Magnetic Circuits and Magnetic Materials 1. Introduction to Magnetic Circuits 2. Flux Linkage, Inductance, and Energy 3. Properties of Magnetic Materials 4. AC Excitation 5. Permanent Magnets 6. Application of Permanent Magnet Materials
49 trang |
Chia sẻ: linhmy2pp | Ngày: 18/03/2022 | Lượt xem: 209 | Lượt tải: 0
Bạn đang xem trước 20 trang tài liệu Bài giảng Electromechanical energy conversion - Chapter I: Magnetic circuits and magnetic materials - Nguyễn Công Phương, để xem tài liệu hoàn chỉnh bạn click vào nút DOWNLOAD ở trên
Nguyễn Công Phương
ELECTROMECHANICAL ENERGY
CONVERSION
Magnetic Circuits
and Magnetic Materials
Contents
I. Magnetic Circuits and Magnetic Materials
II. Electromechanical Energy Conversion
Principles
III. Introduction to Rotating Machines
IV. Synchronous Machines
V. Polyphase Induction Machines
VI. DC Machines
VII.Variable – Reluctance Machines and Stepping
Motors
VIII.Single and Two – Phase Motors
IX. Speed and Torque Control
sites.google.com/site/ncpdhbkhn 2
Magnetic Circuits
and Magnetic Materials
1. Introduction to Magnetic Circuits
2. Flux Linkage, Inductance, and Energy
3. Properties of Magnetic Materials
4. AC Excitation
5. Permanent Magnets
6. Application of Permanent Magnet Materials
sites.google.com/site/ncpdhbkhn 3
Introduction to Magnetic Circuits
(1)
HLJS.d= . d
∫ ∫ S
BS.d = 0
∫S
BH= µ
• H: magnetic field intensity, A/m
• J: current density, A/m2
• B: magnetic flux density, Wb/m2 (T)
• µ: permeability, H/m
sites.google.com/site/ncpdhbkhn 4
Introduction to Magnetic Circuits
(2)
6
4
2
0
-2
-4
-6
1
0.5 1
0.5
0
0
-0.5
= -0.5
HLJS.d . d -1 -1
∫ ∫ S
sites.google.com/site/ncpdhbkhn 5
Introduction to Magnetic Circuits
(3)
HLJS.d= . d
∫ ∫ S
sites.google.com/site/ncpdhbkhn 6
Introduction to Magnetic Circuits
(4)
BS.d = 0
∫S
sites.google.com/site/ncpdhbkhn 7
Introduction to Magnetic Circuits
Magnetic flux Φ (5) Mean core length, lc
i
Cross-sectional area, S
+ c
Air gap, length g,
– permeability µ0,
area S
Winding, N turns g
Magnetic core permeability µ
µ >> µ0 : the magnetic flux is confined
almost entirely to the core
F = Ni : magnetomotive force (mmf)
sites.google.com/site/ncpdhbkhn 8
Introduction to Magnetic Circuits
Magnetic flux Φ (6) Mean core length, lc
i
Cross-sectional area, S
+ c
Air gap, length g,
– permeability µ0,
