String of 5 insulators on 100 kV, ratio 0.1

Assumption: 100 kV is the line voltage of a three-phase line, so the voltage across each string is the phase voltage.

Data

• Phase voltage $V = 100/\sqrt3 = 57.735$ kV
• Ratio of shunt to self-capacitance $m = 0.1$
• Discs numbered 1 (tower end) to 5 (line end)

Voltage distribution

Using $V_{k+1} = V_k + m(V_1+\dots+V_k)$:

$$\begin{aligned} V_2 &= 1.1V_1\\ V_3 &= 1.1V_1 + 0.1(2.1V_1) = 1.31V_1\\ V_4 &= 1.31V_1 + 0.1(3.41V_1) = 1.651V_1\\ V_5 &= 1.651V_1 + 0.1(5.061V_1) = 2.1571V_1 \end{aligned}$$ $$V = 7.2181V_1 \;\Rightarrow\; V_1 = \frac{57.735}{7.2181} = 7.999\ \text{kV}$$

String efficiency

$$\eta = \frac{V}{5V_5} = \frac{57.735}{5\times 17.254} = 66.92\ \%$$

Answer: The disc voltages are 8.00, 8.80, 10.48, 13.21 and 17.25 kV, and the string efficiency is 66.92 %. If 100 kV is instead taken as the voltage across the string, every disc voltage scales by $\sqrt3$ (to 13.85, 15.24, 18.15, 22.87 and 29.88 kV), and the efficiency stays at 66.92 %.