School of Electrical and Computer Engineering
Applied Electronics and Electrical
Machines
(ELEC 365)
Fall 2015
Induction Machine
1
AC Machines
 Key educational goals:
Develop the basic principle of operation of a AC machine. Classify the different
types of ac machines and evaluate their characteristics.
 Reading/Preparatory activities for class
i)Textbook:
Chapter 17: 17.1, 17.2
ii) Power-point file:Induction_machine
2
AC Machines
 Questions to guide your reading and to think about ahead of time.
1. What are the advantages and disadvantages of AC machines related to DC
machines?
3. What are the different hardware components of an AC machine?
4. What are the different types of AC machines?
5. What is rotating magnetic field and what is slip? Does slip exist for all AC
motors?
6. What is operating principle of a AC machine?
3
Introduction
THE 3 PHASE INDUCTION MACHINE
• 55% of world's generated energy →rotating machines.
• >90% of this →induction machines.
• The induction machine consumes more of world’s generated
electricity than any other piece of electrical equipment.
Power Range
• 100-500W small fans.
• 1-50kW fans, pumps, conveyors, escalators.
• 500kW water pumping, coal cutting.
• 1MW high speed train motor.
• 10MW warship/cruise ship motor.
4
Induction machine
• Stator has 3 windings AA’, BB’, CC’ wound 120°apart in space.
• Stator windings connected to 3-phase mains at ωe= (2π) 60Hz mains.
• Fed by 3-phase currents 120°apart in time to create rotating magnetic field
• Rotor has NO windings.
• It has a cage of Aluminium bars; currents will be induced in it.
5
Induction motor
•
Most popular motor today in the low and medium horsepower range.
•
Very simple and robust in construction.
•
Speed easily controllable using V/f or Field Oriented Controllers.
•
Have replaced DC Motors in areas where traditional DC Motors cannot be
used such as mining or explosive environments.
•
Of two types depending on motor construction: Squirrel Cage or Wound
Rotor (Slip Ring).
Disadvantages:
• Speed control difficult.(??)
• Starting torque is inferior to that of dc shunt motor.
• Speed decreases with load (similar to dc shunt motor).
• Most of them run with a lagging power factor.
6
Construction
Stator showing
enclosure
with winding
End Bells
Rotor with bearing
7
Construction
 Consists of (a) Stator and (b) Rotor (decides squirrel cage or wound rotor).
 Stator consists of a 3-phase winding (located in slots) and fed from a 3phase supply.
 It is wound for a definite number of poles [the number of poles P, produced
in the rotating field is P = 2n where n is the number of stator
slots/pole/phase], the exact number of poles being determined by the
requirements of speed.
8
Two pole concentrated winding
9
Squirrel Cage Rotor
 Rotor consists of a cylindrical laminated core with parallel slots carrying heavy
bars of copper, aluminum or alloys.
 No external connections are possible for the rotor
10
Squirrel cage rotor
 The conductors are not arranged parallel to the axis of the rotor, but at an
angle, this is done to reduce torque vibrations and noise.
Squirrel cage rotor
11
Slip Ring Rotor (wounded rotor)
• The rotor contains a set of three-phase coils that are placed in slots.
• The connections from rotor are brought out using slip rings that are rotating
with the rotor and carbon brushes that are static.
12
Slip-ring close-up
Slip-ring induction motor with a close-up of the slip-ring
13
Slip-ring rotor
Possible applications for wound-rotor machines include
 Speed control of very large machines (multi-MW)
 Hoists, Pumps, Fans and Blowers, Conveyors, …
 Doubly-fed induction generation (used in some wind turbines)
Advantages:
 Starting torque of an induction motor can be decreased by increasing the
rotor resistance.
 Torque speed characteristic can be modified.
Disadvantage:
 More expensive
 Less rugged
14
Mathematical expression for RMF
20
Mathematical expression for RMF
21
Mathematical expression for RMF
s 

P/2
rad . / sec .
