Transcript Mechanics of wind turbine - Department of Civil, Architectural and

Short Summary of the Mechanics
of Wind Turbine
Korn Saran-Yasoontorn
Department of Civil Engineering
University of Texas at Austin
8/7/02
1
Summarized from

Wind energy explained: theory, design and
application./ Manwell, J. F. / Chichester / 2002

Wind turbine technology: fundamental concepts
of wind turbine engineering. / New York / 1994

Wind energy conversion systems/ Freris, L.L./
Prentice Hall/ 1990

Wind turbine engineering design/ Eggleston,
D.M. and Stoddard, F.S./ New York/ 1987

Introduction to wind turbine engineering/
Wortman, A.J./ Butterworth Publishers/ 1983
2
Rotor
Hub
Simple model of
Micon 65/13
Drive train
Yaw system
Column
1D steady
wind flow
(12m/s)
3
Fundamental Concepts
Mass flow rate = Av
Energy per unit volume = 1/2 v2
Power = rate of change of energy
= force * velocity
= (Av) 1/2 v2 = 1/2 Av3
Dynamic pressure = force/area
= power/vA
=1/2 v2
4
Actuator Disk Model with no
wake rotation
Assumptions:
i.
Homogeneous, incompressible, steady wind
ii.
Uniform flow velocity at disk (uniform thrust)
iii.
Homogenous disk
iv.
Non-rotating disk
U1
upstream
U2
stream tube boundary
U3
U4
rotor disk
downstream
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Conservation of Linear Momentum
dp
 F  dt

T   m(u 4  u1 )

T  m(u1  u 4 )
where T is the thrust acting uniformly on the disk
(rotor) which can be written as a function of the
change of pressure as follow
T  A( p2  p3 )
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Bernoulli’s Equation (energy conserved)
1 2
1 2
p1  u1  p2  u 2
2
2
1 2
1 2
p3  u3  p4  u4
2
2
Relate above equations and define the axial induction factor, a
as
we obtain
(u1  u2 )
a
u1
1
2
T  Au 1 [4a (1  a )]
2
7
Power output of the turbine is defined as the
thrust times the velocity at the disk. Hence
1
P  Au13 [4a (1  a ) 2 ]
2
1
T  Au 12 [4a (1  a )]
2
Wind turbine rotor performance is usually
characterized by its power and thrust coefficients
C p  P /(1 / 2 Au 3 )  4a (1  a ) 2
CT  T //(1 / 2 Au 3 )  4a (1  a )
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Notice that

Wind velocity at the rotor plane is
always less than the free-stream
velocity when power is being absorbed.

This model assumes no wake rotation,
i.e. no energy wasted in kinetic energy
of a twirling wake.

The geometry of the blades does not
involve the calculations.
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
If the axial induction factor of the
rotor is founded, one can simply
calculate for the thrust and power
output.

An ideal turbine generates maximum
power. After some manipulations, one
can find that the axial induction
factor, a, for the ideal turbine is 1/3.

Even with the best rotor design, it is
not possible to extract more than
about 60 percent of the kinetic energy
in the wind
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upstream
disk
u1
u2
Wind Velocity
downstream
u4
Total Pressure
1/2u2
Dynamic Pressure
p3
Static Pressure
p0
p0
p2
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Actuator Disk Model with wake
rotation
U1
U2
U3
U4
rotor disk
U(1-a)
U
U(1-2a)
dr
r
The thrust distribution is circumferentially uniform.
(infinite number of blades)
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Conservation of Linear momentum
dp
 F  dt

dT (r )   m(u4  u1 )
1 2
dT  4a(1  a) u 2rdr
2
Conservation of Angular Momentum

dQ  (d m)(r )( r )  u2 2rdr (r 2 )
1
dQ  4a(1  a) ur 2 2rdr
2
13
Bernoulli’s Equation (energy conserved)
1
( p2  p3 )   (   )r 2
2
1
dT  ( p2  p3 )dA  [  (   )r 2 ]2rdr
2
Define the angular induction factor a’ as a 
Hence,

