Download Slide 1

CHAPTER 3
ELECTROMECHANICAL ENERGY
CONVERSION
1
Electrical energy is the most popular form of
energy, because:
1. it can be transmitted easily for long distance, at high
efficiency and reasonable cost.
2. It can be converted easily to other forms of energy such as
sound, light, heat or mechanical energy.
2
Hidro power station, Kenyir
Terengganu
Power consumers,
JB Johor
3
Electrical energy
Sound energy
Loud speaker
Electrical energy
Light energy
Lamp
Heat energy
Electrical energy
Kettle
4
Electromechanical energy conversion
device:
converts electrical energy into mechanical energy
or
converts mechanical energy into electrical energy.
5
There are various electromechanical conversion devices
may categorized as under:
a. Small motion
- telephone receivers, loud speakers, microphones
b. Limited mechanical motion
- electromagnets, relays, moving-iron instruments,
moving-coil instruments, actuators
c. Continuous energy conversion
- motors, generators
6
Principle of Energy Conversion
According to the principle of conservation of energy, energy
can neither be created nor destroyed,
it can merely be converted from one form into another.
The total energy in a system is therefore constant.
7
Energy conversion in
electromechanical system
In an energy conversion device, out of the total input energy,
some energy is converted into the required form, some
energy is stored and the rest is dissipated.
It is possible to write an equation describing energy
conversion in electromechanical system:
Electrical
energy
from
source
=
Mechanical
energy to
load
+
Increase of
field
energy
+
Energy
converted
to heat
(losses)
3.1
8
Electrical
energy
from
source
Mechanical
energy to
load
=
+
Increase of
field
energy
+
Energy
converted
to heat
3.1
(losses)
The last term on the right-hand side of Eq. 3.1 (the losses)
may be divided into three parts:
Energy
converted
to heat
=
Resistance
losses
+
Friction and
windage
losses
+
Field losses
3.2
(losses)
Then substitution from Eq. 3.2 in Eq. 3.1 yields
9
Electrical
energy from
source minus
resistance
losses
=
Mechanical
energy to
load plus
friction and
windage
losses
+
Increase of
magnetic
coupling field
energy plus
core losses
3.3
Now consider an electromechanical system (actuator)
illustrated in Fig. 3.1.
10
Bahagian
tak boleh
Fixed
steel
core
gerak
gu
SW
v
R

i
Moveable
steel
Bahagian boleh
armature
gerak
Fm
e
x
dx
Figure 3.1
11
At any instant, the emf e induced in the coil by the
change in the flux linkage  is
d
e
dt
volt
3.4
Consider now a differential time interval dt, during which
the current in the coil is changing and the armature is
moving.
12
Therefore, the differential energy transferred in
time dt from the electric source to the coupling
field is given by the energy output of the source
minus the resistance loss:
dWe  vidt  Ri dt
2
 (v  Ri )idt
 eidt

dWe  eidt
Joule
3.5
13
The coupling field forms an energy
storage to which energy supplied by the
electric system. At the same time,
energy is released from the coupling
field to the mechanical system.
The rate of release energy is not
necessarily equal at any instant to the
rate of supply of energy to the field, so
that the amount of energy stored in the
coupling field may vary.
14
It’s like a pipe system in our house.
Water tank
The water out from the tap will make water flow into the
storage tank from the supply.
15
It’s like a pipe system in our house.
Water tank
The water out from the tap will make water flow into the
storage tank from the supply.
16
In time dt, let dWf be the energy
supplied to the field and either stored or
dissipated. Let dWm be the energy
converted to mechanical form, useful or
as loss, in the same time, dt.
Then, by the principle of conservation of
energy, the following equation may be
written for the field:
dWe  dWm  dW f
3.6
17
Field Energy
To obtain an expression for for dWf of
Eq. 3.6 in terms of the system
variables, it is first necessary to find
an expression for the energy stored in
the magnetic field for any position of
the armature. The armature will
therefore be clamped at some value
of air-gap length g so that no
mechanical output can be produced.
dWm = 0
3.6
18
Field Energy (continue…..)
If switch SW in Fig. 3.1 is now closed, the current
will rise to a value v/R, and the flux will be
established in the magnetic system. Let the
relationship between coil flux linkage  and the
current i for the chosen air-gap length be that
shown in Fig. 3.2



