Environmental Engineering Reference
In-Depth Information
the stator. The larger the applied braking torque the higher the slip, the larger are the
induced EMFs and resulting rotor currents, and the stronger the interaction between the
two RMFs to produce an electrical torque equal and opposite to the braking torque. The
induction motor therefore exhibits a small decrease in speed with increments in braking
torque.
An ideal induction machine could be imagined to operate at zero slip. This is equivalent
to the 'idling' state of the fi xed speed synchronous machine. The vital difference, however,
in the induction machine is that motoring or generating torques will be accompanied by a
decrease or increase of speed below or above synchronous respectively. For generating, Equa-
tion (4.16) now gives a negative slip. The induction machine will move seamlessly from the
motoring into the generation mode as the external torque changes from a decelerating to an
accelerating type. Indeed, in small wind turbines, it is very common to fi nd that the induction
generator was originally designed as a motor and has been employed as a generator without
any modifi cation.
Worked Example 4.2
A six-pole 50 Hz induction motor runs at 4% slip at a certain load. Calculate the synchronous
speed, the rotor speed, the frequency of the rotor currents, the speed of the rotor RMF with
respect to the rotor and the speed of the rotor RMF with respect to the stator.
Model answer
The synchronous speed from Equation (4.4) is
N
s
=
f
/
p
= 50/3 rev/s = 50
×
60/3 =
1000 rev/min
The rotor speed from Equation (4.16) is (1
−
s
)
N
s
= (1
−
0.04)
×
1000 = 960 rev/s
The frequency of the rotor currents are:
f
r
=
sf
= 0.04
×
50 = 2 Hz
The speed of the rotor RMF with respect to the rotor:
Nf
=×
60
p
=×
2
60 3
=
40
rev min
rr
r
The speed of the rotor RMF with respect to the stator:
NN
r
+= +=
960
40
1000
rr
(i.e. the rotor and stator RMFs rotate together)
4.4.2 The Induction Machine Equivalent Circuit
The induction machine can be viewed as a transformer with a rotating secondary. Imagine
an induction machine with its rotor mechanically locked, i.e. at standstill. The stator RMF
will be rotating at
s
with respect to the rotor and inducing in each phase the voltage
E
2
at
mains frequency
f
. The current that fl ows in each phase will be
ω
E
E
2
2
I
=
=
(4.19)
2
Rj
+
2
π
fL RjX
+
2
2
2
2
where
R
2
and
L
2
are the effective per-phase resistance and inductance of the rotor winding
and
X
2
is the rotor reactance at mains frequency. At standstill the slip
s
= 1 and the rotor
voltages and currents are of the stator frequency
f
. At any other rotor speed, the slip is
s
, the
