INDUCTION MOTOR
- In the most common types of AC motor,
electrical power is not conducted to the rotor
directly, the rotor receives its power inductively
in exactly the same way as the secondary of a
transformer receives its power.
- Of all the AC Motors, the polyphase Induction
Motor is the one which is extensively used for
various kinds of industrial drives.
In fact, induction
motor can be
treated as a
rotating
transformer i.e.
one in which
primary winding is
stationary but the
secondary is free
to rotate.
ADVANTAGES OF INDUCTION MOTOR
It has very simple and extremely rugged,
almost unbreakable construction.
2. Its cost is low and it is very reliable.
3. It has sufficiently high efficiency.
4. It requires minimum of maintenance.
5. It starts up from rest and needs no extra
starting motor and has not be synchronized.
1.
DISADVANTAGES OF INDUCTION MOTOR
Its speed cannot be varied without sacrificing
some of its efficiency.
2. Just like a DC shunt motor, its speed decreases
with increase in load.
3. Its starting torque is somewhat inferior to that of
a DC shunt motor.
1.
STATOR
- the stationary part of the induction motor,
consists of a cylindrical laminated slotted core
that is placed in a frame.
ROTOR
- there are two types of construction for the
rotor of an induction motor; the squirrel cage
and wound rotor.
A. SQUIRREL CAGE TYPE ROTOR
- This type of rotor
consists of a cylindrical
laminated
core
with
parallel slots for carrying
the rotor conductors. The
rotor bars are electrically
welded or bolted to two
heavy
and
stoutcircuiting-end-rings.
B. WOUND ROTOR TYPE
- This type of rotor
provided with 3-phase,
double layer, distributed
winding consists of coils
as used in alternators. The
rotor is wound for as
many poles as
the
number of stator poles
and is always wound
three phase.
STATOR WINDING
- the type of winding used in the stator and
rotor of a polyphase induction motor is exactly
the same as that alternators.
REVOLVING FIELD
- the speed of revolving field is inversely
proportional to the no. of pairs of poles and the
speed of the revolving field is directly
proportional to frequency.
If the foregoing two statements are
combined for what is called the
synchronous speed the revolving field
speed in revolution per second can be
written.
Nsyn = rpmsyn =
𝟏𝟐𝟎𝒇
𝑷
PROBLEM 1:
Calculate the synchronous speed of an
8 pole induction motor when applied with
power from a.) 60 cycle source, b.) 50
cycle source, c.) 25 cycle source.
SOLUTION:
rpmsyn =
a. rpmsyn =
𝟏𝟐𝟎𝒇
𝑷
𝟏𝟐𝟎(𝟔𝟎)
𝟖
b. rpmsyn =
rpmsyn = 900rpm
c. rpmsyn =
𝟏𝟐𝟎(𝟓𝟎)
𝟖
rpmsyn = 750rpm
𝟏𝟐𝟎(𝟐𝟓)
𝟖
rpmsyn = 375rpm
PROBLEM 2:
Calculate the synchronous speed of 60cycle induction motor having a.) 4 poles,
b.) 6 poles, c.) 10 poles?
SOLUTION:
rpmsyn =
a. rpmsyn =
𝟏𝟐𝟎𝒇
𝑷
𝟏𝟐𝟎(𝟔𝟎)
𝟒
b. rpmsyn =
rpmsyn = 1800rpm
c. rpmsyn =
𝟏𝟐𝟎(𝟔𝟎)
𝟔
rpmsyn = 1200rpm
𝟏𝟐𝟎(𝟔𝟎)
𝟏𝟎
rpmsyn = 720rpm
SLIP AND ROTOR SPEED
In practice, the rotor never succeeds in
catching up with the stator field. If it really did
so, then there would be no relative between
the two, hence no rotor emf, no rotor current
and so no torque to maintain rotation. That is
why rotor runs at a speed which is always less
than the speed of the stator field. The
difference in speeds depends upon the load
of the motor.
