Bài giảng Electromechanical energy conversion - Chapter II: Electromechanical Energy Conversion Principles - Nguyễn Công Phương
1. Forces and Torques in Magnetic Field Systems
2. Energy Balance
3. Energy in Singly – Excited Magnetic Field
Systems
4. Determination of Magnetic Force and Torque
from Energy and Coenergy
5. Multiply – Excited Magnetic Field Systems
6. Forces and Torques in Systems with Permanent
Magnets
7. Dynamic Equations
8. Analytical Techniques
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Nguyễn Công Phương
ELECTROMECHANICAL ENERGY
CONVERSION
Electromechanical – Energy –
Conversion Principles
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
Electromechanical – Energy –
Conversion Principles
1. Forces and Torques in Magnetic Field Systems
2. Energy Balance
3. Energy in Singly – Excited Magnetic Field
Systems
4. Determination of Magnetic Force and Torque
from Energy and Coenergy
5. Multiply – Excited Magnetic Field Systems
6. Forces and Torques in Systems with Permanent
Magnets
7. Dynamic Equations
8. Analytical Techniques
sites.google.com/site/ncpdhbkhn 3
Forces and Torques in Magnetic
Field Systems (1)
Fq() E v B
v
FE q
B
Fq() v B
F
Fv () v B
J v
FJBv
sites.google.com/site/ncpdhbkhn 4
Forces and Torques in Magnetic
Ex. Field Systems (2) B0 yˆ ˆ
A nonmagnetic rotor contains a single – turn coil, it
is in a uniform magnetic field. The rotor is of radius
R and of length l. Find the θ – directed torque as a
function of α?
I
FJBv
xˆ
FJB S 1( ) I
IB
Fin IB0 l sin
Fout IB0 l sin
T(2 R ) F 2 RIB0 l sin (Nm)
sites.google.com/site/ncpdhbkhn 5
Forces and Torques in Magnetic
Field Systems (3)
i f fld
Lossless magnetic
,e x
energy storage system
Electrical Mechanical
terminal terminal
i
1 Magnetic core
Winding f
resistance x fld
v e
Movable
magnetic
Lossless winding plunger
sites.google.com/site/ncpdhbkhn 6
Forces and Torques in Magnetic
Field Systems (4)
i f fld
Lossless magnetic
,e x
energy storage system
Electrical Mechanical
terminal terminal
dW dx
eifld f
dtfld dt
d
e
dt
dWfld id f fld dx
sites.google.com/site/ncpdhbkhn 7
Electromechanical – Energy –
Conversion Principles
1. Forces and Torques in Magnetic Field Systems
2. Energy Balance
3. Energy in Singly – Excited Magnetic Field
Systems
4. Determination of Magnetic Force and Torque
from Energy and Coenergy
5. Multiply – Excited Magnetic Field Systems
6. Forces and Torques in Systems with Permanent
Magnets
7. Dynamic Equations
8. Analytical Techniques
sites.google.com/site/ncpdhbkhn 8
Energy Balance
Energy is neither created or destroyed, it is merely changed in form
dWelectrical eidt dW mechanical dW field
sites.google.com/site/ncpdhbkhn 9
Electromechanical – Energy –
Conversion Principles
1. Forces and Torques in Magnetic Field Systems
2. Energy Balance
3. Energy in Singly – Excited Magnetic Field
Systems
4. Determination of Magnetic Force and Torque
from Energy and Coenergy
5. Multiply – Excited Magnetic Field Systems
6. Forces and Torques in Systems with Permanent
Magnets
7. Dynamic Equations
8. Analytical Techniques
sites.google.com/site/ncpdhbkhn 10
Energy in Singly – Excited
Magnetic Field Systems (1)
x
Mechanical
i source
L() x i f
fld
+
1 –
2 ,e Massless
Wfld Li
2 magnetic
armature
1 2 Lossless
W fld coil
2L ( x ) Magnetic core
sites.google.com/site/ncpdhbkhn 11
Energy in Singly – Excited
Ex. Magnetic Field Systems (2)
A relay is made of infinitely – permeable magnetic
