Magnetic Materials and Magnetic Circuit Analysis

Transcription

1 Chapter 7. Magneti Materials and Magneti Ciruit Analysis Topis to over: 1) Core Losses 2) Ciruit Model of Magneti Cores 3) A Simple Magneti Ciruit 4) Magneti Ciruital Laws 5) Ciruit Model of Permanent Magnets 6) Calulation of Indutane, EMF, and Magneti Energy Introdution In general, magneti materials an be lassified as magnetially "soft" and "hard" materials. Soft materials are normally used as the magneti ore materials for indutors, transformers, atuators and rotating mahines, in whih the magneti fields vary frequently, whereas hard materials, or permanent magnets, are used to replae magnetization oils for generating stati magneti fields in devies suh as eletri motors and atuators. The - relationships and hysteresis loops have been disussed earlier. In this hapter, we are going to examine the power losses in a soft magneti ore under an alternating magnetization, and further develop an eletrial iruit model of a magneti ore with a oil. For performane predition of eletromagneti devies, magneti field analysis is required. Analytial magneti field analysis by the Maxwell s equations, however, has been shown very diffiult for engineering problems owing to the fat that most pratial devies are of ompliated strutures. Powerful numerial methods, suh as the finite differene and finite element methods, are out of the sope of this subjet. In this hapter, we introdue a simple method of magneti iruit analysis based on an analogy to d eletrial iruits. Soft Magneti Materials under Alternating Exitations Core Losses Core losses our in magneti ores of ferromagneti materials under alternating magneti field exitations. The diagram on the right hand side plots the alternating ore losses of M-36, mm steel sheet against the exitation frequeny. In this setion, we will disuss the mehanisms and predition of alternating ore losses. As the external magneti field varies at a very low rate periodially, as mentioned earlier, due to the effets of magneti domain wall motion the - Alternating ore loss of M36, mm steel sheetat different exitation frequenies

2 relationship is a hysteresis loop. The area enlosed by the loop is a power loss known as the hysteresis loss, and an be alulated by Physt = d (W/m3 /yle) or (J/m 3 ) For magneti materials ommonly used in the onstrution of eletri mahines an approximate relation is P n = C f (1.5 < n < 2.5) (W/kg) hyst h p where C h is a onstant determined by the nature of the ferromagneti material, f the frequeny of exitation, and p the peak value of the flux density. Example: A - loop for a type of eletri steel sheet is shown in the diagram below. Determine approximately the hysteresis loss per yle in a torus of 300 mm mean diameter and a square ross setion of mm. Solution: The are of eah square in the diagram represents (0.1 T) (25 A/m) = 2.5 (Wb/m 2 ) (A/m) = 2.5 VsA/m 3 = 2.5 J/m 3 If a square that is more than half within the loop is regarded as totally enlosed, and one that is more than half outside is disregarded, then the area of the loop is = 215 J/m 3 The volume of the torus is π = m 3 Energy loss in the torus per yle is thus = J ysteresis loop of M36 steel sheet Page 7-2

3 When the exitation field varies quikly, by the Faraday's law, an eletromotive fore (emf) and hene a urrent will be indued in the ondutor linking the field. Sine most ferromagneti materials are also ondutors, eddy urrents will be indued as the exitation field varies, and hene a power loss known as eddy urrent loss will be aused by the indued eddy urrents. The resultant - or λ-i loop will be fatter due to the effet of eddy urrents, as illustrated in the diagram below. Under a sinusoidal magneti exitation, the average eddy urrent loss in a magneti ore an be expressed by ( ) P = C f eddy e p 2 (W/kg) where C e is a onstant determined by the nature of the ferromagneti material and the dimensions of the ore. Sine the eddy urrent loss is aused by the indued eddy urrents in a magneti ore., an effetive way to redue the eddy urrent loss is to inrease the resistivity of the material. This an Relationship between flux linkage and exitation urrent when eddy urrent is inluded (dashed line loop), where the solid line loop is the pure hysteresis obtained by d exitation be ahieved by adding Si in steel. owever, too muh silion would make the steel brittle. Commonly used eletrial steels ontain 3% silion. Another effetive way to redue the eddy urrent loss is to use laminations of eletrial steels. These eletrial steel sheets are oated with eletri insulation, whih breaks the eddy urrent path, as illustrated in the diagram below. Eddy urrents in a laminated toroidal ore The above formulation for eddy urrent loss is obtained under the assumption of global eddy urrent as illustrated shematially in figure (a) of the following diagram. This is inorret for materials with magneti domains. When the exitation field varies, the domain walls move aordingly and loal eddy urrents are indued by the flutuation of the loal flux density aused by the domain wall motion as illustrated in figure (b) of the diagram below. The total eddy urrent aused by the loal eddy urrents is in general higher than that predited by the formulation under the global eddy urrent assumption. The differene is known as the exess loss. Sine it is very diffiult to Page 7-3

