05

621.313 534.44

A.A. AFANASJEV A.G. BABAK A.V. NIKOLAEV

THE ANALYTICAL APPROACHES TO CALCULATION OF ELECTRIC MACHINES ON THE BASIS OF THE ANNULAR DOMAINS

BOUNDARY PROBLEMS SOLUTION BY THE METHOD OF FOURIER VARIABLES

SEPARATION

ABSTRACT The air gap, with double-sided stairstep jaggies being two-downlink area, can be conformally imaged as circular ring. Traditionally this procedure executes approximately with the help of Carter factors. Cross sections of the stator and rotor cores also have the form of rings.

The rotating magnetic field in the locations of conductors (in slots, windows between poles) can be presented by the sum of potential and some adding (easily calculated) fields. The localisation of an adding

Prof. A.A. AFANASJEV, D. Sc. ph. (+8352) 451880, the fax (+8352) 209488

e-mail: [email protected]

A.G. BABAK ph. (+095) 775-0748, the fax (095) 775-0749,

e-mail: [email protected]

Chuvash state university Faculty of control and computer science in technical systems,

Moscow circular 15, 428015. Cheboksary, Russia,

A.V. NIKOLAEV ph. (+ 8352) 620114, e-mail: [email protected]

Chuvash state university

PRACE INSTYTUTU ELEKTROTECHNIKI, zeszyt 220, 2004

A.A. Afanasjev, A.G. Babak, A.V. Nikolaev 42

field in the location of conductors requires arrangement on a part of boundary of each slot (at the bottom or in the wedge part) infinitely thin current layers, the joint currents of them are equal to full currents of slots.

Usage of scalar magnetic potentials (SMP) of current layers allows to receive the analytical solution of Dirichlet problem for circular rings of three mentioned environments (air gap, magnetic stator core and rotor) assuming its magnetic permeabilities are constant.

For the sake of greater generality of the problem we assume that the core of the rotor is performed from massive steel. The selection of this material does the mathematical model more universal, as at transition to high enough values of magnetic permeability and resistivity the massive environment gains properties of laminated steel.

SMP of current layers of stator windings and rotor are submitted by Fourier series.

The parameters of an electromagnetic field in a solid rotor (radial and tangential component of flux density, current density and power loss) are obtained on the basis of the analytical solution of Bessel differential equation for a vector magnetic potential by a method of variables separation and are submitted by appropriate Bessel and Kelvin functions.

With reference to an asynchronous motor with a solid rotor the performance data are calculated and compared with the test results. The changes of magnetic permeability in steel of a rotor are taken into account by splitting it on concentric rings, in each of which the permeability is constant.

1. FORMULATION

The rotating magnetic field in the locations of conductors (in slots, windows between poles) can be presented by the sum of potential pH and

some adding 0H (easily calculated) magnetic fields. The localisation of adding field in the location of conductors requires arrangement on a part of each slot boundary (at the bottom or in the wedge part) infinitely thin current layers with a line density 0H , the joint currents of its are equal to full currents of slots.

The scalar magnetic potential (SMP) of a current layer j-го of a slot will be defined by expression

( ) dlHIlUl

jjj ∫−=0

0/2п τ&&&

The analytical approaches to calculation of electric machines ... 43

where jIп – full current of a slot,

l – variable of integration on slot boundary, read out from edges of a current layer,

joH τ – projection of vector of a adding field to a direction of integration.

At summation SMP of current layers of all slots

( ) ( )∑=∑j

j lUlU

we will receive resultant relation, the geometrical site of which is dictated by convenience of calculation of magnetic field. For example, it is possible to suppose the field is fixed on a circle of an air gap passing on stator teeth edges.

In the beginning for simplification of a problem we neglect stairstep jaggies of stator and rotor cores (it is considered they are smooth; in subsequent, at refinement of the solution we shall come to conformal mapping of an air gap with bilateral stairstep jaggies on a circular ring)1) and we assume the length of the machine infinitely large (edge effects missed).

For the sake of greater generality we assume that the rotor of the

machine is performed from massive steel. The selection of this material does a mathematical model by more universal, as at transition to high enough values of magnetic permeability and resistivity on an axis z the massive environment gains properties of laminated steel core.

The cross section of the machine is shown in Fig. 1. By digits 1, 2 and 3

are marked accordingly: the massive ferromagnetic rotor, air gap, stator core, magnetic permeability which in the beginning is considered as infinitely large.

Usage SMP of current layers allows receiving the analytical solution the Dirichlet problem for circular rings of three mentioned environments at constancy of its magnetic permeabilities.

1) As a first approximation for estimated value of a smooth air gap it is possible to accept value

of a geometrical air gap δ (distance between circles passing on heads of fingers of a stator and a rotor), multiplied on air gap factor (Carter factor).


r

Fig. 1. Cross section of an electric machine with a solid rotor

2. CALCULATION OF A MAGNETIC FIELD IN AN AIR GAP

On the stator core surface with the bolded arcs are shown current layers

of phase A slots. Their scalar magnetic potential (SMP) can be presented by a trigonometric series

( ) ( ) ϕ=ϕ ∑∞

=

narUn

nA cos,1

222 (1)

where ( ) ( ) tFa nn ω= cos22 (2)

( )

mnw

n Ipn

kwF

π22 −= (3)

where mI – amplitude of a phase current of frequency ω ; nwkw, – number of orbits and winding factor of a phase;

2 1

3


p – number of pairs of poles of a statoric winding, ϕ – a ngular coordinate of a view point in a magnetic field ( мpϕϕ = , here мϕ – angular coordinate of the same view point, but in mechanical (geometrical) coordinate system). Summarising (1) with SMP two other phases

( ) ( ) ⎟⎠⎞

⎜⎝⎛ π

−ϕ⎟⎠⎞

⎜⎝⎛ π

−ω=ϕ ∑∞

= 32cos

32cos,

1

222 ntFrU

nnB (4)

( ) ( ) ⎟⎠⎞

⎜⎝⎛ π

+ϕ⎟⎠⎞

⎜⎝⎛ π

+ω=ϕ ∑∞

= 32cos

32cos,

1

222 ntFrU

nnC (5)

we have

( ) ( ) ( ) ( ) ( ) ( ) ϕϕωϕωϕ nj

nn

ntj

nn

nn eFeFntFrU mmm ∑∑∑

∞

=

∞

=

∞

=

⇔⇔=1

2

1

2

1

222 2

323cos

23,

(6)

where the upper sign corresponding to harmonics numbers

( ),...13,7,1,...,2,1,016 ==+= nkkn

the lower sign – ( ) .,...17,11,5,...,3,2,116 ==−= nkkn 1)

On other side of air area 2 (with radius 1r ) current of phase A and the eddy currents of ferromagnetic rotor will induce SMP in the form

( ) ( ) ( )∑∞

=

ϕ+ϕ=ϕ1

1112 sincos,

nnnA nbnarU && (7)

where ( ) 11 , nn ba && – unknown factors.

