Ijeart021104 by International Journal of Engineering & Technical Research

International Journal of Engineering and Advanced Research Technology (IJEART) ISSN: 2454-9290, Volume-2, Issue-10, October 2016

Capacitive dividers on the high voltage lines Haroun Abba Labane, Alphonse Omboua 

Abstract— In the developing countries, several villages are crossed by lines with high voltage yet the populations do not profit from the electricity which passes above their heads. The classical solution for the transformation of the very high voltage into low voltage for the profit of the rural populations is expensive and is not economically profitable for the distributors of electrical energy. The voltage dividers already used like transformers in the stations, can be resized for the extraction of small quantities along high voltage lines. This single-phase technique, would plaid in favor of the reasonable costs. On the work of transporting, the high voltage line would assume the role of distributor of electrical energy to the rural populations. This article is dedicated to the detailed research of the system, for the control of the parameters of the choice equipment, for an easy dimensioning in rural electrification.

II. SCHEMA OF THE PRINCIPLE CAPACITIVE DIVIDER Line HV I C1

L Transfo MV/LV

I1 I2 F

C2 F

Ground

Fig.2: Schema of the capacitive divider

Index Terms— Extracting, Energy, capacitive dividers, high voltage lines.

The equivalent schema reduced to the primary transformer is: Line HV V1

I. INTRODUCTION PRESENTATION OF A CAPACITIVE VOLTAGE DIVIDER

xf2

VΦ

I2 V2

For this kind of power supply, the return of the electric power being done by the ground. So, the level of the voltage of transport is reduced to a level of distribution then allowing the use of a MV/LV distribution transformer to pass from the average voltage level to that of the use. One distinguishes 3 parts associated on three voltage levels: high, medium and low.

xf1

Zn xm

Groun

Fig.3: Schema of the system reduced to the primary MV/LV transformer Notations Z1 ,Z2: respectively impedances of capacity C1 and C2 Z2: impedance of capacity C2 Zn: impedance of the load to be supplied ZPO: impedance of the compensation coil ZF: impedance of the shock absorber filter Z: equivalent impedance of the network subjected to the medium voltage V2 R1: Resistance of winding of the primary transformer R2: Resistance of winding of the second transformer Rm: Resistor losses by iron in the magnetic circuit of the transformer Rp: Equivalent resistance of the transformer (reduce to the primary) XF1: Reactance of winding of the primary transformer XF2: Reactance of winding of the secondary transformer Xm: Reactance of magnetizing the MV/LV transformer V1 ,V2: respectively voltages of poles capacity C1 and C2 V: Single Voltage of HV Line Un: Nominal voltage of the load Zn Sn: Apparent power of the load cosφ: Factor of power of the load Zw = Z-Zpo α: Real part of Zw

Fig. 1: Capacitive voltage divider - Photo Hydro Quebec 225 kV/20 kV - The high voltage is composed by the line and the capacitor C1 for coupling. - The medium voltage is composed by the capacitor C2, the compensation coil, the damping filter and the primary transformer MV/LV. -The Low voltage includes the second transformer and the load impedance Zn.

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Capacitive dividers on the high voltage lines β: Imaginary part of Zw F: Filter shock absorber k: Transformation ratio I: The primary current through C1 I1: The current in the branch MT I2: The current through the capacitor C2 L: Inductance coil of compensation of agreement coil T: MV/LV transformer

VΦ: single voltage transmission line on which the draw is made. I

Fig.5: Simplified representation of the system U is the voltage of the line was for phase-ground voltage U V = 3

Role of the filter shock absorber: The filter shock absorber (F) is at the same time used to filter harmonics on the side distribution. III. CALCULATION OF THE IMPEDANCE Z SUBJECTED TO THE MV VOLTAGE V2

Ohm's

In load, the current value IO is negligible that we can remove the branch of magnetizing and the voltage V2 is applied to the following circuit: Z

V

Role of the compensation coil: The compensation coil plays a role in the compensation of the reactive energy that the capacitors produce.

ωL

C1 I

Z1 

1 jC1

law

gives V1  Z1 I

; Z2 

1 jC 2

; so

and V  V1  V2

V2  V  V1  V  Z1I . __

Call Z éq  Z 1  Z 2 // Z that is to say __

Z éq  Z 1 

Z2Z Z2  Z

[Ω] (2)

Z éq : Represents the equivalent impedance of the entire system connected to the line HV.

