Anna University
B.E / B.Tech DEGREE EXAMINATION, APRIL/MAY 2010
SIXTH SEMESTER
ELECTRICAL AND ELECTRONICS ENGINEERING
EE1353 HIGH VOLTAGE ENGINEERING
(REGULATIONS 2007)
Time: Three Hours
Maximum: 100 Marks
Answer ALL Questions
Part – A – (10 X 2 = 20 Marks)
1. What is isokeraunic level?
It is the number of days in which thunderstorm is recorded in a year at that particular location. It is given by,
2. What is insulation coordination?
The process of bringing the insulation strengths of electrical equipment and buses into the proper relationship with expected overvoltages and with the characteristics of the insulating media and surge protective devices to obtain an acceptable risk of failure
3. Define Paschen’s law.
The breakdown voltage of a uniform field gap is a unique function of the product of p, the gas pressure and d, the electrode gap, for a particular gas or for a given electrode material.
V= f(pd)
4.What are pure liquid dielectrics?
Pure liquids are those which are chemically pure and do not contain any other impurity even in traces of 1 in 109, and are structurally simple.
Examples of such simple pure liquids are n-hexane (C6H14). n-heptane (C7H16) and other paraffin hydrocarbons.
5. For a cascaded voltage multiplier circuit, with f=50Hz, C=0.1uF, Vmax=50kV and I=5mA, obtain the optimum no. of stages.
Maximum Voltage, Vmax=50kV=50X103V
6.Draw cascaded voltage doubler circuit.
7. What are the advantages of Generating Voltmeter?
Advantages:
· No source loading by the meter.
· No direct connection to high voltage electrodes.
· Scale is linear and extension of range is easy.
· Convenient instrument.
8. What is skin depth?
The depth to which electromagnetic radiation can penetrate a conducting surface decreases as the conductivity and the oscillation frequency increase.
9. What is disruptive discharge voltage?
The voltage which produces the loss of dielectric strength of insulation is called as disruptive discharge voltage.
10. How radio frequency noise id measured?
The noise generated in the radio frequency band as a result of corona or partial discharges in high voltage power apparatus may be measured
i. By the radio frequency line to ground voltage known as the radio influence voltage or RIV, and
ii. As an interfering field by means of an antenna known as the radiated radio interference voltage or RI.
PART – B
11.a. Explain the mechanism of lightning strokes. (10)
· Electric field required is 30 kV/cm peak.
· The current in the streamer is of the order of 100 amperes and the speed of the streamer is 0.16 m/μsec.
· Pilot streamer - 50m in length and are accomplished in about a microsecond
· Once the stepped leader has made contact with the earth it is believed that a power return stroke.
· The current varies between 1000 amps and 200,000 amps and the speed is about 10% that of light.
· This charged cell will try to neutralize through this ionized path. This streamer is known as dart leader. The velocity of the dart leader is about 3% of the velocity of light. The effect of the dart leader is much more severe than that of the return stroke.
· The discharge current in the return streamer is relatively very large but as it lasts only for a few microseconds the energy contained in the streamer is small and hence this streamer is known as cold lightning stroke whereas the dart leader is known as hot lightning stroke because even though the current in this leader is relatively smaller but it lasts for some milliseconds and therefore the energy contained in this leader is relatively larger.
11.b. How lightning is modeled mathematically?
Let I0 – Current of lightning stroke.
Z0 – Source impedance
Z – Impedance of an Object
The Voltage across the object, V=IZ
Source impedance: Estimated to be about 1000 to 3000W
Objects Considered:
· Transmission Lines or Towers – Surge impedances less than 500 W
· OH lines: 300-500W
· Ground wires: 100-150W
· 
Towers: 10-50W)
Therefore,
where, I0 – Current of lightning stroke and Impedance of a line
· If the lightning current as low as 10,000A strikes on a transmission line having 400W surge impedance, then the over voltage may be 4000kV.
· This voltage is heavy enough to cause flashover on the insulators.
· If a direct stroke occurs over an unshielded line, the current wave tries to divide into two and travels in both the directions.
· Hence the effective surge impedance of the line as seen by the wave is Z0/2. Therefore the overvoltage caused as above may be only 2000kV.
12.a.Explain various methods to control switching overvoltages.(10)
Following are some of the techniques currently in use to control the magnitudes of switching surges:
a.Resistor Switching:
The initial amplitude of the energization surge when a preinsertion resistor of value R is used will be only Z0/(R+Z0) of that reached in the absence of the resistor, where Zo is the surge impedance of the line.