area S
Winding, N turns g
Magnetic core permeability µ
Φ = BS.d F= Ni = HL. d
∫S ∫
≈ ≈
Bc const Hc const
→ Φ = → =
cB c S c F Hc l c
sites.google.com/site/ncpdhbkhn 9
Introduction to Magnetic Circuits
Magnetic flux Φ (7) Mean core length, lc
i
Cross-sectional area, S
+ c
Air gap, length g,
– permeability µ0,
area S
Winding, N turns g
Magnetic core permeability µ
= = = +
FNi HL. d Hl Hl B Bg
∫ cc gg →F =c l + g
=µ = µ µc µ
BHBHc c; g0 g 0
Φ = Φ = Φ = Φ = Φ
cBS cc; g BS gg ; c g
l g
→F = Φc +
µ µ
Sc0 S g
sites.google.com/site/ncpdhbkhn 10
Introduction to Magnetic Circuits
(8)
l g
F = Φc +
µ µ
Sc0 S g → = Φ +
FRR(c g )
l g
c =RR; =
µc µ g
Sc0 S g
FFF
→ Φ = = = = PF
+ l g total
RRRc gc + total
µ µ
Sc0 S g
sites.google.com/site/ncpdhbkhn 11
Introduction to Magnetic Circuits
Φ lc
Ex. 1 (9)
Given Bc = 1T, lc = 40cm, g = 0.05cm, i Sc
2
N = 1000 turns, Sc = Sg = 16cm , +
µ = 65000 µ . Find: g,
0 µ ,
a) The reluctances R & R ? 0
c g – S
b) The flux Φ? g
c) The current i? N turns
µ
l 40× 10−2 A.turns
R =c = = 3061
c µ π ×−7 × − 4
Sc 65000(4 10 )(16 10 ) Wb
g 0.0510× −2 A.turns
R == =×2.49 10 5
g µ π ×−7 × − 4
0Sg (4 10)(1610) Wb
Φ= =××−4 =× − 4
Bc S c 1 16 10 16 10 Wb
F Φ(R + R ) 16× 10−4 (3061 + 2.49 × 10 5 )
i ===c g = 0.41A
NN 1000
sites.google.com/site/ncpdhbkhn 12
Introduction to Magnetic Circuits
Φ lc
Ex. 2 (10)
2
Given Φ = 0.141 mWb, Sc = Sg = 4cm , i Sc
lc = 44cm, g = 0.02cm, N = 400 turns. +
The core is made of nickel – iron alloy. g,
µ ,
Find the current i? 0
– S
− g
Φ0.141 × 10 3
B = = = 0.35T N turns
c −4 µ
Sc 4× 10
→ =
Hc 850 A/m
B (T)
sites.google.com/site/ncpdhbkhn H (A/m) 13
Introduction to Magnetic Circuits
Φ lc
Ex. 2 (11)
2
Given Φ = 0.141 mWb, Sc = Sg = 4cm , i Sc
lc = 44cm, g = 0.2cm, N = 400 turns. +
The core is made of nickel – iron alloy. g,
µ ,
Find the current i? 0
– S
− g
Φ0.141 × 10 3
B = = = 0.35T N turns
c −4 µ
Sc 4× 10
→ =
Hc 850 A/m
−
Φ0.141 × 103 B 0.35
B== =→==0.35T H g =× 2.7910A/m5
g −4 g µ − 7
Sg 410×0 410π ×
= + =× + ×××5− 3 =
F Hlcc Hl gg 850 0.44 2.79 10 2 10 931A.turns
F 931
→=i = = 2.33 A
N 400
sites.google.com/site/ncpdhbkhn 14
Introduction to Magnetic Circuits
Ex. 3 (12) g
2
Given i = 10 A, Sg = 1500 cm , g = 2 cm,
N = 500 turns. Rotor and stator are made i
of iron (infinite permeability). Find the
flux of the air gap?