Ns=120f/P rpm
Synchronous speed
22
Speed for different poles
Rotating field set up by stator currents rotates at synch speed.
fe
We =2Π(fe)
P (Poles)
≈Ws (rad/s)
Ws (rpm)
60
376
2
360
3600
60
376
4
180
1800
60
376
6
120
1200
60
376
8
90
900
60
376
10
72
720
60
376
12
60
600
25
Torque Production in an Induction Motor
 When rotor is stationary, the rotating flux induces an e.m.f. in the rotor.
Frequency of induced e.m.f. is same as stator (transformer action). Since
the rotor conductors form a closed circuit, rotor currents are produced
whose direction, as given by Lenz’s law, is same as to oppose the very
cause producing it. In this case, the cause which produces the rotor
current is the relative velocity between the rotating flux of the stator and
the stationary rotor conductors. Hence, to reduce the relative speed, the
rotor starts running in the same direction as that of the flux and tries to
catch up with the rotating flux.
26
Slip in Induction Motor
 Slip: Rotor never succeeds in catching up with stator field. If it really did so,
then there would be no relative speed between the two, therefore no rotor
e.m.f. and so no torque to maintain the rotation.
 The parameter slip ‘s’ is a measure of this relative speed difference (in
percentage)
s
Ns  N
Ns
s  

s
where Ns, s are the speeds of the RMF in RPM and rad/sec respectively and N,
m are the actual speeds of the motor in RPM and rad/sec respectively
27
Slip in Induction Motor
Rotor Speed = N = (1 - s) 𝑁s
The rotating field produced by the rotor rotates at a corresponding synchronous
speed Nr, relative to the rotor such that Nr=120fr/P = 120(sf)/P = sNs
S=1 for N=0 (stationary); s=0 for N=Ns,
But the speed of rotor with respect to the stator is
N = (1 - s) 𝑁s
Therefore, speed of rotor field w.r.t. stator (absolute speed) = 𝑁 ′ 𝑠
= 𝑆𝑁𝑠 + 1 − 𝑆 𝑁𝑠 =𝑁𝑠
The speed of rotor field w.r.t. stator field = 𝑁𝑠 - 𝑁𝑠 = 0.
28
Slip in Induction Motor
Rotating magnetic field caused by stator:
Synchronous Speed = 𝑁s
Rotor Speed = N = (1 - s) 𝑁s
Rotating field produced by the rotor Nr = sNs
Speed of rotor field w.r.t. stator = 𝑵′ 𝒔 = 𝑺𝑵𝒔 + 𝟏 − 𝑺 𝑵𝒔 = 𝑵𝒔
29
Example 1
The stator of a three-phase, induction motor has 3 slots per pole per phase. If the
supply frequency is 60 Hz, Calculate
(a) The number of stator poles produced and total number of slots on the stator.
(b) Speed of the rotating stator flux (or magnetic field).
(c) If the motor has speed of 1164 rpm on full load, calculate (i) slip and (ii)
frequency of rotor currents.
(d) The speed of rotor when slip s = 0.04.
31
Example 1
(a) The number of stator poles produced and total number of slots on the stator.
Number of stator poles produced, P = 2n = 2 x 3 = 6 poles.
Total number of slots = (3 slots/pole/phase) x (6 poles) x (3 phases) = 54.
(b) Speed of the rotating stator flux (or magnetic field).
(c) If the motor has speed of 1164 rpm on full load, calculate (i) slip and (ii)
frequency of rotor currents.
32
Example 1
(c) If the motor has speed of 1164 rpm on full load, calculate (ii) frequency of rotor
currents.
(ii) Frequency of rotor current, fr = sf = 0.03 × 60 = 1.8 Hz.
(d) The speed of rotor when slip s = 0.04.