When the resistor is shorted at the end of the preinsertion period, another surge will develop. If R is too small, control of the first surge becomes ineffective; if it is too large, the second surge becomes dangerous. An optimal value of R would normally be a fraction of Z0, and depends on transmission-line length
b. Phase-Controlled Closure
By properly timing of the closing of the circuit breaker poles, the resulting switching overvoltage can be greatly reduced.
c. Use of Shunt Reactors
Shunt reactors are used on many high-voltage transmission lines as a means of shunt compensation to improve the performance of the line, which would otherwise draw large capacitive currents from the supply. They have the additional advantage of reducing energization surge magnitudes. This is accomplished mainly by the reduction in temporary overvoltages, as will be seen in the next section.
d. Drainage of Trapped Charges
An effective way to reduce the trapped charges during the lead time before reclosing is by temporary insertion of resistors to ground or in series with shunt reactors and removing before the closure of the switches.
12.b. Give brief note on protection of transmission lines using surge diverters.(6)
· Surge diverters are devices that provide low resistance paths for overvoltages through an alternate ground path.
· Spark gap inside the diverter acts as a fast acting switch while non-linear diverter elements provide the low impedance ground path.
· 
The arrester voltage at its terminal when connected to a line of surge impedance Z to ground, is given as
where,
Z - Line surge impedance,
R - Resistance of the non-linear element,
r - Ground to earth resistance, and
u(t) - Surge voltage.
· The Thevenin’s equivalent circuit for the diverter is shown in Fig.
· S is open for voltages less than the sparkover voltage of the surge diverter VS.
· It is closed only for voltage magnitudes greater than VS.
· The closing of the switch is represented by injecting a voltage cancellation wave having a negative amplitude equal to the potential difference between the voltage that appears when the switch is open Voc, and the voltage developed across the impedance of the device after the switch is closed.
· ZTH is the impedance of the system viewed from the terminals of the protective device.
13.a.What is an electrical avalanche? How do avalanche give rise to an electrical breakdown in case of Townsend’s type of discharge?
13.b.How vacuum breakdown occurs according to particle exchange mechanism?
14. Explain various theories which explain breakdown in commercial liquid dielectrics. (16).
15.a.How high frequency high voltage is generated in test laboratories?
High frequency high voltages are required for rectifier d.c. power supplies. Also, for testing electrical apparatus for switching surges, high frequency high voltage damped oscillators are needed which need high voltage high frequency transformers. The advantages of these high frequency transformers are:
i) the absence of iron core in transformers and hence saving in cost and size,
ii) pure since wave output,
iii) slow build-up of voltage over a cycles and hence no damage due to switching surges, and
iv) uniform distribution of voltage across the winding coils due to subdivision of coil stack into a number of units.
The commonly used high frequency resonant transformer is the Tesla coil, which is a doubly tuned resonant circuit shown schematically in Fig. 6.13a.
The primary voltage rating is 10 kV and the secondary may be rated to as high as 500 to 1000 kV. The primary is fed form a.d.c. or a.c. supply through the condenser 
A spark gap G connected across the primary is triggered at the desired voltage 
which induces a high self excitation in the secondary. The primary and the secondary windings 
are wound on an insulated former with no core (air cored) and are immersed in oil. The windings are tuned to a frequency of 10 to 100kHz by means of the condensers 
The output voltage 
is a function of the parameters L1, L2, C1, C2, and the mutual inductances M. Usually, the windings resistance will be small and contribute only for damping of the oscillations.
15.b.An Impulse generator has 12 capacitors of 0.12uF and 200kV rating. The wave front and wave tail resistances are 1.25kW and 4kW respectively. If the load capacitance including that of test object is 10000pF, find the wave front and wave tail times and the peak voltage of impulse wave produced.
16.a.Explain the methods of generating switching surges in laboratories. (8).
Generally, for a given impulse generator of Fig, the generator capacitance C1 and load capacitance C2 will be fixed depending on the design of the generator and the test object. Hence, the desires waveshape is obtained by controlling R1 and R2. The following approximate analysis is used to calculate the wave front and wave tail times.
The resistance R2 will be large. Hence, the simplified circuit shown in Fig. 4.16b is used for wave front time calculation. Taking the circuit inductance to be negligible during charging, C1 charges the load capacitance C2 through R1. Then the time taken for charging is approximately three times the time constant of the circuit and is given by
where 
If 
is given in ohms and 
in microfarads, t1 is obtained in microseconds.
