Three phase induction motors

Chapter Three
Poly phase Induction Machines
By Yimam A.
May 12, 2022

Chapter 3
Outline
1 Introduction
2 Review of 3 phase Induction Machine
3 Equivalent circuit and Phasor diagram
4 Torque and Efficiency of 3 phase induction motor
5 Induction Motor Torque-Speed Characteristics
6 Computation and circle diagram
By Yimam A. Chapter 3 May 12, 2022 2 / 93

Chapter 3
Learning Objectives
At the end of this chapter the students should be able to understand:
✒ Understand the construction and working principle of an induction motor.
✒ Understand the concept of rotor slip and its relationship to rotor frequency.
✒ Understand and know how to use the equivalent circuit of an induction motor.
✒ Understand power flows and the power flow diagram of an induction motor.
✒ How to use the equation for the torque-speed characteristic curve for different rotor
design
✒ How to compute performance of 3 phase induction motor using circle diagram

Chapter 3
Introduction
Introduction
Induction Motors transform electrical energy into mechanical energy.
Induction motors are popularly used in the industry, in fact, 90% of industrial motors
are induction motors.
By induction motor, we mean that the stator windings induces a current which flow in
the rotor conductors, like a transformer.
Induction motors are used worldwide in many residential, commercial, industrial, and
utility applications. It can be part of a pump or fan, or connected to some other form
of mechanical equipment such as a winder, conveyor, or mixer.

Chapter 3
Review of 3 phase Induction Machine
The two basic parts of an AC motor are the stator and rotor.
The stator and the rotor are electrical circuits that perform as electromagnets.
The stator consists of stator frame, core and stator winding.
The rotor is the rotating part of the electromagnetic circuit.It can be found in two
types:
Squirrel cage rotor
Wound(slip ring) rotor
However, the most common type of rotor is the squirrel cage rotor.

Chapter 3
Cont...
(a) Rotor (b) Stator

Chapter 3
Working Principle of 3 Phase Induction Motor
When three phase supply is given to the three phase stator winding of the induction
motor, a rotating magnetic field is developed around the stator which rotates at syn-
chronous speed.
This rotating magnetic field passes through the air gap and cuts the rotor conductors
which were stationary.
Due to the relative speed between the stationary rotor conductors and the rotating
magnetic field, an emf is induced in the rotor conductors.
As the rotor conductors are short circuited, current starts flowing through it.
As these current carrying rotor conductors are placed in the magnetic field produced
by the stator, they experiences a mechanical force i.e. torque which moves the rotor in
the same direction as that of the rotating magnetic field.

Chapter 3
The Development of Induced Torque in an Induction Motor
A three-phase set of voltages has been applied to the stator, and a three-phase set of
stator currents is flowing.
These currents produce a magnetic field Bs which is rotating in a clockwise or coun-
terclockwise direction.
The speed of the magnetic field’s rotation is given by
ns =
120f
P
Where f is the system frequency applied to the stator in hertz
P is the number of poles in the machine.
ns is called the synchronous speed in rpm

Chapter 3
Cont...
This rotating magnetic field Bs passes over the rotor bars and induces a voltage in
them.
The voltage induced in a given rotor bar is given by the equation
eind = (V × B).l
where v= Velocity of the bar relative to the magnetic field
B= Magnetic flux density vector
l= length of conductor in the magnetic field
It is the relative motion of the rotor compared to the stator magnetic field that produces
induced voltage in a rotor bar.

Chapter 3
Cont...
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 in the rotor windings.
The rotor current flow produces a rotor magnetic field BR.
A torque is produced as a result of the interaction of those two magnetic fields
τind = kBR × BS
Where τind is the induced torque and
k is a constant representing the construction of the machine.
BR and BS are the magnetic flux densities of the rotor and the stator respectively

Chapter 3
Induction Motor Speed
At what speed will the induction motor run?
Can the induction motor 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.
The induction motor will always run at a speed lower than the synchronous speed.

Chapter 3
Rotor Slip
The voltage induced in a rotor bar of an induction motor depends on the speed of the
rotor relative to the magnetic fields.
The difference between the motor speed and the synchronous speed is called the Slip
speed.
nslip = ns − nm
Where nslip is slip speed of the machine
ns is speed of the rotating magnetic field
nm is mechanical shaft speed of the motor
In normal operation both the rotor and stator magnetic fields BR and BS rotate to-
gether at synchronous speed ns, while the rotor itself turns at a slower speed.

