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KV
WIND ENERGY
Keith Vaugh BEng (AERO) MEng
KV
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
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KV(Lutyens & Tarbuck 2000)
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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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.
KV
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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It can be shown that the power, PD, a wind turbine delivers is proportional to the cube of the wind velocity
PD= 162712ρu3Aη
WIND MEASUREMENT
KV
It can be shown that the power, PD, a wind turbine delivers is proportional to the cube of the wind velocity
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
KV
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;
KV
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;
lnzz0
⎛
⎝⎜⎞
⎠⎟⎡
⎣⎢⎢
⎤
⎦⎥⎥
−1
z - the height of turbine
KV
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;
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
KV
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
KV(Boyle, G. 2004)
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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
KV
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;
Vol = A ×D
AREA
D
Air parcel
KV
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;
Vol = A ×D
Vol = πd2
4×D or Vol = πr 2 ×D
AREA
D
Air parcel
KV
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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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
Transposing this formula to get it in terms of m, yields;
m = ρ × Vol
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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
Transposing this formula to get it in terms of m, yields;
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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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
Transposing this formula to get it in terms of m, yields;
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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Substituting expressions into original formula for kinetic energy
KE = 12mu2
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Substituting expressions into original formula for kinetic energy
Substituting for m;
KE = 12mu2
KE = 12× ρ × Vol( )× u2
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Substituting expressions into original formula for kinetic energy
Substituting for m;
KE = 12mu2
KE = 12× ρ × Vol( )× u2
Vol can be replaced;
KE = 12× ρ × πd2
4
⎛
⎝⎜⎞
⎠⎟×D
⎛
⎝⎜⎞
⎠⎟× u2
KV
Substituting expressions into original formula for kinetic energy
Substituting for m;
KE = 12mu2
KE = 12× ρ × Vol( )× u2
Vol can be replaced;
KE = 12× ρ × πd2
4
⎛
⎝⎜⎞
⎠⎟×D
⎛
⎝⎜⎞
⎠⎟× u2
D can be replaced;
KE = 12× ρ × πd2
4
⎛
⎝⎜⎞
⎠⎟× v × T
⎛
⎝⎜⎞
⎠⎟× u2
KV
Substituting expressions into original formula for kinetic energy
Substituting for m;
KE = 12mu2
KE = 12× ρ × Vol( )× u2
Vol can be replaced;
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
KV
Let us now consider the Power that can be achieved
Substituting expressions into original formula for kinetic energy
Substituting for m;
KE = 12mu2
KE = 12× ρ × Vol( )× u2
Vol can be replaced;
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
KV
Dividing by T
Power = 12× ρ × πd2
4
⎛
⎝⎜⎞
⎠⎟⎛
⎝⎜⎞
⎠⎟× u3
KV
Dividing by T
if we divide the Power by the cross-sectional area (A) of the parcel, then we are left with the expression;
Power = 12× ρ × πd2
4
⎛
⎝⎜⎞
⎠⎟⎛
⎝⎜⎞
⎠⎟× u3
Power
πd2
4
⎛⎝⎜
⎞⎠⎟
= 12× ρ( )× u3
KV
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
Power = 12× ρ × πd2
4
⎛
⎝⎜⎞
⎠⎟⎛
⎝⎜⎞
⎠⎟× u3
Power
πd2
4
⎛⎝⎜
⎞⎠⎟
= 12× ρ( )× u3
KV
MOMENTUM BALANCE ACROSS
ROTOR BLADES
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Applied rotation
MOMENTUM BALANCE ACROSS
ROTOR BLADES
KV
u1
Applied rotation
MOMENTUM BALANCE ACROSS
ROTOR BLADES
KV
Undisturbed air flow at u1
u1
Applied rotation
MOMENTUM BALANCE ACROSS
ROTOR BLADES
KV
①
Undisturbed air flow at u1
u1
Applied rotation
MOMENTUM BALANCE ACROSS
ROTOR BLADES
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① ②
Undisturbed air flow at u1
u1
Applied rotation
MOMENTUM BALANCE ACROSS
ROTOR BLADES
KV
① ② ③
Undisturbed air flow at u1
u1
Applied rotation
MOMENTUM BALANCE ACROSS
ROTOR BLADES
KV
④① ② ③
Undisturbed air flow at u1
u1
Applied rotation
MOMENTUM BALANCE ACROSS
ROTOR BLADES
KV
④① ② ③
Undisturbed air flow at u1
u1
u4
Applied rotation
MOMENTUM BALANCE ACROSS
ROTOR BLADES
KV
④① ② ③
Slipstream outer edge velocity
differential u4 > u1
Undisturbed air flow at u1
u1
u4
Applied rotation
MOMENTUM BALANCE ACROSS
ROTOR BLADES
KV
④① ② ③
Slipstream area reduces in flow
direction
Slipstream outer edge velocity
differential u4 > u1
Undisturbed air flow at u1
u1
u4
Applied rotation
(a) Air flow over a rotating aircraft propeller showing flow acceleration and slipstream boundary
