Previous lectures Today’s goal - IEA - Lund University · Previous lectures • Geometry + medium...
Transcript of Previous lectures Today’s goal - IEA - Lund University · Previous lectures • Geometry + medium...
Industrial Electrical Engineering and AutomationLund University, Sweden
L8: Electric circuits
The MMF and field distribution of an AC winding
eien20vt18
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Previous lectures
• Geometry + medium = a circuit
– Soft magnetic core– Permanent magnets
• Energy conversion = a balance
– Energy vs torque• Intermediate magnetic field
– Facilitate flux linkage– Prevent magnetisation
losses A
snA
s rdABBrdAtT0
1
MV
HC dVBdHWT
0
irBlNiT
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Today’s goal
• Scope on AC 3-phase machines
• Winding configuration for the stator
• Winding factor• Insulation system
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Electromagnet at generator operation
• Excitation -> establish H-field
• Core -> guide the B-field
• Coil -> link by inducted E-field
• Winding -> add e
I H
dAtBEdl
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Placement of coils A
• Distribution• Coil width(pitch)
• Skewing• …
ψa
Ψ=N·Φ=L·I
N·I=Aw·J·Kf
coil
Ψm
ia
θmech
ω
x
magnet
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Placement of coils B
• Concentrated (non overlapping coils)• Distributed (overlapping coils)• TARGET: n-phase symmetric winding
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3-phase system
• Excitation flux Ψm– Rotates and links (projects)
• Spatial distribution of coils– Phase-A ej0·⅔π = a0
– Phase-B ej1·⅔π = a1
– Phase-C ej2·⅔π = a2
• Coordinate frames αβ, dq, xy• Space vector
3
43
2
j
c
j
ba eekj
ψmψa
ψb
ψc
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Sinusoidally-fed PM motor
• N-phase system: Nph=3• P-pole excitation: Np=2 (4,6..)• S-slot stator: Ns={3,6,9,12, …}• Sinusoidal distribution
tPiDNk
tPrPK
dd
rKdrK
iPNktP
eis
s
espis
s
sp
ississsp
sspespsp
2cosˆ6
2cos
2)(
)(1)()()(
ˆ423
2sin)(
1
1
2
0
111
s
s
F
FF
FFF
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Number of coils vs poles
15 20 25 30 35
15
20
25
30
35
number of poles, Np []
num
ber o
f tee
th, N
t []
0.25
0.25
0.25
0.5
0.5
0.5
0.6
0.6
0.6
0.6
0.7
0.7
0.7
0.7
0.7
0.7
0.8
0.8
0.8
0.8
0.8
0.8
0.8
0.85
0.85
0.85
0.85
0.85
0.85
0.85
0.9
0.9
0.9
0.9
0.9
0.9
0.9
0.9
0.9
0.95
0.95
0.95
0.95
0.95
0.95
0.95
0.95
0.95
0.95
0.99
0.99
0.95
0.99
0.2
0.2
0.2
0.3
0.3
0.3
0.3
0.4
0.4
0.4
0.5
0.5
0.5
0.6
0.6
0.7
0.7
0.8
0.9
• Distribution of concentrated coils – distributedconcentrated winding
1
6
11
oo
xxooxx
ooxxoo
xxoo x x
ooxx
123456789101112q=1.00 Kw =1.00
Ns=12 Np=12
1
6
11
oxoo
xoxxoxoxoo
xoxx
ox ox oo xoxx
ox1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
q=1.25 Kw =0.95
Ns=15 Np=12
1
6
11
16
oxox
oxoxoxoxox
oxoxoxox
ox ox ox ox oxox
ox1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
q=1.50 Kw =0.87
Ns=18 Np=12
1
6
11
ox
ooxoxx
oxooxo
xxo x o o
xox
x
1
2
3
45
6
7
8
9
1011
12
q=0.86 Kw=0.93
Ns=12 Np=14
1
6
11
ooxx
oxooxxooxx
oxoo
xx oo xx oxoo
x
x
12
3
45 6
7 8
9101112
13
1415q=1.07 Kw=0.95
Ns=15 Np=14
1
6
11
16
oxoo
xoxoxxoxoxoo
xoxoxx
ox ox oo xoxo
xxox
1
2
3
4
5
6
7
8
9
10
1112
13
14
15
16
17
18
q=1.29 Kw=0.90
Ns=18 Np=14
1
6
11
ox
oxoxox
oxoxox
oxo x o x
oxo
x
1
2
3
