Blade Element Theory for Wind Turbines - Small Wind Turbines: Analysis, Design, and Application - page 46

Because wind turbines operate at high values of k, typically in the range of
7-10, Xr at the tip is about ten times greater than U 1 . At the hub, Xr is nearly zero,
so that h P must vary significantly with radius to maintain the angle of attack, a,at
reasonable values to avoid flow separation. Chapter 4 shows that efficient opera-
tion of wind turbines occurs over a limited range of a.
The basic assumption is that the lift and drag acting on the blade element are the
same as those on an aerofoil of the same section, angle of attack, and effective
velocity.

From the definitions of the lift and drag coefficients, C l and C d

respectively:

LIFT ¼ 1 DRAG ¼ 1

2 qU T C l c 2 qU T C d c

and ð 3 : 9 Þ

where c is the chord; its precise definition will be given in Chap. 4 along with
further

information on the coefficients. It is next necessary to resolve the

lift and drag into the circumferential and axial components of interest to

the wind turbine designer. For an N-bladed turbine, the total thrust on

N blade elements is

dT ¼ 1

Þ dr ¼ 1

2 qU T cN C l cos / þ C d sin / 2 qU T cNC a dr

ð ð 3 : 10 Þ

where C a = C l cos / ? C d sin / and the torque due to the circumferential

force is

dQ ¼ 1 Þ rdr ¼ 1

2 qU T cN C l sin / C d cos / 2 qU T cNC a 0 rdr

ð ð 3 : 11 Þ

where C a 0 ¼ C l sin / C d cos /.
Equations 3.10 and 3.11 are the basic blade element equations. A number of
modifications to them have been proposed in the century following their original
development for propeller analysis; some of these will be considered in Chap. 5
following the discussion of aerofoil lift and drag in Chap. 4 .

Example 3.1 Suppose a turbine is operating at its maximum efficiency where
a = 1/3 and a 0 is negligible. If k = 7, estimate the h P necessary to achieve
a ¼ 6 at r = R ¼ 0 : 25 and r = R ¼ 1 :

Answer for r/R = 0.25 In the blade element diagram of Fig. 3.2 , U 1 /U 0 = 2/3 and
k r ¼ rX = U 0 ¼ 1 : 75. Now / ¼ tan 1 Þ ¼ 20 : 85 ; so that, from Eq. 3.7a , b ,

ð 2 = 3k ðÞ

h P = 14.85.
Answer for r/R = 1.0 In the blade element diagram of Fig. 3.2 , U 1 /U 0 = 2/3,
k = RX/U 0 = 7.0, so that / = 5.44. From Eq. 3.7a and 3.7b , h P =-0.56.
These answers show the necessity for a significant variation in twist along a
well-designed wind turbine blade.