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Wind Energy Conversion Systems April 21-22, 2003 K Sudhakar Centre for Aerospace Systems Design & Engineering Department of Aerospace Engineering

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Presentation on theme: "Wind Energy Conversion Systems April 21-22, 2003 K Sudhakar Centre for Aerospace Systems Design & Engineering Department of Aerospace Engineering"— Presentation transcript:

1 Wind Energy Conversion Systems April 21-22, 2003 K Sudhakar Centre for Aerospace Systems Design & Engineering Department of Aerospace Engineering http://www.casde.iitb.ac.in/~sudhakar

2 Horizontal Axis WECS Energy extraction at a plane normal to wind stream. Rotor plane - a disc

3 Aerodynamics of Wind Turbines Aerodynamics Forces and Moments on a body in relative motion with respect to air Topics of intense study aerospace vehicles, road vehicles, civil structures, wind turbines, etc.

4 Atmosphere International Standard Atmosphere –Sea level pressure = 101325 Pa –Sea level temperature= 288.16 K (IRA 303.16) –Sea level density= 1.226 kg/m^3 (IRA 1.164) –dt/dh= -0.0065 K/m –p/p SL = (t/t SL ) 5.2579 Planetary boundary layer extends to 2000m V(50 m) / V(20 m) = 1.3 city = 1.2 grassy = 1.1 smooth

5 Bernoulli Equation p + 0.5  V 2 = constant Incompressible flows; along a streamline,.. A 1, V 1 A 2, V 2 Internal flows: Conservation of mass;  A V = constant If  is constant, A 1 V 1 = A 2 V 2

6 Actuator Disc Theory A  V  p  A d V d A 1 V 1 p  pd-pd- pd+pd+ A  V  = A d V d =A 1 V 1 ; mass flow rate, m =  A d V d P = 0.5 m (V  2 - V 1 2 ) = 0.5  A d V d (V  2 - V 1 2 ) T = m (V  - V 1 ) =  A d V d (V  - V 1 ) = A d ( p d - - p d + ) p d - - p d + =  V d (V  - V 1 )

7 Actuator Disc Theory A  V  p  A d V d A 1 V 1 p  pd-pd- pd+pd+ p  + 0.5  V  2 = p d - + 0.5  V d 2 p  + 0.5  V 1 2 = p d + + 0.5  V d 2 p d - - p d + = 0.5  (V  2 - V 1 2 ) =  V d (V  - V 1 ) V d = 0.5 (V  + V 1 ) ; V d = V  ( 1 - a); V 1 = V  ( 1 - 2 a) P = 0.5  A d V d (V  2 - V 1 2 ) = 0.5  A d V d 2 V d (V  - V 1 ) =  A d V d 2 (V  - V 1 ) =  A d V  2 (1 - a) 2 2aV  = 2  A d V  3 a (1 - a) 2

8 Actuator Disc Theory P = 2  A d V  3 a (1 - a) 2 Non-dimensional quantities, C P = P / (0.5  A d V  3 ) ; C Q = Q/ (0.5  A d R V  2 ) C T = T/ (0.5  A d V  2 ) ; = r  / V  C P = 4 a (1 - a) 2 ; C T = 4 a (1 - a) dC P /da = 0  a = 1/3 C P-max = 16/27 ; C T @ C P-max = 8/9  a = 1/3 C T-max = 1 ; C P @ C T-max = 1/2  a = 1/2

9 Rotor & Blades Energy extraction through cranking of a rotor Cranking torque supplied by air steam Forces / moments applied by air stream? Blade element theory of rotors?

10 Aerodynamics Aerodynamics - Forces and Moments on a body in relative motion with respect to air VV F M PoPo * P 1

11 Forces & Moments Basic Mechanisms –Force due to normal pressure, p = - p ds n –Force due to tangential stress,  =  ds (   n = 0) u y VV ds n r MRP

12 Drag & Lift D - Drag is along V  L - Lift is the force in the harnessed direction How to maximise L/D VV drag

13 Drag For steam lined shapes D f >> D P For bluff bodies D P >> D f Pressure drag, D P Skin friction drag, D f

14 Streamlining! Equal Drag Bodies 1 mm dia wire Airfoil of chord 150 mm

15 Wind Turbine Typical Vertical Axis WECS - Rotor with n-blades Cranked by airflow. Cranking torque? Tower loads VV ,Q r 

16 Wind Turbine Rotor How to compute Q = Torque, T = Tower load VV drag Lift

17 Why non-dimensional Coefficients With dimensional values –At each ( , , , V , a, c) measure L, D, M –Many tests required With non-dimensional coefficients –At desired Re, M,  and V  –for each  measure L, D, M –Convert to C L, C D, C M –At any other  and V  compute L, D M

18 Airfoil Characteristics h t VV C Camber line h(x)  0 camber  symmetric airfoil (h/c)max and (x/c) @ (h/C)max (t/c)max and (x/c) @ (t/c)max Leading edge radius

19 Airfoil Characteristics C L  = dC L /d  = 2  rad -1 = 0.11 deg -1 C Lo = f (h/c) max  i = f(h/c) max C M = constant = f(h/c) max  CLCL  CDCD  CMCM Moment Ref Pt = 0.25 c 13 o ii stall Special airfoils for wind turbines with high t/c @ low Re  SERI / NREL

20 Cranking Torque? Air cranks rotor  equal, opposite reaction on air Rotor angular velocity,  Torque on rotor Q , Q  Angular velocity of air downstream of rotor,  = 2a’  Angular velocity at rotor mid-plane, 0.5  = a’  a’- circumferential inflow

21 Cranking Torque?  = 2a’  , Q  r dr

22 Flow velocities VV a V  r  r a’  W    =  -  C L, C D = f (  ) CLCL CDCD C x = C L Sin  - C D Cos  = C L Sin  ( 1 -  Cot  ) C T = C L Cos  +C D Sin  = C L Cos  ( 1+  Tan  )

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28 CPCP 16/27 Betz C pi - Energy extraction is through cranking

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