MOMENTUM BALANCE ACROSS
ROTOR BLADES
KV
④① ② ③
Slipstream area reduces in flow
direction
Slipstream outer edge velocity
differential u4 > u1
Undisturbed air flow at u1
u1
u1u4
Applied rotation
(a) Air flow over a rotating aircraft propeller showing flow acceleration and slipstream boundary
MOMENTUM BALANCE ACROSS
ROTOR BLADES
KV
④① ② ③
Slipstream area reduces in flow
direction
Slipstream outer edge velocity
differential u4 > u1
Undisturbed air flow at u1
u1
u1u4
Applied rotation
Air flow driven rotation
(a) Air flow over a rotating aircraft propeller showing flow acceleration and slipstream boundary
MOMENTUM BALANCE ACROSS
ROTOR BLADES
KV
④① ② ③
Slipstream area reduces in flow
direction
Slipstream outer edge velocity
differential u4 > u1
Undisturbed air flow at u1
u1
Undisturbed air flow at u1
u1u4
Applied rotation
Air flow driven rotation
(a) Air flow over a rotating aircraft propeller showing flow acceleration and slipstream boundary
MOMENTUM BALANCE ACROSS
ROTOR BLADES
KV
④① ② ③ ① ② ③
Slipstream area reduces in flow
direction
Slipstream outer edge velocity
differential u4 > u1
Undisturbed air flow at u1
u1
Undisturbed air flow at u1
u1u4
Applied rotation
Air flow driven rotation
(a) Air flow over a rotating aircraft propeller showing flow acceleration and slipstream boundary
MOMENTUM BALANCE ACROSS
ROTOR BLADES
KV
④① ② ③ ④① ② ③
Slipstream area reduces in flow
direction
Slipstream outer edge velocity
differential u4 > u1
Undisturbed air flow at u1
u1
Undisturbed air flow at u1
u1u4
Applied rotation
Air flow driven rotation
(a) Air flow over a rotating aircraft propeller showing flow acceleration and slipstream boundary
MOMENTUM BALANCE ACROSS
ROTOR BLADES
KV
④① ② ③ ④① ② ③
Slipstream area reduces in flow
direction
Slipstream outer edge velocity
differential u4 > u1
Undisturbed air flow at u1
u1
Undisturbed air flow at u1
u1u4
u4
Applied rotation
Air flow driven rotation
(a) Air flow over a rotating aircraft propeller showing flow acceleration and slipstream boundary
MOMENTUM BALANCE ACROSS
ROTOR BLADES
KV
④① ② ③ ④① ② ③
Slipstream area reduces in flow
direction
Slipstream outer edge velocity
differential u4 > u1
Undisturbed air flow at u1
u1
Undisturbed air flow at u1
u1
Slipstream outer edge velocity
differential u4 < u1
u4
u4
Applied rotation
Air flow driven rotation
(a) Air flow over a rotating aircraft propeller showing flow acceleration and slipstream boundary
MOMENTUM BALANCE ACROSS
ROTOR BLADES
KV
④① ② ③ ④① ② ③
Slipstream area reduces in flow
direction
Slipstream outer edge velocity
differential u4 > u1
Undisturbed air flow at u1
u1
Undisturbed air flow at u1
u1
Slipstream outer edge velocity
differential u4 < u1
Slipstream area increases in flow direction
u4
u4
Applied rotation
Air flow driven rotation
(a) Air flow over a rotating aircraft propeller showing flow acceleration and slipstream boundary
(b) Air flow over a wind turbine showing flow deceleration and slipstream boundary
MOMENTUM BALANCE ACROSS
ROTOR BLADES
KV
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;
KV
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;
p1+ 1
2ρu
12 = p
2+ 1
2ρu
22 and p
3+ 1
2ρu
32 = p
4+ 1
2ρu
42
KV
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;
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( )
KV
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;
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( )
KV
Substituting for p2 - p3 into the thrust formula
KV
Substituting for p2 - p3 into the thrust formula
T = A p2− p
3( )T = A 1
2ρ u
12 − u
42( )⎛
⎝⎜⎞⎠⎟
T = 12Aρ u
12 − u
42( )
KV
Substituting for p2 - p3 into the thrust formula
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( )
KV
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
Substituting for p2 - p3 into the thrust formula
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( )
KV
The rotor power is the the product of the thrust, T and the velocity at the rotor, u2
KV
The rotor power is the the product of the thrust, T and the velocity at the rotor, u2
Pr= 12Aρ u
12 − u
42( )u2
KV
The rotor power is the the product of the thrust, T and the velocity at the rotor, u2
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
KV
which reduces to
The rotor power is the the product of the thrust, T and the velocity at the rotor, u2
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
KV
For a frictionless system, the maximum power coefficient can be expressed as
KV
For a frictionless system, the maximum power coefficient can be expressed as
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
⎛
⎝⎜⎞
⎠⎟
KV
For a frictionless system, the maximum power coefficient can be expressed as
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
⎛
⎝⎜⎞
⎠⎟The maximum power coefficient is obtained for dCPdur
= 0
KV
For a frictionless system, the maximum power coefficient can be expressed as
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
⎛
⎝⎜⎞
