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Keith Vaugh
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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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KV 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)
KV Idealized winds generated by pressure gradient and Coriolis Force. Actual wind patterns owing to land mass distribution.. (Lutyens & Tarbuck 2000)
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KV Geostrophic wind/Prevailing wind Storms TYPES OF WIND
KV Geostrophic wind/Prevailing wind Storms TYPES OF WIND Local winds/Sea breezes Mountain wind/Valley wind
KV Geostrophic wind/Prevailing wind Storms TYPES OF WIND Local winds/Sea breezes Mountain wind/Valley wind
KV Inter annual - Longer than 1 year variations Annual - Seasonal or monthly variation Diurnal - Daily variation Short term - Turbulence and gusts WIND VARIATION WITH TIME
KV The Griggs - Putnam Index of deformity Wind Atlas Anemometers (measurement) SITE ASSESSMENT
KV TYPES OF WIND
KV Anemometers
KV 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
KV It can be shown that the power, PD, a wind turbine delivers is proportional to the cube of the wind velocity PD = 16 27 1 2 ρu3 Aη 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 = 16 27 1 2 ρu3 Aη The mean power output over a period 0 toT is proportional to the cube of the mean cubic wind velocity, ū u ≡ 1 T u3 dt 0 T ∫ ⎛ ⎝⎜ ⎞ ⎠⎟ 1 3 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; ln z z0 ⎛ ⎝⎜ ⎞ ⎠⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ −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; ln z z0 ⎛ ⎝⎜ ⎞ ⎠⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ −1 Terrain z0 (m) Urban Area 3 - 0.4 Farmland 0.3 - 0.002 Open sea 0.001 - 0.0001 z - 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
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 p u( )= u σ 2 e − 1 2 u σ ⎛ ⎝⎜ ⎞ ⎠⎟ 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 = 2 π u2
KV Rayleigh frequency distribution for a mean wind speed of 8 m⋅s-1 Rayleigh distribution Probabilityofoccurance(%) Wind speed, u, m/s Mean speed 8 m⋅s-1
KV Height(m) Wind Speed (m⋅s-1)
KV Height(m) Wind Speed (m⋅s-1) A common describer of the dependence of u on height z is u z( )= us z zs ⎛ ⎝⎜ ⎞ ⎠⎟ αs where; zs is height at which u is measured to be us, typically 10 m αs is the wind shear coefficient, which is dependent on the terrain
KV Rotor diameter Slipstream Tower Nacelle Blade Hub WIND TURBINE CONFIGURATION
KV (Boyle, G. 2004)
KV 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
KV 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 = 1 2 mu2 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
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; 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 = πr2 × 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 = m Vol
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 = m Vol Transposing this formula to get it in terms of m, yields; m = ρ × Vol
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 = m Vol 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 = D T
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 = m Vol 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 = D T Transposing this formula D = u × T
KV Substituting expressions into original formula for kinetic energy
KV Substituting expressions into original formula for kinetic energy KE = 1 2 mu2
KV Substituting expressions into original formula for kinetic energy Substituting for m; KE = 1 2 mu2 KE = 1 2 × ρ × Vol( )× u2
KV Substituting expressions into original formula for kinetic energy Substituting for m; KE = 1 2 mu2 KE = 1 2 × ρ × Vol( )× u2 Vol can be replaced; KE = 1 2 × ρ × πd2 4 ⎛ ⎝⎜ ⎞ ⎠⎟ × D ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × u2
KV Substituting expressions into original formula for kinetic energy Substituting for m; KE = 1 2 mu2 KE = 1 2 × ρ × Vol( )× u2 Vol can be replaced; KE = 1 2 × ρ × πd2 4 ⎛ ⎝⎜ ⎞ ⎠⎟ × D ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × u2 D can be replaced; KE = 1 2 × ρ × πd2 4 ⎛ ⎝⎜ ⎞ ⎠⎟ × v × T ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × u2
KV Substituting expressions into original formula for kinetic energy Substituting for m; KE = 1 2 mu2 KE = 1 2 × ρ × Vol( )× u2 Vol can be replaced; KE = 1 2 × ρ × πd2 4 ⎛ ⎝⎜ ⎞ ⎠⎟ × D ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × u2 D can be replaced; KE = 1 2 × ρ × πd2 4 ⎛ ⎝⎜ ⎞ ⎠⎟ × v × T ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × u2 Rewriting KE = 1 2 × 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 = 1 2 mu2 KE = 1 2 × ρ × Vol( )× u2 Vol can be replaced; KE = 1 2 × ρ × πd2 4 ⎛ ⎝⎜ ⎞ ⎠⎟ × D ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × u2 D can be replaced; KE = 1 2 × ρ × πd2 4 ⎛ ⎝⎜ ⎞ ⎠⎟ × v × T ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × u2 Rewriting KE = 1 2 × ρ × πd2 4 ⎛ ⎝⎜ ⎞ ⎠⎟ × T ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × u3 Power = KE T Power = Energy Time Power = 1 2 × ρ × πd2 4 ⎛ ⎝⎜ ⎞ ⎠⎟ × T ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × u3 T
KV Dividing by T Power = 1 2 × 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 = 1 2 × ρ × πd2 4 ⎛ ⎝⎜ ⎞ ⎠⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × u3 Power πd2 4 ⎛ ⎝⎜ ⎞ ⎠⎟ = 1 2 × 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 = 1 2 × ρ × πd2 4 ⎛ ⎝⎜ ⎞ ⎠⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × u3 Power πd2 4 ⎛ ⎝⎜ ⎞ ⎠⎟ = 1 2 × ρ( )× u3
KV MOMENTUM BALANCE ACROSS ROTOR BLADES
KV 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
KV 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 2 ρu1 2 = p2 + 1 2 ρu2 2 and p3 + 1 2 ρu3 2 = p4 + 1 2 ρu4 2
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 2 ρu1 2 = p2 + 1 2 ρu2 2 and p3 + 1 2 ρu3 2 = p4 + 1 2 ρu4 2 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 − p3 = 1 2 ρ u1 2 − u4 2 ( )
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 2 ρu1 2 = p2 + 1 2 ρu2 2 and p3 + 1 2 ρu3 2 = p4 + 1 2 ρu4 2 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 − p3 = 1 2 ρ u1 2 − u4 2 ( ) The thrust can be expressed as the sum of the pressures on either side of the rotor disc T = A p2 − p3( )
KV Substituting for p2 - p3 into the thrust formula
KV Substituting for p2 - p3 into the thrust formula T = A p2 − p3( ) T = A 1 2 ρ u1 2 − u4 2 ( )⎛ ⎝⎜ ⎞ ⎠⎟ T = 1 2 Aρ u1 2 − u4 2 ( )
KV Substituting for p2 - p3 into the thrust formula T = A p2 − p3( ) T = A 1 2 ρ u1 2 − u4 2 ( )⎛ ⎝⎜ ⎞ ⎠⎟ T = 1 2 Aρ u1 2 − u4 2 ( ) The thrust can also be expressed as T = !m u1 − u4( )
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 = u3 = u1 + u4( ) 2 Substituting for p2 - p3 into the thrust formula T = A p2 − p3( ) T = A 1 2 ρ u1 2 − u4 2 ( )⎛ ⎝⎜ ⎞ ⎠⎟ T = 1 2 Aρ u1 2 − u4 2 ( ) The thrust can also be expressed as T = !m u1 − u4( )
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 = 1 2 Aρ u1 2 − u4 2 ( )u2
KV The rotor power is the the product of the thrust, T and the velocity at the rotor, u2 Pr = 1 2 Aρ u1 2 − u4 2 ( )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 = 1 2 Aρ u1 2 − u4 2 ( )u2 1 2 Aρu1 3
KV which reduces to The rotor power is the the product of the thrust, T and the velocity at the rotor, u2 Pr = 1 2 Aρ u1 2 − u4 2 ( )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 = 1 2 Aρ u1 2 − u4 2 ( )u2 1 2 Aρu1 3 CP = 1 2 u1 + u4( ) u1 2 − u4 2 ( ) u1 3
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 = 1 2 1+ ur − ur 2 − ur 3 ( ) CP = 1 2 u1 + u4( ) u1 2 − u4 2 ( ) u1 3 = 1 2 u1 u1 2 − u4 2 ( )+ u4 u1 2 − u4 2 ( ) u1 3 = 1 2 u1 3 − u1 u4 2 + u4 u1 2 − u4 3 u1 3 = 1 2 u1 3 u1 3 − u4 2 u1 2 + u4 u1 − u4 3 u1 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟
KV For a frictionless system, the maximum power coefficient can be expressed as where, ur = u4/u1CP = 1 2 1+ ur − ur 2 − ur 3 ( ) CP = 1 2 u1 + u4( ) u1 2 − u4 2 ( ) u1 3 = 1 2 u1 u1 2 − u4 2 ( )+ u4 u1 2 − u4 2 ( ) u1 3 = 1 2 u1 3 − u1 u4 2 + u4 u1 2 − u4 3 u1 3 = 1 2 u1 3 u1 3 − u4 2 u1 2 + u4 u1 − u4 3 u1 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ The maximum power coefficient is obtained for dCP dur = 0
KV For a frictionless system, the maximum power coefficient can be expressed as where, ur = u4/u1CP = 1 2 1+ ur − ur 2 − ur 3 ( ) CP = 1 2 u1 + u4( ) u1 2 − u4 2 ( ) u1 3 = 1 2 u1 u1 2 − u4 2 ( )+ u4 u1 2 − u4 2 ( ) u1 3 = 1 2 u1 3 − u1 u4 2 + u4 u1 2 − u4 3 u1 3 = 1 2 u1 3 u1 3 − u4 2 u1 2 + u4 u1 − u4 3 u1 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ The maximum power coefficient is obtained for dCP dur = 0 dCP dur = 1 2 1− 2ur − 3ur 2 ( )= 0 ur = 1 3 1 2 1− 2ur − 3ur 2 ( )= 0 0.5 −1ur −1.5ur 2 = 0 1.5ur 2 +1ur − 0.5 = 0 rearranged apply −b ± b2 − 4ac 2a to solve for ur
KV For a frictionless system, the maximum power coefficient can be expressed as where, ur = u4/u1CP = 1 2 1+ ur − ur 2 − ur 3 ( ) CP = 1 2 u1 + u4( ) u1 2 − u4 2 ( ) u1 3 = 1 2 u1 u1 2 − u4 2 ( )+ u4 u1 2 − u4 2 ( ) u1 3 = 1 2 u1 3 − u1 u4 2 + u4 u1 2 − u4 3 u1 3 = 1 2 u1 3 u1 3 − u4 2 u1 2 + u4 u1 − u4 3 u1 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ The maximum power coefficient is obtained for dCP dur = 0 dCP dur = 1 2 1− 2ur − 3ur 2 ( )= 0 ur = 1 3 1 2 1− 2ur − 3ur 2 ( )= 0 0.5 −1ur −1.5ur 2 = 0 1.5ur 2 +1ur − 0.5 = 0 rearranged apply −b ± b2 − 4ac 2a to solve for ur Therefore substituting back into the efficiency formula CP = 1 2 1+ 1 3 − 1 9 − 1 27 ⎛ ⎝⎜ ⎞ ⎠⎟ = 16 27 = 59.3% Betz or Lanchester-Betz limit
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 = CP 1 2 ρu3 Aη 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 = CP 1 2 ρu3 Aη lift drag = 1 2 ρCL u3 A 1 2 ρCD u3 A WIND TURBINE BLADE DESIGN
KV 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 ri rmaxΩ riΩ
KV
KV Plane of Rotation
KV riΩ Plane of Rotation riΩ -Air velocity component
KV riΩ Plane of Rotation u1 riΩ -Air velocity component u1 - wind speed
KV riΩ Plane of Rotation uT,i u1 riΩ -Air velocity component u1 - wind speed uT, i - The total velocity vector
KV riΩ Plane of Rotation uT,i u1 α riΩ -Air velocity component α - Angle of attack u1 - wind speed uT, i - The total velocity vector
KV riΩ Plane of Rotation uT,i u1 Φα riΩ -Air velocity component α - Angle of attack u1 - wind speed uT, i - The total velocity vector
KV riΩ Plane of Rotation 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
KV riΩ Plane of Rotation 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
KV riΩ Plane of Rotation 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
KV Plane of Rotation FL FR FD Φ
KV FD FLcosΦ FL sin Φ FDsinΦ FD cos Φ Resolved components: FL Axial Tangential The tangential and axial loads are established by resolving the vector diagrams for lift, drag and resultant lift Plane of Rotation FL FR FD Φ
KV FT = FL sinΦ − FD cosΦ FA = FL cosΦ − FD sinΦ FD FLcosΦ FL sin Φ FDsinΦ FD cos Φ Resolved components: FL Axial 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 of Rotation FL FR FD Φ
KV 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.
KV Ω Φ at rmax < Φ at r1 rmaxΩ > r1Ω
KV Blade radius, rmax Axis ofrotation Ω Φ at rmax < Φ at r1 rmaxΩ > r1Ω
KV Blade radius, rmax Axis ofrotation Ω Φ at rmax < Φ at r1 rmaxΩ > r1Ωu1 u1 u1 u1 u1
KV ⑤ rmax = 1.0 Blade radius, rmax Axis ofrotation Ω Φ at rmax < Φ at r1 rmaxΩ > r1Ωu1 u1 u1 u1 u1
KV ⑤ rmax = 1.0 Blade radius, rmax Axis ofrotation Ω Φ at rmax < Φ at r1 rmaxΩ > r1Ωu1 u1 u1 u1 u1 rmaxΩ
KV ⑤ rmax = 1.0 Blade radius, rmax Axis ofrotation Ω Φ at rmax < Φ at r1 rmaxΩ > r1Ωu1 u1 u1 u1 u1 rmaxΩ uT,max
KV ⑤ rmax = 1.0 Blade radius, rmax Axis ofrotation Ω Φ at rmax < Φ at r1 rmaxΩ > r1Ωu1 u1 u1 u1 u1 rmaxΩ Φ uT,max
KV ① r1 = 0.2 ⑤ rmax = 1.0 Blade radius, rmax Axis ofrotation Ω Φ at rmax < Φ at r1 rmaxΩ > r1Ωu1 u1 u1 u1 u1 r1Ω rmaxΩ Φ uT,max uT,1
KV ① r1 = 0.2 ② r2 = 0.4 ③ r3 = 0.6 ④ r4 = 0.8 ⑤ rmax = 1.0 Blade radius, rmax Axis ofrotation Ω Φ at rmax < Φ at r1 rmaxΩ > r1Ωu1 u1 u1 u1 u1 r1Ω r2Ω r3Ω r4Ω rmaxΩ Φ uT,max uT,4 uT,3 uT,2 uT,1
KV Plane of Rotation uT,tip u1 FL FR FD Φ α FD FR FL rtipΩ rrootΩ Plane of Rotation Φ u1 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
KV Tip Speed Ratio, λ, a dimensionless term, describes the relationship between the rotational speed of the blades and the wind driving the turbine.
