Voltage Distribution of Suspension Insulator

Voltage Distribution of Suspension Insulator.
Overhead line to the mechanical design is considered that the important issue of the suspension and the Voltage Insulator Division of the disk.
Most of the disk and the metal pin comprising insulators Porcelain. Insulator strings, so all across the capacitor and the pin - to - pin and the pin - to - Earth Capacitance influence. Voltage Insulator strings are equally active in the Division as well as a string of calculated predominates Capacitance. The pin - to - Earth (shunt)Capacitance  because the string can not be everywhere at alls voltage equally. But what kind of steps required to improve the efficiency of the string it all needs to know.
Efficient string: -
N-number of units with a unit spark over voltage (s.o.v.)  and over-voltage N times the spark Efficiency Ratio is a string. The N - number of units s.o.v. means that the maximum lateral line units.
Efficient string η = N - number of units s.o.v.
                                    N × one units s.o.v.
            = String of S.O.V. / N × line units s.o.v.(maximum voltage)
            = (V / NVN × 100)
Efficiency of the entire unit if the strings above the string Equals the spark over voltage (S.O.V.) is out. Efficiency is an important factor for the overhead line as a string of design work.
Efficient string upgrade:-
Insulator metal parts comprising units of the string, as well as mutual Capacitance Cross - arm, etc., exist in the shunt Capacitance. The voltage per unit Capacitance effected  imposed equally on the whole string can not exist. It can be seen, most reactions are line voltage unit. The decrease  Cross - most are low voltage arm near the top of the unit. This significantly reduces the efficiency of the string Efficiency.
Systematic way, therefore each unit is equal to the voltage is raised through a string of Efficiency. String Efficiency is the development of methods, namely: -
1. By using longer cross-arms.
2. By grading the insulator units.
3. By using guard ring.

1. By conducting glaze method.
   5. By conducting glaze method.
   Method of voltage grading in suspension insulators:
   Different dimension of this process are different units chain (string) in the sorting or grading (grading) is, so that the maximum is Capacitance of minimum  unit. Since the amount of voltage to current flow Capacitance vary. (V = i / ωC), line voltage will reduce the unit. On the other hand, other units of voltage increase. Thus, the grading of the Division and the voltage per unit to upgrade the skills of the string.
   Now,
   C = top of the unit mutual Capacitance
   C1 = per unit Earth Capacitance. If the ratio m of the C1 and C,
   C1 = m C
   C2, C3, C4 , or 3^rd and 4^th  units of mutual Capacitance.

V = voltage across each unit.

now
a and b point to karsoff’s Low the current formula,
I2 = I1 + Ia
ωC2 V = ω CV + ω m C V
C2 = C (1 + m) ............................. (i)
Or
I3 = I2 + Ib
ω C3 V = ω C2 V + 2ω m C V
C3 = C2 + 2 m C = C (1 + m) + 2m C = 3mC
C3 = C (1 + 3m) ........................... (ii)
C4 = C (1 + 6m) ............................ (iii)
Thus, a four-unit Insulator string of top-line unit to the unit replaced Capacitance ratio 1: (1 + m): (1 + 3m): (1 + 6m) will be equal to the per unit voltage.
Solved problems on string efficiency and voltage distribution: -
1. An overhead line suspension Insulator consists of 4 units . Transport and land are the differences between the 66 kV splendor. If the land in each insulators mutual Capacitance and Capacitance ratio 5: 1, the voltage across the string to determine the efficiency and the per unit tax.

Answer: - provided
N = 4, V = 66 kv
m = C1 / C
  = 1/5 = 0.2
String efficiency of η =?
V1, V2, V3 & V4 =?
V2 = V1 (1 + m)
    = V1 (1 + 0.2)
  = 1.2 V1
V3 = V1 (m² + 3m + 1)
    = V1 [(0.2) ² + 3 × 0.2 + 1]
    = 1.64 V1
V4 = V1 (m³ + 5m ² + 6m + 1)
         = [(0.2) ³ + 5 × (0.2) ² + 6 × 0.2 + 1]
        = 2.408 V1
= V1 + V2 + V3 + V4 = V1 (1 + 1.2 + 1.64 + 2.408) = V
Or
6.248 V1 = 66
V1 = 66 / 6.248
= 10.563kV
V2 = 1.2 × 10.563 = 12.676 kV
V3 = 1.64 × 10.563 = 17.324 kV
V4 = 2.408 × 10.563 = 25.4367 kV

Efficient string η = N - number of units s.o.v.
                                    N × one units s.o.v.

                  % η = V / 4 × V4 × 100
                          = 66/4 × 25.4367 × 100
                          = 64.87% Ans:
2. Insulator with a 4 - units set and the 2^nd or 1^st  Units, 10 kv and 12 kv, respectively, when the supply voltage  and string to determine efficiency. Ans: - N = 4

V1 = 10 kv V2 = 12 kv
VL =? η =?
V2 = V1 (1 + m)
1 + m = V2 / V1 = 12/10 = 1.2
m = 1.2 - 1 = 0.2
V3 = V1 (m³ + 3m +1)
= 1.0 [(0.2) ² + 3 × 0.2 + 1]
= 10 × 1.64
= 16.4 kv
V4 = V1 (m³ + 5m ² + 6m + 1)
= 10 [(0.2) ³ + 5 × (0.2) ² + 6 × 0.2 + 1]
= 10 × 2.408
= 24.08 kv
V = V1 + V2 + V3 + V4
10 + 12 + 16.4 + 24.08
= 62.48 kv
Line Voltage VL = √ 3  × V
= √ 3 × 62.48
108.22 kv
Efficient string η = N - number of units s.o.v.
                                    N × one units s.o.v.

                    % η = 62.48 / 24.08 × 100
                   = 64.86% (Ans).
3. A 3-φ overhead line in 3 Insulator a set of (string) is. If the voltage at the top of the 1^st  unit and 2^nd units across Voltage with in 10 kv, 12 kv respectively is but the tax :- 1) m 2) Line Voltage 3) Efficient string.

Ans :- 3-φ Line,
N = 3
V1 = 10 kv,
V2 = 12kv, m = ? VL = ? η = ?
We know that
V2 = V1 (1 + m)
1 + m = V2 / V1 = 12/10 = 1.2
 earth and mutual Capacitance ratio
m = 1.2 - 1 = 0.2
V3 = V1 (m³ + 3m +1) = 1.64 V1
 = 1.64 × 10 = 16.4 kv
V = V1 + V2 + V3
= 10 + 12 + 16.4 = 38.4 kv
Line Voltage VL = √ 3 × V = 66.5 kv
String Efficiency = V / NV3 × 100
 = 38.4 / 3 × 16.4 × 100
= 78.05% (Ans).