ENGINEERING TRIPOS PART IB PAPER 8 ELECTIVE () Mechanical Engineering for Renewable Energy Systems Dr. Digby Symons Wind Turbine Blade Design Student Handout
CONTENTS 1 Introduction... 3 Wind Turbine Blade Aerodynamics... 4.1 Aerofoil Aerodynamics... 4. Wind Turbine Blade Kinematics... 7 3 Blade Element Momentum Theory... 1 3.1 Momentum changes... 1 3. Blade forces... 13 3.3 Induction factors... 14 3.4 Iterative procedure... 14 4 Blade Loading... 19 4.1 Aerodynamic Loading... 19 4. Centrifugal Loading... 1 4.3 Self Weight loading... 4.4 Combined Loading... 3 4.5 Storm Loading... 4 More detailed coverage of the material in this handout can be found in various books, e.g. Aerodynamics of Wind Turbines, Hansen M.O.L. 000
1 INTRODUCTION 1.1.1 Aim Preliminary design of a wind turbine 1.1. Wind turbine type Horizontal axis wind turbine (HAWT) with 3 blade upwind rotor the Danish concept : 1.1.3 Load cases We will consider two load cases: 1) Normal operation continuous loading Aerodynamic, centrifugal and self-weight loading ) Extreme wind loading storm loading with rotor stopped 3
WIND TURBINE BLADE AERODYNAMICS.1 Aerofoil Aerodynamics.1.1 Lift, drag and angle of attack α V c.1. Lift and drag coefficients Define non-dimensional lift and drag coefficients 4
.1.3 Variation of lift and drag coefficients with angle of attack How does lift and drag vary with angle of attack α? C L, C D α Stall:.1.4 Application of D theory to wind turbines Tip leakage means flow is not purely two dimensional Wind turbine blades are spinning with an angular velocity ω The angle of attack depends on the relative wind velocity direction. 5
.1.5 Example aerofoil shape used in wind turbines Lift and drag coefficients for the NACA 001 symmetric aerofoil (Miley, 198) 6
. Wind Turbine Blade Kinematics..1 Blade rotation θ ϖ r Relative wind α θ V 7
.. Wake rotation R Axial velocity V 0 0 V0(1 a) V0(1 a) Tangential velocity 0 a = axial induction factor a = angular induction factor Rotor plane 8
..3 Annular control volume Wake rotates in the opposite sense to the blade rotation ω r R ϖ V V (1 ) 0 0 a V0(1 a)..4 Wind and blade velocities Induced wind velocities seen by blade + blade motion V0(1 a) ϖ r a Local twist angle of blade = θ 9
..5 Blade relative motion and lift and drag forces V0 (1 a) θ Local angle of attack = α Relative wind speed Vrel has direction φ = α + θ where V rel sinφ = V0 (1 a) and V rel cosφ = ωr(1 + a ) F L and FD are aligned to the direction of Vrel Obtain C and C for α = φ θ from table or graph for aerofoil used L D 10
..6 Resolve forces into normal and tangential directions V rel F L F D We can resolve lift and drag forces into forces normal and tangential to the rotor plane: We can normalize these forces to obtain force coefficients: Hence: C N = CL cos φ + CD sinφ CT = CL sin φ CD cosφ 11
3 BLADE ELEMENT MOMENTUM THEORY Split the blade up along its length into elements. Use momentum theory to equate the momentum changes in the air flowing through the turbine with the forces acting upon the blades. Pressure distribution along curved streamlines enclosing the wake does not give an axial force component. (For proof see one-dimensional momentum theory, e.g. Hansen) 3.1 Momentum changes r R V V (1 ) 0 ϖ 0 a ϖ r a V0(1 a) ϖr a Thrust from the rotor plane on the annular control volume is δ N δ N = m& V u ) = πrρu( V u ) δr ( 0 1 0 1 Torque from rotor plane on this control volume is δ T δt = m& ru θ = m & ϖr a 1
3. Blade forces Now equate the momentum changes in the flow to the forces on the blades: 3..1 Normal forces δ N = 0 4π rρv a(1 a) δr = Therefore: 1 V a = Bρ 0 (1 ) ccnδr sin φ 1 (1 a) 4 πra = B cc N sin φ Define the rotor solidity: Hence: 3.. Tangential forces δ T = 3 4π r ρv0(1 a) ωa δr = Therefore: 1 V a rω a = rbρ 0 (1 ) (1 + ) cctδr sinφ cosφ 1 (1 + a ) 4 πr a = B cct sinφ cosφ Use the rotor solidity σ : 13
3.3 Induction factors These equations can be rearranged to give the axial and angular induction factors as a function of the flow angle. Axial induction factor: Angular induction factor: However, recall that the flow angleφ is given by: (1 a) V tanφ = 0 (1 + a ) ωr Because the flow angle φ depends on the induction factors a and a' these equations must be solved iteratively. 3.4 Iterative procedure Choose blade aerofoil section. Define blade twist angle θ and chord length c as a function of radius r. Define operating wind speed V and rotor angular velocity ω. 0 For a particular annular control volume of radius r : 1. Make initial choice for a and a, typically a = a = 0.. Calculate the flow angle φ. 3. Calculate the local angle of attack α = φ θ. 4. Find C and C for α from table or graph for the aerofoil used. L D 5. Calculate CN and 6. Calculate a and a. C T. 7. If a and a have changed by more than a certain tolerance return to step. 8. Calculate the local forces on the blades. 14
