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Maximum voltage of 3-disc string, 17.5 kV per unit

Short answer

The maximum phase voltage the string can withstand is 44.24 kV (76.63 kV line voltage). String efficiency is 84.27%.

Determine the maximum voltage that the string of suspension insulators having 3 no. of discs can withstand if the maximum voltage per unit is 17.5 kV. It is given that shunt capacitance between each joint and metal work is 12.5% of the capacitance of each disc.

The line-end disc carries the largest voltage, so the string is limited when that disc reaches 17.5 kV.

Data

Shunt capacitance C1=0.125CC_1 = 0.125C, so m=0.125m = 0.125. Number the discs 1 (tower end) to 3 (line end).

Voltage distribution

At junction A:

V2=V1+mV1=1.125V1V_2 = V_1 + mV_1 = 1.125V_1

At junction B:

V3=V2+m(V1+V2)=1.125V1+0.125(2.125V1)=1.3906V1V_3 = V_2 + m(V_1+V_2) = 1.125V_1 + 0.125(2.125V_1) = 1.3906V_1

The line-end disc has the highest voltage. Setting V3=17.5V_3 = 17.5 kV:

V1=17.51.3906=12.584 kV,V2=1.125×12.584=14.157 kVV_1 = \frac{17.5}{1.3906} = 12.584\ \text{kV},\qquad V_2 = 1.125\times12.584 = 14.157\ \text{kV}

Maximum voltage of the string

V=V1+V2+V3=12.584+14.157+17.5=44.24 kVV = V_1+V_2+V_3 = 12.584 + 14.157 + 17.5 = 44.24\ \text{kV}

This is the phase (line-to-earth) voltage. The corresponding three-phase line voltage is:

VL=3×44.24=76.63 kVV_L = \sqrt3\times 44.24 = 76.63\ \text{kV}

String efficiency

η=V3V3=44.243×17.5=84.27 %\eta = \frac{V}{3V_3} = \frac{44.24}{3\times17.5} = 84.27\ \%

Answer: The string can withstand 44.24 kV to earth (a line voltage of 76.63 kV), with a string efficiency of 84.27 %.

✓ Answer checked by Claude Opus 5.5 (high effort)

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