Airgap MMF from Double Layer
Lap Winding
Airgap MMF for a Full Pitch Winding (1)
Ni
2 Fa = Ni ⇒ Fa =
2
Fa is the MMF of one half of the magnetic circuit.
Airgap MMF for a Full Pitch Winding (2)
Airgap MMF for a Full Pitch Winding (3)
3π / 2
Ni
2  π / 2 Ni

θ
θ
θ
θ
+
−
Fa1 =
d
d
cos
(
)
cos
a
a
a
a
∫π / 2 2
2π  ∫−π / 2 2

4N
=  i
π2
4N
⇒ Fa1 =  i cos θ a
π2
Airgap MMF for a Full Pitch Winding (4)
Extended to multi-pole case
( 3π / 2 ) /( P / 2 )
P
Ni
P
2
 (π / 2 ) /( P / 2 ) Ni

θ
θ
θ
θ
Fa1 =
d
d
cos(
)
(
)
cos(
)
+
−
a
a
a
a
∫(π / 2) /( P / 2) 2
2π /( P / 2)  ∫−(π / 2 ) /( P / 2 ) 2
2
2

3π / 2
Ni
2  π / 2 Ni

θ
θ
θ
θ
d
d
cos(
)
(
)
cos(
)
=
+
−
ae
ae
ae
ae 
∫π / 2 2
2π  ∫−π / 2 2

4N
=  i
π2
4N
P 
⇒ Fa1 =  i cos θ a 
π2
2 
Define
θ ae =
P
θa
2
Harmonics of MMF for a Full Pitch Winding
Assume
Fa =
⇒ Fah
∑ Fah
h =1, 3, 5...
Fah = Fah cos(hθ ae ) = Fah cos(h
P
θa )
2
3π /2
2  π /2 Ni
Ni

θ
θ
θ
θ
cos(
h
)
d
(
)
cos(
h
)
d
+
−
ae
ae
ae
ae 
∫π /2 2
2π  ∫−π /2 2



0,
for h = 2, 4,6,8,...
 π

sin  h 
4N
4Ni
 2  i =
= =

  , for h 1,5,9,13,17,...
h
π  2 
π
 2 h

 4Ni
3,7,11,15,19, ...
 −   , for h =
2
h
π
 

Fractional Pitch Winding (1)
Full Pitch
Fractional Pitch
π
Nc turns per coil
N=2Nc turns
a axis
ρm
N=2Nc turns
Nc turns per coil
a axis
Fractional Pitch Winding (2)
Fa from conductors 2 and 3
N ci
2
−
N ci
2
−
ρ π
2 2
ρ
2
3 4
ρ
π+
2
ρ
π−
2
π
2π
ρ
2π −
2
+
θ ae
1 2
Fa from conductors 1 and 4
π
ρ
π−
2
2
N ci
2
Ni
− c
2
π ρ
− −
ρ 2 2
−π +
ρ
π
Fa
N ci
− N ci
ρ
2
−
2π −
ρ
2
π−ρ
ρ
ρ π
2π −
ρ
2
2
2 2
2π
θ ae
⇓
π−
−π +
N=2Nc turns
Nc turns per coil
ρ
π+
2
2
2
ρ
2
π+
π
ρm
ρ
2
2π
θ ae
a axis
Fractional Pitch Winding (3)
Fa
π−ρ
N ci
−π +
− N ci
ρ
2
−
Fa =
ρ
ρ
2
2
π−
ρ
2
π+
π
ρ
2
∑F
h =1, 3, 5...
2π −
ρ
2
2π
θ ae
ah
Fah = Fah cos(hθ ae ) = Fah cos(h
P
θa )
2
π +ρ / 2
2  ρ /2
⇒ Fah =
(− N c i ) cos(hθ ae )dθ ae 
N c i cos(hθ ae )dθ ae + ∫
∫

π −ρ / 2
2π  − ρ / 2
 ρ
pitch factor for
sin  h 
4Nc
 2 i
hth harmonic
=
π
h
 ρ
k ph = sin  h 
 2
the
Distributed Winding (1)
Full Pitch
Distributed Winding (2)
Fractional Pitch
3γ m
2
(exaggerated end turns)
γm
2
θa
a axis
ρm γ
m
θa
a axis
q=4
γ =
P
γm
2
q=2
q coils per group
Distributed Winding (3)
3γ m
2
γm
2
Fa =
=
=
q −1
q −1
F υ = ∑ ∑F
∑
υ
υ
= 0 ,1...
a,
q −1
q −1


