HIGH VOLTAGE TECHNOLOGY

QUESTION AND ANSWER

PREPARED BY

ASSOC. PROF. DR. ZOLKAFLE BIN BUNTAT

MAY 2013

CHAPTER 1

ELECTRIC FIELDS

(QUESTION & ANSWER)

Question 1

a. There are two types of field distribution, known as homogeneous and non-homogeneous

field. What are the differences of both field distributions? State the electrodes-gap

configuration to simulate the homogeneous and non-homogeneous field.

Answer:

i. Homogeneous field

E is the same throughout the field region.

Uniform or approximate uniform field distributions exists between two infinite

parallel plates, or 2 spheres of equal diameters with gap spacing < sphere

radius.

“Profiled” parallel plates of finite sizes are also used to simulate

homogeneous field.

ii. Nonhomogeneous field

E is different at different points in the field region.

In the absence of space charges, E usually obtains the maximum value at the

surface of the conductor which has the smallest radius of curvature –

nonhomogeneous and asymmetrical.

Most of the practical HV components have nonhomogeneous and

asymmetrical field distribution..

In some gaps – will produce nonhomogeneous fields and symmetrical, e.g.

rod-rod or sphere-sphere (large distance between spheres) gaps.

HV electrode has higher E than the grounded electrode.

b. Experimental analogue is one of the methods for determining the potential distribution.

Briefly describe any two of the experimental analogs used for space-charge-free fields.

Answer:

i. Electrolytic Tank

Widely used for decades.

Equipotential boundaries are represented in the tank by specially form sheets

of metal.

Example, a single dielectric problem such as a three-core cable may be

represented by electrolytes of different conductivities separated by special

partitions.

ii. Semiconducting Paper Analog

Less accurate but attractively simple alternative to the electrolyte.

Errors from this method result from the non-homogeneity of the paper

resistivity.

Errors also dependence on the ambient humidity and the contact resistance

to the electrodes.

iii. Resistive Mesh Analog

The continuous field is replacing by a discrete set of points as depicted by a

mesh of resistors.

The used of discrete resistors introduce an error arising from the finite mesh

analysis.

This error may be reduced by reducing the mesh size.

c. A high voltage DC transmission line rated at 132 kV peak traverses a location where a

road shall be constructed below it as shown in Figure Q1. The metallic walls L1 and L2

are energized at 20 kV peak and 80Kv respectively, and each standing on the insulator

made of polycarbonate. Use the Finite Difference Method to determine the potential at

point 2, 4 and 6. Limit the iteration process to two only.

Figure Q1

Answer:

1st iteration

V 1=132+0+0+204

=38kV

V 2=132+0+0+384

=42.5kV

V 3=132+80+0+42.54

=63.63 kV

V 4=38+0+0+204

=14.5kV

V 5=42.5+0+0+14.54

=14.25kV

V 6=63.63+80+0+14.254

=39.47 kV

2nd iteration

V 1=132+42.5+14.5+204

=52.25kV

V 2=132+63.63+14.25+52.254

=85.53 kV

V 3=132+80+39.47+65.534

=79.25kV

V 4=52.25+14.25+0+204

=21.63kV

V 5=65.53+39.47+0+21.634

=31.66kV

V 6=79.25+80+0+31.664

=47.73kV

The potential at point 2, 4 and 6 after 2 iterations are;

V2 = 65.53 kV, V4 = 21.63 kV, V6 = 47.73 kV

Question 2

a. Briefly describe any two (2) of the followings:

i. Field enhancement factor

ii. ‘Medium High Voltage’ (MHV), ‘High voltage’ (HV), ‘Extra high Voltage’ (EHV)

and ‘Ultra high voltage’ (UHV)

iii. Three applications of high voltages excluding those in the generation,

transmission and distribution of electrical energy

Answer:

i. Whereas any designer of the high voltage apparatus must have a complete

knowledge of the electric field distribution, for a user of the system the knowledge

of the maximum value of the electric field

Emaxto which the insulation is likely to be subjected and the location of such

maximum gradient point is generally sufficient. Consequently, the concept of

field enhancement factor or simply field factor f is of considerable use. This

factor is defined as

f=EmaxEav

( for homogenousdieletric medium)

Where Eavthe average is field in the gap and is equal to the applied potential/gap

separation between the electrodes

Emax=f .V /d

Field utilization factor μ f=1f

(larger μ f represents a more compact equipment)

ii. The following voltage classification describes the meanings:

Voltage class Voltage range

Medium high voltage (MHV) 1kV<V=<70kV

High Voltage (HV) 110kV<V=<230kV

Extra high voltage (EHV) 275 kV<V=<800kV

Ultra high voltage (UHV) 1000kV=<V

iii. Any three of the followings:

ESPs for the removal of dust from flue gases

Atomization of liquids, paint spraying and pesticide spraying

Ozone generation for water and sewage treatment

X-ray generators and particles accelerators

High power lasers and ion beams

Plasma sources for semiconductor manufacture

Superconducting magnet coils

b. There are several properties of a dielectric which are of practical importance for an

engineer. Name five (5) most important properties of a dielectric and briefly describe

each of them.

Answer:

Any five of the followings:

i. DC conductivity

σ= JE=1

ρ

Dependent on material purity, T and E

σ (T )=Ae− EkT

Due to polarization effects, σ also depends upon time of application of the

stress

ii. Dielectric permittivity

εr=C /Co

Dependent on T, f and molecular structure of the insulating material

iii. Complex permittivity

Parallel RC model R represents the lossy part (electronic and ionic conductivity, dipole orientation and space charge polarization, etc.) capacitance in the presence of the dielectric I= jωCoV (εr− jεr tan δ)= jωCoVε∗

ℜ=εr

ℑ=εrtanδωCV 2tanδ

tanδ=σ /ωεoεr

iv. Loss angle

See (iii)

v. Dissipation factor

Similar to tan delta

vi. Polarization

Most electron are bound

Electrostatic forces create some level of polarization forming dipoles

It is this electronic polarization which results in relative permittivity of more

than 1 for most dielectric materials

Atomic polarization

Organic substances, e.g. Polymers

Interfacial polarization

vii. Dielectric strength

The maximum value of applied electric field at which a dielectric material is stressed in a homogenous(hence uniform?) field electrode system, breaks down and loses its insulating property

Dependent on purity, time and method of voltage application, type of stress, other experimental and environmental parameters.c. A 50kV voltage is applied to a square-shaped structure made from stainless steel,

except its base plate which is insulated from the rest of the structure and is grounded.

Taking a cross section of the structure and using a grid with sixteen equal squares

(giving nine points with unknown voltage), determine the voltages at all nine points after

one iteration.

Answer:

50kV

1 2 3

6 5 4

7 8 9

0kV

V 1=50+50+0+04

=25.00 kV

V 2=25+50+0+04

=18.75kV

V 3=50+50+18.75+04

=29.69kV

50kV 50kV

V 4=29.69+50+0+04

=19.92kV

V 5=18.75+19.92+0+04

=9.67kV

V 6=25+50+9.67+04

=21.17kV

V 7=21.17+50+0+04

=17.79kV

V 8=9.67+17.79+0+04

=6.87kV

V 9=19.92+6.87+50+04

=19.20kV

Node 1st iteration (kV)

1 25.00

2 18.75

3 29.69

4 19.92

5 9.67

6 21.17

7 17.79

8 6.87

9 19.20

QUESTION 3

a. The build-up of high currents in a gas breakdown is due to the process of ionization in

which electrons and ions are created from neutral atom or molecules. Explain how the

ionization process occurs prior to gas breakdown phenomena.

ANSWER:

When a high voltage is applied between the two electrodes immersed in a gaseous

medium, the gas becomes a conductor and an electrical breakdown occurs. The

process that responsible for the breakdown of a gas is called ionization. This process

initially liberates an electron from a gas molecule with the simultaneous production of a

positive ion. The generations of new electrons are from ionization by collision, photo-

ionization and the secondary ionization process. Under high voltage stress, a few of the

electrons produced at the cathode due to the certain process will produce positive ions

and additional electrons. The process repeats itself and hence increases in the electron

current.

b. In an experiment using certain gas, it was found that a steady state current of 600 µA

flowed through the plane electrode separated by a distance of 0.5 cm when a voltage of

10 kV is applied. Determine the Townsend’s first iteration coefficient if a current of 60 µA

flow when the distance of separation is reduced to 0.1 cm and the field is kept constant

at the previous value.

If the breakdown occurred when the gap distance was increased to 0.9 cm, what is the

value of Townsend’s secondary ionization coefficient?

ANSWER:

Since the field is kept constant (i.e., if distance of separation is reduced, the voltage is

also reduced by the same ratio so that V/d is kept constant)

I=I o eαx

Substituting two different set of values:

600=I oe0.5α and 60=I0 e0.1α

=> 10=e0.4α

Therefore, α =5.76 ionizing collisions/cm

The breakdown criterion is given by:

1−γ (eαx−1 )=0

Therefore the Townsend’s secondary ionization for the breakdown to be occurred at gap

distance 0.9 cm is:

γ (eαx−1 )=1

γ= 1

(e¿¿ αx−1)=1

(e¿¿ (5.76 )(0.9)−1)=1

177.4¿¿

¿5.64 x 10−1 cm−1

c. In SF6 gas, the effective ionization coefficient is given by:

αp

= 27.7[ Ep ] – 2460

Where α is the effective ionization coefficient incm−1, E is the electric field strength in

kV/cm and p is the pressure (referred to 20 °C) in bar. Breakdown may be predicted

using streamer criterion∫0

d

α .dx=18, where d is the length of the electrodes gap in cm.

Estimate the length of a uniform field gap that will just hold off a steady voltage of 100 kV

in SF6 at 4 bar and 60 °C.

ANSWER:

Given

αp=27.7

Ep−2460

Therefore,

α=27.7 E – 2460 p;

Multiply both sides with d , gives

αd=27.7 Ed−2460 pd

¿27.7V−2460 pd

pd=27.7V−αd2460

(1)

Given; ∫0

d

αdx=18 ;=¿αd=18(2)

The normalized pressure of 4 bar at 60°C is;

pn=40001013

.293

273+60=3.47 ¿

From (1) and (2);

d=27.7V−αd2460 pn

=27.7 (100 )−18

2460(3.47)=0.32cm

QUESTION 4

a. Describe briefly the reasons for electric stress being considered as the main contributor

to the breakdown of insulation. The description should be based on the principle of

insulation breakdown.

ANSWER:

Breakdown criteria for gas:

γ (eαx−1 )=1

Since αp=f ( E

p )∧γ=g (Ep)

Then at breakdown, α ds=pds f (E s

p )=p ds f (V s

pd)

» g( V s

pds)¿

The ionization of gases is related to

αp=f ( E

p )

I.e. α is dependent on E which is electric field. The rate of gas ionization is

dependent on the energy and velocity of free electrons, whereas electron energy and

velocity are dependent on the electric field applied to gas medium.

b. Describe briefly the element of electric stress optimization in the case where a solid

cylindrical insulator is sandwiched between a circular electrode and ground as shown in

Fig. Q4b. The description should put more emphasis on the tangential field distribution

and method to achieve it.

Figure Q4 b)

ANSWER:

The cylindrical shape insulator sandwiched between the plane electrode and ground

will experience non-optimized tangential field considering it from ground plane.

Whereas, with an insulator of profile shown in Fig. 1(b) with dotted line will provide

optimized field distribution.

c. Use the iteration method to find the finite difference approximation to the potential points

1, 2, 3 and 8 of the Fig.Q4c (Limit the iteration up to 2 only). The nodal voltage follows

the sequence as shown in Fig.Q4c that is the node number is 1,2,3,4,5,6,7 and 8

respectively where R₁ = 10 kΩ, R₂ = 30 kΩ, R₃= 20 kΩ, R₄ = 10 kΩ, Eₐ = 400V and EB =

600V

Figure Q4.c)

ANSWER:

V aa=R1

R1−R2

.Ea=10kΩ

(10−30)kΩ.400=100V

V bb=R3

R3−R4

. Eb=10kΩ

(10−20)kΩ.600=200V

Iteration 1:

V 1=100+200+0+0

4=75V

V 2=200+75+0+0

4=68.8V

V 3=200+200+0+68.8

4=117.2V

V 4=100+100+0+0

4=50V

V 5=75+0+0+50

4=31.3V

V 6=68.8+0+31.3+0

4=25V

V 7=117.2+0+25+0

4=35.5V

V 8=200+200+35.5+0

4=108.9V

Iteration 2:

V 1=100+200+68.8+31.3

4=100V

V 2=200+117.2+25+100

4=110.6V

V 3=200+200+35.5+110.6

4=136.5V

V 4=100+31.3+0+100

4=57.8V

V 5=100+25+57.8+0

4=45.7V

V 6=110.6+35.5+0+45.7

4=48.4V

V 7=136.5+108.9+0+48

4=73.4V

V 8=200+200+73.4+0

4=118.4 V

Therefore:

V 1=100V ,V 2=110.6V ,V 3=136.5V ,V 8=118.4 V

QUESTION 5

a. State the most useful equation which can be used directly to solve electric field problem

using Finite Difference Method?

ANSWER:

V o=14[V 1+V 2+V 3+V 4 ]

Where V 1+V 0 , V 2+V 0 , V 3+0 , V 4+V 0 are equidistance .

b. State the differences between electrostatic field and electromagnetic field?

ANSWER:

Electrostatic field Electric Field

i) Static charge i) Time varying current

ii) Coulomb’s Law ii) Maxwell’s Law

c. A high voltage DC transmission line rated at 132 kV peak traverses a location where a

road shall be constructed below as shown in Fig.Q5. The metallic walls L1 and L2 are

respectively energized at 20 kV peak and each standing on an insulator made of

polycarbonate, Use the Finite Difference Method to determine the potential value of point

5. Imaginary meshes are constructed below the transmission line each of size 10 meters

by 10 meters. Determine the potential at point 4 if the metallic wall L1 is shorted to the

ground. Limit the iteration process to 2 only.

Fig Q5.(c)

ANSWER:

Iteration 1:

V 1=0+132+20+0

4=38.0kV

V 2=0+132+38+0

4=42.5kV

V 3=20+132+42.5+0

4=48.63kV

V 4=0+38+20+0

4=14.5kV

V 5=0+42.5+14.5+0

4=14.3 kV

V 6=20+20.72+14.3+0

4=49.3kV

Iteration 2:

V 1'=42.5+132+20+14.5

4=52.3kV

V 2'=162.6+132+38+14.3

4=86.7kV

V 3'=20+132+61.78+49.3

4=58.68kV

V 4' =52.3+14.3+0+20

4=21.7 kV

V 5'=49.3+86+21.7+0

4=26.03kV

V 6'=20+61.0+26.03+0

4=26.06 kV

∴Wall L1 will falloff .

Iteration 1:

V 1=0+132+20+0

4=38.0kV

V 2=0+33+0+132

4=41.3kV

V 3=20+132+41.3+0

4=48.3kV

V 4=0+38+20+0

4=14.5kV

V 5=0+33+0+0

4=8.3 kV

V 6=20+48.3+12.4+0

4=20.2kV

Iteration 2:

V 1'=41.3+132+0+8.3

4=45.4kV

V 2'=43.3+132+45.4+12.4

4=59.5kV

V 3'=20+132+59.5+20.2

4=58.0 kV

V 4' =12.4+45.4+0+0

4=14.5kV

V 5'=20.2+59.5+14.5+0

4=23.6kV

V 6'=20+58.0+23.6+0

4=25.4 kV

Question 6

a. Give two advantages for the provision of the electric field stress control in high voltage

equipment.

ANSWER:

i. Insulation life is adversely affected by an increase in the operating stress values.

ii. Increase the efficiency of HV equipment.

iii. Can understand the failure mechanisms. The knowledge of electric field within

the insulation is essential since it is the intensity of electric field that determines

the onset of breakdown in dielectrics.

iv. The clearances of various HV components can only be determined if the

insulation behaviour under various stress as well as the field distributions within

the components are known.

b. Briefly describe the use of high voltage system in the following applications:

i. Removing industrial flue gases or dust particles floating in air from steel mill

chimneys

ii. Spraying pesticide to agriculture plantation

ANSWER:

i. By applying high voltage power supply to the electrode, corona activity will take

place creating ion pairs, which some ions will be positively charge will stick to the

flue gases or particles. Then passage towards the coherent cause some of them

to be drawn to collecting electrode.

ii. Due to the intense field at the tip of the nozzle, the emitting droplets of pesticide

are broken down to smaller and almost equal sizes. Thus, in affect increase the

coulombs force acting on the tiny droplets of spray and make it higher than

gravitational and inertial forces. This electrically changed force union of pesticide

has high attraction towards the leaves of the plants and ensures safeness.

Average on both sides of the leaves.

c. Use the iteration method to find the finite difference approximation to the potentials at

points 1, 2, 3, 9 and 15 of the system in Figure Q1. The nodal voltage follows the

sequence as shown in Figure Q1 that is node number is 1, 2, 3, 4, 5, 6, 7 up to 16

respectively.

Figure Q1

ANSWER:

The potential on 1, 5, 6, 12, 13, 14, 15, 16, and 17 are as follows.

Node Number Potential (V)

1 100

5 200

6 200

12 100

13 0

14 0

15 0

16 0

17 0

Potential to be estimated are 2, 3, 4, 7, 8, 9, 10, 11.

Iteration 1:

p1=200+0+0+100

4=75V

p2=200+0+0+75

4=69V

p3=200+200+0+69

4=117.3V

p4=200+200+0+0

4=100V

p5=117.3+100+0+0

4=54.3V

p6=69+54.3+0+0

4=30.83V

p7=75+30.83+0+0

4=26.5V

p8=100+100+26.5+0

4=56.63V

Iteration 2:

p1=200+69+26.5+100

4=98.9V

p2=200+117.3+30.83+98.9

4=111.76 V

p3=200+200+54.3+111.76

4=141.5V

p4=200+200+0+54.3

4=113.8 V

p5=141.5+113.8+0+30.82

4=71.53V

p6=111.76+71.53+0+26.5

4=52.4V

p7=98.9+52.4+0+56.63

4=52V

p8=100+52+0+100

4=63V

So: Potential at 1 = 100 VPotential at 2 = 98.90 VPotential at 3 = 111.76 VPotential at 9 = 52.4 VPotential at 15 = 0 V

QUESTION 7

a. Give three reasons for finding the electric field distribution in high equipment.

ANSWER:

Insulation life is adversely affected by an increase in the ooperating stress

values.

Increase the efficiency of HV equipment.

Can understand the failure mechanisms. The knowledge of electric field within the

insulation is essential since it is the intensity of electric field that determines the onset of

breakdown in dielectrics

b. Define the basic field equations

i. If there is no space charge in the dielectric medium

ii. If there is space charge in the dielectric medium

ANSWER:

i. ∇ .E= ρε

E=−∇V

∇ 2V=− ρε

ii. ∇ .E= ρε

E=−∇V

∇ 2V=− ρε

c. Use the iteration method to find the finite difference approximation to the potentials at

points 1 and 4 of the system in Figure Q1 (Limit the iteration up to 3). Take note that

nodal voltage should be in proper sequence that is node 1, 2, 3, 4, 5, 6 and 7 not

otherwise.

