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Self and Mutual Inductances forSynchronous Machine with
Round Rotor
Double Layer Lap Winding
on Stator
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Cross Section Diagram
a axis
baxis
caxis
qaxis
qmqa
qd
daxis
mad qqq
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Stator WindingFractional Pitch
a axis
aq
q=2
rm gm
m
Pgg
2
a axis
2
mg
aq
q=4
(exaggerated end turns)
qcoils per group
2
3 mg
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Self and Mutual Inductances (1)
ia(t)
ib(t)
ic(t)
a axis
daxis
qaqm
mad qqq qd
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Self and Mutual Inductances (2)
cos( )
2cos( )
3
2cos( )3
aa bb cc ls A
ab bc ca s
f lf mf
af sf me
bf sf me
cf sf me
L L L L L
L L L M
L L L
L L
L L
L L
q
q
q
mme
Pqq
2
Linear Model
Balanced Winding
lsL is leakage inductance of armature phase A winding which is about10% of the maximum self inductance.
is leakage inductance of field winding.lfL
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Flux Linkage (1)
( )
( )
( )
[
a aa a ab b ac c af f
aa a s b c af f
b ba a bb b bc c bf f
aa b s a c bf f
c ca a cb b cc c cf f
aa c s a b cf f
f f f af a bf b cf c
f f sf
L i L i L i L i
L i M i i L i
L i L i L i L i
L i M i i L i
L i L i L i L i
L i M i i L i
L i L i L i L i
L i L i
2 2
cos cos( ) cos( )]3 3
a me b me c mei i
q q q
At steady state, fi is DC.
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Flux Linkage (2)
f
c
b
a
fmesfmesfmesf
mesfaass
mesfsaas
mesfssaa
f
c
b
a
i
i
i
i
LLLL
LLMM
LMLM
LMML
)3
2cos()
3
2cos(cos
)3
2cos(
)3
2cos(
cos
q
q
q
q
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Flux Linkage (3)
sAlssaas
mecmeb
measffff
mefsfcs
fcfcsaac
mefsfbs
fbfbsaab
mefsfas
fafasaaa
MLLMLL
ii
iLiL
iLiL
iLiML
iLiL
iLiML
iLiL
iLiML
)]3
2cos()
3
2cos(
cos[
)
3
2cos(
)(
)3
2cos(
)(
)cos(
)(
q
q
q
q
q
q
0 cba iiiY connected without neutral return or balanced Dconnected :
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Flux Linkage (4)
f
c
b
a
fmesfmesfmesf
mesfs
mesfs
mesfs
f
c
b
a
i
i
i
i
LLLL
LL
LL
LL
)3
2cos()
3
2cos(cos
)3
2cos(00
)3
2cos(00
cos00
q
q
q
q
0
cba iiiWhen
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Flux Linkage in Phase A Winding
There are a total ofPgroups. These groups may be connected in series, or parallel,
or partly series and partly parallel.
cospkwcsgroupsa kqNPP Then:
csa qNPN
waa kNN
cosa a pk N
number of series turns per phase per circuit
CPP sAssume:
Note:
effective number of seriesturns per phase
per circuit on armature winding
number of parallel circuitsC
)2
cos( q apkP
BB
P
DlBpkpk
2
pk a pkN
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Self Inductance of Stator Winding
If we apply current in Phase A winding, then the magnetic field for fundamentalharmonic is:
aa
a
eff
a
Pi
P
N
gB q
2cos
4 0
Now, we can calculate flux in Phase A winding from its own current.
Following the formula derived in Notes Flux Linkage in Phase Winding
pkaapkaaa NN ,,)0cos(
P
DlB pkapka
,
,
2 a
a
eff
pka iP
N
gB
4 0,
aa
eff
aa
eff
aa i
P
N
g
Dli
P
N
gP
DlN
2
00842
2
08
P
N
g
Dl
iL a
effa
aA
0
where
aa bb cc ls AL L L L L
This equation is true no matter how those P groupsof windings are connected. Note iais phase A terminal current.Na is effective number of turns connected in series per phase.
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Mutual Inductance between
Stator Windings
If we apply current in Phase B winding, then the magnetic field is:
3
2
2cos
4 0 q
ab
a
eff
b
Pi
P
N
gB
Now, we can calculate flux linkage in Phase A winding from Phase B current.
)3
2cos(| ,windingBPhasefrom pkbaa N
P
DlB pkbpkb
,
,
2 b
a
eff
pkb iP
N
gB
4 0,
ba
eff
ba
eff
aa iP
N
g
Dli
P
N
gP
DlN
2
00442
2
1
2
42
0 Aa
effb
a
s
L
P
N
g
Dl
iM
where
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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:
Now, we can calculate flux in Phase A winding from field current.
mepkfaa N q cos
| ,windingfieldfrom
P
DlB p kfpkf
,
,
2
fme
fa
eff
f
f
eff
meaa iP
NN
g
DliP
N
gP
DlN q
q cos
842cos
2
00
)2
cos(4 0
df
f
eff
f
Pi
P
N
gB q
f
f
eff
pkf iP
N
gB
4 0,
mesfme
fa
efff
aaf L
P
NN
g
Dl
iL qq
coscos
82
0
mad qqq
where
mme
P
qq 2Define:
20
8
P
NN
g
DlL fa
eff
sf
where
meq
fwff NkN Effective number
of turns on fieldwinding.
