CHAPTER 3 LEAKAGE REACTANCE CALCULATIONS

Transcription

1 CHAPTER 3 LEAKAGE REACTANCE CALCULATIONS 3.1 Introduction In Chapter 2, while considering the fundamental energy conversion process in a L~I, the attention was necessarily focussed on the active region of the machine. Inevitably, even in the idealised theoretical model, there exist some leakage fields that link either the stator or rotor circuits only. The leakage reactances assume importance when the applied voltage and power factor characteristics of the current-fed machine are examined, or when the machine is run from a constant-voltage source. There has not been any significant contribution in this as~ect of LIMs except the paper by Allin et al 30 in which the leakage reactances of a laboratory LIM model is experimentally obtained. It is probable that questions regarding the leakage reactances were pushed into the background as the emphasis in LIM research so far has all along been on the current-fed mode of operation 10, In addition, exhaustive work had been previously done on the leakage reactances of different types of conventional round rotor induction motors (Alger, P.L. 31 ), and it is quite possible that some of the formulae could be modified and used for

2 38 the LIMs. Yet it is considered desirable to explore the possibility of developing analytical expressions for the leakage reactances of LIMs keeping the geometry of the machine more or less intact for the following reasons: 1. The stator winding overhang leakage reactance of the longitudinal-flux LIM is relatively large on account of the larger pole pitch as shown by Laithwaite 1,3. 2. A topological cylindrical equivalent of the doublesided LIM does not exist conceptually, since only the outer surface of the round rotor is available for induction. 3. It is recognised that the sheet secondary construction of the LIM facilitates a simplified field theoretic formulation of the leakage reactance problem. The elements of leakage reactance that are generally taken into consideration for a conventional induction motor can be broadly classified as Primary winding overhang leakage reactance 2. Primary slot leakage reactance 3. Secondary leakage reactance 4. Air-gap leakage reactance Some of the physical features characterising the round rotor induction motor and giving rise to the above leakage paths are not present in the type of linear induction motors studied in this investigation.

3 39 Specifically, the aluminium sheet rotor construction results in a low secondary leakage reactance even for thick plates as shown by Laithwaite 1 Analogous to the terminology of conventional induction motors, the airgap leakage reactance is brought about by the differential, zig-zag and peripheral leakage fluxes. Again the unslotted sheet rotor configuration permits ampere-turn balance of the primary currents in the opposite gap surface at every point, thereby eliminating the differential fluxes. For the same reason, the zig-zag leakage fluxes also vanish almost entirely. It may be noted that in the previous chapter the performance of the linear induction motor is predicted by solving for the vector potential along the stator surface so that the airgap leakage flux is automatically included in the analysis of the active region. Consequently the important components of the leakage flux that have to be, separately accounted for are the coil end-turn leakage and the primary slot leakage fields Stator Overhang Leakage Reactance In general, the coil end configuration of the stator overhang region along with the geometry of the end ring assembly gives rise to a complicated three-dimensional pattern of end leakage flux that is very difficult to determine exactly. One redeeming feature, however, is the

4 truely leakage nature of the overhang flux so that it can 40 be treated independent of the main field. In fact, as Alger 31 has pointed out, the calculation can be made as if the stator core were removed and the winding projections on either side brought together with the flux paths entirely in air. Hence for the sake of analysis it is assumed that the overhang field is unaffected by the active,region and rotor currents. This implies that the longitudinal end effects and transverse edge effects of the rotor can be neglected, affording a simplified treatmente Alger 31 has shown that for a double layer winding with diamond shaped coils, the overhang currents can be resolved into the axial (z) and peripheral (x) components, which obey sinusoidel variation in both x and z directions. A three-dimensional solution for the leakage field is obtained by representing the winding overhang by a thin continuous current sheet on either side of the stator core level with the stator-airgap interface. Also, in the following analysis, only the space fundamental of the stator current distribution is considered, while the field quantities have sinusoidal time variation at supply frequency ~ as in eqn ~ Referring to fig. 3.1 which shows the winding disposition along with the co-ordinate axes, the fundamental

5 current distribution in the stator due to the three phase double layer winding is given by (Alger 31 ) 41 h = 2 6. cos ~ x amp/m where ~ is the peak value of the linear current density per layer of the primary and is given by 6. = 3j2~T1 P l amp/m where I is the primary phase current (r.m.s.) and.~ is the pole pitch. Consider a point P in the overhang region with coordinates P (x,y,z). By Alger 31 there is an outgoing current 1 0 and a return current 1 r at this point the linear densities of which are given by ( ' z = cos ~ x -2' Zo ) and I 6 cos ( p x + 't.. z = --r ) r 2 0 where i' is the coil span in radians. The transverse (z) and longitudinal (x) current density components are given by = = 2 6. cos f3 x cos 1\ z