For discharging or tail time, the capacitance 
may be considered to be in parallel and discharging occurs
. Hence, the time for 50% discharge is approximately given by
These formulae for 
hold good for the equivalent circuits are shown in Fig. 6.15b and c. For the circuit given in Fig. 6.15d, R is to be taken as 2 R1. With the approximate formulae, the wave front and wave tail times can be estimated to within 
for the standard impulse waves.
16.b.What is the importance of controlled tripping of impulse generators? How is it done using trigatron gap? (8)
In large impulse generators, the spark gaps are generally sphere gaps or gaps formed by hemispherical electrodes. The gaps are arranged such that sparking of one gap results in automatic sparking of other gaps as overvoltage is impressed on the other. In order to have consistency in sparking, irradiation from an ultra-violet lamp is provided from the bottom to all the gaps.
The three electrode gap requires larger space and an elaborate construction. Now-a-days a trigatron gap shown in Fig is used, and this requires much smaller voltage for operation compared to the three electrode gap. A trigatron gap consists of a high voltage spherical electrode of suitable size, an earthed main electrode of spherical shape, and a trigger electrode through the main electrode. The trigger electrode is a metal rod with an annular clearance of about 1 mm fitted into the main electrode through a bushing. The trigatron is connected to a pulse circuit as shown in Fig. Tripping of the impulse generator is effected by a trip pulse which produces a spark between the trigger electrode and the earthed sphere. Due to space charge effects and distortion of the field in the main gap, sparkover of the main gap occurs. The trigatron gap is polarity sensitive and a proper polarity pulse should be applied for correct operation.
17. Explain the working principle and operation of Electrostatic Voltmeters.(16).
18.a.What are Capacitive voltage dividers? Explain various capacitance voltage dividers used to measure impulse voltage upto 2MV.(10)
18.b.Explain the methods of measuring high DC currents.(6)
19.a.What are the impulse tests done on insulators? Explain. (8)
19.b.Explain the synthetic testing of circuit breakers. (8).
à Heavy current at low voltage is applied
à Recovery voltage is simulated by high voltage, small current source
à Procedure:
i. Auxiliary breaker 3 and test circuit breaker T closed, making switch 1 is closed. \ Current flows through test CB.
ii. At time t0, the test CB begins to open and the master breaker 1 becomes to clear the gen circuit.
iii. At time t1, just before zero of the gen current, the trigger gap 6 closes and high frequency current from capacitance Cv flows through the arc of the gap
iv. At time t2, gen current is zero. Master CB 1 is opened
v. The current from will flow through test CB and full voltage will be available
vi. At the instant of breaking, the source is disconnected and high voltage is supplied by auxiliary CB 4
20.a.What are the tests done on cables? How samples are prepared? Explain any two test. (10)
20.b.Explain long duration impulse current test and operation duty cycle test on surge diverters. (6)
High Current Impulse Test
à Impulse current wave of 4/10µS is applied to pro-rated arrester in the range of 3 to 12kV.
à Test is repeated for 2 times
à Arrester is allowed to cool to room temperature
The unit is said to pass the test if
i. The power frequency sparkover voltage before and after the test does not differ by more than 10%
ii. The voltage and current waveforms of the diverter do not differ in the 2 applications
iii. The non linear resistance elements do not show any puncture or flashover
Operating Duty Cycle Test
· This test is conducted on pro-rated units of diverters and gives better closeness to actual conditions.
· The diverter is kept energized at its rated power frequency supply voltage.
· The rated impulse current wave is applied first at a phase angle of about 30° from the a.c. voltage zero. If the power frequency follow-on current is not established, the angle at which current wave is applied is advanced in steps of 10° up to 90° or the peak position of the supply voltage wave till the follow-on current is established.
· In the course of application of the current wave, if the power frequency voltage is reduced during the flow of current, it can be compensated up to a maximum of 10% of the overvoltage.
· During the follow-on current period, the peak voltage across the diverter should be less than or equal to the rated peak voltage.
· Twenty applications of the impulse current at the selected points on the voltage wave are made in four groups. The time interval between each application is about 1 min, and between successive groups it is about half an hour.
· The arrester is said to have passed the test, if
i. The average power frequency sparkover voltage before and after the test does not differ by more than 10%.
ii. the residual voltage at the rated current docs not vary by more than 10%,
iii. the follow-on power frequency current is interrupted each time, and
iv. no significant change, signs of flashover, or puncture occurs to the pro-rated unit.
