WIND ENERGY SYSTEMS - Cd3wd

Chapter 6—Asynchronous Generators 6–18
The inductance is
The power per phase is
L s = X s
ω = 5.9
377 =15.65 × 10−3 H
10, 000
P e = = 3333 W/phase
3
At resonance, the inductive reactance and the capacitive reactance cancel, so V a = E a . The resistance
R a is
The current I a is given by
R a = V 2
a
P e
= (150)2
3333 =6.75 Ω
I a = V a
R a
= 150
6.75 =22.22 A
At 40 Hz, the circuit is no longer resonant. We want to use Eq. 18 to find the power but we need
k e first. It can be determined from Eq. 15 and rated conditions as
The total power is then
k e = E ω = 150
377 =0.398
P tot = 3P e =
3(0.398) 2 [2π(40)] 2 (6.75)
(6.75) 2 +[2π(40)(15.65 × 10 −3 ) − 1/(2π(40)(450 × 10 −6 ))] 2
202, 600
=
45.56 + 24.10 = 2910 W
If the power followed the ideal cubic curve, at 40 Hz the total power should be
( ) 3 40
P tot,ideal =10, 000 = 2963 W
60
We can see that the resonant circuit causes the actual power to follow the ideal variation rather closely
over this frequency range.
4 AC GENERATORS
The ac generator that is normally used for supplying synchronous power to the electric utility
can also be used in an asynchronous mode[14]. This machine was discussed in the previous
Wind Energy Systems by Dr. Gary L. Johnson November 21, 2001