ANSWER:

Iteration 1

For node 1:

V 1=0+0+0+0

4=0V

For node 2:

V 2=0+0+50+0

4=12.5V

7 6

5 4 3

21

V = 0

50V

Not connected

Not connected

For node 3:

V 3=0+50+50+50

4=37.5V

For node 4:

V 4=0+12.5+37.5+0

4=12.5V

For node 5:

V 5=0+0+12.5+0

4=3.13V

For node 6:

V 6=0+12.5+50+50

4=28.1V

For node 7:

V 7=0+3.13+28.1+0

4=7.8V

Iteration 2

V 1=0+0+12.5+0

4=3.13V

V 2=3.9+0+50+12.5

4=16.9V

V 3=12.5+50+50+50

4=40.62V

V 4=3.13+16.6+40.63+28.1

4=22.12V

V 5=0+3.9+22.12+7.8

4=8.44V

V 6=7.8+22.12+50+50

4=32.5V

V 7=0+8.46+32.5+0

4=10.24 V

Iteration 3:

V 1=0+0+16.6+8.46

4=6.27V

V 2=6.27+0+50+22.12

4=19.6V

V 3=22.12+50+50+50

4=12.5V

V 4=8.46+19.6+43.03+32.5

4=25.9V

V 5=0+6.27+25.9+10.24

4=10.6V

V 6=10.24+25.9+50+50

4=34.04V

V 7=0+10.6+34+0

4=11.15V

Summary of node potential

V 1=6.27 ,3.13 ,0

V 2=19.6 ,16.9 ,12.5

V 3=12.5 , 40.62,37.5

V 4=25.9 ,22.12 ,12.5

V 5=10.6 ,8.44 ,3.13

V 6=34.04 ,32.5 ,28.1

V 7=11.15 ,10.24 ,7.8

CHAPTER 1

INTRODUCTION TO

HIGH VOLTAGE TECHNOLOGY

(QUESTION AND ANSWER)

QUESTION 1

a. There are two type of field distribution, known as homogeneous and non-

homogeneous field. What are differences of both field distributions? State the

electrodes-gap configuration to simulate the homogeneous and non-homogeneous

field.

ANSWER:

Homogeneous field:

E is the same throughout the field region.

Uniform or approximate uniform field distributions exist between 2 infinite parallel

plates, or 2 spheres of equal diameters with gap spacing < sphere radius.

“Profiled” parallel plates of finite sizes are also used to simulate homogeneous

fields.

Non-homogeneous field:

E is different at different points in the field region.

In the absence of space charges, E usually obtains the maximum value at the

surface of the conductor which has the smallest radius of curvature non-

homogeneous and asymmetrical.

Most of the practical HV components have non-homogeneous and asymmetrical

field distribution.

In some gaps – will produce non-homogeneous fields and symmetrical, e.g. rod-

rod or sphere-sphere (large distance between spheres) gaps.

HV electrode has higher E than the grounded electrode.

b. Experimental analog is one of the methods for determining the potential distribution.

Briefly describe any two of the experimental analogs used for space-charge-free

fields.

ANSWER:

1. Electrolytic Tank

Widely used for decades.

Equipotential boundaries are represented in the tank by specially formed

sheets of metal.

Example, a single dielectric problem such as a three-core cable may be

represented by electrolytes of different conductivities separated by special

partitions.

Fig: Electrolytic tank model of a three-core cable represented at the

instant when one core is at zero voltage, the same as the sheath

2. Semiconducting Paper Analog

Less accurate but attractively simple alternative to the electrolyte.

Errors in this method result from the nonhomogeneity of the paper

resistivity.

Errors also dependence on the ambient humidity and the contact

resistance to the electrodes.

Fig: Field plot between two spheres with skanks,

as plotted by a semiconducting paper model

3. Resistive-Mesh Analog

The continuous field is replaced by a discrete set of points as depicted by

a mesh of resistors.

The used of discrete resistors introduce an error arising from the finite

mesh analysis.

This error may be reduced by reducing the mesh size.

Fig: Resistive mesh analog of the field pattern

between two electrodes.

The potential at node 0;

V 0− (V 1+V 2+V 3+V 4 )÷ 4

Due to discretization, simulation of electrostatic fields in the vicinity of

curved surfaces of the electrodes is bound to be of reduced accuracy.

c. A high voltage DC transmission line rated at 132 kV peak traverses a location where

a road shall be constructed below it as shown in figure Q1. The metallic walls L1 and

L2 are energized at 20 kV peak and 80 kV respectively, and each standing on the

insulator made of polycarbonate. Use the Finite Difference Method to determine the

potential at point 2, 4 and 6. Limit the iteration process to two only.

Power line

12 3

4 5 6

Figure Q1

ANSWER:

1st iteration

V 1=132+0+0+204

=38kV

V 2=132+0+0+384

=42.5kV

V 3=132+80+0+42.54

=63.63kV

V 4=38+0+0+204

=14.5kV

V 5=42.5+0+0+14.54

=14.25kV

V 6=63.63+80+0+14.254

=39.47

L1

L2

Ground

2nd iteration

V 1=132+42.5+14.5+204

=52.25kV

V 2=132+38+63.63+14.254

=61.97 kV

V 3=132+80+39.47+65.534

=79.25kV

V 4=52.25+14.25+0+204

=21.63kV

V 5=65.53+39.47+0+21.634

=31.66kV

V 6=79.25+80+0+31.664

=47.73kV

The potential at point 2, 4 and 6 after 2 iterations are;

V2 = 61.97 kV, V4 = 21.63 kV, V6 = 47.73 kV

QUESTION 2

a. Describe briefly, with the aid of suitable diagrams, equations and/or examples, where

appropriate, the avalanche process in the breakdown phenomenon of gaseous

dielectrics.

ANSWER:

The avalanche process is one of the processes which occur in the breakdown of

gaseous dielectrics and is based on the generation of successive ionizing collisions

leading to an avalanche. Suppose a free electron exists (caused by some external

effect such as radio-activity or cosmic radiation) in a gas where an electric field

exists. If the field strength is sufficiently high, then it is likely ionize a gas molecule by

simple collision resulting in 2 free electrons and a positive ion. These 2 electrons will

be able to cause further ionization by collision leading in general to 4 electrons and 3

positive ions. The process is cumulative, and the number of free electrons will go on

increasing as they continue to move under the action of the electric field. The swarm

of electrons and positive ions produced in this way is called an electron avalanche. In

the space of a few millimeters, it may grow until it contains many millions of

electrons.

b. Show that the breakdown criterion in gas according to Paschen’s Law is given by:

g (V s / pds ) exp [ pds . f (V s/ pds ) ]−1 =1

where,

ds – gap distance at sparkover voltage

p – pressure

Vs – sparkover voltage

F & g – different functions

ANSWER:

By neglecting attachment, breakdown criterion

γ (ead−1 )=1 ………….. (1)

Since (Paschen’s Law), α / p=f (E / p ) and γ=g (E/ p), where f∧g significant

different function.

At breakdown, α ds=pdf (E s/ p )

¿ pds f (V s/ pds ) ………. (2)

And γ=g (E s/ p )=g (V s/ pds) ……… (3)

Substitute (2) & (3) into (1) gives;

g (V s / pds ) exp [ pds . f (V s/ pds ) ]−1 =1

c. The following data are given for two parallel while the electric field stress, E is kept

constant.

i. I = 1.2I0 when d = 0.5 cm

ii. I = 1.6I0 when d = 1.3 cm

iii. I = 2.3I0 when d = 2 cm

Where I0 is the initial current and d is distance between the plates.

Find the values of the Townsend Primary and Secondary coefficients, α and γ

ANSWER:

Using equation, I=I oeαd

For d = 0.5 cm, 1.2 I o=I o e0.5α∨α 1d 1=ln1.2=0.182α 1=0.38 /cm

For d = 1.3 cm, 1.6 I o=I oe1.3α∨α 2d2=ln 1.6=0.47α 1=0.36 /cm

For d = 2 cm, 2.3 I o=I oe2α∨α 3 d3=ln 2.3=0.83α 1=0.42 /cm (This suggests

that for this gap γ starts to be active).

The value of γ can be found from the equation;

I=Ioexp (αd)

1−γ [exp (αd )−¿1 ]¿

2.3= e (2 ×0.36 )

1−γ (e0.72−1)= 2.0544

1−1.0554 γ

γ=0.101/cm

QUESTION 3

a. Briefly describe any two (2) of the followings:

i. Field enhancement factor:

ii. ‘Medium high Voltage’ (MHV), ‘High Voltage’ (HV), ‘Extra High Voltage’ (EHV)

and ‘Ultra High Voltage’ (UHV)

iii. Three applications of high voltages excluding those in the generation,

transmission and distribution of electrical energy

ANSWER

i. Whereas any designer of the high voltage apparatus must have a complete

knowledge of the electric field distribution, for a user of the system the knowledge

of the maximum value of the electric field Emax to which the insulation is likely to

be subjected and the location of such a maximum gradient point is generally

sufficient. Consequently, the concept of field enhancement factor or simply field

factor (f) is of considerable use. This factor I defined as

F= E maxE av

(for homogeneous dielectric medium)

Where E av is the average field in the gap and is equal to the applied

potential/gap separation between the electrodes.

E max = f.Vd

Field utilization factor µf = 1f

(larger µf represents a more compact equipment)

ii. The following voltage classification describes the meanings:

Voltage Class Voltage Range

Medium high voltage (MHV) 1kV < V =< 70 kV

High Voltage (HV) 110kV < V =< 230 kV

Extra high voltage (EHV) 275kV < V =< 800 kV

Ultra high voltage (UHV) 1000kV =< V

iii. Any three of the followings:

i. ESPs for the removal of dust from flue gases

ii. Atomization of liquids, paint spraying and pesticide spraying

iii. Ozone generation for water and sewage treatment

iv. X-ray generators and particle accelerators

v. High-power lasers and ion beams

vi. Plasma sources for semiconductor manufacture

vii. Superconducting magnet coils.

b. There are several properties of dielectric which are of practical importance for an

engineer. Name five (5) most important properties of a dielectric and briefly describe

each of them

ANSWER

i. DC conductivity

σ= JE=1

ρ

dependent on material purity,T and E

σ (T )=Ae –EkT

due to polarization effects, σ also depends upon time of application of the

stress

ii. Dielectric permittivity

ε r= CC o

Dependent on T , f and molecular structure of the insulating material

iii. Complex permittivity

Parallel RC model

R represents the lossy part (electronic and ionic conductivity, dipole

orientation and space charge polarization, etc. Capacitance in the

presence of the dielectric

I= jωC ˳V (εᵣ− jεᵣ tan δ)= jωC ˳Vε

Rₑ=ε ᵣ

ℑ=εᵣ tan δ ωCV 2 tan δ

tan δ is dependent on f , E∧¿ T

Power loss = ωC ˳V 2 εᵣ tan δ

tan δ=¿σ /ωε ˳εᵣ¿

iv. Loss angle

see (iii)

v. Dissipation factor

Similar to tan δ

vi. Polarization

Most electrons are bound

Electrostatic forces create some level of polarization forming dipoles

It is this electronic polarization which results in relative permittivity of

more than 1 for most dielectric materials.

Atomic polarization

Organic substances, e.g. Polymers

Interfacial polarization

vii. Dielectric strength

The maximum value of applied electric field at which a dielectric material

is stressed in a homogeneous (hence uniform) field electrode system,

breakdown and loses its insulating property

Dependent on purity, time and method of voltage application, type of

stress, other experimental an environmental parameters.

c. A 50 kV AC voltage is applied to a square-shaped structure made from stainless

steel, except its base plate which is insulated from the rest of the structure and is

grounded. Taking a cross section of the structure and using a grid with sixteen equal

squares (giving nine points with unknown voltage), determine the voltages at all nine

points after one iteration

50kV

1 2 3

65 4

7 8 9

0kV

V 1=50+50+0+04

=25.00 kV

V 2=25+50+0+04

=18.75kV

V 3=50+50+18.75+04

=29.69kV

V 4=29.69+50+0+04

=19.92kV

50kV

50kV

V 5=18.75+19.92+0+04

=9.67kV

V 6=25+50+9.67+04

=21.17kV

V 7=21.17+50+0+04

=17.79kV

V 8=9.67+17.79+0+04

=6.87kV

V 9=19.92+6.87+50+04

=19.20kV

Node 1st iteration (kV)

1 25.00

2 18.75

3 29.69

4 19.92

5 9.67

6 21.17

7 17.79

8 6.87

9 19.20

QUESTION 4

a. The build-up of high current in a breakdown is due to the process of ionization in

which electrons and ions are created from neutral atoms or molecules. Explain how

the ionization process occurs prior to gas breakdown phenomena.

ANSWER:

When a high voltage is applied between the two electrodes immersed in a gaseous

medium, the gas becomes a conductor and an electrical breakdown occurs. The

process that responsible for the breakdown of a gas is called ionization. This process

initially liberates an electron from a gas molecule with the simultaneous production of

a positive ion.

The generations of new electrons are from ionization by collision, photo-ionization

and the secondary ionization process. Under high voltage stress, a few of the

electrons produced at the cathode due to the certain process will produce positive

ions and additional electrons. The process repeats itself and hence increases in the

electron current.

b. The ionization coefficient α/p as function of field strength E and gas pressure p is

given by the following threshold equation:

(αp )= fˌ( Ep )

By using the Townsend’s breakdown criterion, show that the breakdown voltage for

uniform field gaps is a function of gap length (d) and gas pressure (p).

ANSWER:

(αp )= fˌ( Ep ) (i)

The Townsend’s breakdown criterion;

γ [exp (αd)−1]=1 (ii)

Substituting (i) Into (ii);

ef (Ep ) pd ¿ 1

Y+1 (iii)

Taking in on both sides of (iii)

f [ Ep ] pd=ln [ 1Y+1]=K (vi)

For uniform field;

E=V b ¿d

Therefore,

ƒ¿b ¿ pd ] ¿K (v)

Or

ƒ¿b ¿ pd¿=K / pd ; V b ¿F (p .d )

Equation (vi) shows that the breakdown voltage of a uniform field gap is a unique

function of the product of gas pressure and the gap length for a particular gas and

electrode material. This relation is known as Paschen’s Law.

c. Fig.Q2 shows the experimental set-up for studying the Townsend discharge. The

experiment is conducted by measuring the current I at the different gap distance, d.

Table Q2 gives the set of observation obtained when studying the conduction and

breakdown in a gas.

i. Determine the initial current, I0.

ii. Calculate the value of the Townsend’s primary and secondary ionization

coefficients.

Table Q2 Townsend’s experimental data

Gap distance, d (mm) 1 2 3 4 5 6 8 10 12 14 16

Current I (pA) 19 21 26 32 40 45 80 106 152 255 430

Fig.Q2 Townsend’s experimental set-up

ANSWER

Gap distance, d (mm)

1 2 3 4 5 6 8 10 12 14 16

Current I (pA) 19 21 26 32 40 45 80 106 152 255 430

In I 2.94 3.043.26

3.47 3.69 3.81 4.38 4.665.02

5.54 6.06

I=I ˳ead

Taking In on both sides of (1);

ln I=ln ead+ ln I ˳

ln I=αd+ln I ˳ , y=mx+c

i. Plot graph ( ln I ) versus (d );

Gap dis-

tance, d

(mm)

1 2 3 4 5 6 8 10 12 14 160

1

2

3

4

5

6

7

2.94 3.04 3.26 3.47 3.69 3.814.38

4.665.02

5.546.06

In I

From the graph, interception © at ln I axis, gives;

(12-4=8) (5-3.5=1.5)

ln I ˳≈ 2.7 , I ˳≈ 14.88 pA

ii) Townsend’s primary ionization coefficient, α ;

Gradient of the graph (m) shows the value of α.

α ≈1.5÷ 8≈ 0.188mmˉˡ

Townsend’s secondary ionization coefficient,

I= I ˳ eαd

1−γ [eαd−1]

Substituting for higher value of;

430= 14.88 e(0.188 ) (16)

1– γ [ e(0.188 ) (16) –1 ]

¿ 301.271−19.25 γ

γ=0.016mmˉˡ

QUESTION 5

The build-up of high currents in a gas breakdown is due to the process of ionization in which

electrons and ions are created from neutral atoms or molecules. Explain how the ionization

process occurs prior to gas breakdown phenomena.

ANSWER:

When a high voltage is applied between the two electrodes immersed in a gaseous medium,

the gases becomes a conductor and an electrical breakdown occurs. The process that

responsible for the breakdown of a gas is called ionization. This process initially liberates

electron from a gas molecule with the simultaneous production of a positive ion. The

generations of new electrons are from ionization by collision, photo-ionization and the

secondary ionization process. Under high voltage stress, a few of the electrons produced at

the cathode due to the certain process will produce positive ions and additional electrons.

The process repeats itself and hence increases in the electron current.

QUESTION 6

In an experiment using a certain gas, it was found that a steady state current of 600µA

flowed through the plane electrode separated by a distance of 0.5cm when a voltage of 10kV

is applied. Determine the Townsend’s first ionization coefficient if a current of 60µA flows

when the distance of separation is reduced to 0.1cm and the field is kept constant at the

previous value.

If the breakdown occurred when the gap distance was increased to 0.9cm, what is the value

of Townsend’s secondary ionization coefficient?

ANSWER:

Since the field is kept constant (i.e., if distance of separation is reduced, the voltage is also

reduced by the same ratio so that Vd

is kept constant)

I=I o eαx

Substituting two different sets of values;

600=I oe0.5α (1) and 60=I oe

0.1α (2)

(1) ÷ (2)

600

60=

I o e0.5α

I o e0.1α

10=e0.4α ∴ α= 5.76 ionizing collisions/cm

The breakdown criterion is given by;

1−γ (eαx−1 )=0

Therefore the Townsend’s secondary ionization for the breakdown to be occurred at gap

distance 0.9cm is;

γ (eαx−1 )=1

γ= 1

(e¿¿ αx−1)=1

(e¿¿5.76(0.9)−1)=1

177.4¿¿

¿5.64 x 10−3 cm−1

QUESTION 7

In SF6 gas, the effective ionization coefficient is given by;

αp=27.7[ Ep ]−2460

Where α is the effective ionization coefficient in cm-1, E is the electric field strength in kV/cm

and p is the pressure (referred to 20 °C) in bar. Breakdown may be predicted using streamer

criterion, ∫0

d

α .dx=18, where d is the length of the electrodes gap in cm. Estimate the length

of a uniform – field gap that will just hold off a steady voltage of 100 kV SF6 at 4 bar and 60

°C.

ANSWER:

αp=27.7[ Ep ]−2460

α=27.7E-2460 p ; αd=27.7 Ed−2460 pd=27.7−2460 pd ;

pd=27.7V−αd2460

(1)

Given;∫0

d

αdx=18 ; αd=18 (2)

The normalized pressure of 4 bar at 60 °C is;

pn=40001013

.293

273+60=3.47 ¿

∴ From (1) and (2)

d=27.7V−αd2460 pn

.=27.7 (100 )−18

2460(3.47)=0.32cm

QUESTION 8

Use the iteration method to find the finite difference approximation to the potentials at points

1,2,3,9 and 15 of the system in figure Q2(c). (Limit the iteration up to 2 only). The nodal

voltage follows the sequence as shown in figure Q2(c) that is node number is 1,2,3,4,5,6,7

up to 16 respectively.