dWf
i1 i2
i
Fig. 3.2
19
Field Energy (continue…..)
Since core loss is being neglected, this will be a
single-valued curve passing through the origin. In
the absence of any mechanical output energy, all of
the electric input energy must be stored in the
magnetic field:
dWe = dWf
3.8
Substitution from Eqs. 3.4 and 3.8 in Eq. 3.5 yields
dWf = dWe = i.edt = id
J
3.9
20
Field Energy (continue…..)
If now v is changed, resulting in a change in current
from i1 to i2, there will be a corresponding change in
flux linkage from 1 to 2 .
The increase in energy stored during the transition
between these two states is
dW f  
2
1
id
J
The area is shown in Fig 3.2. When the flux
linkage is increased from zero to , the total
energy stored in the field is
3.10
21
Field Energy (continue…..)
Wf  

0
id 
J
3.11
This integral represents the area between the –i
characteristic and the –axis, the entire shaded area
of Fig. 3.2.
If it is assumed that there is no leakage flux, so
that all flux  in the magnetic system links all N
turns of the coil, then
 = N Wb
3.12
22
Field Energy (continue…..)
From Eqs. 3.9 and 3.12,
dWf = id = Nid = F d
J
3.13
where
F = Ni
A
3.14
F is mmf (magneto-motive force)
The characteristic of Fig. 3.3 can also represent
the relationship between  and F .
23


2
1
dWf
F1 i1 iF22
i
F
24
Field Energy (continue…..)
If the reluctance of the air gap forms a large part of
the total reluctance of the magnetic system, then
that of the steel may be neglected and the –i
characteristic becomes the straight line through the
origin shown in Fig. 3.3. For this system,
 = Li
Wb
3.15
Where L is the inductance of the coil.
Substitution in Eq. 3.11 gives the energy Wf in
several useful forms:
Wf  

0

2
Li 2 i
d 


L
2L
2
2
J
3.16
25



dW f
i1 i 2
Fig. 3.3
i
26
Field Energy (continue…..)
If the reluctance of the magnetic system (that is, of
the air gap) as seen from the coil is S, then F = S ,
and from Eq. 3.13,

Wf  
0
S 2 F 2
F d 

2
2S
J
3.17
If A is the cross-section area of the core and l = 2g
is the total length of air gap in a flux path, then
from Eq. 3.16,
i F 1
Wf 

 HBlA
2
2
2
J
J
3.18
27
1
1
1 B2
BH  0 H 2 
lA 2 Energy
2
2 0
Field
(continue…..)
wf 
Wf

Where B is the flux density in the air gaps. Since
B/H=µ0 and lA is the total gap volume, it follows
from Eq. 3.18 that the energy density in the air gaps
is
1
1
1 B2
2
wf 
 BH  0 H 
lA 2
2
2 0
Wf
J/m3
3.19
Equations 3.16, 3.17 and 3.19 represent three
different ways of expressing the field energy.
J
28
Example 3.1 The core and armature dimensions of the
actuator of Fig. 3.1 are shown in Fig. 3.4. Both parts are
made of mild steel, whose magnetization curve is given in
Fig. 3.5. Given la = 160 mm, lb = 80 mm. The coil has 2000
turns. Leakage flux and fringing may be neglected. The
armature is fixed, so that the length of the air gas, lu= 9
mm, and a direct current is passed through the coil,
producing a flux density of 0.8 T in the air gap.
a) Determine the required coil current.
b) Determine the energy stored in the air gap.
c) Determine the energy stored in the steel.
d) Determine the total field energy.
29
la
20 mm
Theketebalan
thickness
= 20 mm
lu
Theangker
armature
lb
20 mm
Fig. 3.4
30
B (Tesla)
2.0
Keluli
Sheetkeping
steel
1.8
1.6
Keluli
tuang
mild steel
1.4
1.2
1.0
0.8
cast tuang
iron
Besi
0.6
0.4
0.2
0
500
1000
1500
Fig. 3.5
2000
2500
3000
H (AT/m)
31
Solution
(a) Area,
A = (20  10-3)(20  10-3) = 4  10-4 m2.
Ni = Htlt + Hulu
lt = 160 + 80 = 240 mm = 240  10-3 m
lu = 2  9 mm = 18 mm = 18  10-3 m
Given Bu = 0.8 T
Bu = Bt = 0.8 T
From Fig. 3.6, magnetic field intensity in the steel is,
Ht = 450 A/m
32
For the air gaps
Bu
0.8
3
Hu 


