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621.313.013.23 The Institution of Electrical EngineersMonograph No. 333May 1959

THE CALCULATION OF THE MAGNETIC FIELD OF ROTATING MACHINESPart 1.The Field of a Tubular CurrentBy P. HA MM ON D, M.A., Associate Member.

(The paper was firstreceived6th October,1958,a ndinrevised form 2ndFebruary, 1959.in May, 1959.) waspublished as an INSTITUTION MONOGRAPHSUMMARY

An outline is given of a method for calculating the magnetic fieldof rotating ma chines. As a first stage in this work the field of atubular current has been calculated, and curves are given for a particularexample of such a current.

A, An AQ ,LIST OF SYMBOLS

AZ = Vector potentials.oifb2- a2.-nig.Velocity of light.Axial length of end-ring.Spacing between centres of successive tubes.HZ Magnetic field streng ths.Current line densities.Maximum values of current line densities.Modified Bessel function and derivative of modifiedBessel functions of the first kind and order p.Modified Bessel function and derivative of modifiedBessel functions of the second kind and order p.Axial length of tube.Self-inductance.A number.Number of turns.Number of pole pairs of current distribution.Radius,co-ordinate.An odd number.Time.A variable.Length, co-ordinate.An angle.\lb 2r2.

a =a 2 =b =c =Id =g =H, H r, HQ ,in /9, i2 =in k>h Ip, Yp =Kp, Kp =

2/ =L =m =N =p =r =s =t =x =za = =7 = R/r.8 = l\r.e = z\r.6 = An angle, co-ordinate.A = Electrical wavelength.

HQ = Magnetic constant.p = A length.a) = Angular frequency.(1) INTRODUCTIONThe continuing increase in the rating of generating plant isforcing manufacturers to work to extremely close tolerances inthe design of a.c. generators. As a result, the magnetic field ofsuch machines is being more closely examined than ever before.In such a programme of examination and research full-scaleand model tests play an important part, but it is also of greatinterest to derive analytical solutions of the fields of certainidealized current distributions. Such solutions can offer generalguidance to the designer, and they are helpful in opening up

Correspondence on Monographs is invited for consideration with a view topublication.Mr. Hammond is in the Department of Engineering, University of Cambridge.

further avenues for experimental investigation. The paper con-tains the first stage of an analytical investigation carried outin the framework of a general research programme into thebehaviour and design of rotating machines.(2) GENERAL CONSIDERATIONS(2.1) Statement of the Problem

An immense amou nt of work has been done since the end of thelast century in examining the ma gnetic field of rotating m achines,but this work has been almost exclusively concentrated on the2-dimensionalfieldof the slot port ion. This region is undoubted lythe most important part of the machine, and a number ofvaluable mathematical tools have been developed to deal withits field. Conformal transformations, relaxation methods andseries solutions of Laplace's equation have all been used withgood effect: in such methods, the a.c. field is described by ascalar potential function, so that a.c. machines are dealt with byextrapolation from the d.c. case. While this is a perfectly satis-factory approach in the derivation of numerical solutions, it isan approach which turns its back on Maxwell's electromagnetictheory and thus sacrifices much of the un derstandin g of problemsinvolving energy transfer by electromagnetic induction. More-over, the methods which are so successful in dealing with the2-dimensional field of the slot p ortion fail w hen they are appliedto the field of the end-windings. Here a 3-dimensional a pproachis essential, but the writer is not aware of any successful attemptat solving this problem. Generally, the end-windings are dis-missed in some remarks of a qualitative nature. Until recently,such a general discussion may have been satisfactory, but it isquite insufficient if information is sought for the design of end-windings which shall give rise to the smallest possible eddy lossesin the core end-plates and associated metal structures.The difficulties in the path of anybody trying to derive ananalytical solution of the magnetic field of an alternator end-winding are considerable. It is true that all the conductors areusually of the same shape, but this shape is not at all amenableto mathematical treatment. To add to the difficulties there areiron masses of various shapes near the conductors. A closeexamination of turbo-alternator end-windings may well give theimpression that a solution of the problem is impossible byanalytical metho ds. Such an impression is strictly correct, butit is not therefore correct to abandon analytical methods:electrical-engineering practice abounds in impossible problems.The general method of attack is to replace the actual physicalstructure by an idealized one which is amenable to analysis.The practical designer can then use the idealized solution togive him general guidance on the problem he has in hand.

(2.2) Proposed Method of AttackFor the alternator end-winding we propose three simpli-fications, namely

(a) The actual conductor currents are to be replaced by currentsheets.(b) The current sheets are to be of simple geometrical shape.(c) There shall be no iron in the vicinity of the current sheets.158 ]

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HAMMOND: THE CALCULATION OF THE MAGNETIC FIELD OF ROTATING MACHINES: PART 1 159The first simplification does not greatly depart from reality, andis equivalent to the simplification which divides the air-gapflux into a series of harmonic components. The magnetic fieldof a suitable arrangemen t of current sheets can correctly describethe magnetic field of the condu ctor cu rrents, except in the imme-diate vicinity of the conducto rs. Mo re specifically, it will besufficient to use a current sheet with simple harmonic variationof current density along it to describe the fundamental spacecomponent of the magnetic field of the conductor currents.The second simplification is imposed at this stage of the work inorder to avoid complications. It is hoped tha t it will be possibleto remove it gradually by the superposition of certain simplestandard current distributions, when these have been analysed.The third simplification is a drastic one. But before the effectof iron can be included in the problem it is in any case necessaryto calculate the magnetizing field. Thus the solutions here setout are incomplete as they stand. It should be kept in mind,however, that not all coils have iron cores. The solutionsderived in the paper do not give a complete description of thefield of rotating machines: they are only the first stage of such awork, but they stand in their own right as the solution of thefield of a certain current distribution and they are likely to findapplicati on in other branches of electrical engineering. It ishoped to include the effect of iron when the magnetizing field ofvarious current distributions has been fully investigated.Whereas the three simplifications cause the present work todepart to some extent from actual engineering practice, there isone aspect of the treatment which is a decided advance onprevious methods. We propose to describe the magnetic fieldin terms of Maxwell's complete theory by means of a vectorpotential and not a scalar potential. Thus the alternatingcurrents will no longer be treated as modifications of directcurrents. The proposed method enables us to deal with alter-nating currents of any frequency, and there are considerableincidental advantages, which are set out in Section 3.2. Thegeneral subject of energy transfer has been discussed in aprevious paper.1 It is hoped to consider it in more detail at alater date when eddy-current losses are to be considered. It isworth noting here that the existence of eddy currents dependson alternating magnetic fields and that the logical approachmust be through the general equations of electromagnetism.This is the approach adopted here.Having now stated the problem in general terms, we proposeto devote the rest of the paper to the detailed discussion of aparticular example.

(3) THE MAGNETIC FIELD OF A TUBULAR CURRENTDISTRIBUTION(3.1) DescriptionThe general symmetry of rotating machines is cylindrical andwe propose to consider cylindrical current sheets. Fig. l( a)shows such a current sheet in which the current flow is axial andis designated iz. The question of continuity of current flow isdiscussed in Sections 3.2 and 3.3, but at this stage the currentsheet should be regarded merely as a current-element, i.e. as anincomplete portion of a current circuit. It will be noted that theline density ofthecurrent varies around the tube. This variationis supposed to be of the form izsinpd and thus to represent awinding with2ppoles.Consider now Fig. 1(6), which shows a circumferentialcurrent sheet ;ealso arranged in2 ppoles. Here the line densityvaries as fecospd .Fig. l(c) shows the radial component of current, ir. Againthe distribution around the tubular sheet is polar and the densityvaries as trsinpd. The current sheet ir is introduced for the

(a)

b)

(c )

Fig.1.Axial(a),circumferential (b )and radial (c) tubularcurrent sheets.sake of completeness, bu t is no t discussed further. It is clearthat any current distribution can be built up by the super-position of the three types of current-element show n.Fig. 2 shows a combination of axial and circumferential


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160 HAMMOND: THE CALCULATION OF THE MAGNETIC FIELD OF ROTATING MACHINES: PAR1 1

Fig. 2.A tubular current made up of an axial section and twoend-rings.current sheets resembling the current distribution in the rotorof a squirrel-cage induction moto r. The axial current sheetrepresents the slot portion and the circumferential sheets theend-rings of the induction moto r. The problem is to find themagnetic field of this postulated current distribution, and wedo not discuss how such a current could be caused to flow in acertain conducting material. For convenience in calculation thecurren t sheets are assumed to be infinitesimally thin. Theradius of the tubular sheets is R, the axial length of the axialcurrent sheet is 2/ and the axial length of each end-ring is2d .Cylindrical co-ordinates r, 9 and z are chosen, the origin ofco-ordinates being on the axis of the tube half-way along it.The positive direction ofz is from left to right and6is thereforemeasured clockwise in a right-handed system of axes.

(3.2) The Field of the End-RingsIt has been stated in Section 2.1 that we are particularlyanxious to calculate the magnetic field of the end-windings. Inthe tubular current distribution shown in Fig. 2 these end-windings take the form of rings, and it would be very desirableto calculate the field of these rings by themselves. If the currentwere of constant strength in the ^-direction, the magnetic fieldcould be described by means of Legendre functions. Sincethere is a variation along the ^-direction [see Figs. 1(6) and 2],the obvious first choice is to use tesseral harmonics, i.e. Legendre

functions which have a ^-variation. However, it was found th ata solution in terms of these functions was not obtainable. Thisis not surprising, when it is remembered that the basis of such asolution is the replacement of the current distribution by amagnetic-dipole layer on the surface of a sphere. The methoduses a scalar magnetic potential and is strictly applicable toproble ms involving only direct-current flow. But isolated end-rings carrying direct currents of varying strengths in the #-direc-tion are a physical impossibility. The mathematical failurearises directly from this physical impossibility. The currenthas to be fed into the end-rings, and it is impossible to obtainthe field of the end-rings by themselves if the current isunidirectional.. This objection falls away if the current is alternating . Thecontinuity of such a current can be assured by the provision ofadditional finite distributions of electric charge such that thedivergence of current is equal to the rate of decrease of charge.

