1

Cálculo de las

Pérdidas Adicionales

en

Bobinas en Folio

Ing. Álvaro Portillo Junio 2010

2

Bobinas en Folio Pérdidas Adicionales debidas al

Campo Magnético de Dispersión Radial

Este cálculo lo realizaremos basados en la Referencia: Prof.Dr. Janusz Turoswki: “Additional losses in foil and bar wound transformers” IEEE PES Winter Meeting and Symposium, New York, January 25-30, 1976, Paper Nº A-76-151-1

MAXOMAX HB µ= OOO HB µ= L

aakR π

δ++−= 211

NIdPHAXC

2=×→→

∫ → NIkL

HR

MAX 2= → RMAX kL

NIH 2=

NIdPHRDC

2=×→→

∫ → NIHaLH OMAX 22 2 =+

RMAXO kNINILHNIHa 2222 2 −=−=

( )NIkHa RO −= 122 2 → ( )

22

12

a

NIkH R

O

−=

3

( )2

2121

22 2

2

2

21

2

2

a

aa

L

NI

L

aa

a

NIk

a

NIH RO π

δπ

δ ++=

++=−=

δπ

++=

21

22aa

aM →

LM

NIHO

2=

O

MAX

R HH

kM

1=

Para dimensiones típicas M resulta comprendido entre 2 y 3.

Basados en el Método de las Imágenes calcularemos la penetración axial del campo magnético de dispersión radial OH en la bobina de lámina suponiendo que la bobina es una pared indefinida de espesor h , (indefinida en el sentido radial x e indefinida en el sentido longitudinal z ).

Ecuaciones de Maxwell: 0=∂∂

+∧∇

→→

t

BE

→→

= HB Oµ 0=∂∂

+∧∇

→→

t

HE Oµ

→→

=∧∇ JH →→

= EJ σ →→

=∧∇ EH σ

Solución del Tipo: ( )→→

= ieyHH tjω ( )→→

= keyEE tjω

Condiciones de Borde: ( ) OO Hh

Hh

yHyH −=

+→+≥−=22

( ) OO Hh

Hh

yHyH +=

−→−≤+=22

4

( )

( ) →

→→→

→

=∂∂

∂∂

∂∂

=∧∇ iedy

ydE

eyEzyx

kji

E tj

tj

ω

ω00

( )

( ) →

→→→

→

−=∂∂

∂∂

∂∂

=∧∇ kedy

ydH

eyHzyx

kji

H tj

tj

ω

ω 00

( )→

→

=∂∂

ieyHjt

H tjωω

0=∂∂

+∧∇

→→

t

HE Oµ →

( ) ( ) 0=+→→

ieyHjiedy

ydE tjO

tj ωω µω

( ) ( ) 0=+ yHj

dy

ydEOµω

→→

=∧∇ EH σ → ( ) ( )

→→

=− keyEkedy

ydH tjtj ωω σ → ( ) ( ) 0=+ yE

dy

ydHσ

( ) ( ) 0=+ yE

dy

ydHσ → ( ) ( )

dy

ydHyE

σ1

−= → ( ) ( )

2

21dy

yHddy

ydEσ

−=

( ) ( ) 0=+ yHj

dy

ydEOµω →

( ) ( ) 01

2

2

=+− yHjdy

yHdOµω

σ

( ) ( ) 02

2

=− yHjdy

yHdO σµω

σµω Ojk =2

( )2

1σµω

σµωσµω OOO jjjk +===

2

σµωα O=

αδ

1=

( )αjk += 1

Solución del tipo: ( ) ykyk eCeCyH −+ += 21

5

Con las condiciones de borde determinamos las constantes de integración 1C y

2C :

O

hk

hk

HeCeCh

H −=+=

+−+

22

212

O

hk

hk

HeCeCh

H +=+=

−+−

22

212

hkhk

hk

hk

O ee

eeHC −+

−+

−+

−=22

1 hkhk

hk

hk

O ee

eeHC −+

−+

−+

=22

2

( ) ykhkhk

hk

hk

Oyk

hkhk

hk

hk

O eee

eeHe

ee

eeHxH −

−+

−+

+−+

−+

−+

+−+

−=2222

( ) ( )ykykhkhk

hk

hk

O eeee

eeHxH −+

−+

−+

−−+

−=22

+

−=−

−+−+−+ 2222

hk

hk

hk

hkhkhk eeeeee

( )22

hk

hk

ykyk

O

ee

eeHxH

−+

−+

−

−−=

( )→→

= jexHH tjω

→

−+

−+→

−

−−= je

ee

eeHH tj

hk

hk

ykyk

Oω

22

→ ( ) →→

= je

hk

ykHH tj

Oω

2sinh

sinh

( ) ( )dy

ydHyE

σ1

−= → ( )2

22

2h

kh

k

ykyk

O

ee

eekHxE

−+

−+

−

+=

σ

( )→→

= kexEE tjω

→

−+

−+→

−

+= ke

ee

eekHE tj

hk

hk

ykyk

Oω

σ2

22

2

6

→→

= EJ σ

→

−+

−+→

−

+= ke

ee

eeHkJ tj

hk

hk

ykyk

Oω

22

→ ( ) →→

= ke

hk

ykHkJ tj

Oω

2sinh

cosh

Simplificaciones:

δα 212

212

>>→>>→>> hhh

k

222

hk

hk

hk

eee+−+

≈−

( ) →−+−→

−+

−+→

+≈

−

+= keeeeHkke

ee

eeHkJ tjykyk

hk

Otj

hk

hk

ykyk

Oωω 2

22

( ) →−+−→

+≈ keeeeHkJ tjykykh

k

Oω2

→

+−

−−→

+≈ keeeHkJ tj

yh

kyh

k

Oω22

22

hy

h+≤≤−

Esta expresión coincide con las fórmula 11 del Paper de Turowski

→

−−→

≈ keeHkJ tjy

hk

Oω2

20

hy +≤≤

( ) 4211

π

δδα

je

jjk =

+=+=

δµδ O

OO BHJ ==1

→

−−→

≈ keeeJJ tjjyh

kω

π42

12 2

0h

y +≤≤

Esta expresión coincide con las fórmulas 12 y 13 del Paper de Turowski

( )

→

−

+−→

−−→

=≈ keeeH

keeHeJ tjy

hjj

Otjy

hk

O

j ωδπ

ωπ

δδ2

1

424 22

→

−−

−−→

−−

−−→

=≈ keeeH

keeeeH

J tj

yh

jyh

Otj

yh

j

yh

jO ω

δπ

δωδδπ

δδ

24222

4 22

7

→

−

−→

−−+

−−≈ ke

yh

jy

h

eB

J tj

yh

O

O ωδ

δπ

δπ

δµ2

4sin2

4cos2

2

Esta es la fórmula 4.111 de la página 154 del libro de Kulkarni Tomando la parte real obtenemos (resultan corrientes en cuadratura):

tjte tj ωωω sincos +=

→

−

−→

−−−

−−≈ kt

yh

ty

h

eB

J

yh

O

O ωδ

πω

δπ

δµδ sin2

4sincos2

4cos2

2

Operando en →

J :

→

−−→

−

−−

−

−≈ keh

yhj

h

yheJJ tjh

yh

ωδ πδ

πδ 42

1sin

42

1cos2 2

1

1

Tendremos además la corriente de carga:

→→

= keJJ tjOL

ω2

La corriente total será:

→

−−→→

−

−−

−

−+= keh

yhj

h

yheJkeJJ tjh

yh

tjOTOT

ωδω πδ

πδ 42

1sin

42

1cos22 2

1

1

→

−−→

−

−−

−

−+= keh

yhj

h

yheJJJ tjh

yh

OTOTωδ π

δπ

δ 421

sin42

1cos2 2

1

1

Esta expresión coincide con la fórmulas 16, 17 y 18 del Paper de Turowski Tomando la parte real obtenemos:

tjte tj ωωω sincos +=

→−−

→

−

−+

−

−+=

−−

−−

kth

yheJt

h

yheJJJ

h

yh

h

yh

eeOTOT ω

πδ

ωπ

δδδ

sin42

1sincos

421

cos22

1

2

1

11

Esta expresión coincide con las fórmulas 17 y 18 del Paper de Turowski

8

→→

= keJJ tjTOTTOT

ω2

−

−−

−

−+=

−−

42

1sin

42

1cos2

1

1

πδ

πδ

δ

h

yhj

h

yheJJJ h

yh

OTOT

Calcularemos ahora la potencia disipada.

Considerando valores eficaces será: ∗×= TOTTOTV JJpσ1

3m

W

−

−−

−−

+= 42

1

2

1

1

πδδ h

yhj

h

yh

OTOT eeJJJ

−=h

yhZ

2

1

δ

−−−+= 4

1

πzj

ZOTOT eeJJJ

−+−∗ += 4

1

πzj

ZOTOT eeJJJ

+

+=×

−+

−

−−

−∗ 41

41

ππzj

ZO

zjZ

OTOTTOT eeJJeeJJJJ

Zzjzj

ZOOTOTTOT eJeeeJJJJJ 22

144

12 −

−+

−−

−∗ +

++=×

ππ

ZZOOTOTTOT eJzeJJJJJ 22

112

4cos2 −−∗ +

−+=×π

( )zzz sincos22

4cos +=

−π

( ) ZZ

OOTOTTOT eJzzeJJJJJ 2211

2 sincos2 −−∗ +++=×

∫∫ ∫ ∫∫∫∫ ∗∗∗ ×=×=×=2

00

2

0 0

221h

TOTTOT

ah

l

TOTTOT

V

TOTTOT dyJJla

dzdydxJJdVJJPσσσ

FOLIOISO eNaaa =−= 2

( )[ ]∫ −− +++=2

0

2211

2 sincos22

h

ZZOO dyeJzzeJJJ

laP

σ

9

−=h

yhZ

21

δ

δdy

dz −= dzdy δ−=

20:

hy → 0

2: →

δh

z

( )[ ]∫ −− +++=δ

σδ 2

0

2211

2 sincos22

h

ZZOO dzeJzzeJJJ

laP

( ) ∫∫ −− +++=δδ

σδ

σδ

σ

2

0

221

2

0

12 2

sincos22

h

Z

h

ZOO dze

laJdzzze

laJJ

hlaJP

( ) [ ]δ

δδδ

2cos1cossincos 22

0

2

0

hezedzzze

hh

Z

h

Z−

−− −=−=+∫

−=

−=

−−−∫ δδ

δ hh

Z

h

Z eedze 12

1

2

1 2

0

22

0

2

−+

−+=

−−δδ

σδ

δσδ

σ

hh

OO ela

Jh

ela

JJhla

JP 12

cos122 2

12

12

−+

−+=

−−δδ δ

δδ

σ

h

O

h

OO e

hJ

Jhe

hJ

JhlaJP 1

2cos1221 2

21212

−+

−+=

−−δδ

δδ

σ

h

O

h

OO e

J

Jhe

hJ

JhlaJP 1

22cos1221 1212

δ>>h →

++≈

OOO J

J

hJ

JhlaJP

2221 112 δ

σ

Esta expresión coincide con la fórmula 20 del Paper de Turowski

δOH

J =1 LM

NIHO

2=

ha

NIJO =

L

h

M

a

NI

ha

LM

NI

NI

haH

J

J O

O δδδ2121 ===

10

+=

+==

L

h

M

a

hL

h

M

a

J

J

hJ

J

RI

Pk

OO

RADADICRADADIC δ

δδ

δ 221

22

22

22 112

+==

LM

ha

LM

a

RI

Pk RADADIC

RADADIC δ2

22

Finalmente resulta la “Fórmula de Turoswki” :

+==

LM

ha

LM

a

RI

Pk RADADIC

RADADIC δ21

42

Calcularemos la profundidad de penetración δ para cobre y aluminio:

O

OO

µωρ

δρµωσµω

α2

22=→==

fπω 2= m

HyO

7104 −×= πµ

Cobre @ ( )Ct º : ( )310

23510210.0 7 t

m+

×=Ω −ρ

Aluminio @ ( )Ct º : ( )300

22510344.0 7 t

m+

×=Ω −ρ

50 Hz 60 Hz ( )mmδ

75ºC 85ºC 75ºC 85ºC

Cobre 10.3 10.5 9.4 9.6

Aluminio 13.2 13.4 12.1 12.3

Para temperaturas comprendidas entre 75ºC y 85ºC será:

Cobre: f

504.10=δ Aluminio:

f

503.13=δ

11

• Fórmula DTDS - PAUWELS:

+=

LM

ha

LM

ak RADADIC δ2

14

Correciones:

• == hL Altura de la Lámina de BT • 6.0=k para Cobre • 9.0=k para Aluminio

kM

a

hM

ak RADADIC

+=

δ21

4

• Fórmula ABB:

+=

LM

ha

LM

ak RADADIC δ2

14

Correciones:

• MM 2→ • =L Altura de la Ventana del Núcleo • =h Altura de la Lámina de BT • =2h Altura de la Bobina de AT

• 9.279.6236 2

2

2 +

−

=h

h

h

hk

12 =→= khh

kLM

ha

LM

ak RADADIC

+=

δ221

2

4

12

“Transformer Engineering – Design and Practice” S.V.Kulkarni – S.A.Kharparde Marcel Dekker Inc. – 2004 4.5.4 Eddy loss in foil windings In small transformers, foil windings (made up of thin sheets of copper or aluminum) are quite popular because of the simplicity obtained in the winding operations. The metal foil and the insulation layer can be wound simultaneously on a special machine and the whole process can be easily mechanized. In foil windings, eddy loss due to axial leakage field is insignificant because of very small thickness of the foil. On the contrary, the radial field results into higher eddy loss at the ends of the foil windings. The eddy loss in a foil winding can be evaluated by analytical or numerical methods. The current density has been obtained as a solution of integral equation (Fredholm type) for a two winding transformer in [20]. For specific transformer dimensions given in the paper, the coefficient of additional loss is 1.046, i.e., the eddy loss is 4.6% of the DC I2R loss. In another approach [21], a boundary-value field problem is solved for magnetic vector potential in cylindrical coordinate system using modified Bessel and Struve functions. For symmetrically placed LV (foil) and HV windings, the additional loss factor reported is 5.8%. For an asymmetrical arrangement of windings this loss increases. In [22], the foil winding is assumed as a vertical section of an infinitely wide and deep conducting plate, which is assumed symmetrically penetrated on both sides by plane electromagnetic waves. The geometry is given in figure 4.10, where h is the height of the foil winding and BX denotes the value of flux density at winding ends.

