EET 4323

Electrical Machines
Chapter –III Induction Machines
Mr. Khalid. A
School of Electrical & Computer Engineering
Haramaya Institute of Technology
Haramaya University

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Outline
Induction Machines
➢Introduction to Induction Machines

➢Construction and Working Principle

➢Types of Induction Motors

➢Equivalent Circuit and Phasor Diagram

➢Performance Analysis

➢Testing and Starting Methods

➢Speed Control Methods

➢Single-phase Induction Motors

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 Class Objectives
Upon completing this chapter, you will
➢ Understand the construction and working principle of 3-phase and single-phase induction motors.
➢ Differentiate between squirrel cage and wound rotor induction motors.
➢ Analyze AC winding types and calculate winding factors (pitch and distribution).
➢ Understand the impact of winding factors on induced EMF and motor performance.
➢ Derive and apply the equivalent circuit of induction motors.
➢ Interpret phasor diagrams under different load conditions.
➢ Construct and analyze the circle diagram for performance parameters.
➢ Conduct no-load and blocked rotor tests for motor evaluation.
➢ Calculate torque, slip, efficiency, and losses from test data.
➢ Describe and compare various motor starting methods (DOL, star-delta, etc.).
➢ Explain speed control methods via stator and rotor side techniques.
➢ Identify industrial and domestic applications of induction motors.
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Introduction
What are Electric Machines?
➢ Electric Machine: A device that converts energy from one form to another form.
➢ Broad classification:
❖ DC Machines
❖ AC Machines

DC vs AC Machines – Key Differences

Feature DC Machines AC Machines

Type of Current Direct Current (unidirectional) Alternating Current (bidirectional)
Construction Commutator and brushes Slip rings (synchronous) or none (induction)
Maintenance High (due to brushes/commutators) Low (especially in squirrel cage)
Speed Control Easy and precise More complex, requires additional devices
Application Areas Electric vehicles, steel mills Fans, pumps, compressors, railways
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Introduction
Synchronous vs Asynchronous (Induction) Machines

Feature Synchronous Machine Asynchronous (Induction) Machine

Rotor Speed Equal to stator rotating field (Ns) Less than Ns (slip is present)
Slip 0% >0% (usually 2–6%)
Starting Needs external means (e.g. DC excitation) Self-starting (in 3-phase)
Construction Complexity More complex (needs excitation) Simpler, especially squirrel cage type
Applications Generators in power plants Motors in industries, households, etc.

Why Focus on Induction Machines? Examples:

•Widely used in industry (over 70% of industrial motors) •Elevators, fans, blowers, washing machines,
•Rugged and reliable air conditioners, conveyor belts.
•No brushes or commutators → low maintenance
•Cost-effective and energy-efficient
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Introduction
In a DC Motor: So, Can a DC Motor be Called a "Conduction
Machine"? Not exactly — here's why:
• Supply is given directly to the armature via
Why Not "Conduction Machine"?
brushes and commutator.
•The term "conduction machine" isn’t standard in
• The stator in a DC motor is typically stationary electrical engineering.
and provides a magnetic field (either via permanent •But if you're referring to the method of energy
magnets or field windings). transfer, yes:
• The armature (rotor) is the part that rotates, not • In a DC motor, electrical energy is supplied
the stator. directly to the rotor (armature) through
physical contact — this is called conduction.
• The commutator and brushes are mechanical
• In contrast, in an induction motor, energy is
devices that maintain electrical contact with the
transferred to the rotor without direct
rotating armature, allowing direct conduction of
electrical contact, using electromagnetic
current
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into the armature windings. Mr. Khalid A. 6
induction.
Introduction
➢ Also known as an asynchronous motor.
➢ Consists of two main parts:
❖ Stator – stationary part connected to AC power supply.
❖ Rotor – rotating part, separated from stator by an air gap.
➢ No direct electrical connection to the rotor.
➢ Rotor current is induced by the stator’s rotating magnetic field.
➢ Energy is transferred through electromagnetic induction
➢ Operates like a rotating transformer:
❖ Stator = Primary winding (stationary)
❖ Rotor = Secondary winding (rotates)
➢ Rotor always rotates at a speed less than the stator's magnetic field speed — this difference is called slip.
➢ If rotor speed equals magnetic field speed, no current is induced, and torque is zero.
➢ This is why it’s called an induction motor — the rotor current is induced, not supplied.
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Introduction
➢ Induction machines are rotating electromechanical energy-converting devices.
➢ Their operating principle is similar to that of all other rotating electrical machines.
•Based on two fundamental electromagnetic laws:
1.Generator Action:– When a conductor moves in a magnetic field, a voltage is induced in the conductor.
2.Motor Action:– When a current-carrying conductor is placed in a magnetic field, it experiences a
mechanical force.
❖ An induction machine is an AC machine that can operate as:
▪ An induction motor (most common use).
▪ An induction generator (used in special cases like wind turbines or braking systems).
➢ In motor mode, electrical energy is converted into mechanical energy.
➢ In generator mode, mechanical energy is converted into electrical energy.

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Applications of Induction Motors
•Induction motors are available in a wide range of sizes and are used in both industrial and domestic settings.
Large Three-Phase Induction Motors Small Single-Phase Induction Motors
(Typically rated in tens or hundreds of horsepower) (Usually fractional horsepower rating)
Used in: Used in:
❖ Pumps and compressors ➢ Household appliances like:
❖ Industrial fans ❖ Juice mixers
❖ Paper mills ❖ Washing machines
❖ Textile mills ❖ Refrigerators
❖ Heavy machinery and processing plants ❖ Fans and blowers
Polyphase (Three-Phase) Induction Motors
Induction generators are used in:
•The most widely used type of AC motor for industrial applications.
❖ Wind energy systems
•Preferred for their: • High efficiency
❖ Regenerative braking etc.
• Robust construction & Low maintenance requirements
•Common
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Advantage & Disadvantage of Induction Machines
Advantages of Induction Motors Disadvantages of Induction Motors
•Simple and rugged construction – especially the squirrel • Speed control is difficult – cannot be
cage type. varied easily without loss of efficiency.
• Low cost and high reliability. • Speed drops with load increase, similar
• High efficiency under typical operating conditions. to a DC shunt motor.
• No commutator or brushes – reduces wear and maintenance • Starting torque is lower compared to DC
needs. shunt motors.
• Operates with a reasonably good power factor.
• Minimal maintenance required.
• Can start from rest without the need for:
• A separate starting motor.
• Synchronization process (unlike synchronous motors).

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Rotating Magnetic Field (RMF)
➢ A Rotating Magnetic Field (RMF) is a magnetic ❖ This means the magnetic poles on the stator are
field that has: not fixed but appear to rotate in space.
❖ Constant strength Proof of RMF Production

❖ Continuously rotating direction ➢ Take a 2-pole, three-phase distributed

How RMF Is Produced winding on the stator.

There are two ways to create RMF: ➢ Apply a balanced three-phase AC supply.

1.By rotating a permanent magnet ➢ Each phase produces an alternating

1. Requires physical rotation of the magnet to magnetic field.

generate the rotating field. ➢ These three fields combine vectorially to

2.By using a 3- ϕ AC supply (used in induction motors) form a single resultant magnetic field.

1. No moving parts required. ➢ This resultant field:

2. When a 3- ϕ AC supply is applied to 3- ϕ ❖ Has constant magnitude

distributed windings on the stator, an RMF is ❖ Rotates smoothly in space around the
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automatically created. stator
Rotating Magnetic Field (RMF)…
Proof for production of rotating magnetic field

ϕ𝑟 = ϕ𝑅 + ϕ𝑌 + ϕ𝐵

3 0 3 0
ϕ𝑅 = ϕ𝑚 sin 𝜔𝑡 At instant 1, 𝜔𝑡 = 00 ϕ𝑟 = 0 + cos 30 + cos 30
2 2
So,
ϕ𝑅 = ϕ𝑚 sin 0 = 0 ϕ𝑟 =1.5 ϕ𝑚
ϕ𝑌 = ϕ𝑚 sin(𝜔𝑡 − 1200 )
So, resultant flux is 1.5 times of the
3
ϕ𝐵 = ϕ𝑚 sin(𝜔𝑡 − 2400 ) ϕ𝑌 = ϕ𝑚 sin( − 1200 ) =− ϕ𝑚
2 maximum flux and at instant 1, it is
Since ϕ𝑅 = ϕ𝐴 , ϕ𝑌 = ϕ𝐵 , & ϕ𝐵 = ϕ𝐶 0) = 3 directed along reference line
ϕ𝐵 = ϕ𝑚Mr.
sin( − 240 ϕ𝑚
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Rotating Magnetic Field (RMF)…
At instant 2, 𝝎𝒕 = 𝟑𝟎𝟎 At instant 3, 𝝎𝒕 = 𝟔𝟎𝟎
ϕ 3
So, ϕ𝑅 = ϕ𝑚 sin( 300 ) = 𝟐𝑚 So, ϕ𝑅 = ϕ𝑚 sin( 600 ) = ϕ
2 𝑚
ϕ𝑌 = ϕ𝑚 sin(300 − 1200 )= ϕ𝑚 sin(−900 )= − ϕ𝑚 3
ϕ𝑌 = ϕ𝑚 sin( − 600 )= − ϕ
2 𝑚
ϕ𝑚
ϕ𝐵 = ϕ𝑚 sin(𝜔𝑡 − 2400 )= ϕ𝑚 sin(−2100 )= ϕ𝐵 = ϕ𝑚 sin( − 1800 )= 0
𝟐

ϕ𝑟 = ϕ𝑅 + ϕ𝑌 + ϕ𝐵

ϕ𝑚 ϕ𝑚 ϕ 𝑟 = ϕ 𝑅 + ϕ𝑌 + ϕ𝐵
ϕ𝑟 = cos 600 + ϕ𝑚 + cos 600
𝟐 𝟐
3 0 3 0
ϕ𝑟 =1.5 ϕ𝑚 ϕ𝑟 = cos 30 + cos 30 +0
2 2