The magnetic field will depend only on the current, and hence itis possible in principle to calculate the magnetic field of theend-rings quite apart from that of the axial current flow.Similarly, it is possible to calculate the magnetic field of theaxial current flow by itself. A scalar magnetic potential is notapplicable to the a.c. case and a vector potential will have to beused. This has already been mentioned in Section 2.2.(3.3) The Form of the Solution

Because of the cylindrical symmetry of the structure, it isconvenient to use Bessel functions in the analysis. The treatmen tis an extension of that given by Moullin,2who employed Besselfunctions in the calculation of the self-inductance of single-layersolenoids. We first obtain the vector potential A of the currentdistribution and thence by the relationship fx0H= curlA themagnetic field strength H.The charge distributions mentioned in Section 3.2 can bedispensed with if the current flow is made continuous. Thus theend-rings could be placed over the ends of the axial current sheetin such a way that the axial current diminished towards the endswhile the circumferential flow increased. A possible distributionof current would be a constant circumferential current densityover the width Id of the end-rings with a corresponding lineardecrease of axial current flow over the distanceId 2g . This means that the solution isexpressed in terms of modified Bessel functions of argumentsubstantially independent of the frequency (see Appendix).


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HAMMOND: THE CALCULATION OF THE MAGNETIC FIELD OF ROTATING MACHINES: PART 1 161The three co mponents of the m agnetic field have been derivedin the Appendix. Rationalized units have been used. Ifr> R, 4 izR2b2 a . H r = cospo sincot

77 p2 sKp(sbr)lp(sbR) sinsb lco ssbz . (1)

4 izR2b . *~H% = s inpo sincotTT r4 izR2b2

Kp(sbr)l'p(sbR) sinsb lc oss bz . (2)s = lcos/>0 sinoot

sK p(sbr)Ip(sbR) sin$Wsin s&z . (3)If r < R,

4 izR2b2 a .Hr cosposincotTT D2 sl'(sbr)K'(sbR) sin56/coss&z . (4)

0 -7

0- 6

0-4H

0- 1

H ro COS 6 SIN i

H e o C -

HZ oc-cos e s^ ^

SIN 6 SIN wtI

0-25 0-5r/R 0 7 5

Fig. 5.Magnetic fields for 2-pole tubular current. a) r > R. 6) r < i?.

4 i ^TT r

4 /H2 =

s\npQ s'xncot00

2 Ip(sbr)Kp(sbR)sinsblcossbz .cos p6sin cot

5 )

2 sIp(sbr)K'(sbR)sin sWsins&z . (6)5 = 1where iz is the maximum line density of axial current, b = ir/g,co is the angular frequency, and s is restricted to odd numbers.It will be noted that the series are convergent everywhere exceptat r = R. HereHr and Hz oscillate finitely except at z = + /,where Hz diverges to infinity. This lack of convergency is aresult of making Id < I and will disappear if this simplificationis abandoned.The above series have been used to calculate the field oftubular currents for which I = 2R and g = 4(/ + R) = 12R ,and p = 1 and 2. The results are shown in Figs. 5 and 6 andmerit close study. Of particular interest is the remarkab ledifference in the internal field between the 2-pole case (p = 1)shown in Fig. 5 and the 4-pole case (p = 2) shown in Fig. 6.It appears thatHr and ffBvary as r 1, whereasHzvaries as / .This follows from the behaviour of the Bessel function Jpnearthe origin of co-ordinates. Thus the stator iron-loss of turbo -alternators is likely to be greater than that of multi-polarmachines because a greater amount of iron surface, particularlyat the end-plates, lies in a strong magnetic field. It will be notedthat Figs. 5 and 6 are given in non-dimensional form.

0 1

\ V - H r COS\ \vH z a COS 2G S IN^ t -^

2G SINoit

SIN 28 SINwt

= = = = =1-5 r / R 2-5

Fig. 6.Magnetic field for 4-pole tubular current.a r> R. b) r

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162 HAMMOND: THE CALCULATION OF THE MAGNETIC FIELD OF ROTATING MACHINES: PART 1It is very important to realize explicitly that the order of theBessel functions is linked to the order of the circular functions.Thuslp, Yp, Kpand Kp' are linked to sin pdand cospd. It is clearthat a change in the number of pole pairs must produce a differentpolar distribution of the magnetic field in the ^-direction, but itis not immediately obvious whether there would be a corre-sponding change in the /--direction. This question is answeredwhen the relationship between Bessel functions and circularfunctions is taken into account. Thus formulae (2)-(6) enable oneto see at aglancehow the num ber of pole pairs affects the magneticfield in both the6-and r-directions.

(3.4) The Limiting Case of an Isolated Tubular CurrentIt was stated in Section 3.3 that the solution obtained appliesstrictly to an infinite succession of tubu lar currents o n a com monaxis. In order to obtain the solution for a single isolated tube,we must makeg very large in comparison with the dimensionsof the tube. In the limit , the infinite series will become infiniteintegrals and we shall have expressions of the type

4 r

cospd sin cot I j3xKp x)Ip yx)s in8 xcosexdxSuch infinite integrals are likely to be exceedingly cumbersomein computation and it is highly desirable to retain the simplicityof the series solutions, especially since the series are rapidlyconvergent, ifg is no t very large. The question arises as to thechoice of g. In effect we must choose g so that the mutualinductance of adjacent tubes is negligible. This provides auseful check, since both the series solution and the integralsolution for the self-inductance of a single-layer solenoid areavailable.2 Fig. 7 shows the self-inductance of solenoids for

0 4

0- 3 0-412

0-268

0-156

0-1

Fig. 7.Self-inductance of single-layer solenoids.whichI = 2R, I = R,and / = $R . It will be seen that the seriessolution in each case tends rapidly to the asymptotic valueobtained by means of an infinite integral. If / = 2R , it is seenthat, even ifg has the smallest possible value ofg = 2/, i.e. ifthere is no gap between successive solenoids, the self-inductanceis already within 18 % of that of an isolated solenoid. Withour value of g = 4(JR -f 0 there is no difference between theseries and the integral solutions. With very short tubes of large

diameter the approach to the asymptotic value is not so rapid,but it is clear that with the choice of a suitable g the seriessolution can be employed without appreciable loss of accuracy.(4) ACKNOWLEDGMENTS

The author wishes to acknowledge the encouragement givento him by Sir Willis Jackson, Director of Research of theMetropolitan-Vickers Electrical Co ., Ltd. Thanks are alsodue to Mr. S. Neville of that Company for many stimulatingdiscussions and to Professor E. B. Moullin for advice throu ghou tthe course of the work.(5) REFERENCES(1) HAMMOND, P.: 'Electromagnetic Energy Transfer', Pro-ceedings I.E.E., Monograph No. 286, February, 195&105C, p . 352).(2) MOULLIN, E. B.: 'The Use of Bessel Functions for Cal-culating the Self-Inductance of Single-Layer Solenoids',ibid., 1949, 96, Part III , p. 133.

(3) MOULLIN, E. B.: 'Radio Aerials' (Clarendon Press, 1949),p. 42 .(4) WATSON, G. N .: 'Theo ry of Bessel Fun ction s' (UniversityPress,C ambridge, 1944), p. 361.(5) MOULLIN, E. B.: 'Radio Aerials' (Clarendon Press, 1949),p. 44.(6) APPENDIX

(6.1) The Vector Potential of a Tubular Current(6.1.1) The Vector Potential of an Axial Current of Line Densityiz sinp a. sino it distributed around the Surface of a Tubeof Radius R.

Refer to Fig. 8 and consider the vector potential at P due to afilament of curr ent at Q. Th e filament is assum ed to be infinitely

p P(r ,e)Fig. 8.Relating to the vector potential of a tubular current.

long and the magnitude of the current varies as cosbzalong thefilament. Then it can be shown3 that= K 0 ( a p ) s i n p a d x . . . 7 )jiio?z/? cosbz sincot

wherep is the distance QP an dwhere a 1 = b2 a2 -

00But Ko (a ' = KJa r)lm(a R) cosm(a - 0) (Ref. 4) (8>m oo

Hence the vector potential Az at P due to the tubular current is.given by2TTA f2irr+ oo ^. = 2 KJa r)lJa R)(cos macos mOIAOi2Rco sbz sincot Jo |_-Ti

+ sinma sin mQ) sinpccda = 2TTKP(CI r) lp(a R) smpd . (9>Hence Az = fjboizRK p(a r)Ip(a R) sin/70 cosbz sin cot . (10)


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HAMMOND: THE CALCULATION OF THE MAGNETIC FIELD OF ROTATING MACHINES: PART 1This is the result for r > R. Ifr R4 . _ . . J2 1

(6.1.3) The Divergence of the Vector Potential.. . i If the current flow is continuous there need be no electric-Az = ~WZR sinpd sin cot -Kp(sbr)Ip(sbR) sinsblcossbz (13) charge distributions; hence the vector potential must have nos~ divergence. The calculation of this divergence thus serves as aand ifrP~\(JLQIQR sin bz sin cot

= *K0(a /o){sin[O + l)a - 0] - sin[(p- l)a - d]}d*= %K0(a p)[sin(p + l)a cos 0 cos(p + l)a sin 0

sin(p l)a cos 0 cos(p l)a sind]da (15)Also

+ 00K0(a p) = 2 Km(a r)I/n(a i?)(cos ma cos m0 + sin ma sin m0)YYl . . . . (16) Also, ifd < g, sinsbd-> s6d

But in order to make the current continuous,i [ izsinpdRdd = 2i&dJo

Therefore -i = 2/W .

J24)

(25)

(26)

Hence-2Ar

= Kp+x (a r)Ip+ x(a R)[cos(p + 1)0sin 0:sin cot- sin(p + 1)0 cos 0] -t- Kp_x(a r)Ip_x(a R)[cos(p - 1)0 sin 0 + sin(p - 1)0 cos 0]

Henceand

. sin{sbd)p-0 sin 6z sin a>/ . (18)

lp+i(a r)Kp+x(a R)] sinpd sin 6z sinc ot . (19)+ I /,+i(a'V)Kp+1 (a i?)] cos/70s in6zsincuf . (20)

(6.2) The Magnetic Field of a Tubular Current(6.2.1) The Axial Component Hz.

27)Consider first an infinitely long tube of current varying as sin bz.Then, for r> R,- fjL0Hz = i/ xo^_ cos/70 sinbz sin co/{[K/,_1(a r)Ip_,(a /?)

