A three-phase 132 kV, 50 Hz transmission line 180 km long consists of three conductors each of diameter 1.2…

Question

A three-phase 132 kV, 50 Hz transmission line 180 km long consists of three conductors each of diameter 1.2 cm and spaced 2.7 m apart in an equilateral triangle formation. The temperature and atmospheric pressure are 30°C and 750 mm of mercury respectively. If the breakdown strength of air is 30 kV/cm (peak) and the irregularity factor is 0.84. Determine the critical disruptive voltage and total corona power loss.

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Here are the solutions to the problems:

: Corona Discharge Calculation

a)

(i) Critical disruptive voltage:

Step 1: Calculate the air density correction factor ( $\delta$ ).
Given temperature $T = 30 \, ^\circ\text{C}$ and pressure $P = 750 \, \text{mm of Hg}$ . We use $25 \, ^\circ\text{C}$ and $760 \, \text{mm of Hg}$ as standard conditions for $g_0$ .

\delta = \frac{P}{760} \times \frac{273 + 25}{273 + T}

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