Φ =F = Ni N turns
+ l g
RRc g c +
µ µ
Sc0 S g
µ → ∞
Niµ S 500× 10(4π × 10−7 )1500 × 10 − 4
→Φ==0 g = 0.024 Wb
2g 2210× × −2
sites.google.com/site/ncpdhbkhn 15
Introduction to Magnetic Circuits
Ex. 4 (13)
l1 l3
i
+ Ф1
l Ф
Ф2 2 3
–
N turns
= = +
F Ni Hl11 Hl 22
=
Hl22 Hl 33
Φ = Φ + Φ
1 2 3
sites.google.com/site/ncpdhbkhn 16
Magnetic Circuits
and Magnetic Materials
1. Introduction to Magnetic Circuits
2. Flux Linkage, Inductance, and Energy
3. Properties of Magnetic Materials
4. AC Excitation
5. Permanent Magnets
6. Application of Permanent Magnet Materials
sites.google.com/site/ncpdhbkhn 17
Flux Linkage, Inductance, and
Energy (1)
d
∫EL.d= − ∫ BS . d
Cdt S
dϕ d λ
e= N =, λ = N ϕ
dt dt
λ
L =
i
ϕ = F
Rtotal
F= Ni
N 2
→L =
Rtotal
sites.google.com/site/ncpdhbkhn 18
Flux Linkage, Inductance, and
Ex. 1 Energy (2)
i Area µ → ∞
Find the inductance of the winding and S1
the flux density of gap 2? +
Area S1
g g Gap1 g g
R=1, R = 2 λ 1 2
1µ 2 µ
01S 02 S −
Gap 2
RR gg N turns
R =12 = 12
total +µ +
R1 R 2 012( gS gS 21 )
λ Nφ φ
L = =
i i φ φ
+ 1 2
Ni
φ = R R
R Ni 1 2
total −
S S
→ =µ 2 1 + 2
LN0
g1 g 2
sites.google.com/site/ncpdhbkhn 19
Flux Linkage, Inductance, and
Ex. 1 Energy (3)
i Area µ → ∞
Find the inductance of the winding and S1
the flux density of gap 2? +
Area S1
λ Gap1 g1 g2
−
φ Gap 2
= 1 N turns
B1
S1
µ
φ =Ni = 0 S 1 Ni φ
1
R g φ φ
1 1 + 1 2
Ni R1 R2
µ Ni −
→ = 0
B1
g1
sites.google.com/site/ncpdhbkhn 20
Flux Linkage, Inductance, and
Φ lc
Ex. 2 Energy (4)
2 i Sc
Given lc = 40cm, Sc = Sg = 16cm , Bc = 1T,
N = 1000 turns, g = 0.05cm, find L for +
g,
a) µ = 65000 µ0, and b) µ = 3000 µ0,
µ0,
lc –
R = Sg
c, a µ N turns
Sc
0.4 A.turns µ
= =3.07 × 10 3
65000(4π × 10−7 )(16 × 10 − 4 ) Wb
0.4 A.turns
R = =6.61 × 10 4
c, b 3000(4π × 10−7 )(16 × 10 − 4 ) Wb
g 0.0510× −2 A.turns
R == =×2.49 10 5
g µ π ×−7 × − 4
0Sg (4 10)(1610) Wb
A.turns A.turns
RRR=+=×2.52105 , RRR =+=× 3.1510 5
total,, a c a g Wbtotalb, cb , g Wb
N21000 2 N 2 1000 2
L== =3.97 H , L == = 3.17 H
a ×5 b × 5
Rtotal, a 2.52 10 R total, b 3.15 10
sites.google.com/site/ncpdhbkhn 21
Flux Linkage, Inductance, and
Energy (5)
4.5
4
3.5
3
2.5
2
Inductance (H) Inductance
1.5
1
0.5
0
0 1 2 3 4 5 6 7 8 9 10
Core relative permeability 4
x 10
sites.google.com/site/ncpdhbkhn 22
Flux Linkage, Inductance, and
Energy (6) Air gap
i
= + i1 2
F Ni11 Ni 22
+ g +
φ =F ≈ F
R+ R R λ λ
g c g 1 2
µ S − −
→φ = + 0 c
(Ni11 Ni 22 )
g N1 turns µ, lc , Sc N2 turns
µS µ S
λ=N φ = N2 0c iNN + 0 c i
11 1g 112 g 2
µS µ S
λ=N φ = NN0c iN + 2 0 c i
22 12g 12 g 2 λ =Li + Li
→ 1 111 122
µS µ S λ = +
LN=20c, LN = 2 0 c (self-inductance) 2Li 211 Li 222
11 1g 22 2 g
µ S
L= NN0 c = L (mutual inductance)
12 12g 21
sites.google.com/site/ncpdhbkhn 23
Flux Linkage, Inductance, and
Energy (7)
dλ
e =
dt
di
e= L
dt
di dL
e= L + i
dt dt
sites.google.com/site/ncpdhbkhn 24
Flux Linkage, Inductance, and
Energy (8)
dλ
p= ie = i
dt
t λ
∆W =2 pdt = 2 id λ
∫ ∫ λ
t1 1
λ λ λ2− λ 2
=2 dλ = 2 1
∫λ
1 LL2
Li 2
λ =0 →W =
1 2
sites.google.com/site/ncpdhbkhn 25
Flux Linkage, Inductance, and
Φ lc
Ex. 3 Energy (9)
2 i Sc
Given lc = 40cm, Sc = Sg = 16cm , Bc = 1T,
N = 1000 turns, g = 0.05cm, µ = 65000 µ . +
0 g,
a) Find the magnetic stored energy?