SLIP AND ROTOR SPEED
The difference between the synchronous
speed Ns and the actual speed N of the rotor
is known as slip. Slip is generally specified in
terms of the synchronous speed.
rpmsyn −rpmrotor
% slip =
x 100
rpmsyn
rpmrotor =
𝟏𝟐𝟎𝒇
𝑷
(𝟏 − 𝒔)
PROBLEM 1:
The rotor speed of a 6-pole, 50
cycle induction motor is 960 rpm.
Calculate the percent slip.
SOLUTION:
rpmsyn =
𝟏𝟐𝟎𝒇
𝑷
=
𝟏𝟐𝟎(𝟓𝟎)
𝟔
rpmsyn = 1000rpm
rpmsyn −rpmrotor
% slip =
x 100
rpmsyn
1000 −960
% slip =
x 100
1000
% slip = 𝟒%
PROBLEM 2:
Calculate the speed of a 60 cycle,
14 pole motor if the slip is 0.05?
SOLUTION:
rpmrotor =
𝟏𝟐𝟎𝒇
𝑷
rpmrotor =
𝟏𝟐𝟎(𝟔𝟎)
𝟏𝟒
(𝟏 − 𝒔)
(𝟏 − 𝟎. 𝟎𝟓)
rpmrotor = 𝟒𝟖𝟖𝒓𝒑𝒎
PROBLEM 3:
A 3-phase 50 cycle 10 pole induction
motor has a full load slip of 0.075. What is
the speed of:
a.) the rotor relative to the revolving field
b.) the revolving field relative to the stator
SOLUTION:
𝟏𝟐𝟎𝒇
a. rpmrotor =
(𝟏 − 𝒔)
𝑷
𝟏𝟐𝟎(𝟓𝟎)
rpmrotor =
(𝟏 − 𝟎. 𝟎𝟕𝟓)
𝟏𝟎
rpmrotor = 555𝒓𝒑𝒎
b. Revolving Field = Ns - NR
Revolving Field = 600 – 555
Revolving Field = 45rpm
GENERATED VOLTAGE AND FREQUENCY IN A ROTOR
- the relative speed of the rotor w/ respect
to the revolving field will affect both the
generated voltage in the rotor and the
frequency of the current in the rotor
conductors.
ER = s x EBR
fR = s x f
where:
ER = generated voltage per phase
EBR = blocked rotor generated voltage
per phase
fR = rotor frequency
s = slip
PROBLEM 1:
A 3 phase 60-cycle 6-pole 220V wound
rotor induction motor has a stator that is
connected in delta and a rotor that is
connected Wye. The rotor has half as many
turns as the stator. For the rotor speed of 1110
rpm. Calculate a.) the slip, b.) the blocked
rotor voltage per phase EBR c.) the rotor
generated voltage per phase ER d.) the rotor
voltage
between
terminals,
e.)
rotor
frequency.
SOLUTION:
a. slip =
𝑵𝒔 −𝑵𝑹
𝑵𝒔
=
slip = 0.075
b. EBR = (220) x
𝟏
𝟐
𝟏𝟐𝟎𝟎 −𝟏𝟏𝟏𝟎
𝟏𝟐𝟎𝟎
e. fR = s x f = (0.075) x (60)
fR = 4.5Hz
EBR = 110V
c. ER = s x EBR = (0.075) x (110V)
ER = 8.25V
d. Volts between rotor terminal = √3 x ER = (√3) x (8.25V)
ER = 14.289V
REVIEW OF WYE CONNECTED (STAR) & DELTA CONNECTED
where:
IL = line current;
VL = line current;
IP = phase current
VP = phase current
ROTOR CURRENT AND POWER
The actual conversion of electrical power
into mechanical power always takes place in
the rotor of a motor. In the induction motor, the
power input to the rotor is not applied directly
ie. conductively, but is transferred across on air
– gap inductively.
IR =
𝒔𝑬𝑩𝑹
𝑹𝑹𝟐+𝒔𝟐 𝑿𝑩𝑹𝟐
where:
RR = rotor resistance; XBR = rotor reactance
IR =
𝑬𝑩𝑹
𝑹𝑹 𝟐
𝟐
+
𝑿𝑩𝑹
𝑺
PROBLEM 2:
Using the data of Problem 1.