636
.
62

10
 o 4  10 7
A/m
(450)( 240  10 3 )  (636.62  10 3 )(18  10 3 )
i
2000
11567.16
=
 5.78
2000
(b)
A
Energy density in the air gaps is
B2
3
w fu 

254
.
65

10
2(4  10 7 )
J/m3
33
Volume of air gaps = length of air gaps  area of air gaps
= 0.018  0.02  0.02
= 7.2  10-6 m3
Energy stored in the air gaps,
Wfu
= the volume of air gaps  wfu
= (7.2  10-6)  254.65  103
= 1.834 Joule.
(c) Energy density in the steel,
w ft 

0.8
HdB
0
34
Energy density in the steel is given by the area enclosed
between the characteristic and the B axis in Fig. 3.6 up to
value of 0.8 T.
wft  ½  0.8  450 = 180 J/m3 (straight-line approximation)
Volume of steel= length of steel  area of steel
= (240  10-3)  (0.02  0.02)
= 9.6  10-5 m3
Energy stored in the steel,
Wft = 9.6  10-5  180 = 0.01728 Joule
(d) Total field energy,
Wf = Wft + Wfu
= 0.01728 + 1.834
= 1.851 Joule.
35
The proportion of field energy stored in the steel is,
therefore, seen to be negligibly.
36
Coenergy
Coenergy, Wf’ is the area enclosed between the -i
characteristic and the i axis of Fig.3.2.

Wf
Wf'
i
Fig. 3.6 Field energy and coenergy
For linear -i characteristic, Wf’ = Wf.
For nonlinear -i characteristic, Wf’ > Wf.
37
Mechanical Energy in a Linear
System
It will be assumed that the armature of the actuator in
Fig. 3.1 may move from position x1 to position x2, as a
result, the length of air gaps is reduced. The –i
characteristics for the two extreme positions of the
armature may be assumed to be the two straight lines
(linear).
38
Bahagian
tak boleh
Fixed
steel
core
gerak
gu
SW
v
R

i
Moveable
steel
Bahagian boleh
armature
gerak
Fm
e
x
dx
Figure 3.1
39
x2
λ
x1
i
40
Mechanical Energy in a Linear System
Consider a very slow armature displacement. It may
assumed that it takes place at essentially constant
current as illustrated in Fig. 3.7 (as d/dt is negligible).
The operational point has changed from a to b.
At the moment of armature movement,
We   eidt  
2
1
id  io (2  1 )
3.20
The change of field energy,
W f  12 io 2  12 io 1
 12 io (2  1 )
3.21
41

x = x2

c

d
b
x = x1
a
o
io
i
Fig. 3.7 Current is fixed
42
Mechanical Energy in a Linear System
From Eq. (3.6),
We  Wm  W f
Wm  We  W f

 io (2  1 )  12 io (2  1 )
 12 io (2  1 )
= ΔWf
= ΔWf’ = the change of coenergy
43
Mechanical Energy in a Linear System
For small change of x or dx,
dWm = dWf’


Fmdx = dWf’
3.21
where
dWm = Fmdx
Fm = mechanical force on moving part (armature)
44
Mechanical Energy in a Linear System
Eq. 3.21 can be written as,
Fm 
W f
x

(i, x)
N
3.22
i = constant
Eq. 3.22 is partial differential since Wf is
function of more than one variable.
45
Mechanical Energy in a Linear System
Consider now a very rapid differential armature
displacement dx. It may be assumed that it takes place at
essentially constant flux linkage o, as illustrated in Fig.
3.8. At the instant, the current is changed from i1 to i2 ,
where i1 > i2.