material, h >> g. Compute the magnetic energy as i
a function of plunger position? + x g
h
l
1 2 – g
Wfld L() x i d
2 Lossless –
N turns coil
S
L() x N 2 0 g
2g d
Magnetic g
flux
x x
Sg l( d x ) ld 1
d
d x g
2
1N0 ld (1 x / d ) 2 x
Wfld i K 1 J
2 2g d
sites.google.com/site/ncpdhbkhn 12
Electromechanical – Energy –
Conversion Principles
1. Forces and Torques in Magnetic Field Systems
2. Energy Balance
3. Energy in Singly – Excited Magnetic Field
Systems
4. Determination of Magnetic Force and Torque
from Energy and Coenergy
5. Multiply – Excited Magnetic Field Systems
6. Forces and Torques in Systems with Permanent
Magnets
7. Dynamic Equations
8. Analytical Techniques
sites.google.com/site/ncpdhbkhn 13
Determination of Magnetic Force and
Torque from Energy and Coenergy (1)
dWfld(,) x id f fld dx
Wfld(,)(,) x W fld x
dWfld (,) x d dx
x
xconst const
Wfld (,) x
i
xconst
W(,) x
f fld
fld x
const
1 2
W(,) x
fld 2L ( x )
12 2 dL ( x )
2
f fld 2 i dL() x
x2 L ( x ) 2 L ( x ) dx f
const fld 2 dx
L() x i
sites.google.com/site/ncpdhbkhn 14
Determination of Magnetic Force and
Torque from Energy and Coenergy (2)
Ex. 1
x (cm) 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0
L (mH) 2.8 2.26 1.78 1.52 1.34 1.26 1.20 1.16 1.13 1.11 1.10
Plot the force as a function of position for a current of 1A.
sites.google.com/site/ncpdhbkhn 15
Determination of Magnetic Force and
Torque from Energy and Coenergy (3)
Ex. 2
x (cm) 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0
L (mH) 2.8 2.26 1.78 1.52 1.34 1.26 1.20 1.16 1.13 1.11 1.10
Plot the force as a function of position for a flux of 2mWb.
sites.google.com/site/ncpdhbkhn 16
Determination of Magnetic Force and
Torque from Energy and Coenergy (4)
dWfld(,) x id f fld dx
dWfld(,) id T fld d
WWfld fld
dW(,) d d
fld
const const
W fld (,)
Tfld
const
1 2
W(,) x
fld 2L ( x )
12 2 dL ( )
Tfld 2 2
2L ( x ) 2 L ( ) d i dL()
const T
fld 2 d
L() i
sites.google.com/site/ncpdhbkhn 17
Determination of Magnetic Force and
Torque from Energy and Coenergy (5)
i
LLL( ) cos(2 )
0 2
Rotor
i2 dL()
T
fld 2 d
Air gap
i2
TL( ) [ 2 sin(2 )]
fld 2 2
sites.google.com/site/ncpdhbkhn 18
Determination of Magnetic Force and
Torque from Energy and Coenergy (6)
Wfld (,)(,) i x i W fld x
dWfld (,)()(,) i x d i dW fld x
d() i id di
dWfld(,) x id f fld dx
dWfld (,) i x di f fld dx
WWfld fld
dW (,) i x di dx
fld i x
xconst i const
Wfld (,) i x
i xconst
W (,) i x
f fld
fld x
iconst
sites.google.com/site/ncpdhbkhn 19
Determination of Magnetic Force and
Torque from Energy and Coenergy (7)
Ex. 3
A relay is made of infinitely – permeable magnetic
material, h >> g. Compute the magnetic energy as i
g
a function of plunger position, if i(x) = I0x/d (A)? + x
h
2 0Sg
L() x N l
2g – g
d
x Lossless –
Sg l( d x ) ld 1
d N turns coil
2
N0 ld(1 x / d )
L() x i2 N 2 l
2g L() x 0
i2 dL() x 2 2g
f fld x
2 dx i() x I0
d
2 2 2
I0 0 N l x
L() x
4g d
sites.google.com/site/ncpdhbkhn 20
Determination of Magnetic Force and
Torque from Energy and Coenergy (8)
()i
0 Energy
W fld
Coenergy
W fld
0 i0 i
Wfld W fld i
1 1 2 1
Ifki : W W i Li2
fld fld 2 2L 2
sites.google.com/site/ncpdhbkhn 21
Determination of Magnetic Force and
Torque from Energy and Coenergy (9)
Ex. 4
Find the torque acting on the rotor as a function of g
the dimensions and the magnetic field in the two r
air gap, suppose that the reluctance of the steel is
negligible (μ → ∞). The axial length is h.