+ K^+^a'V^+^a /?)] + a r[K'p_x(a r)\p_x(a R)K'p+i(a r)Ip+x(a R)} ]} 28)


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164 HAMMOND: THE CALCULATION O FTHE MAGNETIC FIELD OFROTATING MAC HINES: PART 1Hence By themethod given inSection 6.2.1, for r>R,- Hz= /e / ? c o s / > 0 s i n 6 z s i n w / [ - a rKp(a r)Ip_ l(a R) AizR2b . Q . v . ..._ a rK(a r)fp+\a R] . (29) ^ e = i y ~ sin/>0sina>/2 Kp(sbr)\p(sbR) sin56/ cossZ>z (2)

H: =iiBf cospd sin bz sinwt2a Kp(a r)IpXa K) . (30) a n d i f r < 7 ? ,4 i?26 If a ~ bandinstead of/esin 6zwe have H=- - sinpd sincot2 Jp(sbr)Kp(sbR) sin56/ cos s6z (5)77 r 5 = 18 ^ 1- re 2 -sin 56/ sin jfo/ sin $6z,andif79= izR\lpd, (6.2.4) The Magnetic Fieldat theSurface of the Tubular Current.A useful check onthe accuracy ofthe expressions for the4 IJPb co$pd s. nw / Ktsbr)usbK) s i n j A / s i nj W mag netic field isgiven by examining whether the magnetic field77 pd s=\ gives thecorrect current distribution byusing theequat ionsin sbz . (31)

T f . . . , , curl^T= / (34)If sin sbd -> sbd, _ Fr om Section 6.2.1.wehave,atrR,4/R b cos/>#sinco t2 sKJsbr)Jp(sbR) sin56/ sin ^6z 4* D2/,i = ' ( 3 ) Hzomide - H2inside = - - cos/>0sin3 ) ^IfrR, From Section 6.2.2 wehave,at /=R,

Hr= - co spdsincot2 sKXsbr)J'(sbR) sinsblcos sbz tr . .. _ / / ., nS'kTT p s = l JJr inside flr outside w / j \ asitshould.From Section 6.2.3 wehave, atr=i?,and if/

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621.313.32281.013.2 The Institution of Electrical EngineersPaper No. 3489SMar. 1961

THE C LCUL TION OF THEM GNETIC FIELD OFROT TING M CHINESp a r t 2.TheField ofTurbo-Generator End-Windings

By D. S. A S H W O R T H , B.A., and P. H A M M O N D , M.A., Member.(Thepaper wasirstreceived 19///May, and in revised form \4th October, 1960. // waspublished in March,1961,andwasread before the

SUPPLY SECTION 12/// April,1961.)SUMMARY

The magnetic field of turbo-generator end-windings is analysed.Particular attention is given to the field at the surface of the stator-core end-plate. The effects of varying the cone angle of the end-winding, the chording angleand the length of the axial portion ofthe end-winding are discussed. The effect of power factor is con-sidered with particular reference to the large fieldsobservedat leadingpower factors. Experimental results on a model winding showreasonable agreement with the computed results.

LIST OF SYMBOLSAr AotA2

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528 ASHWORTH ANDHAMMOND: THE CALCULATION OF THE MAGNETIC FIELD OF ROTATING MACHINESpropose, therefore, to proceed more cautiously by consideringfirst the fieldofthe w indings in air.We shall deal first with the stator w inding, then w ith the roto rwinding and finally with the fielddue to both stator and rotorwindings acting together.

(3) STATOR END-WINDINGS(3.1) Description of Winding

The stator windingsof modern turbo-generators are madeupfrom a large number of identical coils. These coils fit intoaxial slotsin an iron coreand thereare asmany coilsasthereare slots. Thus there are two conductors, or coil sides,in eachslot and the winding is described as a 2-layer winding. Thestraight slot portion ofthe coils isknownas the active length,and it is here that the electromotive force is induced. Thewindingis arranged in pole groups and thereare generallytwosuch groups in a turbo-generator winding. The end-windingsconvey the current from each slot to another, whichis approxi-mately one pole pitch away. It is usually desirable to makethe coil pitch slightly shorter than a pole pitch. This iscalled'short-pitching' or 'short-chording' the coil, and reference ismade to thechording angle, which willbe somewhat less thana pole pitch.

The end-windings are shaped to lie on the surface of atruncated cone. A complete winding isshown in Fig. 1. The

Fig. 1.Stator end-winding.track of each conductor on the surface of the cone must besuch thatit isat aconstant distance from the track of an adjacentconductor. This is achieved by following an involute of acircle along the cone surface. Fig. 2 shows two views of atypical coil end; (a)is an axial view and (6)a developed viewonthe surface ofthe cone.

(3.2) Simplified Representation of the WindingThe complicated geometry of the individual coils makes itwell-nigh impossible to describe them analytically. We there-fore represent the complete winding by a simplified equivalentcurrent distribution . Fig. 3 (a) shows sucha simplified arrange-ment. The current is supposed to be distributed along thin'current sheets', which liealong the centre-lines of the actualconductors. We thus have a cylindrical current sheet repre-senting the slot portionand a conical current sheet representingthe end-winding. The radius,p,of the cylinder, the length,L,

Fig. 2.Typicalinvolute coil-end.(o) Axial section, (b)Development on coneX.

SURFACE OFEND-PLATE

D 1 2 34

tt

(C )

Fig. 3.Simplified representation of end-winding.of the cone and thecone semi-angle, j8,canbe found fromthedimensions of the actual machine.The equivalent current on the surface of the current sheetsmust everywhere be the sum of the actual currents in the twolayers of the winding. The equivalent curren t will vary withthe magnitude of the actual current and with the chording


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ASHWORTH AND HAMMOND: THE CALCULATION OF THE MAGNETIC FIELD OF ROTATING MACHINES 529angle, 2/x. The length,L , of the cone will be related topand j8.L has been assumed proportional to p, and variation in p canthus be treated as a mere scale effect. The independe nt variablesare the cone angle, 2j8, and the chording angle, 2fi. Thedependence of the magnetic field on these two variables is dis-cussed in Section 5.The stator end-winding is thus represented by current sheetsin the shape of a cylinder and a cone. Since the use of conicalco-ordinates is difficult we make the further simplification ofreplacing th e 'con ical' current sheet by a series of steps as shownin Fig. 3(6). On the conical curren t sheet the current will flowboth along the generators of the cone and perpendicularly tothem. In the equivalent representation of Fig. 3(6) we supposethat the current flows axially and circumferentially on thecylindrical sheets and that it flows radially on the annulardiscs.

The conical current sheet of Fig. 3(a) can thus be replacedby a stepped current sheet as in Fig. 3(6). The steps can becombined into a num ber of tubes and discs as shown in Fig. 3(c).It will be seen tha t th e steps numbered 1, 2, 3 and 4 in Fig. 3(6)have been combined into a single disc and a single tube inFig. 3(c). This involves the expansion of steps1and 2 and thecontraction of steps 3 and 4. The errors introduced by thesechanges will tend to cancel out.In Fig. 3(c) a final representation of three tubes and discswas chosen to account for the conical end-winding. Clearly itwould be possible to improve the accuracy of the representationby increasing the number of tubes and discs. The results of thispaper were obtained by using three tubes and three discs. Inone case, however, this result was compared with that obtainedby using six tubes and six discs. The maximum difference inthe field calculated was only 2 %. This amply justifies the useof the simpler representation by three tubes and discs.The magnitude of the current varies along the cone. Thisvariation depends on the cone dimensions and on the chordingangle of the winding. A method of calculating the current atany point on the cone is given in Section 11.1. The currentsin the equivalent tubes and discs of Fig. 3(c) are taken as thecomponents of current on the cone at the mid-points of thetubes and discs.It should be noted that the current sheets represent the currentin all three phases of the actual stator winding. By means ofFourier series it can be shown that the equivalent current can beexpressed as a series of rotating harmonics of which the funda-mental is strongly dominant. This equivalent current can bedescribed by the expression i= icos(cot pa), where i is theline density of current and a is the angle measured around thestator.The three components of current flow are represented inFig. 4. In order to show the angular variation around thestator a 6-pole arrangement is shown. Turbo-generators have,of course, two poles in general, but the treatment is applicableto any number of poles.

(4) METHOD OF CALCULATING MAGNETIC FIELD(4.1) Use of the Vector Potential

We approach the calculation of the magnetic field by firstfinding the vector potential A of the current distribution. Thevector potential of a current-element is given by the relationship8A = juo/8//r, and it will be seen that the vector potential isparallel to the current from which it is derived. If the currenthas a simple geometry, so has the vector potential; for instance,an axial current gives rise to an axial vector potential. It isthus relatively easy to integrate 8A and the magnetic field canthen be obtained from the relationship curl A =fx0H.

Fig. 4.Cylindrical current sheets.a) Axial current sheet, (b ) Circumferential current sheet,c) Radial current sheet.

An alternative approach would be to use the Biot-Savart law8H = Ibl XrjArrr3without mentioning the vector po tential.However, the resulting integration is likely to be very trouble-some, because the m agnetic field is at right angles to th e current-element and to the radius vector drawn from the current-elementto the field point.There is another advantage in the use of the vector potentialin that the method can be applied to high-frequency problemsby using a delayed current. The Biot-Savart law, on the otherhand, while quite satisfactory at power frequencies, is inapplic-able to the calculation of magnetic fields at high frequencies.Thus the method adopted by us is of very general applicationand does not apply only to the problem of turbo-generatorsdiscussed in this paper.(4.2) Method of Computation

The method is set out in detail in Part 1 of the work.1Cylindrical co-ordinates are used and the three components ofthe magnetic field are expressed as infinite series of products ofmodified Bessel functions and circular functions. The relevantexpressions are derived in Section 11.2. In Par t 1 the com-putations were carried out by hand, but the present work wasdone by means of an electronic digital computer.Examination of the expressions for the magnetic field revealsthat the variation of field with d is sinusoidal in the same wayas the postulated current distribution. This must be so becausethe system is linear. Thus harmon ics in the magnetic field mustbe due to harmonics in the current distribution. If harmonicfields are likely to be of interest, they can be obtained by thesuperposition of a harmonic current on the fundamental com-ponent of current.The expressions for the magnetic field also show that themagnitude of the field is unchanged by altering the scale of thelinear dimensions as long as the current line density (ampere


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530 ASHWORTH AND HAMMOND: THE CALCULATION OF THE MAGNETIC FIELD OF ROTATING MACHINESconductors pe r metre) isunchanged. For a 2-layer winding wecan describe the line density in a single layer by an expressioni -fscos (cot pot). The value of ts will be determined bytheactual conductor current an d willbeindependent of chordingangle. The results given in Figs. 5-10 are calculated for ts = 1

0- 4

0- 3

0-2

0-1

-0-1

-0Q.