⎠⎟The maximum power coefficient is obtained for dCPdur
= 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
KV
For a frictionless system, the maximum power coefficient can be expressed as
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
⎛
⎝⎜⎞
⎠⎟The maximum power coefficient is obtained for dCPdur
= 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
KV
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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riΩ Plane ofRotation
u1riΩ -Air velocity component
u1 - wind speed
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riΩ Plane ofRotation
uT, i u1riΩ -Air velocity component
u1 - wind speed
uT, i - The total velocity vector
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riΩ Plane ofRotation
uT, i u1
α
riΩ -Air velocity component
α - Angle of attack
u1 - wind speed
uT, i - The total velocity vector
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riΩ Plane ofRotation
uT, i u1
Φα
riΩ -Air velocity component
α - Angle of attack
u1 - wind speed
uT, i - The total velocity vector
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riΩ Plane ofRotation
uT, i u1
FL
Φα
FL - Lift force
riΩ -Air velocity component
α - Angle of attack
u1 - wind speed
uT, i - The total velocity vector
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riΩ Plane ofRotation
uT, i u1
FL
FD
Φα
FL - Lift force
FD - Drag force
riΩ -Air velocity component
α - Angle of attack
u1 - wind speed
uT, i - The total velocity vector
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riΩ Plane ofRotation
uT, i u1
FL
FR
FD
Φα
FL - Lift force
FD - Drag force
FR - Resultant lift force
riΩ -Air velocity component
α - Angle of attack
u1 - wind speed
uT, i - The total velocity vector
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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
Φ at rmax < Φ at r1
rmaxΩ > r1Ω
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Blade r
adius, rmax
Axis of rotation
Ω
Φ at rmax < Φ at r1
rmaxΩ > r1Ω u1
u1
u1
u1
u1
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⑤rmax = 1.0
Blade r
adius, rmax
Axis of rotation
Ω
Φ at rmax < Φ at r1
rmaxΩ > r1Ω u1
u1
u1
u1
u1
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⑤rmax = 1.0
Blade r
adius, rmax
Axis of rotation
Ω
Φ at rmax < Φ at r1
rmaxΩ > r1Ω u1
u1
u1
u1
u1
r maxΩ
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⑤rmax = 1.0
Blade r
adius, rmax
Axis of rotation
Ω
Φ at rmax < Φ at r1
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
Φ at rmax < Φ at r1
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
Φ at rmax < Φ at r1
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
Φ at rmax < Φ at r1
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
KV
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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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;
dA = 2πdr
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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;
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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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;
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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The total thrust developed from n blades on the annular area dA equals the rate of change of momentum;
dTn= ρu
122πrdr When the Betz condition is satisfied
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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 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
KV
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
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
Plane ofRotation
Φα
Plane ofRotation
Φ
(a) Turbine tip
(b) Turbine root
KV
The tip speed ratio,
W = 8πR sinΦ3λnC
L
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
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
KV
The time, t for one revolution of the blade tip for a turbine of length R
t = 2πRrΩ
The number of revolutions, nRPM per minute RPM
nRPM
= 60t
KV
The time, t for one revolution of the blade tip for a turbine of length R
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
KV
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
KV
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
Angle (º)
KV
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
Angle (º)
Radius (m) Width (m) Twist (Φ - α) Wind Φ
KV
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
Angle (º)
Radius (m) Width (m) Twist (Φ - α) Wind Φ40 1.2046 -0.5454 4.855
KV
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
Angle (º)
Radius (m) Width (m) Twist (Φ - α) Wind Φ40 1.2046 -0.5454 4.855
20 2.3837 4.2405 9.640
KV
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
Angle (º)
Radius (m) Width (m) Twist (Φ - α) Wind Φ40 1.2046 -0.5454 4.855
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
KV
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