Calculate the rotor current if RR = 0.1
ohms and XBR = 0.5 ohms.
SOLUTION:
IR =
IR =
𝑬𝑩𝑹
𝑹𝑹 𝟐
+ 𝑿𝑩𝑹𝟐
𝑺
𝟏𝟏𝟎
𝟎.𝟏 𝟐
𝟐
+(𝟎.𝟓)
𝟎.𝟎𝟕𝟓
IR = 𝟕𝟕. 𝟐𝟒𝟕𝑨
The total power delivered to the rotor power
per phase consists of two parts: 1.) the power
that causes a copper loss and 2.) the electric
power that is converted into the mechanical
power on a per is phase basis.
Rotor Power Input = Rotor Copper Loss + Rotor Power Developed
RPI = RCL + RPD
2
IR x
𝑹𝑹
𝒔
𝟐
= 𝑰𝑹 𝑹𝑹 +
𝟏−𝑺
𝑰𝑹 𝑹𝑹 ( )
𝑺
𝟐
Rotor Power Input = Rotor Copper Loss + Rotor Power Developed
RPI = RCL + RPD
where:
2
RPI = IR x
𝑹𝑹
𝒔
RCL = 𝑰𝑹𝟐 𝑹𝑹 = 𝑹𝑷𝑰 𝒙 𝒔
RPD = 𝑰𝑹𝟐 𝑹𝑹
𝟏−𝑺
𝑺
RPD = RPI (𝟏 − 𝒔)
=
𝑰 𝑹 𝟐 𝑹𝑹
𝒔
(𝟏 − 𝒔)
PROBLEM 3:
Using the ff. data of Problem 1 and 2.
Calculate a.) the Rotor Power Input (RPI),
b.) Rotor Copper Loss (RCL), c.) Rotor
Power Developed (RPD) by the rotor in
watts, d.) Rotor Power Developed (RPD)
by the rotor in horsepower.
SOLUTION:
a. RPI =
IR2
RPI = 3
x
𝑹𝑹
𝒔
(77.5)2
x
𝟎.𝟏Ω
𝟎.𝟎𝟕𝟓
RPI = 24,025W
b. RCL = 𝑰𝑹𝟐 𝑹𝑹 = 𝑹𝑷𝑰 𝒙 𝒔
RCL = 𝟐𝟒, 𝟎𝟐𝟓 (𝟎. 𝟎𝟕𝟓)
RCL = 1,801.875W
SOLUTION:
c. RPD
RPD =
𝑰𝑹𝟐 𝑹𝑹
= 𝒔 (𝟏
𝟑(𝟕𝟕.𝟓)𝟐 (𝟎.𝟏)
𝟎.𝟎𝟕𝟓
− 𝒔)
(𝟏 − 𝟎. 𝟎𝟕𝟓)
RPD = 𝟐𝟐, 𝟐𝟐𝟑. 𝟏𝟐𝟑𝑾
d. RPD =
𝟐𝟐,𝟐𝟐𝟑.𝟏𝟐𝟑𝑾
𝟕𝟒𝟔𝑾
RPD = 𝟐𝟗. 𝟕𝟖𝟗𝒉𝒐𝒓𝒔𝒆𝒑𝒐𝒘𝒆𝒓
ROTOR TORQUE
- the force tending to produce rotation.
It is usually expressed in pound-feet (lb-ft)
except in the case of very small motors
given in terms of ounce-inches (oz-in).
T = 7.04
𝑹𝑷𝑰
𝑵𝑺
PROBLEM 1:
Calculate the torque developed by the
motor given the values of IR = 77.5, s =
0.75, RR = 0.1Ω at 1200 synchronous
speed.
SOLUTION:
RPI =
IR2
RPI = 3
x
𝑹𝑹
𝒔
(77.5)2
x
𝟎.𝟏Ω
𝟎.𝟎𝟕𝟓
RPI = 24,025W
T = 7.04
𝑹𝑷𝑰
𝑵𝑺
T = 7.04
𝟐𝟒,𝟎𝟐𝟓𝑾
𝟏𝟐𝟎𝟎
T = 140.947 lb-ft