- 0 3

/

A

A\\\ \\V

A

A

0-4 0-8 1-2 1-6 2-0r/pFig. 5.Magneticfieldof end-winding only.

and are plo tted against a dimensionless co-ordinate (i.e. the ratioof two radii). These curves are thus applicable to machines ofany size. To obtain the magneticfieldfor any particular machineit is merely necessary that the field obtained from the curvesbe multiplied by the actual value of f,.(5) COMPUTED RESULTS FOR STATOR WINDING

In Part 1 of the work the field of a simplified winding wasexamined for both a 2-pole and a 4-pole machine. It wasshown that the magnetic field of a 2-pole machine of a certaincurrent loading is always greater than the field of a similarmachine with more poles. This result can be seen intuitivelysince the fields of alternate poles will tend to cancel out at adistance. Mathematically the result follows from the behaviourof the Bessel functions. The orde r of the functions is relatedto the number of poles. The fact tha t a 2-pole field is strongerthan a multi-pole one is in fact well known in practice . Straylosses are far more troublesome in steam-turbine-driven gene-rators than in water-turbine-driven generators, which run moreslowly and have a larger number of poles.Our computations have been confined to the case of 2-polemachines and only the fundamental of field considered. Itwas found to be convenient to use a machine rated at6 5MW.The method is, of course, competent to deal with any number ofpoles and with machines of any rating and dimensions.

(5.1) Magnetic Field at Surface of Core End-Plate due toStator End-Winding in AirThe stator end-winding is represented by tubes and discs asshown in Fig. 3(0- The tubes and discs at A, B and C repre-sent the conical portion of the end-winding. The tube at Drepresents the cylindrical portio n. By use of the formulae ofSection 11.2 the components of the field due to each tube anddisc are found at any radius. The total field is then obtained bysuperposition.

The values to be substituted in the formulae are the linecurrent densities( ir, i$, i2),the radii of the tubes, which are alsothe mean radii of the discs,R, the annular widths of the discs,AT?, the lengths of the tubes, 21,and the distances, z, of thecore end-plate from the mid-points of the tubes. The computerprogramme was arranged to accept these values and to sum thefields of the individual tubes and discs, thus giving the field ofthe complete winding.All these values can be calculated from the machine para-meters by use of Section 11.1. The parameters required, andtheir values for the particular machine chosen, are shown inTable 1.

Table 1PARAMETERS OF STATOR WINDING OF 6-5MW GENERATOR

2fi = 150L = 0-302mp = 0-442mft= 30Z = 0-llm[see Fig.3(o)].For this machine the number of pole pairs, p, is 1. Asalready explained in Section 4.2, ts was taken as unity. Theconstant, g, which determines the convergence of the infiniteseries (see Section 3.4 of Part 1) was taken as g 5m. Theseparameters result in Table 2.

Table 2EQUIVALENT CURRENT SHEETS FOR STATOR END-WINDING OF6-5MW GENERATORTube or disc

ABcD

iz = - irA/m0-8400-5200-1680 966

'6A/ m1-021-752-160

Rm0-4670-5180-5680-442

m0050400504005040

/m0-04370-04370043700550

zm0-1570-2410-3290 055

The resulting magnetic fields are plotted in Fig. 5. The stepin i/y at r/p = 1 must be proportio nal to th e axial current atthat place and affords a useful check. Anothe r check is thatH r andH must be numerically equal on the axis of the tubularcurrent sheets.For the 6-5MW machine considered here the air-gap is atr/p =% and the stator end-shield at rip 2- 1. The full-loadconductor current is 806 amp (r.m.s.). There are 36 slots and4 conductors per slot. Thus the current line density per layerisis = 5-7 x 104A/m . The values of Hn Hand Hz in Fig. 5must therefore be multiplied by 5-7 x 104 to give the end-winding field at full-load cu rrent. The multiplying constant ofa modern hydrogen-cooled machine might well be about threetimes this figure.

(5.2) Magnetic Field due to Stator End-Winding and SlotPortion in AirThe stator winding can be divided into a slot-portion withan end-winding at each end. It is reasonable to ignore thedistant end-winding when we are computing the field on a planecorresponding to the surface of the core end-plate. We must,however, consider the contribution of the slot portio n. Thissection of the winding can be represented by a tubular currentsheet carrying axial current. In our example of a 6-5 M Wgenerator the values for computing the field of a slot portion ofaxial length, 21,of one metre are as shown in Table 3.


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A SH W O RTH AND H A M M O N D : TH E CALCULATION O F T H E MAGNETIC FIELD O F ROTATING MACHINES 531Table 3

E QUI VAL E NT C UR R E NT SHE E T FO R ST AT OR SL OT POR T I ON O F6 - 5 M W GENERATOR

/ - - fr

A/ m0-966fe

A/m0Rm0-442

&Rm0

/m0- 5

zm- 0 - 5

The distance z is negative because the slot portion lies on theother side of the plane of the core end-plate. It was found th atthe magnitude of the fieldwas very insensitive to a change of axiallength oftheslot portio n, as long as this length was considerable.In particular a length of 2 m gave the same field as a lengthof l m . This shows that it must be substantially correct toneglect the d istant end-winding.The total field at the end-plate due to the complete statorwinding in air can thus be found by adding the fieldof a1m-longtube to the field of a single end-winding. This total field hasbeen plotted in Fig. 6. It is of interest to note that HT is the

0 -6

0-5

0-4

0- 3

0-2

0 1

-0-1

-0-2

- 0 3

-0-4

/

.

/

JV

\

V ,

\vV\ VvIvV

0- 4 0-8 1-2 1-6 2 0

Fig. 6.Magneticfieldof end-winding and slot portion.same as in Fig. 5, since the axial current of the slot-portioncannot give rise to axial field.

(5.3) Variation of Magnetic Field with Cone Angle ofStator End-WindingThe variation of field for a range of cone semi-angle, j8, from10 to 90 is shown in Figs. 7-9. The machine parameters,including cone length, L, and the length, Z, remain constant.The length of the straight portion of conductor was taken as1m within the core. It should be noted tha t the curves for a

0-6

0- 5

0- 4

/ '/, /

11 t

\

1\\\ \\;NL V\ \

0 0-4 0-8 V2 1-6 2-0r pFig. 7.fir due tocomp lete winding at different coneangles.

0- 6

0-5

0-4

0-3

0 2

0-1

-0-1

-0-2

-0-3 -

- 0 - 4

O 0 4 0-8 1-2 1-6 20

Fig.8.j?edue to complete winding at different cone angles.cone semi-angle of 90 have been obtained by extrapolation.The method of calculation adopted in the paper cannot beapplied to this cone angle without modification since fe wouldbe infinite.Figs. 7-9 throw an interesting light on the variation of strayloss with cone angle. The loss will depend on the magnitude ofthe field and on th e surface area of the end-plate. The area ata particular radius is proportional to the radius, and hence thefield components at large radii will contribute m ore to thelosses than those at small radii. The inner radius of the core

^ 7 030*10*

\

9 0 *

\


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532 ASHW ORTH AND HA MM OND : THE CALCULATION OF THE MAGNETIC FIELD OF ROTATING MACHINES0 2

0-3

0 - 2

0- 1

-0-1

- 0 2

-

4

-

i

ft*''/ >

^ \\i I

^ .1 10- 4 0 - 8 1-2 V 8 2 0

Fig.9.Hsat different cone angles.end-plate is at about rip = . The mass of the core end-plateis situated at r/p > 1 2. At such radii HQis substantiallyindependent of cone angle, but both Hr and Hz increase withincrease of cone angle. Furth ermo re, the maxima ofH r and Hzoccur at larger radii as the cone angle is increased. Thus agreater cone angle would be expected togivegreater losses, whichis in accordance with test results . (But see also Section 5.5.)

(5.4) Variation of Magnetic Field with Chording Angle ofStator WindingThe line current density due to each layer of the stato r winding

is if, cos(cot a). Hence the line density due to both layersis tssinp\i cos(cot a) , the coil pitch being 2/Lt. The varia-tion of the magnetic field with \i will depend on the magnitudeof sinJU and also on the variation withfj,oftn ?ea nd tz. It wasfound that the field varied dominantly as sin fi. Thus for anyparticular value of the current line density the field is inde-pendent of chording angle. This is an importa nt conclusionbecause it eliminates one of the two variables in the design of aconical end-winding.(5.5) Variation of Magnetic Field with Distance of InclinedPart of End-Winding from Core End-PlateFig. 10 shows the variation of magnetic field,Hz, as the para-meterZ is changed, i.e. as th e axial length of the end-winding ischanged. He and Hr are practically unchanged, because theyinclude the effect of a considerable length of slot current. H2is entirely due to the inclined portion of the end-winding and isconsiderably reduced as this portion is moved away from thecore end-plate.It should be noted that in Section 5.3 we considered a changeof cone angle with constant Z. In a practical design, however,Z will have to be longer for lower values of cone angle in orderto maintain insulation clearances. Thus the curves ofFigs. 7-9and Fig. 10 should be considered together. The decrease instray loss observed at smaller cone angles will be partly due tothe larger values of Z.

(6) ROTOR WINDING(6.1) Description of Rotor Winding

The rotor winding of a typical turbo-generator is shown inFig. 11. The winding consists of an axial slot portion of straight

E 0- 1

-0-1

-

/ ^

I

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ASHWORTH AND H A M M O N D : THE CALCULATION OF THE MAGNETIC FIELD OF ROTATING MACHINES 533andThe rotor winding can thus be represented by a single tube ofaxial current for the slot po rtion and a number of tubes for the

xSJ

POLE CENTRE-LINE

Fig. 12.Development of rotor windingon the surface of acylinder.Top.A ctual winding. Bottom.Sup posed current distribution.

