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# 1 - principles of electrical machines and power electronics p_c_sen

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```However, the air gap re-
mains unsaturated, since the B-H curve for the air medium is linear (J.t is
constant).
A magnetic circuit having two or more media-such as the magnetic core
and air gap in Fig. 1.8-is known as a composite structure. For the purpose
of analysis, a magnetic equivalent circuit can be derived for the composite
structure.
Let us consider the simple composite structure of Fig. 1.9a. The driving
force in this magnetic circuit is the mmf, F = Ni, and the core medium and
the air gap medium can be represented by their corresponding reluctances.
The equivalent magnetic circuit is shown in Fig. 1.9b.
ffi =l
g f.toAg
<I>= Ni
ffic + ffig
Ni = Hclc + Hglg
where lc is the mean length of the core
lg is the length of the air gap
The flux densities are
B = <l>c
c Ac
( 1.15)
(1.16)
(1.17)
(1.18)
( 1.19)
(1.20)
In the air gap the magnetic flux lines bulge outward somewhat, as shown
in Fig. 1.1 0; this is known as fringing of the flux. The effect of the fringing
<I>
ll + ffic N Tg Ni
9kg
(a) (b)
FIGURE 1.9 Composite structure. (a) Magnetic
core with air gap. (b) Magnetic equivalent
circuit.
8 chapter 1 Magnetic Circuits
FIG{Tl?,E 1.10 Fringing flux.
is to increase the cross-
fringing effect can be n
sectional areas of the '
EXAMPLE 1.1
:tional area of the air gap. For small air
'rected. If the fringing effect is neglected,
·e and the air gap are the same and the
Ag =Ac
<I> B =B =-
g c Ac
ps the
cross-
.re
Figure E 1.1 represents the magnetic circuit of a primitive relay. The coil
has 500 turns and the mean core path is lc = 360 mm. When the air gap
lengths are 1.5 mm each, a flux density of 0.8 tesla is required to actuate
the relay. The cc is cast steel.
(a) Find the ct ;ent in the coil.
(b) Compute + e values of permeability and relative permeability of the
core.
(c) If the · -sap is zero, find the current in the coil for the same flux
dens; 8 T) in the core.
Solutior
(a) The
den
the,
N
p is small and so fringing can be neglected. Hence the flux
s the same in both air ga·&quot;1 2nd core. From the B-H curve o~
c steel core (Fig. 1. 7).
FIGURE E1.1 N = 500 turns, lc =
36 em.
-
I
I
Magnetic Circuits 9
For
Be= 0.8 T, He= 510 At/m
mmf Fe= Hele = 510 X 0.36 = 184 At
For the air gap,
mmf Fg = Hg2lg = :: 2lg = 4:i~- 7 X 2 X 1.5 X 10-3
= 1910At
Total mmf required:
F =Fe+ Fg = 184 + 1910 = 2094At
Current required:
. F 2094
z = N = SOO = 4.19 amps
Note that although the air gap is very small compared to the length
of the core (lg = 1.5 mm, lc = 360 mm), most of the mmf is used at
the air gap.
(b) Permeability of core:
(c)
- Be - 0.8 - 1 57 X o-3 J.-te-He-510-. 1
Relative permeability of core:
= f.-te = 1.57 X 10-3 = 1250 f.-tr }-to 47T10-7
F = Hcle = 510 X 0.36 = 184 At
i = ~~~ = 0.368 A
Note that if the air gap is not present, a much smaller current is
required to establish the same flux density in the magnetic circuit. \u2022
EXAMPLE 1.2
Consider the magnetic system of Example 1.1. If the coil current is 4 amps
when each air gap length is 1 mm, find the flux density in the air gap.
Solution
In Example 1.1, the flux density was given and so it was easy to find the
magnetic intensity and finally the mmf. In this example, current (or mmf)
is given and we have to find the flux density. The B-H characteristic for the
10 chapter 1 Magnetic Circuits
air gap is linear, whereas that of the core is nonlinear. We need nonlinear magnetic circuit analysis to find out the flux density. Two methods will be discussed.
1. Load line method. For a magnetic circuit with core length le and air gap length lg,
Rearranging,
( 1.21)
This is in the form y = mx + c, which represents a straight line. This straight line (also called the load line) can be plotted on the B-H curve of the core. The slope is
l 360 m = -1-Lof = -47rl0-7 l = -2.26 X 10-4
g
The intersection on the B axis is
_ Nif.Lo _ 500 X 4 X 47710-7 _ 1 256 l c--1-- 2xto-3 -. tesa g
The load line intersects the B-H curve (Fig. El.2) at B = 1.08 tesla. Another method of constructing the load line is as follows: If all mmf acts on the air gap (i.e., He = 0) the air gap flux density is
Ni Bg = T f.Lo = 1.256 T
g
This value of Bg is the intersection of the load line on the B axis. If all mmf acts on the core (i.e., Bg = 0),
Ni 500 X 4 He= T = 36 X 10_2 = 5556 At/m e
This value of He is the intersection of the load line on the H axis.