KV Tip Speed Ratio, λ, a dimensionless term, describes the relationship between the rotational speed of the blades and the wind driving the turbine. λ = rΩ u1 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 = u1 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;
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; dA = 2πdr
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; 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
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; 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 = 2π nΩ Time for one blade to reach position of next blade n - number of blades Ω -Annular speed of the turbine
KV The total thrust developed from n blades on the annular area dA equals the rate of change of momentum; dTn = ρu1 2 2πrdr
KV The total thrust developed from n blades on the annular area dA equals the rate of change of momentum; dTn = ρu1 2 2πrdr When the Betz condition is satisfied
KV The total thrust developed from n blades on the annular area dA equals the rate of change of momentum; dTn = ρu1 2 2πrdr 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 = 1 2 CL ρuT, i 2 Wdr
KV Wind speed uT, i = u1 sinΦ Plane of Rotation uT,i u1 FL FR FD Φ α riΩ
KV Wind speed uT, i = u1 sinΦ Equating the thrust to the sum of the components of lift and substitute uT,i Plane of Rotation uT,i u1 FL FR FD Φ α riΩ dTn = ρu1 2 2πrdr = ndLcosΦ = 1 2 nCL u1 2 ρWdr cosΦ sin2 Φ
KV Wind speed uT, i = u1 sinΦ Equating the thrust to the sum of the components of lift and substitute uT,i Plane of Rotation uT,i u1 FL FR FD Φ α riΩ dTn = ρu1 2 2πrdr = ndLcosΦ = 1 2 nCL u1 2 ρWdr cosΦ sin2 Φ For the Betz condition to be satisfied the width at radius, r equals W = 4πr tanΦsinΦ nCL
KV The tip speed ratio,λ = rΩ u1 Plane of Rotation Φ α Plane of Rotation Φ (a) Turbine tip (b) Turbine root
KV The tip speed ratio,λ = rΩ u1 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 = u1 rΩ = 2R 3rΩλ Plane of Rotation Φ α Plane of Rotation Φ (a) Turbine tip (b) Turbine root
KV The tip speed ratio, W = 8πRsinΦ 3λnCL λ = rΩ u1 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 = u1 rΩ = 2R 3rΩλ Substituting for tanΦ As sinΦ increases and r decreases, the width of the blade increases from time to root. Plane of Rotation Φ α Plane of Rotation Φ (a) Turbine tip (b) Turbine root
KV (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
KV The time, t for one revolution of the blade tip for a turbine of length R t = 2πR rΩ
KV The time, t for one revolution of the blade tip for a turbine of length R t = 2πR rΩ The number of revolutions, nRPM per minute RPM nRPM = 60 t
KV The time, t for one revolution of the blade tip for a turbine of length R t = 2πR rΩ The number of revolutions, nRPM per minute RPM nRPM = 60 t The tip speed rΩ and the tip speed ratio, λ are rΩ = 2πR t = 2πRnRPM 60 = 2π 40m( )15RPM( ) 60 = 62.8m s λ = rΩ u1 = 62.8m s 8m s = 7.85
KV Given; tanΦ = u1 rΩ = 2R 3rΩλ → Φtip = tan−1 2 3λ = 4.85°
KV Given; tanΦ = u1 rΩ = 2R 3rΩλ → Φtip = tan−1 2 3λ = 4.85° Substituting for Φtip into width equation yields the width at the tip W = 8πRsinΦ 3λnCL = 8π 40( ) sin4.85°( ) 3 7.85( ) 3( )1( ) =1.2m
KV Given; tanΦ = u1 rΩ = 2R 3rΩλ → Φtip = tan−1 2 3λ = 4.85° Substituting for Φtip into width equation yields the width at the tip W = 8πRsinΦ 3λnCL = 8π 40( ) sin4.85°( ) 3 7.85( ) 3( )1( ) =1.2m Angle (º)
KV Given; tanΦ = u1 rΩ = 2R 3rΩλ → Φtip = tan−1 2 3λ = 4.85° Substituting for Φtip into width equation yields the width at the tip W = 8πRsinΦ 3λnCL = 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 = 2R 3rΩλ → Φtip = tan−1 2 3λ = 4.85° Substituting for Φtip into width equation yields the width at the tip W = 8πRsinΦ 3λnCL = 8π 40( ) sin4.85°( ) 3 7.85( ) 3( )1( ) =1.2m Angle (º) Radius (m) Width (m) Twist (Φ - α) Wind Φ 40 1.2046 -0.5454 4.855
KV Given; tanΦ = u1 rΩ = 2R 3rΩλ → Φtip = tan−1 2 3λ = 4.85° Substituting for Φtip into width equation yields the width at the tip W = 8πRsinΦ 3λnCL = 8π 40( ) sin4.85°( ) 3 7.85( ) 3( )1( ) =1.2m Angle (º) Radius (m) Width (m) Twist (Φ - α) Wind Φ 40 1.2046 -0.5454 4.855 20 2.3837 4.2405 9.640
KV Given; tanΦ = u1 rΩ = 2R 3rΩλ → Φtip = tan−1 2 3λ = 4.85° Substituting for Φtip into width equation yields the width at the tip W = 8πRsinΦ 3λnCL = 8π 40( ) sin4.85°( ) 3 7.85( ) 3( )1( ) =1.2m Angle (º) Radius (m) Width (m) Twist (Φ - α) Wind Φ 40 1.2046 -0.5454 4.855 20 2.3837 4.2405 9.640 10 4.5787 13.3641 18.764
KV CP - λ curve for a high tip speed ration wind turbine (Andrews & Jelley, 2007) DEPENDENCE OF CP ON λ
KV Output power versus wind speed (Andrews & Jelley, 20007)
KV TYPICAL WIND TURBINE CONFIGURATION Image source: http://www.popsci.com/content/next-gen-wind-turbine-examined
KV POWER OUTPUT OF A WIND TURBINE The power in the wind, Pw at a given site where: u(z) = wind speed at hub height p(u) = wind frequency distribution Pw = 1 2 ρAu3 = 1 2 ρA 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 height p(u) = wind frequency distribution Pw = 1 2 ρAu3 = 1 2 ρA u z( ){ } 3 p u( )∫ du The average output power Po of a turbine Po = η 1 2 ρA CP λ( ) u z( ){ } 3 p u( )∫ du
KV 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
KV 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
KV 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 Press Boyle, G. (2004) Renewable Energy: Power for a sustainable future, second edition, Oxford University Press Çengel,Y.,Turner, R., Cimbala, J. (2008) Fundamentals of thermal fluid sciences, Third edition, McGraw Hill Da Rosa,A.V. (2009) Fundamentals of renewable energy processes, second edition,Academic Press Douglas, J., Gasiorek, J., Swaffield, J., Jack, L. (2005) Fluid mechanics, fifth edition, Pearson Education Lutyens, F., K. and Tarbuck, E. J. (2000) The Atmosphere:An introduction to meteorology, eighth edition, Prentice Hall Manwell, 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

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Wind Energy Lecture slides

  • 1. KV WIND ENERGY Keith Vaugh BEng (AERO) MEng
  • 2. 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
  • 3. KV
  • 4. KV
  • 5. KV 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
  • 6. KV
  • 7. KV
  • 8. KV
  • 9. KV
  • 12. KV Idealized winds generated by pressure gradient and Coriolis Force. Actual wind patterns owing to land mass distribution.. (Lutyens & Tarbuck 2000)
  • 13. KV
  • 14. KV
  • 16. KV Geostrophic wind/Prevailing wind Storms TYPES OF WIND Local winds/Sea breezes Mountain wind/Valley wind
  • 17. KV Geostrophic wind/Prevailing wind Storms TYPES OF WIND Local winds/Sea breezes Mountain wind/Valley wind
  • 18. KV Inter annual - Longer than 1 year variations Annual - Seasonal or monthly variation Diurnal - Daily variation Short term - Turbulence and gusts WIND VARIATION WITH TIME
  • 19. KV The Griggs - Putnam Index of deformity Wind Atlas Anemometers (measurement) SITE ASSESSMENT
  • 22. KV 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.
  • 23. KV It can be shown that the power, PD, a wind turbine delivers is proportional to the cube of the wind velocity WIND MEASUREMENT
  • 24. KV It can be shown that the power, PD, a wind turbine delivers is proportional to the cube of the wind velocity PD = 16 27 1 2 ρu3 Aη WIND MEASUREMENT
  • 25. KV It can be shown that the power, PD, a wind turbine delivers is proportional to the cube of the wind velocity PD = 16 27 1 2 ρu3 Aη The mean power output over a period 0 toT is proportional to the cube of the mean cubic wind velocity, ū u ≡ 1 T u3 dt 0 T ∫ ⎛ ⎝⎜ ⎞ ⎠⎟ 1 3 WIND MEASUREMENT
  • 26. 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;
  • 27. 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; ln z z0 ⎛ ⎝⎜ ⎞ ⎠⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ −1 z - the height of turbine
  • 28. 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; ln z z0 ⎛ ⎝⎜ ⎞ ⎠⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ −1 Terrain z0 (m) Urban Area 3 - 0.4 Farmland 0.3 - 0.002 Open sea 0.001 - 0.0001 z - the height of turbine
  • 29. 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
  • 30. 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 p u( )= u σ 2 e − 1 2 u σ ⎛ ⎝⎜ ⎞ ⎠⎟ 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 = 2 π u2
  • 31. KV Rayleigh frequency distribution for a mean wind speed of 8 m⋅s-1 Rayleigh distribution Probabilityofoccurance(%) Wind speed, u, m/s Mean speed 8 m⋅s-1
  • 33. KV Height(m) Wind Speed (m⋅s-1) A common describer of the dependence of u on height z is u z( )= us z zs ⎛ ⎝⎜ ⎞ ⎠⎟ αs where; zs is height at which u is measured to be us, typically 10 m αs is the wind shear coefficient, which is dependent on the terrain
  • 36. KV 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
  • 37. KV 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 = 1 2 mu2 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
  • 38. 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; AREA D Air parcel
  • 39. 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
  • 40. 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 = πr2 × D AREA D Air parcel
  • 41. 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 = m Vol
  • 42. 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 = m Vol Transposing this formula to get it in terms of m, yields; m = ρ × Vol
  • 43. 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 = m Vol 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 = D T
  • 44. 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 = m Vol 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 = D T Transposing this formula D = u × T
  • 45. KV Substituting expressions into original formula for kinetic energy
  • 46. KV Substituting expressions into original formula for kinetic energy KE = 1 2 mu2
  • 47. KV Substituting expressions into original formula for kinetic energy Substituting for m; KE = 1 2 mu2 KE = 1 2 × ρ × Vol( )× u2
  • 48. KV Substituting expressions into original formula for kinetic energy Substituting for m; KE = 1 2 mu2 KE = 1 2 × ρ × Vol( )× u2 Vol can be replaced; KE = 1 2 × ρ × πd2 4 ⎛ ⎝⎜ ⎞ ⎠⎟ × D ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × u2
  • 49. KV Substituting expressions into original formula for kinetic energy Substituting for m; KE = 1 2 mu2 KE = 1 2 × ρ × Vol( )× u2 Vol can be replaced; KE = 1 2 × ρ × πd2 4 ⎛ ⎝⎜ ⎞ ⎠⎟ × D ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × u2 D can be replaced; KE = 1 2 × ρ × πd2 4 ⎛ ⎝⎜ ⎞ ⎠⎟ × v × T ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × u2
  • 50. KV Substituting expressions into original formula for kinetic energy Substituting for m; KE = 1 2 mu2 KE = 1 2 × ρ × Vol( )× u2 Vol can be replaced; KE = 1 2 × ρ × πd2 4 ⎛ ⎝⎜ ⎞ ⎠⎟ × D ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × u2 D can be replaced; KE = 1 2 × ρ × πd2 4 ⎛ ⎝⎜ ⎞ ⎠⎟ × v × T ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × u2 Rewriting KE = 1 2 × ρ × πd2 4 ⎛ ⎝⎜ ⎞ ⎠⎟ × T ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × u3
  • 51. KV Let us now consider the Power that can be achieved Substituting expressions into original formula for kinetic energy Substituting for m; KE = 1 2 mu2 KE = 1 2 × ρ × Vol( )× u2 Vol can be replaced; KE = 1 2 × ρ × πd2 4 ⎛ ⎝⎜ ⎞ ⎠⎟ × D ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × u2 D can be replaced; KE = 1 2 × ρ × πd2 4 ⎛ ⎝⎜ ⎞ ⎠⎟ × v × T ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × u2 Rewriting KE = 1 2 × ρ × πd2 4 ⎛ ⎝⎜ ⎞ ⎠⎟ × T ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × u3 Power = KE T Power = Energy Time Power = 1 2 × ρ × πd2 4 ⎛ ⎝⎜ ⎞ ⎠⎟ × T ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × u3 T
  • 52. KV Dividing by T Power = 1 2 × ρ × πd2 4 ⎛ ⎝⎜ ⎞ ⎠⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × u3
  • 53. 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 = 1 2 × ρ × πd2 4 ⎛ ⎝⎜ ⎞ ⎠⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × u3 Power πd2 4 ⎛ ⎝⎜ ⎞ ⎠⎟ = 1 2 × ρ( )× u3
  • 54. 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 = 1 2 × ρ × πd2 4 ⎛ ⎝⎜ ⎞ ⎠⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × u3 Power πd2 4 ⎛ ⎝⎜ ⎞ ⎠⎟ = 1 2 × ρ( )× u3
  • 58. KV Undisturbed air flow at u1 u1 Applied rotation MOMENTUM BALANCE ACROSS ROTOR BLADES
  • 59. KV ① Undisturbed air flow at u1 u1 Applied rotation MOMENTUM BALANCE ACROSS ROTOR BLADES
  • 60. KV ① ② Undisturbed air flow at u1 u1 Applied rotation MOMENTUM BALANCE ACROSS ROTOR BLADES
  • 61. KV ① ② ③ Undisturbed air flow at u1 u1 Applied rotation MOMENTUM BALANCE ACROSS ROTOR BLADES
  • 62. KV ④① ② ③ Undisturbed air flow at u1 u1 Applied rotation MOMENTUM BALANCE ACROSS ROTOR BLADES
  • 63. KV ④① ② ③ Undisturbed air flow at u1 u1 u4 Applied rotation MOMENTUM BALANCE ACROSS ROTOR BLADES
  • 64. 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
  • 65. 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
  • 66. 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
  • 67. 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
  • 68. 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
  • 69. 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
  • 70. 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
  • 71. 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
  • 72. 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
  • 73. 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
  • 74. 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;
  • 75. 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 2 ρu1 2 = p2 + 1 2 ρu2 2 and p3 + 1 2 ρu3 2 = p4 + 1 2 ρu4 2
  • 76. 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 2 ρu1 2 = p2 + 1 2 ρu2 2 and p3 + 1 2 ρu3 2 = p4 + 1 2 ρu4 2 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 − p3 = 1 2 ρ u1 2 − u4 2 ( )
  • 77. 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 2 ρu1 2 = p2 + 1 2 ρu2 2 and p3 + 1 2 ρu3 2 = p4 + 1 2 ρu4 2 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 − p3 = 1 2 ρ u1 2 − u4 2 ( ) The thrust can be expressed as the sum of the pressures on either side of the rotor disc T = A p2 − p3( )
  • 78. KV Substituting for p2 - p3 into the thrust formula
  • 79. KV Substituting for p2 - p3 into the thrust formula T = A p2 − p3( ) T = A 1 2 ρ u1 2 − u4 2 ( )⎛ ⎝⎜ ⎞ ⎠⎟ T = 1 2 Aρ u1 2 − u4 2 ( )
  • 80. KV Substituting for p2 - p3 into the thrust formula T = A p2 − p3( ) T = A 1 2 ρ u1 2 − u4 2 ( )⎛ ⎝⎜ ⎞ ⎠⎟ T = 1 2 Aρ u1 2 − u4 2 ( ) The thrust can also be expressed as T = !m u1 − u4( )
  • 81. 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 = u3 = u1 + u4( ) 2 Substituting for p2 - p3 into the thrust formula T = A p2 − p3( ) T = A 1 2 ρ u1 2 − u4 2 ( )⎛ ⎝⎜ ⎞ ⎠⎟ T = 1 2 Aρ u1 2 − u4 2 ( ) The thrust can also be expressed as T = !m u1 − u4( )
  • 82. KV The rotor power is the the product of the thrust, T and the velocity at the rotor, u2
  • 83. KV The rotor power is the the product of the thrust, T and the velocity at the rotor, u2 Pr = 1 2 Aρ u1 2 − u4 2 ( )u2
  • 84. KV The rotor power is the the product of the thrust, T and the velocity at the rotor, u2 Pr = 1 2 Aρ u1 2 − u4 2 ( )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 = 1 2 Aρ u1 2 − u4 2 ( )u2 1 2 Aρu1 3
  • 85. KV which reduces to The rotor power is the the product of the thrust, T and the velocity at the rotor, u2 Pr = 1 2 Aρ u1 2 − u4 2 ( )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 = 1 2 Aρ u1 2 − u4 2 ( )u2 1 2 Aρu1 3 CP = 1 2 u1 + u4( ) u1 2 − u4 2 ( ) u1 3
  • 86. KV For a frictionless system, the maximum power coefficient can be expressed as
  • 87. KV For a frictionless system, the maximum power coefficient can be expressed as where, ur = u4/u1CP = 1 2 1+ ur − ur 2 − ur 3 ( ) CP = 1 2 u1 + u4( ) u1 2 − u4 2 ( ) u1 3 = 1 2 u1 u1 2 − u4 2 ( )+ u4 u1 2 − u4 2 ( ) u1 3 = 1 2 u1 3 − u1 u4 2 + u4 u1 2 − u4 3 u1 3 = 1 2 u1 3 u1 3 − u4 2 u1 2 + u4 u1 − u4 3 u1 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟
  • 88. KV For a frictionless system, the maximum power coefficient can be expressed as where, ur = u4/u1CP = 1 2 1+ ur − ur 2 − ur 3 ( ) CP = 1 2 u1 + u4( ) u1 2 − u4 2 ( ) u1 3 = 1 2 u1 u1 2 − u4 2 ( )+ u4 u1 2 − u4 2 ( ) u1 3 = 1 2 u1 3 − u1 u4 2 + u4 u1 2 − u4 3 u1 3 = 1 2 u1 3 u1 3 − u4 2 u1 2 + u4 u1 − u4 3 u1 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ The maximum power coefficient is obtained for dCP dur = 0
  • 89. KV For a frictionless system, the maximum power coefficient can be expressed as where, ur = u4/u1CP = 1 2 1+ ur − ur 2 − ur 3 ( ) CP = 1 2 u1 + u4( ) u1 2 − u4 2 ( ) u1 3 = 1 2 u1 u1 2 − u4 2 ( )+ u4 u1 2 − u4 2 ( ) u1 3 = 1 2 u1 3 − u1 u4 2 + u4 u1 2 − u4 3 u1 3 = 1 2 u1 3 u1 3 − u4 2 u1 2 + u4 u1 − u4 3 u1 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ The maximum power coefficient is obtained for dCP dur = 0 dCP dur = 1 2 1− 2ur − 3ur 2 ( )= 0 ur = 1 3 1 2 1− 2ur − 3ur 2 ( )= 0 0.5 −1ur −1.5ur 2 = 0 1.5ur 2 +1ur − 0.5 = 0 rearranged apply −b ± b2 − 4ac 2a to solve for ur