-PLANE OF FIELD PLOT

A

z

a

zL_-

c

Fig. 13.Simplified representation'of rotor winding.end sections. We use two such tubes, shown in Fig. 13 as tubesA and B. The magni tude of current in these tubes is taken asthaton the rotor at their mid-points.

(6.3) Computed Results for Rotor WindingF o r the generator considered in the paper the relevant ro tordimensions are as shown in Table4 .

Table 4PARAMETERS OF R O T O R W I N D I N G S OF 6 5 M W GENERATOR

6= l - 2 r a dPR = 0-242mLR =0-222mZR = - 0 0 1 5 m(see Figs. 12 and 13).

Table 5EQUIVALENT CURRENT SHEETS FOR R O T O R W I N D I N G OF 65 MW

GE NE R AT OR

ABC

rA/m

0 3180-8401 000

aA/m1 430-9260

m

0-2420-2420-242

m000

/m

005500550-500

m01500040

0 515

An axial length of 1 metre has been taken, as for the statorwinding, in order to represent the slot portion of the winding.

The resulting values for the magnetic field of the rotor areshown in Fig. 14. In order to facilitate comparison of statorand rotor fields the magnetic fields in Fig. 14 are plotted againstrip, wherep is the radius of the stator current sheet. It will be

0 - 8

0 - 6

0 - 4

0- 2

-O-2

-0-4 . s-

/

j/v/ A\

Hr

k**. mm

0-4 0 - 8 2 - 0

Fig. 14.Field of rotor winding.

0 - 3

0 - 2

0 -1

-0 -1

- 0 - 2

- 0 - 3

//

//

\

///

i

\

0 -4 0 - 8 r p 1-2 1-6 2- 0The negative sign inZR indicates that the rotor end-winding F i g . i5 ._ V a r i a t i on ofHz with distanceof rotor end-winding fromprojects inside the stat or core end- plate. These dimen sions core end-plate,give values for the tubes A, B and C of Fig. 13 as shown inTable 5. z - o - r a s m .Z = -0015m.= OO35m.= 0085 m.


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534 ASHWORTH AND HAMMOND: THE CALCULATION OF THE MAGNETIC FIELD OF ROTATING MACHINESnoted that there is a discontinuity in Hz as well as in HQ. Thisarises from the fact that the plane of the field plot intersects therotor winding where it carries not only axial but also circum-ferential current.The machine considered here has a rotor current of 265ampat full load and rated power factor. There are 168 turns per pole.This corresponds to a fundamental current line density, iR) of1-8 X 105A/m . The ordinates in Fig. 14 must therefore bemultiplied by this number to give the full-load field of the roto r.(6.3.1) Variation of Magnetic FieldwithDistance of Rotor End-Windingfrom Core End-Plate.

H r and H$ are practically unchanged with variation of Z.This is to be expected, because they both depend on the longslot portion. Hz is, however, considerably reduced as Z isincreased. This is shown in Fig. 15.(7) ADDITION OF STATOR AND ROTOR FIELDS ATVARIOUS POWER FACTORS

The magnetic-field components of the stator winding shownin Fig. 6 can be combined with the magnetic field of the rotorwinding shown in Fig. 14. It is of particular interest to examinethe resultant field at various power factors, since turbo-generators have particularly high stray losses at leading powerfactors. These losses can cause serious overheating.For a machine operating at constant apparent power it iseasy to find the stator and rotor fields at any power factor andto determine the phase angle between these field s. The resultantfield can then be obtained by vector addition. This was donefor the generator considered in the paper operating at its fullrating of 9-2M VA . The results are plotted in Fig. 16.It will be noted from Fig. 16 that at small radii a laggingpower factor results in a stronger field. But it is the field atlarger radii that is of particular interest, because it is thiswhich causes the eddy-current loss in the core end-plate. Over

most of th e region of the end-plate (i.e.rjp> 0 8 it is a leadingpower factor which gives the stronger field. This is consistentwith test results. Fig. 17 shows the effect of power factor on thecomponents of field in the region of the bottom of the statorslots (i.e. r/p = I-2). These results should be compared withFig. 6 of Reference 5, which shows the variation of temperaturerise with power factor. This experimental curve is of the sameshape as the curves ofHrand ife in Fig. 17. In seeking to reducestray losses, designers should therefore pay particular attention

x1O

(b )

Fig. 16.Field due to combined stator and roto r windings at different power factors and constant apparent power.(a) Radial component, (b) Circumferential com ponen t, (c) Axial com ponen t.


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ASHWORTH AND H A M M O N D : THE CALCULATION OF THE MAGNETIC FFELD OF ROTATING MACHINES 535x1 0

ao

16

12 .

y

/ \ \ .

i

^

^ . . Hz

AH P

AHe

0- 4 0-8 10 0-8 0-4L A G L E A DPOWER FACTOR

xiO108

-. 1 w

AH r -AHeAH z

0- 4LA G 0 8 K ) 0-8POWER FACTOR(6)0-4LEAD

Fig. 17.Variation of total field with power factor. A ta r a d i u s c o r r e s p o n d i n g t ot h e b o t t o m o ft h e s t a t o r s l o t s : r/ p = 1 2 . b) A ta l a r g e r a d i u s : r / p = 1 - 8 .

to areduction ofHrand H$. Theaxial component, Hz, seemsto be ofless importance atleading power factors.It is at first sight strange that thefield should bestronger atleading power factors, when the rotor current issmall. It must,however, beborne inmind that theresultant field ofFig. 17 isobtained by vector additionofthe roto r fieldand thestator field.The angle between the vectorsissuch tha t this results effectivelyin an arithmetical subtraction. Moreover the radial field, Hnmust beconstant at the air-gap radius because themachine isoperating at constant apparent power. This explains thecrossing-overof thecurves in Fig. 16, sothat the field beyondthe air-gap radius changes in the opposite manner to thatwithinit.

(8) EXPERIMENTAL RESULTS WITH MODEL WINDINGA modelof thestator windingof the65MW machine usedin these calculations wasconstructed to a scale of 1 : 3. Asmall 3-dimensional search coilwasused, connected to avalvevoltmeter to explore thefieldof themodel winding. Thefieldcomponents were measured on a plane corresponding to thesurface of the core end-plate. These measured values areplotted inFig.18 and forcomparison the calculated valuesarealso given. It willbe seen that there is reasonable agreement.The gapinthe measured curves is duetothe fact that the windinginterfered withthe search coil inthis region.

(9) ACKNOWLEDGMENTSThe authors wish to acknowledge the encouragement givento them by SirWillis Jackson, DirectorofResearchand Educa-tion, A.E.I. (Manchester) Ltd. Their thanks are also due to

Fig. 18.Comparisonofcalculatedandmeasured valuesofstator field.Measured.Calculated.

Mr. S. Neville,Mr. E. W.Consterdineand especiallyto Mrs.M.Phillips, whodeveloped thecomputer programme,all ofwhomare with A.E.I. (Manchester)Ltd.(10) REFERENCES(1) HAMMOND, P.: TheCalculation of the Magnetic Field ofRotating Machines. Part 1.The Field of a TubularCurrent', Proceedings I.E.E., Monograph No. 333, May,1959 (106 C , p. 158).(2) HAMMOND,P .: 'Electricand Magnetic Images',ibid.. Mono-graph No .379, May,1960 (107 C,p. 306).

(3) SMITH, R. T.: End Component of Armature LeakageReactance of Round-Rotor Generators', Transactions oftheAmerican I.E.E., 1958,77,PartIII, p. 636.(4) HONSINGER, V. B.: 'Theory of End-Winding LeakageReactance',ibid.,1959,78, PartHI, p. 417.(5) MASON, T. H., AYLETT, P. D., and BIRCH, F. H.: Turbo-Generator Performance under Exceptional OperatingConditions', Proceedings I.E.E., Paper No. 2846S,January, 1959 (106 A,p. 357).

(11) APPENDICES(11.1) Distribution of Current on a Conical End-Winding

Fig. 19shows adevelopmentof theinvolute end-windingonthe surface of acone. The winding ismadeup from anumberof separate coilsasshowninFig.2, thecoil nose being neglected.Consider the current at a general point P whose position is


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536 ASHWORTH AND HAMMOND: THE CALCULATION OF THE MAGNETIC FIELD OF ROTATING MACHINESThen the current in PQ is

iPQ = isin9\ Isin/? [a (A + Xfi)]where A= (TT/2/?) fju.

Similarly the current in PR isiPR = Isin/?[a + (A + x/x)]

Thus the componentkof the current at P is given byk = hR c o sy I'PQc o s y

= 2 / cosy cos pa sin/?(A + x/x)= t cos pa. cos p(l x)fj, . . . . (1)

where fe is a constant.Similarly i z = i z s i n p a s i n p l X [ M . . . . 2 )

Eqns. (1) and (2) give the current for different positions of P.The condition for continuity of current over half a pole

pitch is

ZERO CURRENTKs BOTTOM LAl ERZERO CURRENTIN TOP LAYER

Fig. 19.Development of end-winding on the surface of a cone.determined by the lengthxL and the angle a. (All angles shownin Fig. 19 excepty and r\are measured in a plane perpendicularto the axis of the machine and are proportional to the arcsshown.)

Two simplifying assumptions are made:(a) The angle y is constant, i.e. the involute has the same shapeas an equi-angular spiral.(6) The angleipis proportional to the distance xL, i.e.ift= X/J..This assumes that the involute has the shape of an Archimedeanspiral.

Fig. 20 compares an involute, plotted in polar co-ordinates,with these two spirals. The spirals are chosen to be close to

ARCHIMEDIAN SPIRALINVOLUTE

EQUIANGULAR SPIRAL

O 20 40 60 80 100 120tj , DEGFig. 20.Comparison of involute with spirals.

the involute over a range of 77 from 25 to 120, which is likelyto be the greatest range met in an actual design of an involutecoil. (The zero of 17 is defined with reference to the involute.)Thus, although the two assumptions are contradictory, they areboth reasonable. In most cases the range of 17will be muchless, and closer-fitting spirals could be chosen. Having justifiedthe two assumptions we need not consider the spirals anyfurther.

