Wind Energy Conversion SystemsApril 21-22, 2003

K Sudhakar

Centre for Aerospace Systems Design & Engineering

Department of Aerospace Engineeringhttp://www.casde.iitb.ac.in/~sudhakar

Horizontal Axis WECS

Energy extraction at a plane normal to wind stream.

Rotor plane - a disc

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.

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/pSL = (t/tSL)5.2579

• Planetary boundary layer extends to 2000mV(50 m) / V(20 m) = 1.3 city

= 1.2 grassy

= 1.1 smooth

Bernoulli Equation

p + 0.5 V2 = constant

Incompressible flows; along a streamline, . .

A1, V1

A2, V2

Internal flows:

Conservation of mass; A V = constant

If is constant, A1 V1 = A2 V2

Actuator Disc Theory

A V

p

A d Vd

A 1 V1

p

pd- pd

+

A V= A d Vd =A1 V1 ; mass flow rate, m = Ad Vd

P = 0.5 m (V2 - V1

2) = 0.5 Ad Vd (V2 - V1

2)

T = m (V- V1) = Ad Vd (V- V1) = Ad ( pd- - pd

+)

pd- - pd

+ = Vd (V- V1)

Actuator Disc Theory

A V

p A d Vd

A 1 V1

p

pd- pd

+

p + 0.5 V2 = pd

- + 0.5 Vd2

p + 0.5 V12 = pd

+ + 0.5 Vd2

pd- - pd

+ = 0.5 (V2 - V1

2 ) = Vd (V- V1)

Vd = 0.5 (V+ V1) ; Vd = V( 1 - a); V1 = V( 1 - 2 a)

P = 0.5 Ad Vd (V2 - V1

2) = 0.5 Ad Vd 2 Vd (V- V1)

= Ad Vd2

(V- V1)

= Ad V2

(1 - a)2 2aV

= 2 Ad V3 a (1 - a)2

Actuator Disc Theory

P = 2 Ad V3 a (1 - a)2

Non-dimensional quantities,

CP = P / (0.5 Ad V

3 ) ; CQ = Q/ (0.5 Ad R V2 )

CT = T/ (0.5 Ad V2 ) ; = r / V

CP = 4 a (1 - a)2 ; CT = 4 a (1 - a)

dCP/da = 0 a = 1/3

CP-max = 16/27 ; CT @ CP-max = 8/9 a = 1/3

CT-max = 1 ; CP @ CT-max = 1/2 a = 1/2

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?

Aerodynamics

Aerodynamics - Forces and Moments on a body in relative motion with respect to air

F)PP(M

FrMM

M

M

M

M;

F

F

F

F

010P

0P1P

z

y

x

z

y

x

V

FM

Po

* P1

Vectors F ,F

Forces & Moments

Basic Mechanisms – Force due to normal pressure, p = - p ds n

– Force due to tangential stress, = ds ( n = 0)

uy

smkg

10x789.1airfor

dydu

5

dsdsn̂pF

FdrMd;dsdsn̂pFd MRP

V

dsn

rMRP

Drag & Lift

• D - Drag is along V

• L - Lift is the force in the harnessed direction

How to maximise L/D

Lift

forceside

drag

z

y

x

F

F

F

F

F

F

FV

drag

Drag

For steam lined shapes Df >> DP

For bluff bodies DP >> Df

dsVdsVn̂pV1

VVF

Drag

Pressure drag, DPSkin friction drag, Df

Streamlining!

Equal Drag Bodies

1 mm dia wire

Airfoil of chord 150 mm

Wind Turbine

Typical Vertical Axis WECS - Rotor with n-blades

Cranked by airflow. Cranking torque?

Tower loadsV

,Q

r

Wind Turbine Rotor

How to compute Q = Torque, T = Tower load

V

dragLift

cSV5.0M

C;SV5.0

DC;

SV5.0L

C

analysislDimensionaa

VM;

cVRe);M(Re,fC

givenaat)c,a,,,V(fL,Lift

2M2D2L

L

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 CL, CD, CM

– At any other and V compute L, D M

Airfoil Characteristics

h

t

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

Airfoil Characteristics

CL = dCL/d = 2 rad-1 = 0.11 deg -1 CLo = f (h/c)max

i = f(h/c)max

CM = constant = f(h/c)max

CL

CD

CM

Moment Ref Pt = 0.25 c

13o i

stall

Special airfoils for wind turbines with high t/c @ low Re SERI / NREL

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

Cranking Torque?

= 2a’

, Q

r

dr

dr)a1('aVr4

r'a2r)a1(Vdrr2

r)0r(Vdrr2dQ

dr)a1(aVr4

a2V)a1(Vdrr2

)VV(Vdrr2dT

Vdrr2md

3

d

2

1d

d

Flow velocities

V

a V

r r a’

W

Sin/)a1(VW

)'a1(r)a1(V

tan

= - CL, CD = f ()

CL

CD

Cx = CLSin - CD Cos = CLSin ( 1 - Cot )

CT = CLCos +CD Sin = CLCos ( 1+ Tan )

)6(dr)Tan1(Sin

CosC)a1(VBcR

21

BCdrcW21

dT

)5(dr)Cot1(SinC

)a1(VBcR21

BrCdrcW21

dQ

)tan1(CosCC

)Cot1(SinCC

2L22

Y

2

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