4
5
6
7
8
9
10
11
12
q=0.75 Kw =0.87
Ns=12 Np=16
1
6
11
ooxx
oxooxxooxx
oxoo
xx oo xx oxoo
x
x
12
3
45 6
7 8
9101112
13
1415
q=0.94 Kw =0.95
Ns=15 Np=16
1
6
11
16
ooxo
xxooxoxxoo
xoxxooxo
xx oo xo xx ooxoxx
1
2
34
5
67
8
9
10
11
1213
14
1516
17
18
q=1.13 Kw =0.95
Ns=18 Np=16
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Winding specification
• Number of turns per phase: Nt=U1/Ψ1ω1
– Voltage and flux at nominal operation point
• Number of turns per coil: Ntc=Nt/integer
• Number of strands– Wire diameter
according to frequency
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AC machine
• Sinusoidally-fed PM motor
• Electromagnetic torque
• Sinusoidal distribution of stator MMF
BlNirNT qpem 5.0
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MMF of a distributed winding
• Focus on stator (Hpm=0,μpm=1)• Amperes circuital law (PQRS)
• H-field uniform in gap (μfe=∞)
• MMF distribution
θ2 θ1
θ
P
Q R
S g
P
S
S
R
R
Q
Q
P
HdlHdlHdlHdlHdlNi
12 rr
S
R
Q
P
gHgHHdlHdlNi
0 gHgHNiF Avo R Design of Electrical Machines 14
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Flux & MMF waveforms
• Electromagnetic force
• The air-gap flux density
• Sinusoidal current density
θ
0 π 2π rotor
stator F(θ), B(θ)
0.5NI B
egmg BB
sinsin4)(
tPirNkK
ddF
rKdrKF
tiNNkF
eis
ss
sp
ississsp
eep
ssp
2cosˆ
26)(
)(1)()()(
sinˆ423)(
1
2
0
1
s
s
ssgmesgmeis
eissgm
iNkBlKBlD
dlrKBF
ˆ32
)()(
1111
2
0
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Full pitch winding
• MMF changes abruptly• A single-phase full-pitch
winding – one slot per pole• Lengthy end-turns• Open for space harmonics
...5sin513sin
31sin
24
sin12
4
INF
oddhhh
INF
tp
tph
θ
0 π 2π rotor
stator F(θ) NtI
½NtI
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Fractional pitch winding
• Coil pitch/Pole pitch is a fractional number
• Ordinarily two layer lap wound type
• MMF waveform becomes more sinusoidal
• Prevent harmonic content
θ
0 π 2π rotor
stator F(θ) ½NtI
½NtI
γ
,5,3,1sin
odd2cos
24
hphp
tph
hFF
hh
INF
h
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Pitch factor
• Coil pitch w
• Pole pitch p
• Pitch factor kph
• w/p ≈0.83 suppress 5th and 7th harmonic
hoddhh
hk
p
w
ph
2sin
2sin
...2
cos
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Distributed windings
• Removal of undesirable harmonics
• If q=2 is not possible, rather than selectingq=1, q can be selectedas a fraction between 1 and 2
θ
0 π 2π rotor
stator F(θ) 1/(2q)NtI
1/(2q)NtI
γ
q=2
,5,3,1sin
odd2
cos
2sin
2sin
24
hphp
tph
hFF
hhqhq
hq
hINF
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Distribution factor
• Number of slots per phase and per pole q
• Harmonic distribution factors for a winding with 600 phase belt
• Slot hamonics hk = 6kq +/- 1 = 2kS/P +/- 1
oddh
qhZq
hZ
hq
hq
k
p
pdh
2sin
2sin
2sin
2sin
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Slot openings
• Slot opening has filtering effect on the MMF
• It is assumed that the MMF varies linearly across the slot
θ
0 π 2π rotor
stator F(θ) NtI
½NtI
χ
,5,3,1sin
odd
2
2sin
12
4
hphp
tph
FF
hh
INF
h
h
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Skewing
• Reduces slot harmonics• Reduces flux variation
that is due to the fringing• Skew angle
θ
0 π 2π rotor
stator F(θ) NtI
½NtI
α
,5,3,1sin
odd
2
2sin
12
4
hphp
tph
FF
hh
INF
h
h
