Lecture Electromechanical energy conversion: Magnetic circuits and magnetic materials - Nguyễn Công Phương

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
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Magnetic Circuits
and Magnetic Materials
1.
2.
3.
4.

5.
6.

Introduction to Magnetic Circuits
Flux Linkage, Inductance, and Energy
Properties of Magnetic Materials
AC Excitation
Permanent Magnets
Application of Permanent Magnet Materials

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Introduction to Magnetic Circuits
(1)

∫ H.dL = ∫

S

∫

S

J.dS

B.dS = 0

B = µH
•
•
•
•

H:
J:
B:
µ:

magnetic field intensity, A/m
current density, A/m2
magnetic flux density, Wb/m2 (T)
permeability, H/m
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Introduction to Magnetic Circuits
(2)
6
4
2
0
-2
-4
-6
1

1

0.5
0.5

0
0

∫ H.dL = ∫

S

-0.5

J.dS

-0.5
-1

-1

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Introduction to Magnetic Circuits
(3)

∫ H.dL = ∫

S

J.dS
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Introduction to Magnetic Circuits
(4)

∫

S

B.dS = 0
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Introduction to Magnetic Circuits
Magnetic flux Φ (5)
Mean core length, l
c

i

+

Cross-sectional area, Sc

–

Air gap, length g,

permeability µ0,
area Sg

Winding, N turns

Magnetic core permeability µ

µ >> µ0 :

the magnetic flux is confined
almost entirely to the core

F = Ni :

magnetomotive force (mmf)
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8


Introduction to Magnetic Circuits
Magnetic flux Φ (6)
Mean core length, l

c

i

+

Cross-sectional area, Sc

–

Air gap, length g,

permeability µ0,
area Sg

Winding, N turns

Magnetic core permeability µ

Φ = ∫ B.dS

F = Ni =

S

B c ≈ const

∫ H.dL
H c ≈ const

→ Φ c = Bc S c
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→ F = H c lc
9


Introduction to Magnetic Circuits
Magnetic flux Φ (7)
Mean core length, l
c

i

+

Cross-sectional area, Sc

–

Air gap, length g,

permeability µ0,
area Sg

Winding, N turns

Magnetic core permeability µ

F = Ni =

∫ H.dL = H l

c c

+ H glg

→F=

Bc

lc +

Bg

g

µ
µ0
Bc = µ H c ;
Bg = µ0 H g
Φ c = Bc Sc ; Φ g = Bg S g ; Φ c = Φ g = Φ
 lc
g
→ F = Φ
+
 µ Sc µ0 S g

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




Introduction to Magnetic Circuits
(8)
 lc
g 
F = Φ
+
 µ Sc µ0 S g 


lc
g
= Rc ;
= Rg
µ Sc
µ0 S g

→ F = Φ ( Rc + Rg )

F
F
F
→Φ=
=

=
= Ptotal F
lc
g
Rc + Rg
R
total
+
µ S c µ0 S g
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Ex. 1

Introduction to Magnetic Circuits
Φ
(9)

Given Bc = 1T, lc = 40cm, g = 0.05cm,
N = 1000 turns, Sc = Sg = 16cm2,
µ = 65000µ0. Find:
a) The reluctances Rc & Rg?
b) The flux Φ?
c) The current i?

lc
Sc

i

+

g,

µ0,
Sg

–
N turns
µ

lc
40 × 10−2
A. turns
Rc =
=
= 3061
−7
−4
µ Sc 65000(4π × 10 )(16 × 10 )
Wb
g
0.05 × 10−2
5 A. turns
Rg =
=
= 2.49 × 10
−7

−4
µ0 S g (4π × 10 )(16 × 10 )
Wb

Φ = Bc Sc = 1 × 16 × 10−4 = 16 × 10−4 Wb
F Φ ( Rc + Rg ) 16 × 10−4 (3061 + 2.49 × 105 )
i= =
=
= 0.41A
N
N
1000
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Ex. 2

Introduction to Magnetic Circuits
Φ
(10)

Given Φ = 0.141 mWb, Sc = Sg = 4cm2,
lc = 44cm, g = 0.02cm, N = 400 turns.
The core is made of nickel – iron alloy.
Find the current i?

Φ 0.141 × 10−3
Bc =

=
= 0.35T
−
4
Sc
4 × 10
→ H c = 850 A/m

lc
Sc

i

+

g,

µ0,
Sg

–
N turns
µ

B (T)

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H (A/m)

13


Ex. 2

Introduction to Magnetic Circuits
Φ
(11)

Given Φ = 0.141 mWb, Sc = Sg = 4cm2,
lc = 44cm, g = 0.2cm, N = 400 turns.
The core is made of nickel – iron alloy.
Find the current i?

Φ 0.141 × 10−3
Bc =
=
= 0.35T
−
4
Sc
4 × 10
→ H c = 850 A/m

lc
Sc

i

+

g,

µ0,
Sg

–
N turns
µ

Bg
Φ 0.141 × 10 −3
0.35
5
Bg =
=
=
0.35T
→
H
=
=
=
2.79
×
10
A/m
g
−
4

−
7
Sg
µ0 4π × 10
4 × 10
F = H c lc + H g l g = 850 × 0.44 + 2.79 × 105 × 2 × 10 −3 = 931A. turns

F 931
→i=
=
= 2.33 A
N 400
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Ex. 3

Introduction to Magnetic Circuits
(12)

Given i = 10 A, Sg = 1500 cm2, g = 2 cm,
N = 500 turns. Rotor and stator are made
of iron (infinite permeability). Find the
flux of the air gap?

