ABOUT

Best Online Coaching for all AE/JE

exam for Electrical Engineering. One
stop solution for all AE/JE Exam

ENGINEERS CAREER
INSTITUTE
BY SURESH SIR
 ENGINEERS CAREER INSTITUTE
P = 8: Ns = 750 RPM
Induction Motor P = 10: Ns = 600 RPM
Approx. 80% to 90% machines are
induction motors, because: Example 1: A 3-ϕ, 6-pole squirrel cage
induction motor runs at 940 RPM. Find
Simple construction the slip.
Easy maintenance
No armature reaction effect Given:
Low cost
P = 6, Nr = 940 RPM, f = 50 Hz

Synchronous speed, Ns = (120 × f) / P =

Operating Principle
(120 × 50) / 6 = 1000 RPM
When a 3-ϕ supply is given to a 3-ϕ
Slip, S = ((Ns - Nr) / Ns) × 100
winding 120° displaced w.r.t. time, due to
this a revolving magnetic field is S = ((1000 - 940) / 1000) × 100 = (60 /
produced which revolves at a constant 1000) × 100 = 6%
speed, called synchronous speed.
Example 2:
Synchronous speed (Ns) is given by:
A 3-ϕ, 4-pole 50 Hz induction motor has a
120𝑓
Ns = 𝑃 slip of 4%. Find the rotor speed.

IM always rotates in the direction of P = 4, S = 4% = 0.04, f = 50 Hz

revolving magnetic field.
Synchronous speed, Ns = (120 × f) / P =
IM can never rotate at synchronous speed
(120 × 50) / 4 = 1500 RPM
because at synchronous speed: EMF,
current & torque = 0 Rotor speed, Nr = (1 - S) × Ns
This motor is also called Asynchronous
Motor Nr = (1 - 0.04) × 1500 = 1440 RPM
IM Also called rotating transformer
Slip at Different Conditions
Slip Speed During starting: Rotor speed, Nr = 0
120𝑓
Synchronous speed (Ns) = Slip, S = (Ns - Nr) / Ns = (Ns - 0) / Ns = 1
𝑃

Slip speed (SNs) = Ns - Nr ⇒ S = 1 (i.e., 100%)

Ns − Nr At running condition: Slip,

Slip (S) = 𝑁𝑠
S = (Ns - Nr)/Ns
At frequency f = 50 Hz
⇒ Slip typically ranges from 1% to 5%
P = 2: Ns = (120 × 50) / 2 = 3000 RPM
P = 4: Ns = (120 × 50) / 4 = 1500 RPM
P = 6: Ns = (120 × 50) / 6 = 1000 RPM

CONTACT NUMBER-: 8077071848

 ENGINEERS CAREER INSTITUTE
Effect of Slip on Different ➢ Design → Quite Difficult
➢ Power factor → Moderate (air
Parameters :- gap)
➢ Power factor → Moderate
1) Rotor Frequency (fr) = sf
(leakage reactance)
2) Rotor EMF: E₂' = SE₂, where E₂ is ➢ Generally preferred slot type:
standstill EMF Semi-closed
➢ For large rating machines: Open
3) Rotor Reactance: X₂' = S X₂ slot is used because:
• Maintenance is easy
4) Rotor Resistance: No change
• Construction is simple
5) Rotor Current or Power Factor:
SE₂) Rotor Construction
I₂' =
(R₁ + jSX₂)

𝑅2
cos(ϕ₂') = Squirrel Cage Rotor
√𝑅2 2 +(𝑠𝑋2) 2

➢ No winding and no terminal.

Construction of Induction Motor ➢ Solid bars are used:
➢ Copper (Cu) for large rating
➢ Stator → Stationary part →
machines
Contains distributed winding.
➢ Aluminum (Al) for small rating
➢ Magnitude of resultant magnetic
machines
field: φr = (3/2) × φm
➢ With the help of end rings, the
solid bars remain short-circuited.
➢ This type of construction is used
Types of Stator Slots
in low starting torque
Open Slot: applications.
➢ In this rotor, slots are slightly
➢ Design → Easy skewed to avoid cogging
➢ With respect to air gap, power problems.
factor → Low ➢ Rotor slots should not be equal or
➢ With respect to leakage reactance, in integral multiple to avoid
power factor → High cogging.

