MEC 3102 – PRODUCTION ENGINEERING I AND

ELECTRICITY & ELECTRONICS II

Mr. Boyd Munkombwe

Engineering Annex Board Room
Cell: (+260)-968315273/50435239/72860920
Email: boyd.munkombwe@unza.zm

MEng in Power Electronics and Motor Control (Southeast; China, 2020)

BEng in Electrical Power System and Machines (UNZA; January, 2017)
 LECTURE 6[1]
Induction Motor
 Introduction
• Three-phase induction motors (IM) are the most common and frequently
encountered machines in industry. The are mostly referred to as the workhorses
of the industry.
• This is due to:
✔ simple design, rugged in construction, low-price, easy maintenance
✔ wide range of power ratings: fractional horsepower up-to 10 MW
✔ run essentially as constant speed from zero to full load
✔ speed is power source frequency dependent
• not easy to have variable speed control. This is because power source has
a constant frequency source of 50 Hz or 60 Hz
• requires a variable-frequency power-electronic drive for optimal speed
control (so as to alter the source frequency)

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Construction
• An induction motor has two main parts
• a stationary stator:
• consisting of a steel frame that supports
a hollow, cylindrical core.
• core, constructed from stacked
laminations (why?), having a number of
evenly spaced slots, providing the space
4.1 Stator of IM
for the stator winding
❖ Stacked lamination reduce Eddy currents,
thereby allowing the stator core to maintain
contant power.

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 • a revolving rotor
• composed of punched laminations, stacked to create a series of rotor
slots, providing space for the rotor winding
• one of two types of rotor windings
• conventional 3-phase windings made of insulated wire (wound-rotor)
» similar to the winding on the stator
• aluminum bus bars shorted together at the ends by two aluminum
rings, forming a squirrel-cage shaped circuit (squirrel-cage)
• Two basic design types depending on the rotor design
• squirrel-cage
• wound-rotor

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 Squirrel cage rotor

Wound rotor

Notice the
slip rings

Fig. 4.2 rotor tyrpes

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Slip rings

Fig. 4.3
Cutaway in a typical
wound-rotor IM.
Notice the brushes
and the slip rings

Brushes
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 Rotating Magnetic Field

Fig. 4.4 rotating field

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Principle of operation
The general principle of operation of IM is similar to that of d.c. motors.
• This rotating magnetic field cuts the rotor windings and produces an induced
voltage in the rotor windings.
• Due to the fact that the rotor windings are short circuited, for both squirrel cage
and wound-rotor, and induced current flows within the rotor windings.
• The rotor current produces another magnetic field opposite to the main field in
order to preserve the m.m.f. balance in the iron core
• A torque is produced as a result of the interaction of those two magnetic fields

Where, Tind is the induced torque and BR and BS are the magnetic flux densities of
the rotor and the stator respectively.

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Induction motor speed
• At what speed will the IM run?
• Can the IM run at the synchronous speed, why?
• If rotor runs at the synchronous speed, which is the same speed of the
rotating magnetic field, then the rotor will appear stationary to the rotating
magnetic field and the rotating magnetic field will not cut the rotor. So, no
induced current will flow in the rotor and no rotor magnetic flux will be
produced so no torque is generated and the rotor speed will fall below the
synchronous speed
• When the speed falls, the rotating magnetic field will cut the rotor windings
and a torque is produced.
• So, the IM will always run at a speed lower than the synchronous speed.
• The difference between the motor speed and the synchronous speed is called
the Slip.

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Where nslip= slip speed
ns = speed of the magnetic field
nr = mechanical shaft speed of the motor

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Examples
1. A 208-V, 10hp, 4-pole, 60 Hz, Y-connected induction motor has a full-load slip
of 5 percent.
i. What is the synchronous speed of this motor?
ii. What is the rotor speed of this motor at rated load?
iii. What is the rotor frequency of this motor at rated load?
iv. What is the shaft torque of this motor at rated load?

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Solution

1.

2.

3.

4.

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2. A 220 V, 3-phase, 2-pole, 50 Hz induction motor is running at a slip 5%. Calculate:
i. The speed of the magnetic fields in rpm.
ii. The speed of the rotor in revolutions per minute.
iii. The slip speed of the rotor.
iv. The rotor frequency in Hertz

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Equivalent Circuit

Torque is produced by the interaction of a stator-bound flux wave and an induced

rotor-bound current wave.
▪ Since the flux and current waves are internal to the machine, it is very difficult to
measure these on load and use them for machine performance calculations.
▪ However, the similarities between the induction motor and the transformer
permit the adaptation of the equivalent circuit model of the transformer to
represent an induction motor.
▪ As such, an electrical equivalent circuit model can be used to predict the torque/
speed characteristics and efficiency of the motor, in terms of machine currents,
voltage, resistances and reactances – instead of flux and current waves.