µ0,
b) Bc = sin(314 t) (T), find e? –
Sg
l A.turns N turns
R =c =3.07 × 10 3
c µ
Sc Wb µ
g A.turns
R = =2.49 × 10 5
g µ
0Sg Wb
N 21000 2
L = = = 3.97 H
+ ×+×3 5
Rc R g 3.07 10 2.49 10
Li 23.97(0.41) 2
W = = = 0.67J
2 2
dλ dN ϕ dB −
e = = = NS c =1000 ×× (16 104 ) × 314cos(314t ) = 502cos(314t )
dt dt c dt
sites.google.com/site/ncpdhbkhn 26
Magnetic Circuits
and Magnetic Materials
1. Introduction to Magnetic Circuits
2. Flux Linkage, Inductance, and Energy
3. Properties of Magnetic Materials
4. AC Excitation
5. Permanent Magnets
6. Application of Permanent Magnet Materials
sites.google.com/site/ncpdhbkhn 27
Properties of Magnetic Materials
(1)
Magnetic materials can be used:
• To constrain and direct magnetic fields in well –
defined paths
• In transformer:
– To maximize the coupling between the windings, and
– To lower the excitation current required for
transformer operation
• In electric machinery:
– To shape the fields to obtain desired torque –
production and electrical terminal characteristics
• The most common: ferromagnetic materials
sites.google.com/site/ncpdhbkhn 28
Properties of Magnetic Materials
(2)
sites.google.com/site/ncpdhbkhn 29
Properties of Magnetic Materials
(2)
sites.google.com/site/ncpdhbkhn 30
Properties of Magnetic Materials
Φ µ lc
Ex. 1 (3)
Given Bc = 0.9T, lc = 40cm, g = 0.05cm, i Sc
2
N = 1000 turns, Sc = Sg = 16cm . Find +
the current i? g,
µ0,
= –
H c 10 A.turns/m Sg
= N turns
Fc H c l c
=10 × 0.4 = 4 A.turns
B
F= Hg = c g
g g µ
0
× × −4
= 1 (16 10 )
4π × 10 −7
= 1273 A.turns
F+ F 4+ 1273
i =c g =
N 1000
=
1.28A sites.google.com/site/ncpdhbkhn 31
Properties of Magnetic Materials
Φ µ lc
Ex. 2 (4)
Given Bc = 2T, lc = 40cm, g = 0.05cm, i Sc
2
N = 1000 turns, Sc = Sg = 16cm . Find +
the current i? g,
µ0,
= –
H c 10000 A.turns/m Sg
= = × N turns
Fc H c l c 10000 0.4
= 4000 A.turns
B
F= Hg = c g
g g µ
0
× × −4
= 1 (16 10 )
4π × 10 −7
= 1273 A.turns
F+ F 4000+ 1273
i =c g =
N 1000
=
5.27A sites.google.com/site/ncpdhbkhn 32
Magnetic Circuits
and Magnetic Materials
1. Introduction to Magnetic Circuits
2. Flux Linkage, Inductance, and Energy
3. Properties of Magnetic Materials
4. AC Excitation
5. Permanent Magnets
6. Application of Permanent Magnet Materials
sites.google.com/site/ncpdhbkhn 33
AC Excitation (1)
ϕ
iϕ ( t ) i1
0 i2
t t t 0 i iϕ
1 2 i2 1
ϕ(t )
e( t )
ϕφω= = ω
(t )max sin tBS max c sin t
d
et()= [ Ntϕωφ ()] = N cos()2 ωπ t = fNBS cos() ω t
dt max max c
π
2 fNBmax S c lc H c, rms
E= = 2π fNB S ; Iϕ =
rms max c ,rms N
2 sites.google.com/site/ncpdhbkhn 34
AC Excitation (2)
lc H c, rms
