Self and Mutual Inductances for
Synchronous Machine with
Round Rotor
Double Layer Lap Winding
on Stator
Cross Section Diagram
b axis
qd
q axis
d axis
qd  qa  qm
qm
qa
a axis
c axis
Stator Winding
Fractional Pitch
3g m
2
(exaggerated end turns)
gm
2
qa
a axis
rm g
m
qa
a axis
q=4
g 
q=2
q coils per group
P
gm
2
Self and Mutual Inductances (1)
ib (t)
qd
qd  qa  qm
d axis
ia (t)
qm
qa
a axis
ic (t)
Self and Mutual Inductances (2)
Linear Model
Balanced Winding
q me
P
 qm
2
Laa  Lbb  Lcc  Lls  LA
Lab  Lbc  Lca  M s
L f  Llf  Lmf
Laf  Lsf cos(q me )
2
Lbf  Lsf cos(q me 
)
3
2
Lcf  Lsf cos(q me 
)
3
Lls is leakage inductance of armature phase A winding which is about
10% of the maximum self inductance.
Llf
is leakage inductance of field winding.
Flux Linkage (1)
a  Laaia  Labib  Lacic  Laf i f
 Laaia  M s (ib  ic )  Laf i f
b  Lbaia  Lbbib  Lbcic  Lbf i f
 Laaib  M s (ia  ic )  Lbf i f
At steady state, i f is DC.
c  Lcaia  Lcbib  Lccic  Lcf i f
 Laaic  M s (ia  ib )  Lcf i f
 f  L f i f  Laf ia  Lbf ib  Lcf ic
2
2
 L f i f  Lsf [ia cos q me  ib cos(q me  )  ic cos(q me 
)]
3
3
Flux Linkage (2)

 a  
  
 b
 c  
  
 f   L
 sf

Laa
Ms
Ms
cos q me
Lsf cos q me 
2   ia 
Laa
Ms
Lsf cos(q me  )   
3  ib
2   
Ms
Laa
Lsf cos(q me  )  ic 
3  
 i f 
2
2
Lsf cos(q me  ) Lsf cos(q me  )
Lf

3
3

Ms
Ms
Flux Linkage (3)
Y connected without neutral return or balanced D connected :
a  ( Laa  M s )ia  Laf i f
 Ls ia  Lsf i f cos(q me )
b  ( Laa  M s )ib  Lbf i f
2
 Ls ib  Lsf i f cos(q me 
)
3
c  ( Laa  M s )ic  Lcf i f
2
)
3
 f  L f i f  Lsf [ia cos q me 
 Ls ic  Lsf i f cos(q me 
2
2
ib cos(q me 
)  ic cos(q me 
)]
3
3
L s  Laa  M s  Lls  LA  M s
ia  ib  ic  0
Flux Linkage (4)
When
ia  ib  ic  0

 a  
  
 b
 c  
  
 f   L
 sf

Ls
0
0
cos q me
Lsf cos q me 
2   ia 
Ls
0
Lsf cos(q me  )   
3  ib
2   
0
Ls
Lsf cos(q me  )  ic 
3  
 i f 
2
2
Lsf cos(q me  ) Lsf cos(q me  )
Lf

3
3

0
0
Flux Linkage in Phase A Winding
There are a total of P groups. These groups may be connected in series, or parallel,
or partly series and partly parallel.
P  PsC
Assume:
Then:
Note:
a  Ps group  Ps qNc kw pk cos 
N a  Ps qNc
C
Nˆ a  N a k w
 a  Nˆ a  pk cos 
number of series turns per phase per circuit
number of parallel circuits
effective number of series turns per phase
per circuit on armature winding
P
B  B pk cos( q a   )
2
2 B pk Dl
 pk  Nˆ a  pk
 pk 
P
Self Inductance of Stator Winding
If we apply current in
harmonic is:
4 0
Ba 
 g eff
Phase A winding, then the magnetic field for fundamental
 Nˆ a 
P
equation is true no matter how those P groups

ia cos q a  This
of
windings
are connected. Note ia is phase A terminal current.
 P 
2


Na is effective number of turns connected in series per phase.


