Page 156 - Wind Energy Handbook

130 AERODYNAMICS OF HORIZONTAL-AXIS WIND TURBINES

For n ¼ 1 the polynomials are
p ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
0
P (v) ¼ v ¼ 1 ì 2 (3:160a)
1
and
1
0 1
Q (iç) ¼ ç tan 1 (3:160b)
1
ç
0
So, on the disc, where ç ¼ 0, Q (i0) ¼ 1.
1
Therefore, the pressure distribution is
p ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
p( ì) ¼ C 0 1 ì 2 (3:161)
1
If the pressure in Equation (3.161) is non-dimensionalized using the free-stream
0
dynamic pressure (1=2)rU 2 1 the value of C can be related to the thrust coefficient
1
by integrating the pressure distribution of Equation (3.161) over the disc area
ð ð ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2ð 1 p
2
2
2
2
2
ðR C T ¼ R C 0 1 ì ì dì dł ¼ ðR C 0
1 1
0 0 3
Therefore,
3
0
C ¼ C T (3:162)
1
2
and so the pressure step across the disc is
3 p ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
p 1 ( ì) ¼ C T 1 ì 2 (3:163)
2
All the remaining polynomials (for m ¼ 0 and odd values of n . 1) produce zero
thrust. To modify the pressure distribution to suit the boundary conditions an
appropriate linear combination of solutions can be added to that of Equation
(3.163).
The application to helicopter rotors leads to a requirement for the pressure and
the radial pressure gradient to be zero at the rotor axis as these conditions corre-
spond to the pressure on actual rotors. The above pressure distribution does not
have zero pressure at the rotor axis and so needs to be combined with at least one
other solution. The second axisymmetric solution, n ¼ 3, is

1 1 p ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
0
2
P (v) ¼ v(5v 3) ¼ 1 ì (2 5ì ) (3:164a)
3 2 2
and
ç 1 5 2 2
2
0
2
0
Q (iç) ¼ (5ç þ 3) tan 1 þ ç þ ,so Q (i0) ¼ (3:164b)
3 3
2 ç 2 3 3