E=2π fNB S ; I ϕ =
rmsmax c , rms N
lc H c, rms
EIϕ =2π fNSB = 2 π fBH () Sl
rms, rms c maxN maxc , rms c c
EIϕ EI ϕ 2π f
P=rms, rms = rms , rms = B H
a ρ ρ maxc , rms
mass Sccc l c
sites.google.com/site/ncpdhbkhn 35
AC Excitation (3)
• The exciting current i supplies the mmf needed to
produce:
– the core flux, and
– the power input associated with the energy in the
magnetic
• This energy:
– Part is dissipated as losses & results in heating of the
core ( RI 2)
– The rest appears as reactive power, it is not dissipated
in the core, but cyclically supplied & absorbed by the
excitation source
sites.google.com/site/ncpdhbkhn 36
AC Excitation (4)
sites.google.com/site/ncpdhbkhn 37
AC Excitation (5)
20 cm
Ex.
The winding is excited with a 60-Hz voltage to 5 cm
produce B = 1.6sin ωt T, the steel occupies 0.94 of the i
3
core cross-sectional area, ρc = 7.65 g/cm . Find: +
a) The applied voltage? e 15 cm
5 cm
b) The peak current?
c) The rms exciting current? –
d) The core loss?
300 turns
B=1.6sin(2π × 60 t ) = 1.6sin 377 t T
dϕ dSB dB
eN= = Nc = NS
dt dtc dt
=300 ×× (25 10−4 ) × 377cos377 t
= 283cos377t V
= → =
Bmax1.6 T H max 70 A.turns/m
l H
I = c max
max N
+++ ×−2 ×
= (15 10 15 10) 10 70
300
= 0.12A sites.google.com/site/ncpdhbkhn 38
AC Excitation (6)
20 cm
Ex.
The winding is excited with a 60-Hz voltage to 5 cm
produce B = 1.6sin ωt T, the steel occupies 0.94 of the i
3
core cross-sectional area, ρc = 7.65 g/cm . Find: +
a) The applied voltage? e 15 cm
5 cm
b) The peak current?
c) The rms exciting current? –
d) The core loss?
300 turns
= → =
Bmax 1.6 T P a 2VA/kg
=×× + + +
Vc (5 5) (15 10 15 10)
= 1250 cm 3
=ρ = ×
Wc c V c 7.65 1250
=9562.5g = 9.56kg
E I
= rmsϕ , rms
Pa
Wc
×
→ =PWa = 2 9.56
Iϕ ,rms
Erms 283/ 2
= 95.57A
sites.google.com/site/ncpdhbkhn 39
AC Excitation (7)
20 cm
Ex.
The winding is excited with a 60-Hz voltage to 5 cm
produce B = 1.6sin ωt T, the steel occupies 0.94 of the i
3
core cross-sectional area, ρc = 7.65 g/cm . Find: +
a) The applied voltage? e 15 cm
5 cm
b) The peak current?
c) The rms exciting current? –
d) The core loss?
300 turns
= → =
Bmax 1.6 T P c 1.4W/kg
=
Pctotal, PW c c
=1.4 × 9.56
= 13.39 W
sites.google.com/site/ncpdhbkhn 40
Magnetic Circuits
and Magnetic Materials
1. Introduction to Magnetic Circuits
2. Flux Linkage, Inductance, and Energy
3. Properties of Magnetic Materials
4. AC Excitation
5. Permanent Magnets
6. Application of Permanent Magnet Materials
sites.google.com/site/ncpdhbkhn 41
Permanent Magnets (1)
sites.google.com/site/ncpdhbkhn 42
Permanent Magnets (2)
Ex.