Chapter 3
Slip
Slip is the relative speed expressed on a per-unit or a percentage basis.
s =
nslip
ns
× 100% =
ns − nm
ns
× 100%
Where s is the slip
This equation can also be expressed in terms of angular velocity ω (radians per second)
as
s =
ωs − ωm
ωs
× 100%
If the rotor runs at synchronous speed, s = 0 if the rotor is stationary, s =1

Chapter 3
Cont...
The mechanical speed of the rotor shaft can be expressed in terms of synchronous speed
and slip.
From the above expression for slip, solving equations for mechanical speed yields
nm = (1 − s)ns or ωm = (1 − s)ωs
These equations are useful in the derivation of induction motor torque and power
relationships.

Chapter 3
Induction Motors and Transformers
Both induction motor and transformer works on the principle of induced voltage.
Transformer: voltage applied to the primary windings produce an induced voltage in
the secondary windings.
Induction motor: voltage applied to the stator windings produce an induced voltage in
the rotor windings.
The difference is that, in the case of the induction motor, the secondary windings can
move.
Due to the rotation of the rotor (the secondary winding of the IM), the induced voltage
in it does not have the same frequency of the stator (the primary) voltage.

Chapter 3
Frequency of rotor
The frequency of the voltage induced in the rotor is given by
fr = sf
Since the slip of the rotor is defined as
s =
ns − nm
ns
fr = (ns − nm)
P
120f
× f ⇒ fr =
P
120
(ns − nm)
Where fr is rotor frequency (Hz) , f is supply frequency(Hz) ,and s is slip

Chapter 3
Cont...
What would be the frequency of the rotor’s induced voltage at any speed nm ?
fr = sf
When the rotor is blocked (s = 1) , the frequency of the induced voltage is equal to
the supply frequency.
If the rotor runs at synchronous speed (s = 0) , the frequency will be zero.

Chapter 3
Torque
While the input to the induction motor is electrical power, its output is mechanical
power.
Any mechanical load applied to the motor shaft will introduce a torque on the motor
shaft. This torque is related to the motor output power and the rotor speed.
τload =
pout
ωm
and ωm =
2πnm
60
rad/s
Another unit used to measure mechanical power is the horse power and
1hp = 746W

Chapter 3
Cont...
Example 1
A 3 phase, 460 V, 100 hp, 60 Hz, 4 pole induction machine delivers rated output power at
a slip of 5 %. Determine the
a) Synchronous speed and motor speed.
b) Speed of the rotating air gap field.
c) Frequency of the rotor circuit.
d) Slip speed (rpm)
e) Speed of the rotor field relative to the
(i) rotor structure , (ii) stator structure and (iii) stator rotating field.
f) Rotor induced voltage at the operating speed, if the stator to rotor turns ratio is 1:0.5.

Chapter 3
Equivalent circuit and Phasor diagram
Equivalent circuit of three phase induction motor
The induction motor is similar to the transformer with the exception that its secondary
windings are free to rotate.
Figure 2: The transformer model of an induction motor, with rotor and stator connected by an
ideal transformer of turns ratio a

Chapter 3
The Rotor Circuit Model
In general, the greater the relative motion between the rotor and the stator magnetic
fields, the greater the resulting rotor voltage and rotor frequency.
When the rotor is locked (or blocked), i.e. s =1, the largest voltage and rotor frequency
are induced in the rotor.Why?
If the rotor rotates at synchronous speed, i.e. s = 0, the induced voltage and frequency
in the rotor will be equal to zero. Why?
ER = sERO
Where ERO is the largest value of the rotor’s induced voltage obtained at s=1(locked
rotor). The same is true for the frequency, i.e.fr = sf

Chapter 3
Cont...
It is known that X = ωL = 2πfL
So, as the frequency of the induced
voltage in the rotor changes, the
reactance of the rotor circuit also
changes.
XR = ωrLR = 2πfrLR
= 2πsfLR = s (2πfLR)
= sXRO
XRO is the rotor reactance at the
supply frequency(at blocked rotor)
Where ER is the induced voltage in
the rotor and RR is the rotor resis-
tance

Chapter 3
Cont...
Now we can calculate the rotor current as
IR =
ER
(RR + jXR)
=
sERo
(RR + jsXRo)
Dividing both the numerator and denominator by s, we get
IR =
ERo

RR
s + jXRo

Where ERo is the induced voltage and XRo is the rotor reactance at blocked rotor
condition (s = 1)

Chapter 3
Cont...
Figure 3: The rotor circuit model with all the frequency (slip) effects concentrated in resistor RR.

