Magnetic Circuits

1
Magnetic Circuits

• Machine conversion Elec. lt-gt Mech.
• Mechanical Electrical Terminal Vars.
• Potential difference, current, power
• factor, freq, speed, torque or force
• relationships between these variables is affected
  by behavior of magnetic Circuit
• A useful concept is the Eq mag. CCT

2
Equivalent Mag. Elect. CCTs

• Coil of N turns wound on plastic torus
• ?H.dl?J .dA (1.1)
• H(2?a)Ni A (1.2)
• HNi/2?aNi/l A/m (1.3)
• llength flux path
• H outside torus Zero
• Bavµo Hav T (1.4)

3
Eq. Mag. Elec. CCTs ..

• FBav ?d2/4 Wb (1.5)
• ?Ni mmf (1.6)
• ?Ni?J. dA?H.dl A (1.7)
• ?H(2?a)B/µo (2?a)
• F/(?d2/4) .2?a/µo A (1.8)
• R?/F A/Wb (1.9)
• R2?a/µo(?d2/4) A/Wb (1.10)
• Reluctance proportional with l, and inversely
  with its S

4
Equivalent Magnetic CCT.

• The CCT is analog
• of resistor in an Elec. CCT.
• Acc. To Faraday's law
• eturn dF/dt V (1.12)
• eNdF/dtd?/dt (1.13)
• Where
• ?flux linkage
• ?N2µ?/2?a . ?d2/4 .i wb

5
Magnetizing Curve

• B Saturates as H increases in Magnetic Core
• B-H curve named Magnetizing Curve
• Low B represents small R, while High B represents
  large R
• While in Electric cct. R is Indep. of I
• In Magnetic Circuit R is Dep. to B
• Cast Steel, Cast Iron, Silicon Sheet Steel
  different B-H curves

6
Magnetic Circuit with Air Gap

• Rotor separated by an air gap
• Cross section of a DC machine shown
• F in air gapf in poles
• Poles made from magnetic core
• To have equal f, air gap associate a higher mmf
• Having high value of B results in core saturation
• While air gap will not saturate as B-H linear

7
Magnetic Circuit with Air Gapcontinued

• Core air gap shown by equivalent Reluctances

8
Mag CCT with Air Gapcontinued
9
Mag CCT with Air Gapcontinued

• lc average length of core
• Lg length of air gap
• Bcfc/Ac and Bgfg/Ag
• In the air gap flux lines slightly fringes
  (extended as shown at corners)
• Effect simulated by increasing equivalent
  cross-sectional
• For small air gaps fringing neglected
• AgAc BgBcf/Ac

10
Example 1.1 of Mag. CCT with air gap

• Mag cct of primitive relay
• Coil N500 turns, lc360mm, g1.5 mm, B0.8 Tesla
• Core is from cast steel
• The current in coil
• Compute µ and µr of core
• If g0 what is current in coil for B0.8 Tesla in
  core

11
Solution of Exp 1.1

• Since gap is small fringing can be ignored
• From curve of fig1.7Bc0.8T, Hc510 At/m
• Fc (mmf)Hc lc 510x0.36184 At
• Fg(mmf)Hg 2lgBg/µ?x2xlg0.8/(4?x10-7)
• x2x1.5x10-31910 At
• FFcFg18419102094 At
• gtiF/N2094/500 4.19 amps
• µcBc/Hc0.8/5101.57x10-3
• µrµc/µ?1.57x10-3/4?x10-71250
• FHclc510x0.36184 At
• i184/5000.368 A

12
Example 1.2

• In Exp 1.1 if coil current 4 amps and g1 mm,
  find Bg?
• Since B is not given directly B-H curve can not
  be employed, two method can be employed
• a-Load line
• NiHg lgHc lcBg/µ? lgHc lc rearranging ?
• BgBc-µ? lc/lg HcNiµ? / lg ym x c
• m-µ? lc/lg-4? 10-7x360/2-2.26x10-4, cNi/lg x
  µ?1.256T
• Load line intersect B-H curve at B1.08 T
• another method of building load line is as
  follows
• if all mmf acts on core (i.e. Hc0)
• Hc0gtBgNi/lg x µ?500x4x4?x10-7/2x10-3
• 1.256 Tesla
• if all mmf acts on the core (i.e. Bg0),
  intersection with H axis is
• HcNi/lc500x4/(36x10-2)5556 At/m