Closed Slot: Slip Ring Induction Motor

(Wound Rotor IM)
➢ Design → Difficult
➢ Power factor → High (air gap) ➢ Rotor winding is essentially star
➢ Power factor → Low (leakage connected.
reactance) ➢ Used in high starting torque
applications.
Semi-Closed Slot:
➢ No problem of crawling.

CONTACT NUMBER-: 8077071848

 ENGINEERS CAREER INSTITUTE
➢ At running condition, external ➢ If R2 changes → Tmax remains
resistance is short-circuited with constant, but slip at which
the help of metal collar maximum torque occurs will
arrangement. changes Sm changes
➢ Starting torque (Tst) is directly ➢ Tst (Starting Torque) → Increases
proportional to external first, then decreases
resistance (Rext). ➢ All the operating region is a stable
region but its converse is not true
Starting Torque and Maximum Torque Conditions for Maximum Torque
➢ If an induction motor starts with
Starting Torque (Tst):
its maximum torque, then Sm = 1.
3 × 60 E2² R2 ➢ If external resistance is added to
Tst = 2πNs R2² + X2² the rotor circuit, then:
(R2 + Rext)
➢ Condition for maximum starting 𝑆𝑚 = X2
torque: R2 = X2 Equivalent Circuit of Induction Motor

Maximum Torque (Tmax):

3 × 60 𝐸2 2
Tmax = 2πNs 2𝑋2

➢ Maximum torque does not depend

on rotor resistance.
Electrical equivalent of mechanical load
➢ Power factor at maximum torque
RL = R2(1/S - 1),
= 1/√2 (lagging)

Running Torque (Tfl): Testing of Induction Motor

3 × 60 sE2² R2
Tfl = 1. 1. No-load test
2πNs R2² +(sX2)²
2. 2. Blocked rotor test
➢ Condition for maximum torque:
R2 = S X2 1. No Load Test
➢ Slip at which maximum torque
occurs: Sm =𝑋2
𝑅 ➢ This test is performed at a rated
2 voltage and rated frequency.
➢ Power factor at maximum torque ➢ In this test, no load is connected
= 1/√2 (lagging) to the shaft, but the shaft is freely
rotating.
➢ In this test, the no-load current
Torque-Slip Characteristics magnitude is 30% to 40% of rated
current.
➢ Low slip region → Straight line ➢ Losses → Mechanical loss + Stator
➢ High slip region → Rectangular core loss (V, f → rated) + Stator
hyperbola copper loss

CONTACT NUMBER-: 8077071848

 ENGINEERS CAREER INSTITUTE
➢ Low power factor (pf) η = Po / Pi

Blocked Rotor Test

➢ This test is equivalent to short
circuit test of Transformer (T/F).
➢ This test is performed at a
reduced voltage and reduced
frequency.
➢ High power factor (pf).

➢ In this test, rated current flows. Rotor Power and Copper

➢ Losses = Rotor copper loss + Loss Equations
Stator copper loss ➢ Rotor Power Input:
Pᵢ = Tg × ω = Tg × (2πNₛ / 60)
Power Flow and Loss Equations in ➢ Rotor Power Output:
Induction Motor Pₒ = Tg × ω = Tg × (2πNᵣ / 60)
➢ Rotor Copper Loss (R.C.L):
➢ Stator input - Stator losses = R.C.L = S Pᵢ
Stator output Rotor Output and Efficiency
➢ Stator output = Air gap power = Derivation
Rotor inpu ➢ ∴ Rotor input - Rotor Copper Loss
(Pg = Rotor input) (RCL) = Rotor output
➢ Rotor input - Rotor copper loss = ➢ Rotor input - S ×Power input =
Rotor output Rotor output
➢ Rotor output = Mechanical power ➢ Rotor Efficiency = Pₒ / Pᵢ = 1 - S
developed
Why we need starter
➢ Pm = Mechanical Power Developed ➢ During starting, the motor will
➢ Rotor output - Friction and draw a huge amount of current
windage loss = Shaft output from the supply, which produces a
➢ Efficiency voltage dip in the supply. This is
➢ Efficiency (η) = Shaft output / not acceptable. So, in order to
Stator input limit this current, a starter is used.
Types of Starters
• DOL Starter
• Star Delta Starter
• Auto Transformer Starter
• Stator Resistance Starter
• Rotor Resistance Starter (only for
slip ring I.M.)

CONTACT NUMBER-: 8077071848