FIGURE Q2 (c)

ANSWER:

The general formula:

V n=V 1+V 2+V 3+V 4

4

Iteration 1

P1 P2 P3

P4P5P6P7P8

P 1=V 2=200+0+0+100

4=75V

P 2=V 3=200+0+0+75

4=68.75V

P 3=V 4=200+200+0+68.75

4=117.2V

P 4=V 7=200+200+0+0

4=100V

P 5=V 8=117.2+100+0+0

4=54.3V

P 6=V 9=68.75+54.3+0+0

4=30.76V

P 7=V 10=75+30.76+0+0

4=26.44 V

P 8=V 11=100+26.44+0+100

4=56.61V

Iteration 2:

P 1=V 2=200+68.75+26.44+100

4=98.8V

P 2=V 3=200+117.2+30.76+98.8

4=111.7 V

P 3=V 4=200+200+54.3+111.7

4=141.5V

P 4=V 7=200+200+0+54.3

4=113.6V

P 5=V 8=141.5+113.6+0+30.70

4=67.97V

P 6=V 9=111.7+67.97+26.44+0

4=51.53V

P 7=V 10=98.8+51.53+0+56.61

4=51.7V

P 8=V 11=100+51.7+0+100

4=62.92V

V 1=100V

P 1=V 2=98.8V

P 2=V 3=111.7 V

P 6=V 9=51.53V

V 15=0V

QUESTION 9

Describe the reasons for electric stress being considered as the main contributor to the

breakdown of insulation. The description should be based on the principle of insulation

breakdown.

ANSWER:

Breakdown criteria of gas

γ (eαx−1 )=1

αp=f [ Ep ]∧γ=g [ Ep ]

Then at breakdown, α ds=pds f [ E s

p ]=pds f [ V s

pd ]

∴g [ V s

pds](e pds f [ V s

pd ]−1)=1

The ionization of gases is related to

αp=f [ Ep ]

i.e. α is dependent on E which is electric field. The rate of gas ionization is dependent on

the energy and velocity of free electrons, whereas electron energy and velocity are

dependent on the electric field applied to the gas medium.

QUESTION 10

Describe briefly the elements of electric stress optimization in the case where a solid

cylindrical insulator is sandwiched between a circular electrode and ground as shown in Fig.

Q1b. The description should put more emphasis on the tangential field distribution and

method to achieve it.

ANSWER:

The cylindrical shape insulator sandwiched between the plane electrode and ground will

experience non-optimized tangential field considering it from ground plane. Whereas, with an

insulator of profile shown in Fig. 1(b) with dotted line will provide optimized field distribution.

QUESTION 11

Optimized profile

Nonoptimized

Distance Z (cm)

0

30

25

20

15

10

8642

Optimized

Nonoptimize

Use the iteration method to find the finite difference approximation to the potential at points

1, 2, 3 and 8 of the system in Fig.Q1c (Limit the iteration up to 2 only). The nodal voltage

follows the sequence as shown in Fig.Q1c that is the node number 1, 2, 3, 4, 5, 6, 7 and 8

respectively where R1 = 10KΩ, R2 = 30KΩ, R3 = 20KΩ, R4=10KΩ, Ea = 400V , Eb = 600V.

ANSWER:

V aa=R1

R1+R2

. Ea=10 K Ω

10 K Ω+30 K Ω.400=100V

Vaa

Vbb

V bb=R4

R4+R3

. Ea=10 K Ω

10 K Ω+20 K Ω.600=200V

Iteration 1

V 1=100+200+0+0

4=75V

V 2=200+75+0+0

4=68.8V

V 3=200+200+0+68.8

4=117.2V

V 4=100+100+0+0

4=50V

V 5=75+0+0+50

4=31.3V

V 6=68.8+0+31.3+0

4=25V

V 7=117.2+0+25+0

4=35.5V

V 8=200+200+35.5+0

4=108.9V

Iteration 2:

V 1=100+200+68.8+31.3

4=100.0V

V 2=200+117.2+25+100

4=110.6V

V 3=200+200+35.5+110.6

4=136.5V

V 4=100+31.3+0+100

4=57.8V

V 5=100+25+57.8+0

4=45.7V

V 6=110.6+35.5+0+26.44+0

4=45.7V

V 7=136.5+108.9+0+48

4=73.4V

V 8=200+200+73.4+100

4=118.4 V

V 1=100V

V 2=110.6 V

V 3=136.5V

V 8=118.4V

QUESTION 12

a. Show that in the process of gas breakdown, the Townsend First Ionization

Coefficient, α is given by;

α=1d

ln [ I t

I o]

where,

d- gap distance

I t- total current

I o- initial current

ANSWER:

Total no. of electron at anode, n(d)=no eαd. At steady state, average current in gap

at distance x,

I−¿(x)= Io eαx ¿ and I +¿(x)=I o(e¿¿αd−eαx)¿¿

Total number of current, I t=I−¿( x)+ I+¿(x) ¿¿ ¿ I o eαd

I t

I o

=eαd

αd=ln [ It

I o]

∴α= 1d

ln [ I t

I o]

b. The following data in table Q12b are given for two parallel plates while the electric

field, E is kept constant.

Table Q12b

Gap distance, d(cm) Ratio of Current and Initial Current, II o

0.5 1.2

1.3 1.6

2.0 2.3

ANSWER:

By using equation, I ¿ I o eαd

For d = 0.5 cm, 1.2 I o=I o e0.5α

or α 1 d1 ln 1.2=0.182 ,∴α1=0.36 /cm

For d = 1.3 cm, 1.6 I o=I o e1.3 α

or α 2 d2 ln 1.6=0.47 ,∴α 2=0.36 /cm

For d = 2 cm, 2.3 I o=I o e2α

or α 3 d3 ln2.3=0.83 ,

∴α3=0.42/cm (This suggest that for this gap γ start to be active)

The value of γ can be found from the equation;

II o

= eαd

1−γ (eαd−1)

2.3= e(2x 0.36)

1−γ (e0.72−1)= 2.0544

1−1.0544 γ

∴ γ=0.101/cm

c. At distance of 22.8mm and pressure 200mm Hg, the breakdown voltage of a uniform

field electrode in air is found to be 19.15Kv. Determine the breakdown voltage if the

secondary ionization coefficient γ is doubled. The values for the ratio of electric field

and pressure, Ep

and ratio of first ionization coefficient and pressure, αp

are given in

Table Q2c.

Table Q2c

Ep

(v/cm mm Hg)αp

(ion pairs/cm mm Hg)

41 0.0196

42 0.0222

ANSWER:

d = 22.8mm = 2.28 cm, p = 200 mm Hg V s = 19.15KV

Ep

= 42 v/cm mm Hg, αp

= 0.0222

Ep

= 41 v/cm mm Hg, αp

= 0.0196

Find V s when γ is doubled?

From secondary Townsend Breakdown Process,

I=I o eαd

1−γ (eαd−1)

and E = V s

d

Breakdown criteria: 1−γ (eαd−1 )=0 or γ eαd=1

E = V s

d =19.15

2.28 = 8.40 KV/cm = 8400V/cm

Ep=8400

200 = 42 v/cm mm Hg

From table;

αp

= 0.0222 ∴ α = 0.0222(200) = 4.44

∴αd=4.44 x2.28=10.12

From breakdown criteria (is doubled, α α ' ),

∴ γ eαd=2 γ eα 'd=1

eαd

eα' d=2

¿ - α ' ) = ln 2

α - α ' = ln22.28

∴α '=4.14

α '

p=4.14

200=0.02068

By Interpolation;

0

αp

0.02222

0.02068

0.0196

Ep42

2

Ep

41

Ep=41+

(0.02068−0.0196)(0.0222−0.0196)

=41.42

∴E=41.42 (200 )=8284 v/cm

∴V s=Ed=8284 x 2.28=18.89 Kv

QUESTION 13

a. State the most useful equation which can be used directly to solve electric field

problem using Finite Difference Method?

ANSWER:

V o=14[V 1+V 2+V 3+V 4 ]

Where V 1+V 0 , V 2+V 0 , V 3+0 , V 4+V 0 are equidistance .

b. State the differences between electrostatic field and electromagnet field?

ANSWER:

Electrostatic Field Electromagnet

a) Static Field a) Time Varying current

b) Coulomb’s Law b) Maxwell’s Law

c. A High Voltage DC transmission line rated at 132 kV peak traverses a location where

a road shall constructed below it as shown in Fig.1. The metallic walls L1 and L2 are

respectively energized at 20kV peak and each standing on an insulator made of

polycarbonate. Use the Finite Difference Method to determine transmission line each

of size 10 meters by 10 meters. Determine the potential at point 4 if metallic wall L1 is

shorted to the ground. Limit the iteration process to 2 only. Fig.1 The general

arrangement of a transmission power line traversing a piece of land.

Fig.1

ANSWER :

Iteration 1:

V 1=0+132+20+0

4=38.0kV

V 2=0+132+38.0+0

4=42.5kV

V 3=20+132+42.5+0

4=48.63kV

V 4=0+38+20+0

4=14.5kV

V 5=0+42.5+14.5+0

4=14.25 kV

V 6=20+48.63+14.25+0

4=20.72kV

Iteration 2:

V 1'=42.5+132+20+14.5

4=52.25kV

V 2'=52.25+132+48.63+14.25

4=61.78kV

V 3'=20+132+61.78+20.72

4=58.63 kV

V 4' =52.25+14.25+0+20

4=21.13 kV

V 5'=61.78+21.13+20.72+0

4=25.91kV

V 6'=20+58.63+25.91+0

4=26.14 kV

∴Wall L1 will falloff .

Iteration 1:

V 1=0+132+0+0

4=33.0kV

V 2=132+33+0+0

4=41.25kV

V 3=132+41.25+20+0

4=48.31kV

V 4=33+0+0+0

4=8.25kV

V 5=41.25+8.25+0+0

4=12.38kV

V 6=48.31+12.38+20+0

4=20.17 kV

Iteration 2:

V 1'=132+0+41.25+8.25

4=45.38kV

V 2'=132+45.38+48.31+12.38

4=59.49kV

V 3'=132+59.49+20+20.17

4=57.92kV

V 4' =45.38+0+12.38+0

4=14.44kV

V 5'=59.49+14.44+20.17+0

4=23.53 kV

V 6'=57.92+23.53+20+0

4=25.36kV

QUESTION 14

Discuss with suitable diagrams the mechanisms which lead to breakdown in liquid insulation.

ANSWER:

Suspended Particle Mechanism

1. Impurities present as fibers or dispersed solid particles

2. Electrostatic force acting on impurities

3. Solid impurities – force directed towards maximum stress

4. Gas impurities – force directed towards areas of lower stress

5. Form a stable chain bridging the gap.

Cavitations & Bubble Mechanism

1. Breakdown strength depends on applied hydrostatic pressure

2. Formation of vapor bubble responsible for breakdown due to

- Gas pockets at electrodes surface

- Electrostatic repulsive forces

3. Gases products by electron collision

4. Vaporization of liquid by corona at sharp points and surface irregularities.

Thermal Mechanism

1. Breakdown under pulse condition

2. High density current pulses give rise to localized heating and formed bubbles

3. Breakdown occurs due to elongation of bubbles to critical size and bridge the gap

4. Breakdown strength depends on pressure and liquid molecular structure.

Stressed Oil Volume Mechanism

1. Breakdown strength is determined by largest possible impurity or weak link.

2. Breakdown strength is inversely proportional to the stressed oil volume

3. Breakdown voltage influence by gas content in the oil, viscosity and the presence of

impurities.

QUESTION 15

Show the breakdown criterion in gas according to Paschen’s Law is given by

g (V s / pds ) exp [ pds . f (V s/ pds ) ]−1 =1

Where,

d s - gap distance at sparkover voltage

p - pressure

V s - sparkover voltage

f∧g - different function

ANSWER:

By neglecting attachment, breakdown criterion

γ (ead−1 )=1 ………….. (1)

Since (Paschen’s Law), α / p=f (E / p ) and γ=g (E/ p), where f∧g significant different

function.

At breakdown, α ds=pdf (E s/ p )

¿ pds f (V s/ pds ) ………. (2)

And γ=g (E s/ p )=g (V s/ pds) ……… (3)

Substitute (2) & (3) into (1) gives;

g (V s / pds ) exp [ pds . f (V s/ pds ) ]−1 =1

QUESTION 16

Breakdown voltage measurement of a uniform field gap in air at 32930 K gave the following

results shown in Table Q16.

Table Q16

pd (bar-cm) E/p at breakdown (kV bar¯¹ cm¯¹)

1.0 30.30

9.0 26.00

Determine the breakdown voltage of a 20 mm gap at a pressure of 3 bars and

temperature of 3000 K .

ANSWER:

V s=Apd+B (pd )12

Ed=Apd+B (pd )12

E=Ap+B ( p/d )12

E / p=A+B/ ( pd )12

From the data given,

30.3=A+B / (1 )12

30.3=A+B …………. (1)

26.0¿ A+B/ (9 )12

26.0=A+0.33 B ………. (2)

From (1) and (2); A=23.55 and B =6.42

For the case of atmospheric air;

V s=23.88 (p /d )+6.42 (p /d )12

p=3 , t=300 K ,d=2cm ,V s=?

QUESTION 17

a. Give two advantages for the provision of the electric field stress control in high voltage

equipment.

ANSWER

Increase the efficiency of high voltage equipment

Insulation life is adversely affected by an increase in the operation stress return.

b. Briefly describe the use of high voltage system in the following application:

i) Removing industrial flue gases or dust particles floating in air from steel mill

chimneys

ANSWER:

By applying high voltage power supply to the electrode, corona activity will

take place creating ion. Some ion will be positively changed which stick to

the gas or particles.

ii) Spraying pesticide to agriculture plantation

ANSWER:

Due to Incense field at the tip of the nozzle the emitting of one particle

broken down to smaller and almost equal sizes.

iii)

c. Use the iteration method to find the finite difference approximate to the potentials at

point 1, 2, 3, 9, and 15 of the system in Fig. Q1 (limit the iteration up to 2 only). The

nodal voltage follows the sequence as shown in Fig. Q1 that is node number is 1, 2,

3, 4, 5, 6, 7 up to 16 respectively.

Figure Q1

ANSWER:

Iteration 1:

p1=100V

p2=200+0+0+100

4=75V

p3=200+0+0+75

4=68.75V

p4=200+200+0+68.75

4=117.19V

p7=200+200+0+0

4=100V

p8=117.19+100+0+0

4=54.29V

p9=68.75+54.29+0+0

4=30.76V

p10=75+0+0+30.76

4=26.44V

p11=100+100+0+26.44

4=56.61V

p15=0V

Iteration 2:

p1=100V

p2=200+100+68.75+26.44

4=98.79V

p3=200+98.79+30.76+117.19

4=111.69 V

p4=200+111.69+54.29+200

4=141.49V

p7=200+54.29+0+200

4=113.57V

p8=141.49+30.76+0+113.57

4=71.45V

p9=11.69+26.44+0+71.45

4=52.39V

p10=98.79+56.61+0+52.39

4=51.95V

p11=100+100+0+51.95

4=62.99 V

p15=0V

Therefore: p1=100V , p2=98.79V , p3=111.69V , p9=52.39V , p15=0V

QUESTION 18

Discuss the processes that lead to ion-generation in a gas breakdown.

ANSWER:

i. Ionization by Electron Impact

Ionization by Electron Impact is the most important process for gas discharge. Kinetic

energy exchanged during collision. Gas atom or molecule becomes excited or ionized

by the energy acquired from the incident atom. Portion of kinetic energy prior to impact

converted to potential energy. Atom or molecule may be ionized by a subsequent with

another slow-moving election.

ii. Photoionization

Result of external radiations. Eg. Cosmic rays, X-rays, Nuclear radiations. Continuous

process produces ions and electrons. It’s capable of penetrating most conventional

walls. It’s also an easy method to produce spark or to ignite combustible mixture with

free electrons. Insulation of high-voltage systems at high attitude is subjected to reduce

air density and increase in ionization by cosmic rays.

iii. Electron Detachment

Electron detached from negative ions in the gas. It’s requires large concentration of

negative ions. Eg. Gas discharged under impulse voltages.

A−h∪⇔ A+e−¿

QUESTION 19

In an experiment to determine the breakdown properties of air, the uniform field electrode is

used. The breakdown process occurs in accordance with Townsend First and Second

Ionization coefficient, α and γ .

At a distance of 22.8 mm and pressure 200 mm Hg, the breakdown voltage is found to be

19.5 kV. Determine the breakdown voltage if the secondary ionization coefficient γ is

doubled. Data’s for the ratio of electric field and pressure, Ep

and ration of first ionization

coefficient, αp

are given in the Table Q19.

Table Q19

Ep

(V/cm mm Hg)αp

(ion pairs/cm mm Hg)

41 0.0196

42 0.0222

ANSWER:

d=22.8mm=22.28cm , p=200mm. Hg ,V s=19.15kV

Ep=42

Vcm

.mm.Hg ;αp=0.0222

Ep=42

Vcm

.mm.Hg ;αp=0.0196

Find V swhen γ is double?

From secondary Townsend Breakdown Process,

I=I o eαd

1−γ [eαd−1]∧E=

V s

d

Breakdown criteria:

1−γ [eαd−1 ]=0∨γ eαd=1

I=V s

d=19.15

2.28=8.4

kVcm=8400

Vm

Ep=8400

200=42

Vcm

.mm.Hg

From Table:

αp=0.00222 , α=0.0222 (200 )=4.44 , αd=4.44 x2.28=10.12

From breakdown criteria (γ is double, α →α ' )

γ eαd=γ eα' d=1

eαd

eα' d=2

(α−α ' )d=ln2

(α−α ' )d= ln22.28

α '=4.14

α 'p=4.14

200=0.02068

By interpolation

Ep=41+

(0.02068−0.0196)0.0222−0.0196

=41.42

E=41.42(200)=8284Vcm

V s=Ed=8284 x 2.28=18.89kV

QUESTION 20

a. Give three reasons for finding the electric field distribution in high voltage equipment.

ANSWER:

1. The use of high voltage in electric power transmission to avoid excessive line

currents which would render the transmission system uneconomical.

2. High voltage is utilized is based on the fact that bodies charged under high

voltage develop an electrostatic force. Applications: cathode-ray tubes,

particle accelerators, xerography, spray painting, and electrostatic

precipitators.

3. High-voltage presence makes use of the ability of high voltage to initiate

ionization in dielectric materials where energy is subsequently released in

controlled quantities. Applications: e.g., ignition in internal combustion

engines, gas-discharge lamps, and ozone generation.

b. Define the basic field equation:-

i) If there is no space charge in the dielectric medium.

ii) If there is space charge in the dielectric medium.

ANSWER:

i) Using the Laplace’s equation if there is no space charge in the dielectric

medium.

∇2V=0

ii) Using Poisson’s equation if the medium has a space charge density.

