ample: -phase, \( 220 \mathrm{kV}, 50 \mathrm{~Hz} \) transmission line consists of 1.5 cm radius tductor spaced 2 meters apart in equilateral triangular formation. temperature is \( 40^{\circ} \mathrm{C} \) and atmospheric pressure is 76 cm , calculate corona loss per km of the line. Take mo \( =0.85 \).

To calculate the corona loss per km of the transmission line, we can use the formula: \[ \text{Corona Loss per km} = \frac{2.5 \times 10^{-3} \times \sqrt{U} \times \sqrt{I} \times \sqrt{R} \times \sqrt{D} \times \sqrt{P} \times \sqrt{m} \times \sqrt{A}}{1000} \] Given: - Phase: -phase - Voltage: \( 220 \, \text{kV} \) - Frequency: \( 50 \, \text{Hz} \) - Radius of conductor: \( 1.5 \, \text{cm} \) - Spacing between conductors: \( 2 \, \text{m} \) - Temperature: \( 40^{\circ} \text{C} \) - Atmospheric pressure: \( 76 \, \text{cm} \) - Dielectric constant: \( m = 0.85 \) We need to calculate the corona loss per km of the line. Let's calculate it step by step: 1. Calculate the radius of the conductor in meters: \[ R = 1.5 \, \text{cm} = 0.015 \, \text{m} \] 2. Calculate the spacing between conductors in meters: \[ D = 2 \, \text{m} \] 3. Calculate the corona loss per km using the formula: \[ \text{Corona Loss per km} = \frac{2.5 \times 10^{-3} \times \sqrt{220} \times \sqrt{50} \times \sqrt{0.015} \times \sqrt{2} \times \sqrt{76} \times \sqrt{0.85} \times \sqrt{1}}{1000} \] Let's calculate the corona loss per km of the line. Calculate the value by following steps: - step0: Calculate: \(\frac{2.5\times 10^{-3}\sqrt{220}\times \sqrt{50}\times \sqrt{0.015}\times \sqrt{2}\times \sqrt{76}\times \sqrt{0.85}\times \sqrt{1}}{1000}\) - step1: Simplify the root: \(\frac{2.5\times 10^{-3}\times 2\sqrt{55}\times \sqrt{50}\times \sqrt{0.015}\times \sqrt{2}\times \sqrt{76}\times \sqrt{0.85}\times \sqrt{1}}{1000}\) - step2: Simplify the root: \(\frac{2.5\times 10^{-3}\times 2\sqrt{55}\times 5\sqrt{2}\times \sqrt{0.015}\times \sqrt{2}\times \sqrt{76}\times \sqrt{0.85}\times \sqrt{1}}{1000}\) - step3: Simplify the root: \(\frac{2.5\times 10^{-3}\times 2\sqrt{55}\times 5\sqrt{2}\times \frac{\sqrt{6}}{20}\sqrt{2}\times \sqrt{76}\times \sqrt{0.85}\times \sqrt{1}}{1000}\) - step4: Simplify the root: \(\frac{2.5\times 10^{-3}\times 2\sqrt{55}\times 5\sqrt{2}\times \frac{\sqrt{6}}{20}\sqrt{2}\times 2\sqrt{19}\times \sqrt{0.85}\times \sqrt{1}}{1000}\) - step5: Simplify the root: \(\frac{2.5\times 10^{-3}\times 2\sqrt{55}\times 5\sqrt{2}\times \frac{\sqrt{6}}{20}\sqrt{2}\times 2\sqrt{19}\times \frac{\sqrt{85}}{10}\sqrt{1}}{1000}\) - step6: Simplify the root: \(\frac{2.5\times 10^{-3}\times 2\sqrt{55}\times 5\sqrt{2}\times \frac{\sqrt{6}}{20}\sqrt{2}\times 2\sqrt{19}\times \frac{\sqrt{85}}{10}\times 1}{1000}\) - step7: Multiply: \(\frac{\frac{\sqrt{510}}{4}\times 10^{-3}\times 2\sqrt{1045}}{1000}\) - step8: Reduce the fraction: \(\frac{\frac{\sqrt{510}}{4}\times 10^{-3}\sqrt{1045}}{500}\) - step9: Factor the expression: \(\frac{\frac{\sqrt{510}}{4}\times 10^{-3}\sqrt{1045}}{100\times 5}\) - step10: Factor the expression: \(\frac{\frac{\sqrt{510}}{4}\times 10^{-3}\sqrt{1045}}{10^{2}\times 5}\) - step11: Reduce the fraction: \(\frac{\frac{\sqrt{510}}{4}\sqrt{1045}}{10^{5}\times 5}\) - step12: Calculate: \(\frac{\frac{5\sqrt{21318}}{4}}{5\times 10^{5}}\) - step13: Multiply by the reciprocal: \(\frac{5\sqrt{21318}}{4}\times \frac{1}{5\times 10^{5}}\) - step14: Reduce the numbers: \(\frac{\sqrt{21318}}{4}\times \frac{1}{10^{5}}\) - step15: Multiply the fractions: \(\frac{\sqrt{21318}}{4\times 10^{5}}\) The corona loss per km of the transmission line is approximately \(0.000365 \, \text{kW/km}\).