The main current is then I  Fig. 4: Summary of the circuit under the MV voltage V2 (a) is the ratio: voltage medium / low voltage transformer. xp and Rp be the reactance and resistance of the transformer reduced to the primary and Z p 0  jL

V2  V  Z1(

V Z éq

V2  V (1 

We have : x p  xf1  a 2 xf 2 ; R p  R1  a 2 R2 ; Z  Z p0 

 Z  )  V 1  1   Z éq 

Z1 ) Z2  Z Z1  Z2  Z

( R p  jx p  a 2 Z n )  Z F ( R p  jx p  a 2 Z n )  Z F ( R p  jx p  a Z n )  Z F

V2  V (

We set ZW 

( R p  jx p  a 2 Z n )  Z F

, now Z1 

knowing that

xy y ,  y x y x y

we have ZW  Z F 

1 jC1

R p  jx p  a 2 Z n  Z F

Knowing that ZW  Z F 

*Ω+

(1)

(Z F ) 2 R p  jx p  a 2 Z n  Z F

Z p 0  jL we have Z  jL  Z w

We set   Re( Z w) and   Im(Z w) ,

1 ) Z1 Z 1  1 Z2 Z

; Z2 

we find V2  V (

(Z F ) 2

(Z F )2 Z  ( Z po  Z F )  R p  jx p  a 2 Z n  Z F

Writing

V Z éq

[V] (3)

1 , jC 2

1 ) C 2 Z1 1  C1 Z

  C1 1 V2  V  1 (C1  C 2 ) 1   j (C1  C 2 ) Z 

     

  V 1  V2  C1  C 2  1 ( ) 1  j  ( C C1 1  C 2 )Z 

     

(4)

The transformation _

ratio k 

we have: Z  Z   2  ( L   ) 2

V V2

C  C2 1 ratio is then: k  1  C1 jC1 Z __

IV. DIVISION RATIO (K) AND OUTPUT VOLTAGE V2 V2: Output voltage divider MV

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International Journal of Engineering and Advanced Research Technology (IJEART) ISSN: 2454-9290, Volume-2, Issue-10, October 2016  C  C2  1 k ( 1 ) 1   C1 jZ (C1  C2 )  

The limited development

(5)

1  1 x

A schema opens Z   , so k  k0 

V V2

(

 (1)

xk ,

k 0

We can write :

C1  C2 C1

(6)

1  C 1 ( 1 ) C2

It has already been established that Z can be written Z    j ( L   ) . Transformation ratio. k

This indicates that the voltage

 (1)

(

k 0

n C 1 1 1  ( )   (1) k 1k 1 C C1  C 2 C 2 C2 k 0 1 1 C2 k

n 1 C    (1) k 1k 1 , so C2 k 1 C2

is in phase with the line

voltage V of the power line that k is pure real. So just what Z    j ( L   ) is imaginary pure, and therefore

  0  Z  j( L   )



n C1  C2  C  1  1  ( ) (1) k ( 1 ) k  jC1C2  C2  1  jC2Z k 1



If we call the C0 capacity equivalent to the set of two capacitors C1 and C2 supposed in series and Z 0 the corresponding impedance, we would have:

*Ω+ (7)

Of the formula of k, we understand that the voltage division ratio depends on Z and therefore much of the load Z n . The

1 1 1 C  C2    1 C0 C1 C 2 C1  C2

ratio k is not constant. The voltage regulation V 2 is necessary

from where C 0 

because V 2 varies with load Z n . C1

[F]

   2  __ __   C1  C1   1     Z éq  Z 0 1    __   C   C   1  jC2 Z  2  2  

(8)

It is known that k0 is a constant strictly greater than 1. The relationship C2  (k 0  1)C1 shows that C2  C1 ; the lowest capacity is one that is directly connected to the line. __

Z

, (

Z1 Z 2  Z1 Z  Z 2 Z , so Z2  Z

jC1  C 2 1   C1  C 2 Z éq

Z éq 

jC1C 2  C1  C 2

(1 

, So

Z éq  Z 0 

)0

jC1  C 2 1   C1  C 2 Z éq

1 1  ( C1  C 2 C 2

*Ω+

(11)

(C1  C 2 ) jC1  C 2

[F] (9)

Z éq whatever Z .