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 Fig. 4.5 Single-phase equivant
circuit of the 3-phase IM

Where, R1 = per-phase resistance of the stator

X1 = per-phase reactance of the stator
Xm = per-phase magnetizing reactance of the IM
Rc = core loss resistance
The airgap of the IM is represented by the ideal transformer model with the primary
being assumed to have N1 turns and the secondary have N2 turns.
The magnitude of the e.m.f. wave induced in the rotor is a function of slip, sE2 − E2
being the rotor e.m.f. at standstill.
R2 = series resistance
X2 = series reactancce
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• The effective reactance, sX2, varies with the frequency of the induced rotor e.m.
fs, X2 being the value when the rotor is stationary (when s = 1).
• As the motor speeds up, the slip is reduced and the effective reactance of the
rotor cage is reduced.
• The equivalent circuit is simplified by making two assumptions:
1. That the stator and rotor windings have the same number of turns per phase.
2. That the airgap flux has a constant amplitude and speed.
As there is an ampere-turn (m.m.f.) balance between stator and rotor:
I 1 N 1 = I2 N 2
Using assumption (1), N 1 = N2, it follows that

(4.4)

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• If φ is the phase difference between I2 and E2, the rotor current and voltage:

(4.5)

and

(4.6)

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• Equation (4.6) suggests the following simplified circuit by referring the rotor
impedance to the stator. This facilitates the calculation of the rotor current referred
to the stator.

Fig. 4.6 Simplified

equivalent circuit of the IM

• Since a large m.m.f. is required to get the working flux across the airgap, the
magnetizing current in an induction motor can be 30% of the full load current
of the machine – and must be included when estimating machine performance.
In a transformer, the magnetizing current is usually only 5% of the full load
current so the magnetizing branch can often be ignored.
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Mechanical power and torque
• If the power losses in the iron core of the stator are neglected, the equivalent
circuit model in fig. 4.7, the real power taken from an a.c. supply is given by:

4.7 Final per-phase equivalent circuit of the induction motor

W/ph (4.7)
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The power dissipated in the resistance of stator and rotor windings is:

W/ph (4.8)

The difference between these two quantities must be the power dissipated in the load, i.
e. the mechanical shaft power Pm:

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This is a per-phase quantity so the total three-phase mechanical power developed is

Equation (4.9) leads to the final and most useful form of the equivalent circuit,
shown in the fig. 4.7 above , which allows the mechanical power and the rotor
losses to be perated.
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For a rotor speed of nr rpm,

Combining equations (4.8), (4.9) and (4.11), the torque is

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Power flow in induction motor
The power flow through the motor and the efficiency can thus be calculated and
are shown in Fig. 4.8.
✔ Friction and windage loss has been included
In the mathematical analysis above, this loss has been included in the
mechanical power developed by the motor

4.8 Real power flow through an induction motor 24

Example:
The 415 V, three-phase, 50 Hz, star-connected induction motor shown in Fig. 4.9 has
the following per-phase equivalent circuit parameters:
R1 = 1Ω, X1 = 3 Ω, Xm = 60 Ω, Rc = 240 Ω, R′2 = 1 Ω, X′2 = 2 Ω.
Using the final per-phase equivalent circuit of the machine, calculate the current
drawn from the supply, and draw a diagram (as Fig. 4.8) showing the flow of real
power through the machine when running with a slip of 5 percent. The friction and
windage loss in the machine is 200 W. Calculate the efficiency of the motor.

Solution:

4.9 Equivalent circuit for the Example

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26
(g) Power flow diagram for the Example

(h) Efficiency of motor,

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• Neglecting the magnetizing component of current in the equivalent circuit of the
induction motor, we can calculate the torque delivered to the mechanical load driven
by the induction motor from equation (4.12).

• Where I1 is given by,

Substituting for the current in the torque equation, we obtain the following expression;

where V1 is the r.m.s. voltage per phase

Equation (4.13) shows the dependency of motor torque on slip (assuming that system
voltage and frequency are constant) and a typical torque–speed curve.
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Torque-speed characteristics
The graph shows what happens in terms of output torque and motor speed when a
motor is started with full voltage applied.
▪ The motor is initially stationary and C
develops locked-rotor torque (point A).
▪ As the motor accelerates, some motor
designs produce a slight dip in torque, A
the lowest point being called the pull-in B
D
or pull-up torque (point B).
▪ As the speed increases further, the
E
torque reaches the highest point on the
curve (point C), the pull-out or
breakdown torque. 4.10 Typical torque-speed
characteristics of induction motor

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▪ Finally, when the motor is loaded to its full-load torque, the motor speed stabilizes
(point D). If the motor is not driving anything, the speed increases to the no-load or
synchronous speed (point E)
Assuming the impedance of the stator to be small for a given supply voltage, then, for
a given a voltage and frequency, the torque equation simplifies to;

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• The value of is typically large compared with the rotor resistance. The
dependency of τm on slip and rotor resistance can be illustrated by an example, in
which = 8 Ω.
• Calculating τm for different values of slip and rotor resistance, the resulting curves
being close to the exact shape of typical torque–speed curves. Note that, when slip
is zero, speed reaches synchronous speed.

4.11 Torgue/slip curves of an

induction motor

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Important conclusions:
i. Increasing the rotor resistance moves the maximum value of torque towards
higher slip values, that is, in the direction of lower speed. Thus, a higher rotor
resistance gives a higher starting torque.
ii. A higher rotor resistance gives a lower on-load efficiency but a higher starting
torque. A lower rotor resistance gives less starting torque but higher on-load
efficiency.
iii. τm is a constant. This can be explained by differentiating equation (4.14) for
torque with respect to slip and equating to zero:

By substituting for back in equation (4.14), it is revealed that τmax is

independent of

Note that is the leakage reactance at standstill and is a constant for a given rotor.
Hence the maximum torque is the same, whatever the value of rotor resistance. 32
END OF LECTURE!

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