chapter 3_magnetic circuit
Post on 02-Mar-2016
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MAGNETIC CIRCUITS
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Chapter 3 : MAGNETIC CIRCUITSLesson Outcome:At the end of the lesson, students should be able to:
Define the basic concept of magnetic flux, magneto-motive force, reluctance, permeability, flux density in magnetic 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 DensityWhen magnetomotif force (mmf) applied to magnetic material, flux will be inducedMagnetic flux () measured in Webers (Wb)Number of flux lines per unit area is called flux density (B) and measured in Tesla (T)Where, = flux in Wb A = area in m2*nsmh...
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PermeabilityThe strength of magnetic will vary in accordance with the core usedThis variation in strength is due to the number of flux lines passing through the core
Where, o = permeabilty of free space (4 x 10-7)r = relative permeabilty of the core*nsmh...
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Amperes Circuital LawSimilarity between electric circuit and magnetic circuit Electric Circuit Magnetic Circuit
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EmfElectromotif Force, EMmfMagnetomotif Force, F = NIRResistance, R RReluctance, R =
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DC
E
R
I
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Kirchoffs Voltage Law in Magnetic CircuitIn electric circuit
In magnetic circuit
Practical equation for mmf drop
Where; H = magnetizing force l = length of section*nsmh...
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ironsteelcobaltNI
All terms are known except magnetizing force (H) for each materialH can be obtained from B-H Curve (if flux or flux density B is known)*nsmh...
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Kirchoffs Current Law in Magnetic CircuitFluxes entering a junction is equal to the sum of fluxes leaving the junctionNIabcaAt junction a:a = b + c*nsmh...
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Example 1The magnetic core below consists of three types of materials which are sheet steel, cast steel and cast iron. The cross sectional areas for 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. Given that the number of turns is 1000 turns. Neglect fringing and magnetic effect. (use o = 4 x 10-7)
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150 mm
150 mm
200 mm
100 mm
400 mm
1 mm
I
N = 1000
Sheet steel
Cast steel
Cast iron
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Example1Given,A = 2 x 10-3 m2, gap = 4 x 10-4 Wb,N = 1000lss = 450 mm, lcs = 300 mm, lci = 649 mm, lg = 1mm
= gap = 4 x 10-4 Wb
For air gap:
Form B-H curve:Cast Steel: At B = 0.2 T, Hcs = 170 At/mCast Iron: At B = 0.2 T, Hci = 400 At/mSheet Steel: At B = 0.2 T, Hss = 40 At/mBy performing KVL at magnetic circuit loop:
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DC
1000 I
(Hl)ss
(Hl)cs
(Hl)ci
(Hl)gap
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Example 2The magnetic circuit shown in Figure Q4b is made of cast steel. The cross-sectional area of all limbs is 5 x 10-4 m2. If the number of turns is 500, determine the current i that will produce a magnetic flux of 1 x 10-4 Wb in the air gap 1. (Neglect magnetic leakage and fringing effect)
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200 mm
170 mm
200 mm
100 mm
100 mm
i
N = 500 turns
1 mm
1 mm
Air gap 1
Air gap 2
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Example 2Given, g1 = 1 x 10-4 Wb A = 5 x 10-4 m2l1 =199 mm, lg1 = 1 mm, l2 = 540 mmlg2 = 1 mm, lT =599 mm
1 = g1 = 1 x 10-4 Wb
For air gap 1:
From B-H curve:At B1 = 0.2 T, H1 = 170 At/mH2l2 = H1l1 + Hg1lg1H2 (540 m) = (170)(199 m) + (159.15 k) (1 m)
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DC
500 i
H2l2
H1l1
Hg1lg1
HTlT
Hg2lg2
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Example 2From B-H curve:At H2 = 377.37 At/m, B2 = 0.62 TFrom B-H curve:At BT = 0.82 T, HT = 525 At/mKVL at right oop:*nsmh...
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