4 alulate the total average eddy urrent loss analytially, by statistial analysis, it was postulated that for most soft magneti materials under a sinusoidal magneti field exitation, the exess loss an be predited by ( ) P = C f ex ex p 32 / (W/kg) where C ex is a onstant determined by the nature of the ferromagneti material. Therefore, the total ore loss an be alulated by Pore = Physt + Peddy + Pex The diagram below illustrates the separation of alternating ore loss of Lyore-140, 0.35 mm nonoriented sheet steel at 1 T. Using the formulas above, the oeffiients of different loss omponents an be obtained by fitting the total ore loss urves. M s M s M s (a) (b) Eddy urrents, (a) lassial model, and (b) domain model Core Loss (J/kg) = 1 T Pex/Freq Peddy/Freq Physt/Freq Frequeny (z) Separation of alternating ore loss of Lyore-140 at =1 T Ciruit Model of Magneti Cores In the equivalent iruit of an eletromagneti devie, the iruit model of the magneti ore is an essential part. Consider a magneti ore with a oil of N turns uniformly wound on it. As illustrated below, under an sinusoidal voltage (flux likage) exitation, Page 7-4

5 the orresponding exitation urrent is nonsinusoidal due to the nonlinear - relationship of the ore. When only the fundamental omponent of the urrent is onsidered, however, the relationship between the phasors of voltage and urrent an be determined by a resistor (equivalent resistane of the ore loss) in parallel of an lossless indutor (self indutane of the oil) as illustrated in the diagram below. Coil of N turns with a magneti ore Ciruit model of magneti ores Exitation urrent orresponding to a sinusoidal voltage exitation Fundamental and third harmoni in the exitation urrent Page 7-5

6 A Simple Magneti Ciruit Consider a simple struture onsisting of a urrent arrying oil of N turns and a magneti ore of mean length l and a ross setional area A as shown in the diagram below. The permeability of the ore material is µ. Assume that the size of the devie and the operation frequeny are suh that the displaement urrent in Maxwell s equations are negligible, and that the permeability of the ore material is very high so that all magneti flux will be onfined within the ore. y Ampere s law, we an write C dl = J da l = Ni S where is the magneti field strength in the ore, and Ni the magnetomotive fore. The magneti flux through the ross setion of the ore an expressed as φ = A where φ is the flux in the ore and the flux density in the ore. The onstitutive equation of the ore material is = µ Therefore, we obtain Ni φ = = l µ A ( ) F R A simple magneti iruit If we take the magneti flux φ as the urrent, the magnetomotive fore F=Ni as the emf of a voltage soure, and R =l /(µ A ) (known as the magneti relutane) as the resistane in the magneti iruit, we have an analog of Ohm s law in eletrial iruit theory. Eletri Ciruit Magneti Ciruit I φ E R F R I = E R φ F = R Page 7-6

7 Magneti Ciruital Laws Consider the magneti iruit in the last setion with an air gap of length l g ut in the middle of a leg as shown in figure (a) in the diagram below. As they ross the air gap, the magneti flux lines bulge outward somewhat as illustrate in figure (b). The effet of the fringing field is to inrease the effetive ross setional area A g of the air gap. y Ampere s law, we an write where and F = Ni = l + l l l g g Aording to Gauss law in magnetis, g g φ l = A l R = = φ µ µ g φg = l A l R g = = φ µ µ o o g g g g we know S da = 0 φ = φ = φ g Therefore, A simple magneti iruit with an air gap ( g) F = R + R φ φ R That is, the above magneti iruit with an air gap is analogous to a series eletri iruit. Further, if we regard l and g l g as the voltage drops aross the relutane of the ore and airgap respetively, the above equation from Ampere s law an be interpreted as an analog to the Kirhhoff s voltage law (KVL) in eletri iruit theory, or R φ = F k k k F Series magneti iruit R g The Kirhhoff s urrent law (KCL) an be derived from the Gauss law in magnetis. Consider a magneti iruit as shown below. When the Gauss law is applied to the T joint in the iruit, we have Page 7-7

