Potential distribution over a suspension insulator string
The figure below shows a 3-disc string of suspension insulator. As each porcelain disc lies in between two metal links, it forms a capacitor. This capacitance is known as self-capacitance or mutual capacitance. Moreover, air capacitance is also present between metal links and the earthed tower. This is known as shunt capacitance. The figure below illustrates the equivalent circuit of a 3-disc suspension insulator (assuming that shunt capacitance is some fraction of self-capacitance i.e shunt capacitance = k * self-capacitance).If there were only mutual capacitances, then the charging current would have been the same through all the discs. In this case, the voltage would have been uniformly distributed across the string, i.e. voltage across each disc would have been the same. But, due to the shut capacitances, charging current is not the same through all the discs.
From the above equivalent circuit, applying Kirchoff's current law to node A,
I2 = I1 + i1
V2ωC = V1ωC + V1ωkC
V2 = V1 + V1k
V2 = (1 + k)V1 ...... eq.(i)
V2ωC = V1ωC + V1ωkC
V2 = V1 + V1k
V2 = (1 + k)V1 ...... eq.(i)
applying Kirchoff's current law to node B,
I3 = I2 + i2
V3ωC = V2ωC + (V2 + V1)ωkC
V3 = V2 + (V1 + V2)k
V3 = kV1 + (1 + k) V2
V3 = kV1 + (1 + k)2 V1 ...... from eq.(i)
V3 = V1 [k + (1 + k)2]
V3 = V1 [k + 1 + 2k + k2]
V3 = V1 (1 + 3k + k2) ...... eq.(ii)
V3ωC = V2ωC + (V2 + V1)ωkC
V3 = V2 + (V1 + V2)k
V3 = kV1 + (1 + k) V2
V3 = kV1 + (1 + k)2 V1 ...... from eq.(i)
V3 = V1 [k + (1 + k)2]
V3 = V1 [k + 1 + 2k + k2]
V3 = V1 (1 + 3k + k2) ...... eq.(ii)
Now, voltage between the conductor and the earther tower is,
V = V1 + V2 + V3
V = V1 + (1 + k)V1 + V1 (1 + 3k + k2)
V = V1 (3 + 4k + k2) ...... eq.(iii)
V = V1 + (1 + k)V1 + V1 (1 + 3k + k2)
V = V1 (3 + 4k + k2) ...... eq.(iii)
from the above equations (i), (ii) & (iii), it is clear that the voltage across the top disc is minimum while voltage across the disc nearest to the conductor is maximum, i.e. V3 = V1 (1 + 3k + k2). As we move towards the cross arm, voltage across the disc goes on decreasing. Due to this non-uniform voltage distribution across the string, the unit nearest to the conductor is under maximum electrical stress and is likely to be punctured.
String efficiency
As explained above, voltage is not uniformly distributed over a suspension insulator string. The disc nearest to the conductor has maximum voltage across it and, hence, it will be under maximum electrical stress. Due to this, the disc nearest to the conductor is likely to be punctured and subsequently, other discs may puncture successively. Therefore, this unequal voltage distribution is undesirable and usually expressed in terms of string efficiency.
The ratio of voltage across the whole string to the product of number of discs and the voltage across the disc nearest to the conductor is called as string efficiency
String efficiency = Voltage across the string / (number of discs X voltage across the disc nearest to the conductor).
Greater the string efficiency, more uniform is the voltage distribution. String efficiency becomes 100% if the voltage across each disc is exactly the same, but this is an ideal case and impossible in practical scenario. However, for DC voltages, insulator capacitances are ineffective and voltage across each unit would be the same. This is why string efficiency for DC system is 100%.
Inequality in voltage distribution increases with the increase in the number of discs in a string. Therefore, shorter strings are more efficient than longer string insulators.