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8/13/2019 Chapter 3 EEE121 Magnetic Circuit_hanim
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8/13/2019 Chapter 3 EEE121 Magnetic Circuit_hanim
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MAGNETIC CIRCUITS
Lesson Outcome:At the end of the lesson, students should be able to:
, -
motive force, reluctance, permeability, flux density in
ma netic circuit.
Describe and apply the concepts of Amperes Circuital
Law in magnetic circuit. Solve/determine the magnetic circuit parameter for a
series and series parallel magnetic circuit.
8/13/2019 Chapter 3 EEE121 Magnetic Circuit_hanim
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Flux Density
When magnetomotif force (mmf) applied tomagne c ma er a , ux w e n uce
Magnetic flux () measured in Webers (Wb)
Number of flux lines per unit area is called fluxdensit B and measured in Tesla T
B Where = flux in Wb
A = area in m2
3nsmh...
8/13/2019 Chapter 3 EEE121 Magnetic Circuit_hanim
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8/13/2019 Chapter 3 EEE121 Magnetic Circuit_hanim
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Similarity between electric circuit and magnetic
c rcu
Electric Circuit Magnetic Circuit
I
N
Emf Electromotif Force, E Mmf Magnetomotif Force,=
R Resistance, R R Reluctance, R =A
l
RNI
REI
5nsmh...
8/13/2019 Chapter 3 EEE121 Magnetic Circuit_hanim
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Circuit
In electric circuit
In magnetic circuit0V
Practical equation for mmf drop0F
HlFWhere; H = magnetizing force
6nsmh...
8/13/2019 Chapter 3 EEE121 Magnetic Circuit_hanim
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iron steel
I
N
0F
)(
)()()(
HlNI
HLHLHlNIironcobaltsteel
-All terms are known except magnetizing force (H) for each material
- -
known)7nsmh...
8/13/2019 Chapter 3 EEE121 Magnetic Circuit_hanim
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Circuit Fluxes entering a junction is equal to the sum of fluxes leaving the
junction
I a c
a
N b
At junction a:
a = b + c8nsmh...
8/13/2019 Chapter 3 EEE121 Magnetic Circuit_hanim
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The magnetic core below consists of three types of materials which, .
the materials are 2 x 10-3 m2. If the flux that pass through the air gap
is 4 x 10-4
Wb, find the value of current I that flows in the windings..
magnetic effect. (use o = 4x 10-7)
9nsmh...
8/13/2019 Chapter 3 EEE121 Magnetic Circuit_hanim
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Given,
For air gap:
HB oA = 2 x 10-3 m2, gap = 4 x 10
-4 Wb,
N = 1000
l = 450 mm l = 300 mm
m/kAt15.159104
2.0H
7gap
lci = 649 mm, lg = 1mmForm B-H curve:
Cast Steel:
At B = 0.2 T, Hcs = 170 At/m
At B = 0.2 T, Hci = 400 At/mSheet Steel:
At B = 0.2 T, Hss = 40 At/m
By performing KVL at magnetic circuit loop:
= gap = 4 x 10-4 Wb
-mmk
mmI
cgapcsss
0)649400()115.159(
)300170()45040(1000
T2.0AB 3-10x2
x
AI
I
48775.0
75.4871000
10nsmh...
8/13/2019 Chapter 3 EEE121 Magnetic Circuit_hanim
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The magnetic circuit shown in Figure Q4b is made of cast steel. The- -4 2 .
is 500, determine the current i that will produce a magnetic flux of 1 x
10-4
Wb in the air gap 1.
11nsmh...
8/13/2019 Chapter 3 EEE121 Magnetic Circuit_hanim
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Given, g1 = 1 x 10
-4 Wb
For air gap 1:
2.0BA = 5 x 10-4 m2
l1 =199 mm, lg1 = 1 mm, l2 = 540 mm
l = 1 mm l =599 mm From B-H curve:
m.104 7
o
1g
At B1 = 0.2 T, H1 = 170 At/m
H2l2 = H1l1 + Hg1lg1
2
1
T
TR
H2 (540 m) = (170)(199 m) + (159.15 k) (1 m)2
R
1R
2gR
1g
mAtH
m
m.mH
/37.357
540
2
2
1 = g1 = 1 x 10-4 Wb
T2.0101
BB4
1
105Ag
12nsmh...
8/13/2019 Chapter 3 EEE121 Magnetic Circuit_hanim
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From B-H curve:
At H2 = 377.37 At/m, B2 = 0.62 T
AB22 TBB
TWb
AB
Tg
TT
82.0
82.0105
101.4
2
4
Wb101.3
10562.04
4
From B-H curve:
= = . ,
m/At54.652104
82.0BH
7
o
2g
2g
T
T
4
44
21
101.3101
KVL at right oop:
lHlHilHlH 0500
.
mkmmmk
mkmImmk
)1)(15.159()199)(170()599)(525()1)(54.652(
0)1)(15.159()199)(170(500)599)(525()1)(54.652(
i 32.2
500
13nsmh...