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Overview
•
•
•
•
28-Principles of AC Machines
Introduction
Rotating Magnetic Field
Magnetic Field Rotational Speed
Synchronous Speed
Text 3.5
ECEGR 450
Electromechanical Energy Conversion
2
Dr. Louie
Introduction
Rotating Magnetic Field
• AC machines rely on a rotating magnetic field
• Stator houses current-carrying conductors
• Three-phase motors are very common
• Requires a three phase source
• Under linear conditions, flux and current will have
similar waveforms
 Stator is the armature
Ia
N
Ib
b’
f
N
Ia  imax sin(t)
b’
Ib  imax sin(t  120 )
S
b
Ic
c
a’
a
c’
Ib
S
b
Ic
AC generator
a
c’
T
c
a’
3
a
b
pi
2pi
theta (rad)
3pi
4pi
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Conceptual Illustration e = 0o
1
b  fmax sin(t  120 )
c  fmax sin(t  120 )
0
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b: -0.5p.u.
a  fmax sin(t)
0
Ic  imax sin(t  120 )
revolving magnetic
field provided by rotor
revolving magnetic
field provided by stator
current (A), Flux ( f)
AC motor
Ia
Conceptual Illustration e = 30o
c
1
c: -0.5p.u.
b: 0p.u.
a
b
c
c: -0.86p.u.
0.5
0.5
0
0
-0.5
-0.5
a: 0.86p.u.
a: 1p.u.
-1
0
-1
0
30 60 90 120 150 180 210 240 270 300 330 360
30 60 90 120 150 180 210 240 270 300 330 360
e
e
r
r
direction of flux through rotor
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direction of flux through rotor
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Conceptual Illustration e = 45o
1
b: 0.255p.u.
a
b
Conceptual Illustration e = 60o
c
1
c: -0.97p.u.
b: 0.5p.u.
a
b
0.5
0.5
0
0
-0.5
-0.5
a: 0.707p.u.
a: 0.5p.u.
-1
0
-1
0
30 60 90 120 150 180 210 240 270 300 330 360
30 60 90 120 150 180 210 240 270 300 330 360
e
e
r
r
direction of flux through rotor
direction of flux through rotor
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1
a
b
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Conceptual Illustration e = 90o
b: 0.86p.u.
c
c: -1p.u.
Conceptual Illustration e = 120o
c
1
a
b
c
c: -0.86p.u.
0.5
0.5
0
0
-0.5
a: 0p.u.
-1
0
30 60 90 120 150 180 210 240 270 300 330 360
What direction will the
flux through the rotor be
when e = 120o?
-0.5
-1
0
30 60 90 120 150 180 210 240 270 300 330 360
e
r
e
direction of flux through rotor
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Conceptual Illustration e = 120o
1
b: 1p.u.
a
b
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Two-Pole Three-Phase Revolving Field
• Coils are spatially separated by 60o
 Do not confuse the spatial direction with the phase of
the flux
• We will analyze how the flux varies with time
• Note: if flux is negative, the direction is opposite as shown
c
c: -0.5p.u.
0.5
0
c
b’
-0.5
notation
a: -0.5p.u.
r
-1
0
30 60 90 120 150 180 210 240 270 300 330 360
e
fx
x
a’
fb
b
direction of flux through rotor
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x’
fc
11
fa
a
reference
c’
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Three-Phase AC Motor
Rotating Magnetic Field
• Want to analyze the net flux as seen by the rotor
• General approach:
c
b’
fc
a’
fb
a
fa
b
reference
 Consider the flux at 0, 60 and 120 degrees in time
 Compute the a, b, c phase flux magnitudes
 Determine resulting flux by adding a, b, c phase
flux
 Generalize results
c’
Note: salient windings are shown.
Large motors use cylindrical windings.
