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

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wind energy conversion systems april 21 22 2003

Wind Energy Conversion SystemsApril 21-22, 2003

K Sudhakar

Centre for Aerospace Systems Design & Engineering

Department of Aerospace Engineering

http://www.casde.iitb.ac.in/~sudhakar

horizontal axis wecs
Horizontal Axis WECS

Energy extraction at a plane normal to wind stream.

Rotor plane - a disc

aerodynamics of wind turbines
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
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 2000m

V(50 m) / V(20 m) = 1.3 city

= 1.2 grassy

= 1.1 smooth

bernoulli equation

A1, V1

A2, V2

Bernoulli Equation

p + 0.5  V2 = constant

Incompressible flows; along a streamline, . .

Internal flows:

Conservation of mass;  A V = constant

If  is constant, A1 V1 =A2 V2

actuator disc theory

A 

V

p 

pd+

pd-

A d

Vd

A 1

V1

p

Actuator Disc Theory

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

P = 0.5 m (V2 - V12) = 0.5  Ad Vd (V2 - V12)

T = m (V- V1) =  Ad Vd (V- V1) = Ad ( pd- - pd+)

pd- - pd+ =  Vd (V- V1)

actuator disc theory1

A 

V

p 

pd+

pd-

A d

Vd

A 1

V1

p

Actuator Disc Theory

p  + 0.5  V2 = pd- + 0.5  Vd2

p  + 0.5  V12 = pd+ + 0.5  Vd2

pd- - pd+ = 0.5  (V2- V12 ) =  Vd (V- V1)

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

P = 0.5  Ad Vd (V2 - V12) = 0.5  Ad Vd 2Vd (V- V1)

=  Ad Vd2(V- V1)

=  Ad V2(1 - a)2 2aV

= 2  Ad V3 a (1 - a)2

actuator disc theory2
Actuator Disc Theory

P = 2  Ad V3 a (1 - a)2

Non-dimensional quantities,

CP= P / (0.5  Ad V3 ) ; CQ = Q/ (0.5  AdR V2 )

CT = T/ (0.5  Ad V2 ) ;  = 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
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

M

F

*P1

V

Po

Aerodynamics

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

forces moments

y

n

u

rMRP

ds

V

Forces & Moments

Basic Mechanisms

  • Force due to normal pressure, p = - p ds n
  • Force due to tangential stress,  =  ds (  n = 0)
drag lift

drag

V

Drag & Lift
  • D - Drag is along V
  • L - Lift is the force in the harnessed direction

How to maximise L/D

slide13

Skin friction drag, Df

Pressure drag, DP

Drag

For steam lined shapes Df >> DP

For bluff bodies DP >> Df

streamlining
Streamlining!

Equal Drag Bodies

Airfoil of chord 150 mm

1 mm dia wire

wind turbine

r 

,Q

Tower loads

V

Wind Turbine

Typical Vertical Axis WECS - Rotor with n-blades

Cranked by airflow. Cranking torque?

wind turbine rotor

Lift

drag

V

Wind Turbine Rotor

How to compute Q = Torque, T = Tower load

why non dimensional coefficients
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

Camber line

t

h

V

C

Airfoil Characteristics

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 characteristics1

stall

Moment Ref Pt = 0.25 c

CL

CM

CD

i

13o

Airfoil Characteristics

i = f(h/c)max

CM = constant = f(h/c)max

CL = dCL/d

= 2  rad-1 = 0.11 deg -1

CLo = f (h/c)max

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

cranking torque

, Q

Cranking Torque?
  • Air cranks rotor  equal, opposite reaction on air
  • Rotor angular velocity, 
  • Torque on rotor Q
  • Angular velocity of air downstream of rotor, = 2a’
  • Angular velocity at rotor mid-plane, 0.5  = a’
  • a’- circumferential inflow
cranking torque1

dr

r

, Q

Cranking Torque?

 = 2a’

flow velocities

CL

CD

Flow velocities

r a’

r 

V

W

a V

 =  - 

CL, CD = f ()

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

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

slide28

Betz

16/27

CP

Cpi - Energy extraction is through cranking