  • 90. KV For a frictionless system, the maximum power coefficient can be expressed as where, ur = u4/u1CP = 1 2 1+ ur − ur 2 − ur 3 ( ) CP = 1 2 u1 + u4( ) u1 2 − u4 2 ( ) u1 3 = 1 2 u1 u1 2 − u4 2 ( )+ u4 u1 2 − u4 2 ( ) u1 3 = 1 2 u1 3 − u1 u4 2 + u4 u1 2 − u4 3 u1 3 = 1 2 u1 3 u1 3 − u4 2 u1 2 + u4 u1 − u4 3 u1 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ The maximum power coefficient is obtained for dCP dur = 0 dCP dur = 1 2 1− 2ur − 3ur 2 ( )= 0 ur = 1 3 1 2 1− 2ur − 3ur 2 ( )= 0 0.5 −1ur −1.5ur 2 = 0 1.5ur 2 +1ur − 0.5 = 0 rearranged apply −b ± b2 − 4ac 2a to solve for ur Therefore substituting back into the efficiency formula CP = 1 2 1+ 1 3 − 1 9 − 1 27 ⎛ ⎝⎜ ⎞ ⎠⎟ = 16 27 = 59.3% Betz or Lanchester-Betz limit
  • 91. 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 = CP 1 2 ρu3 Aη WIND TURBINE BLADE DESIGN
  • 92. 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 = CP 1 2 ρu3 Aη lift drag = 1 2 ρCL u3 A 1 2 ρCD u3 A WIND TURBINE BLADE DESIGN
  • 93. KV 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 ri rmaxΩ riΩ
  • 94. KV
  • 96. KV riΩ Plane of Rotation riΩ -Air velocity component
  • 97. KV riΩ Plane of Rotation u1 riΩ -Air velocity component u1 - wind speed
  • 98. KV riΩ Plane of Rotation uT,i u1 riΩ -Air velocity component u1 - wind speed uT, i - The total velocity vector
  • 99. KV riΩ Plane of Rotation uT,i u1 α riΩ -Air velocity component α - Angle of attack u1 - wind speed uT, i - The total velocity vector
  • 100. KV riΩ Plane of Rotation uT,i u1 Φα riΩ -Air velocity component α - Angle of attack u1 - wind speed uT, i - The total velocity vector
  • 101. KV riΩ Plane of Rotation 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
  • 102. KV riΩ Plane of Rotation 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
  • 103. KV riΩ Plane of Rotation 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
  • 105. KV FD FLcosΦ FL sin Φ FDsinΦ FD cos Φ Resolved components: FL Axial Tangential The tangential and axial loads are established by resolving the vector diagrams for lift, drag and resultant lift Plane of Rotation FL FR FD Φ
  • 106. KV FT = FL sinΦ − FD cosΦ FA = FL cosΦ − FD sinΦ FD FLcosΦ FL sin Φ FDsinΦ FD cos Φ Resolved components: FL Axial 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 of Rotation FL FR FD Φ
  • 107. KV 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.
  • 108. KV Ω Φ at rmax < Φ at r1 rmaxΩ > r1Ω
  • 109. KV Blade radius, rmax Axis ofrotation Ω Φ at rmax < Φ at r1 rmaxΩ > r1Ω
  • 110. KV Blade radius, rmax Axis ofrotation Ω Φ at rmax < Φ at r1 rmaxΩ > r1Ωu1 u1 u1 u1 u1
  • 111. KV ⑤ rmax = 1.0 Blade radius, rmax Axis ofrotation Ω Φ at rmax < Φ at r1 rmaxΩ > r1Ωu1 u1 u1 u1 u1
  • 112. KV ⑤ rmax = 1.0 Blade radius, rmax Axis ofrotation Ω Φ at rmax < Φ at r1 rmaxΩ > r1Ωu1 u1 u1 u1 u1 rmaxΩ
  • 113. KV ⑤ rmax = 1.0 Blade radius, rmax Axis ofrotation Ω Φ at rmax < Φ at r1 rmaxΩ > r1Ωu1 u1 u1 u1 u1 rmaxΩ uT,max
  • 114. KV ⑤ rmax = 1.0 Blade radius, rmax Axis ofrotation Ω Φ at rmax < Φ at r1 rmaxΩ > r1Ωu1 u1 u1 u1 u1 rmaxΩ Φ uT,max
  • 115. KV ① r1 = 0.2 ⑤ rmax = 1.0 Blade radius, rmax Axis ofrotation Ω Φ at rmax < Φ at r1 rmaxΩ > r1Ωu1 u1 u1 u1 u1 r1Ω rmaxΩ Φ uT,max uT,1
  • 116. KV ① r1 = 0.2 ② r2 = 0.4 ③ r3 = 0.6 ④ r4 = 0.8 ⑤ rmax = 1.0 Blade radius, rmax Axis ofrotation Ω Φ at rmax < Φ at r1 rmaxΩ > r1Ωu1 u1 u1 u1 u1 r1Ω r2Ω r3Ω r4Ω rmaxΩ Φ uT,max uT,4 uT,3 uT,2 uT,1
  • 117. KV Plane of Rotation uT,tip u1 FL FR FD Φ α FD FR FL rtipΩ rrootΩ Plane of Rotation Φ u1 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
  • 118. KV Tip Speed Ratio, λ, a dimensionless term, describes the relationship between the rotational speed of the blades and the wind driving the turbine.
  • 119. KV Tip Speed Ratio, λ, a dimensionless term, describes the relationship between the rotational speed of the blades and the wind driving the turbine. λ = rΩ u1 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 = u1 rΩ ≡ 2R 3rΩλ
  • 120. 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;
  • 121. 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; dA = 2πdr
  • 122. 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; 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
  • 123. 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; 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 = 2π nΩ Time for one blade to reach position of next blade n - number of blades Ω -Annular speed of the turbine
  • 124. KV The total thrust developed from n blades on the annular area dA equals the rate of change of momentum; dTn = ρu1 2 2πrdr
  • 125. KV The total thrust developed from n blades on the annular area dA equals the rate of change of momentum; dTn = ρu1 2 2πrdr When the Betz condition is satisfied
  • 126. KV The total thrust developed from n blades on the annular area dA equals the rate of change of momentum; dTn = ρu1 2 2πrdr 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 = 1 2 CL ρuT, i 2 Wdr
  • 127. KV Wind speed uT, i = u1 sinΦ Plane of Rotation uT,i u1 FL FR FD Φ α riΩ
  • 128. KV Wind speed uT, i = u1 sinΦ Equating the thrust to the sum of the components of lift and substitute uT,i Plane of Rotation uT,i u1 FL FR FD Φ α riΩ dTn = ρu1 2 2πrdr = ndLcosΦ = 1 2 nCL u1 2 ρWdr cosΦ sin2 Φ
  • 129. KV Wind speed uT, i = u1 sinΦ Equating the thrust to the sum of the components of lift and substitute uT,i Plane of Rotation uT,i u1 FL FR FD Φ α riΩ dTn = ρu1 2 2πrdr = ndLcosΦ = 1 2 nCL u1 2 ρWdr cosΦ sin2 Φ For the Betz condition to be satisfied the width at radius, r equals W = 4πr tanΦsinΦ nCL
  • 130. KV The tip speed ratio,λ = rΩ u1 Plane of Rotation Φ α Plane of Rotation Φ (a) Turbine tip (b) Turbine root
  • 131. KV The tip speed ratio,λ = rΩ u1 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 = u1 rΩ = 2R 3rΩλ Plane of Rotation Φ α Plane of Rotation Φ (a) Turbine tip (b) Turbine root
  • 132. KV The tip speed ratio, W = 8πRsinΦ 3λnCL λ = rΩ u1 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 = u1 rΩ = 2R 3rΩλ Substituting for tanΦ As sinΦ increases and r decreases, the width of the blade increases from time to root. Plane of Rotation Φ α Plane of Rotation Φ (a) Turbine tip (b) Turbine root
  • 133. KV (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
  • 134. KV The time, t for one revolution of the blade tip for a turbine of length R t = 2πR rΩ
  • 135. KV The time, t for one revolution of the blade tip for a turbine of length R t = 2πR rΩ The number of revolutions, nRPM per minute RPM nRPM = 60 t
  • 136. KV The time, t for one revolution of the blade tip for a turbine of length R t = 2πR rΩ The number of revolutions, nRPM per minute RPM nRPM = 60 t The tip speed rΩ and the tip speed ratio, λ are rΩ = 2πR t = 2πRnRPM 60 = 2π 40m( )15RPM( ) 60 = 62.8m s λ = rΩ u1 = 62.8m s 8m s = 7.85
  • 138. KV Given; tanΦ = u1 rΩ = 2R 3rΩλ → Φtip = tan−1 2 3λ = 4.85° Substituting for Φtip into width equation yields the width at the tip W = 8πRsinΦ 3λnCL = 8π 40( ) sin4.85°( ) 3 7.85( ) 3( )1( ) =1.2m
  • 139. KV Given; tanΦ = u1 rΩ = 2R 3rΩλ → Φtip = tan−1 2 3λ = 4.85° Substituting for Φtip into width equation yields the width at the tip W = 8πRsinΦ 3λnCL = 8π 40( ) sin4.85°( ) 3 7.85( ) 3( )1( ) =1.2m Angle (º)
  • 140. KV Given; tanΦ = u1 rΩ = 2R 3rΩλ → Φtip = tan−1 2 3λ = 4.85° Substituting for Φtip into width equation yields the width at the tip W = 8πRsinΦ 3λnCL = 8π 40( ) sin4.85°( ) 3 7.85( ) 3( )1( ) =1.2m Angle (º) Radius (m) Width (m) Twist (Φ - α) Wind Φ
  • 141. KV Given; tanΦ = u1 rΩ = 2R 3rΩλ → Φtip = tan−1 2 3λ = 4.85° Substituting for Φtip into width equation yields the width at the tip W = 8πRsinΦ 3λnCL = 8π 40( ) sin4.85°( ) 3 7.85( ) 3( )1( ) =1.2m Angle (º) Radius (m) Width (m) Twist (Φ - α) Wind Φ 40 1.2046 -0.5454 4.855
  • 142. KV Given; tanΦ = u1 rΩ = 2R 3rΩλ → Φtip = tan−1 2 3λ = 4.85° Substituting for Φtip into width equation yields the width at the tip W = 8πRsinΦ 3λnCL = 8π 40( ) sin4.85°( ) 3 7.85( ) 3( )1( ) =1.2m Angle (º) Radius (m) Width (m) Twist (Φ - α) Wind Φ 40 1.2046 -0.5454 4.855 20 2.3837 4.2405 9.640
  • 143. KV Given; tanΦ = u1 rΩ = 2R 3rΩλ → Φtip = tan−1 2 3λ = 4.85° Substituting for Φtip into width equation yields the width at the tip W = 8πRsinΦ 3λnCL = 8π 40( ) sin4.85°( ) 3 7.85( ) 3( )1( ) =1.2m Angle (º) Radius (m) Width (m) Twist (Φ - α) Wind Φ 40 1.2046 -0.5454 4.855 20 2.3837 4.2405 9.640 10 4.5787 13.3641 18.764
  • 144. KV CP - λ curve for a high tip speed ration wind turbine (Andrews & Jelley, 2007) DEPENDENCE OF CP ON λ
  • 145. KV Output power versus wind speed (Andrews & Jelley, 20007)
  • 146. KV TYPICAL WIND TURBINE CONFIGURATION Image source: http://www.popsci.com/content/next-gen-wind-turbine-examined
  • 147. KV POWER OUTPUT OF A WIND TURBINE The power in the wind, Pw at a given site where: u(z) = wind speed at hub height p(u) = wind frequency distribution Pw = 1 2 ρAu3 = 1 2 ρA u z( ){ } 3 p u( )∫ du
  • 148. KV POWER OUTPUT OF A WIND TURBINE The power in the wind, Pw at a given site where: u(z) = wind speed at hub height p(u) = wind frequency distribution Pw = 1 2 ρAu3 = 1 2 ρA u z( ){ } 3 p u( )∫ du The average output power Po of a turbine Po = η 1 2 ρA CP λ( ) u z( ){ } 3 p u( )∫ du
  • 149. KV 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
  • 150. KV 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
  • 151. KV 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
  • 152. Andrews, J., Jelley, N., (2007) Energy science: principles, technologies and impacts, Oxford University Press Boyle, G. (2004) Renewable Energy: Power for a sustainable future, second edition, Oxford University Press Çengel,Y.,Turner, R., Cimbala, J. (2008) Fundamentals of thermal fluid sciences, Third edition, McGraw Hill Da Rosa,A.V. (2009) Fundamentals of renewable energy processes, second edition,Academic Press Douglas, J., Gasiorek, J., Swaffield, J., Jack, L. (2005) Fluid mechanics, fifth edition, Pearson Education Lutyens, F., K. and Tarbuck, E. J. (2000) The Atmosphere:An introduction to meteorology, eighth edition, Prentice Hall Manwell, 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

Editor's Notes

  1. \n
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  3. Power extracted from the wind has been utilised for millennia (wind propelled boats along the Nile as early as 5000 B.C., China was using windmills to pump water in 200 B.C.). Prior to the industrial revolution and the introduction of the steam engine, windmills were a common device used for pumping water and for grinding grain. The first modern wind turbines appeared in the 1940&amp;#x2019;s and technologically evolved particularly in times of oil crises. This technology evolution continues with wind turbines become more cost effective and efficient. That being said, there still remains great scope to further advance these technological advancements. \n\nBefore delving into the technological aspects of wind turbines it is essential that comprehensive understanding of wind characteristics and the influence of land and or the build environment can have on site selection.\n \n
  4. The original source of wind energy is solar radiation. The radiation is absorbed by both land and sea, which in turn heats the surrounding air. The wind is generated as a result of different materials absorbing the solar radiation at different rates, causing temperature gradients to arise which in turn causes convection and pressure changes. \n\nAt any given time or location the wind is variable. Where either high or low wind speeds occur are determined by the rotational effect of the earth.\n\n
  5. Consider a simple, single cell atmospheric convection in a non-rotating Earth.&amp;#xA0; &quot;Single cell&quot; being either a single cell north or south of the equator.\nImagine the earth as a non-rotating sphere with uniform smooth surface characteristics. The sun heats the equatorial regions more than the polar regions. In response two large convection cells develop. \nIntermediate model: the earth rotates.&amp;#xA0; As expected, air traveling southward from the north pole will be deflected to the right. Air traveling northward from the south pole will be deflected left.\nHowever, considering the actual winds, after averaging them over a long period of time, we do not observe this type of motion.&amp;#xA0; In the 1920&amp;#x2019;s a new conceptual model was devised that had three cells instead of the single Hadley cell.&amp;#xA0; These three cells better represent the typical wind flow around the globe.\nImage source: Figure 7.5 in The Atmosphere, 8th edition, Lutgens and Tarbuck, 8th edition, 2001\n
  6. Global winds shape the Earth&apos;s climate, determining - which areas are tropical, desert, or temperate.\n\nThe Sun heats the Earth most intensely in the tropical zone around the equator. The heated air rises, cools, and then dumps its moisture as rain. That&apos;s why there are rain forests in the tropics.\n\nThe now drier air is forced by the continuously rising equatorial air to move towards the temperate latitudes on either side of the equator. At roughly 30&amp;#xB0; N and S, the &quot;horse latitudes&quot; - it can move no further due to the Earth&amp;#x2019;s rotation, and settles to the surface. \n\nAs the air sinks, it compresses and warms, creating hot, rain-free conditions. This circulation pattern, called a Hadley cell, is why the deserts of the world are located just poleward of the tropics, to the north and south.\n\nHorse Latitudes Around 30&amp;#xB0;N we see a region of subsiding (sinking) air.&amp;#xA0; Sinking air is typically dry and free of substantial precipitation. Many of the major desert regions of the northern hemisphere are found near 30&amp;#xB0; latitude.&amp;#xA0; E.g., Sahara, Middle East, SW United States.\nDoldrums Located near the equator, the doldrums are where the trade winds meet and where the pressure gradient decreases creating very little winds.&amp;#xA0; That&apos;s why sailors find it difficult to cross the equator and why weather systems in the one hemisphere rarely cross into the other hemisphere.&amp;#xA0; The doldrums are also called the intertropical convergence zone (ITCZ).\nImage source: Figure 7.5 in The Atmosphere, 8th edition, Lutgens and Tarbuck, 8th edition, 2001\n
  7. These give rise to the trade winds and westerlies. Trade winds occur between 0 and 30 degrees latitude, westerlies lie between 30 and 60 degrees - where Ireland lines. \n\nThe trade winds are so named as they carried the Spanish and Portuguese conquerors west to the Americas and they then returned using the westerlies to bring them back east with their heavily laden ships.\n\nCoriolis Force - Once air has been set in motion by the pressure gradient force, it undergoes an apparent deflection from its path, as seen by an observer on the earth. This apparent deflection is called the &quot;Coriolis force&quot; and is a result of the earth&apos;s rotation. \nImage source: Figure 7.8 in The Atmosphere, 8th edition, Lutgens and Tarbuck, 8th edition, 2001\n
  8. Owing to the tilt of the Earth&apos;s axis in orbit, the ITCZ will shift north and south.&amp;#xA0; It will shift to the south in January and north in July.\nThis shift in the wind directions owing to a northward or southward shift in the ITCZ results in the monsoons.&amp;#xA0; Monsoons are wind systems that exhibit a pronounced seasonal reversal in direction.&amp;#xA0; The best known monsoon is found in India and southeast Asia.\nWinter -- Flow is predominantly off the continent keeping the continent dry.\nSummer -- Flow is predominantly off the oceans keeping the continent wet.\nMonsoons happen not only in southeast Asia and India, but also in North America.&amp;#xA0; They are responsible for the increased rainfall in the southwest US during the summer months and the very dry conditions during the winter months.\n\nSource: Figure 7.9 in The Atmosphere, 8th edition, Lutgens and Tarbuck, 8th edition, 2001\n
  9. Owing to the tilt of the Earth&apos;s axis in orbit, the ITCZ will shift north and south.&amp;#xA0; It will shift to the south in January and north in July.\nThis shift in the wind directions owing to a northward or southward shift in the ITCZ results in the monsoons.&amp;#xA0; Monsoons are wind systems that exhibit a pronounced seasonal reversal in direction.&amp;#xA0; The best known monsoon is found in India and southeast Asia.\nWinter -- Flow is predominantly off the continent keeping the continent dry.\nSummer -- Flow is predominantly off the oceans keeping the continent wet.\nMonsoons happen not only in southeast Asia and India, but also in North America.&amp;#xA0; They are responsible for the increased rainfall in the southwest US during the summer months and the very dry conditions during the winter months.\n\nSource: Figure 7.9 in The Atmosphere, 8th edition, Lutgens and Tarbuck, 8th edition, 2001\n
