Disadvantages of Corona

41.29. Disadvantages of Corona

1. There is a definite dissipation of power although it is not so important except under abnor- mal weather conditions like storms etc.

2. Corrosion due to ozone formation.

3. The current drawn by the line due to corona losses is non-sinusoidal in character, hence it causes non-sinusoidal drop in the line which may cause some interference with neighbouring communication circuits due to electromagnetic and electrostatic induction. Such a shape of corona current tends to introduce a large third harmonic component.

However, it has been found that corona works as a safety valve for surges.

4. Particularly intense corona effects are observed at a working voltage of 35 kV or higher. Hence, designs have to be made to avoid any corona on the bus-bars of substations rated for

35 kV and higher voltages during their normal operation. Corona discharge round bus-bars is extremely undesirable because the intense ionization of the air reduces its dielectric strength, makes it easier for the flashover to occur in the insulators and between phases particularly when the surfaces concerned are dirty or soiled with other deposits. The ozone produced due to corona discharge aggressively attacks the metallic components in the substations and switchgear, covering them with oxides. Moreover, the crackling sound of the corona discharge in a substation masks other sounds like light crackling noise due to arcing in a loose contact, the sound of an impending breakdown or creepage discharge in the equip- ment, the rattling noise due to the loosening of steel in a transformer core etc. The timely detection of such sounds is very important if any serious breakdown is to be avoided.

1644 Electrical Technology

Example 41.33. Find the disruptive critical voltage for a transmission line having : conductor spacing = 1 m ;

conductor (stranded) radius = 1 cm barometric pressure = 76 cm of Hg ;

temperature = 40ºC

Air break-down potential gradient (at 76 cm of Hg and at 25ºC) = 21.1 kV (r.m.s.)/cm.

(Electric Power Systems-I, Gujarat Univ.) Solution.

V C = 2.3 m 0 g 0 r log 10 D/r kV/phase

Here,

g 0 = 21.1 kV (r.m.s.)/cm ; m 0 = 0.85 (assumed)

δ = 3.92 × 76/(273 + 40) = 0.952 ; log 10 D/r = log 10 100/1 = 2 ∴

V c = 2.3 × 0.85 × 21.1 × 0.952 × 1 × 2 = 78.54 kV (r.m.s.)/phase Line value

= 78.54 × √ 3= 136 kV (r.m.s.)/phase

Example 41.34. Find the disruptive critical and visual corona voltage of a grid-line operating at 132 kV.

= 3.81 m temperature

conductor dia

= 1.9 cm

conductor spacing

barometric pressure = 73.7 cm conductor surface factor : fine weather

= 44ºC

rough weather

(Electrical Power Systems-III, Gujarat Univ.) Solution.

V c = 48.8 m 0 δ r log 10 D/r kV/phase

Here,

m 0 = 0.8 ; δ = 3.92 × 73.7/(273 + 44) = 0.91

log 10 381/1.9 = log 10 200.4 = 2.302

V c = 48.8 × 0.8 × 0.91 × 1.9 × 2.302 = 155.3 kV/phase

V v = 48.8 m v δ r   1 + 0.3 

 log 10 D/r kV/phase

är 

Here, m v = 0.66 ; δ = 0.91 ; ä r = 0.91 × 1.9 = 1.314 ∴

V v 0.3 = 48.8 × 0.66 × 0.91 × 1.9 1 +

2.302 = 157.5 kV/phase

Example 41.35. A certain 3-phase equilateral transmission line has a total corona loss of

53 kW at 106 kV and a loss of 98 kW at 110.9 kV. What is the disruptive critical voltage between lines? What is the corona loss at 113 kV?

(Electrical Power Systems-II, Gujarat Univ.)

Solution. As seen from Art. 41.28, the total corona loss for three phases is given by

) × 10 kW/km Other things being equal, P 2 ∝ (V − V

241 ( f 25)

c )  2  ∴ 2 53 ∝ 106

− V c  ∝ (61.2 − V c )

— 1st case

— 2nd case

98 2 (64 − V

or V c =

54.2 kV/km

(61.2 − V c )  2 113 

Similarly, 2 W ∝  − V

c ∝ (65.2 − V c )

A.C. Transmission and Distribution

(65.2 − V c )

∴ W = 123.4 kW

98 2 (64 2 − V c ) (64 54.2) −

Example 41.36. A 3-phase, 50-Hz, 220-kV transmission line consists of conductors of 1.2 cm radius spaced 2 metres at the corners of an equilateral triangle. Calculate the corona power loss per km of the line at a temperature of 20ºC and barometric pressure of 72.2 cm. Take the surface factors of the conductor as 0.96.

(Electrical Power-II, Bangalore Univ.)

Solution. As seen from Art. 41.28, corona loss per phase is

( f + 25)

P = 241

(/) rD . (V − V c ) × 10 kW/km/phase

V c = 48.8 m 0 δ r log 10 D/r = 48.8 × 0.96 × 0.966 × 1.2 × log 10

= 120.66 kV/phase

V = 220/ 3 = 127 kV/phase

2 − = 5 241 × 75 × 1.2 × (127 120.66) − × 10 = 0.579 kW/km/phase

Total loss for 3 phase = 3 × 0.579 = 1.737 kW/km