WIND ENERGY SYSTEMS - Cd3wd

Chapter 5—Electrical Network 5–29
where p is the number of poles.
n s = 7200
p
r/min (48)
Figure 17: Equivalent circuit for one phase of a three-phase induction motor
The total losses in the motor are given by
P loss = 3|V A| 2
+3|I 1 | 2 R 1 +3|I 2 | 2 R 2 W (49)
R m
The first term is the loss due to eddy currents, hysteresis, windage, and friction. The second
term is the winding loss (copper loss) in the stator conductors and the third term is the
winding loss in the rotor. The factor of 3 is necessary because of the three phases.
The power delivered to the resistance at the right end of Fig. 17 is
P m,1 = |I 2| 2 R 2 (1 − s)
W/phase (50)
s
The power P m,1 is not actually dissipated as heat inside the motor but is delivered to a load
as mechanical power. The total three-phase power delivered to this load is
P m = 3|I 2| 2 R 2 (1 − s)
s
W (51)
To analyze the circuit in Fig. 17, we first need to find the impedance Z in which is seen by
the voltage V 1 . We can define the impedance of the right hand branch as
Z 2 = R 2 + jX 2 + R 2(1 − s)
s
The impedance of the shunt branch is
= R 2
s + jX 2 Ω (52)
Z m =
jX mR m
R m + jX m
Ω (53)
Wind Energy Systems by Dr. Gary L. Johnson November 21, 2001