1) For the sake of simplicity of the analysis we assume that the stator winding has an integer of

slots on a pole and phase. In this case miss even harmonics in the formulas (2),(5),(6).


SMP, assigned by the formulas (1) and (7), allows finding the solution the Dirichlet problem for an annular domain 2 [1]

( ) ( ) ϕ++ϕ+=ϕ −−∞

=∑ nDrBrnCrArrU n

nn

nn

nn

n

nA sincos)(,

12 (8)

where the factors nnnn DCBA ,,, are determined by the formulas

( ) ( )( ) ( )( ) ( )( ) ( )

( ) ( ) ( )( ) ( )( ) ( ) ( )( ) ( ) ⎪

⎪

⎭

⎪⎪

⎬

⎫

−−=

−−=

−−=

−−=

./

,/

,/

,/

21

22

21

1221

21

22

21

1221

21

22

11

22

21

22

11

22

nnn

nn

nnn

nnn

nn

nnn

nnn

nn

nn

nnn

nn

nn

rrbrbrrrD

rrararrrC

rrbrbrB

rrararA

&&

&&

&&

&&

(9)

From expressions (8) and (9) follows the relation for radial component of magnetic field (MF) intensity on boundary of areas 1 and 2

( ) ( ) ( ) ( )( )( ) ( )( ) ϕνν

ϕννϕϕ

nbbn

naanrr

rUrH

nnnn

nnnnn

rr

ArA

sin

cos1,,

22

11

1

22

11

1

212

1

++

++=∂

∂−= ∑

∞

==

&&

(10)

where

( ) ( )nnnnn rrrr 2

12

22

22

11 / −+=ν

( ) ( )nnn

n rrrr 21

22212 /2 −−=ν

Taking into account operation of two other stator windings phases (B and

С), the formulas (7) and (10) can be given in the form

( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ϕ∞

=

ϕω∞

=

∞

=

∑

∑∑

⇔

⇔⇔ϕω−ϕω=ϕ

nj

nn

ntj

nnn

nn

eF

eFntFntFrU

m

m

&

&m&m&

1

1

1

112

1

1112

23

23sincos

23,

(11)


( ) ( ) ( )( ) ( )

( ) ( )( ) ϕ∞

=

ϕω∞

=

∑

∑

ν+ν⇔

⇔ν+ν=ϕ

nj

nnnnn

ntj

nnnnnr

eFFnr

eFFnr

rH

m

m

&&

&&

1

22

11

1

1

22

11

112

23

23,

(12)

where

( ) ( ) ( ) ( )

( ) ( ) ( ) ⎪⎭

⎪⎬⎫

+=

==

12

11

1

12

111

1 ,sin,cos

nnn

nnnn

FjFF

tFbtFa

&&&

&&& ωω

From the formula (11) we can receive expression for tangential component of MF intensity

( ) ( ) ( ) ( ) ( )∑∑∞

=

ϕ∞

=

ϕωϕ ±⇔±=

ϕ∂ϕ∂

−=ϕ1

1

11

1

11

1212 2

323,,

n

njn

n

ntjn eFn

rjeFn

rj

rrUrH mm &&

(13)

3. CALCULATION OF A MAGNETIC FIELD IN A FERROMAGNETIC ROTOR

We draw attention now to the ferromagnetic rotor – environment 1, rotating with speed Ω (in a direction of a first harmonic MF). Assuming const=1μ

1) and entering a vector magnetic potential

111 rot AH =μ

we have from the first Maxwell equation, written in fixed coordinate system

1) Below this condition will concern only to rather thin elementary concentric annular domains,

into which will be divided the massive ferromagnetic cylinder of a rotor. On boundaries of these rings the magnetic permeability will change jumps. Inside rings the environment will be linear, i.e. its parameters μ and γ are saved invariable. The monotone change of these parameters will be watched at infinitely large number of elementary concentric rings.


11rotrot δμ=A (14) where

[ ]⎟⎟⎠

⎞⎜⎜⎝

⎛+

∂∂

−−== 111

111 rotrad AvtAUqE γγδ (15)

is a current density in steel rotor (here 1v – running speed of a view point of the ferromagnetic environment).

From the vector analysis it is known

AAA 211 divqradrotrot ∇−=

Selecting for calibration of the vector potential the Culomb condition

0div 1 =A

And neglecting strength of an electrical (Culomb) field inside the steel array

0qrad 1 =U

from an equation (14) we shall have

[ ] 0111

12 =⎟⎟

⎠

⎞⎜⎜⎝

⎛+

∂∂

−γμ+∇ BvtAA

At harmonically varying MF it is fair

012 =Ω−−∇ zr eBrAjA &&& γμγμω (16)

As considered MF is two-dimensional parallel, thus in cylindrical coordinate

system

01rot 1111 =⎟⎟

⎠

⎞⎜⎜⎝

⎛

∂

∂−

∂

∂==

ϕϕ r

zzA

rAr

rBA


From here follows

zzr eAAAA === ϕ 111 ,0

⎪⎪

⎭

⎪⎪

⎬

⎫

∂

∂−=⎟⎟

⎠

⎞⎜⎜⎝

⎛∂

∂−

∂

∂−==

∂

∂=⎟⎟

⎠

⎞⎜⎜⎝

⎛∂

∂−

∂

∂==

rA

zA

rA

BA

Arz

ArAr

BA

zrz

zzrr

11111

11111

rot

,11rot

ϕϕ

ϕ

ϕϕ (17)

Therefore equation (16) can be written

0111

121

2

21

21

2

=−∂

∂Ω−

∂

∂+

∂

∂+

∂

∂z

zzzz AjAA

rrA

rr

A&

&&&&γμω

ϕγμ

ϕ (18)

We shall solve it by a method of variables separation, supposing

( ) ( )ϕΦ= nrRA z1

& (19)

And, if to select

( ) ϕ=ϕΦ njen m

then after the substitution (19) in (18) we receive an ordinary differential equation

( ) ( ) ( ) ( ) 02222

22 =+−++ rRrSkn

rdrdRr

rdrRdr n

where

γμωjk −=2

nSnpSn )1(1 1−=Ω

= mm

ωω

where

ωΩ−= pS 11 – slip of a rotor for first harmonic of a magnetic field;

The upper sign (-) corresponds to values 16 += kn ( ,...)13,7,1,...;1,0 == nk ; The lower sign (+) – to ,...).17,11,5,...;3,2,1(16 ==−= nkkn


Its solution is known [2]

( ) ( ) ( )rSkYCrSkJCrR nnnn

nnn 21

1 += ∑∞

= (20)

where ( ) ( )rSkYrSkJ nnn , – cylindrical Bessel functions of n order, accordingly

of the first and second kind; nn CC 21 , – arbitrary constants.