  10. The geostrophic winds are largely driven by temperature differences, and thus pressure differences, and are not very much influenced by the surface of the earth. The geostrophic wind is found at altitudes above 1000 metres (3300 ft.) above ground level. The geostrophic wind speed may be measured using weather balloons. Land masses are heated by the sun more quickly than the sea in the daytime. The air rises, flows out to the sea, and creates a low pressure at ground level which attracts the cool air from the sea. This is called a sea breeze. At nightfall there is often a period of calm when land and sea temperatures are equal. At night the wind blows in the opposite direction. The land breeze at night generally has lower wind speeds, because the temperature difference between land and sea is smaller at night. \n\nOne example is the valley wind which originates on south-facing slopes (north-facing in the southern hemisphere). When the slopes and the neighbouring air are heated the density of the air decreases, and the air ascends towards the top following the surface of the slope. At night the wind direction is reversed, and turns into a downslope wind. If the valley floor is sloped, the air may move down or up the valley, as a canyon wind. If the valley is constricted this can further increase the wind speed.\n\nWinds flowing down the leeward sides of mountains can be quite powerful: Examples are the Foehn in the Alps in Europe, the Chinook in the Rocky Mountains, and the Zonda in the Andes. \nExamples of other local wind systems are the Mistral flowing down the Rhone valley into the Mediterranean Sea, the Scirocco, a southerly wind from Sahara blowing into the Mediterranean sea. \n
  11. Interannual &amp;#x2013;longer than 1 year variations - can have a large effect on the overall performance of a wind farm during its lifetime. Meteorologists reckon it takes 30 years of data to determine long term values and 5 years data is needed to arrive at a reliable wind speed for a site. However 1 years data is sufficient to predict long term seasonal mean wind speeds within 10% and 90% confidence. Up to 25% variation can occur in inter annual wind speeds\n\nAnnual Significant variation in seasonal or monthly averaged wind speeds are common through out the world &amp;#x201C;march &amp;#x2013; in like a lion and out like a lamb&amp;#x201D;... In Ireland the winter is much windier than the summer\n\nDiurnal &amp;#x2013; daily time scale - sea breezes and valley winds are an example of these . Generally the diurnal variation is much greater in the summer than in the winter &amp;#x2013; due to solar radiation.\n\nShort term variations include turbulence and gusts, any wind speeds that have a period between less than one second to 10 minutes and have a stochastic nature are considered to be turbulent. A gust is a discrete event within a turbulent air flow, and has measureable characteristics such as amplitude, rise time, max gust variation and lapse time\n
  12. The Griggs-Putnam Index of Deformity is an additional useful tool to help determine the potential of a wind site. The idea is to observe the area&amp;#x2019;s vegetation. A trees shape, especially conifers or evergreens, is often influenced by winds. Strong winds can permanently deform the trees. This deformity in trees is known as &amp;#x201C;flagging&amp;#x201D;. Flagging is usually more pronounced for single, isolated trees with some height. The Griggs-Putnam diagram, like the Wind Resource Maps, can offer a rough estimate of the wind in your area. The more information that you can obtain from the various sources, the greater degree of accuracy you will have in determining your wind speed and your potential power output.\n\nThe Griggs Putnam index should be used with a degree of caution, don&amp;#x2019;t just depend on one tree, make sure there are several used in the survey. Conifers give better indications that broadleaf trees. Absence of deformation doesn&amp;#x2019;t necessarily rule a site out of contention \n
  13. Wind atlas or wind resource maps are available for Europe and most Western regions. In the case of Ireland SEAI was instrumental in its development. Ireland&amp;#x2019;s wind atlas is available online through SEAI&amp;#x2019;s website. A variety of alternative options can be overlaid on the map including the national road network, topography, conservations or ares of special interest, local or electoral regions, as well as all current wind farms, and wind speeds both off shore and on shore at 50, 75 and 100 m.\n
  14. &amp;#x201C;Data from the wind monitoring site is essential for determining the viability of the project and, particularly, for assessing financial viability. Problems with the quality of wind data can lead to significant difficulties in obtaining financing. The importance of paying attention to this cannot be over-stated. It is hard to overemphasise how easy it is to acquire bad data. A significant effort is required to ensure good data.&amp;#x201D; - IWEA best practice guidelines 2008 state:\n\nThe best way of measuring wind speeds at a prospective wind turbine site is to fit an anemometer to the top of a mast which has the same height as the expected hub height of the wind turbine to be used. This way one avoids the uncertainty involved in recalculating the wind speeds to a different height. \n\nBy fitting the anemometer to the top of the mast one minimises the disturbances of airflows from the mast itself. If anemometers are placed on the side of the mast it is essential to place them in the prevailing wind direction in order to minimise the wind shade from the tower. \n\nPlanning and Development Regulations 2008 (S.I. No. 235 of 2008), state that for; The erection of a mast for mapping meteorological conditions.\n1. No such mast shall be erected for a period exceeding 15 months in any 24 month period.\n2. The total mast height shall not exceed 80 metres.\n3. The mast shall be a distance of not less than: (a) the total structure height plus: (i) 5 metres from any party boundary, (ii) 20 metres from any non-electrical overhead cables, (iii) 20 metres from any 38kV electricity distribution lines, (iv) 30 metres from the centreline of any electricity transmission line of 110kV or more. (b) 5 kilometres from the nearest airport oraerodrome, 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.\n4. Not more than one such mast shall be erected within the site.\n5. All mast components shall have a matt, nonreflective finish and the blade shall be made of material that does not deflect telecommunications signals.\n6. No sign, advertisement or object, not required for the functioning or safety of the mast shall be attached to or exhibited on the mast.\n
  15. Interannual &amp;#x2013;longer than 1 year variations - can have a large effect on the overall performance of a wind farm during its lifetime. Meteorologists reckon it takes 30 years of data to determine long term values and 5 years data is needed to arrive at a reliable wind speed for a site. However 1 years data is sufficient to predict long term seasonal mean wind speeds within 10% and 90% confidence. Up to 25% variation can occur in inter annual wind speeds\n\nAnnual Significant variation in seasonal or monthly averaged wind speeds are common through out the world &amp;#x201C;march &amp;#x2013; in like a lion and out like a lamb&amp;#x201D;... In Ireland the winter is much windier than the summer\n\nDiurnal &amp;#x2013; daily time scale - sea breezes and valley winds are an example of these . Generally the diurnal variation is much greater in the summer than in the winter &amp;#x2013; due to solar radiation.\n\nShort term variations include turbulence and gusts, any wind speeds that have a period between less than one second to 10 minutes and have a stochastic nature are considered to be turbulent. A gust is a discrete event within a turbulent air flow, and has measureable characteristics such as amplitude, rise time, max gust variation and lapse time\n
  16. where; \n&amp;#xBD;&amp;#x3C1;u3 is the power density in the wind, 16&amp;#x2044;27 &amp;#xBD;&amp;#x3C1;u3 is the available power density of the wind, A is the swept area and &amp;#x3B7; is the efficiency of the wind turbine\n\nThe normal anemometer averages &amp;#x16B; are not appropriate readings for the siting of wind turbines, e.g. consider a wind that is constantly blowing at m&amp;#x22C5;s-1, therefore it&amp;#x2019;s average wind speed is 10 m&amp;#x22C5;s-1. This wind will carry energy proportional to u3 = 1000. Now consider a wind of 50 m&amp;#x22C5;s-1 which is blowing for 20% of the time and remains calm for the intervals. This wind speed will also have a mean velocity of 10 m&amp;#x22C5;s-1 but the energy it carries is proportional to 0.2 &amp;#xD7; 503 = 25000, or 25 times more than the initial consideration. In the first case the average wind speed, &amp;#x16B; = 10 m&amp;#x22C5;s-1, while in the second case the average wind speed, &amp;#x16B; = 29.2 m&amp;#x22C5;s-1.\n\nThe quantity &amp;#x16B; can be measured directly by dedicated instrumentation but is more conveniently derived form anemometers equipped to process it and store for later use. \n\n\n
  17. where; \n&amp;#xBD;&amp;#x3C1;u3 is the power density in the wind, 16&amp;#x2044;27 &amp;#xBD;&amp;#x3C1;u3 is the available power density of the wind, A is the swept area and &amp;#x3B7; is the efficiency of the wind turbine\n\nThe normal anemometer averages &amp;#x16B; are not appropriate readings for the siting of wind turbines, e.g. consider a wind that is constantly blowing at m&amp;#x22C5;s-1, therefore it&amp;#x2019;s average wind speed is 10 m&amp;#x22C5;s-1. This wind will carry energy proportional to u3 = 1000. Now consider a wind of 50 m&amp;#x22C5;s-1 which is blowing for 20% of the time and remains calm for the intervals. This wind speed will also have a mean velocity of 10 m&amp;#x22C5;s-1 but the energy it carries is proportional to 0.2 &amp;#xD7; 503 = 25000, or 25 times more than the initial consideration. In the first case the average wind speed, &amp;#x16B; = 10 m&amp;#x22C5;s-1, while in the second case the average wind speed, &amp;#x16B; = 29.2 m&amp;#x22C5;s-1.\n\nThe quantity &amp;#x16B; can be measured directly by dedicated instrumentation but is more conveniently derived form anemometers equipped to process it and store for later use. \n\n\n
  18. where; \n&amp;#xBD;&amp;#x3C1;u3 is the power density in the wind, 16&amp;#x2044;27 &amp;#xBD;&amp;#x3C1;u3 is the available power density of the wind, A is the swept area and &amp;#x3B7; is the efficiency of the wind turbine\n\nThe normal anemometer averages &amp;#x16B; are not appropriate readings for the siting of wind turbines, e.g. consider a wind that is constantly blowing at m&amp;#x22C5;s-1, therefore it&amp;#x2019;s average wind speed is 10 m&amp;#x22C5;s-1. This wind will carry energy proportional to u3 = 1000. Now consider a wind of 50 m&amp;#x22C5;s-1 which is blowing for 20% of the time and remains calm for the intervals. This wind speed will also have a mean velocity of 10 m&amp;#x22C5;s-1 but the energy it carries is proportional to 0.2 &amp;#xD7; 503 = 25000, or 25 times more than the initial consideration. In the first case the average wind speed, &amp;#x16B; = 10 m&amp;#x22C5;s-1, while in the second case the average wind speed, &amp;#x16B; = 29.2 m&amp;#x22C5;s-1.\n\nThe quantity &amp;#x16B; can be measured directly by dedicated instrumentation but is more conveniently derived form anemometers equipped to process it and store for later use. \n\n\n
  19. For any given location the wind speed will vary considerably with time, thereby affecting the amount of power in the wind and the loading experienced by the wind turbine. Changes in the weather will affect the wind speed over periods of days, while gusts will affect it over periods of minutes. In general, averages over a 10 min interval are used to define steady wind speed. Shorter term fluctuations about this interval are quantified by the turbulence intensity IT, which is defined as the ratio of the standard deviation, &amp;#x3C3;T of the wind speed to the steady wind speed. As the steady wind speed increases, &amp;#x3C3;T increases and IT is found to be depend on the terrain and height. \n\nIf the exact behaviour of the wind could be predicted, then a wind turbine or indeed a wind farm could be optimally designed to match the local conditions. unfortunately, wind varies considerably throughout the day, the season and the year. Even if precise data is available for the site, there is no guarantee the each year will match with that data. In the event where precise longterm data is not available, it is possible to use statistical information to make and educated guesstimation of the wind behaviour. \n
  20. For any given location the wind speed will vary considerably with time, thereby affecting the amount of power in the wind and the loading experienced by the wind turbine. Changes in the weather will affect the wind speed over periods of days, while gusts will affect it over periods of minutes. In general, averages over a 10 min interval are used to define steady wind speed. Shorter term fluctuations about this interval are quantified by the turbulence intensity IT, which is defined as the ratio of the standard deviation, &amp;#x3C3;T of the wind speed to the steady wind speed. As the steady wind speed increases, &amp;#x3C3;T increases and IT is found to be depend on the terrain and height. \n\nIf the exact behaviour of the wind could be predicted, then a wind turbine or indeed a wind farm could be optimally designed to match the local conditions. unfortunately, wind varies considerably throughout the day, the season and the year. Even if precise data is available for the site, there is no guarantee the each year will match with that data. In the event where precise longterm data is not available, it is possible to use statistical information to make and educated guesstimation of the wind behaviour. \n
  21. For any given location the wind speed will vary considerably with time, thereby affecting the amount of power in the wind and the loading experienced by the wind turbine. Changes in the weather will affect the wind speed over periods of days, while gusts will affect it over periods of minutes. In general, averages over a 10 min interval are used to define steady wind speed. Shorter term fluctuations about this interval are quantified by the turbulence intensity IT, which is defined as the ratio of the standard deviation, &amp;#x3C3;T of the wind speed to the steady wind speed. As the steady wind speed increases, &amp;#x3C3;T increases and IT is found to be depend on the terrain and height. \n\nIf the exact behaviour of the wind could be predicted, then a wind turbine or indeed a wind farm could be optimally designed to match the local conditions. unfortunately, wind varies considerably throughout the day, the season and the year. Even if precise data is available for the site, there is no guarantee the each year will match with that data. In the event where precise longterm data is not available, it is possible to use statistical information to make and educated guesstimation of the wind behaviour. \n
  22. \n
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  26. The wind speed u will be zero at ground level, however increases significantly as altitude increases. Its rate of change decreases with height as the frictional forces decrease. \n\n&amp;#x3B1;s can show a large variation over a 24 hour period i.e. &lt; 0.15 during the day to &gt; 0.5 at night. This diurnal variation occurs because during night time the surface temperature drops as the ground losses heat by radiation giving a stable atmosphere with the lower part cooler than the upper part of the atmosphere. The air is not mixed and wind shear can be high. After sunrise the ground is heated by the sun and warms the air in contact, which then rises causing mixing and reduced wind shear (Andrews and Jelley, 2007)\n\nThe terrain is characterised by a surface roughness parameter z0. An approximate parametrisation for the dependence of the wind shear coefficient &amp;#x3B1;s and z0 for steady wind speeds lying between 6 and 10 m&amp;#x22C5;s-1 at a height of 10 m is &amp;#x3B1;s = &amp;#xBD;(z0/10)0.2 (Andrews and Jelley, 2007)\n