Now, we can calculate flux in Phase A winding from its own current.
Following the formula derived in Notes Flux Linkage in Phase Winding
a  Nˆ a  a, pk cos(0)  Nˆ a  a, pk
where
 a , pk 
2 Dl 4 0  Nˆ a
ˆ

a  N a
P  g eff  P
2 Ba , pk Dl
P
Ba , pk
4 0  Nˆ a 
 ia

 g eff  P 
2

80 Dl  Nˆ a 
ia 

 ia


 g eff  P 

a 80 Dl  Nˆ a
 LA 

ia
 g eff  P
Laa  Lbb  Lcc  Lls  LA




2
 0
Mutual Inductance between
Stator Windings
If we apply current in Phase B winding, then the magnetic field is:
4 0
Bb 
 g eff
 Nˆ a

 P


P
2 
ib cos q a 


3 
2

Now, we can calculate flux linkage in Phase A winding from Phase B current.
where
2
ˆ
a |from Phase B winding N a  b, pk cos( )
3
2 Bb, pk Dl B  4 0  Nˆ a i
b , pk
 b, pk 
 P b

g
eff 

P
2
 40 Dl  Nˆ a 
 1 ˆ 2 Dl 4 0  Nˆ a 

ib 

 ib
a 
Na



2
P  g eff  P 
 g eff  P 
2
a  40 Dl  Nˆ a 
LA
 Ms 



ib
 g eff  P 
2
Mutual Inductance between Stator and
Rotor Field Winding
If we apply current in rotor field winding, then when rotor is moving,
the magnetic field in airgap from rotor field winding is: q  q  q
40  Nˆ f
Bf 
g eff  P
d
a

i f cos( P q d ) Define: Nˆ  k N
f
wf
f

2

Now, we can calculate flux in Phase A winding from field current.
a |from field winding Nˆ a  f , pk cos q me
m
q me 
Effective number
of turns on field
winding.
  q me
 Nˆ f 

i f
B f , pk
where
 f , pk 
 P 
P


 Nˆ f 
 Nˆ a Nˆ f 

8

2
Dl
4
Dl
0
0

i f 

 cos q mei f
a  Nˆ a cos q me
2
P  g eff  P 
 g eff  P 
a 80 Dl  Nˆ a Nˆ f 
 Laf 

cos q me  Lsf cos q me
2


if
 g eff  P 
2 B f , pk Dl
where
40

g eff
80 Dl  Nˆ a Nˆ f
Lsf 
 g eff  P 2




P
qm
2
Self Inductance of Rotor Field Winding
For the magnetic field from rotor field winding is:
40
Bf 
g eff
 Nˆ f

 P


i f cos( P q d )

2

Now, we can calculate flux in field winding by integrating on q d .
 f  Nˆ f  f , pk
where
 Nˆ f 

i f
 f , pk
 P 


ˆ 
ˆ


N
N

8

2
Dl
4
Dl
f
0 
0
i f 
 f
 f  k f Nˆ f
P  g eff  P 
 g eff  P
2
ˆ


 f 80 Dl  N f 
 Lmf 

if
 g eff  P 
40
2 B f , pk Dl B

f , pk

g eff
P
L f  Llf  Lmf
2

 if


Steady State Analysis of
Round Rotor Machine
Terminal Voltage for Round Rotor Motor
 va   Rs
v   0
 b
 vc   0
  
v f   0
When

 a  
  
 b
 c  
  
 f   L
 sf

0
Rs
0
0
0
0
Rs
0
0   ia 
 a 
0   ib  d  b 

0   ic  dt  c 
 
 
R f  i f 
 f 
ia  ib  ic  0
Ls
0
0
cos q me
Lsf cos q me 
2   ia 
Ls
0
Lsf cos(q me  )   
3  ib
2   
0
Ls
Lsf cos(q me  )  ic 
3  
 i f 
2
2
Lsf cos(q me  ) Lsf cos(q me  )
Lf

3
3

0
0
Round Rotor Motor at Steady State
va  Rsia 
d a
dt
a  Lsia  Lsf cos q mei f
dq me
 e q me  et  r
At steady state
dt
di
va  Rsia  Ls a  Lsf I f e sin et  r 
dt


j  r  


dia
2  je t

 Rsia  Ls
 Re  Lsf I f e e
e 


dt




j



r

je t
je t
2

v

Re
V
e
ia  Re I Ae
Let
E A  Lsf I f e e
a




 V  Rs I A  jX s I A  E A
X s  e Ls
80 Dl  Nˆ a Nˆ f
Lsf 
 g eff  P 2





Round Rotor Generator at Steady State
Motor
Generator
V  Rs I A  jX s I A  E A
V   Rs I A  jX s I A  E A
d a
dt
a  Ls (ia )  Lsf cosqmei f
va   Rsia 


j  r  


dia
2  je t

va   Rsia  Ls
 Re  Lsf I f e e
e 


dt


EA
va  Re V e jet
ia  Re I Ae jet
80 Dl  Nˆ a Nˆ f 
Lsf 
 g eff  P 2 




Open Circuit Voltage (1)
Assume the armature windings are open circuit, the magnetic field in the air gap
comes from the field winding only.
P
P
Bg |from field winding  B f , pk cos(et  q a  r )  B f , pk cos( q a  et  r )
2
2
2 B f , pk Dl