2
1 2
dT  4a(1  a) u 2rdr
2
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Equating the thrust on an annular element derived
from the conservation of linear momentum and the
Bernoulli’s equation gives
a(1  a)
 2r
a(1  a)
where r  r / R,   R u
For an ideal turbine that produces maximum power
output,
a 
1  3a
4a  1
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In summary
1
dT  4a(1  a) u 2 2rdr
2
1
dQ  4a(1  a ) ur 2 2rdr
2
a(1  a)
 2r
a(1  a)
Notice that

the geometry of the blades still does not involve the
calculations.

if the turbine is assumed to be ideal generating
maximum power, one can find a and a’ in each section.

once a and a’ are founded, the total thrust and rotor
torque can be determined by integration along the
blade spanwise.
16
Blade Element Theory
R
dr
r
blade element
rotor blade
Blade geometry is considered in this part and
we may use this to calculate the induction
factors that relates the thrust and rotor
torque.
17
Lift and Drag Forces
FL
Ωr
urel
(top view)
FD
u
1
2
FL  C L ( Aurel )
2
1
2
FD  C D ( Aurel )
2
Note that CL and CD vary with cross section
18
Typical Variation of Aerofoil
Coefficients
flow separation
Cl
Values of Coefficients
1.0
Cd
0.5
0.0
-10
0
angle of attack (degrees)
90
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Relative Velocity
Wind velocity at the rotor blade is u(1-a) in
horizontal direction. Also, the wind rotates
with the angular velocity of ω/2 (=Ωa’) while
the angular velocity of the rotor is Ω in the
opposite direction.
FL
Ωr(1+a’)
urel
FD
u(1-a)
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Blade Geometry
dFL
dFN
dFD
Ωr(1+a’)
θp
φ
dFT
a
urel
u(1-a)
21
From blade geometry, one simply obtains the
following relations.
 p a
1 a
tan  
(1  a)r
u (1  a )
u rel 
sin 
1 2
dFL  C L u rel
cdr
2
1 2
dFD  C D u rel cdr
2
dFN  dFL cos   dFD sin 
dFT  dFL sin   dFD cos 
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Finally, the total normal force on the
section and torque due to the tangential
force operating at a distance, r, from
the center are
1 2
dFN  B urel (C L cos   C D sin  )cdr
2
1 2
dQ  B urel (C L sin   C D cos  )crdr
2
23
Since the forces and moments derived
from momentum theory (actuator model)
and blade element theory must be equal,
1 2
dT  u 4a (1  a)rdr  B urel (C L cos   C D sin  )cdr
2
1 2
3
dQ  4a(1  a ) ur dr  B urel (C L sin   C D cos  )crdr
2
2
from momentum theory
blade element theory
24
2r (cos   r sin  )
C L  4 sin 
Bc (sin   r cos  )
One can solve for C and α at each section by
using this equation and the empirical C vs α
curves. Once both parameters are known, a
and a’ at the section can be determined
from
a (1  a)  BcC L cos  /(8r sin 2  )
a (1  a)  BcC L /(8rr sin  )
25
Iterative solution for a and a’
1.
2.
3.
4.
5.
6.
Guess values of a and a’
Calculate φ
Calculate angle of attack, α
Calculate Cl and Cd
Update a and a’
Check if a > 0.5
(In the case of turbulent
wake this analysis may lead
to a lack of convergence to
a solution)
26
Note that to keep the lift and drag coefficients,
and thus the angle of attack, constant over the
spanwise of the blade, it is necessary to twist the
blade along the length. This however may increase
the complexity of their manufacture.
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Tip loss factor
The tip loss factor allows for the
velocities and forces not being
circumferentially uniform due to the
rotor having a finite number of blades.
The Prandtl tip loss factor can be
express as
  r
2
F  (2 /  ) arccos(exp 0.51   B 1   )
  R

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