Chapter 3
Cont...
Now as we managed to solve the induced voltage and different frequency problems, we
can combine the stator and rotor circuits in one equivalent circuit.
Figure 4: The per-phase equivalent circuit of an induction motor

Chapter 3
Cont...
Where
X2 = a2
XRo
R2 = a2
RR
I2 =
IR
a
E1 = aERo
a =
NS
NR

Chapter 3
Approximate equivalent circuit
(a) (b)
Figure 5: Approximate equivalent circuit

Chapter 3
IEEE-Recommended Equivalent Circuit
In the induction machine, because of its air gap, the exciting current Iϕ is high of the
order of 30 to 50 % of the full-load current. The leakage reactance X1 is also high.
The IEEE recommends that in such a situation, the magnetizing reactance XM not be
moved to the machine terminals, but be retained at its appropriate place.
The resistance Rc is however omitted and the core loss is lumped with the windage
and friction losses.
This equivalent circuit( figure 6) is to be preferred for situations in which the induced
voltage E1 differs appreciably from the terminal voltage V1 .

Chapter 3
Cont...
Figure 6: IEEE-Recommended Equivalent Circuit

Chapter 3
Phasor Diagram of 3 phase Induction motor
Let us take the mutual flux ϕ between the stator and rotor windings as the reference
phasor.This flux induces emf’s in the stator and rotor.
These emf’s under running conditions are E1 and sE2 respectively lagging ϕ by 90◦.
Since the rotor winding is a short circuit, the voltage sE2 sets up the rotor current I2
lagging behind by an angle ϕ2 , and sE2 will be equal to rotor impedance drop.
ϕ2 = tan−1

sX2
R2

sE2 = I2 (R2 + jsX2)

Chapter 3
Phasor diagram of induction motor on load condition
Io = Iϕ = No load current
V1 = I1R1 + I1X2 − E1
sE2 = I2R2 + sI2X2

Chapter 3
Cont...
The current flowing in the stator winding i.e., stator current is
I1 = IO + I′
2
Where
IO = No load current of the induction motor.
I′
2 = Rotor current referred to the stator.
The phasor sum of E1, I1R1 and I1X2 gives the applied voltage V1.
cos ϕ1 is the p.f. of the stator and cos ϕ2 is the p.f. of the rotor circuit.

Chapter 3
Torque and Efficiency of 3 phase induction motor
Losses and the power flow diagram
Power Losses in Induction Machines
Copper losses
Copper loss in the stator (PSCL) = (I1)
2
R1
Copper loss in the rotor (PRCL) = (I2)
2
R2
Core loss (Pcore)
Mechanical power loss due to friction and windage
How this power flow in the induction motor?
The relationship between the input electrical power and the output mechanical power of
this motor is shown in the power flow diagram.

Chapter 3
Cont...
Figure 7: Power flow diagram of an induction motor

Chapter 3
Cont...
The input power to an induction motor Pin in the form of three-phase electric voltages
and currents.
The first losses encountered in the machine are I2R losses in the stator windings is
stator copper loss ( PSCL). Then some amount of power is lost as hysteresis and eddy
currents in the stator (PCore).
The power remaining at this point is transferred to the rotor of the machine across the
air gap between the stator and rotor. This power is called the air gap power PAG of
the machine.
After the power is transferred to the rotor, some of it is lost as I2R losses(rotor copper
loss (PRCL))
The rest is converted from electrical to mechanical form (Pconv).

Chapter 3
Cont...
Finally, friction and windage losses (PFW ) and stray losses Pmisc are subtracted.
The remaining power is the output of the motor Pout.
The higher the speed of an induction motor, the higher its friction, windage, and stray
losses.
The higher the speed of the motor (up to ns), the lower its core losses.
These three categories of losses are sometimes lumped together and called rotational losses.
PAG : PRCL : Pconv
1: s: 1 − s
Pin PSCL Pcore PAG PRCL Pconv Pfw Pstray Pout

Chapter 3
Power and Torque in an Induction Motor
The input current to a phase of the
motor can be found by
I1 =
Vϕ
Zeq
Zeq = R1 + jX1 +
1
GC − jBM + 1
R2
S
+jX2
Pin =
√
3VLIL cos θ = 3VphIph cos θ
PSCL = 3I2
1 R1
PAG = Pin − (PSCL + Pcore)
Pconv = PAG − PRCL
Pout = Pconv − (Pf+w + Pstray)
Pcore = 3E2
1GC
PRCL = 3I2
2 R2
τind =
Pconv
ωm
PAG = Pconv + PRCL

