A maior rede de estudos do Brasil

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

Pré-visualização | Página 21 de 50

constant during the motion. The operating point has therefore moved up-
ward from point a to b as shown in Fig. 3.5a. During the motion, 
dWe = J ei dt = r: i dA. = area abed 
dWr = area Obe - Oad 
dWm = dWe- dWr 
= area abed + area Oad - area Obe 
=area Oab 
(3.23) 
(3.24) 
(3.25) 
If the motion has occurred under constant-current conditions, the mechani-
cal work done is represented by the shaded area (Fig. 3.5a), which, in fact, 
is the increase in the coenergy. 
dWm =dWi 
If fm is the mechanical force causing the differential displacement dx, 
fmdx = dWm = dWi 
fm = f!Wf(i,x)l 
ax i = constant 
X= Xz 
~) ~) 
(3.26) 
X = Xz 
FIGURE 3.5 Locus of the operating point for motion in system of Fig. 3.2. (a) At 
constant current. (b) At constant flux linkage. 
Mechanical Force in the Electromagnetic System 103 
Let us now consider that the movement has occurred very quickly. It may 
be assumed that during the motion the flux linkage has remained essentially 
constant, as shown in Fig. 3.5b. It can be shown that during the motion the 
mechanical work done is represented by the shaded area Oap, which, in fact, 
is the decrease in the field energy. Therefore, 
fm dx = dWm = -dWr 
t; = _ awr(A,x)l 
m ax A = constant 
(3.27) 
Note that for the rapid motion the electrical input is zero (i d,.\ = 0) because 
flux linkage has remained constant and the mechanical output energy has 
been supplied entirely by the field energy. 
In the limit when the differential displacement dx is small, the areas Oab 
in Fig. 3.5a and Oap in Fig. 3.5b will be essentially the same. Therefore the 
force computed from Eqs. 3.26 and 3.27 will be the same. 
EXAMPLE 3.2 
The A-i relationship for an electromagnetic system is given by 
. _ ( Ag ) 2 
l- 0.09 
which is valid for the limits 0 < i < 4 A and 3 < g < 10 em. For current 
i = 3 A and air gap length g = 5 em, find the mechanical force on the moving 
part, using energy and coenergy of the field. 
Solution 
The A-i relationship is nonlinear. From the A-i relationship 
,.\ = 0.09i 112 
g 
The coenergy of the system is 
From Eq. 3.26 
WI = ' A di = ' . z di f . f. o 09"1/2 0 0 g 
0.09 2 .312 • I 
= ---z JOU es g 3 
fm = aWi(i, g) I 
ag i ~ constant 
= -0.09 X ~i 312 ~~ 3 g i = consta:1t 
104 chapter 3 Electromechanical Energy Conversion 
For g = 0.05 m and i = 3 A, 
fm = -0.09 X~ X 3312 X 0.~52 N · m 
= -124.7N·m 
The energy of the system is 
fA. fA ( Ag )
2 
Wr = 0 z dA = 0 0.09 dA 
From Eq. 3.27 
r; = _ aWr(A, g) I 
m dg A ~ constant 
3 X 0.092 
For g = 0.05 m and i = 3 A, 
0.09 X 3112 
A= 0.05 = 3.12 Wb-turn 
and 
{; = _ 3.123 X 2 X 0.05 
m 3 X 0.092 
= -124.7N·m 
The forces calculated on the basis of energy and coenergy functions are the 
same, as they should be. The selection of the energy or coenergy function 
as the basis for calculation is a matter of convenience, depending on the 
variables given. The negative sign for the force indicates that the force acts 
in such a direction as to decrease the air gap length. • 
3.3.1 LINEAR SYSTEM 
Consider the electromagnetic system of Fig. 3.2. If the reluctance of the 
magnetic core path is negligible compared to that of the air gap path, the 
A-i relation becomes linear. For this idealized system 
A= L(x)i (3.28) 
where L(x) is the inductance of the coil, whose value depends on the air 
gap length. The field energy is 
Wr = f idA (3.29) 
Mechanical Force in the Electromagnetic System 105 
From Eqs. 3.28 and 3.29 
From Eqs. 3.27 and 3.30 
a ( A2 ) I t; = -- -
m ax 2L(x) A~ constant 
For a linear system 
From Eqs. 3.26, and 3.33 
A2 dL(x) 
----
2L2(x) dx 
_ 1 . 2 dL(x) 
-zl --
dx 
Wr = W[ = !L(x)i2 
a I fm =- (!L(x)i2) ax i = constant 
_ 1 . 2 dL(x) 
-zl --
dx 
(3.30) 
(3.31) 
(3.32) 
(3.33) 
(3.34) 
Equations 3.32 and 3.34 show that the same expressions are obtained for 
force whether analysis is based on energy or coenergy functions. For the 