(

t


)
 f , pk 
From Notes Flux Linkage in Phase Winding
e
r
P
 a |from field winding Nˆ a  f , pk cos   Nˆ a  f , pk coset  r 
E A (t ) 
d a |from field winding
dt
At steady state, we have
d  f , pk
ˆ
ˆ
  N a  f , pk e sin(et   r )  N a cos(et   r )
dt
d f , pk
dt
0
E A (t )   Nˆ a  f , pke sin(et  r )

 Nˆ a  f , pke cos(et  r  )
2
Phasor of EA(t) is:
EA  e Nˆ a  f , pk e

j ( r  )
2
Open Circuit Voltage (2)
E A  e Lsf I f e
8 Dl
Lsf  0
 g eff

j (r  )
2
 Nˆ a Nˆ f

2
P

EA  e Nˆ a  f , pk e



 f , pk 
4 0
Br 
 g eff
2 Br , pk Dl
 Nˆ f

 P
P

P
I
cos(
q a  q me )
 f
2

We can find out that they are the same.
E A, rms
 f , pk  Nˆ a  f , pk


 Nˆ a Nˆ f 
j (r  )
j ( r  )
2
2
I
e



e


e f , pk
2
 f
P


Dl  Nˆ a Nˆ f 
1
 8 2 f e 0
e f , pk

 I f 
2
g eff  P 
2
80 Dl
E A  e
 g eff

j ( r  )
2
Volt-Second Balance
E A,rms  2 f e Nˆ a  f , pk  4.44 f e Nˆ a  f , pk
 f , pk
E A,rms
or:
 2 Nˆ a  f , pk  4.44 Nˆ a  f , pk
fe
Example: if a 60Hz generator is to be operated at 50 Hz, then the
operating voltage must be derated to 50/60 of its original value.
Induced Phase Voltage
For the net magnetic field, at steady state:
P
Bnet  B pk cos(et  q a  net )
2
 a ,net  Nˆ a  pk cos et  net 
 pk 
2 B pk Dl
P
The steady state phase voltage in armature phase A winding is:
V (t ) 
da

  Nˆ a  pke sin(et  net )  Nˆ a  pke cos(et  net  )
dt
2

j ( net  )
2
ˆ
V  e N a  pk e
Phasor
The rms phase voltage is
V ,rms  2 f e Nˆ a  pk  4.44 f e Nˆ a  pk
 pk
Y Connection (Generator)
D Connection (Generator)
Example 1
For a 3 phase, 4 pole, 24 slot, 5/6 pitch machine with double layer lap
winding, the peak magnetic filed intensity in the airgap is 0.45 T. There
is one slot skew. The machine shaft speed is 8000 rpm. The stator
inner diameter is 0.5 m. The machine length is 0.3 m. There are 10
turns per coil. All the turns are connected in series. The three phase
coils are Y connected.
(1)What is the rms phase voltage of the machine?
(2)What is the rms terminal voltage of the machine?
V ,rms  2 f e Nˆ a  pk  4.44 f e Nˆ a  pk
 pk 
 pk
2 B pk Dl
P
P4
Nˆ a  N a k w
N a  PqN c
q
S
mP
ACmachine1.m
Example 2
For a simple 2 pole, 3 phase, Y connected machine (single layer
winding) shown in the figure, the peak magnetic field intensity in the
airgap is 0.2T. There is no skew. The machine shaft speed is 3600 rpm.
The stator inner diameter is 0.5 m. The machine length is 0.3 m. There
are 15 turns in the coil.
(1)What is the rms phase voltage of the machine?
(2)What is the rms terminal voltage of the machine?
V ,rms  2 f e Nˆ a  pk  4.44 f e Nˆ a  pk
 pk 
 pk
2 B pk Dl
P
P2
For this example:
Nˆ a  N c
(Note: This is single layer winding.)
ACmachine2.m
Voltage and Speed Regulation
Voltage regulation:
VR 
Vnl  V fl
V fl
 100%
Speed regulation:
SR 
Or:
nnl  n fl
n fl
 100%
 nl   fl
SR 
 100%
 fl
AC Machine Efficiency
Pout

 100%
Pin
Pout  Pin  Ploss
AC Machine Loss Mechanism
1.
2.
3.
4.
Electrical or copper losses (I2R losses)
Core losses
Mechanical losses
Stray or miscellaneous losses
Electrical or Copper Loss
Stator Copper Loss (SCL):
PSCL  3I A2 Rs
Rotor Copper Loss (RCL):
PRCL  I F2 RF
Core, Mechanical and Stray Losses
AC Generator Power Flow
power converted from mechanical to electrical
Pconv  Temm
AC Motor Power Flow
power converted from electrical to mechanical
Pconv  Temm