F
h
θ
γ
υγ
cos
(
)
+
−
∑ ∑ ah  ae 2


υ = 0 ,1... h =1, 3, 5...
∑ Fah
h =1, 3, 5...
θa
= 0 ,1... h =1, 3, 5...
ah ,υ
q −1
q −1


h
θ
γ
υγ
cos
(
)
+
−
∑  ae 2

υ = 0 ,1...
where
a axis
Fah =
Let
4Nc i
k
π h ph
q −1
ξ h = hθ ae +
hγ
2
Distributed Winding (4)
 jξ h 1 − e − jqhγ 
 jξ h q −1 − jυhγ 
cos(ξ h − υhγ ) = ∑ Re[e
] = Re e ∑ e
∑
 = Re e 1 − e − jhγ 
υ =0
υ =0
υ =0




qhγ 

sin(
)
q −1
 j ( hθ ae + 2 hγ ) e − jqhγ / 2 (e jqhγ / 2 − e − jqhγ / 2 ) 
 jhθ ae
2
Re
= Re e
e
=



jhγ / 2
− jhγ / 2
− jhγ / 2
γ
h
(
)
−
e
e
e



sin( ) 
2 

qhγ
sin(
)
2 = qk cos(hθ )
= q cos(hθ ae )
dh
ae
hγ
q sin( )
2
qhγ
sin(
)
2
The distribution factor for hth harmonic is defined as k dh =
hγ
q sin( )
2
q −1
q −1
j (ξ h −υhγ )
Distributed Winding (5)
q −1
q −1


cos
(
)
h
θ
γ
υγ
+
−
∑
∑
ae


2
υ 0,1...
h 1,3,5...
=
=
4Nc i
= ∑
k ph  qkdh cos ( hθ ae ) 
h
h =1,3,5... π
= ∑ Fˆ cos ( hθ )
=
Fa
Fah
h =1,3,5...
Define:
ah
ae
N ph = PqN c
ia = Ci
N ph
Na =
C
number of turns per phase
C is the number of parallel circuits per phase
number of turns connected in series per phase per circuit
4  N a k ph k dh  ia
ˆ

⇒ Fah = 
P
π
h
Effect of Winding Skew (1)
When the winding skew exists, for the hth harmonic, cos(hθ ae )
becomes a function of z: cos h P θ + ζ m z 
 ζ z
 a

θa =
ρm
2
+
ζ mz
r
+
q −1
γ m − υγ m
2
ia
rθ a
θa = −
a axis
ρm
2
 2
+
ζ mz
r
+
q −1
γ m − υγ m
2
r 
m
υ = 0,1,2  ( q − 1)
ζm
B
0
ia
r
r
l
ζm
α sm
l
r
z
α sm =
If we say that the stator is skewed n slots, then
α sm = nγ m
α sm
ζ ml
r
Effect of Winding Skew (2)
The average of the above along z:
α sm =
ζ ml
r
α s = α sm
l
P
2
ζm
α sm
r
where
k sh =
sin( hα s / 2)
hα s / 2
 P
ζ m z 
1 l/2
cos
h
θ
+
 a
 dz

∫
l
/
2
−
l
r 
 2
1
=
sin( h
ζ m zP
+ hθ ae ) l−/l2/ 2
lζ m P
2r
2r
αs
αs 
1 
=
sin( hθ ae + h ) − sin( hθ ae − h )

2
2 
hζ 
sin( hα s / 2)
cos(hθ ae ) = k sh cos(hθ ae )
=
hα s / 2
h
winding skew factor for the
hth harmonic
Summary – Stator MMF of Phase A
for Double Layer Lap Winding
Fa =
where
∑ Fah
h =1, 3, 5...
Fah = Fh ia cos(hθ ae ) = Fh ia cos(h
P
θa )
2
4  Nˆ a  1

