‫بسم هللا الرحمن الرحيم‬
‫رب اشرح لى صدرى‬
‫ويسر لى أمر‬
‫واحلل عقدة من لسانى‬
‫يفقهوا قولى‬
‫صدق هللا العظيم‬
Chapter 11
Overhead Line
Insulators part 2
pp. 370-389
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
Prof. Dr. Magdi El-Saadawi
4
4/17/2013
Prof. Dr. Magdi El-Saadawi
5
• 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
4/17/2013
Prof. Dr. Magdi El-Saadawi
6
• 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 = V 1 + 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.
4/17/2013
Prof. Dr. Magdi El-Saadawi
7
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
4/17/2013
Line voltage
Prof. Dr. Magdi El-Saadawi
8
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
4/17/2013
Prof. Dr. Magdi El-Saadawi
9
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
4/17/2013
Prof. Dr. Magdi El-Saadawi
10
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
• Where
V: Voltage across the insulator string, (phase Volt)
4/17/2013
Prof. Dr. Magdi El-Saadawi
11
For m < 1
V 3 > V2 > V 1
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
Prof. Dr. Magdi El-Saadawi
12
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.
4/17/2013
Prof. Dr. Magdi El-Saadawi
13
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%
4/17/2013
Prof. Dr. Magdi El-Saadawi
14
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
4/17/2013
Prof. Dr. Magdi El-Saadawi
15
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.
4/17/2013
Prof. Dr. Magdi El-Saadawi
16
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%
4/17/2013
Prof. Dr. Magdi El-Saadawi
17
Improvement of String Efficiency
Methods 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.
4/17/2013
Prof. Dr. Magdi El-Saadawi
18
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.
4/17/2013
Prof. Dr. Magdi El-Saadawi
19
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.
4/17/2013
Prof. Dr. Magdi El-Saadawi
20
Improvement of String Efficiency
3- Static Shielding (Guard Ring): ( ‫)حلقة الحماية‬
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.
4/17/2013
Prof. Dr. Magdi El-Saadawi
21
Guard ring serves two purposes:
- Equalizing the voltage drop across each insulator
unit.
- protects the insulator against flash over.
4/17/2013
Prof. Dr. Magdi El-Saadawi
22
pp. 389
4/17/2013
Prof. Dr. Magdi El-Saadawi
23
4/17/2013
Prof. Dr. Magdi El-Saadawi
24
4/17/2013
Prof. Dr. Magdi El-Saadawi
25
4/17/2013
Prof. Dr. Magdi El-Saadawi
26
4/17/2013
Prof. Dr. Magdi El-Saadawi
27
4/17/2013
Prof. Dr. Magdi El-Saadawi
28
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.
4/17/2013
Prof. Dr. Magdi El-Saadawi
29
4/17/2013
Prof. Dr. Magdi El-Saadawi
30
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
4/17/2013
Prof. Dr. Magdi El-Saadawi
31
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.
4/17/2013
Prof. Dr. Magdi El-Saadawi
32