We have :

b) Calculation of the current that passes through C1 We already know that

1 ) jZ (C1  C 2 ) 1 (1  ) jC 2Z

(1 

As C1 is small compared to C2, to see what limit will tend __

Zéq  Z0 lim C

In this case Z has no more influence and it is as if the system is idle. So if C2 >> C1 it Z   is as if it is, as if the vacuum system.

) jC 2Z j (1  ) Z (C1  C 2 )

C1 0 C2

Z éq  Z 0

Z 2 Z

Z éq 

*Ω+ (10)

Conclusion: As C2 to C1 is too large then so

a) The Study of impedance Z éq Z2 __

;

Development to order 2 gives:

C2  (k0  1)C1

C1  C 2 C1  C 2

1 j C 0

Z0 

The relationship k 0  C1  C 2 gives

C1 k ) C2 k

and so

 C1  C2  1 ) 1  , C1 j  Z ( C  C ) 1 2  

The result Z éq  Z 1 

1 , with x  1 being 1 x

(1 

) jC 2Z j (1  ) Z (C1  C 2 )

I 

V

gives

Z éq

with 1 ) with C 1 1 C2

C1  0 C2

1 ) jC2Z I   j (C1  C2 ) (1  ) Z (C1  C2 )

jC1C2V

(1 

[A] (12)

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Capacitive dividers on the high voltage lines

Let us now expressed

A empty (transformer MV/LV not connected) we have Z  I  I0 

C1C 2V

[A]

C1  C 2

(Z F )

Z  (Z po  Z F ) 

transformation ratio k  (13)

k (

U2 R p  jx p  Z F  a n e j Sn This expression of the current (13) can also be written: (

jC1C2V (C1  C2 )

)(

 C1  C 2  1 ) 1   C1 jZ (C1  C 2 )  

(14)

which gives: 1 

C1 little toward C2 we have (1 

) jC2Z I ( )( ) 1 (C1  C2 ) (1  ) jC2Z jC1C2V

jC1C 2 V

I 

1 C jC 2Z (1  1 ) C2

A

C1  C 2

(15)

V. REACTIVE POWER

Without the compensation coil, the power Qc supplied by the capacitors is: QC  X 2 ( I 2 ) 2  X1 ( I ) 2

[VAR]

V   (

has already been (1 

) jC 2Z ) 1 (1  ) jZ (C1  C 2 )

QC 

jC 2Z 1 1 jZ (C1  C 2 )

V2 2 CC  )  X1 ( 1 2 ) 2 V X2 C1  C2

by replacing X1=

 C1  C2  1 ) 1  V2 C1 j  Z C 2 

because : (21)

1 1 and X2 = , C1 C2 2

[VAR] (17)

2 QC  C2  C1 k  1 2  (V2 )

[F]

(22)

V2: MV voltage supplied to the MV / LV transformer and

The calculation V  gives: _

(20)

2 2 we find : QC  C2V2  C1V2 k  1 ;

 C1  C2  1 ) 1  V2 C1 j  Z ( C  C ) 1 2  

V   (

C1  C 2 )V2 (k  1) C2

2 V2 2  X 1 (C1 ) 2 V2 2 k  1 X2

Calculation: V  : It had already been shown V  (

(19)

2 V2 2 CC  C C  X1 ( 1 2 ) 2  ( 1 2 ) 2V2 2 k  1 X2 C1  C2 C2

so QC  X 2 (

C1  1 so C1  C 2

and the expression V CC  QC  X 2 ( 2 ) 2  X 1 ( 1 2 ) 2 V  X2 C1  C 2

we set  

C1  C 2 )V2 (k  1) so C2

(16)

Now I 2  V2 while I the expression of the main current

1

C1 C1  C 2

C1 C k (C  C2 ) )V2 (1  1  1 ) C1  C2 C2 C2

V   (

(C1  C 2 )

k

C1 C  C2 C1k )V2 (1  ( 1 )(  1)) C1  C2 C2 (C1  C2 )