∇ .∇V=∇2 V

¿−PE0

c. Use the iteration method to find the finite diffence approximation to the potentials at

point 1 and 4 of the system in fig. Q1 (limit the iteration up to 3 only). Take node that

nodal voltage should be in proper sequence that is node 1, 2, 3, 4, 5, 6 and 7 not

otherwise.

ANSWER:

Iteration 1:

V 1=0+0+0+0

4=0V

V 2=0+0+50+0

4=12.5V

V 3=0+50+50+50

4=37.5V

V 4=10+12.5+37.5+0

4=12.5V

V 5=0+0+12.5+0

4=3.13V

V 6=0+12.5+50+50

4=28.1V

V 7=0+3.13+28.1+0

4=7.8V

Iteration 2:

V 1=0+0+12.5+3.13

4=3.9V

V 2=3.9+0+50+12.5

4=16.6V

V 3=12.5+50+50+50

4=40.62V

V 4=3.12+16.6+40.63+28.1

4=22.12V

V 5=0+3.9+22.12+7.8

4=8.46V

V 6=7.8+22.12+50+50

4=32.5V

V 7=0+8.46+32.5+0

4=10.24 V

Iteration 3 :

V 1=0+0+16.6+8.46

4=6.27V

V 2=6.27+0+50+22.12

4=19.6V

V 3=22.12+50+50+50

4=43.03V

V 4=8.46+19.6+43.03+32.5

4=25.9V

V 5=0+6.27+25.9+10.24

4=25.9V

V 6=10.24+25.9+50+50

4=34.04V

V 7=0+10.6+34+0

4=11.15V

Therefore:

V 1=6.27V ,V 4=25.9V

QUESTION 21

a. Describe the secondary process which can follow an electron avalanches and how

these process may be identified. Show that discharge current in a multi avalanche

Townsend process in a non-attaching gas is given by

I = I 0 exp(αd)

1−γ [exp (αd−1 )]

ANSWER:

The electrical breakdown of a gas is brought about by various processes of

ionization. These are gas processes involving the collision of electrons, ions and

photons with gas molecules and electrode processes which take place at or near the

electrode surface. When a pair of electrodes is immersed in a gas and a voltage

applied across them, the current – voltage characteristic of figure below is observed.

At low voltage the observed current is due to collection of free charge carriers in the

gap and as the voltage is increased a level is reach at which the free electrons gain

enough energy to ionize. Electrons produced may cause further ionization so that an

electron avalanche is generated. Ionization is the process by which an electron

removed from an atom, leaving the atom with a net positive charge. The probability of

ionization due to the electrons will depend on the number of collisions made per unit

distance with coefficient α. α is referred as the primary ionization coefficient which is

number of ionizing collisions per electrons per cm travel.

With the primary ionization alone the discharged is not self – sustaining. If the source

of initial electrons is removed, the current I fall to zero. This suggests that processes

other than the simple α- process are occurring. The additional current is produced by

secondary emission processes. A secondary ionization coefficient γ is defined as the

number of secondary electrons produced at the cathode per electron produced in the

gap.

These processes for secondary – electron liberation can be identified by;

i) Positive – ion , γ i – ions do not have enough energy to ionize gas molecules directly

but may release electrons on colliding with the cathode surface.

ii) Photon, γ p – a proportion of the collisions in the gap cause excitation of neutral –

gas molecules which on return to the ground state may emit photons which release

electrons by photoemission.

iii) Metastables, γ m – metastable molecules may diffuse to the cathode and release

electrons.

One or more secondary mechanism may exist giving a total secondary effect

describe by

γ = γ i + γ p + γ m

Let no = the number of initial electrons at cathode

n 0 ‘ = the number of initial secondary

no “ = the total emission including secondary

i.e no “ = no + no ‘

At x, n(x) = no “ exp(αx)

The total number of new electrons produced, n(d) = no “¿].

If γ electrons are produced at the cathode per ionizing collision in the gap, then

n0 ‘ = γn0” ¿]

Thus , no “ = no + γn0” ¿]

no “ = n0

1−γ [exp (αd−1 )]

Therefore n(d) = no “ exp(αd) = n0exp (αd)

1−γ [exp (αd−1 )]

Under steady state conditions , I = I 0 exp (αd)

1−γ [exp (αd−1 )]

b. What is meant by ‘time lag to breakdown’ and describe how it may be influenced and

exploited.

ANSWER:

On the application of a voltage, a certain time elapses before actual breakdown

occurs even though the applied voltage may be much more than sufficient to cause

breakdown. In considering the time lag observed between the application of a voltage

sufficient to cause breakdown and the actual breakdown, the two basic processes of

concern are the appearance of avalanche initiating electrons and the temporal

growth of current after the criterion for static breakdown is satisfied.

In time – resolved studies. A step function voltage pulse is applied to the gap. The

time to breakdown then comprises;

i) Statistical time lag, ts elapsing prior to the appearance of an electron to initiate the

primary avalanche.

ii) Formative time lag’, tf required for the current builds up by secondary processes.

Analysis of formative time lags can yield information on the relative contributions of

the various secondary processes. The shortest formative times would be expected

with the photon secondary mechanism when the secondary cathode photoelectrons

are produced during the avalanche – crossing time. In general the formative time lag

is a function of the pulse amplitude, pressure and gap spacing.

c. Describe with diagrams the principle breakdown mechanisms which can occur in

solid dielectrics and identify their order of occurrence on a stress-time diagram.

ANSWER:

In almost all electrical equipment, solid insulating materials are used to separate

conductors at different potentials. Failure of the insulation occurs if a conducting or

partially conducting path is established between these conductors. This can occur

either over the surface of the insulating materials or through the body of the material.

When the discharge part occurs across the surface of the material, this is known as

“surface tracking” or “surface flashover”. When breakdown occurs through the body

of the insulating materials, the damage is totally irreparable and the insulation must

be replaced.

It is generally accepted that there are seven ways in which solid insulation can:

1) Electrochemical failure

2) Discharges in cavities within the insulation

3) Breakdown in initiated by spark penetration

4) Electromechanical failure

5) Ambience discharges

6) Thermal breakdown

7) Intrinsic breakdown

The order of the occurrence of the above mechanism can be illustrated on a stress-time

diagram as below.

1) Electrochemical breakdown

In electrical component design for use at low voltage and frequencies, electrochemical

damage is more probable than other types of failure. The deterioration is cause by irons

liberated at the electrodes by conducting current. The damage is dependent both on the

current and reaction with the insulating material. The effect can always be reduced by

reducing the leakage current.

2) Discharges in cavities

Solid insulating material often contains small cavity of gas in which the applied stress is

considerably greater than in the solid material. This can be understood by considering

the equivalent circuit shown below.

If as is almost inevitable, the stress in the gas phase exceeds the breakdown stress for

the gas, then partial breakdown will take place within the cavity this causes thermal and

chemical degradation of the solid dielectric around the boundaries of the cavity and over

a period of time this can lead to a failure of the dielectric.

3) Breakdown initiated by spark penetration on ambient discharge

In practice, the electrodes are never perfectly embedded in the solid insulation and the

dielectric is stressed in conjunction with one or more materials. If one of these materials

has a lower dielectric strength than the solid dielectric then the measured breakdown

voltage will be influence more by that weak medium than by the solid.

A local breakdown of the dielectric at the tips of the discharge is therefore likely and

complete breakdown is the result of such breakdown channels formed in the solid.

4) Electromechanical failure

This types of failure is more applicable to less rigid forms of insulating materials such as

rubber, pvc etc. if we consider two electrodes supported apart by an elasticity material

and a voltage is applied , an attracting electrostatic force will be established between the

plates. This will cause them to move together against the natural elasticity of the

dielectric material because the plates move together, the electrostatic force between

them increases and this will caused a further reduction in spacing. At a certain applied

voltage Vs, an unstable situation will be set up and dielectric material will be collapsed.

5) Thermal breakdown

When a field applied to a dielectric at room temperature the conduction current is

normally very low but its value increases rapidly with temperature. Correspondingly as

current increases so the heat generated within the dielectric increases the temperature

rises further. As the temperature of the dielectric increases, thermal dissipation from its

surface occurs and the final resulting temperature depends upon the heat dissipated in

relation to the heat generated.

6) Intrinsic breakdown

It is considered to be the breakdown mechanisms which take place in the absences of all

other no mechanism know of failure. Intrinsic breakdown theories generally involve

electronic process and it was natural in view of the success of Townsend avalanche

theory in gasses.

QUESTION 22

Measurement of breakdown voltages in a uniform-field spark gap in air gave the results as

shown in Table below

Gap spacing

(mm)Pressure (Bar)

Temperature

(°C)

Breakdown

Voltage, Vs

(kV)

2.5 1.03 30 0.91

27 1.185 15 88.38

Using the expression derived from Paschen’s law, Vs=A (pd )+B (√ pd ), determine;

i. The relative air density ρ referred to standard atmospheric conditions of 1013

25 mbar and 20°C.

ANSWER:

d=0 .25cm , ρ=1034 mbar , t=30° C,Vs =0 .90kV

d=2.7cm , ρ=1180mbar , t=15° C,Vs=88 .56kV

ρ20=10341013

⋅273+20273+30

=10341018

⋅293303

=0.99

ρ20=11801013

⋅298288

ii. The value of constants A and B

ANSWER:

V ll g/m3=V 9+V 9⋅¿ ¿∆V

=V 9(1+∆V )

=0.90 [1+0 .002(2) ]=0.91kV

V ll g/m3=V 12(1−∆V )

=88. 56 [1−0 . 002(1) ]=88. 38kV

V s1=Aρ1d1+B√ ρ1d1

V s2=Aρ2d2+B√ ρ2 d2 ________ χ

=1 .21

ρ1d1=0 .99(0 .25)=0 .25

√ ρ1d1=0 .5

ρ2 d2=1 .21(2 .7)=3 .27

√ ρ2d2=1 .81Dari _______ χ0 .91=A (0.25 )+B (0.5 )________ 188 .38=A (3 .27 )+B (1 .81) ______2

[88 .380 .91 ]=[3 .27 1 .81

0 .25 0 .5 ][ AB ]A=[88 . 38 1 . 81

0 . 91 0 . 53 . 27 1 . 810 . 25 0 . 5

]

B=[ 3 . 27 88 .380 . 25 0 . 913 . 27 1 .810 . 25 0 .5

]

iii. The breakdown voltage of a 3 cm gap spacing at a pressure of 3000 mbar and a

temperature of 20°C.

ANSWER:

d=3cm, ρ=3000mbar,t=20 °C

√ ρd=2 .98

∴V s=35. 75(8 . 88 )−16 . 07(2 .98 )

=88. 38(0 .5)−1. 81(0 .91)3 .27(0 . 5)−1.81(0 . 25)

=44 .19−1 .651 .64−0 .45

=42. 541 .19=35 . 75

=3 .27(0 .91)−88 .38 (0.25 )1 .19

=2 .98−22.101 .19=−16 .07

ρ20=30001013

×293293

=2 .96

ρd=2 .96(3 )=8 .88

=317 . 46−47 .89=269 . 57kV

QUESTION 23

a. In a strongly in homogeneous field, external partial discharges occur at electrodes of

small radius of curvature when a definite voltage is exceeded. These are referred to

as corona discharges and depending upon the voltage amplitude, they result in a

larger number of charge pulses of very short duration. With the aid of the diagram

where appropriate,

i. Define the terms corona

ANSWER:

The term corona is used to describe the discharge phenomena which occur at

highly – stressed electrodes prior to the complete breakdown of the gap

between the electrodes. It is a partial discharge in air around a sharp point or

thin wire in a strong, non-uniform field. It is characterized by a visible glow, an

audible noise, radio interference, chemical effect such as production of ozone

and loss of electrical power. It occurs whenever the local voltage gradient

exceeds the ionization value of the air and depends on the air density,

humidity and in outdoor situations whether it is fair weather or raining and

also on the roughness of the conductor surface.

ii. Types of corona and how it occurs

ANSWER:

Corona can be classified into;

1) DC Corona

a) Positive corona(anode)

When the highly-stressed electrode is the anode, the following corona

modes are observed as the voltage is increased.

Onset streamers

Also known as ‘burst’ pulses, these are intermittent, filamentary

discharges which propagates only a short distance from the highly-

stressed electrodes.

Hermstein glow

As the voltage increased, the intermittent streamer discharges give

way to a steady glow discharges. This transition occurs when a large

enough negative – iron space is generated near the anode to give a

quasi-uniform field in that region.

Breakdown streamers

Eventually, the shielding effect of the glow discharge is not able to

prevent the formation of large streamers which propagate well into the

gap.

b) Negative corona (cathode)

When the highly-stressed electrode is negative, three modes are

again observed.

Trichel pulses

These differ from the burst pulses in that their magnitude and

repetition frequency are both very regular.

Cathode glow

As the voltage is raised a critical Trichel pulses repetition frequency is

reached and the repetitive discharge is replaced by a steady cathode

glow.

Negative streamers

These discharges are usually known as negative feathers to avoid

confusion with positive streamers discharges. They develop out of the

glow mode and a long rise time compared with other pulsating

coronas.

2) AC corona

With an alternating voltage applied, the same basic corona types will

appear although their characteristic maybe altered to an extent which

depends on the gap length:

Small gap (d ¿ 100cm)

Here, the irons generated in any of the corona modes above are able

to cross the gap during one half cycle of the voltage. Space charges

will therefore not persist from one half cycles to the next and the

corona modes will therefore be similar to those for direct voltages,

although all three modes maybe observed in one half cycle.

Large gaps (d > 1m)

For those gaps, space charge can persists from one half cycles to the

next and can have an effect on the corona modes observed. The

usual effect is to enhance the positive-glow phase. Further, the

negative streamers are never observed in ac-stressed gaps, since its

onset potential is higher than the positive polarity breakdown voltage.

Breakdown always occurs on the positive half cycle.

3) Transmission line corona

The above description of the types of corona discharge referred

particularly to the point-plain gap where there is a single side for

discharge to occur. On a transmission line, corona may occur anywhere

the line and the average corona currents will be much higher.

iii. The problem which are created by corona discharges on high voltage

transmission lines.

ANSWER:

The problems which are created by corona discharges on high voltage

transmission lines are:

Power losses

The power losses depend upon the maximum gradient for which the

line is designed. For a single conductor, this occurs at the conductor

surface. For a given cross-section of conductor required for current

carrying capacity, the maximum stress maybe reduced by using

bundled conductors in which 2, 4 or 6 wire assembly is used.

Radio interference

Radio interference is caused only by the pulse corona modes and only

Trichel pulses and positive streamers are of interest. The positive

streamers usually have shorter rise time than the Trichel pulses and

greater amplitude, so that the rate of change of current greater and

their RI effect is therefore greater. These corona discharges caused

radiation of electromagnetic waves.

Audible noise

Recent studies on EHV and UHV lines indicate that audible noise

maybe a problem where such lines pass near inhabited areas.

Difficulties arise in monitoring such noise levels as the ‘apparent

noise’ is a nonlinear function of frequency. Measuring instruments

have thus been developed which has similar respond to their human

ears and level has been set based limit at which most people find the

noise objectionable.

QUESTION 24

a. Discuss three (3) mechanism of solid insulation breakdown

ANSWER:

Insulation breakdown:

1. Ionic insulator or intrinsic

2. Electromechanical insulation

3. Thermal insulation

4. Chemical insulation

5. Treeing and tracking insulation

6. Internal discharge.

i. Ionic and intrinsic insulation

When the voltage is higher than that for a long time.

Electric power is determined by the intrinsic strength.

Depends on the presence of free electrons that move through solid

lattice.

These electrons produce a current flow.

ii. Electromechanical insulation

Less rigid insulation material. (Rubber, PVC)

Electrostatic forces that act exceed the mechanical strength of solid.

Mechanical damage will occur.

iii. Thermal insulation

Leakage current to flow when electrical stresses imposed.

Solid temperature increases when going process heating.

Heat is transfer to the surrounding insulation through conduction and

radiation process.

Limiting the maximum thickness of a solid.

iv. Chemical insulation

Chemical changes by continuous electrical stress through reaction with air

and gas.

Reaction that occurs – oxidation, hydrolysis, chemical reaction. Can be

minimised by carefully inspect the materials.

v. Treeing and tracking insulation

Two effects when the old electrical stress;

-presence of conductive paths across the surface effects of spark caused

the leakage current through the path-producing sparks.

Dissemination channels sparks during the tracking process in the form of

branches called treeing.

vi. Internal discharge insulation

Cavity containing air in the solid insulation

Field in the cavity is larger than the field of insulation

Breakdown exists in the cavity.

b. Show the breakdown criterion in gas according to Paschen’s Law is given by:

g(V s/ p ds)e[pd sf ( V s

pd s)]−1 =1

Where

d s - gap distance at sparkover voltage

ρ - pressure

V s - sparkover voltage

f∧g - different function

ANSWER:

By neglecting attachment, breakdown criterion

γ (ead−1 )=1 ………….. (1)

Since (Paschen’s Law), α / p=f (E / p ) and γ=g (E/ p), where f∧g significant

different function.

At breakdown, α ds=pdf (E s/ p )

¿ pds f (V s/ pds ) ………. (2)

And γ=g (E s/ p )=g (V s/ pds) ……… (3)

Substitute (2) & (3) into (1) gives;

g (V s / pds ) exp [ pds . f (V s/ pds ) ]−1 =1

c. In nitrogen gas, the static breakdown voltage V sof a uniform field gap maybe as

express as,

V s=A ( ρd )+B(√ ρd)

Where A and B a constant, ρ is the gas pressure in torr referred to a temperature of

20° C and d is the gap length in cm.

A 1 cm uniform field gap in nitrogen at 760 torr and 25° C is found to breakdown at a

voltage of 33.3kV . The pressure then reduced an after a period of stabilization, the

temperature and pressure are measured 30° C and 500 torr respectively. The

breakdown voltage is found to be reduced to 21.9kV . If the pressure further

reduced to 350 torr while the temperature of the closed vessel is raise to 60° C and

the gap distance increase to 2cm, determined the breakdown voltage.

ANSWER:

V s1=A ρ1 d1+ β√(ρ1d1)

ρ1=760 torr , t1=25 ° C ,d1=1cm,V s 1=33. 3kV

Corrected pressure to standard temperature of 20° C

ρ1=760 (273+20 )

273+25=747.25 torr ,√(ρ1 d1)=27.34

∴33 .3=A (747 .25 )(1)+B(27 .34 )(1 )

=747 . 25 A+27 .34 B . . .. .. . .. .. . .. .. . .. .. .. . .. .. (1 )

V s2=A ρ2 d2+ β√( ρ2d2) ρ2=500 torr ,t2=30° C ,d2=1cm ,V s2=21. 9kV

Corrected pressure to standard temperature of 20° C

ρ2=500 (273+20 )

273+30=483.50 torr ,√¿¿

∴21 .9=A (483 .50 )(1)+B(21 .99)(1 )

=483 .50 A+21 .99 B . .. .. . .. .. . .. .. . .. .. . .. .. . ..(2 )

(1)×21,99 :732 .27=16432 .03 A+601. 21B . .. .. . .. .. . .. .. .. . .. .(3)(2)×27 .34 : 598.75=13218. 89 A+601 . 21B .. .. . .. .. .. . .. .. . .. ..( 4 )(3 )−( 4 ):133 . 52=3213 .14 A∴A=0. 042andB=0 . 072

V s3=Aρ3 d3+B√( ρ3d3 ) ρ3=350 torr ,t3=60 °C ,d3=2cm ,V s 3=?