  27. The wind speed u will be zero at ground level, however increases significantly as altitude increases. Its rate of change decreases with height as the frictional forces decrease. \n\n&amp;#x3B1;s can show a large variation over a 24 hour period i.e. &lt; 0.15 during the day to &gt; 0.5 at night. This diurnal variation occurs because during night time the surface temperature drops as the ground losses heat by radiation giving a stable atmosphere with the lower part cooler than the upper part of the atmosphere. The air is not mixed and wind shear can be high. After sunrise the ground is heated by the sun and warms the air in contact, which then rises causing mixing and reduced wind shear (Andrews and Jelley, 2007)\n\nThe terrain is characterised by a surface roughness parameter z0. An approximate parametrisation for the dependence of the wind shear coefficient &amp;#x3B1;s and z0 for steady wind speeds lying between 6 and 10 m&amp;#x22C5;s-1 at a height of 10 m is &amp;#x3B1;s = &amp;#xBD;(z0/10)0.2 (Andrews and Jelley, 2007)\n
  28. The wind speed u will be zero at ground level, however increases significantly as altitude increases. Its rate of change decreases with height as the frictional forces decrease. \n\n&amp;#x3B1;s can show a large variation over a 24 hour period i.e. &lt; 0.15 during the day to &gt; 0.5 at night. This diurnal variation occurs because during night time the surface temperature drops as the ground losses heat by radiation giving a stable atmosphere with the lower part cooler than the upper part of the atmosphere. The air is not mixed and wind shear can be high. After sunrise the ground is heated by the sun and warms the air in contact, which then rises causing mixing and reduced wind shear (Andrews and Jelley, 2007)\n\nThe terrain is characterised by a surface roughness parameter z0. An approximate parametrisation for the dependence of the wind shear coefficient &amp;#x3B1;s and z0 for steady wind speeds lying between 6 and 10 m&amp;#x22C5;s-1 at a height of 10 m is &amp;#x3B1;s = &amp;#xBD;(z0/10)0.2 (Andrews and Jelley, 2007)\n
  29. Wind turbines are either a horizontal or vertical axis configuration (HAWT or VAWT respectively). HAWT are by far the most common given that these are generally more efficient that VAWT&amp;#x2019;s. HAWT&amp;#x2019;s are typically two or three blade design, however single blade and four or more blade designs are also available. The main components are the blade and hub assembly, the nacelle which houses the gearbox, pitch control and electrical generator, and the tower. The power generated by the generator is transferred to an on-site transformer located on the ground which ensures the supply is compatible with the network voltage. Weight and strength are of critical importance particularly for the blades. Fibreglass reinforced polyester is the preferred choice given these design considerations. \n\nVAWT&amp;#x2019;s do have some major advantages over HAWT&amp;#x2019;s. These allow the generator and all associated electronics, transformers and control systems to be located at ground level thereby allowing easy access. As these components are located at ground level, the overall costs are reduced significantly. However, as VAWT&amp;#x2019;s are less efficient that HAWT&amp;#x2019;s, the obvious initial investment most be considered over the life and the payback period of the installation. \n
  30. A wide variety of horizontal and vertical axis wind turbines have emerged.\n\nBefore delving into the technological aspects of wind turbines it is essential that a comprehensive understanding of the sources of the wind, its characteristics, measurement and the influence that land and or the build environment can have on site selection has been attained. \n \n
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  42. \n
  43. \n
  44. \n
  45. \n
  46. \n
  47. \n
  48. \n
  49. \n
  50. \n
  51. \n
  52. \n
  53. \n
  54. \n
  55. \n
  56. \n
  57. The momentum balance across the rotor blades can be established from the velocity differential found the the slip stream of a rotating propeller and a wind turbine. \n\nFigure (a) is representative of an aircraft. Torque is applied to the propeller causing it to generate thrust as it rotates. The airflow is accelerated through the propeller with the velocity increasing from u2 to u3. The downstream airflow is also accelerated to a velocity u4 and its cross-sectional area is reduced as illustrated. As the airflow crosses the propeller, its pressure increases, however it must return to its original value downstream of the propeller. The diameter of the propeller and its rotational speed would be known, therefore it the momentum balance can be achieved with ease. \n\nFigure (b) represents a wind turbine positioned in an airstream. The purpose of the turbine is to extract energy from the airstream and as consequence the the airflow decelerates and its cross sectional area increases as it passes through the turbine. \n\nThe application of the momentum equation in either instance is the same. \n
  58. The momentum balance across the rotor blades can be established from the velocity differential found the the slip stream of a rotating propeller and a wind turbine. \n\nFigure (a) is representative of an aircraft. Torque is applied to the propeller causing it to generate thrust as it rotates. The airflow is accelerated through the propeller with the velocity increasing from u2 to u3. The downstream airflow is also accelerated to a velocity u4 and its cross-sectional area is reduced as illustrated. As the airflow crosses the propeller, its pressure increases, however it must return to its original value downstream of the propeller. The diameter of the propeller and its rotational speed would be known, therefore it the momentum balance can be achieved with ease. \n\nFigure (b) represents a wind turbine positioned in an airstream. The purpose of the turbine is to extract energy from the airstream and as consequence the the airflow decelerates and its cross sectional area increases as it passes through the turbine. \n\nThe application of the momentum equation in either instance is the same. \n
  59. The momentum balance across the rotor blades can be established from the velocity differential found the the slip stream of a rotating propeller and a wind turbine. \n\nFigure (a) is representative of an aircraft. Torque is applied to the propeller causing it to generate thrust as it rotates. The airflow is accelerated through the propeller with the velocity increasing from u2 to u3. The downstream airflow is also accelerated to a velocity u4 and its cross-sectional area is reduced as illustrated. As the airflow crosses the propeller, its pressure increases, however it must return to its original value downstream of the propeller. The diameter of the propeller and its rotational speed would be known, therefore it the momentum balance can be achieved with ease. \n\nFigure (b) represents a wind turbine positioned in an airstream. The purpose of the turbine is to extract energy from the airstream and as consequence the the airflow decelerates and its cross sectional area increases as it passes through the turbine. \n\nThe application of the momentum equation in either instance is the same. \n
  60. The momentum balance across the rotor blades can be established from the velocity differential found the the slip stream of a rotating propeller and a wind turbine. \n\nFigure (a) is representative of an aircraft. Torque is applied to the propeller causing it to generate thrust as it rotates. The airflow is accelerated through the propeller with the velocity increasing from u2 to u3. The downstream airflow is also accelerated to a velocity u4 and its cross-sectional area is reduced as illustrated. As the airflow crosses the propeller, its pressure increases, however it must return to its original value downstream of the propeller. The diameter of the propeller and its rotational speed would be known, therefore it the momentum balance can be achieved with ease. \n\nFigure (b) represents a wind turbine positioned in an airstream. The purpose of the turbine is to extract energy from the airstream and as consequence the the airflow decelerates and its cross sectional area increases as it passes through the turbine. \n\nThe application of the momentum equation in either instance is the same. \n
  61. The momentum balance across the rotor blades can be established from the velocity differential found the the slip stream of a rotating propeller and a wind turbine. \n\nFigure (a) is representative of an aircraft. Torque is applied to the propeller causing it to generate thrust as it rotates. The airflow is accelerated through the propeller with the velocity increasing from u2 to u3. The downstream airflow is also accelerated to a velocity u4 and its cross-sectional area is reduced as illustrated. As the airflow crosses the propeller, its pressure increases, however it must return to its original value downstream of the propeller. The diameter of the propeller and its rotational speed would be known, therefore it the momentum balance can be achieved with ease. \n\nFigure (b) represents a wind turbine positioned in an airstream. The purpose of the turbine is to extract energy from the airstream and as consequence the the airflow decelerates and its cross sectional area increases as it passes through the turbine. \n\nThe application of the momentum equation in either instance is the same. \n
  62. The momentum balance across the rotor blades can be established from the velocity differential found the the slip stream of a rotating propeller and a wind turbine. \n\nFigure (a) is representative of an aircraft. Torque is applied to the propeller causing it to generate thrust as it rotates. The airflow is accelerated through the propeller with the velocity increasing from u2 to u3. The downstream airflow is also accelerated to a velocity u4 and its cross-sectional area is reduced as illustrated. As the airflow crosses the propeller, its pressure increases, however it must return to its original value downstream of the propeller. The diameter of the propeller and its rotational speed would be known, therefore it the momentum balance can be achieved with ease. \n\nFigure (b) represents a wind turbine positioned in an airstream. The purpose of the turbine is to extract energy from the airstream and as consequence the the airflow decelerates and its cross sectional area increases as it passes through the turbine. \n\nThe application of the momentum equation in either instance is the same. \n
  63. The momentum balance across the rotor blades can be established from the velocity differential found the the slip stream of a rotating propeller and a wind turbine. \n\nFigure (a) is representative of an aircraft. Torque is applied to the propeller causing it to generate thrust as it rotates. The airflow is accelerated through the propeller with the velocity increasing from u2 to u3. The downstream airflow is also accelerated to a velocity u4 and its cross-sectional area is reduced as illustrated. As the airflow crosses the propeller, its pressure increases, however it must return to its original value downstream of the propeller. The diameter of the propeller and its rotational speed would be known, therefore it the momentum balance can be achieved with ease. \n\nFigure (b) represents a wind turbine positioned in an airstream. The purpose of the turbine is to extract energy from the airstream and as consequence the the airflow decelerates and its cross sectional area increases as it passes through the turbine. \n\nThe application of the momentum equation in either instance is the same. \n
  64. The momentum balance across the rotor blades can be established from the velocity differential found the the slip stream of a rotating propeller and a wind turbine. \n\nFigure (a) is representative of an aircraft. Torque is applied to the propeller causing it to generate thrust as it rotates. The airflow is accelerated through the propeller with the velocity increasing from u2 to u3. The downstream airflow is also accelerated to a velocity u4 and its cross-sectional area is reduced as illustrated. As the airflow crosses the propeller, its pressure increases, however it must return to its original value downstream of the propeller. The diameter of the propeller and its rotational speed would be known, therefore it the momentum balance can be achieved with ease. \n\nFigure (b) represents a wind turbine positioned in an airstream. The purpose of the turbine is to extract energy from the airstream and as consequence the the airflow decelerates and its cross sectional area increases as it passes through the turbine. \n\nThe application of the momentum equation in either instance is the same. \n
  65. The momentum balance across the rotor blades can be established from the velocity differential found the the slip stream of a rotating propeller and a wind turbine. \n\nFigure (a) is representative of an aircraft. Torque is applied to the propeller causing it to generate thrust as it rotates. The airflow is accelerated through the propeller with the velocity increasing from u2 to u3. The downstream airflow is also accelerated to a velocity u4 and its cross-sectional area is reduced as illustrated. As the airflow crosses the propeller, its pressure increases, however it must return to its original value downstream of the propeller. The diameter of the propeller and its rotational speed would be known, therefore it the momentum balance can be achieved with ease. \n\nFigure (b) represents a wind turbine positioned in an airstream. The purpose of the turbine is to extract energy from the airstream and as consequence the the airflow decelerates and its cross sectional area increases as it passes through the turbine. \n\nThe application of the momentum equation in either instance is the same. \n
  66. The momentum balance across the rotor blades can be established from the velocity differential found the the slip stream of a rotating propeller and a wind turbine. \n\nFigure (a) is representative of an aircraft. Torque is applied to the propeller causing it to generate thrust as it rotates. The airflow is accelerated through the propeller with the velocity increasing from u2 to u3. The downstream airflow is also accelerated to a velocity u4 and its cross-sectional area is reduced as illustrated. As the airflow crosses the propeller, its pressure increases, however it must return to its original value downstream of the propeller. The diameter of the propeller and its rotational speed would be known, therefore it the momentum balance can be achieved with ease. \n\nFigure (b) represents a wind turbine positioned in an airstream. The purpose of the turbine is to extract energy from the airstream and as consequence the the airflow decelerates and its cross sectional area increases as it passes through the turbine. \n\nThe application of the momentum equation in either instance is the same. \n
  67. The momentum balance across the rotor blades can be established from the velocity differential found the the slip stream of a rotating propeller and a wind turbine. \n\nFigure (a) is representative of an aircraft. Torque is applied to the propeller causing it to generate thrust as it rotates. The airflow is accelerated through the propeller with the velocity increasing from u2 to u3. The downstream airflow is also accelerated to a velocity u4 and its cross-sectional area is reduced as illustrated. As the airflow crosses the propeller, its pressure increases, however it must return to its original value downstream of the propeller. The diameter of the propeller and its rotational speed would be known, therefore it the momentum balance can be achieved with ease. \n\nFigure (b) represents a wind turbine positioned in an airstream. The purpose of the turbine is to extract energy from the airstream and as consequence the the airflow decelerates and its cross sectional area increases as it passes through the turbine. \n\nThe application of the momentum equation in either instance is the same. \n
  68. The momentum balance across the rotor blades can be established from the velocity differential found the the slip stream of a rotating propeller and a wind turbine. \n\nFigure (a) is representative of an aircraft. Torque is applied to the propeller causing it to generate thrust as it rotates. The airflow is accelerated through the propeller with the velocity increasing from u2 to u3. The downstream airflow is also accelerated to a velocity u4 and its cross-sectional area is reduced as illustrated. As the airflow crosses the propeller, its pressure increases, however it must return to its original value downstream of the propeller. The diameter of the propeller and its rotational speed would be known, therefore it the momentum balance can be achieved with ease. \n\nFigure (b) represents a wind turbine positioned in an airstream. The purpose of the turbine is to extract energy from the airstream and as consequence the the airflow decelerates and its cross sectional area increases as it passes through the turbine. \n\nThe application of the momentum equation in either instance is the same. \n
