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Unformatted text preview: frequency f 2 of the induced voltage and
current in the rotor circuit will correspond to this slip rpm, because
this is the relative speed between the rotating field and the rotor
winding. Thus, from Equation (4.9): f2 = p
120 (ns − n ) = p
120 sns = sf1 (4.11) ThreePhase Induction Machine 227 This rotor circuit frequency f 2 is also called slip frequency.
The voltage induced in the rotor circuit at slip s is: E2 s = 4.44 f 2 N 2φ p KW 2 = 4.44 sf1 N 2φ p KW 2 = sE2 (4.12)
Where E2 is the induced voltage in the rotor circuit at standstill,
that is, at the stator frequency f1 .
The induced currents in the threephase rotor windings also
produce a rotating field. Its speed (rpm) n2 with respect to rotor is: n2 = 120 f 2 120 sf1
=
= sns
p
p (4.13) Because the rotor itself is rotating at n rpm, the induced rotor
field rotates in the air gap at speed n + n2 = (1  s)ns + sns = ns
rpm. Therefore, both the stator field and the induced rotor field
rotate in the air gap at the same synchronous speed ns. The stator
magnetic field and the rotor magnetic field are therefore stationary
with respect to each other. The interaction between these two
fields can be considered to produce the torque. As the magnetic
fields tend to align, the stator magnetic field can be visualized as
dragging the rotor magnetic field. 228 Chapter Four
Example 4.1 A 3phase, 460 V, 100 hp, 60 Hz, fourpole
induction machine delivers rated output power at a slip of 0.05.
Determine the:
(a) Synchronous speed and motor speed.
(b) Speed of the rotating air gap field.
(c) Frequency of the rotor circuit.
(d) Slip rpm.
(e) Speed of the rotor field relative to the (i) rotor structure. (ii)
Stator structure. (iii) Stator rotating field.
(f) Rotor induced voltage at the operating speed, if the
statortorotor turns ratio is 1 : 0.5.
Solution:
(a) n s = 120 f
p = 120 * 60
= 1800 rpm ,
4 n = (1 − s )ns = (1 − 0.05) * 1800 = 1710 rpm
(b) 1800 rpm (same as synchronous speed)
(c) f2 = sf1 = 0.05 x 60 = 3 Hz.
(d) slip rpm = s ns = 0.05 * 1800 = 90 rpm
(e) (i) 90 rpm (ii) 1800 rpm (iii) 0 rpm (f) Assume that the induced voltage in the stator winding is the
same as the applied voltage. Now, E2 s = sE2 = s N2
460
E1 = 0.05 * 0.5 *
= 6.64V / Phase
N1
3 229 ThreePhase Induction Machine 4.6 Equivalent Circuit of the Induction Motor
In this section we develop the equivalent circuit from basic
principles. We then analyze the characteristics of a lowpower and
highpower motor and observe their basic differences.
Finally, we develop the equivalent circuit of an asynchronous
generator and determine its properties under load.
A 3phase woundrotor induction motor is very similar in
construction to a 3phase transformer. Thus, the motor has 3
identical primary windings and 3 identical secondary windings one
set for each phase. On account of the perfect symmetry, we can
consider a single primary winding and a single secondary winding
in analyzing the behavior of the motor.
When the motor is at standstill, it acts exactly like a
conventional transformer, and so its equivalent circuit (Fig.4.6) is
the same as that of a transformer, previously developed. Fig.4.6 Equivalent circuit of a woundrotor induction motor at
standstill. 230 Chapter Four
In the case of a conventional 3phase transformer, we would be
justified in removing the magnetizing branch composed of jX m and Rm because the exciting current I o is negligible
compared to the load current I P . However, in a motor this is no
longer true: I o may be as high as 40 percent of I P because of the
air gap. Consequently, we cannot eliminate the magnetizing
branch.
The stator and rotor winding can be represented as shown in
Fig.4.7 (a) and (b), where,
V1 = perphase terminal voltage.
R1 = perphase stator winding resistance.
L1 = perphase stator leakage inductance.
E1 = perphase induced voltage in the stator winding
Lm = perphase stator magnetizing inductance
Rc = perphase stator core loss resistance. E2 = Perphase induced voltage in rotor at standstill (i.e., at stator
frequency f,) R2 = Perphase rotor circuit resistance
L2 = Perphase rotor leakage inductance
Note that there is no difference in form between this equivalent
circuit and that of the transformer primary winding. The difference ThreePhase Induction Machine 231 lies only in the magnitude of the parameters. For example, the
excitation current I o is considerably larger in the induction
machine because of the air gap. In induction machines it is as high
a: 30 to 50 percent of the rated current, depending on the motor
size, whereas it is only 1 to 5 percent in transformers. Moreover, the
leakage reactance X 1 is larger because of the air gap and also
because the stator and rotor windings are distributed along the
periphery of the air gap rather than concentrated on a core, as in the
transformer. 232 Chapter Four Fig.4.7 Development of the induction machine equivalent circuit
Note that the rotor circuit frequency and current are: f 2 and I 2 . I2 = sE2
R2 + jsX 2 (4.14) T...
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This document was uploaded on 03/12/2014 for the course ENGINEERIN electrical at University of Manchester.
 Spring '14

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