Taking into account, that ( ) ∞=0nY , it is necessary to accept 02 =nC . In

outcome we shall have

( ) ϕnjnn

nnz erSkJCA m& ∑

∞

=

=1

11 (21)

where the constants nC1 will be retrieved from boundary conditions of a problem.

Using (17) we have

( ) ϕ∞

=∑= nj

nn

nnr erSkJCnrjB mm&

111 (22)

( ) ϕ∞

=ϕ ∑ ′−= nj

nn

nnn erSkJCSkB m&1

11 (23)

where ( ) xJxJ nn ∂∂=′ / .

On boundary of areas 1 and 2 ( )1rr = we have

ϕϕ 2121 , HHBB rr&&&& == (24)

To draw attention to the equations (12) and (13) for rH2 and ϕ2H , from

equations (24) we shall receive a set of equations for finding of unknowns nC1

and ( )1nF&

( ) ( ) ( )( )( ) ( )

⎪⎪⎭

⎪⎪⎬

⎫

±=′−

+=

1

111

1

22

11

011

23

,2

3

nnnnn

nnnnnnn

FnrjrSkJCSk

FFrSkJCj

&

&&m

μ

ννμ


Its solution will be

( )

( ) ( )111011

2201

122

3

rSkJrSkrSkJn

FnjC

nnnnnn

nnn

νμ+μ

νμμ±=

& (25)

( )

( )

( )( ) n

n

nnn

nn

nn

rSkJrSk

rSkJrn

FF

2

1

11

1

10

1

21

ν

ν+

′μμ

−=&

& (26)

Considering known equalities

( ) ( ) ( )( ) ( ) ( )[ ]⎭

⎬⎫

+=−=′

+−

+−

xJxJxxJnxJxJxJ

nnn

nnn

11

11

2,2

(27)

it is possible to give the formulas (25), (26) in form

( )

( ) ( )[ ]11211111

221

1

3

rSkJarSkJarSk

FnjC

nnnnnnnn

nnn

+− +ν

νμ±=

& (28)

( )

( )

( ) ( )( ) ( ) n

n

nnnn

nnnn

n

nn

rSkJrSkJ

rSkJrSkJF

F

2

1

1111

1111

20

1

21

ν

ν+

−

+

νμμ

−=

+−

+−

&& (29)

where

nna

10

11 1

νμμ

+= , n

na10

12 1

νμμ

+−=

After a substitution of expression (28) for a factor nC1 in the formulas

(22), (23) for radial and tangential components of a magnetic flux density in the steel cylinder we shall have

( ) ( ) ( )[ ]

( ) ( )∑∞

= +−

ϕ+−

+

+

ν

νμ=

1 112111

112

1

2

1

11 2

3n nnnnnn

njnnnnn

n

nr

rSkJarSkJa

erSkJrSkJFnr

Bm&

& (30)


( ) ( ) ( )[ ]( ) ( )∑

∞

= +−

ϕ+−

ϕ +

−ν

νμ=

1 12111

112

1

2

1

11 2

3n nnnnnn

njnnnnn

n

n

SkJarSkJaerSkJrSkJFn

rjB

m&m& (31)

On other side of an air gap (in the environment 2) – on a circle 2rr = –

– we have according to the formulas (8), (11), (12), (29)

ϕνμ

ϕ jn

nnnnnn

nnnnnnn

nnr e

rqjJarqjJa

rqjJarqjJaFn

rrB m&&

)()(

)(~)(~

23

),(1

23

1212

3

11

12

3

1212

3

11)2(

11

2

022

+−

+−∞

= +

−= ∑ (32)

∑∞

=

±=1

)2(

2

022 2

3),(

n

jnn eFn

rj

rB ϕϕ

μϕ m&& (33)

where

nn

nn aa 1

2

1

21

~ −⎟⎟⎠

⎞⎜⎜⎝

⎛=

νν

, nn

nn aa 2

2

1

22

~ +⎟⎟⎠

⎞⎜⎜⎝

⎛=

νν .

The obtained formulas were used with reference to an asynchronous

motor (AM) with a solid rotor executed on the basis mass-produced three-phase AM type 4А200М 2УЗ, with the following data:

37н2 =P kW; 2943н =n rpm; 220нф =U V; 70н =I A; 06,120н =M Nm; 89,0cos =ϕ ; 90,0=η ; 361 =z ; 6=q ; 60=W ; 0652,0=R Ω (0,0207 rel.u.);

314=ω rad/s; 7105,0 ⋅=γ (Ω.m)-1; 971 =r mm; 9,972 =r mm; 9,0=δ mm; 130=l mm; material of rotor – steel 3.

In Figs. 2-3 the peak values of the first and seventh harmonics of a magnetic flux density on a surface of a solid rotor in a function of slip for the given motor, calculated on the formulas (30), (31)1), are shown at 0100μμμ == e and

var== eμμ (here eμ - magnetic permeability of steel on a rotor surface).

1) Cylindrical, Bessel functions of the first kinds with complex arguments:

0, 4343 >==−= njj

nnn SxereqrSjrSk ππγωμ

0, 44 <=== njj


where

nn Sq ωμγ= , rqx n=

expresses through Kelvin functions:

( ) ( )xijxrxeJ nnj

n bebe43 +=π

( ) ( ) nxijxrxeJ nnj

n ππ cosbebe4 −=


a) b)

Fig. 2. Amplitude values of radial (a) and tangential (b) components of first harmonic of a magnetic flux density on a solid rotor surface in function of the rotor slip concerning a first harmonic of a magnetic field (Curve 1: 0100μμμ == e ;

- 2: eμμ = accordingly to magnetisation curve of steel 3)

4. CURRENT DENSITY IN A FERROMAGNETIC ROTOR

In accordance with expressions (15), (16) complex amplitudes of a current density in a rotor will be

rzz BrAj 111&&& Ω−−= γγωδ (34)

After a substitution in (34) formulas (21) and (22) accordingly for zA1

& and

rB1& we shall have

( ) ϕ∞

=∑ωγ−=δ nj

nnnn

nz erSkJCSj m&&1

11

Taking into account expressions (27), (28) we shall have

( ) ( ) ( )[ ]( ) ( )

ϕ∞

= +−

+−∑+

+

ν

νγμω±=δ nj

n nnnnnn

nnnnnn

n

nz e

rSkJarSkJa

rSkJrSkJFSrr m

&&

1 112111

112

1

2

11 2

3


a) b)

Fig. 3. Amplitude values of the seventh harmonics – radial (a) and tangential (b) components of a magnetic flux density on the solid rotor surface in a function of slip concerning the first harmonic of a magnetic field at 0100μμμ == e

Subsequently, our interest will be the square of the module of current density

for n harmonic. From last formula, skiping indexes 1z, we have

( )

( ) ( )( ) ( ) ( ) ( )