ϕ𝑟 =1.5 ϕ𝑚

So, resultant flux is again 1.5 times of the maximum flux and
at instant 2 & 3, it is directed at angle of 300 & 600 from
reference respectively, in clock-wise direction.
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Rotating Magnetic Field (RMF)…
➢ So, from the above phasor diagrams, we see that a
rotating magnetic field of constant magnitude (1.5
ϕ𝑚 ) is setup, when a 3- ϕ balance supply is given
to 3- ϕ distributed windings.
➢ The speed at which the magnetic field rotates is
known as synchronous speed, denoted by N𝑠

120𝑓
N𝑠 = Where
𝑃
N𝑠 = synchronous speed
f =supply frequency
P = Number of poles of stator winding

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Construction of 3-ϕ Induction motor
➢ An induction motor consists essentially of two main parts: The stator consists of the following main parts:
A. Stator B. Rotor 1. Stator Frame
2. Stator Core
3. Stator Windings

A. Stator:
•The stator is the stationary part of an induction motor.
•It is responsible for producing a rotating magnetic field when
supplied with 3-phase AC power.
•This rotating magnetic field induces an EMF in the rotor by
mutual induction, which leads to the production of torque and
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motor rotation.
Construction of 3-ϕ Induction motor
1. Stator Frame:
2. Stator Core:
•The outer shell of the stator assembly, providing
•Made up of thin laminated silicon steel stampings to reduce
mechanical strength and protection.
eddy current losses.
•Mechanical Support:
•These laminations are slotted to hold the stator windings.
• Holds the stator core and windings firmly in
•Provides a low-reluctance path for the magnetic field.
place.
3. Stator Windings:
•Protection:
•Consist of copper or aluminum conductors, placed inside the
• Shields internal parts from dust, moisture, and
slots of the stator core.
physical damage.
•Connected to a 3-phase AC supply.
•Heat Dissipation:
•When energized, the windings produce a rotating magnetic
• Transfers heat to the surroundings, often
field of constant magnitude at synchronous speed.
enhanced by external cooling fins.
•Mounting Base:
• Includes provisions like feet or flanges for
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Construction of 3-ϕ Induction motor
➢ The number of poles (P) in the stator is determined
Air Gap Considerations:
by the winding configuration, where:
•The air gap between the stator and rotor should be as
P=2n
small as practically possible.
n = number of stator slots per pole per phase.
• A small air gap:
The speed of the rotating magnetic field (synchronous • Reduces leakage flux
speed N𝑠 ​) depends on the number of poles and is • Improves magnetic coupling
given by: • Enhances operating power factor
120𝑓
N𝑠 =
𝑃
Greater the number of poles, lesser the speed, and
vice versa.
•The rotating magnetic flux crosses the air gap and
induces EMF in the rotor conductors.
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Construction of 3-ϕ Induction motor
Construction:
B. Rotor:
•The rotor consists of a cylindrical laminated core made of
➢ The rotor is the rotating part of an induction
silicon steel to minimize eddy current losses.
motor.
•Slots are cut on the outer periphery of the rotor core.
➢ It receives power from the stator by
•These slots contain rotor bars, made of aluminum, copper, or
electromagnetic induction and produces torque
brass.
that drives mechanical loads.
•The rotor bars are not placed exactly parallel to the shaft —
➢ There are two main types of rotors used in
they are skewed at a small angle. This design:
induction motors:
• Reduces magnetic locking (cogging) between stator
I. Squirrel-Cage Rotor:
and rotor teeth.
➢ The squirrel-cage rotor is the most commonly
• Ensures smoother operation.
used type of rotor in induction motors.
• Helps reduce humming noise during rotation.
➢ Motors employing this type are called squirrel-
•The rotor bars are short-circuited at both ends by end rings,
cage induction motors.
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forming a closed electrical circuit that resembles a 18
squirrel cage.
Construction of 3-ϕ Induction motor
•High Efficiency and Reliability:
Key Features:
• Very popular for industrial and commercial
•Simple and Rugged Design:
applications.
• Extremely robust and reliable.
• About 90% of induction motors use the squirrel-cage
• Can withstand harsh operating conditions.
rotor due to its advantages.
•Permanently Short-Circuited Rotor Bars:
Limitations:
• No brushes or slip rings.
•Fixed rotor resistance:
• No external resistance can be added to the
• Cannot adjust starting characteristics.
rotor circuit.
• Offers lower starting torque compared to wound
• This limits control over starting torque and
rotor motors.
starting current.
Applications:
•Low Maintenance:
•Fans, blowers, pumps, compressors, conveyors, and general-
• Since there are no sliding contacts, the motor is
purpose industrial machinery where constant speed and
almost maintenance-free.
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A. performance are required. 19
Construction of 3-ϕ Induction motor
In Summary:
Rotor slots are skewed to:
Rotor slots are skewed mainly to reduce
1.Reduce magnetic locking (cogging):
cogging, smooth torque, lower noise, and
1. Prevents alignment of stator and rotor teeth, avoiding
minimize harmonic effects — leading to a
magnetic locking that can stop the rotor from starting
quieter and more efficient motor.
smoothly.
2.Ensure smooth and quiet operation:
1. Skewing helps distribute the torque more evenly, which
reduces torque pulsations, vibration, and noise during
motor operation.
3.Minimize harmonic effects:
1. Reduces the influence of slot harmonics, which
improves efficiency and reduces electromagnetic noise
and additional losses.
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Construction of 3-ϕ Induction motor
II. Phase-Wound Rotor (Wound Rotor / Slip-Ring Rotor):
➢ The phase-wound rotor is used in slip-ring •One end of each phase is internally connected in star (most
induction motors, which are typically employed in commonly).
applications requiring high starting torque or •The other ends of the three windings are brought out and
adjustable speed. connected to three insulated slip rings, which are mounted on
➢ This type of rotor has a construction similar to the the rotor shaft.
stator. •Carbon brushes rest on these slip rings to allow external
Construction: electrical connection.
•The rotor core is cylindrical and laminated, similar to the
External Circuit:
squirrel-cage rotor.
➢ The slip rings are connected to an external 3-phase
•It carries a three-phase winding, either star- or delta-
star-connected rheostat (resistance bank).
connected, and is distributed across the rotor slots.
➢ This setup allows variable resistance to be
•The winding is wound for the same number of poles as the
introduced into the rotor circuit during starting.
stator to ensure proper magnetic coupling.
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Construction of 3-ϕ Induction motor
Key Features and Advantages:
•High Starting Torque:
• Additional external resistance improves starting torque and reduces starting current.
•Speed Control:
• The inserted rotor resistance also enables better control over speed-torque characteristics.
•Smooth Acceleration:
• Torque can be maintained at a nearly constant level during acceleration.

Normal Operation:
•Once the motor reaches near full speed:
• The slip rings are short-circuited automatically using a metal collar or shorting device.
• This bypasses the external resistance, allowing the rotor to function similar to a squirrel-cage rotor
during steady-state operation.

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Construction of 3-ϕ Induction motor
Applications:
•Cranes, elevators, hoists, conveyors, compressors, and other heavy-load machinery requiring:
•High starting torque
•Variable speed
•Frequent starts and stops

Limitations:
•Higher cost and complexity due to the presence of slip rings, brushes, and external resistors.
•Requires more maintenance (brush wear, dust collection, etc.).
•Larger physical size than squirrel-cage motors of the same rating.

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Construction of 3-ϕ Induction motor

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Construction of 3-ϕ Induction motor

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Construction of 3-ϕ Induction motor
Comparison: Squirrel-Cage Rotor vs Wound Rotor (Slip-Ring)
Feature Squirrel-Cage Rotor Wound Rotor (Slip-Ring)
Simple, rugged, with copper/aluminum Complex, with 3-phase winding
Construction
bars short-circuited by end rings connected to slip rings
Rotor Windings Conducting bars (no actual windings) 3-phase winding, like stator
No (bars are permanently short-
External Electrical Access Yes (through slip rings and brushes)
circuited)
High (can add external resistance during
Starting Torque Low to moderate
start)
Speed Control Very limited Good (via external rotor resistance)
Requires regular maintenance (due to
Maintenance Very low (no brushes or slip rings)
brushes/slip rings)
Cost Less expensive More expensive
Slightly lower (due to rotor resistance
Efficiency High
and friction losses)
Widely used in fixed-speed and general Used for high torque and variable-speed
Usage
applications applications
Fans, pumps, blowers, compressors, Cranes, elevators, hoists, heavy-duty
Applications
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Working Principle of induction motor
1. Three-Phase Supply to Stator Windings 3. EMF is Induced in Rotor Conductors
➢ The stator has a three-phase winding. ➢ According to Faraday’s Law, when a conductor
➢ When connected to a three-phase AC power supply, experiences a changing magnetic flux, an electromotive
the windings produce a rotating magnetic field force (EMF) is induced in it.
(RMF). ➢ Since the rotor conductors are part of a closed circuit (e.g.,
➢ This RMF rotates at synchronous speed (Ns), which squirrel-cage bars shorted by end rings), this EMF causes
depends on supply frequency and number of poles: a current to flow in the rotor.

2. RMF Cuts Stationary Rotor Conductors

•The rotor is initially stationary.
•The rotating magnetic field sweeps across the air gap and
cuts the stationary rotor conductors.
•This causes relative motion between the magnetic field and
the rotor.
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Working Principle of induction motor…
4. Rotor Current Produces Its Own Magnetic Field 6. Rotor Starts Moving in Same Direction as RMF
•The rotor current produces its own magnetic field. •The rotor always rotates in the same direction as
•According to Lenz’s Law, this field opposes the the rotating magnetic field to reduce relative motion
cause of its production — i.e., the relative motion (slip).
between stator field and rotor. •The greater the slip (difference between stator

5. Torque is Produced on the Rotor field speed and rotor speed), the larger the induced

•Now the rotor is carrying current and is placed in a EMF and torque.

rotating magnetic field. 7. Total Torque = Sum of Torque on All Conductors

•According to the Lorentz force law (or Left-Hand •If there are n rotor conductors, the net torque is the

Rule), this interaction causes a force (torque) on the sum of torque on each conductor:

rotor conductors. •T𝑇𝑜𝑡𝑎𝑙 = 𝑇1 + 𝑇2 + T3 + ⋯ T𝑛

•This torque causes the rotor to start rotating. •This net torque causes the rotor to accelerate and rotate
continuously.
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Working Principle of induction motor…
8. Rotor Reaches Steady Speed (Slip ≠ 0)
•As the rotor speeds up, the relative speed between rotor and stator field decreases, which reduces the
induced EMF and current.
•Eventually, the rotor reaches a constant speed where the electromagnetic torque balances the load torque.
•The rotor never reaches synchronous speed in an induction motor (except in special cases like synchronous
motors), so a small slip (S) always exists:

Conclusion:
The rotor of an induction motor rotates because the rotating magnetic field of the stator induces current in
the rotor, which then interacts with the magnetic field to produce torque, causing the rotor to rotate in the same
direction as the stator field.