  69. The momentum balance across the rotor blades can be established from the velocity differential found the the slip stream of a rotating propeller and a wind turbine. \n\nFigure (a) is representative of an aircraft. Torque is applied to the propeller causing it to generate thrust as it rotates. The airflow is accelerated through the propeller with the velocity increasing from u2 to u3. The downstream airflow is also accelerated to a velocity u4 and its cross-sectional area is reduced as illustrated. As the airflow crosses the propeller, its pressure increases, however it must return to its original value downstream of the propeller. The diameter of the propeller and its rotational speed would be known, therefore it the momentum balance can be achieved with ease. \n\nFigure (b) represents a wind turbine positioned in an airstream. The purpose of the turbine is to extract energy from the airstream and as consequence the the airflow decelerates and its cross sectional area increases as it passes through the turbine. \n\nThe application of the momentum equation in either instance is the same. \n
  70. The momentum balance across the rotor blades can be established from the velocity differential found the the slip stream of a rotating propeller and a wind turbine. \n\nFigure (a) is representative of an aircraft. Torque is applied to the propeller causing it to generate thrust as it rotates. The airflow is accelerated through the propeller with the velocity increasing from u2 to u3. The downstream airflow is also accelerated to a velocity u4 and its cross-sectional area is reduced as illustrated. As the airflow crosses the propeller, its pressure increases, however it must return to its original value downstream of the propeller. The diameter of the propeller and its rotational speed would be known, therefore it the momentum balance can be achieved with ease. \n\nFigure (b) represents a wind turbine positioned in an airstream. The purpose of the turbine is to extract energy from the airstream and as consequence the the airflow decelerates and its cross sectional area increases as it passes through the turbine. \n\nThe application of the momentum equation in either instance is the same. \n
  71. The momentum balance across the rotor blades can be established from the velocity differential found the the slip stream of a rotating propeller and a wind turbine. \n\nFigure (a) is representative of an aircraft. Torque is applied to the propeller causing it to generate thrust as it rotates. The airflow is accelerated through the propeller with the velocity increasing from u2 to u3. The downstream airflow is also accelerated to a velocity u4 and its cross-sectional area is reduced as illustrated. As the airflow crosses the propeller, its pressure increases, however it must return to its original value downstream of the propeller. The diameter of the propeller and its rotational speed would be known, therefore it the momentum balance can be achieved with ease. \n\nFigure (b) represents a wind turbine positioned in an airstream. The purpose of the turbine is to extract energy from the airstream and as consequence the the airflow decelerates and its cross sectional area increases as it passes through the turbine. \n\nThe application of the momentum equation in either instance is the same. \n
  72. The momentum balance across the rotor blades can be established from the velocity differential found the the slip stream of a rotating propeller and a wind turbine. \n\nFigure (a) is representative of an aircraft. Torque is applied to the propeller causing it to generate thrust as it rotates. The airflow is accelerated through the propeller with the velocity increasing from u2 to u3. The downstream airflow is also accelerated to a velocity u4 and its cross-sectional area is reduced as illustrated. As the airflow crosses the propeller, its pressure increases, however it must return to its original value downstream of the propeller. The diameter of the propeller and its rotational speed would be known, therefore it the momentum balance can be achieved with ease. \n\nFigure (b) represents a wind turbine positioned in an airstream. The purpose of the turbine is to extract energy from the airstream and as consequence the the airflow decelerates and its cross sectional area increases as it passes through the turbine. \n\nThe application of the momentum equation in either instance is the same. \n
  73. The momentum balance across the rotor blades can be established from the velocity differential found the the slip stream of a rotating propeller and a wind turbine. \n\nFigure (a) is representative of an aircraft. Torque is applied to the propeller causing it to generate thrust as it rotates. The airflow is accelerated through the propeller with the velocity increasing from u2 to u3. The downstream airflow is also accelerated to a velocity u4 and its cross-sectional area is reduced as illustrated. As the airflow crosses the propeller, its pressure increases, however it must return to its original value downstream of the propeller. The diameter of the propeller and its rotational speed would be known, therefore it the momentum balance can be achieved with ease. \n\nFigure (b) represents a wind turbine positioned in an airstream. The purpose of the turbine is to extract energy from the airstream and as consequence the the airflow decelerates and its cross sectional area increases as it passes through the turbine. \n\nThe application of the momentum equation in either instance is the same. \n
  74. The momentum balance across the rotor blades can be established from the velocity differential found the the slip stream of a rotating propeller and a wind turbine. \n\nFigure (a) is representative of an aircraft. Torque is applied to the propeller causing it to generate thrust as it rotates. The airflow is accelerated through the propeller with the velocity increasing from u2 to u3. The downstream airflow is also accelerated to a velocity u4 and its cross-sectional area is reduced as illustrated. As the airflow crosses the propeller, its pressure increases, however it must return to its original value downstream of the propeller. The diameter of the propeller and its rotational speed would be known, therefore it the momentum balance can be achieved with ease. \n\nFigure (b) represents a wind turbine positioned in an airstream. The purpose of the turbine is to extract energy from the airstream and as consequence the the airflow decelerates and its cross sectional area increases as it passes through the turbine. \n\nThe application of the momentum equation in either instance is the same. \n
  75. The momentum balance across the rotor blades can be established from the velocity differential found the the slip stream of a rotating propeller and a wind turbine. \n\nFigure (a) is representative of an aircraft. Torque is applied to the propeller causing it to generate thrust as it rotates. The airflow is accelerated through the propeller with the velocity increasing from u2 to u3. The downstream airflow is also accelerated to a velocity u4 and its cross-sectional area is reduced as illustrated. As the airflow crosses the propeller, its pressure increases, however it must return to its original value downstream of the propeller. The diameter of the propeller and its rotational speed would be known, therefore it the momentum balance can be achieved with ease. \n\nFigure (b) represents a wind turbine positioned in an airstream. The purpose of the turbine is to extract energy from the airstream and as consequence the the airflow decelerates and its cross sectional area increases as it passes through the turbine. \n\nThe application of the momentum equation in either instance is the same. \n
  76. The momentum balance across the rotor blades can be established from the velocity differential found the the slip stream of a rotating propeller and a wind turbine. \n\nFigure (a) is representative of an aircraft. Torque is applied to the propeller causing it to generate thrust as it rotates. The airflow is accelerated through the propeller with the velocity increasing from u2 to u3. The downstream airflow is also accelerated to a velocity u4 and its cross-sectional area is reduced as illustrated. As the airflow crosses the propeller, its pressure increases, however it must return to its original value downstream of the propeller. The diameter of the propeller and its rotational speed would be known, therefore it the momentum balance can be achieved with ease. \n\nFigure (b) represents a wind turbine positioned in an airstream. The purpose of the turbine is to extract energy from the airstream and as consequence the the airflow decelerates and its cross sectional area increases as it passes through the turbine. \n\nThe application of the momentum equation in either instance is the same. \n
  77. The momentum balance across the rotor blades can be established from the velocity differential found the the slip stream of a rotating propeller and a wind turbine. \n\nFigure (a) is representative of an aircraft. Torque is applied to the propeller causing it to generate thrust as it rotates. The airflow is accelerated through the propeller with the velocity increasing from u2 to u3. The downstream airflow is also accelerated to a velocity u4 and its cross-sectional area is reduced as illustrated. As the airflow crosses the propeller, its pressure increases, however it must return to its original value downstream of the propeller. The diameter of the propeller and its rotational speed would be known, therefore it the momentum balance can be achieved with ease. \n\nFigure (b) represents a wind turbine positioned in an airstream. The purpose of the turbine is to extract energy from the airstream and as consequence the the airflow decelerates and its cross sectional area increases as it passes through the turbine. \n\nThe application of the momentum equation in either instance is the same. \n
  78. The momentum balance across the rotor blades can be established from the velocity differential found the the slip stream of a rotating propeller and a wind turbine. \n\nFigure (a) is representative of an aircraft. Torque is applied to the propeller causing it to generate thrust as it rotates. The airflow is accelerated through the propeller with the velocity increasing from u2 to u3. The downstream airflow is also accelerated to a velocity u4 and its cross-sectional area is reduced as illustrated. As the airflow crosses the propeller, its pressure increases, however it must return to its original value downstream of the propeller. The diameter of the propeller and its rotational speed would be known, therefore it the momentum balance can be achieved with ease. \n\nFigure (b) represents a wind turbine positioned in an airstream. The purpose of the turbine is to extract energy from the airstream and as consequence the the airflow decelerates and its cross sectional area increases as it passes through the turbine. \n\nThe application of the momentum equation in either instance is the same. \n
  79. \n
  80. \n
  81. \n
  82. \n
  83. \n
  84. &amp;#x1E41; is the mass flow rate which is the product of the volumetric flow rate and density. In this case the volumetric flow rate is the throughflow, i.e. Aiui, where i = 1 to 4\n\n\n\n
  85. &amp;#x1E41; is the mass flow rate which is the product of the volumetric flow rate and density. In this case the volumetric flow rate is the throughflow, i.e. Aiui, where i = 1 to 4\n\n\n\n
  86. &amp;#x1E41; is the mass flow rate which is the product of the volumetric flow rate and density. In this case the volumetric flow rate is the throughflow, i.e. Aiui, where i = 1 to 4\n\n\n\n
  87. &amp;#x1E41; is the mass flow rate which is the product of the volumetric flow rate and density. In this case the volumetric flow rate is the throughflow, i.e. Aiui, where i = 1 to 4\n\n\n\n
  88. &amp;#x1E41; is the mass flow rate which is the product of the volumetric flow rate and density. In this case the volumetric flow rate is the throughflow, i.e. Aiui, where i = 1 to 4\n\n\n\n
  89. For the efficiency recall the derivation on slides 22 - 26, and that u2 = &amp;#xBD;(u1 + u4)\n\n
  90. For the efficiency recall the derivation on slides 22 - 26, and that u2 = &amp;#xBD;(u1 + u4)\n\n
  91. For the efficiency recall the derivation on slides 22 - 26, and that u2 = &amp;#xBD;(u1 + u4)\n\n
  92. For the efficiency recall the derivation on slides 22 - 26, and that u2 = &amp;#xBD;(u1 + u4)\n\n
  93. For the efficiency recall the derivation on slides 22 - 26, and that u2 = &amp;#xBD;(u1 + u4)\n\n
  94. This analysis illustrates that the maximum fraction of the power in the wind that can be theoretically extracted is approximately 60%.\nThis does not take into account the aerodynamic losses or the mechanical and electrical power losses. If a wind turbine is to be maximised, these additional losses must be minimised.\n
  95. This analysis illustrates that the maximum fraction of the power in the wind that can be theoretically extracted is approximately 60%.\nThis does not take into account the aerodynamic losses or the mechanical and electrical power losses. If a wind turbine is to be maximised, these additional losses must be minimised.\n
  96. This analysis illustrates that the maximum fraction of the power in the wind that can be theoretically extracted is approximately 60%.\nThis does not take into account the aerodynamic losses or the mechanical and electrical power losses. If a wind turbine is to be maximised, these additional losses must be minimised.\n
  97. This analysis illustrates that the maximum fraction of the power in the wind that can be theoretically extracted is approximately 60%.\nThis does not take into account the aerodynamic losses or the mechanical and electrical power losses. If a wind turbine is to be maximised, these additional losses must be minimised.\n
  98. This analysis illustrates that the maximum fraction of the power in the wind that can be theoretically extracted is approximately 60%.\nThis does not take into account the aerodynamic losses or the mechanical and electrical power losses. If a wind turbine is to be maximised, these additional losses must be minimised.\n
  99. This analysis illustrates that the maximum fraction of the power in the wind that can be theoretically extracted is approximately 60%.\nThis does not take into account the aerodynamic losses or the mechanical and electrical power losses. If a wind turbine is to be maximised, these additional losses must be minimised.\n
  100. This analysis illustrates that the maximum fraction of the power in the wind that can be theoretically extracted is approximately 60%.\nThis does not take into account the aerodynamic losses or the mechanical and electrical power losses. If a wind turbine is to be maximised, these additional losses must be minimised.\n
  101. This analysis illustrates that the maximum fraction of the power in the wind that can be theoretically extracted is approximately 60%.\nThis does not take into account the aerodynamic losses or the mechanical and electrical power losses. If a wind turbine is to be maximised, these additional losses must be minimised.\n
  102. This analysis illustrates that the maximum fraction of the power in the wind that can be theoretically extracted is approximately 60%.\nThis does not take into account the aerodynamic losses or the mechanical and electrical power losses. If a wind turbine is to be maximised, these additional losses must be minimised.\n
  103. This analysis illustrates that the maximum fraction of the power in the wind that can be theoretically extracted is approximately 60%.\nThis does not take into account the aerodynamic losses or the mechanical and electrical power losses. If a wind turbine is to be maximised, these additional losses must be minimised.\n
  104. It is imperative that power coefficient be maximised during all operating conditions (except those above the rated wind speed) to ensure that the maximum power is extracted from the wind. The power coefficient can be maximised by maximising the the turbine blades lift:drag ratio, &amp;#xBD;&amp;#x3C1;CLu2A: &amp;#xBD;&amp;#x3C1;CDu2A\n
  105. It is imperative that power coefficient be maximised during all operating conditions (except those above the rated wind speed) to ensure that the maximum power is extracted from the wind. The power coefficient can be maximised by maximising the the turbine blades lift:drag ratio, &amp;#xBD;&amp;#x3C1;CLu2A: &amp;#xBD;&amp;#x3C1;CDu2A\n