0
dWf
o
i2
i1
Fig. 3.8 Flux linkage is fixed
i
46
Mechanical Energy in a Linear System
Refer to Fig. 3.8, the change of field energy is
dW f  12 o i2  12 o i1
 12 o (i2  i1 )
3.23
Since  does not change, no emf is induced in the
coil , and

dWe = 0
3.24
 From Eq.3.6,
47
Mechanical Energy in a Linear System
dWe  dWm  dW f
-Fmdx = dWf

3.25
 Fm dx  12 o (i2  i1 )
3.26
= the change of field energy
Eq. 3.26 can be written as,
Fm  
W f
x
( , x )
 = constant
Since the electrical input energy is zero, the mechanical output
energy has been supplied entirely by the coupling field.
3.27
48
Mechanical Energy in a Linear System
For a linear electromagnetic system,
 = L(x) i
3.28
where
L(x) = the inductance of the coil which dependent on length of
the air gaps.
From Eqs. 3.11 and 3.28,
Wf 

 id  
0

0

L( x )
d 
2
3.29
2 L( x )
L( x ) 2 i 2 1

 2 L( x)i 2
2 L( x )

W f  W f  12 L( x)i 2
3.30
3.31
49
Mechanical Energy in a Linear System
From Eqs. 3.22 and 3.31,
Fm 
W f
x

(i, x)
 1
 ( 2 L( x)i 2 )
x
1 2 dL ( x)
 i
2
dx

1 2 dL( x)
Fm  i
2
dx
i = constant
i = constant
3.32
3.33
50
Mechanical Energy in a Linear System
From Fig. 3.1 (for linear system),
Ni  H u 2 g
Bu

2g
o
3.34
From Eq. 3.18
2
Bu
Wf = volume of air gaps 
2 o
2
Bu
 Au 2 g 
2o
3.35
where Au = cross section area of air gap
51
Mechanical Energy in a Linear System
From Fig. 3.1, it is seen that a positive displacement
dx will correspond to a reduction dg in the air gap
length. Thus,
dx =  dg m
3.36
From Eqs. 3.27, 3.35 and 3.36 yield,
2
Bu 
 
 Au 2 g 

Fm 

g 
2o 
2

Bu
Fm  2 Au
2 o
3.37
where
2Au = The total cross-section area of air gaps
52
Mechanical Energy in a Linear System
 The force per unit area of air gaps, fm is
2
Bu
fm 
2 o
N/m2
3.38
53
Example 3.2
An electromagnet system is shown in Fig. 3.9.
i
N
lu
Fig. 3.9: linear system
Given that N = 600, i = 3 A, cross section area of air gap
is 5 cm2 and air gap length is 1.5 mm. By neglecting core
reluctance, leakage flux and fringing effects, find:
(a) Force between the electromagnetic surfaces.
(b) Energy stored in the air gap.
54
Solution
(a) The total cross-section area of air gap = Au, Eq. 3.37
becomes,
2
Bu
Fm  Au
2 o
3.39
For linear system,
Ni  H u lu 
 Bu 
 o Ni
Bu lu
o
3.40
lu
55
Substitution from Eq. 3.40 in Eq. 3.39 yields

Fm 
Au  o N 2 i 2
2lu
2
(5  10 4 )( 4  10 7 )(600) 2 (3) 2

2(1.5  10 3 ) 2
= 452.39 N
(b) Since the system is linear, the entire field energy is
stored in the air gap,
56
2
Bu
W f  volume of air gap 
2o
2
Bu
 l u  Au 
2 o
= lu  Fm
= (1.5  10-3)  452.39 Nm
= 0.6789 Nm
= 0.6789 Joule
57
Example 3.3
Electromagnet system in Fig. 3.10 has cross-section
area 25 cm2. The coil has 350 turns and 5 ohm
resistance. Magnetic core reluctance, fringing effects
and leakage flux can be neglected. If the length of air
gap is 4 mm and a 110 V DC supply is connected to the
coil, find
(a) Stored field energy
(b) Lifting force
lu
Fig. 3.10
58
Solution
Coil current,
110
i
 22
5
A
Since the electromagnet system is linear, core reluctance is
neglected,
Bu

Ni  H u lu 
lu

Bu 
 o Ni
o
2lu
(4  10 7 )(350)( 22)

