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Rotating Magnetic Field (Chap 3)
Rotating Magnetic Field
• In the DC machine previously studied, the rotor motion was
produced through the interaction of two stationary magnetic
fields combined with commutator action.
• In AC machines, rotating motion is produced via rotating
magnetic fields. We begin our study of rotating magnetic
fields by considering the mmf distribution of a single N- turn
coil carrying a current.
• Consider the air gap flux shown in the following figure. Notice
how the flux direction reverses on either side of the stator
winding.
Rotating Magnetic Field
Rotating Magnetic Field
• Notice that as additional stator windings are added the flux in the
air gap produces a closer approximation to a sinusoid. We should
now be able to convince ourselves that it is possible to create a
nearly sinusoidal mmf distribution in the air gap. In any event the
fundamental component of the mmf is given by
=
4
∙ ()
4 − − − − !
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Rotating Magnetic Field
Rotating Magnetic Field
If the current is = " cos a sinusoid, then
& = '() cos cos()
*+
where '() = ".
,-
Now, since cos . cos / = &⁄0{cos(A-B) + cos(A+B)}
& = '() {&⁄0 cos − 3 + &⁄0 cos( + )}
& = 5 + 6
5 travels to the right 6 travels to the left
& = '() {&⁄0 cos − 3 + &⁄0 cos( + )}
Rotating MMF
Rotating MMF
Big Picture Result
Flux distribution in 3 phase machine
• The mmf of a single phase winding excited by an
alternating current can be resolved into two traveling
waves.
• In a 3 phase machine, the individual phase windings
are displace by 120°. Thus 3 spatial sinusoidal mmf
waves separated by 120° are also produced.
• Phasors
& = '() {&⁄0 cos − 3 + &⁄0 cos( + )}
& = 5 + 6
5
&
6
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Rotating MMF
Flux distribution in 3 phase machine
If each phase winding is supplied by an alternating
current forming a 3Φ balanced set
( = "
3
7 = "
3 − 120°
< = "
3 − 240°
Then, , = (& + 7& + <&
, = ( 5 + ( 6 + 7 5 + 7 6 + < 5 + < 6
Rotating MMF
Flux distribution in 3 phase machine
where from the previous single phase case
( 6 = &⁄0 '() cos + 3 and ( 5 = &⁄0 '() cos − 3
7 6 = &⁄0 '() cos + 3 − 240 and 7 5 = &⁄0 '() cos − 3
< 6 = &⁄0 '() cos + 3 + 240 and < 5 = &⁄0 '() cos − 3
, = {( 6 + 7 6 + < 6 } + {( 5 + 7 5 + < 5 }
, = = 0 + {32 '() cos − 3 }
, = 32 '() cos − 3
Rotating MMF
Flux distribution in 3 phase machine
, = 32 '() cos( − )
Rotating MMF
Flux distribution in 3 phase machine
• The angular velocity of the wave is 3 = 2
for a ‘P’ pole machine the rotational speed is
0
-
3' = 3 @(A⁄BC<
A single positive travelling wave!!
• (, )is a sinusoidal function of θ. At a fixed
time (, )describes a sinusoid in space (around
the air gap)
• The angle ωt provides for motion of the entire
wave around the air gap from pole to pole at a
angular velocity ω =2. (where f is the electrical
frequency)
=
3' x
60s
1min
120
1rev
2
= 2π
2πrads
120
= 60
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Phasor analysis of 3 φ MMF
, = ( + 7 + <
= '() 3 + '() cos − 120 cos(3 − 120) +
'() cos( + 120)cos(3 + 120)
At t = 0,
-1/2
, 0 = '() + '() cos( − 120)cos(−120) +
'() cos( − 120)cos(−120)
-1/2
, 0 = '() −
PQRS
cos
0
− 120 −
PQRS
cos(
0
Phasor analysis of 3 φ MMF
, 0 = ('() < 0) − (
PQRS
, π/3 = (
( , π/3 =
0
< 0) + (
U⁄ 0 '()
PQRS
0
1
2
−1
< 60°
U
3/2Fmax
Atalatertime3 = 2/3=120° ,
, 23 = '() 120 +
'() cos − 120 cos(120 − 120) +
'() cos( + 120)cos(120 + 120)
Phasors
, 2π/3 = (−1/2'() < 0) + ('() < −120)+
(−1/2'() < +120)
( , 2π/3 = U⁄0 '() < 120°
Fb
120°
60°
60°60°
3/2Fmax
Fc
Fb
< 120)
Phasor analysis of 3 φ MMF
< −120) + (−'() < −120)
Fc
PQRS
0
+ 120)
1
2
'() cos − 120 cos(60 − 120) +
'() cos( + 120)cos(60 + 120)
Phasors
< −120) −(
, 0 = 0 '() < 0°
Phasor analysis of 3 φ MMF
At a later time 3 = /3= 60° ,
, 3 = '() 60 +
PQRS
0
Fa
Fa
4