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Rotating Magnetic Field
Rotating Magnetic Field
Let t = 0. The magnitudes are:
• Maximum flux occurs in the following sequence
| a | fmax sin(t)  0
 a, c’, b, a’, c, b’ and so on
 same relative ordering of coils around stator
a
a’
b
| b | fmax sin(t  120 ) | 
b
| c | fmax sin(t  120 ) 
c
current (A), Flux ( f)
a
b’
current (A), Flux ( f)
c
0
c’
0
pi
c’
2pi
theta (rad)
3pi
a’
b’
3
f
2 max
pi
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At t =
3pi
4pi
16
3
f
2 max
| b | fmax sin(t  120 ) | 
3
f |
2 max
| c | fmax sin(t  120 )  0
b’
b  
c 
3
f 240
2 max
3
f 120
2 max
c’
current (A), Flux ( f)
a
a  00
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b
c
0
0
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2pi
theta (rad)
60o:
| a | fmax sin(t) 
 3
3
f 240 
f 120  1.5fm90
2 m
2 m
b
c
Rotating Magnetic Field
r  a  b  c
a
b
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Rotating Magnetic Field
a’
a
0
0
• At t = 0 the flux is as shown
• The resulting flux,  r, is found through vector addition
c
3
f |
2 max
4pi
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=0+
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pi
2pi
theta (rad)
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3pi
4pi
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Rotating Magnetic Field
Rotating Magnetic Field
• Let t = 120o
r  a  b  c
=
3
f
2 max
| b | fmax sin(t  120 )  0
| a | fmax sin(t) 
3
 3
f 0 
f 240  0  1.5fm30
2 m
2 m
| c | fmax sin(t  120 ) | 
b’
a
a 
a
a’
b
3
f 0
2 max
current (A), Flux ( f)
c
3
f 240
2 max
c  0120
b  
c’
c
2pi
theta (rad)
3pi
4pi
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Dr. Louie
Rotating Magnetic Field
Rotating Magnetic Field
• At t = 60o the flux is as shown
• The resulting flux is found through vector addition
t = 0o
r  a  b  c
c
3
3
f 0  0  
f 120  1.5fm  30
2 m
2 m
c
pi
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b
0
0
=
3
f |
2 max
b’
t = 60o
a
a’
3
f 0
2 max
b  0240
c
b’
t = 120o
b’
c
b’
a a’
a’
a
a 
a
a’
c  
b
b
3
f 120
2 max
c’
c’
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b
c’
b
c’
resulting flux vector rotates CW in time
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Rotating Magnetic Field
Rotating Magnetic Field
 Resulting flux magnitude is constant
 Direction of the resulting flux rotates with time
 120o phase shift in the time domain has shifted the
spatial orientation of the flux 120o
 To make the field rotate in the opposite direction
(counter clockwise) switch any two phases (e.g. b
and c phases)
Conceptually like rotating magnets around the
periphery
N
S
Observations:
r
N
S
t = 0o
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t = 60o
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t = 120o
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Magnetic Field Rotational Speed
Magnetic Field Rotational Speed
• For a two pole motor one full rotation of the
magnetic field occurs after one complete
electrical cycle
• How does a 4-pole motor affect the rotational
speed of the magnetic field?
• Coils separated by 30 degrees
 a, c’, b, a’, c, b’ ordering is preserved
• Four poles, examine one pole-pair
a’
c
b’
a’
b
t = 60o
c
b
c’
b’
b’
c’
a
a
c
a’
m =
a’
c
c’
b’
b’
c’
b
c
45o
a’
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• For a 4 pole-motor one full rotation of the
magnetic field requires two complete electrical
cycles
P
• To generalize: Ts  2 T
Note: do not confuse “T” for period, with “T” for torque.
b
m = 15o
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Magnetic Field Rotational Speed
• Also
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 Ts: period of the flux rotation (s)
 T: period of the AC waveform (s)
 P: number of poles
a
a
b
Magnetic Field Rotational Speed
60 degrees time shift resulted in spatial rotation of
30 degrees
t = 0o
a’
S
Magnetic Field Rotational Speed
N
c’
c
25
S
N
a
a
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S
N
b’
S
N
b
c’
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Example
1
2f
ns 

Ts
P
Write Ns, the synchronous speed in revolutions
per minute (RPM) and radians per second (s) as
a function of the number of poles and frequency f
 ns: speed of the revolving field (revolutions/s)
 f: frequency of the AC waveform (Hz)
• ns is known as the synchronous speed
Note: this and previous equations relate frequency
of applied source with rotation of magnetic field, not the
actual rotation of the rotor.
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Example
Summary
An 6-pole AC motor is connected to 50 Hz source.
What is the synchronous speed of the motor in
rpm?
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• Magnetic field rotates with constant magnitude
• The resulting flux is 0.5n times the single phase
flux, where n is the number of phases
• The synchronous speed is inversely proportional
to the number of poles and proportional to the
frequency of the applied source
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