Corrected pressure to standard temperature of 20° C

ρ3=350(273+20)273+60

=307 . 96 torr ,√( ρ3 d3)=24 . 82

∴V s3=(307. 96 )(2 )( 0.042 )+(24 .82)(0 . 072)

=27 . 66kV

QUESTION 25

a. Briefly describe the definition of high voltage and the classification of voltage levels.

ANSWER:

Definition of high voltage:

IEC 1970 – A high voltage is voltage being greater than 1000V for alternating and

greater than 1200 for direct.

Voltage Classification:

Low Voltage (LV) : 220V to 1000V (Europe) or 120V to 660V (US)

Medium Voltage (MV) : 5kV to 66kV (Europe) or 2.4kV to 69kV (US)

High Voltage (HV) : 110kV to 220kV (Europe) or 115kV to 230kV (US)

Extra High Voltage (EHV) : 275kV to 800kV (Europe) or 287kV to 765kV (US)

Ultra High Voltage (UHV) : above 1000kV

b. Show that in the process of gas breakdown, the Townsend First Ionization

Coefficient, α is given by ;

α=1d

ln [ I t

I o]

where, d - gap distance

It - total current

Io - initial current

ANSWER:

Total no. of electrons at anode, n (d )=no exp (αd )

At steady state, average current in gap at distance x,

L ( x )=I oexp (αx) and I ( x )=I t ¿

Total number of current , I t=L ( x )+ It (x )=I o exp (αd)

I t

I o

=exp (αd)

Ad=ln(I t /I o¿)¿

Thus, α=1d

× ln( I t / I o¿)¿

c. At a distance of 22.8 mm and pressure 200 mm Hg, the breakdown voltage of a

uniform field electrode in air is found to be 19.15 kV. Determine the breakdown

voltage if the secondary ionization coefficient ϒ is doubled. The values for the ratio

of electric field and pressure , E/p and ratio of first ionization coefficient , α/p are

given in Table Q4.

Table Q4

E/p (V/cm mm Hg) α/p ( ion pairs/cm mm Hg0

41 0.0196

42 0.0222

ANSWER:

d = 22.8mm = 2.28cm, p = 200 mmHg Vs = 19.15 kV

E/p = 42 V/cm mmHg α/p = 0.0222

E/p = 41 V/cm mmHg α/p = 0.0196

Find Vs when ϒ is doubled?

From secondary Townsend Breakdown process,

I=I o expαd

1−γ [exp (αd )−1 ]∧E=

V s

d

Breakdown criteria: 1−γ [exp (αd )−1 ]=0∨γ exp (αd )=1

E=V s

d=19.15

2.28=8.40kV /cm=8400V /cm

Ep=8400

200=42v /cmmmHg

From Table;

αp=0.0222 ,thus α=0.0222 (200 )=4.44

Thus αd=4.44 ×2.28=10.12

From breakdown criteria (ϒ is doubled , α α’)

Thus γ exp (αd )=2 γ exp (α ' d )=1

exp (αd)exp (α' d)

=2

(α−α' )d=ln2

α−α '= ln22.28

Thus, α '=4.14

A '

p= 4.14

200=0.02068

By interpolation;

Ep=41+0.02068−0.0196¿ ¿

(0.0222−0.0196)=41.42

Thus E=41.42 (200 )=8284v /cm

Thus V s=Ed=8284×2.28=18.89kV

QUESTION 26

Describe the secondary processes which can lead to on electron avalanche and how these

processes may be identified. Show that the discharge current in a multi avalanche.

Townsend process in a non-attaching gas is given by;

I=I o ead

1−γ [e(ad−1) ]

Where,

I o – initial current

α – first Townsend ionization coefficient

γ – second Townsend ionization coefficient

d – gap distance in cm

ANSWER

The electrical breakdown of a gas is brought about by various processes of ionization. These

are gas processes involving the collision of electron, ions and photons with gas molecules

and electrode processes which take place at or near the electrode surface. When a pair of

electrodes is immersed in a gas and a voltage applied across them the current-voltage

characteristics of figure below is observed.

At low voltage the observed current is due to collection of free charge carriers in the gap and

as the voltage is increased a level is reach at which the free electrons gain enough energy to

ionize. Electrons produced may cause further ionization so that an electron avalanche is

generated. Ionization is the process by which an electron is removed from an atom, leaving

the atom with a net positive charge. The probability of ionization due to the electrons will

depend on the number of collisions made per unit distance with coefficient α, α is referred as

the primary ionization coefficient which is the number of ionizing collisions per electron per

cm travel. With the primary ionization alone the discharged is not self-sustaining. If the

source of initial electrons is removed, the current I fall the zero. This suggests that processes

other than the simple α-process are occurring. The additional current is produced by

secondary emission processes. A secondary ionization coefficient, Ƴ is defined as the

number of secondary electrons produced at the cathode per electron produced in the gap.

These processes for secondary-electron liberation can be identified by;

i. Positive-ion, γᵢ – ions do not have enough energy to ionize gas molecules directly but

may release electrons on colliding with the cathode surface.

ii. Photon, γ p– a proportion of the collisions in the gap cause excitation of neutral-gas

molecules which on return to the ground state may emit photons which release

electrons by photoemission.

iii. Metastables, γm - metastable molecules may diffuse to the cathode and release

electrons.

One or more secondary mechanisms may exist giving a total secondary effect described by

γ=γi+γ p+γm

Let no = the number of initial electrons at cathode

no ' = the number secondaries

no = the total emission including secondaries.

i.e. no=no+no '

At x, n ( x )=no' ' exp (αx )

The total number of new electrons produced, n (d )=no ' ' [exp (αd−1 ) ].

If γ electrons are produced at the cathode per ionizing collision in the gap,

then no' =γ n0 ' ' [exp (αd−1 ) ].

Thus, no' '=no+γ no ' ' [exp (αd−1 ) ]

no ' '=no

1−γ [exp (αd−1 ) ]

∴n (d )=no' ' exp (αd )=

no exp (αd )1−γ [exp (αd−1 ) ]

Under steady state condition,

I=I o exp (αd )

1−γ [exp (αd−1 ) ]

QUESTION 27

Measurement of breakdown voltages in a uniform-field spark gap in air gave the

results as shown in Table below.

Gap spacing (mm) Pressure (Bar) Temperature (°C) Breakdown

Voltage, Vs (kV)

2.5 1.03 30 0.91

27 1.185 15 88.38

Using the expression derived from Paschen’s law, Vs=A (pd )+B (√ pd ), determine;

i. The relative air density ρ referred to standard atmospheric conditions of 1013-

25mbar and 20°C.

ii. The value of constants A and B

iii. The breakdown voltage of a 3 cm gap spacing at a pressure of 3000 mbar

and a temperature of 20°C.

ANSWER

i. d1=0.25cm, p=1034 mBar , t=30° c ,V s=0.90 w

ρ20=ρ

1013∙273+20273+t

¿ 10341013

∙293303

=0.99

d2=2.7 cm, p=1180 mBar , t=15 °c ,V s=88.56w

ρ21=ρ

1013∙273+20273+t

¿ 11801013

∙293288

=1.19

ii .V s 1=A ρ1 d1+B√ ρ1 d1

V s 2=A ρ2d2+B√ ρ2d2

ρ1 d1=0.99 (0.25 )=0.25

√ ρ1 d1=0.5

ρ2 d2=1.19 (2.7 )=3.27

√ ρ2 d2=1.79

Dari X:

0.91=A (0.25 )+B (0.5 )⟶1

X

88.38=A (3.21 )+B (1.79 )⟶2

[88.380.91 ]=[3.27 1.81

0.25 0.5 ][AB ] R1 ⟷ R2

A=[ 88.38 1.810.91 0.53.27 1.810.25 0.5

]=88.38 (0.5 )−1.81 (0.91 )3.27 (0.5 )−1.81 (0.25 )

¿ 44.19−1.651.64−0.45

=42.541.19

=35.75

B=|3.27 88.38

0.5 0.91 ||3.27 1.810.25 0.5 |

=3.27 (0.91 )−88.38 (0.25 )

1.19

¿ 2.98−22.101.19

=−16.07

iii. Given: d=3cm, ρ=3000mBar, t=20

ρ20=30001013

×293293

=2.96

ρd=2.96 (3 )=8.88

√ ρd=2.98

∴V s=35.75 (8.88 )−16.07 (2.98 )

¿317.46−47.89

¿269.57kV

QUESTION 28

Briefly explain the difference between High voltage and Extra High Voltage equipment in

terms of high impulse voltage testing requirements.

ANSWER:

The high voltage range is from about 50kV to about 300 kV. In this range the lightning over

voltage are much higher than that due to switching. Up to about 300kV experiments

indicates that the highest voltage stress arises from lightning. From transmission system

above 300kV the switching overvoltage increase in importance so that at about 500kV the

point has been reached where they are equipment to that of lightning overvoltage. Based on

above facts, impulse tents on high voltage equipment are usually specified for the case of

extra high voltage equipment, in addition to lightning wave shape (1.2/50μ) the switching

impulse (eg. 250/2500μs) wave shape are also specified.

QUESTION 29

The ionization coefficient of the electron α and its dependence upon the field strength E can

be described by the formula.

αp=Ae−B E

P

Where A and B are empirical constants and p is pressure. Using the criterion for Townsend

breakdown in a homogeneous field, and assuming Ƴ is constants, show that the breakdown

voltage Vs can be expressed as.

Vs=Bpd

¿( A . pdk )

Where d is the gap distance and k is a constant related to Ƴ.

ANSWER:

The criterion for breakdown is given by

γ αd≅ 1

eαd A≅ 1γ

Since γ is constant

eαd≅ k1

Where k 1 is constant

ed=ln k 1

ed=k

Where k is constant

α= kd

Into

αρ=A e−B

ρE

kρd=A e−B

ρE

Substituting E=V s

d

kρd=A e

−BρdV s

kAρd

=e−B

ρdV

s

¿¿

ln ¿¿

ln ¿¿

V s=Bρd

ln ¿¿¿

QUESTION 30

Paschen Law can be described by the equation.

Vs=Bpd

¿ (C . pd )

Where B and C are constant and p is gas density

The following data were obtained from experiment

T (º0) P (mbar) D (cm) Vs (kV)

25 1009 2 58.9

28 1015 3 88

Calculate the breakdown voltage for the following condition.

T=300 , p=1020mbar=1020m ,x

ANSWER:

Vs=Bpd

¿ (C . pd )

T 1=25 ºC T 2=28 ºC

P1=1009mbar P21=1015mbar

d=2cm d2=3cm

V s1=58.9 kV V s2=88.0kV

Substituting:

V s1=B ρ1d1

ln ¿¿¿

ρ1=ρ1

1013×

293(273+T )

¿ 10091013

×293

(273+25)=0.97934

S.T.P = 1013mbar = 760, T = 20ºC

ρ2=ρ2

1013×

293(273+T )

¿ 10151013

×293

(273+28)=0.975344

V 1 :58.9=β(0.979× 2)

ln (C ×0.97934× 2)V ¿kV , d∈cm

0.033254 β=ln 1.95868C …………………….. (1)

V 2 :88.0=β(0.975344×3)

ln (C ×0.975344 ×3)

0.022167 β=ln2.926032C ……………………. (2)

(1) – (2)

0.011087 β=ln1.95868C

2.926032 lnC

¿ ln 0.669398

β=−36.20244

Into (1)

0.033254 (−36.20244)=ln 1.95868C

exp (−1.203876)=1.95868C

C=−36.20244

V s=−36.20244 ρdln (0.15318 ρd )

;V ¿kV , d∈cm

QUESTION 31

a (i) Define the breakdown of electrical insulation

ANSWER:

Breakdown on electrical insulation is the maximum voltage that can be supported by an

insulating material.

(ii) List all mechanism of breakdown in solid and liquid insulation

ANSWER:

Intrisi sik

electrodynamic

heat

chemical and electrochemical

trace and tracking

internal-discharge

b) Explain the Townsend mechanism of breakdown in gas medium and prove that the

criterion of breakdown is given by (all symbols have their usual meaning): γ eαd=1

ANSWER:

Townsend mechanism of breakdown in gas medium used to describe the process in gas.

The mechanism of breakdown is blown at low pressure conditions with the small distance

between the electrodes. Voltage applied to the electrodes until breakdown happen. The

breakdown process begins with form of the first subsequent calculation of the coefficient and

with voltage change the secondary breakdown will be created, the breakdown process is

then occurs.

Current equations taking into account of the secondary generator.

c) Breakdown voltage measurement of a uniform field pressurized gas is shown in Table

Q4.

Table Q4

Distance (mm) 20.3 28.8

Pressure (mbar) 500 700

Temperature (°C) 20 25

Breakdown voltage (kV) 31.6 53

Assume the Paschen curve for ρd>5mm can be represented by the equation (all symbols

have their usual meanings)

V s=A( ρd)+β √( ρd)

Determine the breakdown voltage for the distance 15mm at a pressure of 250 mbar and

temperature 25°C.

ANSWER:

V s=A( ρd)+β √( ρd)

Given; d1=20.3mm ,P1=500mbar , t1=20 ,V 1=31.6 kV

ρ1=P1

1013(273+20 )

273+ t=0.4936

31.6=A ×0.4936+B√0.4936×20.3=10.02+B × 3.169 ………..(1)

d2=28.8mm ,P1=700mbar , t2=29 ,V 2=53 kV

ρ2=P2

1013(273+20 )273+29

=0.6794

s3=A × 0.6794× 28.8+B√0.6794× 28.8=19.57 A+4.424 B ………..(2)

Persamaan (1) X 19.57

618.412=196.09 A+61.94B ………..(3)

Persamaan (2) X 10.02

531.06=196.09 A+44.33 B ………..(4)

Persamaan (3) – persamaan (4)

837.352=17.61B)

B=4.96

Dari persamaan (1) dan

A=(31.6−15.7)

10.02=1.587

d3=15mm ,P1=250mbar , t2=25 ,

ρ3=P3

1013(273+20 )273+25

=0.2427

V 3=1.587× 0.2427×15+4.96√0.6794×1=15.24kV

QUESTION 32

a. Explain what is meant by partial discharge in high voltage engineering.

ANSWER:

Partial discharge (PD) is a localized dielectric breakdown of a small portion of a solid

or fluid electrical insulation system under high voltage stress, which does not bridge

the space between two conductors. While a corona discharge is usually revealed by

a relatively steady glow or brush discharge in air, partial discharges within solid

insulation system are not visible.

PD can occur in a gaseous, liquid or solid insulating medium. It often starts within

gas voids, such as voids in solid epoxy insulation or bubbles in transformer oil.

Protracted partial discharge can erode solid insulation and eventually lead to

breakdown of insulation.

b. What are the effects to high voltage apparatus if the partial discharge level is greater

than the maximum allowed.

ANSWER:

Insulation of the HV power equipment gradually degrades inside the insulator due to

cumulative effect of electrical, chemical and thermal stress. Due to the high voltage

stress the weak zone inside the insulator causes the partial discharge (PD) which is

known as local electrical breakdown. As a result the insulation properties of such

materials are enormously degrades its quality due to the PD.

Question 33

(a)

(i)

Write short discussion on the followings:

Paschen’sLaw and its significance in gas breakdown phenomenona;

ANSWER:

At very low pressures, deviations from the Paschen’s law are observed when the

breakdown mechanism is not influenced by the properties of gas but depends on the

purity and property of the electrodes.

(ii) Thermal breakdown of solid insulators;

ANSWER:

Conduction current flows – heats up the specimen and the temperature rises.

Heat generated transfers to the surrounding medium by conduction and radiation

Breakdown occurs when heat generated > heat dissipated.

Heat generated is proportional to the frequency – thermal breakdown is more serious at

high frequency.

Thermal breakdown stresses are lower under a.c. condition then d.c.

Thermal instability in solid dielectrics

CHAPTER 2

(QUESTION AND ANSWER)

SOLUTION TUTORIAL 2

ANSWER :

1. Describe briefly, with the aid of diagrams, equations and/or examples, where

appropriate, the avalanche process in the breakdown phenomenon of gaseous

dielectrics.

Answer:

The avalanche process is one of the processes which occurs in the breakdown of

gaseous dielectrics and is based on the generation of successive ionizing collisions

leading to an avalanche. Suppose a free electron exists (caused by some external

effect such as radio-activity or cosmic radiation) in a gas where an electric field exists.

If the field strength is sufficiently high, then it is likely to ionize a gas molecule by

simple collision resulting in 2 free electrons and a positive ion. These 2 electrons will

be able to cause further ionization by collision leading in general to 4 electrons and 3

positive ions. The process is cumulative, and the number of free electrons will go on

increasing as they continue to move under the action of the electric field. The swarm of

electrons and positive ions produced in this way is called an electron avalanche. In the

space of few millimetres, it may grow until it contains many millions of electrons.

2. Show that the breakdown criterion in gas according to Paschen’s Law is given

by:

g( V s

pds)e[

pd s . f ( V s

p ds)]−1=1

Where ds – gap distance at spark over voltage

p – pressure

Vs - spark over voltage

f & g - different functions

Answer:

By neglecting attachment, breakdown criterion is :

γ (ead−1 )=1 .....................(1)

Since Paschen’s Law, αp=f ( E

p) and γ=g( E

p )Where f & g signify different function.

At breakdown, α ds=pd . f ( Es

p )

= pds . f ( V s

pds) ............................(2)

and γ=g( E s

p )=γ=g( V s

pd s)...............(3)

Substitute (2) & (3) into (1) gives:

g( V s

pds)e[

pd s . f ( V s

p ds)]−1=1 proved

3. The following data are given for two parallel plates while the electric field stress,

E is kept constant.

i) I = 1.2 Io when d = 0.5 cm

ii) I = 1.6 Io when d = 1.3 cm

iii) I = 2.3 Io when d = 2.0 cm

where Io is the initial current and d is the distance between the plates. Find the

values of the Townsend Primary and Secondary coefficients, α and γ.

Answer:

Using the equation I=I o ead

For d = 0.5cm,

1.2 I o=I o e0.5 a or

α 1 d1=ln 1.2=0.182

α 1=0.36 /cm

For d = 1.3cm,

1.6 I o=I o e1.3a or

α 2 d2=ln 1.6=0.47

α 2=0.36 /cm

For d = 2.0cm,

2.3 I o=I o e2a or

α 3 d3=ln 2.3=0.83

α 3=0.42/cm α3

(this suggests that for this gap γ starts to be active)

The value of γ can be found from the equation:

II o

= eαd

1−γ (eαd−1 )

2.3= e (2 )(0.42)

1−γ (e0.84−1 )= 2.316

1−1.316 γ

γ=0.0528/c

4. The build-up of high currents in a breakdown is due to the process of

ionization in which electrons and ions are created from neutral atoms or

molecules. Explain how the ionization process occurs prior to gas breakdown

phenomena.