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Slot opening factor
• Skew angle α• Slot width angle χ• Filtering effect β
odd
2
2sin
2
2sin
hZ
Z
k
p
ph
h
h
h
h
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Winding factor
• Distribution factor kdh
• Pitch factor kph
• Slot opening factor koh
• Skewing factor ksh
• Winding factor is ratio of actual winding MMF to the full-pitched winding MMF
• A machine with a high polenumber AND a high numberof slots per pole and phaseis impractical to build
• The number of slots MUST be a multiple of the numberof phases – symmetry.
odd2
4 hkkkkhINF shhphdh
tph
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Specify 3-phase winding
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nFractional slot winding example : I
• 10-pole machine• Select number of slots
– q=1 -> 10*3*1 = 30 slots– q=2 -> 10*3*2 = 60 slots– Pick 42 slots -> q = 42/(10*3) = 7/5 – q=slots per pole per phase.
• q=7/5 implies 1 or 2 slots occupied per pole• Coil arrangements: 2,1,1,2,1 OR 1,1,2,2,1 slots per
pole per phase for 5 poles.• The other 5 poles are copies...
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Fractional slot winding example : II• Pick pitch – same for all coils• Phase shift between coils =
360/42*p/2= 42 6/7 °• Suggest coil span 4 slots ->
pitch angle = 180-4* 42 6/7 ° = 8 4/7 °
• Pitch factorkp1 = cos(8 4/7°/2)=0.997
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
•Winding sketch:
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Fractional slot winding example : III1
2
3
4
56
7
8
9 10
11
12
1314
15
16
1718
19
20
21
The phase coils forms an equivalent 60 degreephase belt, distributedover 7 slots:
956.0
790
2171sin7
9021
71sin
2sin
2sin
qhZq
hZ
k
p
pdh
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Fractional slot winding example : IV1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
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nFractional slot winding example : V
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End-turns
• Single-layer concentric and equal coils
• Double layer distributed winding with concentric coils and concentrated winding
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Double layer diamond winding
• Animation 1: Increasing coil span
• Animation 2: increasing q=Ns/(Np*Nph)
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Diamond windings in Ansys
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Winding configuration
• Slot content– Insulation system– Electrical strands
• Insulation system– Phase/overhang insulation (a)– Ground insulation (b)– Turn insulation (c)
• Insulation tests– Phase to phase (1)– Phase to ground (2)– Turn to turn (3)
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Conductor fill factor
• Coating increases the diameter of copper strand by 5% to 7%
• Different layers of main insulation
• Round conductor in rectangular slot Kf=π/4
• Max range 60-70%
hins r
hs
hwht
bs1
hss
iscuis
isispispsf
sssinsinsinsis
sins
ins
wts
ddAdNdNNK
rbhhrbhhhrA
brh
hhhrb
rhhhh
/25.0
105.02225.05.0
22
22
222
211
1112
11
11
1
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Thermal limits
• Insulation lifetime is shortened radically if temperature exceeds the limit and that is due to accelerated oxidation process in the insulation material.