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Working Principle of induction motor…
1. The synchronous speed depends on supply
3. What If Rotor Reaches 𝑁𝑆 ​?
120𝑓
frequency f and number of poles P: N𝑠 =
𝑃 •If the rotor speed equals synchronous speed (i.e., 𝑁𝑟 = 𝑁𝑆 ​):
2. Rotor Tries to Catch Up • There will be no relative motion between the rotor and
•The RMF cuts the rotor conductors and induces an RMF.
e.m.f. due to relative motion. • Therefore, no induced e.m.f. in the rotor.
•This induced e.m.f. produces current in the rotor, • Without e.m.f., there is no rotor current, and hence no
which interacts with the stator’s magnetic field to torque.
produce torque, causing the rotor to rotate. • The motor would stop producing torque, and the

4. So Rotor Always Runs Below 𝑁𝑆 rotor would slow down again.

•To maintain torque production, the rotor must always rotate at a 5. Slip and Slip Speed
speed less than synchronous speed: 𝑁𝑟 < 𝑁𝑆 •The difference between synchronous speed and rotor
•This operating speed is called sub-synchronous speed. speed is called slip speed:
•That’s why an induction motor is also called an Slip speed = 𝑁𝑆 − 𝑁𝑟
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asynchronous motor.
Working Principle of induction motor…
6. Role of Slip in Torque Production
Slip (S) is usually expressed as a percentage:
•Slip creates the relative speed necessary to:
𝑁𝑆 −𝑁𝑟
%S= *100
𝑁𝑆 • Induce e.m.f in rotor
S* 𝑁𝑆 = 𝑁𝑆 − 𝑁𝑟 • Generate rotor current
𝑁𝑟 = 𝑁𝑆 - S* 𝑁𝑆 • Produce torque
𝑁𝑟 = 𝑁𝑆 (1-S) •More slip = higher torque, especially during starting.
Condition Rotor Speed (𝑁𝑟 ) Slip (s) Notes
At standstill (starting) 0 1 or 100% Maximum slip; full relative motion
Running normally < 𝑁𝑠 0.01 to 0.05 Typical slip range: 1% – 5%
At synchronous speed = 𝑁𝑠 0 Not possible for an induction motor
Conclusion:
➢ The rotor must slip behind the stator’s rotating magnetic field to generate torque.
➢ If it were to catch up completely, all torque production would cease.
➢ That’s why induction motors always run at a speed
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less than 31
synchronous speed:-making them asynchronous machine
Effect of Slip on Rotor Parameters
➢ In a transformer, the frequency of the induced e.m.f in the secondary winding is the same as the frequency
of the voltage applied to the primary winding.
➢ In an induction motor:
❖ At start-up, the rotor speed 𝑁𝑟 =0, so the slip S=1.
❖ Under this condition, the frequency of the induced e.m.f in the rotor is the same as the stator supply
frequency.
➢ As the motor starts rotating and gains speed:
❖ The rotor speed 𝑁𝑟 >0, so slip S<1.
❖ The frequency of the rotor induced e.m.f decreases and is no longer equal to the stator frequency.
➢ Therefore, slip affects the rotor frequency, and other rotor parameters are also affected by slip.
I. Rotor frequency
II. Magnitude of rotor induced e.m.f
III. Rotor reactance
IV. Rotor power factor &
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V. Rotor current
Effect of Slip on Rotor frequency
At Start-up (Standstill Condition) During Running Condition (Motor Picks Up Speed)
➢ Rotor speed 𝑁𝑟 =0, so slip S=1. ➢ Rotor rotates at speed 𝑁𝑟 , so slip S<1.
➢ The rotor is stationary, and the rotating magnetic ➢ The relative speed between the rotor and the R.M.F.
field (R.M.F.) of the stator is moving at becomes:
synchronous speed 𝑁𝑆 . ➢ 𝑁𝑆 − 𝑁𝑟 ⇒ S𝑁𝑆
➢ Hence, the relative motion between the R.M.F. and ➢ Since e.m.f is induced due to cutting of magnetic flux, and
the rotor is maximum. the rate of flux cutting depends on relative speed, the
➢ This causes the maximum e.m.f to be induced in magnitude of induced e.m.f decreases.
the rotor conductors. ➢ As the relative speed decreases, the frequency of induced
➢ The frequency of this induced e.m.f in the rotor is: e.m.f in rotor also decreases.
𝑓𝑟 = 𝑓𝑆 ​ (same as the stator or supply frequency)

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Effect of Slip on Rotor frequency
Rotor and Stator have Same Number of Poles (P)
➢ The rotor is wound for the same number of poles (P) as the stator.
➢ In running condition, the rotor e.m.f frequency 𝑓𝑟 ​ is given by the same type of relation used for synchronous
speed:
𝑁𝑆 −𝑁𝑟 𝑃 𝑁𝑆 −𝑁𝑟
𝑓𝑟 = where S= ⇒ 𝑁𝑆 − 𝑁𝑟 = S* 𝑁𝑆
120 𝑁𝑆

substituting into the above equation gives

S∗ 𝑁𝑆𝑃 120𝑓𝑆
𝑓𝑟 = where N𝑠 =
120 𝑃

S∗120𝑓
𝑃
𝑃
𝑓𝑟 = ⇒ 𝑓𝑟 = 𝑆𝑓𝑆
120

which confirms again that rotor frequency is proportional to slip.

Slip is usually very small, typically: S=0.01 to 0.05 or 1 to 5%
6/10/2025 Mr. Khalid A. 34
Effect of Slip on Magnitude of Rotor induced E.M.F
At Standstill (Starting Condition) During Running Condition

➢ Rotor speed 𝑁𝑟 = 0, so slip S=1. •As the rotor starts rotating at speed 𝑁𝑟 , slip S<1.

➢ Relative speed between rotor and rotating •The relative speed between RMF and rotor becomes:

magnetic field (RMF) is maximum, i.e. 𝑁𝑆 − 𝑁𝑟 ⇒ S* 𝑁𝑆

𝑁𝑆 − 𝑁𝑟 ⇒ 𝑁𝑆 where 𝑁𝑟 = 0 •Since induced e.m.f depends on the rate of cutting of

➢ Therefore, the maximum e.m.f is induced in the flux (i.e., relative speed), the rotor e.m.f decreases

rotor windings. proportionally as speed increases.

➢ Let this maximum e.m.f per phase be: •Let the rotor e.m.f during running condition be:

𝐸2 =Rotor induced e.m.f per phase at standstill 𝐸2𝑟 =Rotor induced e.m.f per phase at running condition
𝐸2 ≈ 𝑁𝑆 while 𝐸2𝑟 ≈ 𝑁𝑆 − 𝑁𝑟 ⇒ S* 𝑁𝑆
Dividing the two proportionality equations

𝐸2𝑟
= S∗𝑁 𝑁𝑆 ⇒ 𝐸2𝑟 =S 𝐸2
𝐸2 𝑆
The magnitude of induced e.m.f in the rotor alsoMr.reduces
6/10/2025 Khalid A. by slip times the magnitude of induced e.m.f
35 at
standstill condition
EXAMPLES…
1. A 4 pole, 3-phase IM is supplied from 50Hz supply. Determine its synchronous speed. On full load, its
speed observed to be 1410 rpm. Calculate its full load slip. Ans =
2. A 4 pole, 3-phase, 50Hz, star connected IM has a full load slip of 4%. Calculate full load speed of the motor.
Ans =
3. A 4 pole, 3-phase, 50Hz IM runs at a speed of 1470 rpm. Find the frequency of induced emf in the rotor
under the condition.
Ans =