system in Fig. 3.2, if the reluctance of the magnetic core path is neglected, 
Bg Ni = H 2g = -2g 
g f.Lo 
From Eq. 3.16, the field energy is 
B2 
Wr = -
2 
g X volume of air gap 
f.Lo 
B2 
= -
2 
g XAg2g 
f.Lo 
where Ag is the cross-sectional area of the air gap. 
From Eqs. 3.27 and 3.36 
a ( B 2 ) fm = - -
2 
g X Ag X 2g 
ag f.Lo 
B2 
= -
2 
g (2Ag) 
f.Lo 
(3.35) 
(3.36) 
(3.37) 
., 
r 
..... 
106 chapter 3 Electromechanical Energy Conversion 
The total cross-sectional area of the air gap is 2Ag. Hence, the force per unit 
area of air gap, called magnetic pressure Fm, is 
EXAMPLE 3.3 
Bz 
F =-g N/m2 
m 2~-to (3.38) 
The magnetic system shown in Fig. E3.3 has the following parameters: 
N = 500 
i=2A 
Width of air gap= 2.0 em 
Depth of air gap = 2.0 em 
Length of air gap = 1 mm 
Neglect the reluctance of the core, the leakage flux, and the fringing flux. 
(a) Determine the force of attraction between both sides of the air gap. 
(b) Determine the energy stored in the air gap. 
Solution 
(a) 
(b) 
N 
~-toNi 
Bg=-z-
g 
= 47710-7(500 X 2)2 X 2.0 X 2.0 X 10-4 
2 X 1 X 1 X 10-6 
= 251.33 N 
FIGURE E3.3 
p 
EXAMPLE 3.4 
Mechanical Force in the Electromagnetic System 107 
Bz 
= -
2 
g XAg X lg 
JLo 
= 251.33 X 10-3 J 
= 0.25133 joules 
The lifting magnetic system shown in Fig. E3.4 has a square cross section 
6 X 6 cm2• The coil has 300 turns and a resistance of 6 ohms. Neglect 
reluctance of the magnetic core and field fringing in the air gap. 
(a) The air gap is initially held at 5 mm and a de source of 120 V is 
connected to the coil. Determine 
(i) The stored field energy. 
(ii) The lifting force. 
(b) The air gap is again held at 5 mm and an ac source of 120 V (rms) at 
60 Hz is connected to the coil. Determine the average value of the lift 
force. 
Solution 
(a) Current in the coil is 
FIGURE E3.4 
i = 120 = 20A 6 
108 chapter 3 Electromechanical Energy Conversion 
Because the reluctance of the magnetic core is neglected, field energy 
in the magnetic core is negligible. All field energy is in the air gaps. 
Field energy is 
B 
Ni=Hl =~lg 
g g f-1-o 
f-1-oNi 
Bg=lg 
47710-7 X 300 X 20 
2 X 5 X 10-3 
= 0. 7 54 tesla 
Bz 
Wr = -
2 
g X volume of air gap 
f-1-o 
2 ~-~~~~- 7 X 2 X 6 X 6 X 5 X 10-7 J 
= 8.1434 J 
From Eq. 3.37 the lift force is 
Bz fm = -
2 
g X air gap area 
f-1-o 
2 ~-~~~~-7 X 2 X 6 X 6 X 10-4 N 
= 1628.7 N 
(b) For ac excitation the impedance of the coil is 
Z =R + jwL 
Inductance of the coil is 
L = N 2 = N 2J1-oAg 
Rg lg 
3002 X 47710-7 X 6 X 6 X 10-4 
2 X 5 X 10-3 
= 40.7 X 10-3 H 
Z = 6 + j377 X 40.7 X 10-3 0, 
= 6 + j15.340 
Current in the coil is 
I = 120 
rms Y(62 + 15.342) 
= 7.29 A 
F 
Rotating Machines 109 
The flux density is 
The flux density is proportional to the current and therefore changes 
sinusoidally with time as shown in Fig. E3.4. The rms value of the flux 
density is 
The lift force is 
}LoNlrms 
Brms = 2g 
= 47710-7 X 300 X 7.29 T 
2 X 5 X 10-3 
= 0.2748 T 
Bz fm = -
2 
g X 2Ag 
}-to 
(3.38a) 
(3.38b) 
The force varies as the square of the flux density as shown in Fig. E3.4. 
Bzl fmlavg = -2 g X 2Ag }-to avg 
B~, . 
= -
2 
X mr gap area 
J.to 
0.27482 X 2 X 6 X 6 X 10-4 
2 X 47710-7 
= 216.3 N 
(3.38c) 
which is almost one-eighth of the lift force obtained with a de supply 
voltage. Lifting magnets are normally operated from de sources. • 
3.4 ROTATING MACHINES 
The production of translational motion in an electromagnetic system has 
been discussed in previous sections. However, most of the energy converters, 
particularly the higher-power ones, produce rotational motion. The essential 
part of a rotating electromagnetic system is shown in Fig. 3.6. The fixed 
part of the magnetic system is called the stator, and the moving part is caUed 
the rotor. The latter is mounted on a shaft and is free to rotate between the 
110 chapter 3 Electromechanical Energy Conversion 
Stator 
i, 
wm{~ 
' 
I ' 
Rotor 
FIGURE 3.6 Basic configuration of a ro-
tating electromagnetic system. 
poles