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- 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
- 10. KV (Lutyens & Tarbuck 2000)
- 11. KV (Lutyens & Tarbuck 2000)
- 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
- 15. KV Geostrophic wind/Prevailing wind Storms TYPES OF WIND
- 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
- 20. KV TYPES OF WIND
- 21. KV Anemometers
- 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
- 32. KV Height(m) Wind Speed (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
- 34. KV Rotor diameter Slipstream Tower Nacelle Blade Hub WIND TURBINE CONFIGURATION
- 35. KV (Boyle, G. 2004)
- 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
- 55. KV MOMENTUM BALANCE ACROSS ROTOR BLADES
- 56. KV Applied rotation MOMENTUM BALANCE ACROSS ROTOR BLADES
- 57. KV u1 Applied rotation MOMENTUM BALANCE ACROSS ROTOR BLADES
- 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
- 95. KV Plane of Rotation
- 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
- 104. KV Plane of Rotation FL FR FD Φ
- 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
- 137. KV Given; tanΦ = u1 rΩ = 2R 3rΩλ → Φtip = tan−1 2 3λ = 4.85°
- 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

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