1.256
1.08
B
0~--~------------------------~~~--H
Ni = 5555.6
lc
FIGURE E1.2
Magnetic Circuits 11
2. Trial-and-error method. The procedure in this method is as follows.
(a) Assume a flux density.
(b) Calculate He (from the B-H curve) and Hg (= Bglp_0).
(c) Calculate Fe ( = Hclc), Fg ( = Hglg), and F ( = Fe + Fg).
(d) Calculate i = FIN.
(e) If i is different from the given current, assume another judicious
value of the flux density. Continue this trial-and-error method until
the calculated value of i is close to 4 amps.
If all mmf acts on the air gap, the flux density is
Ni B = T p_0 = 1.256 T
g
Obviously, the flux density will be less than this value. The procedure
is illustrated in the following table.
B
1.1
1.08
800
785
EXAMPLE 1.3
8.7535 X 105
8.59435 X 105
288
282
1750.7
1718.87
F
2038.7
2000.87
i
4.08
4.0
\u2022
In the magnetic circuit of Fig. El.3a, the relative permeability of the ferro-
magnetic material is 1200. Neglect magnetic leakage and fringing. All dimen-
sions are in centimeters, and the magnetic material has a square cross-
sectional area. Determine the air gap flux, the air gap flux density, and the
magnetic field intensity in the air gap.
Solution
The mean magnetic paths of the fluxes are shown by dashed lines in Fig.
El.3a. The equivalent magnetic circuit is shown in Fig. El.3b.
F1 = N 111 = 500 X 10 = 5000 At
F2 = N 212 = 500 X 10 = 5000 At
P-e = 1200p_0 = 1200 X 47710-7
(fJt _ f bafe
bafe-A
P-c c
3 X 52 X 10-2
1200 X 47710- 7 X 4 X 10-4
= 2.58 X 106 At/Wb
l
12 chapter 1 Magnetic Circuits
10 A
NJ
500 turns
\Jtbafe
(a)
\Jtbcde
+
N2
500 turns
+
F1 = 5000 At F2 = 5000 At
(b)
FIGURE E1.3
From symmetry
5 x w-3
4rr1o-7 x 2 x 2 x w-4
= 9.94 X 106 At/Wb
CZft _ Z be( core) be(core) - A
f-tc c
51.5 x 10-2
1200 X 4rr10- 7 X 4 X 10-4
= 0.82 X 106 At/Wb
The loop equations are
<P1(ezftbafe + ezftbe + ezftg) + <P2(ezftbe + ezftg) = F1
<P1(ezftbe + ezftg) + <Piezftbcde + ezftbe + ezftg) = F2
Magnetic Circuits 13
or
or
<.1>1(13.34 X 106) + <l>il0.76 X 106) = 5000
<.1>1(10.76 X 106) + <l>i13.34 X 106) = 5000
<l>1 = <l>z = 2.067 X 10-4 Wb
The air gap flux is
<l>g = <l>1 + <l>z = 4.134 X 10-4 Wb
The air gap flux density is ·
B = <I>g = 4.134 X w-4 = 1.034 T
g Ag 4 X 10-4
The magnetic intensity in the air gap is
H = Bg = l.034 = 0.822 X 106 At/m \u2022
g /-to 47T10-7
1.1.6 INDUCTANCE
A coil wound on a magnetic core, such as that shown in Fig. 1.11a, is
frequently used in electric circuits. This coil may be represented by an ideal
circuit element, called inductance, which is defined as the flux linkage of
the coil per ampere of its current.
Flux linkage
Inductance
A= N<l>
L=~
l
(1.22)
(1.23)
From Eqs. 1.22, 1.23, 1.11, 1.3, and 1.14,
I A /
Val f
i ~
N4 ~~ ~--~
(a)
L _ N<I> _ NBA _ N ~-tHA ---------
i i i
N~-tHA
Hl!N
(b)
L
( 1.24)
FIGURE 1.11 Inductance of a
coil-core assembly. (a) Coil-core
assembly. (b) Equivalent induc-
tance.
:I
!
14 chapter 1 Magnetic Circuits
(1.25)
Equation 1.24 defines inductance in terms of physical dimensions, such as
cross-sectional area and length of core, whereas Eq. 1.25 defines inductance in terms of the reluctance of the magnetic path. Note that inductance varies
as the square of the number of turns. The coil-core assembly of Fig. 1.11a is represented in an electric circuit by an ideal inductance as shown in Fig. l.llb.
EXAMPLE 1.4```
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