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Self Inductance of Rotor Field Winding
For the magnetic field from rotor field winding is:
Now, we can calculate flux in field winding by integrating on .
pkfff
N,
P
DlB pkfpkf
,
,
2
f
f
eff
f
f
eff
fff iP
N
g
Dli
P
N
gP
DlNk
2
00842
)2
cos(
4 0df
f
eff
fPi
P
N
gB q
f
f
eff
pkf iP
N
gB
4 0,
2
08
P
N
g
Dl
iL
f
efff
f
mf
where
dq
mflff LLL
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Steady State Analysis ofRound Rotor Machine
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Terminal Voltage for Round Rotor Motor
f
c
b
a
f
c
b
a
f
s
s
s
f
c
b
a
dt
d
i
i
i
i
R
R
R
R
v
v
v
v
000
000
000
000
0
cba iiiWhen
f
c
b
a
fmesfmesfmesf
mesfs
mesfs
mesfs
f
c
b
a
i
i
i
i
LLLL
LL
LL
LL
)3
2cos()
3
2cos(cos
)3
2cos(00
)3
2cos(00
cos00
q
q
q
q
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Round Rotor Motor at Steady State
aa s a
dv R i
dt
cosa s a sf me f L i L i q
At steady state me e r tq
2
sin
Rer
e
aa s a s sf f e e r
j
j tas a s sf f e
div R i L L I t
dt
diR i L L I e edt
Let Re ej tav e V Re ej ta Ai e I 2rj
A sf f eL I e
E
s A s A AR jX V I I E
2
08
P
NN
g
DlL
fa
eff
sf
mee
d
dt
q
s e sX L
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Round Rotor Generator at Steady State
Motor Generator
s A s A AR jX V I I Es A s A AR jX V I I E
aa s a
dv R i
dt
( ) cosa s a sf me f L i L i q
2Rer
e
jj ta
a s a s sf f e
div R i L L I e e
dt
Re ej tav e V Re ej ta Ai e IAE
2
08
P
NN
g
Dl
L
fa
effsf
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Open Circuit Voltage (1)
from field winding ,
,
| ( ) sin( ) cos( )a f pkA a f pk e e r a e r
d dE t N t N t
dt dt
repkfapkfaa tNN coscos| ,,windingfieldfrom
from field winding , ,| cos( ) cos( )2 2
g f pk e a r f pk a e rP PB B t B t q q
Assume the armature windings are open circuit, the magnetic field in the air gapcomes from the field winding only.
)( ret From Notes Flux Linkage in Phase WindingP
DlB pkfpkf
,
,
2
At steady state, we have 0,
dt
d pkf
)
2
cos(
)sin()(
,
,
reepkfa
reepkfaA
tN
tNtE( )
2,
rj
e a f pk N e
A
E
Phasor ofEA(t) is:
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Open Circuit Voltage (2)
0
2
8 a fsf
eff
N NDlL
g P
,
,
2r pk
f pk
B Dl
P
0
4 cos( )2
fr f a me
eff
N PB Ig P q q
( )2,
rj
e a f pk N e
AE)2(
rj
fsfeA eILE
We can find out that they are the same.
( ) ( )0 2 2
,2 8 r rj ja f
A e f e f pk
eff
N NDl I e eg P
E
, 0 ,2
18 2
2
a f
A rms e f e f pk
eff
N NDlE f I
g P
, ,
f pk a f pkN
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Volt-Second Balance
, , ,
,
, ,
2 4.44
or: 2 4.44
A rms e a f pk e a f pk
A rms
a f pk a f pk
e
E f N f N
EN N
f
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.
,f pk
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Induced Phase Voltage
Therms phase voltage is, 2 4.44rms e a pk e a pk V f N f N
, cosa net a pk e net N t
For the net magnetic field, at steady state:
P
DlBpkpk
2
The steady state phase voltage in armature phase A winding is:
)2
cos()sin()( neteepkaneteepkaa tNtN
dtdtV
( )2
netj
e a pk N e
VPhasor
)2
cos( netaepknet PtBB q
pk
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Y Connection (Generator)
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Connection (Generator)
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Example 1
4P
2 pkpk
B Dl
P
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. Thereis one slot skew. The machine shaft speed is 8000 rpm. The statorinner diameter is 0.5 m. The machine length is 0.3 m. There are 10turns per coil. All the turns are connected in series. The three phasecoils are Y connected.
(1)What is the rms phase voltage of the machine?(2)What is the rms terminal voltage of the machine?
, 2 4.44rms e a pk e a pk V f N f N
a a wN N k
a cN PqN
Sq
mP
ACmachine1.m
pk
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Example 2
2P
2 pkpk
B Dl
P
For this example: a cN N
(Note: This is single layer winding.)
For a simple 2 pole, 3 phase, Y connected machine (single layer
winding) shown in the figure, the peak magnetic field intensity in theairgap 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. Thereare 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?
, 2 4.44rms e a pk e a pk V f N f N
ACmachine2.m
pk
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Voltage and Speed Regulation
%100
fl
flnl
V
VVVR
Voltage regulation:
%100
fl
flnl
n
nn
SR
Speed regulation:
%100
fl
flnlSR
Or:
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AC Machine Efficiency
100%out
in
P
P
out in lossP P P
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AC Machine Loss Mechanism
1. Electrical or copper losses (I2Rlosses)2. Core losses
3. Mechanical losses4. Stray or miscellaneous losses
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Electrical or Copper Loss
Stator Copper Loss (SCL):
Rotor Copper Loss (RCL):
23SCL A sP I R
2
RCL F FP I R
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Core, Mechanical and Stray Losses
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AC Generator Power Flow
power converted from mechanical to electrical
conv em mP T
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AC Motor Power Flow
power converted from electrical to mechanical
conv em mP T
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