6/10/2025 Mr. Khalid A. 36

Effect of Slip on Rotor Resistance and Reactance
Types of Rotor Construction
Rotor Winding Parameters
1. Squirrel Cage Rotor
•The rotor winding has two main electrical properties:
➢ Rotor resistance 𝑅2 is very small
• Resistance → 𝑅2
➢ No access to rotor windings (closed bars)
• Inductance → represented by reactance 𝑋2
➢ External resistance cannot be added
➢ For analysis, 𝑅2 ​ is often neglected
2. Slip Ring Rotor (Wound Rotor)
➢ Rotor has winding ends connected to slip rings
➢ External resistances can be added through slip rings to
control:
•Starting torque
•Rotor current &
•Speed control
6/10/2025 Mr. Khalid A. 37
Rotor resistance in this case is variable
Effect of Slip on Rotor Resistance and Reactance…
Resistance and Reactance Behavior
General Case (Applicable to Both Types)
•Rotor resistance 𝑅2 is independent of frequency, so it
Let: 𝑅2 = Rotor resistance per phase on standstill
remains constant at: 𝑅2 (same in both standstill and running)
𝑋2 =Rotor reactance per phase on standstill
•Rotor reactance 𝑋2 depends on frequency, and since rotor
Now at standstill,
frequency in running is: 𝑓𝑟 = S𝑓𝑠
𝑓𝑟 = 𝑓𝑠 hence if The rotor reactance during running becomes:
𝐿2 𝑖𝑠 𝑡ℎ𝑒 𝑖𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑟𝑜𝑡𝑜𝑟 𝑝𝑒𝑟 𝑝ℎ𝑎𝑠𝑒, 𝑿𝟐𝒓 = S 𝑿𝟐
𝑋2 = 2π 𝑓𝑟 𝐿2 ⇒ 2π𝑓𝑠 𝐿2 Ω/phase 𝒁𝟐 = Rotor impedance on standstill (𝑵𝒓 ) condition
𝑅2 = Rotor resistance in Ω/phase 𝒁𝟐 = 𝑹𝟐 + 𝒋𝑿𝟐 Ω/phase
While Now in running conditions,
𝑓𝑟 = S𝑓𝑠 hence, 𝒁𝟐 = 𝑹𝟐 2 + 𝑿𝟐 2 Ω/phase … . . 𝑖𝑛 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒
𝑋2𝑟 = 2π 𝑓𝑟 𝐿2 ⇒ 2π𝑆𝑓𝑠 𝐿2 ⇒ 𝑆. (2π𝑓𝑠 𝐿2 )
While 𝒁𝟐𝒓 = rotor impedance in running conditions
𝑿𝟐𝒓 = S 𝑿𝟐
𝒁𝟐𝒓 = 𝑹𝟐 + 𝒋𝑿𝟐𝒓 ⇒ 𝑹𝟐 + 𝒋𝑺 ∗ 𝑿𝟐
Where 𝑋2𝑟 = rotor reactance in running conditions
6/10/2025 Mr. Khalid A. 38
𝒁𝟐𝒓 = 𝑹𝟐 2 + (𝑺 ∗ 𝑿𝟐 )2 Ω/phase … . . 𝑖𝑛 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒
Effect of Slip on Rotor Power factor
From rotor impedance, we can write the expression for the power factor of rotor at standstill and also in running
conditions
➢ The impedance triangle on standstill rotor condition is show in figure below from it we ca write
cos ∅2 = Rotor power factor on standstill
𝑅2 𝑅2
cos ∅2 = =
𝑍2
𝑹𝟐 2 + 𝑿𝟐 2
➢ The impedance in running condition become 𝒁𝟐𝒓 and the corresponding impedance triangle is shown in the
figure below. From it we can write
cos ∅2𝑟 = Rotor power factor in running condition
𝑅2 𝑅2
cos ∅2𝑟 = =
𝑍2𝑟
𝑹𝟐 2 +(𝑺∗𝑿𝟐 )2

Key point: as rotor winding is inductive the rotor p.f is always lagging in nature
6/10/2025 Mr. Khalid A. 39
Effect of Slip on Rotor Current
➢ In the running conditions,
Let 𝐼2 = 𝑟𝑜𝑡𝑜𝑟 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑝𝑒𝑟 𝑝ℎ𝑎𝑠𝑒 𝑜𝑛 𝑠𝑡𝑎𝑛𝑑𝑠𝑡𝑖𝑙𝑙 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛
➢ 𝑍2 , 𝐸2 & 𝐼2 changes to 𝑍2𝑟 , 𝐸2𝑟 & 𝐼2𝑟 , respectively.
➢ The magnitude of 𝐼2 dependence on magnitude of 𝐸2
➢ The equivalent rotor circuit on running condition is
and impedance per phase
𝐸2 𝑝𝑒𝑟 𝑝ℎ𝑎𝑠𝑒 shown in figure below. 𝐼2𝑟 = rotor current in phase
𝐼2 = 𝐴
𝑍2 𝑝𝑒𝑟 𝑝ℎ𝑎𝑠𝑒
𝐸2
Substituting expression of 𝑍2 𝑤𝑒 𝑔𝑒𝑡, 𝐼2 = 𝐴
2 2
𝑹𝟐 +𝑿𝟐

➢ The equivalent rotor circuit on standstill is shown in the

➢ The value of slip depends on speed which in turn
figure below
depends on load on motor hence 𝑋2𝑟 is show variable in
➢ The ∅2 𝑖𝑠 𝑡ℎ𝑒 angle between 𝐸2 and 𝐼2 which
the equivalent circuit. From the equivalent circuit we
determine rotor p.f on standstill.
𝐸 𝑝𝑒𝑟 𝑝ℎ𝑎𝑠𝑒 𝑆𝐸2
can write 𝐼2𝑟 = 2𝑟 =
𝑍2𝑟 𝑝𝑒𝑟 𝑝ℎ𝑎𝑠𝑒
𝑹𝟐 2 + (𝑆 ∗ 𝑿𝟐 )2

➢ ∅2 is the angle between , 𝐸2𝑟 and 𝐼2𝑟 which decides p.f

6/10/2025 Mr. Khalid A. in running condition. 40
Induction Motor as a Transformer
1. Basic Similarity: Electromagnetic Induction 2. Frequency of Induced E.M.F.

•In both the transformer and induction motor, power Aspect Transformer Induction Motor
Primary frequency 𝑓𝑠 (same as supply) f (same as supply)
is transferred from the primary to the secondary by
Secondary 𝑓𝑟 = S𝑓𝑠 (varies
mutual induction. 𝑓𝑠 (always same)
frequency with slip S)
•In a transformer: At standstill (start) — S=1⇒𝑓𝑟 = 𝑓𝑠
• Power is transferred between two stationary During running — S<1⇒𝑓𝑟 < 𝑓𝑠
windings. •In a transformer, the primary and secondary always operate

•In an induction motor: at the same frequency.

• Power is transferred from the stator •In an induction motor, only the stator frequency is fixed at

(primary) to the rotor (secondary), but the supply frequency.

rotor rotates. The rotor frequency changes depending on slip, which in turn
depends on load.

6/10/2025 Mr. Khalid A. 41

Induction Motor as a Transformer…
3. Nature of Energy Transfer
Aspect Transformer Induction Motor
Energy in secondary Entirely electrical Part electrical, part mechanical
Conversion No energy conversion Electrical energy is converted to mechanical
Output Electrical power output Mechanical power output (torque & speed)
•In a transformer, both input and output are electrical.
•In an induction motor, electrical power from the rotor is partly converted to mechanical power, which
drives the load.
4. Conclusion: Why IM is Called a Transformer
•Unlike a transformer, the rotor rotates, and
An induction motor is often called a "rotating transformer" because:
this leads to:
•The stator acts like the primary winding of a transformer.
• Variable frequency in the rotor.
•The rotor acts like the secondary winding.
• Conversion of part of the energy into
•Power is transferred through mutual induction via the air gap.
mechanical form.
6/10/2025 Mr. Khalid A. 42
Induction Motor as a Transformer…
Summary

Transformer Induction Motor

------------------------ ----------------------------
Stator (Primary) Stator (Primary)
↓ ↓
Mutual Induction Mutual Induction (Air Gap)
↓ ↓
Secondary (Stationary) Rotor (Rotating)
↓ ↓
Electrical Output (f = const) Electrical + Mechanical Output (𝑓𝑟 = S𝑓𝑠 )

6/10/2025 Mr. Khalid A. 43

Examples
1. For a 4 pole, 3-phase, 50Hz IM ratio of stator to rotor turn is 2. On a certain load, its speed
is observed to be 1455 rpm when connected to 415 V supply. Calculate
I. Frequency of rotor emf in running condition Ans=1.5 Hz
II. Magnitude of induced emf in standstill Ans=
III. Magnitude of induced emf in the rotor under running condition. Assume star connected stator.
Ans=

6/10/2025 Mr. Khalid A. 44

Torque Equation
➢ Torque in induction motor depends on the Let’s

following parameters,

➢ The torque developed by the motor at the instant of

starting is called starting torque.
➢ In some cases, it is greater then the normal running
torque,
6/10/2025 whereas in some other cases it is somewhat
Mr. Khalid less.
A. 45
Torque Equation…
Let’s

Condition for Maximum Starting Torque

➢ It can be proved that starting torque is maximum when rotor resistance equals rotor reactance

⇒
6/10/2025 Mr. Khalid A. 46
Torque Equation…
Effect of change in supply voltage on starting torque

Note:- Starting torque is very sensitive to the change in

supply voltage as it vary with the square of the supplied
voltage

6/10/2025 Mr. Khalid A. 47

Torque Equation…
Torque under running conditions

6/10/2025 Mr. Khalid A. 48

 Torque Equation…
Condition for maximum Torque under running conditions
➢ The torque of a rotor under running conditions is
➢ Hence, Torque under running conditions is
maximum at that value of slip S which makes

➢ The condition for maximum torque may be obtain by rotor reactance equal to rotor resistance

differentiating the above expression with respect to slip S

and putting it equal to zero. however, it is simpler to put ➢ In the other equation given in above normal

Y=1/T and then differentiate it. torque equation,

6/10/2025 Mr. Khalid A. 49

Torque Equation…
Note:-
➢ Maximum torque is independent of rotor resistance, however, the slip at which maximum torque
occurs is dependent on rotor resistance,.
➢ Maximum torque varies inversely as stand still rotor reactance.
➢ Maximum torque varies directly as the square of applied voltage.
➢ To get maximum torque at starting, rotor resistance must be equal to rotor reactance per phase

6/10/2025 Mr. Khalid A. 50

Torque-Slip Characteristics
➢ As the induction motor is loaded from no-load to full load, it’s speed decreases hence slip increases.
Due to the increased load motor has to produce more torque to satisfy load demand.
➢ The behavior of motor can be easily judged by sketching a curve obtained by plotting torque produced
against slip of induction motor.

6/10/2025 Mr. Khalid A. 51

Torque-Slip Characteristics…

6/10/2025 Mr. Khalid A. 52

Induction Machines Modes
Induction Motor Operation Based on Slip 2. Plugging or Braking Mode (s > 1):
1. Motoring Mode (0 < s < 1): ➢ Occurs when the rotor rotates in the direction
➢ This is the normal operating region of an opposite to that of the stator's rotating field.
induction motor. ➢ Achieved in practice by interchanging any two stator
➢ The rotor rotates in the same direction as the terminals, which reverses the phase sequence and
rotating magnetic field produced by the stator. thereby the direction of the magnetic field.
➢ The motor draws electrical power from the ➢ The motor then experiences a reverse (counter)
supply and delivers mechanical power to the torque, causing it to decelerate quickly.
load. ➢ This method of stopping the motor is known as
➢ Both rotor speed and torque act in the same plugging.
direction. ➢ Important precaution: After the motor stops, the
➢ The machine behaves as a motor. stator must be disconnected from the supply to
prevent the rotor from accelerating in the reverse
6/10/2025 Mr. Khalid A. 53
direction.
 Induction Machines Modes …
Induction Generator Operation (s < 0)
➢ An induction machine operates as a generator when the slip becomes negative (i.e., s < 0).
➢ Negative slip means that the rotor speed is greater than the synchronous speed.
➢ In this condition, the machine is mechanically driven (e.g., by a turbine or engine), and it begins to deliver
electrical power to the grid or load.
➢ The mechanical energy supplied to the rotor is converted into electrical energy, which is delivered
through the stator windings.
➢ In this region, the torque-slip characteristic reverses, as the machine is now acting as a generator rather
than a motor.