It has already been shown in Section 3.2 that the current ineach layer of the winding varies sinusoidally with the angle.Let the current in each layer at the intersection of a planeSRQT with the winding (Fig. 19) be tsinpOand let the originsof dbe at O( and O 2for the top and bottom layers, respectively.

TT/2/J r7t / 2pi>sin/?/zsin/>ada = izR sin pa sinp(l x)fxdaJo+ QLCOS8cos/?(l x)fxdxJo

Henceisps inp\x. 1ZRs inp(l x)/x, IQ LC OSJ3r.4 ^ r [ s i

This must be true for all values ofx. Hence

and

(3)(4)

Eqns. (l)-(4) determine the currents along the generators ofthe cone,iz, and perpendicular to the generators, z"e. If the coneis replaced by a stepped arrangement of tubes and discs, theaxial current will be iz and the circumferential current / e. Theradial current, / will be equal in magnitude to i2, but, beingdirected inwards, it must be taken as negative.

If the end-winding is represented byN tubes and discs, it canreadily be shown by reference to Fig. 21 that

* = 2nR = p +x.Lsinj5

Ai? = L sin 8N

z = Z + xLcos 6(11.2) Magnetic Field of Tubular Current Sheets

Referring to Fig. 4, we consider three types of currentdistribution:

(a) Axial current of line density, i2sinpa sin cor.(b)Circumferential current of line density fe cos/?a sincot.(c) Radial current of line densitytrsin pa sincot.

Currents (a) and (b) are distributed over the surfaces of tubeseach of radius R and length 21 . Current (c) is distributed onthin annular discs each of mean radiusRand radial depth Ai?.


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ASHW ORTH AND HAM MO ND: THE CALCULATION OF THE MAGNETIC FIELD OF ROTATING MACH INES 537Using eqns. (17) and (18) of Part 1we have the vector potentialof a circumferential current given by

2Ar = 77and

1s i n w ' 2 -[K p | l(sbr)lp+ i{sbR)Sm .l S- K p _ {(sbr)Ip_ i(sbR)] sin sb lc ossb : . (11)

PLANE OFFIELD PLOT

Fig. 21.Relating to the determination of current on a cone.These single isolated current sheets are replaced by an infinitesuccession of equal and oppositely directed coaxial currentsheets spaced at a distance g such that mutual effects will benegligible. The currents can then be expressed by Fou rierseries.For currents (a) and (b),

2 1 = -fjboiBR cos pd sin cot 2 ~[ Kp+ l(sbr)Ip+ l(sbR)77 j _ l S+ K p _i(sbr)Ip_ i(sbR)] si nsb lc ossbz . (12)

whence the magnetic field of a circumferential current is given by2b H r = %R cos pd sin cot 2 [Kp+ 1(sbr)Ip+ l(sbR)77 512b t

+ Kp_ i(sbr)lp^ iisbRj] sinsb lsin s&z . (13)= t$R sinpd sin co t 2 [Kp+ iCsr)Ip+1 CyZ>i?)

7 7 .v = 1- K_ sin . (14)r, = - * W 90 sin cor 2 Kp(sbr)Ip(sbR)sin 56/ cosJ6Z

. . . . (15)The vector potential of a radial current can be obtained in asimilar manner:

bAr =4 1i : e = f : e 2 - s i n ^ 6 / c o s s b z . . . . 5 )7T ' 5 = 1 5

sinp8 sincot+ Kp _

For current (c),

where 5 is od d. 5 = 1

(6)

p p . (16)b ^ e = -fxotrRAR cos pd sin ojt 2 [Kp+^jfcOIp+iCjA/J)- K p . , ( ^ r ) I p _ x(sbR)] co s J6 Z . (17)

whence the magnetic field of a radial current is given byb2 H r= irRAR cospd sin

^(sbr) {sbR)] sin sbz . (18)sincot 2

K p sin*6z . (19)Fig. 22.Relating to the vector potential of a tubular current.

Refer now to Fig. 22 and consider the case when r>R.Using eqn. (13) of Part 1of the paper,1the vector potential at Pdue to an axial tubular current is4 1Az = -uotzR sinpdsincot2 -KD (^r)L(^ i?) sinsblcossbz . (7)77 Sm.\S y

Using the relationfi0 H = curlA,we have# z = 0 (8)

Inhnh2 = - ?rA/?cos/70sino>/ 2 KJsbr)JJsbR) cos *6z . (20)5 = 1The total magnetic field can be obtained by adding the com-ponents. If we write cos(cot p6) for sinpdsincotwe obtainthe following expressions for r>R:

sin(cot pd)4b = IQ R2 Kp(sbr)l'(sbR) si n sb lco s sbz

4o 1r = - fz i? cos/>0 sin co t 2 -KJsbr)lB(sbR) sin77T S77T . (9)

46- 2tAR 2

7 7 5 = 1He

co s sbz . (21)

46f,i?sinpd sin cof 2 K'(sbr)IJsbR) sin.s6/cos 56z7 7 5 - 1

co s(cot pd)4b . _. (10) 7 7 5 = 1


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538 ASHW ORTH AND H A M M O N D : THE CALCULATION OF THE MAGNETIC FIELD OF ROTATING MACH INES+ 2-kR [Kp+ l(sbr)Jp+ l(sbR)

+ Kp_1(sbr)Jp_ l(sbR)] sinsb lsinsbzr-hR [Kp+ l(sbr)lp+ l(sbR)

Kp_ l(sbr)lp_i(sbR)] sinsb lsinsbzirRAR s[Kpi_tsbr)Ip+ l(sbR)J T =

-'- Kr sinsbz (22)ft 3

sin (cotnr* KB(j*/-)I-(d>/?) sin 56/ cossbz+p p

b-i,RAR - K p_i(sbr)lp_ i(sbR)] sin j&r . (23)

The expressions for the mag netic field when r < R can beobtained by transposing ran dRin the argument of the Besselfunctions of eqns. (21)(23).[The discussion on the above paper will be found on page549.]

621.313.322 81.013.2 The Institution ofElectrical EngineersPaper No. 3490 SMar. 1961

THE MAGNETIC FIELD OFTHE END-WINDINGS OFTURBO-GENERATORSBy P. J. LAWRENSON, M .Sc, Graduate.

{The paper was first received 19///May,and in revised form \4thOctober,1960. / /was publishedin March, 1961,and wasread beforetheSUPPLY SECTION 12th April, 1961.)

SUMMARYAn investigation of the magnetic field generated by the currentsinthe stator end-windings of a 3-phase 2-pole turbo-generatorisdescribed.It includesan examination of the influence ofwinding design, namelythe effects of cone an gle, straight projection and coil pitch, of theinfluence of the air-gap,and alsoof the permeability, or eddy-currentscreening, of the core end-plate.The simpleand accurate method which isused to calculate the fieldof the stator currents, an d which can be usedfor a circuitofany shape,is described. The effect of the air-gap is represented bya fictitiousconductor,and the influenceof the end-plate isaccounted for by useof the method of images. Results are presented for the magnitudeand phase of the fundamental and harmonic waves of the rotatingfields on the end-plateand rotor-coil retaining-ring surfaces.From the resultsthe following points appear:{a )The predominant loss-producing component of thefieldat theend-plate surface arises from the fringing at the end of the air-gap.(/>) In order to reduce loss-producing co mponents of the fieldatthe end-platedue to the windingthe cone angle should bemade smalland the straight projection large.(r) The useof a non-magnetic end-plate reduces the axial comp onentof the field.(d )The use of eddy-cu rrent screening, whilst eliminating the axialcompo nents, increases the radial components.{e) Near the winding and the air-gap, the field contains a highproportion of harmonic components which induce loss in therotor-coil retaining ring. They increase in magnitude with increases in thestraight projectionof the winding and the air-gap m agnetomotive force.(/) Changes in coil pitch do not significantly influence either thefundamental or the harmonicsoftheend field.

(1) INTRODUCTIONThe need for careful investigation of the effects of the magneticfields generated by the currents in the end-windings of t u rbo-

generators (and other large electrical machines) has long beenMr. Lawrenson was formerly with A.E.I. (Ma nchester) Ltd., and is now in theElectrical Engineering Dept., Leeds University.

appreciated, 1 3 an d dur ing the last 10-15 years, because of theenormous increasesin electrical loadings which have been made,this need has become m ore urgent . The stray fields havetw oparticularly impo rtant effects. First , they produ ce, in all con-ducting parts of the end structure, eddy-curren t and powerlosses which no t only represent a large part of t he total lossin th e machine but also are liable to cause dam age ow ing tooverheating in localized reg ions; ex aminations of eddy-currenteffects in certain parts of the machine , e.g. in the stator wind-ing1* 4> 5 o r in the end-plate teeth,6 have been described in manypapers. Secondly, the stray fields result in the development,on the end-winding itself, of forces which can also cause severedamage during short-circuit conditions; this subject , too, hasbeen discussed in a number of papers.7"9

The complete analysisof the magnetic field in the end regionsof the machine is an extrem ely difficult one. N ot only is thefield gen erated by avery complicated p attern of primary currentsbut also it is mod ified by boundaries which are of verycom-plicated shape, which have varying values of permeabili ty,and which are the seat of modifying eddy curren ts. Con-sideration of these features in dicates th at no exact analyticalo rnumerical solution fo r the field is possible a t present o r fore-seeablein the future. Nevertheless, by ma king certa in simplify-ing assumptions with regard to the form of the enclosingboundaries , it is possible to calculate a considerable amoun tof information which can provide useful guidance to designers.This includes, for example, a knowledgeof the general distribu-tion of fieldin the end space and of the forces experiencedbythe winding.References 5, 7, 8 and 9 aretypical of previous theoreticalwork on the end-field problem an d in all of them the boundarysurfaces, when considered, are represented as being plane. Theassumpt ions made in these works with regard to the formof thewindings vary more widely and, in general , the representationsof th e winding shapes have been inadeq uate; only the one


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621.313.1.013.23 The Institution of Electrical EngineersMonograph No. 514 SApr. 1962

THE CALCULATION OF THE MAGNETIC FIELD OF ROTATING MACHINESPart 3.Eddy Currents Induced in a Solid Slab by a Circular Current Loop

By P. HAMMOND, M.A., Member.{Thepaperwa s firstreceived19th October,1961,and in revised form 25thJanuary,1962.in April,1962.) It waspublished as an INSTITUTION MONOGRAPH

SUMMARYThe paper contains a detailed analysis of the eddy currents induced ina semi-infinite slab by a circular current outside the slab. Particularattention is paid to the magnetic field outside the slab, and the con-ditions are explained under which the method of images can be safelyapplied. Th e effective impe dance of the slab is calculated, and th eresults are compared with recent experimental work. The applicationof the work to the end-fields of rotating machines is discussed, and it

is shown that the diameter of the end-windings and the degree ofsaturation of the end-plates have a profound effect on the eddy-currentbehaviour.

