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String efficiency: 33 kV, 3 discs, 10%

Short answer

With m = 0.1 and phase voltage 33/√3 = 19.053 kV, disc voltages are 5.59 kV (tower), 6.15 kV, 7.32 kV (line). String efficiency = 19.053/(3×7.319) = 86.77%.

An insulator string for 33 kV overhead line has 3 discs. The shunt capacitance between each joint and metal work is 10% of the capacitance of each disc. Find the voltage across the different discs and string efficiency.

Data

  • Phase voltage: V=33/3=19.053V = 33/\sqrt3 = 19.053 kV
  • m=C1/C=0.1m = C_1/C = 0.1
  • Discs numbered 1 (tower end) to 3 (line end)
 Tower ─[C]─ A ─[C]─ B ─[C]─ Line (19.05 kV)
             |       |
           0.1C    0.1C  to tower

Junction A

ωCV2=ωCV1+ω(0.1C)V1  ⇒  V2=1.1V1\omega CV_2 = \omega CV_1 + \omega(0.1C)V_1 \;\Rightarrow\; V_2 = 1.1V_1

Junction B

V3=V2+0.1(V1+V2)=1.1V1+0.21V1=1.31V1V_3 = V_2 + 0.1(V_1 + V_2) = 1.1V_1 + 0.21V_1 = 1.31V_1

Disc voltages

V=V1(1+1.1+1.31)=3.41V1  ⇒  V1=19.0533.41=5.587 kVV = V_1(1 + 1.1 + 1.31) = 3.41V_1 \;\Rightarrow\; V_1 = \frac{19.053}{3.41} = 5.587\ \text{kV}
DiscVoltage (kV)% of VV
1 (tower end)5.58729.33 %
26.14632.26 %
3 (line end)7.31938.42 %

String efficiency

η=V3V3=19.0533×7.319=86.77 %\eta = \frac{V}{3V_3} = \frac{19.053}{3\times 7.319} = 86.77\ \%

Answer: The disc voltages are 5.59 kV, 6.15 kV and 7.32 kV, and the string efficiency is 86.77 %.

✓ Answer checked by Claude Opus 5.5 (high effort)

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