⎥⎥⎦

⎤

⎢⎢⎣

⎡+

⎥⎥⎦

⎤

⎢⎢⎣

⎡+

×

×γμω⎟⎟⎠

⎞⎜⎜⎝

⎛

ν

ν=δδ=δ

∗+

∗−+−

∗

∗

111

21111

1

211

21

2

2

1

2

1

2 223

rSkJaa

rSkJrSkJaa

rSkJa

rSkJrSkJ

SFrn

nnn

nnnnn

n

nnnn

nnnn

nnn

nnnn

&&&

(35)

Cylindrical functions with considered complex argument

rqjrSjrSk nnn2/3=γμω−=

where

nn Sq γμω=

expresses through Kelvin functions


( ) ( ) ( ) njnnnnnnn ebrqbeijrqberrqjI β=+=2/3

where

( ) ( )rqbeirqberb nnnnn22 += ,

( )( )rqber

rqbei

nn

nnn arctq=β

In outcome the formula for a square of a module of a current density (35) will receive a form

( ) ( )( ) ⎟

⎟⎠

⎞⎜⎜⎝

⎛γμω⎟

⎟⎠

⎞⎜⎜⎝

⎛

ν

ν=δ

1

22

2

1

2

1

2 3rqDrqbSF

rn

nn

nnnn

n

nn

&& (36)

where

( ) ( ) ( ) ( ) ( ) ( )⎥⎥

⎦

⎤

⎢⎢

⎣

⎡−+⎟

⎟⎠

⎞⎜⎜⎝

⎛+= −++−+− 111111

1

22

111

21

21

211 cos2 nnnnnn

n

nnn

n

nnnnnn rqbrqb

aa

rqbaa

rqbarqD ββ

From obtained formulas in Figs. 4-5 are shown the main and higher

harmonics amplitudes of the current density on the solid rotor surface in function of rotor slip.

Fig. 4. Amplitude of a first harmonic of a current density on a solid rotor surface in function of slip (curve 1:

0100μμμ == e ; - 2: eμμ = is on a magnetization curve of steel 3)


a) b)

Fig. 5. Amplitudes of fifth (а) and seventh (b) harmonics of a current density on the solid rotor surface in function of slip at 0100μμμ == e

5. LOSSES IN A FERROMAGNETIC ROTOR

The losses in the steel cylinder from n harmonic of eddy currents in accordance with (36) are

( )

( ) ( ) ( )( ) drrrqbeirqberrqDSlF

rn

drrlddrrlP

r

nnnnnn

nn

n

n

r

nnn

r

nn

×+μωπ

⎟⎟⎠

⎞⎜⎜⎝

⎛

ν

ν=

=δδγπ

=ϕδδγ

=

∫

∫∫ ∫ ∗∗π

1

11

0

22

1

2

2

1

2

1

00

2

0

3

2

&

&

(37)

where l – active length of the rotor.

The integral in this expression is named as Lommel one. His value is known [3]. Finally

( )( )

( ) ( ) ( ) ( )[ ]11111111

2

2

1

2

129

rqbeirqberrqbeirqber

FrqDSln

P

nnnnnnnn

nn

n

nn

nn

+−−+ −⋅

⋅⎟⎟⎠

⎞⎜⎜⎝

⎛= &

ννμωπ

(38)


Using this formula the losses in a solid rotor from main and higher harmonics of an eddy current in a function of slip were calculated. The graphs of these relations are shown in Figs. 6-7.

Fig. 6. Losses in a solid rotor from a first harmonic of eddy current in a function of slip (curve 1:

0100μμμ == e ; 2 - eμμ = is on a magnetisation curve of steel 3)

a) b)

Fig. 7. Losses in a solid rotor from a fifth (а) and seventh (b) harmonics of an eddy current in a function of slip at 0100μμμ == e

The depth of penetration in the steel array n of a harmonics of a magnetic

field can be defined according to the formula

ωμγnn S2=Δ (39)


The graph ( )111 SΔ=Δ for a first harmonic of a magnetic field is presented in Fig. 8.

Fig. 8. Depth of penetration in the steel array of a rotor of a first harmonic magnetic fields in a function of slip (curve 1: 0100μμμ == e ; - 2: eμμ = is on a magnetization curve of steel 3)

6. THE ELECTROMAGNETIC TORQUE CALCULATION OF PERFORMANCE DATA

As is known, the area density of electromagnetic forces which are

operational in a tangent direction to a cylindrical surface of rotor will be equal [4]

( ) εϕϕ +==122п rrr HBT

where 0>ε – small constant;

r – radius of a circle flanking to a rotor. On this circle (at 0→ε ) radial component of a magnetic flux density

( )rB2 and tangential component of the magnetic fields intensity ( )τ2H will be

equal

( ) ( ) ( )∑∞

=

=1

1max12 cos,n

nrr ntrBrB ϕωϕ m& (40)


( ) ( ) ( )∑∞

=⎟⎠⎞

⎜⎝⎛ +=

11max12 2

cos,n

n ntrHrH πϕωϕ ϕϕ m& (41)

where

( )( ) ( ) ( )

( ) ( )( ) nrjn

r

nnnnnn

nnnn

n

nnnr

erB

rqjJarqjJarqjJrqjJ

rFn

rB

α

ννμ

1max

12/3

1212/3

11

12/3

112/3

1

11

221

1max 23

&

&&

=

=++

×=+−

+−

( ) ( )( ) ( ) ( )

( ) ( )( ) ( ) )2(

1max

12/3

1212/3

11

12/3

112/3

1

11

22

1max 23

παϕ

ϕ

ϕ

νν

+

+−

+−

=

=+−

=

njn

nnnnnn

nnnn

n

nnn

erH

rqjJarqjJarqjJrqjJ

rFnj

rH

&

&m&

In accordance with the theory of tensions for the electromagnetic torque of an electric machine will be fair the expression

( ) ( )

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )mnrm

n m

nrm

nrm

n m

nr

r

rHrBlrdmt

ntrHrBlr

drHrBlrdTlrM

ϕϕϕ

π

ϕ

ϕ

ππ

ϕ

ααπϕαπϕω

αϕω

ϕϕϕϕ

−=⎟⎠⎞

⎜⎝⎛ ++×

×+=

=×==

∑ ∑

∫∑∑

∫∫

∞

=

∞

=

∞

=

∞

=

sin2

cos

cos

,,

1max11 1

max2

1

2

01max1

1 1max

21

121

2

02

21

2

0п

21

m

m

(42)

With the help of this formula the electromagnetic torques from first and higher harmonics of a magnetic field in a function of slip for the mentioned above tested asynchronous motor with a solid rotor were determined. The outcomes of calculations are adduced in Fig. 9-10.

Being repelled the formula (32) for radial component of a magnetic flux

density it is easy to define the stator current of the asynchronous machine with a solid rotor.


