ميحرلا نمحرلا الله مسب - saadawi1 the voltage distribution of an insulator of 3 units, if the maximum voltage of each unit is 17 kV, and the capacitance to ground

بسم هللا الرحمن الرحيم

رب اشرح لى صدرى ويسر لى أمر

واحلل عقدة من لسانى يفقهوا قولى

صدق هللا العظيم

Chapter 11

pp. 370-389

Overhead Line Insulators part 2

Potential Distribution over a String of Suspension Insulators p. 376

• The voltage across the discs of the string is not uniformly distributed. This is because of the capacitances formed between the metal parts of the insulators and tower structure.

• These capacitances could be made negligibly small by increasing the distance between the insulators and the tower structure which requires larger lengths bigger size of the towers and hence it becomes uneconomical.

4/17/2013 Prof. Dr. Magdi El-Saadawi 3


4/17/2013 5Prof. Dr. Magdi El-Saadawi

• In practice the insulators are not very far from the tower structure and hence these capacitances affect the voltage distribution across the string


• Since the insulator discs are identical, each disc is represented by its mutual capacity C

• A shunt capacitance C1 is the capacitance of metal part of the insulator disc to the tower structure (ground). So that C1 = m C• Where m is the ratio between the capacitance to

ground to the mutual (unit) capacitance

• Let V be the operating voltage and V1,V2,V3

and V4 the voltage drops across the units starting from cross towards the power conductor.

V = V1 + V2 + V3 + V4

• The objective is to find out the voltage across each disc as a multiple of the operating voltage and to compare the voltage across each unit.


C : capacitance of each insulator unit.

C1 = m C : capacitance to ground

is the capacitance of metal part of the insulator

unit to the tower (m<1).

V1, V2, V3 the voltage across each unit starting from

the cross arm towards the power conductor.

V = V1 + V2 +V3 Line voltage


At point P:

I2 = I1 +Ic1

ωC.V2 = ωC.V1 + ωmC.V1

V2 = (1+m).V1

At Point Q:

I3 = I2 + Ic2

ωC.V3 = ωC.V2 + ωmC.(V1+V2)

V3=m.V1 +(1+m).V2

=(m +(1+m)2).V1

V3 = (1+3m +m2).V1


The voltage across the units cam also given by:

n= 1 V2 = (1+ m). V1

n=2 V3 = (1+ 3 m +m2). V1

n=3 V4 = V3.(1+ 6 m +5 m2 + m3). V1

n=4 V5= V4.(1+ 10 m + 15 m2 + 7m3 + m4 ). V1


To Find the voltage across each disk

• For 3 disc string:

• V = V1 + V2 + V3

• V = V1 + V1 (1+ m) + V3 (1+ 3m + m2)

• V = V1 (3 + 4 m + m2)

• Then you can find V1 , V2 and V3

• WhereV: Voltage across the insulator string, (phase Volt)


For m < 1 V3 > V2 > V1

Insulator (String) Efficiency: %η = (V/n.Vmax) x 100

V: Voltage across the insulator string, (phase Volt)

n: number of insulator units.

Vmax: Voltage across the insulator unit near to the

power line (for n = 3, Vmax = V3).4/17/2013 12Prof. Dr. Magdi El-Saadawi

Example (1)

Find the voltage distribution of an insulator of

3 units, if the maximum voltage of each unit

is 17 kV, and the capacitance to ground is

20% of unit capacitance, also find the

insulator efficiency.


V3 = 17 kV, m=20% = 0.2

V2 = (1+m).V1 = 1.2 V1

V3 = (1+ 3 m +m2 ). V1 =1.64 V1

V1 = 17/1.64 = 10.36 kV

V2 = 1.2x10.36 = 12.43 kV

V= V1 + V2 + V3 = 39.8 kV

Insulator efficiency = (39.8/3x17)x100=78.03%


In general, voltage across the units can given by:

Vn+1 = Vn.(1+m) + (V1+V2+……+ Vn-1).m

n= 1 V2 = (1+m). V1

n=2 V3 = V2.(1+m) + V1.m

n=3 V4 = V3.(1+m) + (V1 +V2).m

n=4 V5= V4.(1+m) + (V1 + V2 +V3).m


Example (2)

An insulator string for 66 kV line has 4 units.

The capacitance to ground is 10% of the

capacitance of each insulator unit. Find the

voltage across each insulator unit and string

efficiency.


V2 = (1+m). V1 = 1.1 V1

V3 = V2.(1+m) + V1.m = 1.31 V1

V4 = V3.(1+m) + (V1 +V2).m =1.651 V1

V1 +V2 +V3 +V4 = = 38.1 kV

V1 (1+1.1 +1.31+1.651)=38.1

V1=7.53 kV, V2= 8.28 kV, V3=9.86 kV, V4= 12.43 kV

String efficiency = (38.1/4x12.43)x100=76.6%


Improvement of String EfficiencyMethods of Equalizing Potential (p.386)

1- Reducing the ground capacitance relative to the

capacitance of insulator unit (reduce m where m = c1/c ):

This can be done by increasing the length of cross arm and

hence taller supporting tower which uneconomical.


Improvement of String Efficiency

2- Grading of insulator units: It can be seen that the

unequal distribution of voltage is due to the leakage current

from the insulator pin to the tower structure. The solution is

to use insulator units with different capacitances.

This requires that unit nearest the cross arm should have

minimum capacitance (maximum Xc) and the capacitance

should increase as we go towards the power line.


This means that in order to carry out unit grading,

units of different types are required. This requires

large stocks of different units which is uneconomical

and impractical.


3- Static Shielding (Guard Ring): ( الحماية حلقة )

This method uses a large metal ring surrounding the bottom

insulator unit and connected to the line. This ring is called a

grading or guard ring which gives a capacitance which will

cancel the charging current of ground capacitance.

Improvement of String Efficiency


Guard ring serves two purposes:

- Equalizing the voltage drop across each insulatorunit.

- protects the insulator against flash over.



pp. 389






Example (3)

A 3-unit insulator string with guard ring. The

capacitance to ground and to guard ring are

25 % and 10 % of the capacitance of each

unit. Determine the voltage distribution and

string efficiency.



At point A:

I1 + Iy = Ix + i1

0.1ωC.(V2+V3) + ωC.V2

= ωC.V1 + 0.25 ωC.V1

1.25 V1 -1.1 V2 -0.1 V3 =0.0

At point B:

I2 + Iz = Iy + i2

0.1ωC.V3 + ωC.V3

= ωC.V2 + 0.25ωC.(V1+V2)

0.25 V1 + 1.25 V2 -1.1V3 =0.0


Also: V1 + V2 + V3 = V

Solve, to get:

V1= 0.295V, V2 = 0.2985V, V3 = 0.406V

η = V/(3x0.406V)x100= 82.1 %

Find the voltage distribution and insulator

efficiency without a guard ring.


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ميحرلا نمحرلا الله مسب - saadawi1 the voltage distribution of an insulator of 3 units, if the maximum voltage of each unit is 17 kV, and the capacitance to ground