3.4.1 Example wind turbine Blade element theory has been applied to an example 4 m diameter wind turbine with the parameters below. Each element has a radial thickness r δ = 1m. Incident wind speed V 0 8 m/s Angular velocity ϖ 30 rpm Blade tip radius R 1 m Tip speed ratio λ = ωr /V0 Number of blades B 3 Air density ρ 1.5 kg/m 3 Blade shape (chord c and twist θ ) are based on the Nordtank NTK 500/41 wind turbine (see Hansen, page 6). Chord c 1.8 1.6 1.4 1. Chord c (m) 1 0.8 0.6 0.4 0. 0 0 4 6 8 10 1 14 16 18 0 Radius r (m) 15
Blade twist angle θ Blade twist angle (deg) 0 18 16 14 1 10 8 6 4 0 0 4 6 8 10 1 14 16 18 0 r (m) 3.4. Results of BEM analysis Axial induction factor a 0.5 0.0 Axial induction factor a 0.15 0.10 0.05 0.00 0 4 6 8 10 1 14 16 18 0 r (m) 16
Angular induction factor a 0.035 0.030 Angular induction factor a' 0.05 0.00 0.015 0.010 0.005 0.000 0 4 6 8 10 1 14 16 18 0 r (m) Flow angle φ and local angle of attack α = φ θ 30 5 0 Angle (deg) 15 10 5 0 0 4 6 8 10 1 14 16 18 0 r (m) 17
Normal FN and tangential F T forces on blade Forces per unit length (N/m) 1100 1000 900 800 700 600 500 400 300 00 100 0 FN (N/m) FT (N/m) 0 4 6 8 10 1 14 16 18 0 r (m) Total power (3 blades) Coefficient of performance Contribution of blade elements to total torque (and therefore power) 10 9 Contribution to total torque (%/m) 8 7 6 5 4 3 1 0 0 4 6 8 10 1 14 16 18 0 r (m) 18
4 BLADE LOADING 4.1 Aerodynamic Loading Once values of a and a have converged the blade loads can be calculated: 1 V a FN ρ 0 (1 ) = cc N sin φ F 1 V a ωr a ρ 0 (1 ) (1 + ) = sinφ cosφ T cc T 4.1.1 Stresses at blade root The normal force F N causes a flapwise bending moment at the root of the blade. M N = R F N rmin ( r rmin ) dr The tangential force F T causes a tangential bending moment at the root of the blade. M T = R F T rmin ( r rmin ) dr For convenience we will neglect the relatively small twist of the blade cross section and assume that these bending moments are aligned with the principal axes of the blade structural cross section. The maximum tensile stress due to aerodynamic loading is therefore given by: 19
4.1. Deflection of blade tip ds F N = dr dm S = dr d v M κ = dr [ EI ]( r) Simplified approach: Split blade into elements. Assume that for each element the loading F N and flexural rigidity EI are constant. Find the shear force and bending moment transferred between each element. Use data book deflection coefficients for each element. Find the cumulative rotations along the blade. Find the cumulative deflections along the blade. 0
4. Centrifugal Loading The large mass of a wind turbine blade and the relatively high angular velocities can give rise to significant centrifugal stresses in the blade. Consider equilibrium of element of blade: df c = m( r) ω r dr σ c = F c ( r) A( r) Simplified method: Split blade up into elements. Assume each element has a constant cross-section 1 F c, n = Fc, n+ 1 + m( rn + 1 rn ) ( rn + 1 + rn ) ω 1 c, n+ 1 + n+ 1 n = F m( r r ) ω 1
4.3 Self Weight loading The bending moment at the blade root due to self weight loading can dominate the stresses at the blade root. Because the turbine is rotating the bending moment is a cyclic load with a frequency of f = ω / π. The maximum self-weight bending moment occurs when a blade is horizontal. Bending moment at root of blade due to self weight M sw = R rmin m r) g( r r ( min ) dr where m(r) is the mass of the blade per unit length. This is a tangential (edge-wise) bending moment and therefore the maximum bending stress due to self-weight is given by: Simplified method: split blade into elements, assume each element has uniform self weight.
4.4 Combined Loading M d M σ N o T max, aero = + ITT I NN σ c = σ F c ( r) A( r) M sw max, sw = I NN b b Operational maximum stress: σ max = Minimum stress at same location: σ min = 3
4.5 Storm Loading 4.5.1 Drag force on blade Blades parked. Extreme wind speed load per unit length V max c F D V max = 50 m/s, c = 1.3m ρv c Re = max = μ Hence C D = 4
4.5. Bending moment Find bending moment at root of blade M = R ( r r r min min ) F D dr σ = y M I = Eκ M σ = I d o I 3 3 b( d o d i ) = 1 4.5.3 Shear stress S = R F r D dr min q = SAc y I q = τt w Note: High solidity rotor (multi bladed) gives excessive forces on tower during extreme wind speeds. Therefore use fewer blades. 5