776 BASIC CONTROL SYSTEMS
0 T
ωs
ωm
Trated
δ = 0
δ = 1
1 > δ 1 > δ 2 > 0
δ 2
δ 1
Natural
Figure 16.1.24Effect of static rotor resistance control on
speed–torque curves.
WithjXmshifted to the stator terminals in Figure 16.1.23 as an approximation, we get
I ̄ 2 ′=
V ̄ 1
[
R 1 +R′b+(R′a/S)
]
+j(Xl 1 +X′l 2 )
(16.1.55)
and
T=
3
ωs
(I 2 ′)^2
(
R′a
S
)
(16.1.56)
Figure 16.1.24 shows the nature of the speed–torque characteristics for different values of the
duty ratioδ.
EXAMPLE 16.1.5
A 440-V, 60-Hz, six-pole, wye-connected, wound-rotor induction motor with a full-load speed of
1170 r/min has the following per-phase parameters referred to the stator:R 1 =R 2 ′= 0. 5 , Xl 1 =
X′l 2 = 2 , Xm= 40 , and a stator-to-rotor turns ratio of 2.5.
The scheme of Figure 16.1.21 is employed for speed control withRd= 0. 02 andR= 1 .
For a speed of 1000 r/min at 1.5 times the rated torque, find the duty ratioδ, neglecting friction
and windage, and using the equivalent circuit withjXmmoved adjacent to the stator terminals.
Solution
Full-load torque without rotor resistance control is
T=
3
ωS
[
V 12 (R′ 2 /S)
(R 1 +R′ 2 /S)^2 +(Xl 1 +X′l 2 )^2
]
WithV 1 = 440 /
√
3 =254 V;ωS=125.7 rad/s, and full-load slip=( 1200 − 1170 )/ 1200 = 0 .025,
then
T=
3
125. 7
[
2542 ( 0. 5 / 0. 025 )
( 0. 5 + 0. 5 / 0. 025 )^2 + 42
]
= 70 .6N·m