String efficiency: 66 kV, 4 discs, 10%

Data

• Line voltage 66 kV, so voltage across the string (phase to earth):

$$V = \frac{66}{\sqrt3} = 38.105\ \text{kV}$$

• Number of discs $n = 4$; shunt-to-self capacitance ratio $m = 0.1$.

Let $V_1$ (top, at cross-arm) to $V_4$ (bottom, at line) be the disc voltages.

Equations at each joint

Current through lower disc = current through upper disc + shunt current to tower:

$$\begin{aligned} V_2 &= V_1(1+m) = 1.1V_1\\ V_3 &= V_2 + m(V_1+V_2) = [1.1 + 0.1(2.1)]V_1 = 1.31V_1\\ V_4 &= V_3 + m(V_1+V_2+V_3) = [1.31 + 0.1(3.41)]V_1 = 1.651V_1 \end{aligned}$$

Solving

$$V = V_1(1 + 1.1 + 1.31 + 1.651) = 5.061V_1 \Rightarrow V_1 = \frac{38.105}{5.061} = 7.529\ \text{kV}$$

String efficiency

$$\eta = \frac{V}{n\times V_4} = \frac{38.105}{4\times12.431}\times100 = 76.6\ \%$$

The disc next to the line carries about 33 % of the string voltage, so it is the most stressed. Guard rings or longer cross-arms would improve the distribution.

Answer: $V_1 = 7.53$, $V_2 = 8.28$, $V_3 = 9.86$, $V_4 = 12.43$ kV; string efficiency = 76.6 %.