The induced eddy current density (with the vertical dimensions specified with respect to the middle of the winding height) is [22]:

−−+

−−=

−

−

δπ

δπ

δµδ

xh

jx

h

eB

J

xh

O

Oe

24

sin2

4cos

2 (4.111)

13

The eddy loss in terms of the peak value of current density is given as

∫=V

ee dVJP2

2

1

σ (4.112)

It is clear from equations 4.111 and 4.112 that

δ

−−

∝∝

xh

ee eJP2

22(4.113)

Hence, the maximum value of eddy loss occurs at y=h/2 (at two ends of the winding) since at this height the exponential term has the maximum value of 1, whereas its value is just 0.0067 at y=(h/2)-2.5×δ, which means that more than 99% of the induced eddy current loss occurs in just 2.5×δ depth from the winding ends. For an aluminum foil winding, whose skin depth is 13.2 mm at 50 Hz, almost all the eddy loss is concentrated in 33 mm from the winding ends. The foil winding eddy current density distribution calculated by equation 4.111 for a typical transformer is shown in figure 4.11.

Thus, the current density at the ends can be about 1.5 to 2 times the uniform current density, resulting into local heating of 2.25 to 4 times that in the middle portion of the winding. Hence, the temperature rise at the foil winding ende should be carefully assessed.

Usually, there is no temperature rise problem in a foil winding since the thermal conductivity of aluminum/copper is quite good, and the winding edges are well exposed to the cooling medium. If a transformer with foil winding is supplying a power electronic load, there is increase in the heating effect at winding ends on account of harmonics [23].

14

With the development in computational facilities, the numerical tools such as FEM can easily analyze the foil winding eddy problem without any simplifications done in analytical formulations. It is actually a three-dimensional problem. The losses in a foil winding calculated in the core-window cross section are lower than that calculated in the region outside the window. The radial flux at winding ends is usually more in the region outside the window, and hence the additional loss factor may be of the order of 10 to 15%. The position of foil winding affects its eddy loss significantly. If the foil winding is the outer winding, the radial flux density at ends reduces resulting in reduction of eddy loss [24]. References: 20. N.Mullineux, J.R.Reed, and I.J.Whyte: “Current distribution in sheet and foil wound transformers”, Proceedings IEE, Vol. 116, No. 1, January 1969, pp. 127–129. 21. M.M.El-Missiry: “Current distribution and leakage impedance of various types of foil-wound transformers”, Proceedings IEE, Vol. 125, No. 10, October 1978, pp. 987–992. 22. J.Turowski: “Additional losses in foil and bar wound transformers”, IEEE PES Winter Meeting and Symposium, New York, January 25–30, 1976, Paper No. A-76–151–1. 23. A.Genon, W.Legros, J.P.Adriaens and A.Nicolet: “Computation of extra joule losses in power transformers”, ICEM 88, pp. 577–581. 24. B.S.Ram: “Loss and current distribution in foil windings of transformers”, Proceedings IEE—Generation, Transmission and Distribution, Vol. 145, No. 6, November 1998, pp. 709–716.

15

“Advanced Calculations of Magnetic Leakage Fields in Transformers” O.W. Andersen

2.2 Sheet winding transformer

When radial flux tries to penetrate a sheet winding, induced eddy currents will be set up to prevent it, according to Lenz's law. As a result, flux lines will be straightened out through the winding and will be almost purely axial, except at the very ends. This is shown in Fig.2 in a sheet winding with six turns.

Fig.2. Sheet winding transformer

Induced currents are phase shifted with respect to the main current in the winding, so that vector potentials must be calculated as complex numbers. Current densities will usually be very high at the winding ends, often several times the average. However, this is concentrated in a small volume, and losses and temperature rise are usually tolerable. Current density distribution can be improved by making the low voltage sheet winding slightly longer than the high voltage wire winding in a transformer such as the one in Fig.2. Current density distribution is calculated in each turn. The distribution is shown in Fig.3 for the inner turn of the sheet winding in Fig.2 from the radial centerline to the end of the winding. Each vertical bar represents the current density calculated in a finite element mesh. The horizontal dashed line represents the current density for a uniform distribution.

16

Fig.3. Current density distribution, inner turn