Question

(a) Briefly explain how corona discharge takes place in overhead transmission line systems
(b) What accounts for the unequipotential distribution along a string of insulators? How can it be practically resolved?
(c) A 3-phase 220 kV 50 Hz transmission line consists of 1.5 cm radius spaced 1.8 m apart in an equilateral triangle formation. If the temperature is 360 C and the atmospheric pressure is 76 cm Hg, and assume the breakdown strength of air to be 32 kV/cm (peak value) and the irregularity factor , determine the effective critical disruptive voltage per phase.

Answer 1

ANSWER TO PART (a)

Air is not a perfect insulator and even under normal condition, the air contains a number of free electrons and ions. Consider two large parallel conducting planes. When an electric gradient is set up between them, the electrons and ions acquire motion by this electric field and they maintain a very small current between the conducting planes. This current is negligible, when the electric field intensity is less than 30 KV/cm. But when the electric field intensity or potential gradient reaches the critical value of about 30 KV/cm, the air in the immediate vicinity of conductors no more remains a dielectric and at this intensity, the ions attain high velocity and on striking another neutral molecule dislodge one or more electrons from the neutral molecule. This produces a new electron and a positive ion which in turn are accelerated and collide with other air molecules to ionize them further. Thus the number of charged particles goes on increasing rapidly. If a uniform field intensity is assumed between the electrodes such conditions are produced everywhere in the gap. As a result of this, the saturation is reached. Therefore, the air becomes conducting, hence a complete electric breakdown occurs and arc is established between the two electrodes.

When an alternating potential difference is applied across two conductors whose spacing is large in comparison with the diameter, then the air surrounding the conductor is subjected to electro-static stresses. This stress or intensity is maximum at the surface of the conductor and then decreases in inverse proportion to the distance from the centre of the conductor. If this potential difference is gradually increased, a point will be reached when a faint luminous glow of violet colour will make its appearance, and at the same time a hissing noise will be heard. This phenomenon is called corona and is accompanied by the formation of ozone, as is indicated by the characteristic odour of this gas. This luminous glow is due to the fact that the atmospheric

air around the conductor becomes conducting due to electrostatic stress. If the potential difference is raised still further, the glow and the noise will increase in intensity until eventually a spark- over will take place. If the conductors are perfectly uniform and smooth, the glow will be uniform along their length, otherwise the rough points of the conductor will appear brighter. With conductors only a short distance apart in comparison with their diameter, the spark-over may take place before there is any luminous glow is observed. If the applied potential difference is D.C. instead of A.C., the positive conductor having a uniform glow while the negative conductor has a more patchy glow and often it is accompanied by streamers if there are any rough places. An important point in connection with corona is that it is accompanied by loss of power, which is dissipated in the form of heat, light, sound and chemical action. In case of A.C. system, current flow due to corona is non-sinusoidal and in practice this non-sinusoidal current and non-sinusoidal voltage drop by corona may be more important than the power loss.

ANSWER TO PART (b)

The voltage distribution across an insulator string is not uniform. The units nearest to the line end are stressed to their maximum allowable value while those near the tower end are considerably under stressed resulting in a waste of insulating material. The string efficiency indicates the extent of this wastage. Though string efficiency can never be made 100 per cent, an improvement in its value is necessary to minimize the wastage 13). Some methods to improve string efficiency are: 1. Using longer cross arms It is clear from the above mathematical expression of string efficiency that the value of string efficiency depends upon the value of k. lesser the value of k, the greater is the string efficiency. As the value of k approaches to zero, the string efficiency approaches to 100%. The value of k can be decreased by reducing the shunt

capacitance. In order to decrease the shunt capacitance, the distance between the insulator string and the tower should be increased, longer cross-arms should be used. However, there is a limit in increasing the length of cross-arms due to economic considerations. 2. By grading the insulators In this method, voltage across each disc can be equalize by using discs with different capacitances. For equalizing the voltage distribution, the top unit of the string must have minimum capacitance, while the disc nearest to the conductor must have maximum capacitance. The insulator discs of different dimensions are so chosen that each disc has a different capacitance. They are arranged in such a way that the capacitance increases progressively towards the bottom. As voltage is inversely proportional to capacitance, this method tends to equalize the voltage distribution across each disc [6]

3. By using a grading ring The potential across each unit in a string can be equalized by using a guard ring which is a metal ring electrically connected to the conductor and surrounding the bottom insulator as shown in Fig. 2.3. The guard ring introduces capacitance between metal fittings and the line conductor. The guard ring is connected in such a way that shunt capacitance currents i1, 12 etc. are equal to metal fitting line capacitance currents i'], i'2 etc. The result is that the same charging current I flow throw each unit of string. Consequently, there will be uniform potential distribution across the units [6].

C= C HE G2 --- с c. i2 C3 Guard Ring Fig. 2.3 Grading ring [6]

ANSWER TO PART(c)