Important Note: An induction generator does not self-excite; it requires an external source of reactive
power (usually from the grid or a capacitor bank) to establish the magnetic field necessary for generating
action.

6/10/2025 Mr. Khalid A. 54

 Induction Machines Modes …
The Fig. below shows the complete torque-slip characteristics showing motoring, generating
and the braking region.

6/10/2025 Mr. Khalid A. 55

Torque Ratios
Torque Comparison in Induction Motors 3. Maximum Torque :
The performance of an induction motor is often 1. The highest torque the motor can develop
assessed by comparing the following types of torques: under any condition, also called breakdown
1.Starting Torque : torque.
1. The torque developed by the motor at the 2. Occurs at a particular slip s < 1, beyond which
moment of starting (when slip s = 1). the torque falls.
2. Critical for determining whether the motor can
Why These Ratios Matter
overcome initial load inertia.
•They help evaluate motor suitability for specific
2.Full-Load Torque :
applications (e.g., whether high starting torque is needed
1. The torque developed by the motor when it
for heavy loads).
operates at rated (full-load) conditions.
•Allow comparison between different motor designs.
2. This is the normal operating torque during
•Aid in predicting motor performance under various
steady-state operation.
loading conditions.
6/10/2025 Mr. Khalid A. 56
Torque Ratios…
Full Load and Maximum Torque Ratio Dividing both numerator and denominator by Torque
Ratio 𝑿𝟐 2 . We get,

6/10/2025 Mr. Khalid A. 57

Torque Ratios…
Starting Torque and Maximum Torque Ratio Dividing both numerator and denominator by Torque
Ratio 𝑿𝟐 2 . We get,

6/10/2025 Mr. Khalid A. 58

Speed-Torque Characteristics
At 𝑁𝑟 = 𝑁𝑠 , the motor stops as it cannot produce any torque, as induction motor cannot rotate at synchronous speed.
• At 𝑁𝑟 = 0, the starting condition, motor produces a torque called starting torque.

6/10/2025 Mr. Khalid A. 59

Effect of Resistance on Torque
It is know that in slip ring induction motor, externally Key Point:-it can be observed that 𝑇𝑚 is independent of
resistance can be added in the rotor. Let us see the effect
of change on rotor resistance of the torque produced. 𝑅2 . 𝐡𝐞𝐧𝐜𝐞 whatever may be the rotor resistance,
maximum torque produced never changes but the slip
and speed at which it occurs depends on 𝑅2

6/10/2025 Mr. Khalid A. 60

Effect of Resistance on Torque …

6/10/2025 Mr. Khalid A. 61

 Effect of Resistance on Torque …
Why Add External Resistance? •Therefore, the added resistance is:
➢ The torque T of an induction motor is maximum • Only present during starting

when the rotor resistance 𝑅2 ​ equals the rotor • Gradually removed as the motor picks up speed

reactance 𝑋2 at a given slip: • Fully cut off during normal running condition
𝑅2 How It’s Done in Practice
𝑇𝑚𝑎𝑥 𝑜𝑐𝑐𝑢𝑟𝑠 𝑤ℎ𝑒𝑛 𝑆 =
𝑍2
At starting (s = 1), this condition becomes: 𝑅2 = 𝑋2 •This method is used only in slip ring (wound rotor) induction
motors, because:
So, by adding external resistance to the rotor to
• The rotor windings are accessible through slip rings.
make 𝑅2 = 𝑋2 ​, the motor can produce maximum
• External resistors can be connected and disconnected as
torque at start.
needed.
Important Points
Why It Can’t Be Used in Squirrel Cage Motors
Why Not Keep the Resistance Always?
•In squirrel cage induction motors:
•If high rotor resistance is kept permanently, it
causes: • The rotor bars are short-circuited at both ends.
• High copper losses due to 𝐼 2 𝑅 • No access to rotor circuit ⇒ can’t add external resistance.
6/10/2025 Mr. Khalid A. 62
• Poor efficiency •Therefore, starting torque is fixed by the rotor design
Examples
1. A 3-phase induction motor having a star-connected rotor has an induced e.m.f. of 80 volts between slip-
rings at standstill on open-circuit. The rotor has a resistance and reactance per phase of 1 Ω and 4 Ω
respectively.
Calculate current/phase and power factor when
I. slip-rings are short-circuited
II. slip-rings are connected to a star-connected rheostat of 3 Ω per phase.

2 A 3-phase, 400-V, star-connected induction motor has a star-connected rotor with a stator to rotor turn
ratio of 6.5. The rotor resistance and standstill reactance per phase are 0.05 ohm and 0.25 ohm
respectively. What should be the value of external resistance per phase to be inserted in the rotor circuit
to obtain maximum torque at starting and what will be rotor starting current with this resistance?

6/10/2025 Mr. Khalid A. 63

Examples
3. A 3-phase, slip-ring, induction motor with star-connected rotor has an induced e.m.f. of 120 volts
between slip-rings at standstill with normal voltage applied to the stator. The rotor winding has a
resistance per phase of 0.3 ohm and standstill leakage reactance per phase of 1.5 ohm. Calculate
I. Rotor current/phase when running short-circuit with 4% slip and
II. The slip and rotor current per phase when the rotor is developing maximum torque

4. A 3-phase induction motor having a 6-pole, star-connected stator winding turns on 240-V, 50Hz supply.
The rotor resistance and standstill reactance are 0.12 ohm and 0.85 ohm per phase. The ratio of stator to
rotor turns is 1.8. Full load slip is 4%. Calculate the developed torque at full load, maximum torque and
speed at maximum torque.

6/10/2025 Mr. Khalid A. 64

Examples
5. A 400V, 4 pole, 3 phase, 50 Hz star connected IM has a rotor resistance and reactance per phase of 0.01
ohm and 0.1 ohm respectively. Determine
I. Starting torque
II. Slip at which maximum torque will occur
III. Speed at which maximum torque will occur
IV. Maximum torque
V. Full load torque if full load slip is 4%
Assume ratio of stator to rotor turns as 4.

6. A 24 pole, 50 Hz star connected IM has a rotor resistance and reactance per phase of 0.016 ohm and
0.265 ohm at standstill respectively. It is achieving its full load torque at a speed of 24 rpm. Calculate

𝑇𝑓𝑙 𝑇𝑠𝑡
the ratio of & .
𝑇𝑚 𝑇𝑚
6/10/2025 Mr. Khalid A. 65
 Power Losses in Induction Motor
Note:-
Power Losses in an Induction Motor
➢ Stator iron losses are significant, since the
➢ Power losses in an induction motor are broadly
stator frequency = supply frequency.
classified into:
➢ Rotor iron losses are very small during normal
I. Constant Losses
operation, because:
These losses do not vary significantly with load. 𝑓𝑟 = 𝑆𝑓𝑆
A. Core (Iron) Losses ❖ Since slip S≪ 1 in normal operation, rotor frequency is
➢ Occur in the stator and rotor cores due to the very low ⇒ negligible iron losses.
alternating magnetic field.
B. Mechanical Losses
➢ Comprise: ❖ Hysteresis loss: due to the reversal
➢ Caused by:
of magnetization.
❖ Friction in bearings
❖ Eddy current loss: due to
❖ Windage (air resistance)
circulating currents in the iron core.
➢ These losses are also constant and independent of load.
➢ Depend on: ❖ Supply frequency
6/10/2025 Mr. Khalid A. 66
❖ Flux density & Material of the core
 Power Losses in Induction Motor…
II. Variable Losses in an Induction Motor 2. Rotor Copper Losses

➢ Variable losses are the losses that change with the ➢ Due to resistance of rotor winding:

load, since they depend on the current flowing in ➢ 𝑃𝑐𝑢−𝑟𝑜𝑡𝑜𝑟 = 3𝐼 2 2 𝑅2 … . 𝑎𝑛𝑎𝑙𝑦𝑠𝑒𝑠 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑒𝑙𝑦

the windings. ​ where:

Main Types of Variable Losses: ❖ 𝐼2𝑟 ​ = rotor current per phase at particular load

1. Stator Copper Losses ❖ 𝑅2 ​ = rotor resistance per phase

➢ Due to resistance of stator winding: •Rotor current also increases with load, so these losses vary

➢ 𝑃𝑐𝑢−𝑠𝑡𝑎𝑡𝑜𝑟 = 𝐼 21 𝑅1 accordingly.
Note: Rotor copper loss is also equal to:
where:
𝑃𝑐𝑢−𝑟𝑜𝑡𝑜𝑟 = S* 𝑃𝑔
❖ 𝐼1 = stator current
❖ 𝑅1 ​ = stator winding resistance ​ where:

•These losses increase with load, since stator current S = slip

increases. 𝑃𝑔 g​ = power transferred to rotor (air gap power)

6/10/2025 Mr. Khalid A. 67
•This means part of the power transferred to the rotor is lost as heat.
Power Losses in Induction Motor…
Why They Are Called Variable Losses Combined Losses Note:
➢ As the load increases, the current in both stator and •In performance calculations, stator copper losses and stator
rotor windings increases. iron losses are often grouped together for a specific load.
➢ So, copper losses (𝐼 2 𝑅) increase. • This helps in estimating total stator losses at that load.
➢ These losses directly affect the efficiency of the • But they remain distinct in nature (iron loss is
motor under load. constant, copper loss is variable).