(

I 2  I 2 and I  I

calculated. I  ( jC1C 2V )  (

C jC 2Z (1  1 ) C2

So this expression reported in (19) gives: V  (

;

 C  C1 1  (1  1 ) k ( )  1 C 2  C1  C 2 jC 2Z 

   C1  C 2  1  k ( ) 1  C1   C1 jC  Z ( 1  ) 2  C 2  

)

jC2Z ) 1 (1  ) C jZ C2 ( 1  1) C2

V

But it had already been established previously that

2 2

(1 

1 as a function of the j Z C 2

k 

V

: complex transformation ratio

This result expresses the reactive power that would produce the capacitive divider from the capacities (C1) and (C2), k is

[V] (18)

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International Journal of Engineering and Advanced Research Technology (IJEART) ISSN: 2454-9290, Volume-2, Issue-10, October 2016 generally a complex number because there is no evidence that tensions V and V 2 are in phase. The value

I 

L2

C1  2

(23)

V2



V V2

QC 

 1)2

V2

V V2

QC  QL ,

 C1  C 2  C1 (

V V2

F

We can write that QC  C1(V )  (C2  C1 )V2 2

V

C1  C 2

, the

(VAR)

C1  C 2

The compensation coil here would be one that would swallow up the power QC

 C1  C2  C1 (

C1V

power QC produced by the capacitors C1 and C2 takes its minimum value C1C 2V 2

average of the two extreme values.  1)2  C2  C1 (

We had to establish that for a voltage V2 

V 2 2

V

C1 ) C2

is as low as it can be considered as the value of QC the

C2  C1 (

1

2 2 Equation (30) becomes V2  X 1 I 0 2 2  ( L ) V2 2

V

C1  4C1

)  (1 

I  I0 

V V     L   C 2  C1 (  1) 2   C 2  C1 (1  ) 2  V2 V2     V

1 jC 2Z 1

1

jC 2Z (1 

being low we see immediately that L

V2 2

V

C1  C 2

is within an interval of length L

V 2 2

C1 value

jC1C 2V

(24)

V2 

C2  L

and C2. The expression QC imposes the choice of a low voltage V2 average risk of injecting too much reactive power in the network and therefore the over-sizing of the compensation coil.

V2 2

now

and Z 2   2  ( L   ) 2

 ( L )



(26)

(C1V ) 2



(C1  C2 ) 2  2  ( L   ) 2

C1  C 2  2  ( L   ) 2





(27)



 2  ( L   ) 2  ( L ) 2  (2 

By adopting the solution of the total compensation, the coil of compensation must be able to absorb reactive power produced by the capacities C1 and C2. This results in the following equation: VAR

 ( L )

C2 (C 1 C 2)  2  ( L   ) 2  LC1

VI. COIL COMPENSATION OF REACTIVE POWER INJECTED BY C1 AND C2

X 2 I 2 2  X 1 I 2  X L (I1 ) 2

C1  C 2

C1  C2

V 2 will be the subject of a choice in relation to the capacity C1

C1  C2

C1V

C1C2V 2

3 is the voltage of the line to HV it is imposed then that

C1C2V 2

LC1 C2 (C1  C2 )

(28)

C1 )(L )   2   2  0 C2 (C1  C2 )

 2  ( L ) 2  2  ( L )   2 

(29)

LC1

(30)

C 2 (C1  C 2 ) Equation of second degree that ( L) will determine the value of the compensation coil.

(25)

Line HV

VII. STUDY OF THE VACUUM SYSTEM Lω

Line HV

C1 I 1

V

Ground

Fig.6: capacitive divider and Coil compensation Fig.7: empty capacitive divider I2 

V2 X2

I1 

V2 Z

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Capacitive dividers on the high voltage lines In this case, assume that Z   and therefore expression k (

 C1  C 2  1 ) 1   gives C1 jZ (C1  C 2 )  

k (

C1  C 2 ) , k becomes a pure real which implies that the C1

Another line of reasoning leads to the same result; indeed, to C C empty the capacity equivalent Céq  1 2 to the assembly C1  C2 of the two is at an impedance Z 0  The relationship V 

voltages V and V2 are in phase; so k 1  1 

C2 C 1  2 C1 C1

(31)