6 I I I I ) / ))J)) / / / / / ' /.//// / / / / / / / / / / /, // / / / / / / P(lI".I) Fig.3.1(a) Winding disposition y y. I X (1)1---'--.-- '--.;..,~;; Fig.3.1 (b) Longitudinal view of stator y I;'.'.,', StatorI 'rame rln. (),/erhan. CU"ren\.h, Fig.3.Hc) End view of stator

7 and Ix = (Io-I) r cote = 2 ~ sin ~ x sin "Z cote (3.6) where "'V 2Z o and e is the angle of bend of the end 43 turn measured from the positive x direction as in fig. 3.1(a). The approach for determining the overhang leakage reactances 25 is essentially to solve for the field distribution generated by the transverse and longitudinal components of the end turn currents by formulation of the appropriate boundary value problems. Expressions for the leakage reactive power flow are derived SUbsequently in terms of the Poynting vector. Due to certain constructional details of the disc drive, discussed in Chapter 4, the two sides of the overhang on either side of the stator core are to be treated slightly differently. vlliile the overhang at the outer periphery of the (curved) stator block is placed entirely in air, that at the inner side has an iron backing ring which is part of the stator frame assembly holding the laminations together. Also for each overhang current sheet there are two regions of field propagation 25 In the following, the airgap region between the stators LS denoted by (1) and the back air region by (2). The field equations in the regions (1) and (2) for the outer overhang in the absence of rotor currents are

8 44 + = 0, o<y < CX) = 0 along y = 0, by symmetry about the xz plane and = + along y = g considering only the transverse component at the moment, with I z given by eqn The general solution of eqn. 3.7 for region (1) consistent with eqn. 3.5 is given by A Z1 (x,y,z) = Ai (y) cos ~ x cos1\ z Substitution of A Z1 in eqn. 3.7 results in the ordinary differential equation = 0 where 11 2 =

9 The solution o~ eqn e 3.11 that satisfies the boundary condition 3.8 is given by 45 :: C1 cosh " Y where C 1 is an arbitrary constant. The normal component of the flux density B Y1 (x,y,z) a :: - bx A Zi (x,y,z) :: C 1 (6- cosh "l y sin p x cos AZ strength The tangential component of the magnetic field H X1 (x,y,z) = - 1 ""0 o - oy A Z1 (x,y,z) Cit) = - sinh'" Y cos f! X COB 1\ Z tj. Now, for the back air region (2), the solution of eqn. 3.7 is of the form A Z2 (x,y,z) = C 2 e- t'\ y cos is x cos A Z which satisfies the physical requirement that AZ2~ 0 as y

10 46 The normal component of the flux density B Y2 (x,y,z) o = - ~ A Z2 (x,y,z) = (3 C 2 e- IJ y sin ft x cos A. z Also the tangential component of the field strength H X2 (x,y,z) 1 = - ~o A Z2 (x,y,z) I') 02 = - - e- '1 y cos f3 x cos ~ z fjo Equating B y1 (x,g,z) and B Y2 (x,g,z) for continuity of normal component of flux density along the plane y = g = C 1 e1) g cosh ~ g By eqn. 3.9 which relates the tangential components H X1 and H X2 at y = g, C 1 ~ - sinh t'l g cos (!> x cos 1\ z = tj o cos?\ z + 2 A cos f3 x cos A Z Eliminating C 2 in the last eqn. by using 3.14 and solving for C 1,

11 47 2 A Po = t') (sinh t') g + cosh') g ) and = 2 A f!~ e ''1 g cosh ~ g t') (sinh ~ g + cosh r') g ) Hence the final expressions for the distribution of the magnetic vector potential in regions (1) and (2) are 25 A Z1 (x,y,z) = 2 A J.Locosh!} Y, cos ~ x cos 1\ z ~ ( sinh 1 g + cosh ~ g ) O<'y< g and A Z2 (x,y,z) = 2 A J.l.cosh r') g e- '1 (y-g) cos ~ x cos A z ~ ( sinh ~ g + cosh ~ g ) y> g The reactive power flow leaving the current sheet per unit normal area into the regions (1) and (2) are given by the outward drawn Poynting vectors 32, and 8 2 evaluated at y =g.

