1 | P a g e
1. Calculate the total thrust and aerodynamics power developed in a 3- blade wind turbine at
a wind velocity of 9 m/s.
The machine specifications are as follows:
Diameter = 9 m
Rotational speed = 100 rpm
Blade length = 4 m
TSR = 5.23
Chord length = 0.45 m
Pitch angle = 5
o
Airfoil section = NACA 23018
Distance from shaft to inner edge = 0.5 m
Solution:
2
1
2
L b L
dF dA C e =
2
1
2
D b D
dF dA C e =
cos sin
T L D
dF dF I dF I = +
sin cos
M L D
dF dF I dF I =
| |
| |
sin cos
sin cos
m L D
L D
dM r dF r dF I dF I
dP dM r dF I dF I e e
= =
= =
2 | P a g e
It will be convenient the airfoil in to four blade elements as shown below
V
= 9 m/s
By Betz theory
v = 2/3 * 9 = 6 m/s
= 1.2 kg/m
3
Area of each blade element (dA
b
) = 0.45 m
2
Rotational speed 100 rpm = 100/60 = 1.66 rps.
The angles b/n Relative wind and Direction of rotation, dF
L
& v
Where is the pitch angle and equal to 5
0
1
4
3 2
1 m 1 m 1 m 1 m
4.5 m
0.5 m
0.5 m
1 m
1 m
1 m
1 m
0.45 m
1
1
1
4
1
2
3
1
6
29.81
2 *1*1.66
6
15.98
2 *2*1.66
6
10.81
2 *3*1.6
tan
tan
ta
6
6
8.15
2 *4*1.
n
t n
66
a
I
I
I
I
t
t
t
t
=
=
=
=
| |
=
|
\ .
| |
=
|
\ .
| |
=
|
\ .
| |
=
|
\ .
i = I -
3 | P a g e
The angles of attack for each of the blade element becomes
i
1
= 24.81
o
i
2
= 10.98
o
i
3
= 5.81
o
i
4
= 3.15
o
From the table for specific Airfoil (NACA 23018)
C
L1
0.95
C
D1
0.0105
C
L2
1.20
C
D2
0.0143
C
L3
0.75
C
D3
0.0092
C
L4
0.46 C
D4
0.0078
The power developed can be calculated as
( )
2
1 1 1 1 1
2 0 0
1
sin cos
2
= 10.49*1*0.5*1.2*0.45*12.1 (0.95*sin 29.81 - 0.0105*cos29.81 )
= 198.2
b L D
dP r dA C I C I
watt
e e =
( )
2
2 2 2 2 2
2 0 0
1
sin cos
2
= 10.49*2*0.5*1.2*0.45*21.8 (1.2*sin15.98 - 0.0143*cos15.98 )
= 886.14
b L D
dP r dA C I C I
watt
e e =
( )
2
3 3 3 3 3
2 0 0
1
sin cos
2
= 10.49*3*0.5*1.2*0.45*32 (0.75*sin10.81 - 0.0092*cos10.81 )
= 1190.52
b L D
dP r dA C I C I
watt
e e =
| |
( )
2
sin cos
1
sin cos
2
L D
b L D
dP dM r dF I dF I
r dA C I C I
e e
e e
= =
=
4 | P a g e
The total power developed in a 3- blade wind turbine is given by
The thrust can be calculated as
( )
2
4 4 4 4 4
2 0 0
1
sin cos
2
= 10.49*4*0.5*1.2*0.45*42.32 (0.46*sin8.15 - 0.0078*cos8.15 )
= 1213.38
b L D
dP r dA C I C I
watt
e e =
( )
1 2 3 4
3
1 0, 46 6 10
Total Power dP dP dP dP
W kW
= + + +
= ~
( )
2
cos sin
1
cos sin
2
T L D
b L D
dF dF I dF I
dA C I C I e
= +
= +
( )
( )
2
1 1 1 1 1
2 0 0
1
cos sin
2
0.5*1.2*0.45*12.1 0.95 cos 29.81 0.0105 sin 29.81
33.96
T b L D
dF dA C I C I
N
e = +
= +
=
( )
( )
2
2 2 2 2 2
2 0 0
1
cos sin
2
0.5*1.2*0.45*21.8 1.2 cos 15.98 0.0143sin 15.98
154.24
T b L D
dF dA C I C I
N
e = +
= +
=
( )
( )
2
3 3 3 3 3
2 0 0
1
cos sin
2
0.5*1.2*0.45*32 0.75 cos 10.81 0.0092sin 10.81
212.94
T b L D
dF dA C I C I
N
e = +
= +
=
5 | P a g e
The total thrust is given by
( )
( )
2
4 4 4 4 4
2 0 0
1
cos sin
2
0.5*1.2*0.45*42.32 0.46 cos 8.15 0.0078sin 8.15
229.96
T b L D
dF dA C I C I
N
e = +
= +
=
( )
1 2 3 3
3
1893.4
T T T T T
F dF dF dF dF
N
= + + +
=