6/10/2025 Mr. Khalid A. 68

Power Losses in Induction Motor…
1. Electrical Power Supplied to Stator 3. Air Gap Power (𝑃𝑎𝑔 )
•The three-phase AC supply is connected to the stator •After subtracting stator copper losses, the remaining power is
winding. transferred across the air gap to the rotor.
➢ This is the net electrical input power to the motor. •Called air gap power or rotor input power:
If: ❖ 𝑉𝐿 ​ = line voltage (volts), 𝑃 =𝑃 −𝑷
𝑎𝑔 𝑖𝑛 𝒔𝒕𝒂𝒕𝒐𝒓−𝒄𝒖
❖ 𝐼𝐿 = line current (amps), 4. Rotor Copper Loss
❖ cos ϕ = power factor, •Some power is again lost in the rotor windings (especially in
Then the input electrical power (𝑃𝑖𝑛 ) is: wound rotor or slip ring motors).
𝑃𝑖𝑛 = 3𝑉𝐿 𝐼𝐿 cos ϕ 𝑷𝒓𝒐𝒕𝒐𝒓−𝒄𝒖 = 𝑺 ∗ 𝑃𝑎𝑔
This is the total real power drawn from the supply.
Where: •S = slip (ratio of difference between synchronous
2. Stator Copper Loss
speed and rotor speed to synchronous speed).
➢ Some power is lost due to resistance in stator windings.
➢ This is called stator copper loss, denoted as:
𝐼1 = phase current
𝑷𝒔𝒕𝒂𝒕𝒐𝒓−𝒄𝒖 = 𝟑𝑰𝟏 𝟐 𝑹𝟏
6/10/2025 Mr. Khalid A.
𝑅1 = stator resistance per phase.
69
Power Losses in Induction Motor…
Summary of Power Flow Stages:
5. Mechanical Power Developed (𝑃𝑚 )
1.Input Electrical Power →
•After rotor copper loss, the remaining power is
2.Stator Losses (Copper Loss) →
converted into mechanical power at the shaft:
3.Air Gap Power →
𝑃𝑚 = (1 − 𝑆) 𝑃𝑎𝑔
4.Rotor Losses (Copper Loss) →
6. Mechanical Losses 5.Mechanical Power Developed →
•Include friction and windage losses (due to bearings, 6.Mechanical Losses →
air drag, etc.). 7.Output Shaft Power
•Let’s denote this as 𝑃𝑚𝑒𝑐ℎ_𝑙𝑜𝑠𝑠 ➢ From the power flow diagram we can define
𝑃𝑚 Mechanical Power Developed
7. Output Power (𝑃𝑜𝑢𝑡 ) 𝑅𝑜𝑡𝑜𝑟 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = =
𝑃𝑎𝑔 air gap power or rotor input power
•Final useful mechanical power available at the
𝑵𝒆𝒕 𝒐𝒖𝒕𝒑𝒖𝒕 𝒂𝒕 𝒔𝒉𝒂𝒇𝒕
shaft: Net motor efficiency=
𝑵𝒆𝒕 𝒆𝒍𝒄𝒕𝒓𝒊𝒄𝒂𝒍 𝒊𝒏𝒑𝒖𝒕 𝒕𝒐 𝒎𝒐𝒕𝒓
𝑃𝑜𝑢𝑡 = 𝑃𝑚 − 𝑃𝑚𝑒𝑐ℎ_𝑙𝑜𝑠𝑠
6/10/2025 Mr. Khalid A. 70
Relationship b/n 𝐏𝟐 , 𝐏𝒄 and 𝐏𝒎
Let’s 𝐏𝟐 = 𝑃𝑎𝑔 , 𝐏𝒄 = 𝐏𝒓𝒐𝒕𝒐𝒓_𝒄𝒖

6/10/2025 Mr. Khalid A. 71

Relationship b/n 𝐏𝟐 , 𝐏𝒄 and 𝐏𝒎

6/10/2025 Mr. Khalid A. 72

Efficiency of an Induction Motor
1. Definition of Efficiency
3. Behavior of Efficiency with Load
The overall efficiency (η) of an induction motor is
➢ No Load:
given by: Where:
❖ Current is small → copper losses are low.

•𝑃𝑜𝑢𝑡 = net mechanical power available at the shaft, ❖ But constant losses are still present.

•𝑃𝑖𝑛 = total electrical power supplied to the motor. ❖ So, efficiency is low at no load.

2. Types of Losses in an Induction Motor ➢ Increasing Load:

Constant Losses (Independent of Load): ❖ Current increases → copper losses increase.

•Core (Iron) Losses ❖ Total output power increases.

•Friction and Windage Losses ❖ Efficiency improves.

These remain roughly the same regardless of the load.

Variable Losses (Increase with Load):
•Stator Copper Losses (∝ 𝐼 21 ​)
•Rotor Copper Losses (∝ 𝐼 2 2 ​)
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As the load increases, current increases, so copper losses increase.
Efficiency of an Induction Motor…
➢ Maximum Efficiency Condition:
Maximum Efficiency occurs when:
Variable losses = Constant losses
➢ Beyond Optimum Load:
• Load increases further → copper losses
dominate.
• Variable losses > Constant losses
• Efficiency begins to decrease.
Key Point Summary
➢ Efficiency improves with load, reaches a peak, then
declines.
➢ Maximum efficiency is achieved when:
Copper losses (variable)=Constant losses (core + friction)
6/10/2025 Mr. Khalid A. 74
Examples
1. A 400 V, 3-phase, 50 Hz, 4-pole, star-connected induction-motor takes a line current of 10 A with 0.86 pf
lagging. Its total stator losses are 5% of the input. Rotor copper losses are 4% of the input to the rotor, and
mechanical losses are 3% of the input of the rotor. Calculate
I. slip and rotor speed,
II. torque developed in the rotor, and
III. shaft-torque.
2. A 400 V, 3-phase, 50 Hz, 4-pole, star-connected induction-motor takes a line current of 10 A with 0.86 p.f
lagging. Its total stator losses are 5% of the input. Rotor copper losses are 4% of the input to the rotor, and
mechanical losses are 3% of the input of the rotor. Calculate
I. slip and rotor speed,
II. torque developed in the rotor, and
III. shaft-torque.
IV. the gross electromagnetic torque.
6/10/2025 Mr. Khalid A. 75
Examples…
3. The useful full load torque of 3-phase, 6-pole, 50-Hz induction motor is 162.84 N-m. The rotor e.m.f is observed
to make 90 cycles per minute. Calculate
I. motor output
II. Cu loss in rotor
III. motor input and
IV. efficiency, if mechanical torque lost in windage and friction is 20.36 Nm and stator losses are 830 W.

4. A 6-pole, 50-Hz, 3-phase, induction motor running on full-load with 4% slip develops a torque of 149.3 N-m at its
pulley rim. The friction and windage losses are 200 W and the stator Cu and iron losses equal 1,620 W. Calculate
I. output power
II. the rotor Cu loss and
III. the efficiency at full-load.

6/10/2025 Mr. Khalid A. 76

Examples…
5. A 3-phase induction motor has a 4-pole, Y-connected stator winding. The motor runs on 50-Hz supply with 200V
between lines. The motor resistance and standstill reactance per phase are 0.1 ohm and 0.9 ohm respectively.
Calculate
I. the total torque at 4% slip
II. the maximum torque
III. the speed at maximum torque if the ratio of the rotor to stator turns is 0.67. Neglect stator impedance.

6/10/2025 Mr. Khalid A. 77

Equivalent Circuit of Induction Motor
➢ We have already seen that the induction motor can be treated as generalized transformer.
➢ Thus stator acts as a primary while the rotor acts as a rotating secondary.
➢ So induction motor can be represented as a transformer as shown below.

6/10/2025 Mr. Khalid A. 78

Equivalent Circuit of Induction Motor…
The equivalent circuit of induction motor thus can be represented as below.

The stator and rotor sides are shown separated by an air gap.
𝐼2𝑟 = 𝑟𝑜𝑡𝑜𝑟 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑖𝑛 𝑟𝑢𝑛𝑛𝑖𝑛𝑔 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠

𝐸2𝑟 𝑆𝐸2
𝐼2𝑟 = =
𝑍2𝑟
𝑅2 2 + (𝑆𝑋2 )2

Note:-It is important to note that as load on the motor changes, the motor speed changes, slip changes and
6/10/2025 Mr. Khalid A. 79
as slip changes the reactance 𝑋2𝑟 . 𝐻𝑒𝑛𝑐𝑒 = 𝑆𝑋2𝑟 𝑖𝑠 𝑠ℎ𝑜𝑤𝑛 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒.
Equivalent Circuit of Induction Motor…
➢ Representation of rotor impedance:- Key Point: thus mechanical load on the motor is represented by
𝐸 𝑆𝐸2
It is known that, 𝐼2𝑟 = 𝑍2𝑟 = 1−𝑆
2𝑟
𝑅2 2 +(𝑆𝑋2 )2 the pure resistance value 𝑅2 ( ) so rotor equivalent can be show
𝑆
𝐸2
= as,
𝑅
( 2 )2 +(𝑋2 )2
𝑆
➢ So it can be assumed that equivalent rotor circuit
in running condition has fixed reactance 𝑋2 , fixed
𝑅2
voltage 𝐸2 but a variable resistance , as
𝑆

indicated in the above equation

𝑅2 𝑅2
Now = 𝑅2 + − 𝑅2
𝑆 𝑆
1 1−𝑆
= 𝑅2 +𝑅2 − 1 = 𝑅2 + 𝑅2 ( )
𝑆 𝑆
𝑅2
So the variable rotor resistance has two parts,
𝑆

1. Rotor resistance 𝑅2 itself which represents copper loss

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1−𝑆 Mr. Khalid A. 80
2. 𝑅2 ( ) which represents load resistance 𝑅𝐿 . So it is electrical equivalent of mechanical load on motor.
Equivalent Circuit of Induction Motor…
Now let us obtain equivalent circuit referred to stator side.
Equivalent circuit referred to stator: The rotor current 𝐼2𝑟 has its reflected component on the stator
side which is 𝐼 ′ 2𝑟
transfer all the rotor parameters to stator,
𝐸2 𝐸2 ′ 𝐾𝑆𝐸2
𝐾 = 𝐸 = 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑖𝑜 𝐸 2 = 𝐾 ′ 𝐼 2𝑟 = 𝐾 𝐼2𝑟 =
1 2
(𝑅2 )2 +(𝑆𝑋2 )2
𝑋
𝑋 ′ 2 = 𝐾22 = 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑒𝑑 𝑟𝑜𝑡𝑜𝑟 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 𝑅 𝑅 1−𝑆 1−𝑆
𝑅
𝑅′ 𝐿 = 𝐾𝐿2 = 𝐾22 ( ) = 𝑅′ 2
𝑆 𝑆
𝑅′ 2 = 𝐾22 = 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑒𝑑 𝑟𝑜𝑡𝑜𝑟 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒
Thus 𝑅′ 𝐿 , is reflected mechanical on stator
So equivalent circuit referred to stator can be shown as in figure below

The resistance 𝑅′ 𝐿 =
1−𝑆
𝑅′ 2 is fictitious
𝑆

resistance representing the

mechanical load on the motor.