Chapter 3
Cont...
PAG = Pin − (PSCL + Pcore)
= Pconv + PRCL = 3I2
2
R2
s
=
PRCL
s
Pconv = PAG − PRCL = 3I2
2
R2 (1 − s)
s
=
PRCL (1 − s)
s
= (1 − s) PAG
τind =
Pconv
ωm
=
(1 − s) PAG
(1 − s) ωs
=
PAG
ωs

Chapter 3
Cont...
The induced torque (τind) in a machine is the torque generated by the internal electrical
to mechanical power conversion.
This torque differs from the torque actually available at the terminals of the motor by
an amount equal to the friction and windage torques in the machine.
This torque is also called the developed torque of the machine.
The power converted from electrical to mechanical form is called converted power,
which is sometimes called developed mechanical power.
If the rotor is not turning, the slip s=1 and the air gap power is entirely consumed in
the rotor. since if the rotor is not turning, the output power (Pout = τloadωm) must be
zero.
The air gap power is the power crossing the gap from the stator circuit to the rotor
circuit. It is equal to the power absorbed in the resistance R2/S.

Chapter 3
Separating the Rotor Copper Losses and the Power Converted
Since air gap power would require R2
S and rotor copper loss require R2 element.
Figure 8: Per-phase equivalent circuit with the rotor copper losses and the power converted to
mechanical form separated into distinct elements

Chapter 3
Cont...
Example 2
A 460-V,25-hp,60 Hz,four-pole, Y-connected induction motor has the following impedances
in ohms per phase referred to the stator circuit.
R1 = 0.641Ω R2 = 0.332Ω X1 = 1.106Ω X2 = 0.464Ω XM = 26.3Ω
The total rotational losses are 1100 W and are assumed to be constant. The core loss is
lumped in with the rotational losses. For a rotor slip of 2.2 % at the rated voltage and
rated frequency, find the motor’s
a) Speed (ns)
b) Stator current
c) Stator power factor
d) PconvPout
e) τindPload
f) Efficiency

Chapter 3
Cont...
Example 3
A 480-V,60 Hz,six-pole,three phase delta-connected induction motor has the following
parametres per phase referred to the stator circuit.
R1 = 0.641Ω R2 = 0.258Ω X1 = 5.07Ω X2 = 0.309Ω XM = 30.74Ω
The total rotational losses are 2450 W and are assumed to be constant. The motor drives
a mechanical load at a speed of 1170 rpm.calculate
a) ns in rpm
b) Slip
c) Line current
d) Input power
e) Air gap power
f) Torque developed
g) Output power(hp)
h) Efficiency

Chapter 3
Induction Motor Torque-Speed Characteristics
Thevenin equivalent circuit
Thevenin’s theorem can be used to transform the network to the left of points ‘a’ and ‘b’
into an equivalent voltage source VTH in series with equivalent impedance RTH +jXTH

Chapter 3
Cont...
VTH = Vϕ
jXM
R1 + j (X1 + XM )
|VTH| = |Vϕ|
XM
R2
1 + (X1 + XM )2
ZTH =
jXM (R1 + jX1)
R1 + j (X1 + XM )
RTH + jXTH = (R1 + jX1) //jXM

Chapter 3
Cont...
The current I2 is given by
I2 =
VTH
ZTH + Z2
=
VTH
RTH + R2
S + jXTH + jX2

Chapter 3
Cont....
Since XM ≫ X1 and XM ≫ R1
VTH ≈ Vϕ
XM
X1 + XM
Because XM ≫ X1 and XM + X1 ≫ R1 ,the equivalent impedance
ZTH =
jXM (R1 + jX1)
R1 + jX1 + jXM
RTH ≈ R1

XM
X1 + XM
2
XTH ≈ X1

Chapter 3
Cont...
I2 =
VTH
ZT
=
VTH
r
RTH + R2
S
2
+ (XTH + X2)2
Then the power converted to mechanical Pconv
Pconv = 3I2
2
R2 (1 − s)
s
The mechanical induced torque (τind) is given by
τind =
Pconv
ωm
=
Pconv
ωs(1 − s)
=
3I2
2
R2
s
ωs
=
PAG
ωs