LIST OF PRINCIPAL SYMBOLSA = Magnetic vector potential .a Radius of current loop.Bn BQ,BZ = Magnetic flux densit ies.b = Distance of current loop from slab.Jr, JQ,JZ Eddy -current densit ies.Fi (#) , F2(A:) = Functions of k.Qx{r\d), G2(r/fl) = Functions of r\a.I, I' Lo op curren t, eddy current.Jo , Jx = Bessel functions of the first kind and orders

zero and unity.k = Variable.Lfc = Complex function of k.M k = Co mplex function of k.p = V ( W / - w 2 / p ) .R Equivalen t resistance of slab.r,8,z = Cylindrical co-ordinates.t = Time.X = Equivalent reactance of slab.Z Equivalen t impedance of slab.8 = -\/ 2pl[jL0urco) = Skin depth of eddy current.IXQ = Primary magnetic constant.*fji r= Relative permeabili ty./j,= /uo/x r = Permeabili ty.p = Resistivity.a> = Angular frequency.

(1) INTRODUCTIONFor some time the author has been grappling with the problemsarising from the magnetic fields surrounding the ends of rotatingmachines. These problem s, which are mos t mark ed in largeturbo-gen erators, are of two kin ds: first there is the determ inationof the magnetic field around the end-windings, which enablesinductances and short-circuit forces to be computed, andsecondly there is the problem of eddy-current losses in the coreend-plates and various other metal structures. These 'stray'losses are highly undesirable, both because they represent acontinuous waste of energy and, more importantly, because they

Comm only known as the permeability of free space.

cause heating, which may set a limit to the output obtainablefrom a machine of given frame size.As a first stage in this investigation, a method has beendeveloped1*2 which enables the designer to compute the mag-netic field of the currents in the end-windings in air. Thenext step would appear to be to estimate the effect of themetallic boundaries, and this should lead to the third step, thedetermination of eddy-current losses. Unfortunately, however,it is not possible to divide the problem in this way. The troubleis that the eddy currents modify the magnetic field outside themetallic bodies in which they flow. The second and third stepsmust be taken together, because it is essential to study theexternal field and the internal eddy-current system at the sametime.Such a study is full of interest and impo rtance. The subjectof eddy currents is, of course, far larger than that of straylosses in machines. Indeed, in many importan t cases such asinduction heating, the eddy-current loss, far from being unde-sirable, has to be as large as possible. It is therefore convenientto avoid, in the first instance, the complication of the machineend-windings and to investigate fully a somewhat simplerproblem, which demonstrates the eddy-current behaviour moreclearly, so that the results can be applied to a wide variety ofelectrical devices. This simpler prob lem, the subject of thepaper, is the investigation of eddy currents induced in a largesolid slab by a circular current parallel to the face of the slab.The investigation throws a great deal of light on the more com-plicated problem of stray losses in rotating machines, and it ishoped that it will also form a useful addition to the literaturedealing with eddy currents.

(2) GENERAL DISCUSSION OF THE PROBLEM(2.1) Physical Considerations

Before attempting a mathematical analysis of the particularconfiguration which forms the subject of our inquiry, we shalllook briefly into the various factors which govern eddy-currentphenomena. Suppose that the physical properties of the systemcan be described by the permeability fx , the resistivity p, theangular frequency , the linear dimension a and the excitingcurrent /. Dimensional analysis shows that the eddy currentswill be governed by the dimensionless group fxoja2/p . Thepower loss, for instance, can be expressed by the equation

Correspondence on Monographs is invited for consideration with a view topublication.Mr. Hammond is in the Department of Engineering, University of Cambridge.

a \ pThus, if /xaja2lp is held constant, the power varies as I2p/a,P/xcoa or Ppy^ifxco/p), and under these circumstances thepower factor also remains consta nt. It is clear that a greatdeal of information can be obtained from relatively few experi-ments, once i t is realized that the behaviour is governed by theexpression (xcoa2/p.The expression -s/^pjfxcS)has the dimension of length, and in

[508]


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HAMMOND: THE CALCULATION OF THE MAGNETIC FIELD OF ROTATING MACHINES 509books dealing with eddy currents3 S = -\/ 2plixo)) is called theskin depth of the mate rial at the angular frequency tu. Thisskin depth is defined with reference to the eddy-current loss in asemi-infinite slab to which is applied a uniform tangential mag-netic field. If the eddy current were uniformly distributedthroughout a skin of depth 8, it would produce the same lossas the actual non-uniform cu rrent in the slab. The dimen-sionless factor |U,aja2/p, which governs the behaviour of theeddy-current system, can be written as 2a2/8 2. Thus thebehaviour is determined by the ratio of the leading dimensionof the specimen to the skin depth.So far, we have discussed the simplest possible system, whichcan be described solely by the five variables fx, p, co,a and /.In discussing the behaviour of the more complicated systemshown in Fig. 1 we shall need to introduce two permeabilities,

V////AI, u

- U- 2a -I

Fig. 1.Circularcoil parallel to the face of a semi-infinite slab.fx { and /x2, for the two regions of space and an additionallinear dimension b. The behaviour of the eddy currentsnow depends on three dimensionless factors b\a, yi\lix2 anda/8 = y/(ju,1coa2/2p). It is well to take a close look a t thesethree factors. The first is an obvious geometrical ratio, but thesecond singles out iron from all other materials and warns usto expect a very different type of behavio ur with iron. It isessential to note explicitly that the permeability not only affectsthe skin depth b ut also forms a dimensionless group independentof the resistivity and the frequency. This is, of course , reason-able, since iron will affect the field even in the static case of zerofrequency. The third dimensionless factor is a reminder thatthe skin depth must be taken into consideration and thatthe eddy-current behaviour depends on the ratio of coil dimen-sion to skin dep th. This is a scale factor which is far lessobvious than the ratiob \aand one that can easily be overlooked.

(2.2) Mathematical ToolsThe magnetic field will depend on the current loop and on theslab with its surface polarity and eddy currents. We shallmake use of the boundary conditions at the surface of the slabby making the n ormal flux density and the tangentialfieldstrengthcontinuous.As is customary in eddy-current problems, we shall limit thediscussion to cases for which the dimensions of the apparatusare small compared with the electrical wavelength (6000 km at50c/s). We can thus neglect the displacement current. Thismeans that the magnetic field outside the slab has no sourcesand could be described by a scalar potential as was done in arecent, helpful paper by Mukherji,4 who discussed the eddy-current system induced by an infinite plane sheet carryingsinusoidal current density.However, the scalar potential is not applicable inside the slab,and it seems preferable to use instead a vector poten tial functionwhich can be applied both inside and outside the slab. Thevector potential is governed by the equations

= 0 outside the slab

A suitable system of co-ordinates is the cylindrical (r, d, z)system, the origin being chosen at the centre of the circularcoil. Since the exciting current is circular, the only componentof A is AQ and we can drop the suffix. Since the current isuniform, there is no variation of A with 6. If the current isharmonic in timewecan write7)A/7)t =jcoAand use the notationA, which indicates that A is a phasor and that the completeSolution, including time dependence, is Si{Aejat ).Outside the slab the equation governing the vector potential is

~hr r ~br r2 ~d z2which has solutions of the form

A =ekzlaJ l(krla)Similarly, the vector potential within the slab has solutionsof the form

A= ewherep2 = \una2\p 2a2/S2.

^kr/a)

(3) FORMULAEThe derivation of the formulae is given in the Appendix.For ready reference the chief results are set out in this Section.(3.1) The Vector Potential

Outside the slab (z b):

2 )(3.2) The Magnetic Flux Density

Outside the s lab:6 = ^2a

kd k

2aand = i within the slab

3)4)

(5)


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510 HAMMOND: THE CALCULATION OF THE MAGNETIC FIELD OF ROTATING MACHINESInside the slab :=WL1 fr a J

(6 )7)

Jo

(3.3) The Eddy-Current Density in the Slab(9 )

(10)The current per unit radius is

r t I B //eaz = - e-kb la1x(k)h{krla)+ jp 2) + k2 + jp2dk (11)

and the total current isf00 [w~ 2 r -

Jn Jfi J n

k 12)(3.4) The Power Loss in the Slab

Let R be the equivalent resistance and X the equivalentreactance of the slab, so that the complex power is given by2R +j X = y J1-2kbla[Jl(k)]2

-jp2)+jp2)][k(jL r 7^dk . (13)-JP2)}(4) GENERAL CONSIDERATION OF THE FORMULAEThe equations for the vector potential in Section 3.1 showclearly the separate effects of the exciting current and the slab.Consider the expression inside the square brackets in eqn. (1).The term e~k^-b)la g j v s \\\t effect of the coil only. The termk\xr - Vtyr + Vih2 + jp2)it is interesting to note the occurrence of two of the dimension-less factors (fxr and p) which were discussed in Section 2.1.The third factor, b\a,occurs in the exponential function outsidethe brackets. The minus sign in the num erator indicates thatfj.rand/? will act in opposition; whereas /x r strengthens the vectorpotential outside the slab,p will tend to reduce it. Moreover,

//

1

ZJ///

\\\\\\\ \ \

*

_

the field will be in phase with / if \ir is dominant but will lagby 180 ifp is dominant.Although it will be necessary to have recourse to a digitalcomputer in order to evaluate the integrals in Sections 3.1-3.4accurately, much general information can be obtained by con-sidering the integrand alone. Why this is possible can readilybe understood by reference to Fig. 2, in which is plotted a

01 2

0-10

0 0 8

5 0 0 6

0 0 4

0 0 2

"0 1 2 3 4 5 6 7 8kFig. 2.Behaviour of part of the integrand occurring in variousformulae.particular case of the function e~kb laJ l(k)J l(krla) which formspart of this integrand. It will be noted th at the functioniszero atk = 0 and negligible for values ofklarger than, say, 6: thus, thelimits of the integration could be from k = i tok = 6 withoutappreciab ly affecting the result. This knowledge is very helpfulin considering the behaviour of the integral while either }xrorpis dominant, i.e. while the magnetic field is controlled either bythe magnetostatic field of the slab or by the field of the eddycurrents.