13
Solution 1.2 continued

• Hc is intersection of load line on H axis
• b-Trial Error
• Assume a flux density, then Hc found from B-H
  curve HgBg/µ?
• Then FcHc lc , FgHg lg gtFFcFg
• Then iF/N if result in a different current,
  another B is chosen continue to reach i4 amps
• i.e. assuming B1.1 gt i4.08 amps
• And B1.08 gt i4 amps

14
Example 1.3

• In mag cct of Fig E1.3 µr of core is 1200 ,
  neglecting mag leakage fringing, dimensions in
  cm, core with square cross section
• Find air gap flux, air gap flux density, Hg
• F1N1xI1500x105000 At
• F2N2xI2500x105000 At
• µc1200µ?1200x4?x10-7
• Rbafelbafe/(µc Ac)3x52x10-2/(1200x4?x10-7x4x10
  -4)
• 2.58x106 At/Wb

15
Fig of example E 1.3

16
Solution Example 3

• Because of symmetry RbcdeRbafe
• Rglg/(µ0 Ag)5x10-3/(4?x2x2x10-4)
• 9.94x106 At/Wb
• Rbe(core)lbe(core)/(µc Ac)
• 51.5x10-2/(1200x4?10-7x4x10- 4)
• 0.82x106 At/Wb
• Loop equations are
• F1(RbafeRbeRg)f2(RbeRg)F1
• F1(RbeRg)f2(RbcdeRbeRg)F2
• Or f1(13.34x106)f2(10.76x106)5000

17
Solution Ex.3 continued

• And f1(10.76x106)f2(13.34x106)5000
• gt f1f22.067x10-4 Wb
• air gap flux is fgf1f24.134x10-4 Wb
• Bgfg/Ag4.134x10-4/(4x10-4)1.034T
• HgBg/µ01.034/(4?10-7)0.822x106 At/m

18
INDUCTANCE

• Coil wound on a mag. Core
• This coil equivalent ideal model is inductance
• Defined as flux linkage per ampere
• Flux linkage ?Nf Inductance L?/I
• LNf/iNBA/iNµHA/(Hl/N)N2/(l/µA)
• LN2/R

19
Example 1-4

• in mag. Cct. Of fig 1.9 N400 turns
• lc50 cm , lg1.0 mm, AcAg15 cm2
• Realitive permeability of core µr3000
• i1.0 A
• Find f and B in air gap, also coil inductance
• Solution
• Rclc/(µr µ0Ac)50x10-2/(3000x4?10-7x15x10-4)8
  8.42x103 At/Wb
• Rglg/(µ0Ag)1x10-3/(4?10-7x15x10-4)530.515x10
  3 At/Wb

20
Ex. 1-4, solution continued

• FNi/(RcRg)
• 400x1.0/(88.42530.515)x103
• Bf/Ag
• 0.6463x10-3/(15x10-4)0.4309 T
• LN2/(RcRg) 4002/(88.42530.515)x103
• 258.52x10-3 H
• or
• L?/iNf/i
• 400x0.6463x10-3/1.0258.52x10-3 H

21
Example 1.5

• Coil of fig 1.4 with N250 turns, wound on a
  silicon steel sheet (toroidal)
• Inner and outer radii are 20 25 cm
• Toroidal core has a circular cross section
• if coil current were 2.5 A find
• a- B at mean radius of toroid
• b- L of coil, assuming B within core uniform
  equal to that of mean radius

22
Solution of Ex. 1.5

• Mean radius1/2(2025)22.5 cm
• HNi/l250x2.5/(2?22.5x10-2)442.3 At/m
• From B-H curve of silicon steel sheet (fig1.7)
  B1.225 T
• Cross-sectional area A?(core radius)2
• ?(25-20)/22x10-4?6.25x10-4 m2
• FBA1.225x?6.25x10-424.04x10-4 Wb
• ?250x24.04x10-40.601 Wb. Turn
• L?/i0.601/2.50.2404 H24.04 mH
• Or can be calculated as follows