Answer:

When a high voltage is applied between two electrodes immersed in a gaseous

medium, the gas becomes a conductor and an electrical breakdown occurs. The

process that is responsible for the breakdown of a gas is called ionization. This

process initially liberates an electron from a gas molecule with the simultaneous

production of a positive ion.

The generations of new electrons are from ionization by collision, photo-ionization and

the secondary ionization process. Under high voltage stress, a few of the

electrons produced at the cathode due to the certain process will produce positive ions

and additional electrons. The process repeats itself and hence increases in the

electron current.

5. The ionization coefficient αp

as a function of field strength E and gas pressure p

is given by the following threshold equations :

αp=f ( E

p)

By using the Townsend’s breakdown criterion, show that the breakdown voltage

for uniform field gaps is a function of gap length (d) and gas pressure (p).

Answer:

Given that αp=f ( E

p) ........(i)

The Townsend’s breakdown criterion :

γ (eαd−1 )=1 ................(ii)

Substitute (i) into (ii) ;

ef (Ep ). pd

=1γ+1 ............(iii)

Taking in on both sides of (iii);

f (Ep ) pd=ln [ 1γ +1]=K .......(iv)

For uniform field E =V b

d; therefore f (V b

pd ) pd=K ......(v)

Or f (V b

pd )= Kpd

∴ V b=fpd .................(vi)

Equation (vi) shows that the breakdown voltage of a uniform field gap is a unique

function of the product of gas pressure and the gap length for a particular gas and

electrode material. This relation is known as Pashen’s Law.

6. Figure Q2 shows the experiment set-up for studying the Townsend discharge.

The experiment is conducted by measuring the current I at the different gap

distance,d. Table Q2 gives the set of observation when studying the conduction

and breakdown in a gas.

Gap

distance

1 2 3 4 5 6 8 10 12 14 16

d (mm)

Current I

(pA)19 21 26 32 40 45 80 106 152 255 430

Table Q2: Townsend experimental data

i) Determine the initial current, Io.

ii) Calculate the values of the Townsend’s primary and secondary

ionization coefficients.

Answer:

Gap

distance

d (mm)

1 2 3 4 5 6 8 10 12 14 16

Current

I (pA)19 21 26 32 40 45 80 106 152 255 430

ln I 2.94 3.04 3.26 3.47 3.69 3.81 4.38 4.66 5.02 5.54 6.06

From equation I=I o ead ........(1)

Taking ln on both sides of (1);

lnI=ln eαd+ ln I o I

lnI=ad+ln I o => y=mx+c

i) Plot the graph (ln I) Vs (d);

∴From the graph, interception (c) at ln I axis gives:

lnIo=≈ 2.7

I o ≈14.88 pA .

ii) Townsend’s primary ionization coefficient, α :

Gradient of the graph (m) shows the value of α;

∴α ≈ 1.5/8≈ 0.188/mm.

Townsend’s secondary ionization coefficient, γ :

I=I oe

αd

1−γ (eαd−1 )

Substitute for higher value for d;

430= 14.88 e (0.188 )(16 )

1−γ (e (0.188 ) (16)−1 )= 301.27

1−19.25 γ

∴ γ=0.016/mm

0 2 4 6 8 10 12 14 160

1

2

3

4

5

6

d (mm)

ln I

7. In an experiment using a certain gas, it was found that a steady current of 600μA

flowed through the plane electrode separated by a distance of 0.5cm when a

voltage of 10kV is applied. Determine the Townsend’s first ionization coefficient

if a current of 60μA flows when the distance of separation is reduced to 0.1cm

and the field is kept constant at the previous value. If the breakdown occurred

when the gap distance was increased to 0.9cm, what is the value of Townsend’s

secondary ionization coefficient?

Answer:

Since the field is kept constant (i.e. if distance of separation is reduced, the voltage

is also reduced by the same ratio so that V/d is kept constant).

I=I o eax

Substituting two different sets of value ;

600=I oe0.5a and 60=I oe

0.1a

60060=

I o e0.5a

I o e0.1a

10=e0.4a

∴α=5.76 ionizing collisions /cm

The breakdown criterion is given by;

1−γ (ead−1 )=0

Therefore the Townsend’s secondary ionization for the breakdown to

be occurred at gap distance of 0.9cm is :

γ (ead−1 )=1

γ= 1

(eax−1)= 1

(e (5.76 )(0.9)−1)= 1

177.4=5.64 x 10−3cm−1

8. In SF6 gas , the effective ionization coefficient is given by ;

αp=27.7( E

p )−2460

where α is the effective coefficient in cm-1, E is the electric field strength in

kV/cm and p is the pressure (referred to 20 )in bar. Breakdown maybe

predicted using streamer criterion∫0

d

α .dx=18, where d is the length of the

electrodes gap in cm. Estimate the length of a uniform-field gap that will just

hold off a steady voltage of 100kV in SF6 at 4 bar and 60.

Answer:

Given that ;

αp=27.7( E

p )−2460

α=27.7E-2460 p

αd=27.7 Ed−2460 p

¿27.7V−2460 pd

pd=27.7V – αd2460

.......(1)

Given ∫0

d

α .dx=18

∴αd=18 ..................(2)

The normalised pressure of 4 Bar at 60 is:

Pn=40001013

.293

273+60=3.47 ¿

From (1) and (2)

d=27.2V−αd2460Pn

=27.7 (100 )−18

2460(3.47)=0.32cm

9. Show that in the process of gas breakdown, the Townsend First Ionization

coefficient, α is given by ;

α=1d

ln( I t

I 0)

where d – gap distance

I t - total current

I 0 - initial current

Answer:

Total no of electrons at anode, n(d) =noead

At steady state, average current in gap at distance x,

I ₋(x)=I o eax

I ₊ ( x )=I o ¿¿]

Total number of current, I t=I−¿ ( x )+ I₊ ( x )=Io ead ¿

αd=ln ( I t

I 0)

α=1d

ln( I t

I 0) ........ Proved

10. The following data in Table Q2b are given for two parallel plates while the

electric field, E is kept constant.

Gap distance, d(cm) Ratio of current and initial current, I/Io

0.5 1.2

1.3 1.6

2.0 2.3

Table Q2b

Find values of α and γ

Answer:

Using the equation

I=I o ead

For d = 0.5cm,

1.2 I o=I o e0.5 a

α 1 d1=ln 1.2=0.182

α 1=0.36 /cm

For d = 1.3cm,

1.6 I o=I 0 e1.3 a∨α 2α 2=ln1.6=0.47

α 2=0.36 /cm

For d = 2.0cm,

2.3 I o=I o e2a∨α3 d3=ln 2.3=0.83

α 3=0.42/cm (This suggests that for this gap γ starts to be active)

The value of γ can be found from the equation:

II o

= eαd

1−γ (eαd−1 )

2.3= e (2 )(0.36)

1−γ (e0.72−1 )= 2.0544

1−1.0544 γ

γ=0.101/cm

11. At a distance of 22.3mm and pressure 200mm Hg, the breakdown voltage of a

uniform field electrode in air is found to be 19.15kV. Determine the breakdown

voltage if the secondary ionization coefficient γ is doubled. The values for the

ratio of electric field and pressure E/p and ratio of first ionization coefficient and

pressure, α/p are given in Table Q2c.

E/p (V/cm mm Hg) α/p (ion pairs/cm mm Hg)

41 0.0196

42 0.0222

Table Q2c

Answer:

Given that ;

d =22.8 mm =2.28 cm, p=200mm Hg Vs = 19.15kV

E/p = 42 V/cm mm Hg, α/p = 0.0222

E/p = 41 V/cm mm Hg, α/p = 0.0196

From secondary Townsend Breakdown Process,

I=I oe

αd

1−γ (eαd−1 )∧E=V s/d

Breakdown criterion 1−γ (eαd−1 )=0∨γ (eαd )=1

E=V s

d=19.15km

2.28cm=8.40kV /cm=8400V /cm

Ep=8400

200=42V /cm mm Hg

From Table;

αp=0.0222

∴α=0.0222 (200 )=4.44

∴αd=4.44×2.28=10.12

From breakdown criterion (γ is doubled, α→α '),

∴ γ (eαd )=2 γ (eα ' d )=1

eαd

eα' d=2

(α−α' )d=ln2

α−α '= ln22.28

∴α '=4.14

α '

p=4.14

200=0.02068

By interpolation;

Ep=41+

(0.02068−0.0196)(0.0222−0.0196)

=41.42

∴E=41.42 (200 )=8284V /cm

∴V s=Ed=8284× 2.28=18.89kV

12. Discuss with suitable diagrams the mechanisms which lead to breakdown in

liquid insulation.

Answer:

Suspended Particle Mechanism

1. Impurities present as fibers or dispersed solid particles;

2. Electrostatic force acting on impurities;

3. Solid impurities –force directed towards maximum stress;

4. Gas impurities –force directed towards areas of low stress;

5. Form a stable chain bridging the gap.

Cavitation and Bubble Mechanism

1. Gas pockets at the surfaces of the electrodes;

2. Electrostatic repulsive forces between space charges which may be sufficient to

overcome the surface tension;

3. Gaseous products due to the dissociation of liquid molecules by electron collisions;

4. Vaporization of the liquid by corona type discharge from sharp points and

irregularities on the electrode surfaces.

Thermal Mechanism

1. High density current pulses give rise to localized heating – lead to the formation of

vapour bubbles;

2. Elongation to a critical size or completely bridges the gap from the formed bubbles;

3. Breakdown strength depends on pressure and molecular structure.

Stressed Oil Volume Mechanism

1. Breakdown strength is determined by largest possible impurity or weak link;

2. Breakdown strength is inversely proportional to the stressed oil volume;

3. Breakdown voltage influenced by gas content in the oil, viscosity and the

presence of impurities.

13. Show that the breakdown criterion in gas according to Paschen’s Law is given

by,

g (V s / pds ) e pdsf (vs / pds )−1=1

Where,

ds – Gap distance at spark over voltage

p – Pressure

Vs – Spark over voltage

F & g – different functions

Answer:

Since

γ (ead−1 )=1−−−−−(1)

Based on Paschen’s law:

ap=f (E/ p )

γ=g (E / p )

At breakdown,

ad s=pd f ( Es/ p )

ad s=pds f (V s/ pds )−−−−−(2)

And

γ=g (E / p )=g (V s/ pds )−−−−−(3)

Substitute (1) & (2) into (3)

Then,

g (V s / pds ) e pdsf (vs / pds )−1=1−−−−−(Proved )

14. Breakdown voltage measurements of a uniform-field gap in air at 293K gave the

following results shown in Table Q2.

pd (bar-cm) E/p at breakdown (kV bar-1cm-1)

1.0 30.30

9.0 26.00

Table Q2

Determine the breakdown voltage of a 20mm gap at a pressure of 3 bars and

temperature of 300K.

Answer:

V s=A pd+B ¿

ED=A pd+A B ¿

E=Ap+B( pd )1/2

∴E=Ap+B (pd )1/2

From the data given;

30.3=A+B /(112)

30.3=A+B−−−−−(1)

26.0=A+B 912

26.0=A+0.33 B−−−−−(2)

From (1) and (2); A = 23.88 and B = 6.42

For the case of atmospheric air;

V s=23.88 (pld )+6.42 pld12

p = 3 Bar, t = 300 K, d = 2cm, V s=?

Corrected pressure to standard temperature of 20ºC,

P=3000 (293 )1013 (300 )

=2.89

∴V s=(23.88 ) (2 ) (2.89 )+6.42 (2.89× 2 )12=153.45kV

15. Discuss the processes that lead to ion-generation in a gas breakdown.

Answer:

Types of processes:

Ionization by collision;

Photo-ionization;

Secondary ionization;

Electron attachment process.

Ionization by collision

The process of liberating an electron from a gas molecule with the

simultaneous production of positive ion.

A free electron collides with a neutral gas molecule and generates new

electron and positive ion.

e−¿+A ε>V i

→

e−¿+ A+ ¿+e−¿¿¿ ¿¿

Photo-ionization

The process of ionization by radiation.

Radiation process; excitation of the atom to a higher energy state, continuous

absorption by direct excitation of the atom, dissociation of diatomic molecule.

hv+A↔

A¿

Secondary ionization

Electron emission due to positive ion impact: a positive ion approaching

cathode can cause emission of electrons from the cathode.

Electron emission due to photons: electron can escape from a metal if there is

enough energy to overcome the surface potential barrier.

Electron emission due to metastable and neutral atoms: electrons can be

ejected from the metal surface by the impact of excited (metastable) atoms.

Electron attachment process

Electrons become attached to atoms or molecules to form negative ions.

Atoms/molecules have vacancies in their outermost shells, therefore have an

affinity for electrons.

Play a very important role in the removal of free electrons from the ionized

gas.

16. Prove that the breakdown criterion in gas according to Townsend’s equation is

given by:

I=Ioexp (αd)

1−ɣ [exp (αd ) ]−1

Where,

α – Townsend’s Primary coefficient

ɣ - Townsend’s Secondary coefficient

ds – Gap distance at sparkover voltage

Answer:

Number of electrons reaching the anode;

nd=no exp(αd)

Average current = the number of electrons travelling/sec

I=I o exp (αd)

α = average number of ionizing collisions made by an electron per cm travel in

the direction of the field. (is called Townsend’s first ionization coefficient)

I o = initial current at the cathode

Let :

n0 = electrons emitted from the cathode

no׳ = number of secondary electrons produced due to secondary (y) processes.

no = total number of electrons leaving the cathode.

Then

no= n rsub o + n rsub o

The total number of electrons n reaching the anode becomes,

n=no exp (αd) = ( n rsub o + n rsub o ׳ ) exp (αd

And,

no γ=׳ [n−(no+no [(׳

Eliminating no ׳

n=no exp (αd) over 1- ɣ left [exp left (αd right ) right ] -1¿ Or

I=Ioexp (αd)

1−ɣ [exp (αd ) ]−1

17. In an experiment to determine the breakdown properties of air, the uniform field

electrode is used. The breakdown process occurs in accordance with Townsend

First and Second Ionization Coefficients, and . At a distance of 22.8mm and

pressure 200mm Hg, the breakdown voltage is found to be 19.5Kv . Determine

the breakdown voltage if the secondary ionization coefficient is doubled.

Data’s for the ratio of electric field & pressure, E/p and ratio of first ionization

coefficient, /p are given in Table Q2.

E/p (V/cm mm Hg) E/p (ion pairs/cm mm Hg)

41 0.019642 0.0222

Table Q2.

d = 22.8mm = 2.28cm,

p = 200mm Hg

V s = 19.15kV

E/p = 42 V /cm mm Hg, /p = 0.0222

E/p = 41 V/cm mm Hg, /p = 0.0196

Find V s when is doubled?

Answer :

From secondary Townsend Breakdown Process,

I=IO exp(αd) and E=V s

I−γ exp(¿ αd)¿

Breakdown criteria: I−γ (exp (αd )−1 )=0 or γ exp (αd )=I

E = V s/d = 19.15/2.28 = 8.40kV/cm = 8400V/cm

E/p = 8400/200 = 42V/cm mm Hg

From Table;

/p = 0.0222 = 0.0222 (200) = 4.44

d = 4.44 2.28 = 10.12

From Breakdown Criteria ( is doubled, ),

γ exp (ad )=2 γ exp (a' d )=I

exp (αd )=2

exp (α' d )=ln2

α−α '=ln2

¿2.28

∴α '=4.14 , α ' / p=4.14 /200=0.02068

By interpolation;

41 E/p 42 0

-0.0042000000000001

0.0154

0.0350000000000001

E / p=¿41+(0.02068−0.0196 )=41.42¿

∴E=41.42 (200 )=8284V /cm

∴V s=Ed=8284 X 2.28=18.89kV .

18. Briefly describe the definition of High Voltage and classification of voltage levels

Answer:

IEC 1970 – A high voltage being greater than 1000 V for alternating and greater than 1200 V for direct.

Voltage Classification

Low Voltage (LV) : 220V to 1000V (Europe) or 120V to 660V (US) Medium Voltage (MV) : 5kV – 66 kV (Europe) or 2.4kV – 69kV (US) High Voltage (HV) : 110kV – 220kV (Europe) or 115kV – 230kV (US) Extra High Voltage (EHV) : 275kV – 800kV (Europe) or 287kV – 765kV (US) Ultra High Voltage (UHV) : above 1000Kv

19. Show that in the process of gas breakdown, the Townsend First Ionization

Coefficient, α is given by;

α=1/d ln 〖(¿ /Io )〗

Where;

d=¿ Gap distance

¿=¿ Total current

Io=¿ Initial current

Answer;

Total no. of electrons at anode, n(d)=n₀e(αd)

At steady stage, average current in gap at distance x,

I( x)=I 0e(αx)∧I+(x)=I 0[e(αd )– e(αx)]

Total number of current, ¿=I(x)+ I+(x )=I 0e(αd)

¿ /I 0=e(αd)

αd=ln (¿/ I 0)

→α=1/d ln(¿ /I 0)

20. Discuss three (3) mechanism of solid insulation breakdown

1) Intrinsic Breakdown.

- When a high voltage is applied for a long

- Electric strength is determined by the strength intrinsic

- Depending on the presence of free electrons that move through

the solid lattice

- Electron current flow produces.

2) Electromechanically Breakdown.

- Less rigid insulation materials (rubber, pvc).

- Electrostatic force acting exceeds the mechanical strength of solid.

- Mechanical damage will occur.

3) Heat Breakdown.

- Leakage current of electricity to flow when stress imposed.

- Solid temperature increases through the heating process.

- The heat is transferred to the surrounding insulation through the

conduction and radiation processes.

- Restrictor to a maximum thickness of a solid.

4) Chemical and Electrochemical Breakdown.

- Chemical changes by continuous electrical stress through the

reaction with air and gas.

- Reactions that occur-oxidation, hydrolysis, chemical reaction.

- Could minimized by carefully inspect materials.

5) Treeing and Tracking Breakdown.

- Two effects will occur if the old electrical stress:

Presence across the surface of the conductor paths.

Fireworks effect during the tracking process in the form of

branches called treeing.

- Distribution channels fireworks during the tracking process in the

form of branches called treeing.

6) Breakdown by internal discharge.

- Cavity containing air in solid form.

- Field in the cavity is larger than the field in the insulation.

- Breakdown exists in the cavity.

21. Show that the breakdown criterion in gas according to Paschen’s Law is given

by:

g(Vs / pds)exp[ pds . f (Vs / pds)] –1 =1

Where,

ds=¿ Gap distance at spark over voltage

p=¿ Pressure

Vs=¿ Spark over voltage

f∧g=¿Different functions

Answer:

By neglecting attachment, breakdown criterion is,

γ (eαd – 1)=1−−−−−−−−−−−−−−−−−−−−−−−−−−−−−(1)

Since (Paschen’s Law), α / p=f (E / p)∧γ=g (E/ p)

Where f∧gsignify different function.

At breakdown,

αds=pd f (Es/ p)

¿ pds f (Vs/ pds)−−−−−−−−−−−−−−(2)

And

γ=g (Es/ p)=g(Vs / pds)−−−−−−−−−−−−(3)

Substitute (2) and (3) into (1) gives;

G(Vs / pds)exp¿ PROVED!