(4.1) Two Limiting Forms of the Solution(4.1.1) Resistance-Limited Eddy Currents.

Ifkfxr > -\/ k2+ jp2), the expressions for the vector poten tiaare simplified as follows.Outside the slab:

. . . . (14)

. . . (15)

XInside the slab:

r0 0=fiol\ e-Eqn. (14) shows th at the magnetic field outside the slab isunaffected by the eddy currents in the slab. The eddy -currentdensity in the slab can be found from eqn. (10) and is given by

The eddy current lags behind the exciting current by 90, as isshown in Fig. 3. This is the usual low-frequency approximationand the behaviour is said to beresistancelimited.The condition kfxr> \/(k2 + jp2) is theoretically impossibleif A: is very small, but as remarked previously, we can take the


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HAMMOND: THE CALCULATION OF THE MAGNETIC FIELD OF ROTATING MACHINESj and the total current is given by5

Fig. 3.Phase relationship of resistance-limited eddy current.lower limit of the integration at k = i , say. It is of interest toexamin e what this means in a practical case. Consider a slabof silicon-iron of resistivity p = 5 x 10~7Q-m and of relativepermeability \ir = 1000. Let the frequency be 50 c/s ; then theskin depth S = -\/(2alp) = 1/200TT. If the radius of the coil isl c m, p2 = 79, kf ir is 1000/4 = 250 and ^{k1 + jp2) =VO/16 +y"79), so that the condition k(xr> \/(k2 + jp2) ismet. We note that in this case the relevant dimensionlessfactors are /xr= 1000 and coil-radius/skin-depth = 6-3.It should also be noted that a coil radius of 1m would makep2 _ 79OOOO and th e inequality would no longer hold . Thusthe resistance-limited condition cannot be applied to the case[jL r = 1000 and coil-radius/skin-depth = 630. Similarly, thiscondition does not hold if the iron is saturated and \ir = 10,even for th e case of the smaller coil of1cm radius. It is essentialto keep both dimensionless factors in view, and the general ruleis that for resistance-limited eddy currents we require a largepermeability and a small ratio of coil radius to skin depth.Non-magnetic materials (/xr = 1) do n ot exhibit the resistance-limited behaviour.(4.1.2) Inductance-Limited Eddy Currents.

If p > k\x.n the expressions for the vector potential aresimplified as follows.Outside the slab:

Inside the slab:j - 0 0

P\/JJo

. (17)

. (18)Examination of eqn. (17) shows that the external magnetic fieldis profoundly affected by the eddy currents in the slab. Thenorm al flux densityBzat the face of the slab tends to zero, andthe tangential flux density Br just outside the slab is nearlydoubled by the eddy currents. In other words, the slab hasbecome almost impermeable to magnetic fields.The eddy cu rrent density in the slab is given by

(19)

20)The line density of the eddy current per unit rad ius is

r i / hd z = - - e-kb laJ l(k)3 l(krla)kdk . o. .'o

I' = f fJtfzdr = - I \ e-kb la1{(k)dk . .

511

21)= - / [ - b/V(b2 + a2)]

Under these conditions the total eddy current is in antiphasewith the exciting current and the surface current densityJs leadsthe total eddy current by 45. These phase relations are illus-trated in Fig. 4. The eddy-current loss will vary as kixr may seem impossible, since kis to be taken for values up to infinity. But we have alreadyseen by reference to Fig. 2 that the upper limit could well beset at k = 6. Thus the condition becomes p > 6/xr. Ifix r = 1000, oi = 1007T, p = 5 x 1 0~7 n-m, this meansp> 6000, i .e . a /S = 4 2 5 0 and a> 6-75 m, which may beimpossib le. If, however, /xr = 10 the condition isa> 67 5cm,and ifweexamine the condition for copper, then for inductance-limited eddy currents a> 42 cm.(4.1.3) Comparison of the Two Limiting Solutions.

From the preceding tw o Sections it follows that the resistance-limited behaviour occurs with unsaturated iron at power fre-quencies, provided that the coil radius is of the order of 1cm.The inductance-limited behaviour occurs with non-magnetic


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512 HAMMOND: THE CALCULATION OF THE MAGNETIC FIELD OF ROTATING MACHINESmaterials at power frequencies, for coil radii of the order of10 cm. It also occurs with saturated iro n and coils of theorder of 1m radius.For small coils with saturated iron or large coils withunsaturated iron, neither of the limiting cases is appropriate.Increasing the frequency of the exciting current is equivalentto increasing the coil radius, and increasing the resistivity isequivalent to decreasing the radius. The exact effect can bedetermined from a consideration of the dimensionless factorP =

(4.2) Eddy Currents and the Theory of ImagesIn examining the magnetic fields at the ends of rotatingmachines, various writers have recently made use of the theoryof images.6'7> 8 This method is very attractive because it makesit possible to assess in a straightforward manner the effect ofsolid boundaries on the magnetic field. There have, however,been serious doubts as to whether the method is applicable toboundaries in which eddy currents are flowing . W e are now in a

position to resolve these doubts.Reference to eqn. (14) shows that, if the resistance-limitedsolution applies, use can be made of the theory of magnetostaticimages, because the eddy currents do n ot influence the magneticfield outside the slab. The field outside the slab is as though/z,= oo, and the surface of the slab can be treated as a magneticequipotential surface.Reference to eqn. (17) shows that, if the inductance-limitedsolution applies, the surface of the slab behaves as if it wereinfinitely conducting, and the magnetic field outside the slab isas if the slab had zero permeability. Use can therefore be madeof the theory of electromagnetic images in infinitely conductingsheets. The image of a coil is of the same sign if the resistance-limited solution applies and of the opposite sign if theinductance-limited solution applies.6 Reference to eqn. (2)shows that the limiting image solutions also apply within theslab.Comparison of eqns. (1) and (14) shows that, in between thetwo limiting forms of the solution, the integrand behaves as ifthe effective permeability were given by the expressionkfxrl\/(k2 +jp2). Fig. 6 illustrates the behaviour of this|l EFFECTIVE/^.

0 01 0-2 0-3 0 4 0 5 0 6 0-7 0-8 0-9 1-0

0-1

0-20-3

0 4

\ f^ 4 5 \ N \

, _

9

/i

k{ xr> \/(k2 + jp2) and p > kpr which make it legitimate touse the image method. If the slab is truly infinite, the appli-cation of the image method does not depend on the geometricalfactor b/a,which is the third of the three dimensionless groupsaffecting the problem. But in engineering practice the slab will,of course, be finite, and it becomes important to investigatehow large it has to be before we can apply a solution whichis based on an infinite surface. Eq n. (20) gives the line densityof eddy current per unit radius for the inductance-limited case,and we can use it to find how quickly the eddy current decreaseswith increasing radius. Fig. 7 shows the computed results,

Fig. 6.Effective complex permeability of the slab.function for different values of p/k. If p tends to zero theeffective permeability tends to \xrand is a real number. Fo r allother values the effective permeability is complex and its phaseangle approaches 45 lagging asp tends to infinity. This limitingphase angle has already been noticed in Section 4.1.2 and is ofcommon occurrence in all eddy-current problems. Itisimportantto realize that the condition of zero permeability is approachedat this phase angle.We have so far discussed the two limiting relationships

- 0 - 2

- 0 - 4

- 0 - 6

- 0 - 8

- 1 0

YK

Fig. 7.Eddy-current flow per unit radius.b/a bjabja

which suggest that a radial distance of three times the coilradius will make the slab appear to be infinite and will makethe image method possible. But if the slab has the same radiusas the coil, the method of images is quite inapp ropriate. Ourequations, although they are not based on the method of images,also apply strictly to slabs of infinite extent only, but they giveinformation about the distribution of eddy current which canbe used to investigate the problem even for finite slabs.(5) COMPUTED RESULTS

(5.1) Magnetic Flux DensityThe magnetic flux density Bz on the axis of the coil at thesurface of the slab (r = 0, z = b) was computed for the valuesb/ a = , [Mr = 1, 10, 1000, p = \, 10, 100. The results aregiven in Table 1 and should be compared with the value of Bxdue to the coil alone, which is Bz =0 358/xo//a. Thus themagnetostatic image solution of a doubled flux density holdsfor /xr = 1000,p = 1 and 10, but begins to break down forfj,r = 1000,p = 100 and for /x, = 10,p = 1.The limiting phase angle of 45 is approached for /x r = 10,p =100 and fxr = 1, p = 10 and 100, bu t the case of negligiblysmallBzholds only for yur = 1, p = 100. This is the case thatcorresponds to zero permeability. Reference to Sections 4.1.1and 4.1.2 shows that the computed results of the completeintegrals bear out the predictions made from consideration ofthe integrand, and the previous discussion is thereby justified.


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HAMMOND: THE CALCULATION OF THE MAGNETIC FIELD O F ROTATING MACHINES 513Table 1

AXIAL FLUX DENSITY AT SURFACE OF SLAB

111101010100010001000

p

110100110100110100

Bzaliior

+0-3390 0630 0060-6450-4270-0630-7150-7110-677

J0 0410 0590 0060 0 1 70-1400 0570-00020 0040-034

(5.2) Penetration of Eddy urrents into the SlabThe penetration of the eddy currents into the slab can beexamined with reference to eqn. (10). Fo r inductance-limitedbehaviour the eddy currents are confined to a surface layer ofvanishingly small depth. Fo r all other types of behaviour theeddy currents penetrate into the slab and the magnitude of thecurrent decreases almost exponentially with the depth. Fig. 8

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Fig. 8.Penetration ofeddy current into the slab.(z-b) = 0-5( 6 ) 0 1shows the complex eddy current density in the slab for the ca

10.1.1.165.6983(2)

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