2gHag Ni
1
Energy density : BH
2 i
1 B2
Energy densityof the core :BH steel 0
2steel steel 2
BHH 2
Energy densityof the air-gap : ag ag 0 ag
2 2
H 2 (Ni )2 h ( r 0.5 g )
W 0 ag [2 gh ( r 0.5 g ) ] 0
ag 2 4g 2
0 (Ni ) h ( r 0.5 g )
W (,) i Tfld
ag 4g
Tfld
iconst
sites.google.com/site/ncpdhbkhn 22
Determination of Magnetic Force and
Torque from Energy and Coenergy (10)
Ex. 5
Find the inductance as a function of θ, then extract g
the expression for the torque acting on the rotor as r
a function of i & θ. The axial length is h.
S
LN() 2 0 g
2g
i
Sg h( r 0.5 g )
N2 h( r 0.5 g )
L() 0
2g
2
0 (Ni ) h ( r 0.5 g )
Tfld
2 4g
i dL()
T
fld 2 d
sites.google.com/site/ncpdhbkhn 23
Electromechanical – Energy –
Conversion Principles
1. Forces and Torques in Magnetic Field Systems
2. Energy Balance
3. Energy in Singly – Excited Magnetic Field
Systems
4. Determination of Magnetic Force and Torque
from Energy and Coenergy
5. Multiply – Excited Magnetic Field Systems
6. Forces and Torques in Systems with Permanent
Magnets
7. Dynamic Equations
8. Analytical Techniques
sites.google.com/site/ncpdhbkhn 24
Multiply – Excited Magnetic
Field Systems (1)
i1 Tfld
1
i Lossless magnetic
2 energy storage system
2
Electrical Mechanical
terminals terminal
dWfld(,)(,,) id T fld d dW fld 1 2 i 1 d 1 i 2 d 2 T fld d
WWWfld fld fld
dWfld (,,)1 2 d 1 d 2 d
1 2 , are const
2, are const 1 , are const 1 2
WWWfld(,,)(,,)(,,)1 2 fld 1 2 fld 1 2
i1 ;; i 2 Tfld
1 2 , are const
2, are const 1 , are const 2 2
sites.google.com/site/ncpdhbkhn 25
Multiply – Excited Magnetic
Field Systems (2)
i1 Tfld
1
i Lossless magnetic
2 energy storage system
2
Electrical ,e Mechanical
terminals 1 1 terminal
1L 11 i 1 L 12 i 2
2L 21 i 1 L 22 i 2
LLLL
i 22 1 12 2 22 1 12 2
1
LLLLD11 22 12 21
LLLL21 1 11 2 21 1 11 2
i2
LLLLD11 22 12 21
12 1 2 L12 ( 0 )
WLLfld (,,)()()10 20 0 11 0 20 22 0 10 10 20
2DDD (0 ) 2 ( 0 ) ( 0 )
sites.google.com/site/ncpdhbkhn 26
Multiply – Excited Magnetic
Field Systems (3)
Wfld (,,) i1 i 2 1 i 1 2 i 2 W fld
dWfld (,,) i1 i 2 1 di 1 2 di 2 T fld d
Wfld (,,) i1 i 2
1
i1
i2 , are const
Wfld (,,) i1 i 2
2
i
1 i1, are const
W (,,) i i
T fld 1 2
fld
i1, i 2 are const
1 1
W (,,)()()() i i L i2 L i 2 L i i
fld 1 22 11 1 2 22 2 12 1 2
W (,,) i2 dL()()() i 2 dL dL
Tfld 1 2 1 11 2 22 i i 12
fld 2d 2 d 1 2 d
i1, i 2 are const
sites.google.com/site/ncpdhbkhn 27
Multiply – Excited Magnetic