33
Three Phase Table
Relationships of balanced three phase circuits
Voltage (V)
Current (A)
Active Power (P) (W)
Star (wye)
𝑉𝐿−𝐿 = 3𝑉𝑝ℎ
𝐼𝐿 = 𝐼𝑝ℎ
3𝑉𝐿−𝐿 𝐼𝐿 cos 𝜑
=3𝑉𝑝ℎ 𝐼𝑝ℎ cos 𝜑
Reactive Power (Q) (VAR) 3𝑉𝐿−𝐿 𝐼𝐿 sin 𝜑
=3𝑉𝑝ℎ 𝐼𝑝ℎ sin 𝜑
Apparent Power(VA)
3𝑉𝐿−𝐿 𝐼𝐿 =3𝑉𝑝ℎ 𝐼𝑝ℎ
Delta
𝑉𝐿−𝐿 = 𝑉𝑝ℎ
𝐼𝐿 = 3𝐼𝑝ℎ
3𝑉𝐿−𝐿 𝐼𝐿 cos 𝜑
=3𝑉𝑝ℎ 𝐼𝑝ℎ cos 𝜑
3𝑉𝐿−𝐿 𝐼𝐿 sin 𝜑
=3𝑉𝑝ℎ 𝐼𝑝ℎ sin 𝜑
3𝑉𝐿−𝐿 𝐼𝐿 =3𝑉𝑝ℎ 𝐼𝑝ℎ
Note: 𝑉𝐿−𝐿 = Line-Line Voltage, 𝐼𝐿 = Line Current, 𝑉𝑝ℎ = Phase voltage, 𝐼𝑝ℎ = Phase
Current, cos 𝜑 = Power Factor, 𝜑 = Power Factor Angle.
34
Equivalent circuit of an induction machine
The equivalent circuit of an induction motor (per-phase basis) can be developed
by treating the induction motor as a transformer with the stator as the primary
winding and the rotor as the secondary winding. From this analogy it follows that
the stator and rotor each have their respective resistances and leakage
reactances. The air-gap in an induction motor, makes its magnetic circuit coupled
relatively poor and the corresponding magnetizing reactance (Xm) will be
relatively smaller as compared that of a transformer. A shunt resistance (Rc) can
be used to represent the hysteresis and eddy-current losses. Since the secondary
(rotor) of an induction motor is rotating, the frequency of the rotor current is
different from the frequency of the stator currents.
35
Rotor Equivalent Circuit (per-phase)
𝑋2𝐵𝑅 = Leakage reactance of rotor at standstill or blocked-rotor
(rotor speed, N = 0; or slip s = 1).
= 𝜔𝑠 𝐿2
Ω
where rotor frequency, 𝜔𝑟 = 𝜔𝑠 rads/sec, stator frequency since s = 1 at
standstill.
Leakage reactance 𝑋2 at any slip s is given by
𝑋2 = 2𝜋𝑓𝑟 𝐿2 = 2𝜋𝑠𝑓 𝐿2 = 𝑠 2𝜋𝑓 𝐿2 = 𝑠𝑋2𝐵𝑅
(1)
f = stator (or supply) frequency in Hz.
36
Rotor Equivalent Circuit (per-phase)
If 𝐸2BR is the voltage induced in the rotor circuit per phase at standstill (s = 1),
then
𝐸2𝐵𝑅 = 4.44(𝑓)∅𝑚 𝑁𝑡
𝑁𝑡 is the number turns in the rotor winding and 𝜙𝑚 is the maximum flux. At any
slip s, the rotor frequency becomes sf. Hence, the voltage induced in the rotor
circuit per phase at a slip s is given by
𝐸2 = 4.44(𝑠𝑓)∅𝑚 𝑁𝑡
∴ 𝐸2 = 𝑠𝐸2𝐵𝑅
(2)
37
Rotor Equivalent Circuit (per-phase)
𝑋2𝐵𝑅 = Leakage reactance of rotor at standstill or blocked-rotor = 𝜔𝑠 𝐿2
(rotor speed, N = 0; or slip s = 1).
The rotor current is given by
Per-phase rotor (phasor) equivalent circuit.