23
Solution of Ex. 1.5continued

• µ of coreB/H1.225/442.3
• Rcorel/(µA)
• 2?22.5x10-2/(1.225/442.3)x?6.25x10- 4
• 2599.64x102 At/Wb
• LN2/R2502/2599.64x1020.2404 H
• 240.4 mH

24
Hysteresis 1-characteristic

• Coil-core assembly of fig 1.12 a
• Assuming core initially is demagnetized
• The resulting B-H characteristic shown
• When H becomes zero still a residual flux density
  is left Br
• Hysteresis loops are very narrow, for such cores
  if hysteresis effect can be ignored, then
• B-H characteristic represented by
  magnetization or saturation curve
• Deltamax,a special ferromagnetic alloys has a
  almost square B-H loop
• Alloys with 50 iron, 50nickel has B-H loop of
  fig 1.13
• Coil wound on Deltamax core used as a switch(very
  low current when unsaturated and large current
  when saturated

25
Hysteresis 2- Losses

• As shown in hysteresis loops of fig 1.12c over a
  cycle(where I of coil slowly varying)
• energy flows to coil-core from source
• Howeverenergy flowing ingtenergy returns
• The net energy flow from source to coil is the
  heat in core
• The loss due to hysteresis called Hysteresis loss
  
• Size of hysteresis loop ? hysteresis loss
• To prove this assume coil has no resistance

26
HysteresisLoss , continued

• Voltage e across the coil eN df/dt
• Energy transfer during t1 to t2 is
• VcoreA l, volume of core
• Power loss due to hysteresis in core
• PhVcore Wb f
• f freq. of variation of i

27
HysteresisLoss , approximate relation

• Steinmetz of G.E. through large no. of experiment
  for machine magnetic materials proposed a
  relation
• Area of B-H loop
• Bmax is the max flux density
• n varies from 1.5 to 2.5, K is a constant
• Therefore the hysteresis power loss
• PhKh (Bmax)n f
• Kh a constant depends on ferromagnetic material
  and core volume

28
EDDY CURRENT LOSS

• Another power loss of mag. Core is due to rapid
  variation of B
• In cross section shown voltage induced and ie
  passes and due to resistance of core
• Peie2 R
• this loss can be reduced
• a- using high resistive core material, few
  Si
• b- using a laminated core
• Eddy current loss PeKeBmax2 f2
• Ke constant depends on material lamination
  thickness which varies from 0.01 to 0.5 mm

29
CORE LOSS

• PcPhPe
• If current I varies slowly eddy loss negligible,
  separation in fig 1.16
• Total core loss determined from dynamic B-H loop
• Using a wattmeter core loss can be measured
• It is not easy to know what portion is eddy
  hysteresis

30
SINUSOIDAL EXCITATION

• Ac electric machine work with ac supply have
  sinusoidal voltage flux fig1.17
• F(t)fmax sin?t
• e(t)N df/dtN fmax ? cos?tEmax cos?t
• ErmsEmax/v2N?fmax /v24.44Nffmax

31
Example 1.6

• A 1 phase 120 V, 60 Hz supply and coil with 200
  turns on core with
• core length100 cm, cross-section area20
  cm2, µr2500
• Find a-expression for B in core
• b-expression for I in coil
• Fmax120/4.44x200x600.002253 Wb
• Bmax0.002253/20x10-41.1265 T
• B1.1265 sin2?60t
• Hmax1.1265/2400x4?10-7358.575 At/m
• imax Hl/N358.575x100x10-2/2001.7928 A,
  i1.79sin?t

32
Example 1.7

• A square wave voltage E100 V f60 Hz applied
  coil on a closed iron core, N500
• Cross section area 0.001 mm2, assume coil no
  resistance
• a- max value of flux sketch V f vs time
• b- max value of E if Blt1.2 Tesla

