Electric Machines
CH 7 Induction Motor
By
Hassan Khalid
Department of Electrical Engineering
7.1 Induction Motor Construction
• There are two main types of rotor
Induction motors
• Cage rotor
• Wound rotor
• The figure to the right is a cage rotor
type induction machine
• It contains series of conducting bars
carved in the face of rotor shorted by
using shorting rings
• A wound rotor has a complete set of
three phase windings similar to
winding on stator and are usually Y
connected. While other end is
connected to slip rings on the rotor
shaft
• The wound rotor have there rotor
current accessible at the stator
brushes
• Disadvantage
• Wound rotor type motor are
expensive as compared to cage
rotor type
• Wound rotor type motor require
more maintenance because
brushes and slip rings wear out
with time
6.2 Basic Induction Motor Concept
• IM operation is same as
that of damper or amor-
tisseur winding
The development of induced torque in IM
• Consider the figure of cage rotor type IM. As the three phase voltage is
applied to the stator, three phase stator current will start to flow.
• The current produces a magnetic field BS which rotates counter
clockwise
• The speed of rotating magnetic field is given by
• Where fe is system frequency and P is number of poles
• The voltage induced is given by
• The velocity of the upper rotor bars relative to the magnetic field is to
the right, so the induced voltage in the upper bars is out of the page,
while the induced voltage in the lower bars is into the page.
• This results in a current flow out of the upper bars and into the lower
bars. However, since the rotor assembly is inductive, the peak rotor
current lags behind the peak rotor voltage
• The rotor current flow produces a rotor magnetic field BR.
• The rotating stator field
Bs induces a voltage in
the rotor bars
• the rotor voltage produces
a rotor current flow,
which lags behind the
voltage because of the
inductance of the rotor
• The rotor current
produces rotor magnetic
field that lags behind by
90o.
• There is a finite upper limit to the motor's speed, however. If the
induction motor 's rotor were turning at synchronous speed, then the
rotor bars would be stationary relative to the magnetic field and there
would be no induced voltage.
• If eind were equal to 0, then there would be no rotor current and no
rotor magnetic field.
• With no rotor magnetic field, the induced torque would be zero, and
the rotor would slow down as a result of friction losses.
• An induction motor can thus speed up to near-synchronous speed, but
it can never exactly reach synchronous speed
The Concept of Rotor Slip
• The voltage induced in rotor bar of and IM depends on the speed of
rotor relative to the magnetic fields
• Since the behavior of induction motor depends on rotor voltage &
current and relative speed
• Two terms defines the rotor & magnetic field
• Slip speed
• slip
• The slip speed is the difference between synchronous speed and rotor
speed
• The term used to define motion is called slip given by
• If s is 0 it means rotor is running at synchronous speed
• If s is 1 it means rotor is stationary
Electric frequency on rotor
• Due to its induced voltage and current IM is some times called as an
rotating transformer
• But unlike transformer its frequency varies
• At nm = 0, s=1 and fr=fm
• At nm = nsyn, s=0 and fr=0
• Therefor
Example : 7.1
• Data: • What is synchronous speed ?
• VT= 208 V
• P = 10 Hp
• Poles = 4
• F = 60 Hz
• Connection = Y • What is rotor speed at rated load?
• S=5%
Continue
• What is a rotor frequency of this motor at rated load?
• What is rated torque of this motor at rated load?
7.3: THE EQUIVALENT CIRCUIT OF AN
INDUCTION MOTOR
• Induction motor is just like a transformer inter of its operation
• It is called singly exited machine
• Because an induction motor does not have an independent field
circuit, its mode will not contain an internal voltage source such as
the internal generated voltage EA in a synchronous machine .
Equivalent model
• Where R1 and X1 are the stator
resistance and reactances respectively
• like any transformer with an iron core,
the flux in the machine is related to the
integral of the applied voltage El .
• The higher reluctance caused by the air
gap means that a higher magnetizing
current is required to obtain a given
flux level.
• Therefore, the magnetizing reactance
XM in the equivalent circuit will have a
much smaller value (or the susceptance
BM will have a much larger value) than
it would in an ordinary transformer
• The primary internal stator voltage El is coupled to the secondary ER by an
ideal transformer with an effective turns ratio aeff
• The effective turns ratio aeff is fairly easy to determine for a wound-rotor
motor- it is basically the ratio of the conductors per phase on the stator to the
conductors per phase on the rotor
• The primary impedances and the magnetization current of the
induction motor are very similar to the corresponding components in a
transformer equivalent circuit.