For amplitude of a first harmonic of radial component of a magnetic flux density it is fair

( ) njnnn

nr eGF

rn

B ανμ 2

12

0)(2 2

3=&

Fig. 9. The electromagnetic torque of a first harmonic of a magnetic field in a function of slip

1S (curve 1: 0100μμμ == e ; - 2: eμμ = is on magnetisation curve of steel 3)

where

24

23

22

21

NNNNGn +

+= ,

here:

( ) ( )1121111~~ rqberarqberaN nnnnnn +− +=

( ) ( )122112~~ rqbeiarqbeiaN nnnn +=

( ) ( )1121113 rqberarqberaN nnnnnn +− +=

( ) ( )1121114 rqbeiarqbeiaN nnnnnn +− +=

γμωnn Sq =

nnn 21 ββα −=


121 /arctq NNn =β

342 /arctq NNn =β

( ) ( ) IkwnpF wnn π/222 −=&

As the magnetic flux of n harmonics of one pole is equal

nlB n

rnτ

π)(

22 && =Φ

then for an effective value of EMF vector of n harmonic of stator winding we have

( ) njnnnwnn eIGQjkfwjE απ −=Φ−= && 2/2

where

( )22106

nwnn kwnp

lQ ωνμπ

=

a) b)

c)

Fig. 10. The electromagnetic torque of fifth (а), seventh (b) and eleventh (c) harmonics of magnetic field in a function of slip S1 (curve 1: 0100μμμ == e ; - 2:

eμμ = is on a magnetisation curve of steel 3)


Considering an equilibrium equation of voltages in one phase of a stator winding circuit

1ZIEU &&& +−=

where

∑=n

nEE &&

let's have finally for a stator current

2

1

2

1 cossin ⎟⎠

⎞⎜⎝

⎛++⎟

⎠

⎞⎜⎝

⎛−

=

∑∑n

nnnn

nnn GQXGQR

UI

αα σ

and his phase angle

∑∑

−

+−=

nnnn

nnnn

GQR

GQX

α

αϕ

σ

sin

costqarc

1

1

The calculation of a stator current defined, as visible, in a function of rotor slip 1S , allows to find according to the offered above formulas all performance data of the asynchronous machine with a solid rotor.

The first harmonic of a magnetic flux density displaces on a surface of a rotor, having amplitude

( ) ( ) ( )[ ] ( ) ( )[ ]21max1

21max1max1

1 rBrBB rr ϕ+= (43)

As the amplitudes of the nearest higher harmonics (5-th and 7-th) of flux density are much less the first one, the magnetic permeability of a surface of a rotor eμμ = can be determined on a level of an effective value of a first harmonic of flux density [5].

On a large part of pole pitch of a rotor surface the magnetic permeability is close to minimum value


( )( )1действ1min rBfe === μμμ (44)

appropriate to effective value of a first harmonic of a magnetic flux density on the rotor surface (43). Therefore, as a first approximation, the magnetic permeability of the rotor surface can be determined by expression (44).

It is more logical to determine the permeability eμ as average value of a curve ( )αμμ ee = :

( )∫==π

ααμπ

μμ0

ср1 deee (45)

At this stage the magnetic permeability of whole rotor mass will be assumed also as identical and equal eμ – as magnetic permeability of a rotor surface. Below it will be considered more accurate calculation of perme-ability using splitting the cylinder of a rotor into concentric annular domains.

The nature of change minμμ =e functions of a rotor slip

is shown in Fig. 11. Here curve 2 is obtained with the help of a magnetisation curve of steel 3 and amplitude of a first harmonic of a magnetic flux density calculated on the formulas (40), (41), (43).

In Fig.12 are shown computational and experimental [6] relations of the torque (a) and a stator current (b) from slip for an asynchronous motor with a solid rotor from steel 3, made on the basis mass-produced electric motor АО2-81-2 of power 40 kW with synchronous speed 3000 rpm. The curves were calculated taking into account the frontal effect of eddy currents short circuiting of a rotor by reduction the electrical conductivity γ of rotor material on followed factor [7]

lk πτ21л +=

Fig. 11. Minimum value of relative magnetic permeability on a surface of the steel array of a rotor in a function of his slip (curve 1:

=0min μμe 100; -2: 0min μμe is on a magnet-ization curve of steel 3)


a) b)

Fig. 12. The electromagnetic torque (а) and current of a stator (b) in a function of slip of an asynchronous motor with a smooth solid rotor from steel 3, made on the basis of asynchronous machine АО2-81-2 (curve 1 corresponds to -1

л6 m)Om( к106 ⋅⋅=γ ; curve - 2:

-1л

6 m)Om( к105 ⋅⋅=γ ; curve- 3: -1л

6 m)Om(к104 ⋅⋅=γ ; 5,2кл = ; o - experimental points

The overstating of the calculated torque values as compared to experimental ones (approximately to 14 %) at this stage of calculation is due to neglecting the magnetic voltage in the stator core (it is assumed ∞=3μ ).

In summary we shall mark, that analytical and numerical methods of calculation of asynchronous machines with a solid rotor many activities are dedicated. The analytical approaches, naturally, are connected to those or diverse assumptions. Let's call these of them, which are assumed in this study:

1. the curvature of magnetic cores and air gap is neglected (the electromagnetic phenomena are considered in rectangular coordinate system) [6];

2. the effect of higher harmonics of magneto-motive force (MMF) on characteristics of machines is not analysed [6, 7, 8, 9];

3. the leakage of magnetic flux in an air gap is rather neglected (magnetic flux coming out of stator core is assumed as equal to magnetic flux entering from an air gap to a rotor core) [7, 9] or its calculation is made approximately [6, 8];

4. the calculation of a magnetic field in steel rotor core is made at given on his surface waves of a magnetic flux density ( δB ) and electric field strength ( zE ) [6, 8];

5. the connection between EMF of a rotor and stator is on the basis of an approximated integral representation of a Faraday law of flux density [6, 8, 9].


7. CALCULATION OF MAGNETIC FIELD AT FINAL VALUE OF MAGNETIC PERMEABILITY OF THE STATOR CORE

In the previous calculations the magnetic permeability of the stator core (environment 3 in Fig. 1) was assumed infinitely large. In this case environment 3 did not exert influence on a magnetic field in an adjacent annular domain – air gap. But such effect takes place in reality, when the magnetic permeability of the core of a stator is finite.

7.1. Calculation of magnetic field at the existence

of eddy currents in the stator core

Supposing the stator core electro-conductive, the calculation of a magnetic field in its can be made similarly to calculation of a magnetic field in a massive steel rotor. Pursuant to the formulas (20), (22), (23) we shall have for the environment 3

( ) ( )[ ] ϕnj

nnnnnr erkYDrkJDn

rjB mm& ∑

∞

=

+=1

32313 (46)

( ) ( )[ ] ϕϕ

nj

nnnnn erkYDrkJDkB m& ∑

∞

=

′+′−=1

323133 (47)

where 333 γμωjk −= , nn DD 21 , - constants defined from boundary conditions.

