-
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because the backward equivalent circuit (for small S) is strongly inductive and
the slip increases notably.
It is therefore not practical to let a fully loaded three-phase IM operate a
long time with one phase open.
Example 7.7 One stator phase is open
Let us consider a three-phase induction motor with the data VL = 220V, f1 = 60
Hz, 2p1 = 4, star stator connection Rs = Rr2 = 1 &!, Xsl = Xrl2 = 2.5 &!, X1m = 75
&!, R1m = " (no core losses). The motor is operating at S = 0.03 when one stator
phase is opened. Calculate: the stator current, torque, and power factor before
one phase is opened and the slip, current, and power factor for same torque with
one phase open.
Solution
We make use of Figure 7.26a.
With balanced voltages, only Zf counts.
Xsl + X1m 75 + 2.5
Va = Zf Ia; C1 = = = 1.033 (7.130)
X1m 75
© 2002 by CRC Press LLC
8
8
Author: Ion Boldea, S.A.Nasar& & & & ..& & & ..
VL
3
Ia=Ib=IcH" =
2
Rr '
ëøR + C1 öø 2
+ (Xsl + C1Xrl')
ìø ÷ø
s
S
íø øø
(7.131)
220
3
= = 3.55A ; cosÕ = 0.989!
2
1
ëø1+1.033 öø 2
+ (2.5 +1.033Å" 2.5)
ìø ÷ø
0.03
íø øø
The very high value of cosÕ indicates that (7.131) does not approximate the
phase of stator current while it correctly calculates the amplitude. Also,
jX1m 75
Ir '= Is = 3.55Å" = 3.199A (7.132)
2
Rr '
1
+ j(X1m + Xrl') ëø öø 2
+ (75 + 2.5)
ìø ÷ø
S
0.03
íø øø
From (7.121),
3Rr ' p1 1 2
Te3 = Ir '2 = 3Å" Å"3.1992 = 5.4338Nm (7.133)
S É1 0.03 2À60
It is not very simple to calculate the slip S for which the IM with one phase
open will produce the torque of (7.133).
To circumvent this difficulty we might give the slip a few values higher
than S = 0.03 and plot torque Te1 versus slip until Te1 = Te3.
Here we take only one slip value, say, S = 0.05 and calculate first the
current Ib from (7.128), then Iaf = Iab from (7.125), and, finally, the torque Te1
from (7.121).
VL VL
Ib= H" =
Zf + Zb Rr '
jX1m Rr '
S
2(Rs + jXsl)+ + + jXrl'
Rr '
2 - S
+ j(X1m + Xrl')
S
(7.134)
220 220
= = = 9.2789A
1 23.709
j75
1
0.05
2(1+ j2.5)+ + + j2.5
1
2 -S
+ j(75 + 2.5)
0.05
20.34
cosÕ1 = = 0.858
23.709
© 2002 by CRC Press LLC
Author: Ion Boldea, S.A.Nasar& & & & ..& & & ..
From (7.125),
Ib 9.2789
Iaf = Iab = = = 5.3635A (7.135)
5.73
3
Figure (7.26) yields
jX1m 5.3635Å" jÅ" 7.5
Irf ' = Iaf = = 5.0258A
Rr ' 1
(7.136)
+ j(X1m + Xrl') + j(75 + 2.5)
S 0.05
Irb'H" Iab = 5.3635A
Now, from (7.121), the torque Te1 is
îø5.02582 5.36352 ùø
3Å" 2 Å"1
Te1 = - = 7.809Nm (7.137)
ïø úø
2À60 0.05 2 - 0.05úø
ïø
ðø ûø
In this particular case, the influence of a backward component on torque has
been small. A great deal of torque is obtained at S = 0.05, but at a notably large
current and smaller power factor. The low values of leakage reactances and the
large value of magnetization reactance explain, in part, the impractically high
power factor calculation. For more correct calculations, the complete equivalent
circuit is to be solved. The phase b current is Ib = 9.2789A for phase a open at S
= 0.05 and only 3.199A for balanced voltages at S = 0.03.
7.15 UNBALANCED ROTOR WINDINGS
Wound rotors may be provided with external three-phase resistances which
may not be balanced. Also, broken bars in the rotor cage lead to unbalanced
rotor windings. This latter case will be treated in Chapter 13 on transients.
Te
Zbr
Zar Zcr
Te Tef B
A
B
A
S
cr 3~
1/2
1 0
ar br
Teb
a.)
Figure 7.32 Induction motor with unbalanced wound rotor winding a.) and
torque / speed curve b.)
© 2002 by CRC Press LLC
Author: Ion Boldea, S.A.Nasar& & & & ..& & & ..
However, the wound rotor unbalanced windings (Figure 7.32) may be
treated here with the method of symmetrical components.
We may start decomposing the rotor phase currents Iar, Ibr and Icr into
symmetrical components:
1
Iarf = (Iar + aIbr + a2 Icr)
3
1
Iar b = (Iar + a2 Ibr + aIcr) (7.138)
3
1
Iar0 = (Iar + Ibr + Icr)= 0
3
Also, we do have
Var = -Zar Iar; Vbr = -Zbr Ibr; Vcr = -Zcr Icr (7.139)
In a first approximation, all rotor currents have the frequency f2 = Sf1 at
steady state. The forward mmf, produced by Iarf, Ibrf, Icrf, interacts as usual with
the stator winding and its equations are
Ir 'f Rr '-Vr 'f = - jSÉ1Èr 'f ; Èr 'f = Lr 'Ir 'f +L1m Isf
(7.140)
Isf Rs - Vs = - jÉ1Èsf ; Èsf = Ls Isf + L1m Ir 'f
The backward mmf component of rotor currents rotates with respect to the
stator at the speed n12 .
f1 f1
n1'= n - S = (1- 2S) (7.141)
p1 p1
So it induces stator emfs of a frequency f12 = f1(1 2S). The stator may be
considered short-circuited for these emfs as the power grid impedance is very
small in relative terms. The equations for the backward component are
Ir 'b Rr '-Vr 'b = - jSÉ1Èr 'b ; Èr 'b = Lr 'Ir 'b +L1m Isb
(7.142)
IsbRs = - j(1- 2S)É1Èsb; Èsb = Ls Isb + L1m Ir 'b
For given slip, rotor external impedances Zar, Zbr, Zcr, motor parameters
stator voltage and frequency, Equation (7.139) and their counterparts for rotor
voltages, (7.140) through (7.142) to determine Irf, Irb, Isf, Isb, Vr2 f, Vr2 b. Note that
(7.140) and (7.142) are, per-phase basis and are thus valid for all three phases in
the rotor and stator.
The torque expression is
* *
Te = 3P1L1m[Imag(Isf Ir 'f )+ Imag(Isb Ir 'b )]= Tef + Teb (7.143)
© 2002 by CRC Press LLC
Author: Ion Boldea, S.A.Nasar& & & & ..& & & ..
The backward component torque is positive (motoring) for 1 2S
> ½ and negative (braking) for S
is motoring. Also for S = ½ Isb = 0 and, thus, the backward torque is zero. This
backward torque is also called monoaxial or Geörge s torque.
The torque/speed curve obtained is given in Figure 7.33. A stable zone, AA2
around S = ½ occurs, and the motor may remain hanged around half ideal no-
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