12 -48 * 51 = E Z1 H X1 () A Z1 (x,g,z) e>a 1 * z1 (x,g,z) = (3.19) ~t ay a A Z2 (x,g,z) * (x,g,z) ~ A and 1 Z2 52 = (J. 20) - 2po "b t ~ Y While 8 2 represents reactive power density that is entirely leakage in nature, part of 51 crosses into the secondary representing a mutual effect 25 The latter component denoted by 5 d can be obtained by replacing d for g in eqn and subtracted out from 51 to give the true leakage25,31~ Consequently the stator leakage is given by s = (3.21) The expressions for 51' 52 and 5 d are given by 2j t-t>ou)o~2 cosh II g sinh '1 g cos 2 ~ x cos 2 1\ z ~ ( sinh ~ g + cosh ~ g )2 (3.22) 2J V.~o D,.2 cosb 2 1] g cos 2 fl x cos 2 i\ z., ( sinh 1] g + cosh II g )2

13 49 ~ { sinh ~ g + cosh ~ g )2 where the negative sign indicates inductive reactive power. ~he above procedure needs a modification for the overhang side provided with an iron back-up ring 25 (fig, ~.1c). \lliile eqns. 3.7 and 3.8 hold without change, A Z2 * 0 in eqn It is now required to solve for A z1 only in the airgap region with the boundary condition = along y = g where I is given by z 3.5 as before. The solution of A z1 satisfying eqns. 3.7, 3.8 and 3.25 is straight forward and is given by A Z1 (x,y,z) = 2 6 eocosh t] y cos " x cos " z 1 sinh Y) g (3.26) The rest of the derivation including correction for the mutual field entering the rotor remains the same as in eqns and Poynting vectors by Si and SJ with S~ Denoting the corresponding = 0, the reactive

14 50 volt-ampere per unit area of overhang is given by 8' =8' - 8 I 1 d where 81. = - 2j tj.o~oa2 cosh rt g sinh '1 g cos 2 tsx cos 2 l\z ~ sinh 2 '1 g (3.28) and 8 f ~ _,gj fjo(.l.)o ~ cosh 11 d sinh tl d cos 2 lex cos2~ z d r, Sinh 2 '1 g Let xi denote the leakage reactance per phase due to the transverse component of the overhang current. Equating the 3 phase reactive power with the surface integral of the Poynting vector over P pole pitches x = P 1 Z "r ~ 0 ~'( 8 + 8' ) dx dz z=o x=o ~ cosh!j g sinh., g - cosh!) d sinh" d + c.o~h2 '1 ~ (., ( sinh t'1 g + cosh yt g ) 2 + cosh'1 g sinh ~ g - cosh., d sinh'1" ~ f' '1 Sinh 2 ') g ~

15 where T' = z z=~ l' o ~ x=o 51 Carrying out the above integration and substituting for 6, in eqn in terms of I, 3 l.\)op'czo(1+kz) T 2 K; P'11:' ~ cosh!) g sinh" g - cosh., d sinhr'\ d + cosh 2 Y') g { ( sinh ~ g + cosh ~ g )2 + cosh!\ g sinh!') g - cosh!) d sinh!' d ~ '1 g "00 2 ~ S1 where k z = sinx t and is zero for full pitch coils. Eqn can be rewritten in the form 25 3 wop'oozo (1+k z ) T 2 K; Py\T ~ sinh r'\ (g-d) cosh,,(g+d) + cosh 2 'J g { cosh 2'1 g + sinh 2 " g + sinh ~ (g-d) cosh 1 (g+d) Sinh~ g ohms

16 52 The analysis for the longitudinal component of overhang current parallels the above derivation except that the variables of interest in the Laplace's equations for regions (1) and (2) are A X1 and A The boundary value x2 problem characterised by magnetic lines of force mapped on the y-z plane, is described by + + = 0, o < y < 00 (3.34) ~Ax1 0 along 0 CY = y = (3.35) and OA x1 oa X2 ~ y = by Ix along Y = g, (3.36) where Ix is given by eqn Equating the normal components of flux density at y = g a A x1 (X,g,z) o Z = The tangential components of the field strength are related by eqn and following the steps leading to eqns and 3.18 one gets 25

17 53 A x1 {x,y,z) = 2 A ~ cosh '1 y s in ~ x sin" z cot e (3. 37) ')( cosh '1g + sinh')g ) and l c~sh ~ g e- ~ (y-g) A X2 {x,y,z) = - 2 A U 1"0 If (cosh 1) g + sinh Y) g ) sin ~x sin 1\ z cot 9 The Poynting vectors 8 1 and 8 2 representing the reactive power density are * 8 1 = E x1 H z1 / 2 and a A X1 (x,g, z) * 1 ~ A X1 (x,g,z) = 2 t-lo at oy () A X2 (x,g, z) ()A* x2 (x,g,z) = - 2 f-lo at ay The final expressions for 8 1, 8 2 and 8 d corresponding to the longitudinal component of overhang current are =

18 J U w A f'"'o cosh Y) g cot e { ~ (cosh r')g + Sinh'1g )2 I S = d. 2 ~ sinh t') d cosh f) d cot -2j Mt;t.Q o A t 2 e ~ '1 (cosh W'\ g + sinh 'lg )2 ~ 2. 2 sin ~ x sin A,Z and S = For the overhang region provided with a back-up ring A X2 = 0 and analogous to eqn. 3.26, the vector potential in the airgap is given by A X1 (X,y,Z) = 2 6. f-iocosh '1 y sin~z cot r'} sinh r; g e The corresponding power density vectors in tune with eqns a~d 3.28 are given by SI = 1