Chapter 3
Cont...
τind =
3
ωs




VTH
r
RTH + R2
s
2
+ (XTH + X2)2




2
×

R2
s

τind =
1
ωs
3V 2
TH

R2
s

RTH + R2
S
2
+ (XTH + X2)2

Chapter 3
Torque-speed characteristics of induction motor

Chapter 3
Comments
1 The induced torque is zero at synchronous speed.
2 The curve is nearly linear between no-load and full load. In this range, the rotor
resistance is much greater than the reactance, so the rotor current, torque increase
linearly with the slip.
3 There is a maximum possible torque that can’t be exceeded. This torque is called
pullout torque or breakdown torque, and is 2 to 3 times the rated full-load torque.
4 The starting torque of the motor is slightly higher than its full-load torque, so the
motor will start carrying any load it can supply at full load.
5 The torque of the motor for a given slip varies as the square of the applied voltage.
6 If the rotor is driven faster than synchronous speed it will run as a generator, converting
mechanical power to electrical power.

Chapter 3
Complete Torque-Speed characteristics curve

Chapter 3
Cont...
7 If the motor is turning backward relative to the direction of the magnetic fields, the
induced torque in the machine will stop the machine very rapidly and will try to rotate
it in the other direction.
Since reversing the direction of magnetic field rotation is simply a matter of switching any
two stator phases, this fact can be used as a way to very rapidly stop an induction motor.
The act of switching two phases in order to stop the motor very rapidly is called plugging.

Chapter 3
Modes of operation of induction motor
The induction machine can be operated in three modes: motoring, generating, and
plugging.
1) Motoring,0 nm nsync, 1 s 0
If the stator terminals are connected to a three-phase supply, the rotor will rotate in
the direction of the stator rotating magnetic field.
This is the natural (or motoring) mode of operation of the induction machine.
The steady state speed nm is less than the synchronous speed ns.
Torque and motion are in the same direction.

Chapter 3
Cont...
2) Generating nm nsync, s 0
In this mode, again torque is positive while speed is negative. However, unlike plugging,
Pconv = (1 − s) PAG
indicates that if the power converted is negative, so is the air gap power.
In this case, power flows from the mechanical system, to the rotor circuit, then across
the air gap to the stator circuit and external electrical system.

Chapter 3
cont...
3) Braking, nm 0, s 1
Torque is positive while speed is negative. Considering the power conversion equation
Pconv = (1 − s) PAG
It can be seen that if the power converted is negative (P = τω) then the airgap power
is positive. i.e. the power is flowing from the stator to the rotor and also into the rotor
from the mechanical system.
This operation is also called plugging.

Chapter 3
Cont...
This mode of operation can be used to quickly stop a machine.
If a motor is travelling forwards it can be stopped by interchanging the connections to
two of the three phases.
Switching two phases has the result of changing the direction of motion of the stator
magnetic field, effectively putting the machine into braking mode in the opposite di-
rection.

Chapter 3
Cont...
If an induction motor is driven at a speed greater than nsync by an external prime
mover, the direction of its induced torque will reverse and it will act as a generator.
As the torque applied to its shaft by the prime mover increases, the amount of power
produced by the induction generator increases.
There is a maximum possible induced torque in the generator mode of operation. This
torque is known as pushover torque of the generator.
If a prime mover applies a torque greater than the pushover torque to the shaft of an
induction generator, the generator will over speed.

Chapter 3
Cont...

Chapter 3
Maximum(Pullout) Torque
Maximum power will be achieved when the magnitude of source impedance matches
the load impedance.
Maximum torque occurs when the power transferred to R2/s is maximum.
This condition occurs when R2/s equals the magnitude of the impedance RTH +j(XTH +
X2)
R2
Smax
=
q
R2
TH + (XTH + X2)2
Smax =
R2
q
R2
TH + (XTH + X2)2

Chapter 3
Cont...
The corresponding maximum torque per phase of an induction motor is
τmax =
1
2ωs


3V 2
TH
RTH +
q
R2
TH + (XTH + X2)2


The slip at maximum torque is directly proportional to the rotor resistance R2 .
The maximum torque is independent of R2 and related to the square of applied voltage.
Rotor resistance can be increased by inserting external resistance in the rotor of a
wound-rotor induction motor.
The value of the maximum torque remains unaffected but the speed at which it occurs
can be controlled.

Chapter 3
Cont...
It is possible to take advantage of
this characteristic of wound rotor in-
duction motors to start very heavy
loads.
If a resistance is inserted into the ro-
tor circuit, the maximum torque can
be adjusted to occur at starting con-
ditions.
Therefore, the maximum possible
torque would be available to start
heavy loads.