• An induction motor equivalent circuit differs from a transformer
equivalent circuit primarily in the effects of varying rotor frequency on
the rotor voltage ER and the rotor impedances RR and jXR
The rotor circuit model
• In an induction motor, when the voltage is applied to the stator
windings, a voltage is induced in the rotor windings of the machine,
• In general, the greater the relative motion between the rotor and the
stator magnetic fields, the greater the resulting rotor voltage and rotor
frequency
• The largest relative motion occurs when the rotor is stationary, called
the locked-rotor or blocked-rotor condition, so the largest voltage and
rotor frequency are induced in the rotor at that condition
• The smallest voltage (0 V) and frequency (0 Hz) occur when the rotor
moves at the same speed as the stator magnetic field, resulting in no
relative motion
• Therefore, if the magnitude of the induced
rotor voltage at locked-rotor conditions is
called ERo, the magnitude of the induced
voltage at any slip will be given by the
equation
• The frequency of induced voltage at any slip
is given by
• The reactance of an induction motor rotor
depends on the inductance of the rotor and
the frequency of the voltage and current in
the rotor. With a rotor inductance of LR, the
rotor reactance is given by
• The rotor current flowing in the circuit is
given by
The Final Equivalent Circuit
• Refer rotor side over stator side
7.4: Power and Torque in Induction Motor
• Since the induction machines are singly exited machines so their power
and torque relations are different from synchronous machines
• In transformer the output is considered for electrical output but in
induction motor the secondary side is shorted out
• The output of induction machine is a mechanical and input is electrical
• As the power is given to the stator
• Stator consumes the current
• The first loss encountered in the machine is stator copper loss is I 2 R (PSCL)
• The second loss is due to the hysteresis and eddy current loss is called core
loss (Pcore)
• The remaining power is transferred to the rotor across the air gap between
rotor and stator. This loss is called air gap loss (PAG)
• The remaining power is transferred to rotor and the copper loss associated
there is rotor copper loss (PRCL)
• Remaining are the stray and windage loss (Pstray)
• Finally mechanical power is delivered.
Example 7.2:
• A 480 V, 60 Hz, 50 hp, three-phase induction motor is drawing 60 A at
0.85 PF lagging. The stator copper losses are 2 kW, and the rotor
copper losses are 700 W. The friction and windage losses are 600 W,
the core losses are 1800 W, and the stray losses are negligible. Find the
following quantities:
(a) The air gap power (PAG)
(b) The power converted (Pconv)
(c) The output power (Pout)
(d) Find the efficiency
Power & Torque in an Induction Machine
• The input current per
phase is given by
• Therefore the stator
copper loss and core
loss is given by
• Air gap power loss
• The only element
that is consuming
the power due to
airgap is R2/s
• Therefore we can
simplify our
equation
• The actual resistive loss
in rotor circuit is given
by
• The actual resistive loss
in rotor circuit is given
by equation
• But as the power
remains same so copper
loss can be expressed as
• After calculating copper loss,
core loss and rotor copper loss
we can find the power
transformed from electrical to
mechanical
• Subtract them from input to
get converted power
7.5: Induction Motor Torque Speed
Characteristics
• An induction motor with cage type
rotor running at a very light load near
to no load condition nearly runs at a
synchronous speed
• The net magnetic field Bnet is produced
due to magnetizing current and is
directly proportional to E1
• If E1 is constant then Bnet stays
constant
• In reality E1 changes as the load
changes and this leads to cause
variation in stator resistances and
reactances R1 and X1
• As the induction motor is on no load there
will be a very small slip and it means that the
relative motion between rotor and magnetic
field is very small and thus the
corresponding frequency is very small
• As the relative motion is small the induced
rotor voltage ER will also be small
• Thus the rotor current IR will also be small
• As slip is small rotor frequency fR will also
be small and rotor reactance XR will also be
small
• As IR is small BR will also be small
• But stator current IS will be large
in order to keep most of Bnet
• As the motor is loaded
• The slip of machine increases and
rotor speed falls
• Now there is more relative motion
between stator magnetic field and
rotor
• Greater relative motion produces more
E1 and thus more rotor current IR will
be induced thus producing high BR
• Now the Slip is high the rotor
frequency fR will be high and the
reactance also increases
• With the increase of BR the rotor torque will also be increased
• By increasing torque a point comes when the load on the shaft is
increased but the factor sin (d) decreases more than BR increases
• At this point by increasing more torque the rotor will be stopped
• This condition is called pullout torque
