chapter 3 eee121 magnetic circuit_hanim

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.


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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...


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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...


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Circuit

In electric circuit

In magnetic circuit0V

Practical equation for mmf drop0F

HlFWhere; H = magnetizing force

6nsmh...


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iron steel

I

N

0F

)(

)()()(

HlNI

HLHLHlNIironcobaltsteel

-All terms are known except magnetizing force (H) for each material

- -

known)7nsmh...


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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...


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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...


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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...


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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...


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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...


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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...

chapter 3 eee121 magnetic circuit_hanim

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