  106. It is imperative that power coefficient be maximised during all operating conditions (except those above the rated wind speed) to ensure that the maximum power is extracted from the wind. The power coefficient can be maximised by maximising the the turbine blades lift:drag ratio, &amp;#xBD;&amp;#x3C1;CLu2A: &amp;#xBD;&amp;#x3C1;CDu2A\n
  107. \n
  108. A wind impacting perpendicularly on a turbine (vector u1) causes a rotational movement of the turbine blades in the plane of rotation. This rotational movement induces an air velocity vector, r&amp;#x3A9;. When these vectors are vectorially added, a relative vector, known as the total velocity vector, uT emerges.\n \n The sectional view of the airfoil illustrates, the velocity vectors, the angle &amp;#x3A6; which is the angle between the total velocity vector and the plane of rotation, the lift, drag and the resultant forces. The drag force acts in the same plane as the total velocity vector, uT, and the lift force is always perpendicular to this. \n\n
  109. A wind impacting perpendicularly on a turbine (vector u1) causes a rotational movement of the turbine blades in the plane of rotation. This rotational movement induces an air velocity vector, r&amp;#x3A9;. When these vectors are vectorially added, a relative vector, known as the total velocity vector, uT emerges.\n \n The sectional view of the airfoil illustrates, the velocity vectors, the angle &amp;#x3A6; which is the angle between the total velocity vector and the plane of rotation, the lift, drag and the resultant forces. The drag force acts in the same plane as the total velocity vector, uT, and the lift force is always perpendicular to this. \n\n
  110. A wind impacting perpendicularly on a turbine (vector u1) causes a rotational movement of the turbine blades in the plane of rotation. This rotational movement induces an air velocity vector, r&amp;#x3A9;. When these vectors are vectorially added, a relative vector, known as the total velocity vector, uT emerges.\n \n The sectional view of the airfoil illustrates, the velocity vectors, the angle &amp;#x3A6; which is the angle between the total velocity vector and the plane of rotation, the lift, drag and the resultant forces. The drag force acts in the same plane as the total velocity vector, uT, and the lift force is always perpendicular to this. \n\n
  111. A wind impacting perpendicularly on a turbine (vector u1) causes a rotational movement of the turbine blades in the plane of rotation. This rotational movement induces an air velocity vector, r&amp;#x3A9;. When these vectors are vectorially added, a relative vector, known as the total velocity vector, uT emerges.\n \n The sectional view of the airfoil illustrates, the velocity vectors, the angle &amp;#x3A6; which is the angle between the total velocity vector and the plane of rotation, the lift, drag and the resultant forces. The drag force acts in the same plane as the total velocity vector, uT, and the lift force is always perpendicular to this. \n\n
  112. A wind impacting perpendicularly on a turbine (vector u1) causes a rotational movement of the turbine blades in the plane of rotation. This rotational movement induces an air velocity vector, r&amp;#x3A9;. When these vectors are vectorially added, a relative vector, known as the total velocity vector, uT emerges.\n \n The sectional view of the airfoil illustrates, the velocity vectors, the angle &amp;#x3A6; which is the angle between the total velocity vector and the plane of rotation, the lift, drag and the resultant forces. The drag force acts in the same plane as the total velocity vector, uT, and the lift force is always perpendicular to this. \n\n
  113. A wind impacting perpendicularly on a turbine (vector u1) causes a rotational movement of the turbine blades in the plane of rotation. This rotational movement induces an air velocity vector, r&amp;#x3A9;. When these vectors are vectorially added, a relative vector, known as the total velocity vector, uT emerges.\n \n The sectional view of the airfoil illustrates, the velocity vectors, the angle &amp;#x3A6; which is the angle between the total velocity vector and the plane of rotation, the lift, drag and the resultant forces. The drag force acts in the same plane as the total velocity vector, uT, and the lift force is always perpendicular to this. \n\n
  114. A wind impacting perpendicularly on a turbine (vector u1) causes a rotational movement of the turbine blades in the plane of rotation. This rotational movement induces an air velocity vector, r&amp;#x3A9;. When these vectors are vectorially added, a relative vector, known as the total velocity vector, uT emerges.\n \n The sectional view of the airfoil illustrates, the velocity vectors, the angle &amp;#x3A6; which is the angle between the total velocity vector and the plane of rotation, the lift, drag and the resultant forces. The drag force acts in the same plane as the total velocity vector, uT, and the lift force is always perpendicular to this. \n\n
  115. A wind impacting perpendicularly on a turbine (vector u1) causes a rotational movement of the turbine blades in the plane of rotation. This rotational movement induces an air velocity vector, r&amp;#x3A9;. When these vectors are vectorially added, a relative vector, known as the total velocity vector, uT emerges.\n \n The sectional view of the airfoil illustrates, the velocity vectors, the angle &amp;#x3A6; which is the angle between the total velocity vector and the plane of rotation, the lift, drag and the resultant forces. The drag force acts in the same plane as the total velocity vector, uT, and the lift force is always perpendicular to this. \n\n
  116. A wind impacting perpendicularly on a turbine (vector u1) causes a rotational movement of the turbine blades in the plane of rotation. This rotational movement induces an air velocity vector, r&amp;#x3A9;. When these vectors are vectorially added, a relative vector, known as the total velocity vector, uT emerges.\n \n The sectional view of the airfoil illustrates, the velocity vectors, the angle &amp;#x3A6; which is the angle between the total velocity vector and the plane of rotation, the lift, drag and the resultant forces. The drag force acts in the same plane as the total velocity vector, uT, and the lift force is always perpendicular to this. \n\n
  117. A wind impacting perpendicularly on a turbine (vector u1) causes a rotational movement of the turbine blades in the plane of rotation. This rotational movement induces an air velocity vector, r&amp;#x3A9;. When these vectors are vectorially added, a relative vector, known as the total velocity vector, uT emerges.\n \n The sectional view of the airfoil illustrates, the velocity vectors, the angle &amp;#x3A6; which is the angle between the total velocity vector and the plane of rotation, the lift, drag and the resultant forces. The drag force acts in the same plane as the total velocity vector, uT, and the lift force is always perpendicular to this. \n\n
  118. A wind impacting perpendicularly on a turbine (vector u1) causes a rotational movement of the turbine blades in the plane of rotation. This rotational movement induces an air velocity vector, r&amp;#x3A9;. When these vectors are vectorially added, a relative vector, known as the total velocity vector, uT emerges.\n \n The sectional view of the airfoil illustrates, the velocity vectors, the angle &amp;#x3A6; which is the angle between the total velocity vector and the plane of rotation, the lift, drag and the resultant forces. The drag force acts in the same plane as the total velocity vector, uT, and the lift force is always perpendicular to this. \n\n
  119. A wind impacting perpendicularly on a turbine (vector u1) causes a rotational movement of the turbine blades in the plane of rotation. This rotational movement induces an air velocity vector, r&amp;#x3A9;. When these vectors are vectorially added, a relative vector, known as the total velocity vector, uT emerges.\n \n The sectional view of the airfoil illustrates, the velocity vectors, the angle &amp;#x3A6; which is the angle between the total velocity vector and the plane of rotation, the lift, drag and the resultant forces. The drag force acts in the same plane as the total velocity vector, uT, and the lift force is always perpendicular to this. \n\n
  120. A wind impacting perpendicularly on a turbine (vector u1) causes a rotational movement of the turbine blades in the plane of rotation. This rotational movement induces an air velocity vector, r&amp;#x3A9;. When these vectors are vectorially added, a relative vector, known as the total velocity vector, uT emerges.\n \n The sectional view of the airfoil illustrates, the velocity vectors, the angle &amp;#x3A6; which is the angle between the total velocity vector and the plane of rotation, the lift, drag and the resultant forces. The drag force acts in the same plane as the total velocity vector, uT, and the lift force is always perpendicular to this. \n\n
  121. A wind impacting perpendicularly on a turbine (vector u1) causes a rotational movement of the turbine blades in the plane of rotation. This rotational movement induces an air velocity vector, r&amp;#x3A9;. When these vectors are vectorially added, a relative vector, known as the total velocity vector, uT emerges.\n \n The sectional view of the airfoil illustrates, the velocity vectors, the angle &amp;#x3A6; which is the angle between the total velocity vector and the plane of rotation, the lift, drag and the resultant forces. The drag force acts in the same plane as the total velocity vector, uT, and the lift force is always perpendicular to this. \n\n
  122. A wind impacting perpendicularly on a turbine (vector u1) causes a rotational movement of the turbine blades in the plane of rotation. This rotational movement induces an air velocity vector, r&amp;#x3A9;. When these vectors are vectorially added, a relative vector, known as the total velocity vector, uT emerges.\n \n The sectional view of the airfoil illustrates, the velocity vectors, the angle &amp;#x3A6; which is the angle between the total velocity vector and the plane of rotation, the lift, drag and the resultant forces. The drag force acts in the same plane as the total velocity vector, uT, and the lift force is always perpendicular to this. \n\n
  123. A wind impacting perpendicularly on a turbine (vector u1) causes a rotational movement of the turbine blades in the plane of rotation. This rotational movement induces an air velocity vector, r&amp;#x3A9;. When these vectors are vectorially added, a relative vector, known as the total velocity vector, uT emerges.\n \n The sectional view of the airfoil illustrates, the velocity vectors, the angle &amp;#x3A6; which is the angle between the total velocity vector and the plane of rotation, the lift, drag and the resultant forces. The drag force acts in the same plane as the total velocity vector, uT, and the lift force is always perpendicular to this. \n\n
  124. The application of simple trigonometric rules to the resulting vector diagram, allows for the establishment of the Tangential, FT and Axial, FA forces.\n \nOptimised wind turbine blade design will result in the maximisation of the torque about the axis of rotation, thereby ensuring that the wind turbine will operate at an optimum level. The thrust and axial forces emphasise the need to maximise the lift:drag ratio.\n
  125. The application of simple trigonometric rules to the resulting vector diagram, allows for the establishment of the Tangential, FT and Axial, FA forces.\n \nOptimised wind turbine blade design will result in the maximisation of the torque about the axis of rotation, thereby ensuring that the wind turbine will operate at an optimum level. The thrust and axial forces emphasise the need to maximise the lift:drag ratio.\n
  126. The application of simple trigonometric rules to the resulting vector diagram, allows for the establishment of the Tangential, FT and Axial, FA forces.\n \nOptimised wind turbine blade design will result in the maximisation of the torque about the axis of rotation, thereby ensuring that the wind turbine will operate at an optimum level. The thrust and axial forces emphasise the need to maximise the lift:drag ratio.\n
  127. The application of simple trigonometric rules to the resulting vector diagram, allows for the establishment of the Tangential, FT and Axial, FA forces.\n \nOptimised wind turbine blade design will result in the maximisation of the torque about the axis of rotation, thereby ensuring that the wind turbine will operate at an optimum level. The thrust and axial forces emphasise the need to maximise the lift:drag ratio.\n
  128. The air velocity component, r&amp;#x3A9; varies longitudinally from root to tip. This arises due to the rotational motion of the blade. As the component varies, the angle of attack will vary at differing measurements of the blade radius. A small angle of attack, &amp;#x3B1; reduces the Lift Coefficient, CL and larger angles of attack results in stall. In both instances, a significant reduction in the Power will result, thereby affecting the aerodynamic efficiency of the wind turbine. \n\nThe extractable power is maximised when the Lift:Drag ratio is maximised. This can be achieved by ensuring that the angle of attack is at the optimum along the longitudinal length of the turbine blade. To achieve this a &amp;#x201C;twist&amp;#x201D; measured from the plane of rotation is introduced into the blade, where the maximum twist exists at the root and the minimum at the tip. \n
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  146. For a given radius, r, &amp;#x3A6; is a function of the tip speed ratio, &amp;#x3BB;. As the radius, r decreases, the angle of the wind, &amp;#x3A6; increases. For an efficient wind turbine, it is essential that the angle of attack, &amp;#x3B1; is optimised. This is achieved by maximising the lift to drag ratio therefore turbine blades are designed with a twist which increases as r decreases. With reducing radius r, an increase in the blade width, W is required to maintain the component of lift, L, and the thrust, T in order to maintain the Betz condition. \n \n\n
  147. For a given radius, r, &amp;#x3A6; is a function of the tip speed ratio, &amp;#x3BB;. As the radius, r decreases, the angle of the wind, &amp;#x3A6; increases. For an efficient wind turbine, it is essential that the angle of attack, &amp;#x3B1; is optimised. This is achieved by maximising the lift to drag ratio therefore turbine blades are designed with a twist which increases as r decreases. With reducing radius r, an increase in the blade width, W is required to maintain the component of lift, L, and the thrust, T in order to maintain the Betz condition. \n \n\n
  148. For a given radius, r, &amp;#x3A6; is a function of the tip speed ratio, &amp;#x3BB;. As the radius, r decreases, the angle of the wind, &amp;#x3A6; increases. For an efficient wind turbine, it is essential that the angle of attack, &amp;#x3B1; is optimised. This is achieved by maximising the lift to drag ratio therefore turbine blades are designed with a twist which increases as r decreases. With reducing radius r, an increase in the blade width, W is required to maintain the component of lift, L, and the thrust, T in order to maintain the Betz condition. \n \n\n
  149. For a given radius, r, &amp;#x3A6; is a function of the tip speed ratio, &amp;#x3BB;. As the radius, r decreases, the angle of the wind, &amp;#x3A6; increases. For an efficient wind turbine, it is essential that the angle of attack, &amp;#x3B1; is optimised. This is achieved by maximising the lift to drag ratio therefore turbine blades are designed with a twist which increases as r decreases. With reducing radius r, an increase in the blade width, W is required to maintain the component of lift, L, and the thrust, T in order to maintain the Betz condition. \n \n\n
  150. For a given radius, r, &amp;#x3A6; is a function of the tip speed ratio, &amp;#x3BB;. As the radius, r decreases, the angle of the wind, &amp;#x3A6; increases. For an efficient wind turbine, it is essential that the angle of attack, &amp;#x3B1; is optimised. This is achieved by maximising the lift to drag ratio therefore turbine blades are designed with a twist which increases as r decreases. With reducing radius r, an increase in the blade width, W is required to maintain the component of lift, L, and the thrust, T in order to maintain the Betz condition. \n \n\n