8 or in general, 3 φ k k = 1 = 0 n φ k k = 1 = 0 aving derived the Ohm s law, KVL and KCL in magneti iruits, we an solve very omplex magneti iruits by applying these Gaussian surfae g 1 g 2 φ φ 2 i 1 1 φ 3 i 2 N 1 g 3 N 2 basi laws. All eletrial d iruit analysis tehniques, suh as mesh analysis and nodal analysis, an also be applied in magneti iruit analysis. For nonlinear magneti iruits where the nonlinear magnetization urves need to be onsidered, the magneti relutane is a funtion of magneti flux sine the permeability is a funtion of the magneti field strength or flux density. Numerial or graphial methods are required to solve nonlinear problems. µ Magneti iruit of T joints Magneti Ciruit Model of Permanent Magnets Permanent magnets are ommonly used to generate magneti fields for eletromehanial energy onversion in a number of eletromagneti devies, suh as atuators, permanent magnet generators and motors. As mentioned earlier, the harateristis of permanent magnets are desribed by demagnetization urves (the part of hysteresis loop in the seond quadrant). The diagram below depits the demagnetization urve of five permanent magnets. It an be seen that the demagnetization urves of some most ommonly used permanent magnets: Neodymium Iron oron (NdFe), Samarium Cobalt, and Cerami 7 are linear. For the onveniene of analysis, we onsider the magnets with linear demagnetization urves first. Consider a piee of permanent magnet of a uniform ross setional area of A m and a length l m. The demagnetization urve of the magnet is a straight line with a oerive fore of and a remanent flux density of r as shown below. The demagnetization urve an be expressed analytially as m r = + = + ( m ) µ m( m ) where µ m = r / is the permeability of the permanent magnet, whih is very lose to µ o, the permeability of free spae. For a NdFe magnet, µ m =1.05µ o. The magneti voltage drop aross the magnet an be expressed as l m m m lm = lm = φ l = Rφ F µ µ A m m m m m m m m Page 7-8

9 Demagnetization urves of permanent magnets l m A m m m 0 r m φ m l m R m= µ m A m ml m F m= l m Magneti iruit model of a magnet with linear demagnetization urve lm where Rm = is the relutane and F m = l m the magnetomotive fore ( voltage µ mam soure ) of the magnet. It should be noted that in the magnet, m and m are in opposite direts. For a magnet with a nonlinear demagnetization urve, the above magneti iruit model is still valid, exept that the magneti permeability beomes m µ m = m + whih is a funtion of the magneti field in the magnet. Notie that m is a negative value sine it is in the opposite diretion of m. The derivation for the magneti iruit model of a nonlinear magnet is illustrated graphially by the diagram below. l m A m m m 0 m r m φ m l m R m= µ m A m ml m F m= l m Magneti iruit model of a magnet with nonlinear demagnetization urve Page 7-9

10 It should also be understood that the operating point ( m, m ) will not move along the nonlinear demagnetization urve if a small (suh that the magnet will not be demagnetized) periodi external magneti field is applied to the magnet. Instead, the operating point will move along a minor loop or simply a straight line (enter line of the minor loop) as illustrated in the diagram on the right hand side. Indutane ex r m ex m 0 m Movement of operating point of a nonlinear magnet under an external field ex Consider a two oil magneti system as shown below. The magneti flux linkage of the two oils an be express as λ1 = λ11 + λ12 and λ = λ + λ where the first subsript indiates the oil of flux linkage and the seond the oil arrying urrent. y defining the self and mutual indutanes of the two oils as L jk jk = λ (j=1,2 and k=1,2) i k where L jk is the self indutane of the jth oil when j=k, the mutual indutane between the jth oil and the kth oil when j k, and L jk = L kj, the flux linkages an be expressed as λ 1 = L 11 i 1 + L 12 i 2 and λ 2 = L 21 i 1 + L 22 i 2 The above definition is also valid for a n oil system. For a linear magneti system, the above alulation an be performed by swithing on one oil while all other oils are swithed off suh that the magneti iruit analysis an be simplified. This is espeially signifiant for a omplex magneti iruit. For a nonlinear magneti system, however, the indutanes an only be alulated by the above definition with all oils swithed on. g 1 g 2 φ 1 A 1 µ A 2 φ 2 i 1 i 2 φ 1 φ 2 R g1 R g2 A 3 N 1 N 2 φ 3 g 3 R g3 F1 F 2 φ 3 µ (a) Magneti iruit of a two oil system (b) Eletromotive Fore When a ondutor of length l moves in a magneti field of flux density at a speed v, the indued eletromotive fore (emf) an be alulated by Page 7-10

