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String of 4 insulators at 66 kV, ratio 10

Short answer

Phase voltage 66/√3 = 38.105 kV, m = 0.1. Using V2=1.1V1 etc., disc voltages (tower to line) are V1=7.53 kV, V2=8.28 kV, V3=9.86 kV, V4=12.43 kV. String efficiency = 38.105/(4×12.43) = 76.64%.

A string of 4 insulators is connected across a 66 kV line. It has a self-capacitance equal to 10 times the pin to earth capacitance. Calculate (i) the distribution of voltage on the insulator discs and (ii) the string efficiency.

Data

  • Phase voltage V=66/3=38.105V = 66/\sqrt3 = 38.105 kV
  • Self-capacitance C=10C1C = 10C_1, so m=C1/C=0.1m = C_1/C = 0.1
  • Discs numbered 1 (tower end) to 4 (line end)

(i) Voltage distribution

At each junction, the current through the lower disc equals the current through the upper disc plus the current to earth:

Vk+1=Vk+m(V1+⋯+Vk)V_{k+1} = V_k + m(V_1 + \dots + V_k) V2=1.1V1V3=V2+0.1(2.1V1)=1.31V1V4=V3+0.1(3.41V1)=1.651V1\begin{aligned} V_2 &= 1.1V_1\\ V_3 &= V_2 + 0.1(2.1V_1) = 1.31V_1\\ V_4 &= V_3 + 0.1(3.41V_1) = 1.651V_1 \end{aligned} V=(1+1.1+1.31+1.651)V1=5.061V1  ⇒  V1=38.1055.061=7.529 kVV = (1 + 1.1 + 1.31 + 1.651)V_1 = 5.061V_1 \;\Rightarrow\; V_1 = \frac{38.105}{5.061} = 7.529\ \text{kV}
DiscVoltage% of VV
1 (tower end)7.529 kV19.76 %
28.282 kV21.73 %
39.863 kV25.88 %
4 (line end)12.431 kV32.62 %

(ii) String efficiency

η=V4V4=38.1054×12.431=76.64 %\eta = \frac{V}{4V_4} = \frac{38.105}{4\times 12.431} = 76.64\ \%

Answer: The disc voltages are 7.53, 8.28, 9.86 and 12.43 kV (tower to line end), and the string efficiency is 76.64 %.

✓ Answer checked by Claude Opus 5.5 (high effort)

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