22. In nitrogen gas the static breakdown voltage Vs of a uniform field gap may be

expressed as,

Vs=A pd+B √ pd

where A and B are constants, p is the gas pressure in torr referred to a

temperature of 20°C and d is the gap length in cm. A 1 cm uniform field gap is

nitrogen at 760 torr and 25°C is found to breakdown at a voltage of 33.3kV. The

pressure is then reduced and after a period of stabilization, the temperature and

pressure are measured as 30°C and 5oo torr respectively. The breakdown

voltage is found to be reduced to 21.9 kV. If the pressure is further reduced to

350 torr while the temperature if the closed vessel is raised to 60°C and the gap

distance is increased to 2 cm, determine the breakdown voltage.

Answer:

Vs1=Ap 1d1+B √ (p1d 1)

p1=760 torr ,

t1 = 25°C,

d1 = 1cm,

Vs1=33.3kV

Corrected pressure to standard temperature of 20°C,

p1=760(273+20)

273+25=747.25 torr ,√ p1d 1=27.34

Hence;

33.33=A (747.25)+B (27.34)(1)

¿747.25 A+27.34 B−−−−−−−−−−−−−(1)

Vs2=Ap 2d2+B √ ( p2d 2)

p2=500 torr ,

t 2=30 °C,

d 2=2cm ,

Vs2=21.9 kV

Corrected pressure to standard temperature of 20°C,

p2=500(273+20)

273+30=483.50torr ,√ p2d 2=21.99

Hence;

21.9=A (483.50)+B(21.99)(1)

¿483.50 A+21.99B−−−−−−−−−−−−−−−(2)

By solving the simultaneous equation of (1) and (2), thus, we get:

A=0.042

B=0.072

Vs3=Ap 3d3+B √ ¿

p3=350 torr ,

t 3=60 °C ,

d 3=2cm ,

Vs3=?

Corrected pressure to standard temperature of 20°C,

p3=350(273+20)

273+60=307.96 torr ,√ p3d 3=24.82

Hence,

Vs3= (307.96 ) (2 ) (0.042 )+ (24.82 ) (0.072 )

¿27.66kV .

23. Discuss with suitable diagrams the mechanisms which lead to breakdown in

vacuum insulation.

Answer:

Mechanism of vacuum breakdown

Particle exchange

A charge particle emits from one electrode – impinges on other electrode –

liberates oppositely charged particles.

Involves electrons, positive ions, photons and the absorbed gas; cumulative

process.

Field emission

Electrons produced at small projections of the cathode;

bombard the anode – causing local rise in temperature – release gases vapours;

Electrons ionize atom of the gas;

Produce secondary electrons.

Field emission

Existence of the pre breakdown current at the sharp point of the cathode surface.

Current cause’s resistive heating at the tip of sharp point – tip melts and

explodes – initiates vacuum discharge

24. In an experiment to determine the breakdown voltage in sulphur hexafluoride

(SF6) gas, a non-uniform field electrode is used. At 1 cm gap distance and

pressure of 700 torr, the breakdown voltage is found to be 35kV. Determine the

breakdown voltage if the gap distance is doubled and the Second Townsend

Coefficient; ɣ is halved at a pressure of 1500 torr. The First Townsend

Coefficient, α is given by the equation:

α / p=6.5 exp [−250(E/ p)−1]cm−1 torr−1

Where, E – electric field strength in V /cm

p – Pressure in torr

Answer:

Given,

d=1cm , p=700 torr ,V b /d=35 KV

α / p=6.5 exp [−250(E/ p)−1]cm−1 torr−1

Ep=V s

pd

¿ 35x 10 3700x 1

¿50v /cm

α / p=6.5 exp [−250/50]cm−1 torr−1

¿6.5exp(5.0)

ɣe ad=1

αd=(a/ p)x ( pd)

¿6.5 exp (5.0 ) x700

¿30.66

ɣ=1/exp (αd)

¿1/exp(30.66)

¿ 12.07 x10 13

¿4.83 x 10−14

When p=1500torr , d 2=2d1=2cm, ɣ 2=ɣ /2

pd=1500 x2

¿3000 torr

ɣ2=4.83 x10−14

2

¿2.42 x10−14

For breakdown criteria,

exp (αd)=1/ ɣ=1/ (2.42x 10−14)=4.13 x10 13

αd=ln (4.13x 10 13 )

¿31.35

αp=αd

pd

¿ 31.353000

¿0.0105

αp=6.5exp[−250( E

p )−1]

¿0.0105

exp [−250( Ep )−1]

¿

αp

6.5

¿ 0.01056.5

¿619.05

Ep= 250

ln 619.058.88V /cm torr

¿ E / p=V s /( pd)

V s=(Ep ) x ( pd )

¿38.88 x3000

¿116.64 KV

25. The ionization coefficient of the electrons α and its dependence upon the field

strength E can be described by the formula.

α / p=A e−BEP

Where A & B are empirical constants and p is pressure. Using the criterion for

Townsend breakdown in a homogeneous field, and assuming γ is constant,

show that the breakdown voltage Vs can be expressed as

V s=Bpd

ln(A.pdk )

Where d is the gap distance and k is a constant related to γ.

Answer:

The criterion for breakdown is given by:

γ eαd=1

eαd=1/γ

Since γ is constant ,

eαd=k 1

Where k1 is aconstant .

∴αd=ln k1

αd=k

Where k is constant .

∴α=k /d…………………………… (1)

Substitute (1)intoα / p=A e−BEP

kpd=A e−BE

P

…………………………… ..(2)

substitute E=V s /d into (2) ,

k / pd=A e-BVspd

k / pd A=e-BVspd

ln (k / pd A)=-BpdV s

V s=Bpd

ln(A.pdk ) ……………………………………… (Prove!)

26. Paschen Law can be describe by the equation

V s=Bpdln (C . pd )

Where B and C are constant and p is gas density. The following data were

obtained from experiments:

T (°0) P (mbar) D (cm) Vs (kV)

25 1009 2 58.9

28 1015 3 88.0

Calculate the breakdown voltage for the following conditions:

T = 300°C, p = 1020 mbar, d= 2 cm

Answer:

Paschen Law:

V s=Bpdln (C . pd )

Given,

T 1=25 °C T 2=28 °C

p1=1009mbar p2=1015mbar

d 1=2cm d 2=3cm

V s1=58.9kV V s 2=88.0kV

Substituting,

V s1=bp1d1 / ln(Cp 1d 1)

ρ 1=p1/1013 x293 /(273+T )

¿1009/1013 x293 /(273+25)

¿0.97934

ρ 2=1015 /1013 x 293/(273+28)

¿0.975344

58.9=B(0.97934× 2)

ln (C × 0.975344×3)

0.033254 B=ln 1.95868C…………………………………. (1)

88.0=B(0.97934× 3)

ln (C × 0.975344× 3)

0.022167 B=ln 2.926032C ……………………………….. (2)

(1) – (2);

0.011087 B= ln1.95868Cln 2.926032C

¿ ln 0.669298

∴B=−36.2024

Into(1);

0.033254 (−36.20244)=ln 1.95868C

exp (−1.203876)=1.95868C

∴C=0.15318

∴V s=−36.20244 ρ d / ln(0.15318 ρd)

¿−36.20244 x1020 x 2/ ln (0.15318 x 1020x 2)

¿12.856 kV

27. Describe the secondary processes which can lead to an electron avalanche and

how these processes may be identified. Show that the discharge current in a

multi avalanche Townsend process is a non-attaching gas is given by:

l=I o eαd

1 - γ [ e(αd-1) ]

Where,

I o– initial current

α - first Townsend ionization coefficient

γ - second Townsend ionization coefficient

d – gap distance in cm

Answer:

The electrical breakdown of gas is brought about by various processes of ionization.

These are gas processes involving the collision of electrons, ions and photons with

gas molecules and electrode processes which take place at or near the electrode

surface. When a pair of electrodes is immersed in a gas and a voltage applied across

them the current voltage characteristics is shown on figure below.

I self-sustained

Non self-sustained breakdown

Io

Ionization Vs Charge collector

At low voltage the observed current is due to collection of free charge carriers in the

gap and as the voltage is increased a level is reach ay which the free electrons gain

enough energy to ionize. Electrons produced may cause further ionization so that an

electron avalanche is generated. Ionization is the process by which an electron is

removed from an atom, leaving the atom with a net positive charge. The probability of

the ionization due to electron will depends on the number of collision made per unit

distance coefficient α, α referred as the primary ionization coefficient which is the

number of ionizing collisions per electron per cm travel. With the primary ionization

alone the discharged is not self-sustaining. If the source of initial electrons is removed,

the current I fall to zero. This suggest that processes other than the simple α-process

are occurring. The additional current is produced by secondary emission process. A

secondary ionization coefficient, γ is defined as the number of secondary electrons

produced at the cathode per electron produced in the gap.

These processes for secondary electron liberation can be identified by;

i) Positive ion γi – ions do not have enough energy to ionize gas molecules

directly but may release electrons on colliding with the cathode surface.

ii) Photon γp – a proportion of the collisions in the gap cause excitation of

neutral gas molecules which on return to the ground state may emit

photons which release electrons bt photoemission.

iii) Metastables,γm– metastables molecules may diffuse to the cathode and

Release electrons’

One or more secondary mechanisms may exist giving a total secondary effect

described by, γ=γ i + γ p+ γm

Let,n0=the numberof initial electronsat cathode

n0 '=the numberof secondary

n0= the total emission including secondary ’ s

i.e.n0= n rsub 0 '+ n rsub 0

At x, n0 ( x )=n0 e ^ ( αx )

The total number of new electrons produced, n (d )=n0 [ e ^ ( αd - 1)

If γ electrons are produced at the cathode per ionizing collision in the gap,

then

n0 '=γ n0 ( e ^ left (αd - 1 right )

Thus, n0= n rsub 0 +γ n rsub 0 [e (αd−1 ) ]

n= n 0 over 1−γ [ e ^ ( αd - 1) ]

∴n (d )=n0 e ^ left (αd right ) = n 0 e ^ ( αd ) over 1−γ [ e ^ ( αd - 1) ]

Under steady state conditions, I=I o eαd

1 - γ [ e(αd-1 )]

28. Measurement of breakdown voltages in a uniform field spark gap in air gave the

result as shown in table Q4.

- +

N0

V Volts

d

d Distance

Gas spacing (mm) Pressure (Bar) Temperature (°C)Breakdown voltage,

Vs (kV)

2.5 1.03 30 0.91

27 1.18 15 88.38

Table Q2c

Using the expression derived from Paschen’sVs=A (pd )+B ¿ lawdetermine;

I. The relative air density ρ referred to standard atmospheric conditions of

1013.25 mbar and 20°C.

II. The value of A and B

III. The breakdown voltage of a 3 cm gap spacing at a pressure of 3000 mbar and a

temperature of 20°C.

Answer:

i) d1 = 0.25 cm, p = 1030 mbar, t = 30 °C, Vs = 0.91kV

ρ20=p

1013×(273+20)(273+t )

¿ 10301013

×(273+20)(273+30)

¿0.99

d 2=2.7 cm, p=1180 mbar , t=15 °C ,V s=88.38 kV

ρ20=11801013

×(273+20)(273+15)

¿1.19

ii) V s1=A ρ 1d1+B ¿………………………………………………(1)

V s 2=A ρ 2d2+B(√ ρ2 d2)……………………………………………… (2)

ρ1 d1=0.99(0.25)=0.25

√ ρ1 d1¿¿=√0.25

¿0.5

ρ 2d2=1.19(2.7)=3.27

√ ρ2 d2=√3.27=1.79

From equation (1) and (2);

0.91=A (0.25)+B(0.5)………………………………………………….(3)

88.38=A (3.27)+B(1.79)………………………………………………(4)

From (3) and (4);

[88.380.91 ]=[3.27 1.79

0.25 0.5 ][ AB ]

A=|[88.38 1.79

0.91 0.5 ]||3.27 1.790.25 0.5 |

=88.38 (0.5 )−1.79 (0.91)3.27 (0.5 )−1.79(0.25)

=35.75

B=|[3.27 88.38

0.25 0.91 ]||3.27 1.790.25 0.5 |

3.27 (0.91 )−88.38 (0.25)3.27 (0.5 )−1.79 (0.25)

=−16.07

iii) d = 3 cm, p = 3000mbar, t = 20°C,

ρ20=p

1013×(273+20)(273+t )

¿ 30001013

×(273+20)(273+20)

¿2.96

ρd=2.96 (3 )

¿8.88

√ ρd=√8.88

¿2.98

∴V s=35.75(8.88)−16.07 (2.98)=269.57k

29. Describe the secondary process which can follow an electron avalanches and how

these processes may be identified. Show that the discharge current in a multi

avalanche Townsend process in non-attaching gas is given by:

I=Ioexp (αd)

1−ɣ [exp (αd ) ]−1

Answer:

When the d.c. voltage is applied and when the voltage is low, the current pulses start

appearing due to electrons and positive ions as shown in Figs.(a) and (b). These

records are obtained when the current is measured using a cathode ray oscillograph.

When the applied voltage is increased, the pulses disappear and an average dc.

current is obtained as shown in Fig.(c) In the initial portion (To), the current increases

slowly but unsteadily with the voltage applied. In the regions TI and T2, the current

increases steadily due to the Townsend mechanism. Beyond T2 the current rises

very sharply, and a spark occurs.

Fig. Current as a function of time

(a) When secondary electrons are produced at the cathode by positive ions.

(b) When secondary electrons are produced by photons at the cathode. ideal,

actual. I(t) is the total current and I+and are electron ion currents. τ-andτ+are the

electron and ion transit times.

Fig. (c) Typical current growth curve in a townsend discharge

Number of electrons reaching the anode;

nd=n0e(αd)

Average current = the number of electrons travelling/sec

I=I oe(αd )

α : average number of ionizing collisions made by an electron per cm travel in the

direction of the field. (is called Townsend’s first ionization coefficient)

I o = initial current at the cathode

Let :

n0: electrons emitted from the cathode

n0’ = number of secondary electrons produced due to secondary (y)

processes.

n0 = total number of electrons leaving the cathode.

Then,

n0= n rsub 0 ' + n rsub 0׳

The total number of electrons n reaching the anode becomes,

n=n0 e ^ ( αd ) = ( n rsub 0 + n rsub 0 ׳ ) e ^ ( αd ) ¿

And

n0 Y=׳ ¿

Eliminating n0 ׳

n=n0 e ^ ( αd ) over 1 - γ left [e ^ ( αd ) right ] - 1 ¿ Or

I= Ioe(αd )

1−γ [e(αd )]−1

30. What is meant by ‘time lag to breakdown’ and describe how it may be influenced

and exploited.

Answer:

Time lags for breakdown:

In practical, the breakdown due to the rapidly changing voltages or impulse voltages,

there is a time difference between the application of a voltage sufficient to cause

breakdown and the occurrence of breakdown itself time lag.

t :time lag.

ts: The time during the voltage applications until a primary electron appears to initiate

the discharge and is known as the statistical time lag.

tf: The time required for the breakdown to develop once initiated and is known as the

formative timelag.

31. Describe with diagrams the principle breakdown mechanisms which can occur

in solid dielectrics and identify their order of occurrence on a stress-time

diagram.

Answer:

Breakdown In Solid Dielectrics

Solid insulating materials are used almost in all electrical equipments, be it an electric

heater or a 500MW generator or a circuit breaker, solid insulation forms an integral

part of all electrical equipments especially when the operating voltages are high. The

solid insulation not only provides insulation to the live parts of the equipment from the

grounded structures, it sometimes provides mechanical support to the equipment.

The processes responsible for the breakdown of gaseous dielectrics are governed by

the rapid growth of current due to emission of electrons from the cathode, ionization of

the gas particles and fast development of avalanche process. When breakdown occurs

the gases regain their dielectric strength very fast, the liquids regain partially and solid

dielectrics lose their strength completely. The breakdown of solid dielectrics not only

depends upon the magnitude of voltage applied but also it is a function of time for

which the voltage is applied. Roughly speaking, the product of the breakdown voltage

and the log of the time required for breakdown is almost a constant.

Vb = 1n tb = constant

Figure. Variation of Vb with time of application

The dielectric strength of solid materials is affected by many factors viz. ambient

temperature, humidity, duration of test, impurities or structural defects whether a.c, d.c.

or impulse voltages are being used, pressure applied to these electrodes etc. The

mechanism of breakdown in solids is again less understood. However, as is said

earlier the time of application plays an important role in breakdown process, for

discussion purposes, it is convenient to divide the time scale of voltage application into

regions in which different mechanisms operate. The various mechanisms are:

(i) Intrinsic Breakdown

(ii) Electromechanical Breakdown

(iii) Breakdown Due to Treeing and Tracking

(iv) Thermal Breakdown

(v) Electrochemical Breakdown

( i) Intrinsic Breakdown

Occurs at a very short duration of HV applied (10-8 s).

Depends on the presence of free electron, which capable of migration thru the lattice of

the dielectric.

Electronic breakdown

Assumed to be electronic in nature.

Initial conduction electrons density assumed to be large – electrons collisions occur.

Electrons gain energy and cross to conduction band. The process is repeated and more

electrons are developed in conduction band.

( ii ) Electromechanical Breakdown

Due to electrostatic compressive forces that exceed mechanical

Compressive strength.

The highest apparent electric stress before breakdown, if the thickness of specimen do

is compressed to d under applied HV;

Vd=√ 2 γ

ε o εr

ln1.67∨ Vd0

=Ea=0.6[ γε0 ε r ]

12

Mechanical instability occurs when d/do = 0.6, and Y : Young’s modulus and

depends on mechanical stress

( iii ) Breakdown Due to Treeing and Tracking

Occurs over a long time of electrical stress; presence of conducting path. Leakage

current phenomena – formation of spark and carbon track.

Treeing occurs due to the erosion of the material at the tips of the spark. Breakdown

channels spread through the insulation – formation of conducting channels.

Surface Tracking – caused by dry-band arcing. Formation of carbon track on the

insulation surface.

Tracking occurs even at low voltages, whereas treeing requires high voltage.

( iv ) Thermal Breakdown

Conduction current flows – heats up the specimen and the temperature rises.

Heat generated transfers to the surrounding medium by conduction and radiation.

Breakdown occurs when heat generated > heat dissipated.

Heat generated is proportional to the frequency – thermal breakdown is more

serious at high frequency.

Thermal breakdown stresses are lower under a.c. condition then d.c.

( v ) Electrochemical Breakdown

Chemical and electrochemical deterioration and breakdown---presence of air and other

gases – Chemical reactions occur.

Oxidation – surface cracks.

Hydrolysis – lose electrical and mechanical properties.

Chemical action – chemical instability – oxidation, cracking and hydrolysis.

Chemical and electrochemical deterioration increases very rapidly with temperature.

32. Measurement of breakdown voltage in a uniform-field spark gap in air gave the

results as shown in Table Q12.