In the environment 2 (air gap) on boundary circle with radius 2rr = we have according with the formulas (8), (11), (12)

( ) ( ) ( )( ) ϕννϕ nj

nnnnnr eFFn

rrH m&& ∑

∞

=Σ+−=

1

21

12

222 2

3, (48)

( ) ( ) ϕϕ ϕ nj

nn eFn

rjrH m&& ∑

∞

=Σ±=

1

2

222 2

3, (49)

where ( ) ( ) ( )222

nnn FFF &&& Δ+=Σ , ( )2nF&Δ – adding stipulated by a magnetic field of the

stator core.


The formulas (46)-(49) contain three unknowns of a factor ( )221 ,, nnn FDD &Δ ,

which are determined from three equations expressing boundary condition

( ) ( )ϕμϕ ,, 22023 rHrB rr&& =

( ) ( ) ( )

ϕϕϕ

μϕ

ϕϕ

∂∂

+=2

2222

3

23 ,,,

rrUrH

rB&

&

( ) 0,33 =ϕrB r

&

From these three equations it is easy to receive expressions for unknowns

of factors

( ) ( ) ( )( )NkrMn

FFrkYnjD

n

nnnnnn

32103

21

123330

1 23

νμμννμμ

+−

+±=

&&

( ) ( ) ( )( )

NkrMnFFrkJnj

Dn

nnnnnn

32103

21

123330

2 23

νμμννμμ

+−

+=

&&m

( )( ) ( )( )

NkrMnFFNrk

Fn

nnnnn

32103

21

122302

νμμννμ

+−

+−=Δ

&&&

where

( ) ( ) ( ) ( )23333323 rkYrkJrkYrkJM nnnn −=

( ) ( ) ( ) ( )23333323 rkYrkJrkYrkJN nnnn ′−′=

7.2. Calculation of a magnetic field if neglected eddy currents in the stator core

The magnetic field in the environment 3 (Fig. 1) can be determined on

the basis of the solution of Neumann boundary problem. In this case we assume given the radial (normal) components of magnetic field intensity on boundaries of area 3:


( ) ( ) ( ) ϕϕϕ nbnarH nn

nAr sin~cos~, 2

1

223 +−= ∑

=

( ) 0,33 =ϕrH Ar

Then the scalar magnetic potential in this area will be equal [1]

( ) ( ) ( ) knrDrBnrCrArU nn

nn

n

nn

nnA ++++= −

∞

=

−∑ ϕϕϕ sin~~cos~~,1

3 (50)

where ( )

( )( )

( )( )

( )( )

( ) ⎪⎪⎪⎪⎪

⎭

⎪⎪⎪⎪⎪

⎬

⎫

−−=

−−=

−−=

−−=

+

+

+

+

,~

~

,~~

,~

~

,~~

22

23

223

12

22

23

223

12

22

23

212

22

23

212

nnn

nn

n

nnn

nn

n

nnn

n

n

nnn

n

n

rrnbrrD

rrnarrC

rrnbrB

rrnarA

(51)

k - constant.

As regards the magnetic field of phase A is even function (Fig.1) we can accept ( ) 0~ 2 =nb . Then we have 0~~ == nn DB .

Meaning that

( ) ( ) tHa nn ωcos~ 22 &=− (52)

and summarise magnetic fields of all three phases, we have

( ) ( ) ( ) ( )∑∑∞

=

∞

=

⇔=1

2

1

223 2

3cos23,

n

njn

nnr eHntHrH ϕϕωϕ m&m&& (53)

After the substitution of expressions (51) in the formula for SMP (50) we have for an inner boundary of area 3


( ) ( ) knan

rrU nn

nA +−= ∑

∞

=

ϕν

ϕ cos~, 2

1

3223

where

( ) ( )nnnnn rrrr 2

22

32

22

33 / −+=ν

Summarising SMP of all three phases and taking into account (52) we

shall receive

( ) ( ) keHn

rrU njn

n

n += ∑∞

=

ϕνϕ m& 2

1

3223 2

3,

Then the tangential component of MF intensity for 2rr = will be equal

( ) ( ) ( ) ϕϕ ν

ϕϕ

ϕ njn

nn eHj

rrU

rH m& 2

13

2

2323 2

3,, ∑

∞

=

±=∂

∂−= (54)

The boundary conditions for boundary between areas 2 and 3 are represented by two equations

( ) ( )ϕμϕμ ,, 233220 rHrH rr =

( ) ( ) ( )ϕϕϕϕ ϕϕ ∂

∂−=

2

222322

,,,r

rUrHrH

Substituting in these the formulas (6), (48), (49), (53), (54), we shall

receive two equations for finding two unknowns ( )2nH& and ( )2

ΣnF& :

( ) ( )( ) ( )2

32

11

22

0nnnnn HFF

rn &&& μννμ

=+− ∑ ; ( ) ( ) ( )2

2

23

2

2nnnn F

rnHF

rn &&& +=Σ ν

Solving above equations, we shall find

( )( ) ( )( )

( )nn

nnnnn r

FFnH

31032

21

1202

ννμμννμ

+

+−=

&&& (55)


( )( ) ( )

nn

nnnnn

FFF

3103

23

13202

ννμμμννμ

+

+−=Σ

&&& (56)

We see, that in an ideal case, when ∞=3μ , we have

( ) ( ) ( )222 ,0 nnn FFH &&& == Σ or ( ) 02 =Δ nF&

8. CONSIDERING THE CHANGE OF MAGNETIC PERMEABILITY IN MASSIVE FERROMAGNETIC ROTOR

The relation of magnetic

permeability to a diving depth of a magnetic field in a body of a rotor is easy for taking into account by splitting his array into elementary segments shaped of concentric rings (Fig. 13). To each i elementary segment there will correspond the value of magnetic permeability equal 1μ

i . On boundaries of elementary

segments the magnetic permeability will jump. Inside an elementary ring the parameters of the environment μi and γi are invariable, i.e.

environment will be homogeneous (linear), and the initial differential equation (18) for a vector potential in it will be linear. Therefore for such elementary segments there is no necessity initially to consider magnetic permeability as a continuous function of space coordinates. That it, apparently, becomes, if number of elementary concentric rings to take infinitely large.

