1 - principles of electrical machines and power electronics p_c_sen
634 pág.

1 - principles of electrical machines and power electronics p_c_sen


DisciplinaConversão Eletromecânica de Energia198 materiais1.954 seguidores
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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 
.,.,.,__ __ "'----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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