F
Ni
Φ=

=
lc
g
Rc + Rg
+
µ S c µ0 S g
µ→∞

g

i

N turns

Ni µ0 S g 500 × 10(4π × 10−7 )1500 × 10 −4
→Φ=
=
= 0.024 Wb
−2
2g
2 × 2 × 10
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Ex. 4

Introduction to Magnetic Circuits
(13)

i

+

l1

Ф1
Ф2

–

l3

l2

Ф3

N turns

 F = Ni = H1l1 + H 2l2

 H 2l2 = H 3l3
Φ = Φ + Φ
 1
2
3
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16

Magnetic Circuits
and Magnetic Materials
1.
2.
3.
4.
5.
6.

Introduction to Magnetic Circuits
Flux Linkage, Inductance, and Energy
Properties of Magnetic Materials
AC Excitation
Permanent Magnets
Application of Permanent Magnet Materials

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Flux Linkage, Inductance, and
Energy (1)
d
∫ C E.dL = − dt ∫S B.dS
dϕ d λ
e=N
=
,

dt
dt
L=

ϕ=

λ = Nϕ

λ
i
F
Rtotal

F = Ni
N2
→L=
Rtotal
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Flux Linkage, Inductance, and
Energy (2)

Ex. 1

Find the inductance of the winding and
the flux density of gap 2?

L=

λ

+

λ
−

g
g
R1 = 1 , R2 = 2
µ0 S1
µ0 S2
Rtotal =

Area
S1

i

Gap1

Area S1

g2

Gap 2

N turns

R1R2
g1 g 2
=
R1 + R2 µ0 ( g1S 2 + g 2 S1 )

g1

µ→∞

φ

Nφ
i
i
Ni
φ=
Rtotal
=

+
Ni

 S1 S2 
→ L = µ0 N  + 
 g1 g 2 

−

φ1

φ2

R1

R2

2

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Ex. 1

Flux Linkage, Inductance, and
Energy (3)

Find the inductance of the winding and
the flux density of gap 2?

B1 =

Area
S1

i

+

λ
−

φ1

Gap1

g1

µ→∞
Area S1

g2

Gap 2

N turns

S1

Ni µ0 S1 Ni
φ1 =
=
R1
g1

φ
+

→ B1 =

µ0 Ni

Ni

−

φ1

φ2

R1

R2

g1

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Flux Linkage, Inductance, and
Φ
Energy (4)

Ex. 2

Given lc = 40cm, Sc = Sg = 16cm2, Bc = 1T,

N = 1000 turns, g = 0.05cm, find L for
a) µ = 65000µ0, and b) µ = 3000µ0,

Rc ,a =

=
Rc ,b =

lc
µ Sc

lc
Sc

i

+

g,

µ0,
Sg

–
N turns

0.4
3 A. turns
=
3.07

×
10
65000(4π × 10 −7 )(16 × 10 −4 )
Wb

µ

0.4
4 A. turns
=
6.61
×
10
3000(4π × 10−7 )(16 × 10 −4 )
Wb

g
0.05 × 10−2
5 A. turns
Rg =
=
=
2.49
×
10
µ0 S g (4π × 10 −7 )(16 × 10 −4 )
Wb
A. turns
A. turns
, Rtotal ,b = Rc ,b + Rg = 3.15 × 105

Wb
Wb
N2
1000 2
N2
1000 2
La =
=
= 3.97 H , Lb =
=
= 3.17 H
Rtotal ,a 2.52 × 105
Rtotal ,b 3.15 × 105

Rtotal ,a = Rc ,a + Rg = 2.52 × 105

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Flux Linkage, Inductance, and
Energy (5)
4.5

4

3.5

Inductance (H)

3

2.5

2

1.5

1

0.5

0

0

1

2

3

4
5
6
Core relative permeabili ty

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7

8

9

10
4

x 10

22


Flux Linkage, Inductance, and
Energy (6)
Air gap
i1

F = N1i1 + N 2i2

+

F
F
φ=
≈
Rg + Rc Rg

λ1 = N1φ = N12

−

µ0 S c

N1 turns

g

µ0 Sc

µ0 Sc

g

, L22 = N 22

L12 = N1 N 2

µ0 S c
g

µ0 S c
g

µ, lc , Sc

N2 turns

i2

g
µS
µS
λ2 = N 2φ = N1 N 2 0 c i1 + N 22 0 c i2
g
g

µ0 S c

g

i1 + N1N 2

+

λ2
−

λ1

→ φ = ( N1i1 + N 2i2 )

L11 = N12

g

i2

(self-inductance)

λ1 = L11i1 + L12i2
→
λ2 = L21i1 + L22i2

= L21 (mutual inductance)
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Flux Linkage, Inductance, and
Energy (7)
dλ
e=
dt
di
e=L
dt
di
dL
e = L +i
dt
dt
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Flux Linkage, Inductance, and

Energy (8)
dλ
p = ie = i
dt
t2

λ2

t1

λ1

∆W = ∫ pdt = ∫ id λ
λ2

λ

λ1

L

=∫

Li
λ1 = 0 → W =
2

dλ =

λ22 − λ12

2L

2

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