Pursuant to the formulas (17), (20) we shall have following expressions for components of complex amplitude n of a harmonics of a magnetic flux density in an elementary segment

Fig. 13. Elementary segments of a rotor in the form of concentric rings with numbers 1, 2, …, m


( ) ( ) ( )[ ]rSkYCrSkJCnrjB n

inn

in

inn

inr

i211 += m& (57)

( ) ( ) ( )[ ]rSkYCrSkJCSkB n

inn

in

inn

in

ini ′+′−= 211ϕ& (58)

where

γμω 1,,,2,1 ii jkmi −== K

For last (m) of a segment shaped circle, it is necessary to accept

02 =nmC

The unknown factors niC1 , n

iC2 and ( )1nF& with the help of the formulas (12),

(13) for elementary segments will be determined from boundary conditions (24). For example, from boundary conditions of an air gap and outside surface

of a rotor the equations follow

( ) ( ) ( ) ( )

( ) ( )220

1101

12

11

11

1

2/3

2/3

nn

nnnnnnnn

Fj

FjrSkYCrSkJC&

&m

νμ

νμ

±=

=+ (59)

( ) ( ) ( ) ( ) 02/3 11

111

11

21

11

11 =±′+′ nnnnnnnn FSkrnjrSkYCrSkJC &μ

(60)

Let's enter into consideration vector of unknowns

[ ]Tmxxxx 221 K=

The components which one will be equal

( ) ( ) ( )121122

1221

1

3222

412

321

211

1

,,,

,,,,,

nmnm

mnm

mnm

mnnnn

FxCxCxC

xCxCxCxCx&

K

====

=====

−−

−−

−


Then from boundary conditions i and (i+1) of elementary segments having a boundary circle of radius 1rr i= , we shall receive equations

12222,1212111 baxaxaxa mm =++ (61)

02222,2222121 =++ mm axaxaxa (62)

02222,121212,1222,121212,12 =+++ +++++++−−+ iiiiiiiiiiii xaxaxaxa (63)

02222,221212,2222,221212,22 =+++ +++++++−−+ iiiiiiiiiiii xaxaxaxa (64)

where

i = 1, 2, ..., (m-1)

( )11

11 rSkJa nn= , ( )11

12 rSkYa nn= , 23 102,1 nm ja νμm=

( )11

21 rSkJa nn′= , ( )11

22 rSkYa nn′= , nm Skrjna 111

12,2 23 μ±=

( )112,12 rSkJa i

ni

nii =−+ , ( )11

12,12 rSkJa in

inii

+++ −= , ( )12,12 rSkYa i

ni

nii =+

( )1

122,12 rSkYa i

ni

nii+

++ −= , ( ) ( )111

12,22 rSkJka in

in

iiii ′= +−+ μ

( ) ( )11

12,22 rSkYka i

ni

nii

ii ′= ++ μ , ( ) ( )( )1

11112,22 rSkJka i

ni

nii

ii++

++ ′−= μ

( ) ( )( )1

11122,22 rSkYka i

ni

nii

ii++

++ ′−= μ ,

0,0 2,22,12 ==− mmmm aa , ( ) ( ) ( )[ ]zYzYzY nnn 1121

+− −=′

The unknown constants of rings will be determined from a vector-matrix equation

bAx = (65)

where

( ) ( )2201 23 nn Fjb &νμ±= ; 0=kb , mk 2 ..., ,3,2=


The matrix A has following frame

[ ]

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−

−−−−−−

=

012,222,232,20000000

012,1222,1232,120000000

000006665646300

000005655545300

000000044434241

000000034333231

2,2000000002221

2,1000000001211

A

mmammammammammamma

aaaa

aaaa

aaaa

aaaamaaamaaa

K

K

KKKKKKKKKKKKKKKKK

K

K

K

K

K

K

The solution of a set of equations (65) allows to determine components of complex amplitude n of a harmonics of a magnetic flux density in ring segments of a rotor according to the formulas (57), (58). Therefore, there are modules of amplitudes of a first harmonic of flux density in these segments

( ) ( )[ ] ( )[ ]21

max121

max11max1 ϕBBB i

rii +=

on which ones (turning towards their effective values [5]) with the help of a magnetisation curve of a rotor material – the magnetic permeability i of a segment 1μ

i will be determined. If to enter the denotations

⎪⎭

⎪⎬⎫

==

==

,, Re

, Re

2222

1111

ni

mni

ni

ni

ni

mni

ni

ni

CJbCa

CJbCa

(66)

,ke2he,ke2he xrxixixr nnnn ππ−==

(67)

where

( )1

11 rSx i

nii −= γμω , , ..., 2, ,1 mi = ( )

110 rr = (68)


it is possible to give amplitude values of radial and tangential harmonics of flux density on outside boundaries of annular domains of a rotor with radiuses ( )

11 ri−

( )mi ..., 2, ,1= in form

( )( )

22

111 n

rn

ri

nr

i DCr

nB += − (69)

( ) 22111 22

1nn

niii

ni DCS

B ϕϕϕ

γμω+= (70)

where expression for functions nC and nD depend on the sign of slip nS 1).

1) 0>nS ( ) ( ) ( ) ( )xixibxrxraxrbxiaxCxC n

in

in

in

in

in

in

in

in

in

in

rn

r hebehebebebe 2211 −++−+−=−=+ (71)

( ) ( ) ( ) ( )xrxrbxixiaxibxraxDxD ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

nr

nr hebehebebebe 2211 −−+−+−==+ (72)

( ) ( ) ( )( ) (

) ()xixixixixr

xrxrxrbxrxrxrxrxi

xixixiaxrxrxixb

xixixrxraxCxC

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

nn

11111

111211111

111211111

11111

hehebebehe

hebebehehebebehe

hebebebebebebei

bebebebe

+−+−+

−+−+−+−+

−+−+−+−

+−+−+

−++−−

−++−++−−+−

−++−+−++−+

+−+−== ϕϕ

(73)

( ) ( ) ( )( ) (

) ()xixixixi

xrxrxrxrbxrxrxrxr

xixixixiaxrxrxixib

xixixrxraxDxD

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

nn

1111

111121111

1111211111

11111

hehebebe

hehebebehehebebe

hehebebebebebebe

bebebebe

+−+−

+−+−+−+−

+−+−+−+−

+−+−+

−++−

−+−−++−−+

++−−+−+−+

+−++−== ϕϕ

(74) 1) Bessel cylindrical functions of the second kinds with complex arguments

0,4343 >==−= njj


0,44 <=== njj

nnn SxereqrSjrSk ππωμγ

where nn Sjq ωμγ=

are expressed through Kelvin functions:

( ) ⎟⎠⎞

⎜⎝⎛ +−+−−= xrxijxixrxeY nnnn

jn beke2beke243

πππ

( ) nxrxijxixrxeY nnnnj

n πππ

π cosbeke2beke24⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ +−−= .