33
Ex 1.7 - Solution

• e N df/dt gt N.?fE.?t
• E constant gt 500(2fmax)Ex1/120
• Fmax100/(100x120)Wb0.833x10-3 Wb
• From fig. gt Bmax1.2 T
• FmaxBmax x A1.2 x 0.0011.2 x 10-3 Wb
• N(2fmax)E x 1/120
• E120x500x2x1.2x10-3144 V

34
Exciting Current

• Current which establish the flux in the core
• The term If if B-H nonlinear, non-sinusoid
• At first consider not to have Hysteresis
• B-H curve ? f-i curve (or the rescaled one)
• Knowing sine shape flux, exciting current
  waveform by help of f-i curve obtained
• The current non-sinusoidal, if1 lags V 90
• Or no loss (since Hysteresis neglected)

35
Considering Hysteresisin Exciting Current
determination

• Now if determined from multi-valued f-i
• Exciting current nonsinusoid nonsymmetric
• It can split 2 component ic in phase with e
  rep. loss, im in ph. With f symmetric

36
PERMANENT MAGNET

• Permanent magnet large B-H loop high Br
• Also high Hc , material iron, cobalt, Nickel and
  called hard iron vs soft iron
• P.M. magnetization
• Assume magnet material
• Initially demag.
• large mmf applied
• removed, remain at Br
• If reverse H1 applied
• To hard iron move to b
• By removing reapply recoil
• Through bc line parallel with xay,
• µrec, for alnico 3-5 µ0

37
Approximation Designof Permanent Magnets

• Per. Mag. of last fig. magnetized to Br-point a
• Removing keeper air gap activated
• Assuming no fringing no leakage, also if no mmf
• required for soft iron
• Amperes circuit law
• Hm lm Hg lg0
• Hm-lg/lm Hg
• Continuity of flux requires fBmAmBgAg
• Also Bgµ? Hg gtBmµ? Ag/Am lm/lg Hm
• This equation a straight line, intersection with
  demag. Curve, at b, is operating point

38
Design of Permanent Magnetscontinued

• Demagnetizing curve shear line
• ? intersection shear line- demagn. Curve result
  in operating values of B and H hard iron, keeper
  removed
• ? if keeper inserted, o.p. moves up recoil line
  bc.
• ? this analysis gt O.P. of a
• permanent magnet with
• air gap, determined by
• demag. Portion B-H loop
• Dimensions of magnet and air gap

39
Permanent Magnet DesignContinued

• From last Eqs gt volume of P.M.
• VmAmlmBgAg/Bm x Hglg/Hm
• Bg2
  Vg/(µ0BmHm)
• Where VgAglg
• To have Bg in gap of volume Vg,
• a min volume of hard iron required if final
  o.p. located such that BmHm is a max
• BmHm known as energy product of hard iron

40
Permanent Magnet Materials

• Alinco (Alloys of Al, Ni, Co)used for for P.M.
• Having a Demagnetization curve fig22
• Br 1.25 Tesla , and Hc -50 kA/mF
• From 1950 Ferrite P.M. also are used, with lower
  B (0.35 T) and higher Coercive force(-350 kA/m)
• Since 1960 rare-earth P.M. material
  developed having high residual B as alinco type,
  with a greater coercivity than ferrite
• Demagnetization for samaritum-cobalt and
  neodymium-boron magnet fig 1.24

41
Example 1.8 P.M.

• P.M. in fig 1.21 made of Alinco 5
• B of 0.8 to be stablished in air gap, keeper
  removed
• Air gap Ag2.5 sq. cm, lg0.4 cm
• O.p. correspond to point with max HmBm
• Or Bm0.95 T , Hm-42 kA/m
• Determine lm, Am of P.M.

42
Solution to example 1.8

• Lmlg/Hm Hglg Bg/(Hmµ0) 0.4x0.01x0.8/(42x1000x4?
  x0.0000001)
• 0.0606 m6.06 cm
• From flux continuity AmBg Ag/Bm
• 0.8x2.5x0.0001/0.952.105 sq. cm