6/10/2025 Mr. Khalid A. 81

 Approximate Equivalent Circuit

Now the resistance 𝑅1 𝑎𝑛𝑑 𝑅′ 2 while reactance 𝑋1 𝑎𝑛𝑑 𝑋 ′ 2 can be combined. So we get,
𝑅1𝑒 =equivalent resistance referred to stator = 𝑅1 + 𝑅′ 2
𝑋1𝑒 = equivalent reactance referred to stator = 𝑋1 + 𝑋 ′ 2
𝑅
𝑅1𝑒 = 𝑅1 + 𝐾22
𝑋
𝑋1𝑒 = 𝑋1 + 𝐾22

while 𝐼ഥ1 = 𝐼ഥ0 + 𝐼 ′ 2𝑟 &

𝐼ഥ0 = 𝐼ഥ𝑐 + 𝐼𝑚

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 Power Equations from Equivalent Circuit

&

Key Point: Remember that in all the equations, all the values are per phase values.
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 Maximum Power Output
➢ Consider the approximate equivalent circuit shown
➢ In this circuit, the exciting current Io is neglected hence the exciting no load branch is not shown.

➢ To obtain maximum output power, differentiate the

equation of total 𝑷𝒐𝒖𝒕 with respect to variable 𝑹′ 𝑳
and equate to zero

&
& cancel it

6/10/2025 Mr. Khalid A. 84

 Maximum Power Output…
➢ Expression for maximum 𝑷𝒐𝒖𝒕 : using the

➔ condition obtained in expression of total

𝑷𝒐𝒖𝒕 , we can get maximum 𝑷𝒐𝒖𝒕
➢ For maximum mechanical power output, the mechanical
load (equivalent resistance) referred to the stator should
match the total leakage impedance of the motor (resistive
+ reactive).

But 𝑹𝟏𝒆 𝟐 + 𝑿𝟏𝒆 𝟐 = 𝒁𝟏𝒆 𝟐

𝑽𝟏 𝟐
(𝑷𝒐𝒖𝒕 )𝒎𝒂𝒙 =𝟑 ∗ 𝒁𝟏𝒆
𝟐𝒁𝟏𝒆 (𝒁𝟏𝒆 + 𝟐𝑹𝟏𝒆 )

6/10/2025 Mr. Khalid A. 85

 Maximum Torque
➢ The Thevenin’s circuit for the above network is
shown in the figure below across the terminal x & y

➢ The condition for maximum torque can be obtain

from maximum power transfer theorem.
(𝑰′ 𝟐𝒓 )𝟐 𝑹′ 𝟐
➢ When 𝑺
is maximum consider the
approximate equivalent circuit of induction motor
as shown
B/c 𝑿𝑴 ≫ 𝑿𝟏 & 𝑿𝑴 + 𝑿𝟏 ≫ 𝑹𝟏 ,
𝒕𝒉𝒆 Thevenin resistance and reactance are
approximately given by

➢ The value of 𝑹𝟎 is assume to be negligible. Hence

the circuit will be reduced as shown
6/10/2025 Mr. Khalid A. 86
 Maximum Torque…
➢ The mechanical torque developed by rotor is maximum if
𝑅′ 2
there is maximum power transfer to the resistor .
𝑆
𝑅′ 2
➢ This takes place when equal to impedance looking back
𝑆
into the supply source.

➢ This is the slip corresponding to the maximum torque. The

maximum torque is given by

❖ From the above expression, it can be seen that the

maximum torque is independent of rotor
6/10/2025 Mr. Khalid A. 87
resistance 𝑅′
 Maximum Torque…
What is a Synchronous Watt?
➢ A synchronous watt is a unit of torque-related
measurement used mainly in electrical machine analysis.
➢ A synchronous watt is defined as the torque required to ➢ Thus torque is directly proportional to the rotor input.

produce 1 watt of mechanical power at synchronous ➢ By defining new unit of torque which is synchronous

speed. watt we can write,

Synchronous Watt in an Induction Motor T ≈ 𝑃2 synchrouns-watts
➢ While the synchronous watt originates from synchronous If torque is given in synchronous-watts then it can be

machines, the concept is still useful in induction motors obtained in N-m as,

for analysis, especially in per-unit systems and torque

calculations.
In an induction motor:
•The motor runs at a speed slightly less than synchronous speed due to slip S.
•But6/10/2025
for analytical simplicity, synchronous speed isMr.used
Khalidas
A. a reference. 88
 Phasor Diagram of Induction Motor
Induction Motor vs Transformer: Phasor Diagram Comparison
➢ Induction motors and transformers are
electromagnetically similar — both operate on the
principle of mutual induction.
➢ That’s why: The phasor diagram of a loaded
induction motor looks similar to that of a loaded
transformer.
➢ But there are key differences due to the nature of
the rotor (secondary) and the type of load.

Why the Phasor Diagram is Similar

Both machines have:
•Magnetizing current 𝐼𝑚 •Reflected secondary current 𝐼 ′ 2
•Core loss component 𝐼𝐶 •Voltage drops due to leakage reactance and resistance
6/10/2025 Mr. Khalid A. 89
•Primary current 𝐼1 So the basic voltage equation for both is:
Phasor Diagram of Induction Motor…

6/10/2025 Mr. Khalid A. 90

 No-Load Test (Running Light Test)
1. Purpose of the No-Load Test 2. How the Test Is Performed

The no-load test is done to: •Motor condition: Induction motor runs without mechanical

•Determine magnetizing branch parameters (core loss load.

resistance 𝑅 , magnetizing reactance 𝑋 ​) of the •Supply: Rated voltage and rated frequency are applied to the
𝐶 𝑚

equivalent circuit. stator terminals.

•Estimate core losses, friction, and windage losses. •Measurements taken (per phase):

•Understand the motor’s behavior without mechanical • 𝑉𝑛𝑙 ​: No-load stator voltage

load. • 𝐼𝑛𝑙 ​: No-load stator current

• 𝑃𝑛𝑙 ​: No-load power input (using a wattmeter)

6/10/2025 Mr. Khalid A. 91

 No-Load Test (Running Light Test)…
𝑟′2
➢ The no-load slip 𝑆𝑛𝑙 ​ is very small, therefore ​ in This behaves like a very high resistance, meaning very little
𝑆𝑛𝑙
current flows in the rotor branch. ​
the approximate equivalent circuit is very large
3. Effect on Equivalent Circuit:
compared to 𝑋𝑚 .​ 𝑟′2
Because ≫ 𝑋𝑚 , the rotor branch becomes negligible, and
What It Means: 𝑆𝑛𝑙
the motor draws current mostly through the magnetizing
1. At No Load: branch: 𝐼𝑛𝑙 ≈ 𝐼𝑚 + 𝐼𝐶
➢ The motor is spinning close to synchronous speed. So, under no-load:
➢ So the slip 𝑆𝑛𝑙 ≈0.001 or even smaller (near 0). •Rotor branch ≈ open (very high impedance)
➢ This means the rotor is not absorbing mechanical •The circuit reduces to:
power; it’s only overcoming friction and windage. •𝑉1 → 𝑅1 + 𝑗𝑋1 →parallel(𝑅𝐶 || 𝑋𝑚 )
2. Referred Rotor Resistance Term: Practical Use: This is why during the no-load test, we assume:
In the induction motor’s equivalent circuit: •Rotor circuit is open
𝑟′2 •Motor behaves like a transformer with only the
′
Rotor branch: ​ ​+j 𝑋 2 magnetizing branch active
𝑆𝑛𝑙
𝑟′2 This simplifies the analysis and allows extraction of:
So at no load: →very large
Mr. Khalid A. •Core loss resistance 𝑅𝐶
𝑆𝑛𝑙
6/10/2025 92
•Magnetizing reactance 𝑋𝑚
No-Load Test (Running Light Test)…

𝑉𝑛𝑙

6/10/2025 Mr. Khalid A. 93

 Blocked-Rotor Test
What Is the Blocked-Rotor Test? How It’s Performed
It’s the induction motor equivalent of the short-circuit 1.Lock the rotor using a mechanical method like:
test in transformers. 1. Hand grip
It is performed with the rotor locked (not rotating), so that: 2. Pulley system
•Slip s=1s 3. Mechanical brake
•The motor behaves like a short-circuited transformer
Purpose of the Test 2.Apply reduced voltage (through a

To determine: Variac/autotransformer) until the stator current equals

•Stator resistance 𝑟1 rated current.