2) 0<nS ( ) ( ) ( ) ( )xixibxrxraxrbxiaxCxC n

in

in

in

in

in

in

in

in

in

in

rn

r hebehebebebe 2211 ++++−==− (75)

( ) ( ) ( ) ( )xrxrbxixiaxibxraxDxD ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

nr

nr hebehebebebe 2211 +++−+==− (76)

( ) ( ) ( )( ) (

) ()xixixixi

xrxrxrxrbxrxrxrxr

xixixixiaxrxrxixib

xixixrxraxCxC

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

nn

1111

111121111

1111211111

11111

hehebebe

hehebebehehebebe

hehebebebebebebe

bebebebe

+−+−

+−+−+−+−

+−+−+−+−

+−+−−

−+−+

+−+−+−+−+

++−+−++−−+

+−+−== ϕϕ

(77)

( ) ( ) ( )( ) (

) ()xixixixi

xrxrxrxrbxrxrxrxr

xixixixiaxrxrxixib

xixixrxraxDxD

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

ni

nn

1111

111121111

1111211111

11111

hehebebe

hehebebehehebebe

hehebebebebebebe

bebebebe

+−+−

+−+−+−+−

+−+−+−+−

+−+−−

+−+−

−−+−++−+−

−+−+−+−+−+

++−−== ϕϕ

(78)

The expressions (57), (58) rather easily allow to calculate a two-layer rotor [10] (first layer – massive cylinder from alloy of iron and copper, the second layer – the laminated core from electric machine’s steel) or to take into account availability of the coopering of the solid rotor surface enriching performance data of an asynchronous motor [6, 8]. In this case for one or several first concentric rings it is necessary to accept

0μμμ == eii и Cu(Al)γγ =i

If there are homogeneous concentric ring elementary segments the formulas (35), (38) for a current density and losses in a rotor will be other. Pursuant to initial expression (34) for a current density and, considering the formula (57) for radial component n of a harmonics of a magnetic flux density for i of a ring, we shall have for complex amplitude of a current density in it

( ) ( ) ( )[ ]rSkYCrSkJCSj i

nnii

nni

nini

211 +−= γωδ& (79)


According to denotations above introduced in (66)-(68), (71), (72), (75), (76) it is fair

( ) ( )[ ] ( ) ( )xDjxCrSkYCrSkJCj n

rn

rn

inn

in

inn

i ±± +=+ 21

where

rqrSx ni

nii == 11 γμω ; the sign (+) corresponds to 0>nS ; the sign (-) – 0<nS .

Therefore, for a module of amplitude n of a harmonics of a current density in

i ring we have

( ) ( ) ( )rqDrqCS ni

nr

ni

nr

nini 22

1±± += ωγδ (80)

The losses in a rotor from n harmonic of a current density should be found

according to the formula

( )[ ] ( ) ( ) ( )[ ]( )

rdrrqDrqCSlrdrlPr

rn

in

rn

in

rm

i

in

r

i

ni

n

i

i∫∑∫−

±±

=

+==1

1

1

122

11

2

0 1

2

γωπγ

δπ (81)

In Figs. 14-16 some parameters of a magnetic field of ring concentric

segments of a solid rotor in a function of slips calculated on the formulas (65), (69), (70) are presented.

Fig. 14. Amplitude of a first harmonic of radial component of a magnetic flux density on outer boundaries of ring concentric segments (m = 12) of a solid rotor as a function of slip


Fig. 15. Amplitude of a first harmonic of radial component of a magnetic flux density on outer boundaries of ring concentric segments (m = 4) of a solid rotor as a function of slip

Fig.16. Magnetic permeability of concentric ring segments (m = 12) of a solid rotor in a function of slip

PRACE INSTYTUTU ELEKTROTECHNIKI, zeszyt 220, 2004


LITERATURE1)

1. Polyanin A.D.: Handbook on linear equations of mathematical Physics.-M.: Physmathlit, 2001, p.576

2. Kamke E.: Handbook on ordinary differential equations.-M.: Nayka, p.1971.-576

3. Watson G.: The theory of Bessel functions, parts 1, 2.-M., 1949.

4. Ivanov-Smoleskij A.W. Electromagnetic forces and transformation of energy in electric machines.-М.: Vyssh. shkola, 1989, p.312

5. Neumann L.R.: A skin effect in ferromagnetic bodies. M.-L.: State publishing house, 1949, p.190

6. Lishenko A. I., Lesnik V.A.: Asynchronous machines with solid rotors. Kiev: Nauk. dumka, 1984, p.168

7. Kucevalov W.M.: Asynchronous and synchronous machines with solid rotors. M.: Energia, 1979, p.161

8. Postnikov I.M.: Designing of electric machines. Kiev: Publishing house of a technical publications. UkrSSR, 1960, p.910

9. Kovach K. P., Rac I.: Transients in AC machines (transl. from German). M. , 1963., p.744

10. Mogilnikov W.S., Olejnikov A.M., Strelnikov A.N.: Asynchronous motors with two-layer rotor. - M.: 1983, p.120

Manuscript submitted 28.10.2004 Reviewed by Jerzy Zadrożny

ANALITYCZNE PODEJŚCIE DO OBLICZANIA MASZYN ELEKTRYCZNYCH W OPARCIU O ROZWIĄZANIE PROBLEMÓW BRZEGOWYCH DOMEN PIERŚCIENIOWYCH

METODĄ ROZDZIAŁU ZMIENNYCH FOURIERA

A.A. AFANASJEV, A.G. BABAK, A.V. NIKOLAEV

STRESZCZENIE Szczelina powietrzna obustronnie uzębiona, będąca obszarem przejścia, może być konformalnie odwzorowana jako pierścień kołowy; tradycyjnie tę procedurę przeprowadza się w przybliżeniu przy pomocy współczynników Cartera. Przekroje po-przeczne rdzeni stojana i wirnika mają również kształt pierścieni.

Wirujące pole magnetyczne w obszarach przewodów (w żłobkach, w oknach między biegunami) może być przedstawione przez sumę pól potencjału i pewnego dodatkowego (łatwego do obliczenia) pola.

1) All in Russian


Umieszczenie dodatkowego pola w miejscu, gdzie są przewody wymaga rozmieszczenia w brzegowej części żłobka (w części przy dnie lub klinie) nieskończenie cienkich warstw prądu tak, aby łączne prądy były równe pełnym prądom żłobków.

Użycie skalarnych magnetycznych potencjałów (SMP) warstw prądu pozwala otrzymać analityczne rozwiązanie problemu Dirichleta dla kołowych pierścieni wspomnianych trzech obszarów (szczelina powietrzna, rdzeń magnetyczny stojana i wirnika) przy założeniu, że przenikalności magnetyczne są stałe.

Dla uzyskania lepszego uogólnienia problemu zakładamy, że rdzeń wirnika jest wykonany z litej stali. Wybranie tego materiału czyni model matematyczny bardziej uniwersalnym, gdyż przy przejściu do wysokich wartości przenikalności magnetycznej i rezystywności, ma-teriał lity uzyskuje własności stali blachowanej.

SMP warstw prądu uzwojeń statora i wirnika przedstawia się za pomocą szeregów Fouriera.

Parametry pola elektromagnetycznego w litym wirniku (promie-niowa i styczna składowa indukcji, gęstość prądu i straty mocy) otrzymuje się na podstawie analitycznego rozwiązania różnicz-kowych równań Bessela dla wektorowego potencjału magnetycznego metodą rozdziału zmiennych i przedstawia się je przez odpowiednie funkcje Bessela i Kelvina.

Obliczone dane dla silnika asynchronicznego z litym wirnikiemi porównano z wynikami prób. Zmiany przenikalności magnetycznej w stali wirnika uwzględniono przez podzielenie go na współosiowe pierścienie o stałej przenikalności w każdym z nich.

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