  151. For a given radius, r, &amp;#x3A6; is a function of the tip speed ratio, &amp;#x3BB;. As the radius, r decreases, the angle of the wind, &amp;#x3A6; increases. For an efficient wind turbine, it is essential that the angle of attack, &amp;#x3B1; is optimised. This is achieved by maximising the lift to drag ratio therefore turbine blades are designed with a twist which increases as r decreases. With reducing radius r, an increase in the blade width, W is required to maintain the component of lift, L, and the thrust, T in order to maintain the Betz condition. \n \n\n
  152. \n
  153. The air velocity component, r&amp;#x3A9; varies longitudinally from root to tip. This arises due to the rotational motion of the blade. As the component varies, the angle of attack will vary at differing measurements of the blade radius. A small angle of attack, &amp;#x3B1; reduces the Lift Coefficient, CL and larger angles of attack results in stall. In both instances, a significant reduction in the Power will result, thereby affecting the aerodynamic efficiency of the wind turbine. \n\nThe extractable power is maximised when the Lift:Drag ratio is maximised. This can be achieved by ensuring that the angle of attack is at the optimum along the longitudinal length of the turbine blade. To achieve this a &amp;#x201C;twist&amp;#x201D; measured from the plane of rotation is introduced into the blade, where the maximum twist exists at the root and the minimum at the tip. \n
  154. The air velocity component, r&amp;#x3A9; varies longitudinally from root to tip. This arises due to the rotational motion of the blade. As the component varies, the angle of attack will vary at differing measurements of the blade radius. A small angle of attack, &amp;#x3B1; reduces the Lift Coefficient, CL and larger angles of attack results in stall. In both instances, a significant reduction in the Power will result, thereby affecting the aerodynamic efficiency of the wind turbine. \n\nThe extractable power is maximised when the Lift:Drag ratio is maximised. This can be achieved by ensuring that the angle of attack is at the optimum along the longitudinal length of the turbine blade. To achieve this a &amp;#x201C;twist&amp;#x201D; measured from the plane of rotation is introduced into the blade, where the maximum twist exists at the root and the minimum at the tip. \n
  155. The air velocity component, r&amp;#x3A9; varies longitudinally from root to tip. This arises due to the rotational motion of the blade. As the component varies, the angle of attack will vary at differing measurements of the blade radius. A small angle of attack, &amp;#x3B1; reduces the Lift Coefficient, CL and larger angles of attack results in stall. In both instances, a significant reduction in the Power will result, thereby affecting the aerodynamic efficiency of the wind turbine. \n\nThe extractable power is maximised when the Lift:Drag ratio is maximised. This can be achieved by ensuring that the angle of attack is at the optimum along the longitudinal length of the turbine blade. To achieve this a &amp;#x201C;twist&amp;#x201D; measured from the plane of rotation is introduced into the blade, where the maximum twist exists at the root and the minimum at the tip. \n
  156. The air velocity component, r&amp;#x3A9; varies longitudinally from root to tip. This arises due to the rotational motion of the blade. As the component varies, the angle of attack will vary at differing measurements of the blade radius. A small angle of attack, &amp;#x3B1; reduces the Lift Coefficient, CL and larger angles of attack results in stall. In both instances, a significant reduction in the Power will result, thereby affecting the aerodynamic efficiency of the wind turbine. \n\nThe extractable power is maximised when the Lift:Drag ratio is maximised. This can be achieved by ensuring that the angle of attack is at the optimum along the longitudinal length of the turbine blade. To achieve this a &amp;#x201C;twist&amp;#x201D; measured from the plane of rotation is introduced into the blade, where the maximum twist exists at the root and the minimum at the tip. \n
  157. The air velocity component, r&amp;#x3A9; varies longitudinally from root to tip. This arises due to the rotational motion of the blade. As the component varies, the angle of attack will vary at differing measurements of the blade radius. A small angle of attack, &amp;#x3B1; reduces the Lift Coefficient, CL and larger angles of attack results in stall. In both instances, a significant reduction in the Power will result, thereby affecting the aerodynamic efficiency of the wind turbine. \n\nThe extractable power is maximised when the Lift:Drag ratio is maximised. This can be achieved by ensuring that the angle of attack is at the optimum along the longitudinal length of the turbine blade. To achieve this a &amp;#x201C;twist&amp;#x201D; measured from the plane of rotation is introduced into the blade, where the maximum twist exists at the root and the minimum at the tip. \n
  158. The air velocity component, r&amp;#x3A9; varies longitudinally from root to tip. This arises due to the rotational motion of the blade. As the component varies, the angle of attack will vary at differing measurements of the blade radius. A small angle of attack, &amp;#x3B1; reduces the Lift Coefficient, CL and larger angles of attack results in stall. In both instances, a significant reduction in the Power will result, thereby affecting the aerodynamic efficiency of the wind turbine. \n\nThe extractable power is maximised when the Lift:Drag ratio is maximised. This can be achieved by ensuring that the angle of attack is at the optimum along the longitudinal length of the turbine blade. To achieve this a &amp;#x201C;twist&amp;#x201D; measured from the plane of rotation is introduced into the blade, where the maximum twist exists at the root and the minimum at the tip. \n
  159. The air velocity component, r&amp;#x3A9; varies longitudinally from root to tip. This arises due to the rotational motion of the blade. As the component varies, the angle of attack will vary at differing measurements of the blade radius. A small angle of attack, &amp;#x3B1; reduces the Lift Coefficient, CL and larger angles of attack results in stall. In both instances, a significant reduction in the Power will result, thereby affecting the aerodynamic efficiency of the wind turbine. \n\nThe extractable power is maximised when the Lift:Drag ratio is maximised. This can be achieved by ensuring that the angle of attack is at the optimum along the longitudinal length of the turbine blade. To achieve this a &amp;#x201C;twist&amp;#x201D; measured from the plane of rotation is introduced into the blade, where the maximum twist exists at the root and the minimum at the tip. \n
  160. The air velocity component, r&amp;#x3A9; varies longitudinally from root to tip. This arises due to the rotational motion of the blade. As the component varies, the angle of attack will vary at differing measurements of the blade radius. A small angle of attack, &amp;#x3B1; reduces the Lift Coefficient, CL and larger angles of attack results in stall. In both instances, a significant reduction in the Power will result, thereby affecting the aerodynamic efficiency of the wind turbine. \n\nThe extractable power is maximised when the Lift:Drag ratio is maximised. This can be achieved by ensuring that the angle of attack is at the optimum along the longitudinal length of the turbine blade. To achieve this a &amp;#x201C;twist&amp;#x201D; measured from the plane of rotation is introduced into the blade, where the maximum twist exists at the root and the minimum at the tip. \n
  161. The air velocity component, r&amp;#x3A9; varies longitudinally from root to tip. This arises due to the rotational motion of the blade. As the component varies, the angle of attack will vary at differing measurements of the blade radius. A small angle of attack, &amp;#x3B1; reduces the Lift Coefficient, CL and larger angles of attack results in stall. In both instances, a significant reduction in the Power will result, thereby affecting the aerodynamic efficiency of the wind turbine. \n\nThe extractable power is maximised when the Lift:Drag ratio is maximised. This can be achieved by ensuring that the angle of attack is at the optimum along the longitudinal length of the turbine blade. To achieve this a &amp;#x201C;twist&amp;#x201D; measured from the plane of rotation is introduced into the blade, where the maximum twist exists at the root and the minimum at the tip. \n
  162. The air velocity component, r&amp;#x3A9; varies longitudinally from root to tip. This arises due to the rotational motion of the blade. As the component varies, the angle of attack will vary at differing measurements of the blade radius. A small angle of attack, &amp;#x3B1; reduces the Lift Coefficient, CL and larger angles of attack results in stall. In both instances, a significant reduction in the Power will result, thereby affecting the aerodynamic efficiency of the wind turbine. \n\nThe extractable power is maximised when the Lift:Drag ratio is maximised. This can be achieved by ensuring that the angle of attack is at the optimum along the longitudinal length of the turbine blade. To achieve this a &amp;#x201C;twist&amp;#x201D; measured from the plane of rotation is introduced into the blade, where the maximum twist exists at the root and the minimum at the tip. \n
  163. The air velocity component, r&amp;#x3A9; varies longitudinally from root to tip. This arises due to the rotational motion of the blade. As the component varies, the angle of attack will vary at differing measurements of the blade radius. A small angle of attack, &amp;#x3B1; reduces the Lift Coefficient, CL and larger angles of attack results in stall. In both instances, a significant reduction in the Power will result, thereby affecting the aerodynamic efficiency of the wind turbine. \n\nThe extractable power is maximised when the Lift:Drag ratio is maximised. This can be achieved by ensuring that the angle of attack is at the optimum along the longitudinal length of the turbine blade. To achieve this a &amp;#x201C;twist&amp;#x201D; measured from the plane of rotation is introduced into the blade, where the maximum twist exists at the root and the minimum at the tip. \n
  164. The air velocity component, r&amp;#x3A9; varies longitudinally from root to tip. This arises due to the rotational motion of the blade. As the component varies, the angle of attack will vary at differing measurements of the blade radius. A small angle of attack, &amp;#x3B1; reduces the Lift Coefficient, CL and larger angles of attack results in stall. In both instances, a significant reduction in the Power will result, thereby affecting the aerodynamic efficiency of the wind turbine. \n\nThe extractable power is maximised when the Lift:Drag ratio is maximised. This can be achieved by ensuring that the angle of attack is at the optimum along the longitudinal length of the turbine blade. To achieve this a &amp;#x201C;twist&amp;#x201D; measured from the plane of rotation is introduced into the blade, where the maximum twist exists at the root and the minimum at the tip. \n
  165. The air velocity component, r&amp;#x3A9; varies longitudinally from root to tip. This arises due to the rotational motion of the blade. As the component varies, the angle of attack will vary at differing measurements of the blade radius. A small angle of attack, &amp;#x3B1; reduces the Lift Coefficient, CL and larger angles of attack results in stall. In both instances, a significant reduction in the Power will result, thereby affecting the aerodynamic efficiency of the wind turbine. \n\nThe extractable power is maximised when the Lift:Drag ratio is maximised. This can be achieved by ensuring that the angle of attack is at the optimum along the longitudinal length of the turbine blade. To achieve this a &amp;#x201C;twist&amp;#x201D; measured from the plane of rotation is introduced into the blade, where the maximum twist exists at the root and the minimum at the tip. \n
  166. A wind turbine will operate optimally when the Tip Speed Ratio, &amp;#x3BB;, is in the range of 7 to 10. The power coefficient, CP, is dependent on wind speed and on the airfoil angle of attack,&amp;#x3B1;. Variable pitch control allows for the angle of attack, &amp;#x3B1; to be altered, thereby ensuring that the optimum CP obtained. In small turbines, which do not have a variable pitch control, the angle of attack cannot be altered. The speed of these turbines will be determined by the prevailing wind and as a consequence the overall power output will be limited. \n
  167. The primary objective in wind turbine design is to maximise the CP at wind speeds below the rated value and not to maintain the maximum CP across all operating conditions. The optimum blade design is only one element. The other element is accurate control. Coupling these two elements together will ensure that power output is maintained. \n\nThe diagram illustrates achieved power output at various wind speeds. The cut in wind speed ucut-in, is the speed at which the blades begin to rotate. This is the speed that must be reached for take-off, i.e. the wind speed required to overcome the torque of the device. Once the ucut-in has been exceed the turbine begins to generate power. This generation typically follows an exponential curve as wind speed increases, until such a value that the urated wind speed has been reach. The control system then engages, changing the angle of attack such that it maintains the power generation at this value as the wind speed increases further. High wind speed poses risks of structural and or electrical damage to wind turbines, therefore it is essential that a provision is included to disengage the wind turbine in such cases. This is achieved through the application of a brake to the gearbox or turbine shaft and engages at the ucut-out speed. \n
  168. There are additional requirements for overspeed protection, particularly when there is a reduction in the turbines electrical load during operation at high tip speed ratios in high winds. \n\nYaw control is the simplest method of achieving power control, i.e the turbine is turned out of the wind direction and its blades are orientated parallel to the wind. The wind vane located above the nacelle provides wind directional information which forms an input to the control system which in turn rotates the turbine via its yaw control mechanism if necessary. \n\nActive pitch control is more common in variable speed turbines. In this case the the turbine is run at constant speed, however the angel of attack is altered to reduce the lift, thereby altering the lift:drag ratio.\n\nImage source: http://www.popsci.com/content/next-gen-wind-turbine-examined\n
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  173. Site selection for the wind farm requires that accurate wind data over a period of time is available. Experience has shown that an average wind speed &gt; 6 m&amp;#x22C4;s-1 indicates a good location. Ideally, mountainous areas (assuming that access is not an issue) and coastal regions are good candidates for locating a wind farm. However, consideration should also be given to losses that may occur in the cabling from wind farm to the electrical grid. \n\nSpacing between wind turbines is a critical issue. Typical spacing should be in the region of 5D &amp;#x2192; 10D (D being diameter) between each turbine, thereby allowing for the wind to have regained speed and avoid turbulence. Greater turblance intensity reduces the spacing but increases the fatigue loading on the turbines. \n
  174. Wind turbines and farms can have an adverse impact on the society and the environment. They can spoil areas of natural beauty and affect wildlife. The land area required for wind farms is another concern, however, the land between each turbine can be used of grazing and other farming needs provided access is available and a small exclusion area around the base of the turbine is provided. \n\nThey also can cause electromagnetic interference which affects radar and telecommunications. Noise is another area of concern, however improvements have been made in the blade design to help reduce this. \n\nWhen a turbine has reached its end of service life, it most be disposed of. Established disposal and recycling procedures can be utilised for the metallic parts, however the disposal of the rotor blades consisting of glass reinforced polymers can pose difficulties. \n\nWind turbines reduce CO2 emissions by providing an alternative to fossil fuels. It is also important to stress however that wind turbines do produce CO2, during there manufacture and disposal. Also the construction of turbines in remote regions can affect local communities, the environment, require specialised construction equipment and most also be connected back to the Grid. \n
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