•Rotor resistance referred to stator 𝑟 ′ 2 3. Record per-phase values:

•Stator leakage reactance 𝑋1 1. Voltage 𝑉𝑏𝑟
•Rotor leakage reactance referred to stator 𝑋 ′ 2 2. Current 𝐼𝑏𝑟 ​ = rated current
We calculate: 3. Input power 𝑃𝑏𝑟
•Total leakage resistance: 𝑟1 + 𝑟 ′ 2
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•Total leakage reactance: 𝑋1 + 𝑋 ′ 2
Blocked-Rotor Test…

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Blocked-Rotor Test…

6/10/2025 Mr. Khalid A. 96

Blocked-Rotor Test…

6/10/2025 Mr. Khalid A. 97

 Separation of friction and windage loss from the no-load Test
What Does the No-Load Power Input Include? Measuring Stator Copper Loss

When the induction motor runs without mechanical •Measure the DC resistance of each stator phase (call it 𝑅𝑑𝑐 ​).

load (no-load test), the total power input 𝑃𝑛𝑙 ​ is used •Because AC resistance is higher than DC (due to skin effect

to supply: and temperature), estimate actual AC resistance as:

1.Stator Copper Loss 𝑟1 ​=(1.1 to 1.3)× 𝑅𝑑𝑐

𝑃𝑐𝑢−𝑛𝑙 = 3𝐼 2 𝑛𝑙 𝑟1 •Then compute copper loss: 𝑃𝑐𝑢 = 3𝐼 2 𝑛𝑙 𝑟1

This is the loss in the stator windings due to their resistance

2. Core (Iron) Loss Estimating Friction and Windage Losses (Mechanical Losses)
This includes hysteresis and eddy current losses in the Why this is done:
stator core. •We want to separate mechanical losses from the total no-load
It depends on voltage and frequency. power, which also includes core and copper losses
3. Friction and Windage Losses
Mechanical losses due to bearing friction and air
6/10/2025 Mr. Khalid A. 98
resistance (windage) while rotating.
 Separation of friction and windage loss from the no-load Test…
How: Key Insight from the Graph:

1.Vary the stator voltage from 125% to about 20% •Extrapolate the curve to zero voltage (or very low voltage).

of rated voltage. •At V=0, core loss and stator copper loss are nearly zero.

2.At each voltage level, record: •The remaining power at this point is almost entirely:

1. Input voltage V Friction and Windage Loss

2. No-load input current I

3. No-load power input P
3.Plot a graph of input power P vs voltage squared. 𝑉 2

Why use 𝑉 2 ?
•Core losses are approximately proportional to 𝑉 2 .
•Stator copper losses are proportional to 𝐼 2 .
•Friction and windage losses are mechanical and remain
approximately
6/10/2025
constant, since the speed changes very little.
Mr. Khalid A. 99
 Separation of friction and windage loss from the no-load Test…
➢ From each of the input-power readings, the corresponding stator ohmic loss is subtracted to obtain the core loss
and friction and windage loss, i.e

Where,
𝑃𝑛𝑙 is the per phase power input
𝐼𝑛𝑙 is the per phase stator current
𝑟1 is the effective per phase stator resistance
➢ When you perform a no-load test on an induction motor and vary the applied voltage, you're doing it to
separate the different components of loss — specifically to isolate the mechanical losses (friction and
windage).

6/10/2025 Mr. Khalid A. 100

 Examples
1. A 10-kW, 400-V, 4-pole delta connected squirrel cage induction motor gave the following test results:
No-load Test: 400-V , 8-A, 250-W
Blocked rotor Test: 90-V, 35-A, 1350-W
The dc resistance of the stator winding per phase measured after the blocked–rotor test is 0.6 Ω. Calculate the
rotational losses and the equivalent circuit parameters.
2. A 110 V, 3-phase star-connected induction motor takes 25 A at a line voltage of 30V with rotor locked. With this
line voltage power input to motor is 440 W and core loss is 40 W. The dc resistance between a pair of stator
terminals is 0.1 ohm. If the ratio of ac to dc resistance is 1.3, find the equivalent leakage reactance/phase of the
motor and the stator and rotor resistance per phase.
3. A 460-V, 25-hp, 60-Hz, four-pole, Y-connected wound-rotor induction motor has the following
impedances in ohms per phase referred to the stator circuit:

A. What is the maximum torque of this motor? At what speed and slip does it occur?
B. What is the starting torque of this motor?
C. When the rotor resistance is doubled, what isMr.
6/10/2025 the speed
Khalid A. at which the maximum torque now occurs?
101
D. What is the new starting torque of the motor?
 Starting of Induction Motors
A. Direct-On-Line (D.O.L.) Starting B. Stator Resistance or Reactor Starting
Working:- Motor is connected directly to full line Working: External resistors or reactors are inserted in series
voltage using a contactor. with the stator windings.
•It starts with maximum voltage, drawing high •They reduce voltage applied to the stator → reduces starting
inrush current. current and torque.
Features:- Starting current: 5–7 × full-load current Features: Lower voltage → lower current and torque
•Starting torque: 1.5–2.5 × full-load torque •Resistance dissipates power as heat
Advantages:-Simple, cheap, and easy to maintain Advantages: Simple control
•High starting torque •Limits starting current
Disadvantages: Disadvantages: Power loss in resistors
•High current surge can damage motor or cause •Starting torque is also reduced
voltage dips •Not efficient for large motors
•Not suitable for large motors or weak supply systems Application:-Used for medium-size motors where D.O.L. is
Application:-
6/10/2025 Small motors (usually < 5 kW) Mr. Khalid A.
not acceptable
102
 Starting of Induction Motors…
C. Autotransformer Starting D. Star-Delta Starting

Working: An autotransformer reduces the voltage during Working: The stator windings are initially connected in

startup, then switches to full voltage. star (Y) → voltage per phase reduced.

•Commonly provides 50%, 65%, or 80% of rated voltage at •After startup, they switch to delta (Δ) for normal operation.

start. Features: Star reduces voltage to 1/√3 (≈58%) → reduces

Features:-Starting current and torque depend on tap setting torque to 1/3

•Example: 50% tap → 25% torque (because torque ∝ voltage²) •After few seconds, it changes to full-voltage delta
Advantages:- Efficient Advantages: Reduces starting current (by ~67%)

•Allows adjustable starting voltage •Inexpensive compared to autotransformer

•Suitable for larger motors (above 10 kW) Disadvantages: Low starting torque

Disadvantages: More complex and expensive •Sudden current jump during star-to-delta switch

•Needs more panel space •Not suitable for high-load startup

Application: Large industrial motors with medium starting Application: Motors > 5 kW where load torque is low
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torque needs during start
 Starting of Induction Motors…
E. Rotor Resistance Starting (Slip-ring Induction Motors only)
Working:
•External resistors are connected to the rotor winding via slip rings.
•Gradually reduced as motor speeds up.
Features:
•Adding rotor resistance increases starting torque and limits current
Advantages:
•High starting torque
•Good control over acceleration
•Current remains within safe limits
Disadvantages:
•Only works with wound rotor (slip-ring) motors
•Requires maintenance of slip rings and brushes
Application:-Cranes,
6/10/2025 elevators, mills, where high torque
Mr. Khalidis
A. required at start 104
 Starting of Induction Motors…
Starting Methods and Motor Type Compatibility Explanation:

Squirrel Cage Wound Rotor (Slip- •Squirrel Cage Motors:

Starting Method
Motor ring) Motor • Most common motor type.
Direct-On-Line • All starting methods that do not require rotor
Yes Yes (rarely used)
(D.O.L.)
access (like adding resistance) are usable.
Stator
Yes Yes
Resistance/Reactance • D.O.L., Star-Delta, Stator Resistance,
Autotransformer
Yes Yes Autotransformer → all can be used.
Starting
•Wound Rotor (Slip-ring) Motors:
Yes (only if
Star-Delta Starting windings are No • Allow access to rotor via slip rings → so rotor
delta-connected) resistance starting is possible.
Rotor Resistance Yes (only slip-ring
No • Can also use D.O.L., stator resistance, and
Starting motors)
autotransformer if external resistance is not
needed.
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 Speed Control of Induction Motors
1. Control from Stator Side
(Works for both squirrel cage and wound rotor motors, unless specified)
➢ Changing Applied Voltage ➢ Constant V/f Control (Variable Frequency Drive -
• Principle: Reduce stator voltage → reduces torque (T ∝ V²) VFD)
• Speed reduces slightly under load due to torque drop •Principle: Change frequency (f) and voltage (V)
• Simple and low cost proportionally to keep V/f constant → maintain flux
•Only effective for lightly loaded motors • Best method for smooth and wide-range speed control
• Used for: Fans, pumps • Efficient, with good torque control
➢ Changing Number of Stator Poles
• Requires electronic drive (VFD), more expensive
(Pole changing method / multi-speed motor)
• Used in: Elevators, conveyors, HVAC systems
120𝑓
• Principle: N𝑠 = ​ → Changing poles (P) changes speed
𝑃

•Provides discrete step speed control (e.g., 2-speed or 4-speed motors)

• Not continuous; requires special winding design
6/10/2025 Mr. Khalid A. 106
• Used in: Machine tools, textile mills
 Speed Control of Induction Motors
2. Control from Rotor Side
(Only applicable to wound rotor (slip-ring) induction motors)
➢ Rotor Rheostat Control ➢ Injecting e.m.f in Rotor Circuit
• Principle: •Principle:
Add external resistors to rotor circuit → increases slip → reduces speed Inject external voltage (with same frequency
•Simple method for low-speed, high-torque needs as slip) into rotor → modifies rotor current and
•Inefficient (power loss in resistors), not used for squirrel cage motors slip → controls speed
•Used in: Cranes, hoists •Precise speed control (can even increase
➢ Cascade Operation speed above synchronous)
•Principle: Two motors (main and auxiliary) are coupled and •Requires frequency-matching power source
connected in cascade (rotor of one connected to stator of the other) (complex)
•Allows step changes in speed •Used in: Slip power recovery systems
•Bulky and complex
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•Used where multi-speed operation is needed
 Applications of Induction Motors
Preferred For:
1. Squirrel-Cage Induction Motors
❖ These are applications with low to moderate starting torque
Characteristics:
requirements and where constant running speed is needed.
•Simple, rugged, and maintenance-free
•Moderate starting torque
•Nearly constant speed under varying load
•Less expensive and compact

Application Reason for Suitability

Fans & Blowers Low starting load, runs at constant speed
Water Pumps Moderate torque, constant speed
Grinders Smooth operation, does not require high starting torque
Lathe Machines Consistent speed, good efficiency
Printing Machines Minimal torque variation, reliable performance
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Drilling Machines Moderate load, low maintenance
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6/10/2025 Mr. Khalid A. 109