Ex. Field Systems (4)
L11 = a + cos2θ, L12 = bcosθ,
L22 = c + ecos2θ. Find Tfld(θ)? i1
+ Tfld
– –
,e +
1 1 2,e 2
i
Tmech 2
i2 dL()()() i 2 dL dL
T1 11 2 22 i i 12
fld 2d 2 d 1 2 d
2 2
i1sin 2 ei 2 sin 2 bi 1 i 2 sin
sites.google.com/site/ncpdhbkhn 28
Multiply – Excited Magnetic
Field Systems (5)
2 2
Tfld i1sin 2 ei 2 sin 2 bi 1 i 2 sin
-3
x 10
4
3
2
1
0
Torque (Nm)Torque
-1
-2
Reluctance torque
-3 Mutual - interaction torque
Total torque
-4
0 1 2 3 4 5 6
(rad)
sites.google.com/site/ncpdhbkhn 29
Multiply – Excited Magnetic
Field Systems (6)
W fld (,,)1 2
Tfld
1, 2 are const
2 2
Wfld (,,) i1 i 2 i1 dL 11()()() i 2 dL 22 dL 12
Tfld i1 i 2
2d 2 d d
i1, i 2 are const
12 1 2
Wfld (,,)()()() i12 i L 11 i 1 L 22 i 2 L 12 i 12 i
2 2
Wfld (,,)1 2 x
f fld
x
1, 2 are const
2 2
Wfld (,,) i1 i 2 x idLx1 11()()() idLx 2 22 dLx 12
ffld i1 i 2
x2 dx 2 dx dx
i1, i 2 are const
12 1 2
Wiixfld (,,)()()()12 Lxi 11 1 Lxi 22 2 Lxii 12 12
2 2
sites.google.com/site/ncpdhbkhn 30
Electromechanical – Energy –
Conversion Principles
1. Forces and Torques in Magnetic Field Systems
2. Energy Balance
3. Energy in Singly – Excited Magnetic Field
Systems
4. Determination of Magnetic Force and Torque
from Energy and Coenergy
5. Multiply – Excited Magnetic Field Systems
6. Forces and Torques in Systems with
Permanent Magnets
7. Dynamic Equations
8. Analytical Techniques
sites.google.com/site/ncpdhbkhn 31
Forces and Torques in Systems
with Permanent Magnets (1)
dWfld (,) i x di f fld dx
x
dWfld (,) i f x f di f f fld dx
Wfld ( i f 0, x )
f
fld x
i f const
i f
x
0
Wfld( i f 0, x ) f ( i f , x ) di f
I f 0 f
Fictitious winding
Nf turns
sites.google.com/site/ncpdhbkhn 32
Forces and Torques in Systems
Ex. 1 with Permanent Magnets (2) Depth D
Find an expression for:
Wm W
a) the coenergy of the system as a function of plunger g
position x? i
f d x
b) the force on the plunger as a function of x?
BHHm R() m c
N f
Nf i f H m d H g x H0 g 0
g0
Fictitious W0
m g 0 BSBSBS m m g g 0 0 winding
BWDBWDBWDm m g g 0 0
R()N f i f H c d
Bm
R x g0 Nf W m D R() N f i f H c d
d Wm
WW f
0 g 0 x g
d W R 0
m
0WWg 0
fNNBSNBWD f m f m m f m m
sites.google.com/site/ncpdhbkhn 33
Forces and Torques in Systems
Ex. 1 with Permanent Magnets (3) Depth D
Find an expression for:
Wm W
a) the coenergy of the system as a function of plunger g
position x? i
f d x
b) the force on the plunger as a function of x?
