

where:u(z) = wind speed at hub heightp(u) = wind frequency distribution

Pw= 12ρAu 3 = 1

2ρA u z( ){ }3 p u( )∫ du

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POWER OUTPUT OF A WIND

TURBINE

The power in the wind, Pw at a given site

where:u(z) = wind speed at hub heightp(u) = wind frequency distribution

Pw= 12ρAu 3 = 1

2ρA u z( ){ }3 p u( )∫ du

The average output power Po of a turbine

Po= η 12ρA C

Pλ( ) u z( ){ }3 p u( )∫ du

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Accurate wind data for a period of time is essential

Mountainous regions and coasts are ideal as well as exposed plains

Wind turbine spacing should be of the order 5D → 10D

Wind farms will experience array loss, i.e. an array of turbines will not produce as much power as if they potentially could

Low wind shear reduces the differential loading on turbine blades, i.e. fatigue loading

WIND FARM’s

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Natural scenery and preservation of wildlife particularly avian

Electromagnetic interference and noise

End of Service Life - recyclability

Embodied energy

Remote regions - access and grid connections

ENVIRONMENTAL IMPACT & PUBLIC

ACCEPTANCE

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Advantages Disadvantages

Prime fuel is free Risk of blade failure (total destruction of installation)

Infinitely renewable Suitable small generators not readily available

Non-polluting unsuitable for urban areas

In Ireland the seasonal variation matches electricity demands

Cost of storage battery or mains converter system

Big generators can be located on remote sites including offshore

Acoustic noise of gearbox and rotor blades

Saves conventional fuels Construction costs of the supporting tower and access roads

Saves the building of conventional generation

Electromagnetic interference due to blade rotation

Diversity in the methods of electricity generation

Environmental objections

Andrews, J., Jelley, N., (2007) Energy science: principles, technologies and impacts, Oxford University PressBoyle, G. (2004) Renewable Energy: Power for a sustainable future, second edition, Oxford University PressÇengel, Y., Turner, R., Cimbala, J. (2008) Fundamentals of thermal fluid sciences, Third edition, McGraw HillDa Rosa, A.V. (2009) Fundamentals of renewable energy processes, second edition, Academic PressDouglas, J., Gasiorek, J., Swaffield, J., Jack, L. (2005) Fluid mechanics, fifth edition, Pearson EducationLutyens, F., K. and Tarbuck, E. J. (2000) The Atmosphere: An introduction to meteorology, eighth edition, Prentice HallManwell, J.F., McGowan, J.G., Rodgers, A.L., (2008) Wind energy explained: theory, design and appication, John Wiley & Sons Ltd.Twidell, J. and Weir, T. (2006) Renewable energy resources, second edition, Oxon: Taylor and Francis

Wind Energy Lecture slides

Education

Transcript of Wind Energy Lecture slides