Chapter 3
Cont..
Example 4
A 460 V, 25hp, 60-Hz, four-pole, Y-connected wound rotor induction motor has the following
impedances in ohms per phase referred to the stator circuit
R1 = 0.641Ω, R2 = 0.332Ω, X1 = 1.106Ω, X2 = 0.464Ω, XM = 26.3Ω
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) If the rotor resistance is doubled, what is the speed at which the maximum torque now
occur?
d) What is the new starting torque of the motor?

Chapter 3
Tests on Three phase Induction Motor
Determination of Motor Parameters
Due to the similarity between the induction motor equivalent circuit and the trans-
former equivalent circuit, same tests are used to determine the values of the motor
parameters.
The equivalent circuit parameters of polyphase induction motors can be determined
from no load test, blocked rotor test and stator winding DC resistance.
DC test: determine the stator resistance R1
No-load test: determine the rotational losses and magnetization current (similar to no-
load test in Transformers).
Locked-rotor test: determine the rotor and stator impedances (similar to short-circuit test
in Transformers).

Chapter 3
DC Test
The purpose of the DC test is to determine R1. A variable DC voltage source is
connected between two stator terminals.
The DC source is adjusted to provide approximately rated stator current, and the
resistance between the two stator leads is determined from the voltmeter and ammeter
readings.

Chapter 3
Cont...
then
RDC =
VDC
IDC
The current flows through two of the windings, so the total resistance in the current
path is 2R1 . Therefore,
2R1 =
VDC
IDC
If the stator is Y-connected, the per phase stator resistance is
R1 =
RDC
2
If the stator is delta connected, the per phase stator resistance is
R1 =
3
2
RDC

Chapter 3
No load Test
The no load test of an induction motor measures the rotational losses of the motor and
provides information about its magnetization current.
1 The motor is allowed to spin freely
2 The only load on the motor is the friction and windage losses, so all Pconv is consumed
by mechanical losses.

Chapter 3
Cont...
3 The slip is very small. At this small slip
R2 (1 − s)
s
≫ R2
R2 (1 − s)
s
≫ X2
The equivalent circuit reduces to

Chapter 3
Cont...
5 Combining RC and RFW we get

Chapter 3
Cont...
6 At the no load conditions, the input power measured by meters must equal the losses
in the motor.
7 The PRCL is negligible because I2 is extremely small because R2(1−s)
s is very large.
8 The input power equals
Pin = PSCL + Pcore + Pfw = 3I2
1 R1 + Prot
9 The equivalent input impedance is thus approximately
|Zeq| =
Vϕ
I1,nl
=
Vϕ
IL
≈ X1 + XM
If X1 can be found, the magnetizing impedance XM will be known

Chapter 3
Locked Rotor (Blocked Rotor) Test
In this test, the rotor is locked or blocked so that it cannot move, a voltage is applied
to the motor, and the resulting voltage, current and power are measured.

Chapter 3
Cont...
Since the rotor is not moving,the slip s = 1 and so the rotor resistance R2/s is just equal
to R2 (quite small value).Since R2 and X2 are so small,almost all the input current will
flow through them,instead of through the larger magnetizing reactance XM
Therefore,the circuit under these conditions looks like a series combination of X1, R1
,X2 and R2

Chapter 3
Cont...
The AC voltage applied to the stator is adjusted so that the current flow is approxi-
mately full load value. The input power to the motor is given by
P =
√
3VLIL cos θ
The locked rotor power factor can be found as
PF = cos θ =
Pin
√
3VLIL
The magnitude of the total impedance
|ZLR| =
Vϕ
I
|ZLR| =
Vϕ
I1
=
VT
√
3IL

Chapter 3
Cont...
|ZLR| = RLR + jX′
LR = |ZLR| cos θ + j|ZLR| sin θ
RLR = R1 + R2 X′
LR = X′
1 + X′
2
Where X′
1 and X′
2 are the stator and rotor reactances at the test frequency respectively.
R2 = RLR − R1
XLR =
frated
ftest
× X′
LR = X1 + X2

Chapter 3
Rules of thumb for dividing rotor and stator circuit reactance
Rotor design X1 X2
Wound rotor 0.5XLR 0.5XLR
Design A 0.5XLR 0.5XLR
Design B 0.4XLR 0.6XLR
Design C 0.3XLR 0.7XLR
Design D 0.5XLR 0.5XLR

Chapter 3
Computation and circle diagram
Circle diagram of three phase induction motor
The circle diagram is a graphical representation of the performances of an induction
motor.
It is very useful to study the performance of an induction motor under all operating
conditions.
Its construction is based on the approximate equivalent circuit.
The data required for the construction of circle can be obtained from no load test,
blocked rotor test and stator resistance test.
From the no-load test, Io and ϕO have been calculated.
From blocked-rotor test, short-circuit current (Isc) and ϕsc have been calculated.
By conducting the no load and blocked rotor tests and by drawing circle diagram,one
can compute all the performance characteristics of the motor on all possible loads.