N W D () N i H d
f m R f f c
f N
x g f
d W R 0
m
0 WWg 0
g0
Fictitious W0
Hc d winding
f0 i f I f 0
N f
2
0
Wm D R() H c d
Wfld() x f di f
I
f 0 x g
2 d W R 0
m WW
0 g 0
W ( i 0, x ) W2 D () H d 2
f fld f m R c
fld x
i f const R x g0
2Wg d W m
WW
0 g 0
sites.google.com/site/ncpdhbkhn 34
Forces and Torques in Systems
with Permanent Magnets (4)
S 1 H d F 0
m e
External
d F 1 BSSHHm R() m c
H e magnetic
m circuit
Fe
RSH c
d
BHHm R() m c
S 2 Ni F H
e
External
2 BS R HS
d Fe magnetic
()Ni equiv 1 2
circuit ()Ni F
equiv e
2 RS
d d
BHm R m
If (Ni )equiv H c d
sites.google.com/site/ncpdhbkhn 35
Forces and Torques in Systems
Ex. 2 with Permanent Magnets (5) Depth D
a) Find the x – directed force on the plunger when the
W W
current in the excitation winding is zero and x = 3 mm? g
b) Find the current in the excitation winding required to i1 d
reduce the plunger force to zero? x
()Ni H d
equiv c N1
g0
d d x g W
RRR ;; 0
mS WD x W D0 WD
R R0 g 0
()Ni equiv
1 2 +
W Li – R
fld 2 1 m
1 ()Ni 2
W equiv
fld 2 RRR +
2 m x 0 – N i Rx
N 1 1
L 1
R
Rtotal g0
sites.google.com/site/ncpdhbkhn 36
Forces and Torques in Systems
Ex. 2 with Permanent Magnets (6) Depth D
a) Find the x – directed force on the plunger when the
W W
current in the excitation winding is zero? g
b) Find the current in the excitation winding required to i1 d
reduce the plunger force to zero? x
2
1 ()Ni equiv N
W 1
fld 2 RRR
m x 0 g
W 0
W () Ni 2 dR
f fld equiv x
fld x() R R R2 dx
iequiv const m x 0
2 ()Ni equiv
+
()Ni equiv
2 – Rm
0WDRRRg() m x 0
+
– R
()Ni equiv N1 i 1 x
(Ni )equiv N1 i 1 0 i1
N1 Rg0
sites.google.com/site/ncpdhbkhn 37
Electromechanical – Energy –
Conversion Principles
1. Forces and Torques in Magnetic Field Systems
2. Energy Balance
3. Energy in Singly – Excited Magnetic Field
Systems
4. Determination of Magnetic Force and Torque
from Energy and Coenergy
5. Multiply – Excited Magnetic Field Systems
6. Forces and Torques in Systems with Permanent
Magnets
7. Dynamic Equations
8. Analytical Techniques
sites.google.com/site/ncpdhbkhn 38
Dynamic Equations (1)
x
K
i
B
v0 f fld
+ R Electromechanical –
– ,e energy – conversion
system M
f0
d
v Ri di dL() x dx
0 dt v Ri L() x i
0 dt dx dt
L() x i
sites.google.com/site/ncpdhbkhn 39
Dynamic Equations (2)
x
K
i
B
v0 f fld
+ R Electromechanical –
– ,e energy – conversion
system M
f
fK K() x x0 0
dx
fD B
dt dx d2 x
ffld K( x x0 ) B M2 f 0 0
d2 x dt dt
f M
M dt2
ffld f K f D f M f0 0
sites.google.com/site/ncpdhbkhn 40
Dynamic Equations (3)
x
K
i
B
v0 f fld
+ R Electromechanical –
– ,e energy – conversion
system M
f0
di dL() x dx
v()() t Ri L x i
0 dt dx dt
dx d2 x
f()()(,) t K x x B M f x i
0 0 dt dt 2 fld
sites.google.com/site/ncpdhbkhn 41
Dynamic Equations (4) l0
Ex.
Extract the dynamic equations of motion of the Spring, K
electromechanical system? Coil
l1
Length of flux path in the direction of field
R
(area of flux path perpendicular to field) a
x
g
R
g1
0 dx
g h
Rg 2
da
0
g1 1 g a x
RRRg1 g 2
d x a da x
0 0 g a
2 2
N0 daN x x
L(), x L Cylindrical
R g a x a x d steel plunger,
2 M
0 daN
L Applied force, ft
g
sites.google.com/site/ncpdhbkhn 42
Dynamic Equations (5) l0
Ex.
