WIND ENERGY SYSTEMS - Cd3wd

Chapter 5—Electrical Network 5–34
For a slip s = 0.025 (operation as a motor), compute I 1 , V A , I m , I 2 , speed in r/min, total output
torque and power, power factor, total three-phase losses, and efficiency.
The applied voltage to neutral is
V 1 = 220 √
3
= 127/0 o
V/phase
Z 2 = R 2
s + jX 2 = 0.14
0.025 + j0.35 = 5.60 + j0.35 = 5.61/3.58o Ω
Z m =
jR mX m
= j(120)(13.2)
R m + jX m 120 + j13.2 = 1584/90o
120.72/6.28 o =13.12/83.72o Ω
Z in = R 1 + jX 1 + Z mZ 2
Z m + Z 2
=0.30 + j0.35 + (13.12/83.72o )(5.61/3.58 o )
13.12/83.72 o +5.61/3.58 o =5.29/27.08o Ω
I 1 = V 1
Z in
= 127/0o
5.29/27.08 o =24.01/ − 27.08o A
V A = V 1 − I 1 (R 1 + jX 1 ) = 127 − 24.01/ − 27.08 o (0.30 + j0.35)
= 116.76 − j4.20 = 116.84/ − 2.06 o V
I m = V A
116.84/ − 2.06o
=
Z m 13.12/83.72 o =8.91/ − 85.78 o A
From Eq. 48 we have
I 2 = V A
116.84/ − 2.06o
=
Z 2 5.61/3.58 o =20.83/ − 5.64 o A
From Eq. 47 the speed is
n s = 7200
6
= 1200 r/min
n =(1− s)n s =(1− 0.025)1200 = 1170 r/min
Wind Energy Systems by Dr. Gary L. Johnson November 21, 2001