(3)
38
Rotor Equivalent Circuit (per-phase)
Equation (3) shows that the rotor equivalent circuit can be drawn in an alternative
way as shown in Figure (a) and (b).
Rotor equivalent circuit redrawn: (a) using equation (3), and (b) separating rotor loss and the
torque component.
39
Per-Phase Equivalent Circuit of Induction Motor
If we add the stator equivalent circuit to Fig(b) in previous slide, we get the
complete equivalent circuit of an induction motor shown.
Per-phase equivalent circuit of an induction motor.
40
Per-Phase Equivalent Circuit of Induction Motor
Referred to Stator-Side
All the parameters on the rotor (secondary) side can be referred to the stator
(primary) side in a way similar to that done in a transformer. Next slide figure
shows such a circuit. Primes indicate that these parameters are transferred from
rotor (secondary) to stator (primary) side.
Note that
41
Per-Phase Equivalent Circuit of Induction Motor
Referred to Stator-Side
where 𝑅′ 2 is the per-phase stand-still (or blocked) rotor resistance referred to the
𝑅 ′ 2 (1−𝑠)
𝑠
stator and
is a dynamic resistance that depends on the rotor speed and
corresponds to the load on the motor.
Per-phase equivalent circuit of an induction motor referred to stator side.
42
Per-Phase Equivalent Circuit of Induction Motor and
power flow
43
INDUCTION MOTOR PERFORMANCE
Efficiency Calculations:
Referring to the equivalent circuit (next slide):
Stator copper loss = 3𝐼1 2 𝑅1
Watts
Therefore, power crossing the air-gap = 𝑃ag−3φ
= 𝑃i −3φ − 3𝐼1 2 𝑅1 − (core loss)
watts
where 𝑃i −3φ is the 3-phase input power and we will neglect core-loss but we
will consider it as part of the mechanical loss.
46
INDUCTION MOTOR PERFORMANCE
Per-phase phasor equivalent circuit of an induction motor referred to stator side.
47
INDUCTION MOTOR PERFORMANCE
Power crossing the air-gap, 𝑃ag−3φ is dissipated in 𝑅′ 2 /s and therefore
watts
(5)
2
Subtracting the rotor resistance loss 3𝐼 ′ 2 𝑅′ 2 from the air-gap power 𝑃ag−3φ ,
we get the developed electromagnetic power, 𝑃d−3φ . Hence
watts
(6)
48
INDUCTION MOTOR PERFORMANCE
From equations (5) and (6):
(7)
The output power is given by 𝑃o−3φ = 𝑃d−3φ − [mechanical (rotational) power
(𝑃mech loss] – [core loss] Watts
49
INDUCTION MOTOR PERFORMANCE
The electromagnetic torque of an induction motor is given by
where mechanical (rotor) speed
and Ω𝑠 = synchronous speed in rads/sec.
50
Torque-speed characteristics
54
Example 2
A certain 30-hp four-pole 440-V-rms 60-Hz three-phase delta-connected induction
motor has
Under load, the machine operates at 1746 rpm and has rotational losses of 900W.
Find the power factor, the line current, the output power, copper losses, output
torque, and efficiency.
55
Example 2
synchronous speed : 1800 rpm
Slip:
56
Example 2
The impedance seen by the source is
The power factor is the cosine of the impedance angle. Because the impedance
is inductive, we know that the power factor is lagging:
power factor = cos(27.59 ) = 88.63% lagging
For a delta-connected machine, the phase voltage is equal to the line voltage,
which is specified to be 440V rms. The phase current is
57
Example 2
Thus, the magnitude of the line current is
The input power is
Next, we compute Vx and Ir :
58
Example 2
The copper losses in the stator and rotor are
59
Example 2
Finally, the developed power is
As a check, we note that
The output power is the developed power minus the rotational loss, given by
60
Example 2
The output power is the developed power minus the rotational loss, given
by
This corresponds to 29.62 hp, so the motor is operating at nearly its rated
load. The output torque is
61
Example 2
The efficiency is:
62