Extract the dynamic equations of motion of the Spring, K
electromechanical system? Coil
l1
x daN 2
L(), x L L 0
a x g a
x
W (,) i x i2 dL i 2 aL
f fld
fld x2 dx 2 ( a x )2 h
iconst
dLi() di dL di dLdx
e L i L i
dt dt dt dt dxdt g a
Cylindrical
x di ai dx d steel plunger,
LL
a x dt() a x2 dt M
Applied force, ft
sites.google.com/site/ncpdhbkhn 43
Dynamic Equations (6) l0
Ex.
Extract the dynamic equations of motion of the Spring, K
electromechanical system? Coil
l1
i2 dL i 2 aL
f
fld 2dx 2 ( a x )2 a
x
x di ai dx
e L L
a x dt() a x2 dt
di dL() x dx h
v()() t Ri L x i
0 dt dx dt
dx d2 x
f()()(,) t K x x B M f x i
0 0 dt dt 2 fld
g a
x di a dx
v() t Ri L L i
0 2 Cylindrical
a x dt() a x dt d
steel plunger,
dx d2 x i 2 aL M
Applied force, f
f0()() t K x l 0 B M 2 2 t
dt dt2 ( a x )
sites.google.com/site/ncpdhbkhn 44
Electromechanical – Energy –
Conversion Principles
1. Forces and Torques in Magnetic Field Systems
2. Energy Balance
3. Energy in Singly – Excited Magnetic Field
Systems
4. Determination of Magnetic Force and Torque
from Energy and Coenergy
5. Multiply – Excited Magnetic Field Systems
6. Forces and Torques in Systems with Permanent
Magnets
7. Dynamic Equations
8. Analytical Techniques
sites.google.com/site/ncpdhbkhn 45
Analytical Techniques (1)
v(0) V
If x di a dx
Ri L, L i
2 V
a x dt() a x dt i
R
x di a dx
v() t Ri L L i
2
a x dt() a x dt
2 2
dx d x i aL
f()() t K x l0 B M 2 2
dt dt2 ( a x )
f 0
If
M 0
dx1 a V 2
B L2 K()() x l0 f x
dt2 ( a x ) R
X B
t dx
0 f() x
sites.google.com/site/ncpdhbkhn 46
Analytical Techniques (2)
v(0) V
If x di a dx
Ri L, L i
2 V
a x dt() a x dt i
R
x di a dx
v() t Ri L L i
2
a x dt() a x dt
2 2
dx d x i aL
f()() t K x l0 B M 2 2
dt dt2 ( a x )
f 0
If
B 0
d2 x1 a V 2
M2 L 2 K()() x l0 f x
dt2 ( a x ) R
dx2 x B
v() x dx
dt M0 f() x
sites.google.com/site/ncpdhbkhn 47
Analytical Techniques (3)
di dx
If 0, 0
dt dt
V0 RI 0
x di a dx 1 L aI 2
v() t Ri L L i 0 K() X l f
2 2 0 0t 0
a x dt() a x dt 2 (a l0 )
2 2
dx d x i aL
f()() t K x l0 B M 2 2
dt dt2 ( a x )
IfiIiff0 ,t t 0 fvVvxXx , t 0 , 0
LX()() xdi LaI i dx
V v R() I i 0 0
0 0 2
a X0 x dt() a X 0 x dt
1L a ( I i )2 dx d 2 x
0 K() X x l B M f f
20 0 2 t 0
2 (a X0 x ) dt dt
sites.google.com/site/ncpdhbkhn 48
Analytical Techniques (4)
LX()() xdi LaI i dx
V v R() I i 0 0
0 0 2
a X0 x dt() a X 0 x dt
1L a ( I i )2 dx d 2 x
0 K() X x l B M f f
20 0 2 t 0
2 (a X0 x ) dt dt
L X di L aI dx
v Ri 0 0
2
a X0 dt() a X 0 dt
L aI dx d2 x L aI 2
0i B M K 0 x f
2 2 3
()()a X0 dt dt a X 0
sites.google.com/site/ncpdhbkhn 49
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