Chapter 3
Tests Required
No-load Test
Run the induction motor on no-load at rated supply voltage.
Observe the supply line voltage Vo, No-load line current Io and no-load power Po.
Phase angle for no-load condition
ϕo =
Po
√
3VoIo
Blocked Rotor Test
Block the rotor firmly and apply a reduced voltage to obtain rated current at the motor
terminals.
Observe the supply line voltage Vsc, No-load line current Isc and no-load power Psc.

Chapter 3
Cont...
Phase angle for blocked rotor condition
ϕsc =
Psc
√
3VscIsc
Current drawn if rated voltage is applied at blocked rotor condition ISN = Isc
Vo
Vsc
Power input at rated voltage and motor in the blocked rotor condition PSN = Psc

Vo
Vsc
2
Resistance Test
By voltmeter ammeter method determine per phase equivalent stator resistance, R1.
If the machine is wound rotor type, find the equivalent rotor resistance R′
2 also after
measuring rotor resistance and required transformations are applied.

Chapter 3
Construction of a Circle Diagram for an Induction Motor
The circle diagram of an induction motor may be drawn by using the data obtained
from (i) stator resistance test, (ii) no-load test and (iii) blocked rotor test.
From stator resistance test, the stator winding resistance per phase R1 is determined.
From no-load test, the phase voltage V

V = VL
√
3

, no-load current IO and no-load
power input PO is measured and from this data no-load power factor angle is deter-
mined.
From blocked rotor test, the corresponding value of short circuit current at normal
stator applied voltage ISN

ISN = V
Vsc
× Isc

and short circuit phase angle ϕsc are
determined.

Chapter 3
Cont...
From this data, a circle diagram for an induction motor is drawn by taking the following
steps:
1 Draw the voltage phasor OV along y-axis.
2 Choose a convenient scale for current and draw a line OO′ = IO at an angle of ϕO with
phasor OV.
3 Draw a line OX perpendicular to voltage phasor OV. Also draw a line O′X from O′
parallel to line OX.
4 From O draw a line OA equal to blocked rotor current ISN corresponding to normal
rated voltage, with the same scale used for IO, lagging behind voltage vector OV by
an angle ϕsc. Then phasor O′A will represent the rotor current referred to stator side
i.e., I′
2. The line O′A is known as output line.

Chapter 3
Cont...
5 It is obvious that both point O′and A will lie on the circle. To determine the centre of
circle C, divide the line O′A equally. Extend the dividing line on both sides and the
point where it meets the line O′B that will be the centre C of the circle (see Fig. 9).
6 Taking C as centre and CO′ as radius, draw a semicircle O′AB.
7 Draw vertical lines O′P and AQ from O′ and A respectively parallel to Y-axis or voltage
vector OV.

Chapter 3
Cont...
Figure 9: Circle diagram for an induction motor

Chapter 3
Cont...
All the vertical distances represent the active or power components of current since
they are in phase with the voltage vector OV.
Accordingly, O′P represents working component of no-load current
(i.e., O′P = OO′ cos ϕO = IO cos ϕO = IW ); otherwise it represents no-load input which
includes core losses, friction and windage losses and very small amount of stator copper
loss.
Similarly, AQ i.e., vertical component of OA is proportional to motor input on short
circuit which includes rotor copper loss, stator copper and fixed losses i.e.,stator iron
loss and mechanical loss.
Where FQ = O′P = Stator iron loss + windage and friction loss
AE = rotor copper loss EF = stator copper loss

Chapter 3
Cont...
The rotor and stator copper losses are represented by the torque line O′E ,and E
located as follows:
i) In case of squirrel cage machines: The stator resistance per phase R1 is determined by
stator resistance test. Then
Rotor copper loss = power input at short circuit − stator copper loss
= PSC − 3I2
SCR1
AE
EF
=
PSC − 3I2
SCR1
3I2
SCR1
ii) In case of phase- wound machines: In this case, the stator and rotor resistances per phase
i.e., R1 and R2 can be measured separately for any value of stator and rotor currents, say
I1 and I2.