µ → ∞
2
g = 0.2cm, lm = 2cm, Sm = Sg = 5 cm , Bg = ? Sm
lm
F=0 = HgHl + →H = − H l Magnetic Air gap, g
g m m gg m m material
µ0, Sg
S
φ = = → = m
BSgg BS mm Bg B m µ → ∞
Sg
= µ
Bg0 H g
S l
→ = − µ g m
Bm0 H m
Sm g
− 5 2
=−4π × 10 7 ×× H
5 0.2 m
= − × −7
126 10 H m
=− → =
Hm40kA/m B m 0.50T
= =
Bg B m 0.60T
sites.google.com/site/ncpdhbkhn 43
Permanent Magnets (3)
µ → ∞
S Sm
= m
Bg B m
Sg Magnetic
lm Air gap, g
material µ , S
H l 0 g
m m = − 1
µ → ∞
Hg g
= µ
Bg0 H g Point of
maximum energy product
S l
→2 =µ m m −
Bg0 ( H m B m )
Sg g
Vol
=µ mag −
0 (Hm B m )
Vol air gap
Vol B 2
→Vol = airgap g
mag µ −
0 (Hm B m )
sites.google.com/site/ncpdhbkhn 44
Magnetic Circuits
and Magnetic Materials
1. Introduction to Magnetic Circuits
2. Flux Linkage, Inductance, and Energy
3. Properties of Magnetic Materials
4. AC Excitation
5. Permanent Magnets
6. Application of Permanent Magnet
Materials
sites.google.com/site/ncpdhbkhn 45
Application of Permanent Magnet
Materials (1)
sites.google.com/site/ncpdhbkhn 46
Application of Permanent Magnet
Materials (2)
µ → ∞ i
Permanent
lm magnetic N
material turns
sites.google.com/site/ncpdhbkhn 47
Application of Permanent Magnet
Ex. Materials (3) Sm
2 2 2 µ → ∞
g = 0.2cm, Sm = 2cm , 2cm ≤ Sg ≤ 4cm . g / 2
a) Find the magnet length such that the system will Sm
operate on a recoil line which intersects the Permanent
x µ → ∞
maximum B – H product point on the magnetization lm magnetic
curve. material
b) Calculate the flux density in the air gap as the
plunger moves back and forth. g / 2
S
B= m B Point of
gS m maximum energy product
g
→ = Sm B m
lm g
H l S−µ H
m m = − 1 g0 m
Hg g
= − =
H2 40kA/m; B 2 1T
m2cm m 2cm
→=×−2 2 1 =
l 0.2 10− 3.98cm
m 24π × 107 × 410 × 4
sites.google.com/site/ncpdhbkhn 48
Application of Permanent Magnet
Ex. Materials (4)
2 2 2 µ → ∞
g = 0.2cm, Sm = 2cm , 2cm ≤ Sg ≤ 4cm . g / 2
a) Find the magnet length such that the system will Sm
operate on a recoil line which intersects the Permanent
x µ → ∞
maximum B – H product point on the magnetization lm magnetic
curve. material
b) Calculate the flux density in the air gap as the
plunger moves back and forth. g / 2
S l Point of
= − µ g m
Bm0 H m maximum energy product
Sm g
→ =−×π −7 4 3.98
B2 4 10 H 2
m 4cm 2 0.2 m 4cm
= − × −5
5 10 H 2
m 4cm
2≤≤ 2 ֏ ≤≤
2cmSg 4cm 1.0T B 1.08T
sites.google.com/site/ncpdhbkhn 49
Các file đính kèm theo tài liệu này:
- bai_giang_electromechanical_energy_conversion_nguyen_cong_ph.pdf