A E B F H0
20
40
60
80
100
120
140
160
180
insulation classes
tem
pera
ture
[
C]
40
60
5
40
75
5
40
80
10
40
105
10
40
125
15ambmaxallow edsafety
140 160 180 200 220 240 260 280102
103
104
105
temperature [C]
ther
mal
life
[h]
Class 200Class 155
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Modeling a winding
• Equivalent thermal conductivity of a winding is given by the filling factor of the conductor strands (copper in this example) and the thermal conductivity of the medium between the conductor strands
inscond
fcondfins
ins
f
cond
f
eff
kkLkLkLL
11
fcondfins
inscondeff kk
1
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Winding insulation and heat transfer
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50
1
2
3
4
5
6
7
8
wire diameter dc, [mm]
rela
tive
thic
knes
s of
rhe
insu
latio
n d i, [
%]
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Winding design
• k1→Fsp()– Winding layout and
resulting MMF• →Ubr
– Dielectric properties and voltage limits
• →hs– Heat dissipation
and temperature limits
42
41
40
39
38
37
36
35
34
33
32
31
30
29
28
27
26
25
24
23
22 21
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
11
2
3
4
5
6
78
91011121314
15
16
17
18
19
20
21
22
23
24
25
26
27
2829
30 31 32 33 3435
36
37
38
39
40
41
42
B()MMF()T()
1 5 9 13 17 21 25 29 33 37 41
Ns=42 Np=10 q=1.40 span=4.0
kd=0.978 kp=0.997 ks=0.955 k1=0.931
Does not consider properly concentratedwindings where q<1
1 5 9 13 17 21 25 29 33 37 41
6
5
43
2
1
1
2
3
4
5
6 B()MMF()T()
1 5
Ns=6 Np=2 q=1.00 span=3.0
kd=1.000 kp=1.000 ks=0.955 k1=0.955
Does not consider properly concentratedwindings where q<1
1 5
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Assignment A4 part 1
-0.05 0 0.05-0.05
-0.04
-0.03
-0.02
-0.01
0
0.01
0.02
0.03
0.04
0.05
gap radiustooth ref nodesyoke ref nodes
0 30 60 90 120 150 180 210 240 270 300 330 360
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
angle , [deg]
Mag
netic
flux
dens
ity in
the
airg
ap B
( ),
[N/m
2 ]
BgnL(), [T]
BgtL(), [T]
Bgn0(), [T]
Bgt0(), [T]
0 2 4 6 8 10 12 14 16 18 200
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
harmonic order, [-]
Mag
netic
flux
den
sity
in th
e ai
rgap
Bg, [
T]
BgnL(), [T]
BgtL(), [T]
Bgn0(), [T]
Bgt0(), [T]
0 30 60 90 120 150 180 210 240 270 300 330 360
-3
-2
-1
0
1
2
3
4
x 105
angle , [deg]
Mag
netic
she
ar s
tress
in th
e ai
rgap
(
), [N
/m2 ]
gnL(), [N/m2]
gtL(), [N/m2]
gn0(), [N/m2]
gt0(), [N/m2]
• con.femm=1• θ=0 el.deg• I(θ),Ψ(θ),
T(θ), B(θ)• Bgm1?• EMF?• T calculation?
– Weighted– Line integral– Estimation
• Phasors?
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Assignment A4 part 2
0 50 100 150 200 250 300 350
-300
-200
-100
0
100
200
300
angle , [deg]
curre
nt I,
[A]
abc
0 50 100 150 200 250 300 350-10
-5
0
5x 10
-3
angle , [deg]
flux
linka
ge
, [V
s]
abc
0 50 100 150 200 250 300 350
3.5
4
4.5
5
5.5
6
angle , [deg]
torq
ue T
, [N
m]
0 50 100 150 200 250 300 350
0.2
0.4
0.6
0.8
1
1.2
angle , [deg]
flux
dens
ity in
the
stat
or c
ore
B, [
T]
Bst1Bst2
Bst3
Bsy1
Bsy2Bsy3
• con.femm=2• pos_end=120• pos_step=60
• I(θ),Ψ(θ), T(θ), B(θ)
• Isx=0, Isy>0 • T>0, ω>0