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# 1 - principles of electrical machines and power electronics p_c_sen

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```heating loss due to current flowing in the
winding of the energy converter. This loss is known as the i2R loss in the
resistance (R) of the winding. The field loss is the core loss due to changing
magnetic field in the magnetic core. The mechanical loss is the friction and
windage loss due to the motion of the moving components. All these losses
95
r
96 chapter 3 Electromechanical Energy Conversion
Electrical
system
Electrical
loss
Coupling
field
t
Field
loss
Mechanical
system
t
Mechanical
loss
FIGURE 3.1 Electromechanical converter system.
Pmech
are converted to heat. The energy balance equation 3.1 can therefore be
written as
Electrical energy = mechanical energy + increase in stored
input from source output + friction field energy +
- resistance loss and windage loss core loss (3.2)
Now consider a differential time interval dt during which an increment
of electrical energy dWe (excluding the i 2R loss) flows to the system. During
this time dt, let dWr be the energy supplied to the field (either stored or lost,
or part stored and part lost) and dW m the energy converted to mechanical
form (in useful form or as loss, or part useful and part as loss). In differential
forms, Eq. 3.2 can be expressed as
(3.3)
Core losses are usually small, and if they are neglected, dWr will represent
the change in the stored field energy. Similarly, if friction and windage losses
can be neglected, then all of dW m will be available as useful mechanical
energy output. Even if these losses cannot be neglected they can be dealt
with separately, as done in other chapters of this book. The losses do not
contribute to the energy conversion process.
3.2 FIELD ENERGY
Consider the electromechanical system of Fig. 3.2. The movable part can
be held in static equilibrium by the spring. Let us assume that the movable
part is held stationary at some air gap and the current is increased from
zero to a value i. Flux will be established in the magnetic system. Obviously,
(3.4)
and from Eqs. 3.3 and 3.4,
dWe=dWr (3.5)
If core loss is neglected, all the incremental electrical energy input is stored
as incremental field energy. Now,
dA
e=-dt (3.6)
Immovable
part
Field Energy 9 7
.,.,.,__ __ &quot;'----11 Reference
position
FIGURE 3.2 Example of an electromechanical system.
dw. = eidt (3.7)
From Eqs. 3.5, 3.6, and 3.7,
dWr =idA (3.8)
The relationship between coil flux linkage A and current i for a particular
air gap length is shown in Fig. 3.3. The incremental field energy dWr is
shown as the crosshatched area in this figure. When the flux linkage is
increased from zero to A, the energy stored in the field is
(3.9)
This integral represents the area between the A axis and the A-i characteris-
tic, the entire area shown shaded in Fig. 3.3. Other useful expressions can
also be derived for the field energy of the magnetic system.
Let
He= magnetic intensity in the core
Hg =magnetic intensity in the air gap
lc = length of the magnetic core material
lg = length of the air gap
FIGURE 3.3 A-i characteristic for the system in
Fig. 3.2 for a particular air gap length.
98 chapter 3 Electromechanical Energy Conversion
Then
Also
A =NcfJ
=NAB
where A is the cross-sectional area of the flux path
B is the flux density, assumed same throughout
From Eqs. 3.9, 3.10, and 3.12,
Wr = I Hclc ~ Hglg NA dB
For the air gap,
H=B
g 1'-o
From Eqs. 3.13 and 3.14,
Wr = I ( Hclc + !/g) A dB
= I (He dB Ale + :
0
dB lgA)
= I He dB X volume of magnetic material
B2
+ -
2
X volume of air gap
1'-o
(3.10)
(3.11)
(3.12)
(3.13)
(3.14)
(3.15)
(3.16)
=wrcXVc+WrgXVg (3.17)
= Wrc + Wrg (3.18)
where Wrc = J He dEc is the energy density in the magnetic material
Wrg = B 2/2f.J-o is the energy density in the air gap
Vc is the volume of the magnetic material
Vg is the volume of the air gap
Wrc is the energy in the magnetic material
Wrg is the energy in the air gap
Normally, energy stored in the air gap (Wrg) is much larger than the energy
stored in the magnetic material (Wrc). In most cases Wrc can be neglected.