Fh = 
π  P  h
Nˆ a = N a k wh
effective number of turns connected
in series per phase per circuit
k wh = k ph k dh k sh
 ρ
k ph = sin  h 
 2
winding factor
qhγ
)
2
=
hγ
q sin( )
2
sin(
k dh
k sh =
sin( hα s / 2)
hα s / 2
Rotating MMF from Three
Phase Winding
Stator MMF of Three Phases (1)
θb
θa
θc
2π
θ be = θ ae −
3
2π
θ ce = θ ae +
3
Stator MMF of Three Phases (2)
Fa =
Fb =
Fc =
P
θa )
2
∑ Fah
Fah = Fh ia cos(hθ ae ) = Fh ia cos(h
∑ Fbh
 
 P
2π 
2π 
Fbh = Fh ib cos hθ ae −
F
i
h
θ
cos
=
−


h b
 2 a
3
3


 
 
∑ Fch
 
 P
2π 
2π 
Fch = Fh ic cos hθ ae +
 = Fh ic cos h θ a +

3
2
3


 
 
h =1, 3, 5...
h =1, 3, 5...
h =1, 3, 5...
Total MMF from stator winding:
Fstator = Fa + Fb + Fc =

P
2hπ 
F
i
i
i
h
(
)
cos
cos(
θa ) +
+
+


∑
h a
b
c
3 
2
h =1, 3, 5...

(ib − ic )sin 2hπ sin(h P θ a )
3
2

Stator MMF of Three Phases (3)
For h = 1, 5, 7, 11 …
cos
2 hπ
1
=−
3
2
2hπ
3
=
3
2
For h = 1, 7, 13 …
sin
For h = 5, 11, 17 …
2 hπ
3
sin
=−
3
2
For h = 3, 9, …
cos
2 hπ
=1
3
sin
2 hπ
=0
3
For Y connected windings without neutral return:
ia + ib + ic = 0
Stator MMF of Three Phases (4)
Fstator
P
P
3
3
(ib − ic )sin(h θ a ) +
= ∑ Fh  ia cos(h θ a ) +
2
2
2
2

h =1, 7 ,13...
P
P
3
3

(
)
F
i
h
i
i
h
cos(
)
sin(
)
θ
θ
−
−
∑
h
a
a
b
c
a 
2
2
2
2


h = 5 ,11,17...
Note that the third harmonic components have been eliminated
if the machine is Y connected without neutral return.
Consider balanced three phase sinusoidal currents:
=
ia i pk cos(ωet + θi )
2π
)
=
ib i pk cos(ωet + θi −
3
2π
)
=
ic i pk cos(ωet + θi +
3
⇒
3
3
=
i pk sin(ωet + θi )
(ib − ic )
2
2
Note: ipk may change with time. It’s
a constant Ipk at steady state.
Stator MMF of Three Phases (5)
Fstator
h
P
3
−
ω
θ a + θi ) +
F
i
t
h
cos(
∑
h pk
e
2
2
=
1,7,13...
h
wave rotating in forward
direction (counterclockwise)
P
3
+
ω
θ a + θi )
F
i
t
h
cos(
∑
h pk
e
2
5,11,17... 2
wave rotating in backward
direction (clockwise)
From:



P
θa
θa 
=


ωet  h θ a ω=
t
ω
t


e 
e 
ω
ω
hP
2
2
/
e
h 



2ω
⇒ ωh =e
hP
angular velocity of the hth harmonic
Physical Understanding of Wave (1)
z
f (t − )
vp
z
f (− )
vp
z
f (t − )
vp
= f (−
z − v pt
vp
)
0
z
z1 = v p t
z
f (t − ) : stands for a wave propagating in + z direction with velocity v p .
vp
“forward wave”
Likewise:
z
g (t + ) : stands for a wave propagating in − z direction with velocity v p .
vp
“backward wave”
Physical Understanding of Wave (2)
Plot u + ( z , t ) = A cos(ωt − kz )
A
0
-A
z
Effective Airgap
g eff = k c g
where the Carter’s coefficient
k c = k cs k cr
For stator
kcs =
τs
2
bs
2bs 
g   bs   


  
τs −
− ln 1 + 
atan
π 
2 g bs   2 g   



approximately
kcs ≈
τs
bs2
τs −
5 g + bs
For rotor
kcr =
τr
2
2br 
br
g   br   


  
τr −
− ln 1 + 
atan
2 g br   2 g   
π 



approximately
kcr ≈
τr
br2
τr −
5 g + br
Airgap Magnetic Field
When the airgap is small, we can assume
F ≈ H g g eff
g
Since Bg = µ 0 H g
r’or
geff
Bg =
µ0
g eff
F
r
ror
ris
r’is