2(4  10 3 )
= 1.2095 Tesla
59
Field energy,
2
Bu
W f  volume of air gap 
2o
1.2095 2
 2  (25  10 )  (4  10 ) 
2  4  10 7
4
3
= 11.6413 Joule
(b) Applying Eq. 3.37 to obtain lifting force,
2
Bu
Fm  the total area of air gaps 
2o
2
1
.
2095
 2  (25  10 4 ) 
2(4  10 7 )
= 2910.33 N
60
Mechanical Energy in a Saturable System
Figure 3.11 shows a diagram illustrating the -i characteristics
for the actuator in Fig. 3.1 when the effect of the ferromagnetic
material is taken into account. It is no longer a linear system
due to saturation of the steel. Wf is smaller than coenergy Wf’.

x + dx
x

dWm
di
o
i
i
Fig. 3.11: At constant flux linkage.
61
Mechanical Energy in a Saturable System
However the areas of this diagram may be interpreted in
exactly the same way as were those of Fig. 3.8 for the
ideal linear system. Field energy is still given by Eq. 3.11.
If an analytical expression is available that gives the
coil current as a function of  and x, then the force on
the armature for a given value of x can readily be
determined. Fig. 3.11 illustrates a differential movement
of the operating point in the -i diagram corresponding
to a differential displacement dx of the armature made
at high speed; that is, at constant flux linkage.
62
Mechanical Energy in a Saturable System
By integrating to obtain an expression for the
area between the -i curve for any x and the
–axis, Wf is obtained as a function of  and x
can be written as
Wf = Wf (,x)
J
3.41
For the movement, the electrical energy
input is zero, since  does not change and
the emf is zero. Consequently,
dWm = dWf (,x)
3.42
63
Mechanical Energy in a Saturable System
and

W f
dWm
 , x 
Fm 

dx
x
Fm  
W f
x
 = constant
 , x 
3.43
 = constant
This corresponds to the expression for a linear
system in Eq. 3.27
64
Mechanical Energy in a Saturable System
More usually, however, it is convenient to express  as a
function of x and i and to employ different approach.
Figure 3.12 illustrates a differential movement of the
operating point in the -i diagram corresponding to a
differential displacement dx of the armature made at low
speed; that is, at constant current. For this displacement,
during flux linkage changes, the emf is not zero, and
therefore dWe is not zero.,

dWe  eidt 

2
1
id
= area defg
dWf = area oef – area odg
3.44
3.45
65
Mechanical Energy in a Saturable System


f

g
x + dx
e
d
x
d
dWf
o
i
i
Fig. 3.12: At constant current
66
Mechanical Energy in a Saturable System
From Eq. 3.6,
dWm = dWe – dWf
= area defg + area odg – area oef
= area ode
3.46
The differential mechanical energy associated
with movement dx is given by the shaded area, it
is equal to the increase of coenergy.
dWm = dWf’
Fmdx = dWm = dWf’

Fm 
W f
dx

i, x 
3.47
i = constant
67
Mechanical Energy in a Saturable System
where

i
W f (i, x)   di
Joule
3.47
0
Wf’ is the function of i and x.
Also since  = NΦ, and i = F /N, substitution in Eq. 3.47
yields the coenergy as a function of mmf and
displacement

F
W f (F , x)   dF
0
3.48
68
Mechanical Energy in a Saturable System
and
Fm 
Wf
x

F , x
3.49
F = constant
69
Example 3.4
The flux linkage and current relationship for an actuator
can be expressed approximately by,
 g 
i

 2 
2
Between the limits 0 < i < 3 A and 3 < g < 9 cm. If the
current is maintained at 2 A, what is the force on the
armature for g = 4 cm?
70
Solution
The -i relationship is nonlinear, and thus the force
must be determined using Eq. 3.11.
Wf 


0
idλ 


0
x 2 3
1 2 3
 x 

x 
  d 
4 3 12
 2 
2
From Eq. 3.43,
Fm  
= 
W f
x
 , x 
 = constant
3 2 x
12
71
For x = 0.04 m and i = 2 A,
1
2
1
2
2i
2 2


 70.71
x
0.04

Wb-turn
70.713  2  0.04
Fm  
12
=  2356.95 N
72