Fluid Mechanics and Thermodynamics of Turbomachinery, 5e

Example 10.2. Determine the radii of the unmixed slipstream at the disc (R2) and
far downstream of the disc (R3) compared with the radius far upstream (R1).
Solution. Continuity requires that
pRc 1 x pRcx pRcx
R R cx cx a R R a
R R cx cx a R R a
R R a a
2
1 2 2
2 3 2
= = 3
( 2
2
1)
= 1 2 = 1 ( 1- ) 2 1
.
= 1 ( 1-
)
( 3
2
1)
= 1 3 = 1 ( 1-2 ) 3 1 = 1 ( 1-2 )
(
0.5
)= ( 1- ) ( 1-2 )
3 2
Choosing a value of a – = 1 – 3, corresponding to the maximum power condition, the radius
ratios are R2/R1 = 1.225, R3/R1 = 1.732 and R3/R2 = 1.414.
Example 10.3. Using the above expressions for an actuator disc, determine the
power output of a HAWT of 30m tip diameter in a steady wind blowing at
(i) 7.5m/s,
(ii) 10m/s.
Assume that the air density is 1.2kg/m 3 and that a – = 1 – 3 .
Solution. Using eqn. (10.10) and substituting a – = 1 – 3 , r = 1.2kg/m 2 and A 2 =p15 2 ,
then
P= 2a A c ( 1-a)
3
r
= 251. 3c
2 x1
2 = ¥ 1. 2 ¥ p15
¥ 1-
3
3
x1
P 1
[ ]
( )
(i) With cx1 = 7.5m/s, P = 106kW.
(ii) With cx1 = 10m/s, P = 251.3kW.
2
2 1
3
2
c
3
x1
These two examples give some indication of the power available in the wind.
P 2+
P 2–
05
05 .
Wind Turbines 335
FIG. 10.7. Schematic of the pressure variation before and after the plane of the
actuator disc.