I0 

V2

 C 2  C1 k  1  C 2  C1 (

I0 led to V  (C1  C2 ) I 0 C éq  C1C2

The same result is confirmed: C1C2V

C  C2 1 ,  1 C éq  C1C 2

C2 2 ) C1

(37)

C1  C2

VIII. APPLICATIONS IN CASE OF THE VILLAGE MOGROUM IN CHAD

QC  C 2V 2 2 (1 

C2 ) C1

(32)

QC   (C1  C 2 )

C2 2 V2 C1

(33)

On the assumption that one has a capacitive divider that provides a medium voltage of 20 kV from the potential line 220 kV project. It was V 

As k  1  C2 , we find

220kV

, we may exercise the load calculation to

write

QC   (C1  C2 )(k  1)V2 2

k0 

(34)

C C1  C 2 V   6,35 and so 2  5,35 C1 U C1

Assuming a load current to the relatively low land of about 0.3 A, we write the equation: either I0  ( QC   C1  C 2 ( k  1) 2 V2 V a) Vacuum 2 Calculation

(35)

The combination of these two equations results in:

It was demonstrated that  __  V 1 V2   C1  C 2 1  ( ) 1  j (C1  C 2 ) Z C1 

C1  5,95nF and C2  31,80nF A vacuum, such a divider would inject reactive power QC  X 1 ( I 0 ) 2  X 2 ( I 0 ) 2

     

 ( X 1  X 2 )(I 0 ) 2 2 1 1 I0  ) C1 C2  Include: QC= 25 kVAR

QC  (

V2 : MT output voltage divider

V = U : Line Voltage HT on which it is drawn off. 3

A vacuum Z   , was then

V20 

(C1  C 2 )

V

(1 



(38)

IX. CONCLUSION

V C1  C 2 ( ) C1

V C1  C2 ( ) C1

jC1C 2V

VAR

V2 0 

The system does not produce directly the low voltage, it requires the incorporation of an intermediate MV/LV transformer to supply the load LV. The maintenance of such system can be in favor of the rural populations of the developing countries.

(36)

b) Calculation of the current that passes through C1 and C2 We know that I 

C1  C 2 )V with   314 rad/s C1  C 2

REFERENCES [1]

)

jC 2Z j (1  ) Z (C1  C 2 )

[2]

[3]

A vacuum Z   is found C1C 2V I  I0  C1  C 2

[4]

BG CHECO – CEGELEC – Postes à couplage capacitif(SCC) pour électrification rurale fiable et rentable. Mars 1995 A. OMBOUA - Thèse de doctorat 2002 - Université de liège : Alimentation de faibles charges directement des lignes à haute tension. PP 47-71. Iliceto F. ‘’lightly Loaded Long HV transmission Lines. Guidelines prepared at the request of the Word Bank‘’ Washington DC,1982-1983. Iliceto F, Cinieri E, Casely-Hayford L., Dokyi G. Operation results of an experimental system and Applications in Ghana, IEEE Transactions on Power Delivery, Vol.N°4, October 1989

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International Journal of Engineering and Advanced Research Technology (IJEART) ISSN: 2454-9290, Volume-2, Issue-10, October 2016 [5] [6] [7]

[8]

Cinieri E., Iliceto F., Dokyi G. Operation Results in Ghana’’. African 1992, Swaziland, sept.1992, paper A-054. TURAN GÖNEN: Electric Power Transmission System Engineering: Analysis and Design. Copyrigt 1988 by John Willey Electrical Safety. A guide to the causes and prevention of electrical Hazards, J.Maxwell Adams. IEE power series 19, The Institution of Electrical Engineers, London, UK, 1994. Ch.S. Walker, ‘’Capacitance, Inductance and Cross-Talk Analysis’’ Artech House, Boston, 1990.

Haroun Abba Labane,: Master's degree 2014 University of Brazzaville and PhD student in engineering sciences of the University of Brazzaville-Congo.

Pr. Alphonse Omboua is an electrical engineer; he received his PhD, from the University of Liege (Belgium) in 2002. He received a post-gradual diploma in Ouagadougou in rural development energies. He is presently lecturer at the University of Brazzaville-Congo, Dept. of electricity. He is expert of rural electrification (decentralized) using photovoltaic systems.