19 55 and S 1 ;:: d 2 " 2 -JuU3A rel 0 2 ~ sinht"\dcosh'1dcot e l t Yj sinh 2 t') g The true leakage reactive power density is now 8 1 ;:: S d Let x 2 denote the leakage reactance per phase of the stator due to the longitudinal component of the overhang current~ Equating the 3 phase reactive power with the surface integral of the Poynting vector over the current sheet, with 8 and SI given by eqns and 3.45, = z P ~ 0 z=o T ~ x=o ( 8 + S r ) dx dz ;:: cot2e~ cosh I) g sinh l) g-cosh!) d sinh!) d+cosh 'J g ~ y) (sinh '1 g + cosh ry g) cosh l) g sinh!) g-cosh!)d sinh f) d ~ I 2 ~ ~ sinh Y). g 7:'. ~ -sin 2 ~ x sin 2 "l\ z dx dz (3.46) x=o

20 56 Evaluating the double integral and substituting for ~ in terms of I from eqn. 3.2, the final expression for x 2 is obtained as 25 3 ~o~ozo(1-kz) T 2 ~ cot 2 e P'lT ~ sinh!) (g-d) cosht) (g+d) + COSh2ng ( cosh 2 t')g + sinh 2 " g + sinh" (g-d) cosh'l (g+d) sinh 2 '1 g ohms. Finally the stator overhang leakage reactance per phase is given by the sum of xi and x Slot Leakage Reactance There are standard formulae available in the literature 3i for calculating the slot leakage reactances. In this section the slot leakage reactance is calculated for the upper and lower coil sides of the double layer winding separately, because of the inherent wnbalance of the stator winding disposition, with half-filled slots at the ends of the stator.

21 57 Consider a single open slot shown in fig. 3.2 with coil sides P and Q carrying currents I p and I respectively in the direction normal to the Q paper. The current I p flowing in coil side P produces a flux density varying linearly along its height hi. The flux density at an elemental height dy, y from the bottom of coil P is obtained by considering the current enclosed by the lines of force. Neglecting the reluctance offered by the iron, the flux density variation is given by b x (y) = tjonp I p Y 2h i w In the region h 2 of the slot, the flux density is constant and is given by fjc)n p I p / 2w. The total flux linkage of the coil P due to its own current is therefore given by + dy where 2a is the width of the stator. The self inductance Lp is now given by setting I p equal to unity in the above equation.

22 h, : 9 4 m m hz =1.4mm h J = 4 mm 59 2 w = mm phases n 1 n2 n s A C n, - upper coil sides n2 - lower co il sides n s - numbe r of slots containing only one phase Slots cor. tom ut u a I c 0 i I sides ~ A 8-8 I B C - 6 I CA - 6. Fi g. 3.2 Slot and coil side details

23 59 The current I Q flowing in coil side Q sets up flux density varying linearly along its height, while in the slot region above Q, the flux density is constant, being given by ~onq I Q / 2w. Hence the total flux linkage of the coil Q due to I Q is given by dy I Q = The self-inductance L Q is now given by putting 1 so that l!.n~ a ~"N: a hi L Q = ( h i +h 2 +h:; ) + (:;.49) w :;w To obtain the mutual inductance MPQ' let Q carry a current I Q The field due to I Q is uniform in the region h 1 and h 2 and is given by While the entire flux in region h 2 links the coil P 9 in full, the linkage in hi is linear in y. Hence the mutual inductance MPQ' due to unity current in Q is equal to given by the flux linkage of P

24 60 ( + (3.50) Let the phase A consists of n 1 upper coil sides and n 2 lower coil sides. Also let n 8 denote the number of slots containing only A phase and the number of slots containing A ajld B phases. The self-leakage reactance of phase A is given by = (3.51) and the mutual reactance between A and B phases is = (3.52) The other elements of the slot reactance matrix, namely x b, x c, x ac and c are similarly obtained. Combining the above with the final results of Section 3.2, the stator leakage reactance matrix is given by [x) (3.53)

25 Summary The problem of calculating the leakage reactances in a double-sided, sheet secondary type Linear Induction Motor is considered in this chapter. The leakage fields assume importance when the complete terminal characterisation of the machine is desired, as discussed in the subsequent chapters. In this sense, the results here complement those of Chapter 2. The three-dimensional field model enables computation of the 3 phase reactive power flow in the coil end winding region and facilitates calculation of the 3 x 3 leakage reactance matrix. It may be noted that the derivation does not employ any empirical formula or factor and the final expressions for the stator overhang and slot leakage reactances can be evaluated using normal design data.