For a linear magnetic system
H =Be
c 1'-c (3.19)
Field Energy 99
Therefore
(3.20)
The field energy of the system of Fig. 3.2 can be obtained by using either
of Eqs. 3.9 and 3.16.
EXAMPLE 3.1
The dimensions of the actuator system of Fig. 3.2 are shown in Fig. E3.1.
The magnetic core is made of cast steel whose B-H characteristic is shown
in Fig. 1.7. The coil has 250 turns, and the coil resistance is 5 ohms. For a
fixed air gap length g = 5 mm, a de source is connected to the coil to produce
a flux density of 1.0 tesla in the air gap.
(a) Find the voltage of the de source.
(b) Find the stored field energy.
Solution
(a) From Fig. 1.7, magnetic field intensity in the core material (cast steel)
for a flux density of 1.0 T is
5 em
5 em
He= 670At/m
Length of flux path in the core is
Zc = 2(10 + 5) + 2(10 + 5) em
= 60cm
The magnetic intensity in the air gap is
Bg 1.0
Hg = J.to = 477 1 o-7 At/m
= 795.8 X 103 At/m
Depth= 10 em
FIGURE E3.1
100 chapter 3 Electromechanical Energy Conversion
The mmf required is
Ni = 670 X 0.6 + 795.8 X 103 X 2 X 5 X 10-3 At
= 402 + 7958
= 8360 At
. _ 8360A
l- 250
= 33.44A
Voltage of the de source is
Vdc = 33.44 X 5 = 167.2 V
(b) Energy density in the core is
II.O d Wrc = 0 H B
This is the energy density given by the area enclosed between the B
axis and the B-H characteristic for cast steel in Fig. 1.7. This area is
Wrc =~X 1 X 670
= 335 11m3
The volume of steel is
Vc = 2(0.05 X 0.10 X 0.20) + 2(0.05 X 0.10 X 0.10)
= 0.003 m 3
The stored energy in the core is
Wrc = 335 X 0.003 J
= 1.005 J
The energy density in the air gap is
- 1.02 J/ 3
Wfg- 2 X 41T X 10-7 m
= 397.9 X 103 J/m3
The volume of the air gap is
Vg = 2(0.05 X 0.10 X 0.005) m 3
= 0.05 X 10-3 m 3
The stored energy in the air gap is
Wrg = 397.9 X 103 X 0.05 X 10-3
= 19.895 joules
1 um 1 B 1 B u oT E c A 1 r a A 1 I
>
Field Energy 101
The total field energy is
Wr = 1.005 + 19.895 J
= 20.9 J
Note that most of the field energy is stored in the air gap. \u2022
3.2.1 ENERGY, COENERGY
The A-i characteristic of an electromagnetic system (such as that shown in
Fig. 3.2) depends on the air gap length and the B-H characteristics of the
magnetic material. These A-i characteristics are shown in Fig. 3.4a for
three values of air gap length. For larger air gap length the characteristic
is essentially linear. The characteristic becomes nonlinear as the air gap
length decreases.
For a particular value of the air gap length, the energy stored in the field
is represented by the area A between the A axis and the A-i characteristic,
as shown in Fig. 3.4b. The areaB between the i axis and the A-i characteristic
is known as the coenergy and is defined as
w; = t A.di (3.21)
This quantity has no physical significance. However, as will be seen later,
it can be used to derive expressions for force (or torque) developed in an
electromagnetic system.
From Fig. 3.4b,
Wi +Wr=Ai (3.22)
Note that Wi > Wr if the A.-i characteristic is nonlinear and Wi = Wr if it is
linear.
(a)
air gap
length
(b)
FIGURE 3.4 (a) A.-i characteristics for different air gap lengths. (b)
Graphical representation of energy and coenergy.
102 chapter 3 Electromechanical Energy Conversion
3.3 MECHANICAL FORCE IN THE
ELECTROMAGNETIC SYSTEM
Consider the system shown in Fig. 3.2. Let the movable part move from one
position (say x = x 1) to another position (x = x 2) so that at the end of the
movement the air gap decreases. The A.-i characteristics of the system for
these two positions are shown in Fig. 3.5. The current (i =viR) will remain
the same at both positions in the steady state. Let the operating points be
a when x = x 1 and b when x = x 2 (Fig. 3.5).
If the movable part has moved slowly, the current has remained essentially```
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