11 e = lv For a oil linking a time varying magneti field, the indued emf an be alulated from the flux linkage of the oil by e k n dλ dλ k kj = = = dt dt n j= 1 j= 1 L kj di j dt (k=1,2, n) Magneti Energy In terms of indutane, the magneti energy stored in an n oil system an be expressed as n n n n n n 1 1 λjk λkj 1 Wf = λ jkij = = Ljkijik 2 2 L 2 j= 1 k = 1 j= 1 k = 1 jk j= 1 k = 1 Exerises 1. Show that the hysteresis energy loss per unit volume per yle due to an AC exitation in an iron ring is equal to the area of the - loop, i.e. d The hysteresis loop for a ertain iron ring is drawn in terms of the flux linkage λ of the exitation oil and the exitation urrent i m to the following sales on the exitation urrent i m axis: 1 m = 500 A on the flux linkage λ axis: 1 m = 100 µwb The area of the hysteresis loop is 50 m 2 and the exitation frequeny is 50 z. Calulate the hysteresis power loss of the ring. Answer: 125 W 2. A oils of 200 turns is wound uniformly over a wooden ring having a mean irumferene of 600 mm and a uniform ross setional area of 500 mm 2. If the urrent through the oil is 4 A, alulate: (a) the magneti field strength, (b) the flux density, and () the total flux Answer: 1333 A/m, T, µwb 3. A mild steel ring having a ross setional area of 500 m 2 and a mean irumferene of 400 mm has a oil of 200 turns wound uniformly around it. Calulate: (a) the relutane of the ring and (b) the urrent required to produe a flux of 800 µwb in the ring. (Given that µ r is about 380). Answer: A/Wb, 6.7 A Page 7-11

12 4. Fig.Q4 shows an iron iruit with a small air gap ut in it. A 6000 turn oil arries a urrent I=20 ma whih sets up a flux within the iron and aross the air gap. If the iron ross setion is m 2, the mean length of flux path in iron is 0.15 m, µ r =800 in iron and air gap length is 0.75 mm, alulate the air gap flux density. It may be assumed that the flux lines flow straight aross the air gap, i.e. air gap ross setion is also mm 2. Answer: 0.16 T 5. A magneti iruit is made of mild steel arranged as in Fig.Q5. The enter limb is wound with 500 turns and has a ross setional area of 800 mm 2. Eah of the outer limbs has a ross setional area of 500 mm 2. The air gap has a length of 1 mm. Calulate the urrent required to set up a flux of 1.3 mwb in the enter limb, assuming no magneti leakage and fringing. The mean lengths of the various magneti paths are shown on the diagram. (Use the given - urve). Answer: 4 A 6. A magneti iruit is made up of steel laminations shaped as in Fig.Q6. The width of the iron is 40 mm and the ore is built up to a depth of 50 mm, of whih 8 perent is taken up by insulation between the laminations. The gap is 2 mm long and the effetive area of the gap is 2500 mm 2. The oil is wound with 800 turns. If the leakage fator (the ratio of the total flux linking the oil over the air gap flux) is 1.2, alulate the magnetizing urrent required to produe a flux of Wb aross the air gap. (Use the given - urve). Answer: 5 A Fig.Q4 Fig.Q5 Fig.Q6 Page 7-12

13 7. It is desired to ahieve a time varying magneti flux density in the air gap of the magneti iruit of Fig.Q7(a) of the form g = 0 + 1sinωt where 0 =0.5 T and 1 =0.25 T. The d field 0 is to be reated by a NdFe permanent magnet, whereas the time varying field is to be reated by a time varying urrent. Assume the permeability of the iron is infinite and neglet the fringing effet. (a) For the air gap dimensions given in Fig.Q7(a), find the magnet length d if the magnet area A m equals the air gap area A g. Fig.Q7(b) gives the demagnetization urve of NdFe permanent magnet. (b) Find the exitation urrent required to ahieve the desired time varying air gap flux density. Answer: 2.64 (mm), i = 5.28sinωt (A) i(t) = d NdFe N turns µ Magnet Area Am g (a) µ Air-gap Area Ag Ag = 500 mm2 g = 4 mm N = 250 turns O (T) r = 0.4π (A/m) (b) Fig.Q7 (a) Magneti iruit of Problem 1, (b) Demagnetization urve of permanent magnet NdFe g 8. Fig.Q8 shows a magneti iruit with air gaps 1 g 2 g1 = g 2 = g 3 = 1 mm and oils N A µ A 1 = 100 turns A and N2 = 200 turns. The ross setional area A of the iruit is 200 mm 2 g 3. Assume the permeability of the ore material approahes infinity and the fringing effet is negligible. µ Calulate: (a) the self and mutual indutanes; (b) the total magneti energy stored in the system, if the urrents in the oils are i1 = i 2 = 1 A; () the mutual indutane between N1 and N 2, if the air gap g 3 is losed. Answer: m, m, m, J, 0 i 1 i 2 N 1 N 2 Fig.Q8 Magneti iruit of Problem 7 Page 7-13