WIND ENERGY SYSTEMS - Cd3wd

Chapter 5—Electrical Network 5–35
The total torque is given by Eq. 62.
T m = 90|I 2| 2 R 2
πn s s
= 90(20.83)2 (0.14)
π(1200)(0.025)
=58.01 N · m/rad
The total mechanical power is then
P m = ω m T m = 2πnT m
60
= 2π(1170)(58.01)
60
= 7107 W
At 746 W/hp, the motor is delivering 9.53 hp to the load. From Eq. 17, the power factor is the cosine
of the angle between the input voltage and current, or in this case,
From Eq. 49 the total three-phase losses are
pf = cos 27.08 o =0.890 lag
P loss = 3(116.84)2
120
The efficiency is given by Eq. 59.
η m =
+3(24.01) 2 (0.30) + 3(20.83) 2 (0.14) = 1042 W
P m
P m + P loss
=
7107
7107 + 1042 =0.872
The efficiency is 87.2 percent, a typical value for induction motors of this size.
Example
For the machine of the previous example, compute the input current, total starting torque, and
total three-phase losses while the machine is being started (while the slip is still essentially unity).
Assume the source is able to maintain rated voltage during the start.
With the slip s = 1, the impedance Z 2 becomes
Z 2 = 0.14
1 + j0.35 = 0.377/68.20o Ω
The shunt impedance Z m remains the same as before. The input impedance is then
Z in =0.30 + j0.35 + 13.12/83.72o (0.377/68.20 o )
13.12/83.72 o +0.377/68.20 o =0.816/57.92o Ω
I 1 =
127/0 o
0.816/57.92 o = 155.58/ − 57.92o A
V A = 127 − 155.58/ − 57.92 o (0.30 + j0.35) = 57.07/ − 10.73 o
V
Wind Energy Systems by Dr. Gary L. Johnson November 21, 2001