Chapter 3
Cont...
AE
EF
=
I2
2 R2
I2
1 R1
=
R2
R1
×

I2
I1
2
=
R2
R1
×
1
K2
Since I1
I2
= K
AE
EF
=
R2/K2
R1
=
R′
2
R1
=
Rotor resistance referred to stator
Stator resistance
Value of K can be determined by voltage ratio test or by using two ammeters on stator
and rotor circuit during blocked rotor test.

Chapter 3
Results Obtainable from Circle Diagram
Let us assume that the motor is drawing a current of I1 ampere at any load.
Draw an arc with radius OL = I1 with O as its centre.
From L, draw a line LM parallel to y-axis, as shown in Fig. 9 which intersects various
lines at points N,K and J.
Then
JM represents fixed losses, JK as stator copper loss,
KN as rotor copper loss, KL as rotor input,
NL as rotor output and LM as total motor input.

Chapter 3
Cont...
Motor input = 3V (LM) watt
Fixed losses = 3V (JM) watt
Stator Cu loss = 3V (JK) watt
Rotor Cu loss = 3V(NK) watt
Total losses = 3V(NM) watt
Mechanical power developed =
3V(NL) watt
Rotor input = 3V(KL) watt
Torque developed = 3V (KL)
ω Nm
Motor efficiency = LN
LM = output
input
Slip = NK
LK = Rotor copper loss
Rotor input
Power factor = LM
OL
Rotor output
Rotor input = 1 − S = N
Ns
= NK
LK
Rotor speed = Ns ×

NK
LK

rpm

Chapter 3
Maximum Quantities
Maximum output: From centre point C drop a perpendicular on output line O′A and
extend it to meet the circle at R.
From R draw a vertical line intersecting the output line O′A at S. Then maximum
output is represent by RS, such that
Maximum output = 3V (RS)watt.
Maximum torque or rotor input: From centre point C drop a perpendicular on torque
line O′E and extend it to meet the circle at U. From U draw a vertical line intersecting
the torque line O′E at W. Then maximum torque is represent by UW, such that
Maximum torque = 3V

UW
ω

Nm

Chapter 3
Cont...
Maximum input power: Draw a vertical line parallel to y-axis passing through the
centre of circle C. It intersects the circle at D and the x-axis at Z.
The point D where the tangent to the circle is horizontal represents the maximum
input power which is proportional to DZ, such that
Maximum power = 3V (DZ)watt.
But the motor will be unstable here since the point D occur beyond the point of
maximum torque U.
However, maximum input indicates the ability of the motor to carry short time over
loads.Usually, the maximum input power is 2 to 3 times the rated power.

Chapter 3
Cont...
Figure 10: Maximum quantities on a circle diagram

Chapter 3
Significance of Some Lines in the Circle Diagram
Input line OX: The vertical distance between any point on the circle and this line
represents the input power and hence called input line.
Output line O′A: The vertical distance between any point on the circle and this line
represents the output power and hence called output line.
Torque line or Air-gap power line O′E: Since KL represents mechanical power devel-
oped in the rotor i.e., air-gap power, the line O′E is called torque line or air-gap power
line.

Chapter 3
Cont...
Example 5
A three-phase, 400 V,induction motor gave the following results:
No-load test:400 V,1200 W,8A
Short circuit test:150 V,3900 W,36 A
The rotor copper loss at standstill is half the total copper loss.Draw the circle diagram,
Determine the full load value of the current,pf and slip when the normal rating is 14
kW.Also,calculate motor input at full load,stator copper loss,rotor copper loss,rotor in-
put, mechanical power output, efficiency of the motor,ratio of maximum torque to full load
torque and maximum power.

Chapter 3
Cont...
Example 6
For a 200 V,50 Hz,3-phase,7.46 kW,slip ring induction motor with a star connected stator
and rotor.The winding ratio of the motor is unity, whereas the stator and rotor resistance
per phase is 0.38 and 0.24 ohm respectively.The following are the test results:
No load test:200 V,7.7 A,874 W
Blocked rotor test:100 V,39.36 A,3743 W
Draw a circle diagram and determine
(i) Starting torque,
(ii) Maximum torque,
(iii) Maximum output
(iv) Slip for maximum torque
(v) Maximum output

Chapter 3
Questions
Thank You!

Three phase induction motors

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