Gap Spacing

(mm)

Pressure

(Bar)

Temperature

(°C)

Humidity

(g/mᵌ)

Breakdown

Voltage,Vs

( kV)

2.5 1.03 30 9 0.90

27 1.18 15 12 88.56

Table Q12

Using the expression derived from Paschen’s law,

Vs=A ( ρd )+B ¿

i) Determine the relative air density preffered to standard

atmospheric conditions of 1013.25 mbar and 20°C.

ii) The value of constant A and B.

iii) The breakdown voltage of a 3-cm gap spacing at a pressure of

3000 mbar and temperature of 20°C.

Answer:

i) d = 0.25cm, p = 1034mbar, t = 30°C, Vs = 0.90W

ρ20=¿ P1013

. 273+20273+t

¿

¿ 10341013

.293303

¿0.99

d = 2.7cm, p = 1180mbar, t = 15°C, Vs = 88.56kV

ρ20=¿ 1180

1013. 298288

¿

¿0.99

ii)

V 11( g

m3 )=V 9+V 9 .∆ V

¿V 9(1+∆ V )

¿0.90(1+0.002 (2 ))

¿0.91 KV

V 11( g

m3 )=V 12(1−∆ V )

¿88.56 (1−0.002 (1 ))

¿88.38 KV

Vs1=A ( ρ 1d1 )+B ¿

Vs2=A ( ρ 2d 2 )+B ¿

ρ 1d1=0.99 (0.25 )=0.25

√ ρ1d 1=0.5

√ ρ 2d2=1.21 (2.7 )=3.27

√ ρ 2d 2=1.81

Hence, from (x)

0.91=A (0.25 )+B (0.5 )………(1)

88.38=A (3.27 )+B (1.81 )…… …(2)

From equation 1 ;

0.91=A (0.25 )+B (0.5) ………………………..(a)

(x)

88.38=A (3.27 )+B(1.81)………………………(b)

(88.380.91 )=(3.27

0.251.810.5 )(AB )

A=|88.38 1.81

0.91 0.5 ||3.27 1.810.25 0.5 |

¿88.38 (0.5 )−1.81(0.91)3.27 (0.5 )−1.81(0.25)

¿ 44.19−1.651.64−0.45

¿35.75

B=|3.27 88.380.25 0.91 ||3.27 1.810.25 0.5 |

¿3.27 (0.91 )−88.38 (0.25)

1.19

¿−16.07

Solve the simultaneous equation, hence we get the value of :

A = 35.75

B = -16.07

iii) The breakdown voltage of 3-cm gap spacing at pressure of 3000 mbar and a

temperature of 20ºC

d = 3cm , p= 3000mbar, t = 20°C

ρ20=30001013

.293293

¿2.96

ρd=2.96 (3 )=8.88

√ ρd=2.98

Hence,

V s=35.75 (8.888 )−16.07 (2.98)

¿317.46−47.89

¿269.57kV

33. For a gap spacing of 3cm in a non-attaching gas, the breakdown voltage was

found to be 20kV. Determine the new breakdown voltage if secondary ionization

coefficient, γ is doubled given that α = 0.02 x 10−7

Answer:

Given d=3cm Vs=20kV α=0.02×10−7

Find Vs when γ is double

I=I 0 e(αd )

1−γ [e(αd )]−1

E=V s

d

From Townsend’s breakdown criterion for non attaching gas

γ [e(αd )−1 ]−1=0 or γ e(αd )=1

E=V s

d=20kV

3 cm=6.667 kV /cm

from Breakdown criteria γ is double α------α 1

γ e(αd )=2 γ e(αd )=1

e(αd )

e(α1 d)=2

(α−α1 )d=ln2

(α−α1 )= ln2d= ln 2

3=0.231

(α 1 )=0.02×10−7−0.231=−0.231

34. In a strongly inhomogeneous field, external partial discharge occur at

electrode of small radius of curvature when a definite voltage is exceeded.These

are referred to as corona discharge and depending upon the voltage,amplitude,

they result in a larger number of charge pulses of very short duration. With the

aid of diagram where appropriate;

i) Define the terms corona.

ii) Types of corona and how it occurs.

iii) The problems which are created by corona discharges on high

voltage transmission lines

Answer:

i) The term corona is used to describe phenomena which occur at highly-

stressed electrodes prior to the complete breakdown of the gap between the

electrodes. It is a partial discharge in air around a sharp point or thin wire in a

strong, non-uniform field. It is characterized by a visible glow, an audible

noise, radio interference, chemical effects such as production of ozone and

loss of electrical power. It occurs whenever the local voltage gradient exceeds

the ionization value of the air and depends on the air density, humidity and

the outdoor situations whether it is fair weather or raining and also the

roughness of the conductor surface.

ii) Corona can be classifieds into:

DC Corona

Positive Corona (anode)

When the highly-stressed electrode is anode, the following corona modes are observed

as the voltage is increased.

Onset Streamers

Also known as ‘burst’ pulses, these are intermittent, filamentary discharges which

propagates only a short distance from the highly-stressed electrodes.

Hermstein glow

As the voltage increased, the intermittent streamer discharge give way to a steady glow

discharges. This transition occurs when a large enough negative-ions space is

generated near to anode to give a quasi-uniform field in that region.

Breakdown Streamers

Eventually, the shielding effect of the glow discharge is not able to prevent the formation

of large streamers which propagate well into the gap.

Negative corona.

When the highly-stressed electrode is negative, three modes are again observed.

-Trichel Pulses

These differ from the burst pulses in that their magnitude and repetition

frequency are both very regular.

-Cathode glow

As the voltage is raised a critical trichel pulse repetition frequency is reached

and the repetitive discharge is replaced by a steady cathode glow.

-Negative streamers

These discharges are usually known as negative feathers to avoid confusion

with positive streamers discharges. They develop out of the glow mode and a

long rise time compared with other pulsating coronas.

1) AC Corona

With an alternating voltage applied, the same basic corona types will appear although

their characteristics may be altered to an extent which depends on the gap length;

- Small gaps (d ≤ 100cm)

Here, the ions generated in any of the corona modes above are able to cross the gap

during one half cycle of the voltage. Space charges will therefore not persist from one

half cycle to the next and the corona modes will therefore be similar to those for direct

voltages, although all three modes may be observed in one half cycle.

- Large gaps (d ≤ 1m)

For those gaps, space charge can persist from one half cycle to the next and can have

an effect on the corona modes observed. The usual effect is to enhance the positive-

glow phase. Further, the negative streamer is never observed in ac-stressed gaps, since

its onset potential is higher than the positive polarity breakdown voltage. Breakdown

always occurs on the positive half cycle.

2) Transmission Line Corona

The above description of the types of corona discharge referred particularly to the

point-plane gap where there is a single site for discharge to occur. On a transmission

line, corona may occur anywhere on the line and the average corona currents will be

much higher.

iii) The problems which are created by corona discharges on high voltage

transmission lines are:

Power losses

The power losses depend upon the maximum gradient for which the line is designed.

For a single conductor, this occur at the conductor surface. For a given cross-section of

conductor repaired for current carrying capacity, the maximum stress may be

reduced by using bundled conductors in which 2, 4 or 6 wire assembly is used.

Radio Interference

Radio interference is caused only by the pulse corona modes and only Trichel pulses

and positive streamers are of interest. The positive streamers usually have shorter

risetimes than the Trichel pulses and greater amplitude, so that the rate of change of

current is greater and their RI effect is therefore greater. These corona discharges

cause radiation of electromagnetic wanes.

Audible noise.

Recent studies on EHV and UHV lines indicate that audible noise may be a problem

where such lines pass near inhabited areas. Difficulties arise in monitoring such

noise levels as the ‘apparent noise’ is a nonlinear function of frequency. Measuring

instruments have thus been developed which have a similar response to the human

ear and levels have been set based limits at which most people find the noise

objectionable.

TAI(t)URRt

β-αTUTA UTTZg∆TZt

TT

35. Figure Q2 (a) shows a schematic diagram of a tiled transmission line tower.

Consider the tower top is struck by the lightning current i(t)and causes voltage

rise to u(t).

Figure Q2 (a)

Show that

u (t )=Z g Z t / (Zg+2Z t )i (t )

where,

Zg, is the surge impedance of the ground wire

Z t, is the surge impedance of the tower

u (t ),is the impulse surge function

∆ T ,is time of surge propagation from tower top to the cross-arm

i (t ), is the current wave function

T T, is time of surge propagation from tower top to the tower footing

T A,is time of surge propagation from tower cross-arm to the tower

footing

RT ,is tower footing resistance

Answer:

TAI(t)URRt

β-αTUTA UTTZg∆TZt

TT

Zeq=Zg Z g Z t

¿Zg

2Z t

¿

Zg

2× Z t

Z g

2+Z t

¿Zg Z t

Zg+2 Z t

u (t )=Zeq i (t )

¿Zg Z t

Zg+2 Z t

i (t )

36. Show whether the following equation is right or wrong (write the detail

derivation in order to prove it).

UTT (t )=u ( t )−αT (1+β)u (t−2T T )−(αT β )u ( t−4T T )+(αT β )2 u¿

Where,

UTT (t ), is the potential distribution on the top of tower

UTA ( t ),is the potential distribution on the top of tower cross-arm

Alpha(α ), is the coefficient of reflection

Beta(β ), is the coefficient of reflection on the tower to

TAI(t)URRt

β-αTUTA UTTZg∆TZt

TT

-αt=α0

UTA β∆T UTT

V1

α 3β 3UTT

α 2β 2UTT

αβUTT

α 3β 2UTT

α2βUTT

αUTTUTT

V2

V3

V4

I(t)

Answer :

UTT=V 1+V 2+V 3+V 4

¿u ( t )+α (1+β )u (t−2T t )+α 2 β (1+β )u (t−4T t )+α 3 β2 (1+ β )u (t−6T t )

¿u ( t )+α (1+β )u (t−2T t )+αβu (t−4 TT )+α 2 β2ut−6T T ¿+...¿

Disconnected

151413

Disconnected Disconnected

5

12

17

67891011

1 432

200 V

16

100 V

Disconnected

151413

Disconnected Disconnected

5

12

17

67891011

1 432

200 V

16

100 V

Replace α=−αt

¿u (t )−α T (1+ β )u (t−2TT )−(αT β )u (t−4T T )+(α T β )2u (t−6T T )+...¿

The equation proves that UTT is right

37. Use the iteration method to find the finite difference approximation to the

potentials at points 1, 2, 3, 9 and 15 of the system in Figure Q2(c). (Limit the

iteration up to 2 only). The nodal voltage follows the sequence as shown in

FigureQ2(c) that is node number is 1, 2, 3, 4, 5, 6, 7 up to 16 respectively.

Answer:

The general formula:

V 0=V 1+V 2+V 3+V 4

4

Iteration 1:

V 1'=200+0+0+100

4=75V

V 2'=200+0+0+75

4=68.75V

V 3'=200+200+0+68.75

4=117.2V

V 4'=200+200+0+0

4=100V

V 5'=117.2+100+0+0

4=54.3V

V 6'=68.75+54.3+0+0

4=30.76V

V 7'=75+30.76+0+0

4=26.44V

V 8'=100+26.44+0+100

4=56.61V

Iteration 2:

V 1' '=200+68.75+26.44+100

4=98.8V

V 2' '=200+117.2+30.76+98.8

4=111.7V

V 3' '=200+200+54.3+111.7

4=141.5V

V 4' '=200+200+0+54.3

4=113.6V

V 5' '=141.5+113.6+0+30.76

4=67.97V

V 6' '=111.7+67.97+26.44+0

4=51.53V

V 7' '=98.8+51.53+0+56.61

4=51.7V

V 8' '=100+51.7+0+100

4=62.92V

Result:

Potentials at point : P1= 100V

Potentials at point : P2=98.8V

Potentials at point : P3=111.7 V

Potentials at point : P9=51.53V

CHAPTER 3

LIGHTNING OVERVOLTAGE

(QUESTION AND ANSWER)

Q1. (a) With the aids of suitable labelled diagrams, discuss three

possible discharge paths that can cause surges on the transmission

line.

(b) A lightning stroke which reachs a peak current of 35 kA in 1 µs

strikes a 20 m tower on a 345 kV transmission line. The line has a

ground wire joining the tops of the towers; its surge impedance is 520

Ω. The tower surge impedance os 90 Ω and the ground footing

resistance is 40 Ω.

Determine whether the line insulations will flashover as a consequence

of the surge, assuming that their impulse flashover strength is 1050 kV.

A coupling factor 0.3 with the phase conductor can be assumed; the

impedance of the stroke channel can be ignored; a wave velocity on the

tower of 2.98 x 108 m/s can be assumed. Show the surge progressions in

the form of Bewley lattice diagram.

SOLUTION:-

(a) 2 leader core

earth wire space charge envelope

3

Tower conductor

Tower footing resistance

Earth place

Figure 1 – Geometry of lightning leader stroke and transmission line

In the first discharge path (1), which is from the leader core of the lightning

stroke to the earth, the capacitance between the leader and earth is

discharged promptly, and the capacitance from the leader head to the earth

wire and the phase conductor are discharged ultimately bt travelling wave

action, so that a voltage is developed across the insulator string. This is

known as the induced voltage due to a lightning stroke to nearby ground. It is

not a significant factor in the lightning performance of systems above about 66

kV, but causes considerable trouble on lower voltage systems.

The second discharge path (2) is between the lightning head and the earth

conductor. It discharged the capacitance between these two. The resulting

travelling wave comes down the tower and, acting through its effective

impedance, raises the potential of the tower top to a point where the

difference in voltage across the insulation is sufficient to cause flashover from

the tower back to the conductor. This is the so-called back-flashover mode.

The third mode of discharge (3) is between the leader core and the phase

conductor. This discharges the capacitance between these two and injects the

main discharge current into the phase condustor, so developing a surge

impedance voltage across the insulator string. At relatively low current, the

insulation strength is exceeded and the discharge path is completed to earth

via the tower. This is the shielding failure or direct stroke to the phase

conductor.

(b)

20m

I

35 kA

α3 Β2

α2

ZT = 90Ω

α1

40Ω 40Ω

1 µs t

Equivalent circuit

Find Zeq :

1

Zeq

= 1ZT

+ 1Zg

+ 1Zg

1

90+ 1

520+ 1

520

Zeq = 66.86 Ω (1 Mark)

Find V surge :

V = Imax x Zeq

= 35 x 103 x 66.86

= 2.34 MV

= u(t) (1 Mark)

Find α and β :

α1 = R−ZT

R+ZT

= 40−9040+90

= -0.385 (1 Mark)

α2 = Z g Zg−ZT

Z g Zg+ZT

35kA ZT

Imax

Zg Zg

= 520520−90520 520+90

= 260−90260+90

= 0.486 (1 Mark)

β2 = 1 + α2

= 1 + 0.486

= 1.486 (1 Mark)

Time taken for wave travel from tower top to tower base

∆t = DistanceVelocity

= 20

2.98x 108 = 0.067 µs = 1 T (1 Mark)

No of travel in 1 µs after the lightning strike

0.067 µs → 1T

1 µs → 1

0.067 = 14.9 T

V1 = u(t)

V2 = α1β2 u(t - 2T)

= (-0.385)(1.486) u(t – 2T)

= -0.572 u(t – 2T)

V3 = α12α2β2 u(t – 4T)

= (-0.385)2(0.486)(1.486) u(t – 4T)

= 0.107 u(t – 4T)

V4 = α13α2

2β2 u(t – 6T)

= (-0.385)3(0.486)2(1.486) u(t – 6T)

= -0.02 u(t – 6T)

V5 = α14α2

3β2 u(t – 8T)

= (0.385)4(0.486)3(1.486) u(t – 8T)

= 0.0037 u(t – 8T)

V6 = α15α2

4β2 u(t – 10T) V7 = α16α2

5β2 u(t – 12T)

= (0.385)5(0.486)4(1.486) u(t – 10T) = (0.385)6(0.486)5(1.486) u(t – 12T)

= -0.0007 u(t – 10T) = 0.00013 u(t – 12T)

V8 = α17α2

6β2 u(t – 14T)

= (0.385)7(0.486)6(1.486) u(t- 14T)

= -0.00002 u(t – 14T)

Vtt = V1 + V2 + V3 + V4 + V5 + V6 + V7 + V8

= u(t) – 0.572 u(t – 2T) + 0.107 u(t – 4T) – 0.02 u(t – 6T) + 0.0037 u(t – 8T)

– 0.0007 u(t – 10T) + 0.00013 u(t – 12T) – 0.00002 u(t – 14T)

(4 Marks)

From graph,

Vpeak = 0.56 u(t) Vphase conductor = 0.3 Vpeak

= 0.56 (2.34 x 106) = 0.3 (1.31 x 106)

= 1.31 MV (1 Mark) = 0.393 MV (1 Marks)

Vflashover = Vpeak – Vphase conductor + Vphase

= 1.31 x 106 – 0.393 x 106 + √2 x345 x 103

√ 3

= 1.12 MV (1 Marks)

Vflashover strength = 1050 kV = 1.05

Therefore flashover occur because flashover ¿ flashover strength

(3 Marks)

Q2. (a) Describe the lightning phenomena based on Simpson’s theory as

shown in Figure Q2a.

Figure 2a

(b) Explain what is meant by the terms ‘T1/T2 Impulse Wave’ and outline the

methode of lightning impulse voltage production in the laboratory.

SOLUTION:-

Q2. (a) According to Simpson’s theory as shown in the diagram below, there are

three essential regions the the cloud to be considered for charge formation.

Below region A, air currents travel about 800 cm/s and no rain drops fall

through. In region A, air velocity is high enough to break the falling rain drops

causing a positive charge spray in the clouds and negative charge in the air.

The spray is blown upwards but as the velocity of air decreases, the positively

charge water drops recombine with the larger drops and fall again. Thus

region A eventually becomes pre-dominately positively charged while region B

above it becomes negatively charged by air currents. In this upper regions in

the cloud the temperature is low (below freezing point) and only ice crystals

exists. The impact of air on these crystals makes them negatively charged,

thus the distribution of the charge within the cloud is shown as in the figure.

Lightning phenomena is based on the atmospheric process that takes place

during tunderstorm where charges are accumulated in the cloud or portion of

the cloud and equal and opposite charges are produced in the earth beneath.

These positive and negative charged become seperated by the heavy air

currents with ice crystals in the upper part and rain in lower parts of the cloud.

As the charges increases the potential between the cloud and the earth

increases and therefore the potential gradient is not uniformly distributed.

When the gradient exceeds the strength of the portion of air across which it is

applied, the air breakdown and a streamer starts from the cloud towards

earth. Lightning stroke may be started with potential of the order of 5 to 20 MV

between the cloud and the earth.

(b) An impulse voltage wave is a unidirectional voltage which rises rapidly to a

maximum voltage and the decays rather more slowly to zero as shown in the

diagram below.

V

100 %

90 %

50 %

10 %

T1 t

T2

The waveshape is generally defined in terms of the time T1 and T2 in

microseconds. T1 is the time taken by the voltage wave to reach its peak

value, i.e from 10% to 90% of the voltage wave. T2 is the total time from the

start of the wave to the instant when it has declined to one half of its peak, i.e

from start of the wave to 50% of the peak during decay.

Q3 (a) Will backflashover occur on a tower that was struck by lightning of

current I(t), of wave-shape with maximum value at 40kA occurring at 1.2µs and

decreased linearly. The surge impedance of tower, tower ground wire and

phase conductors are 200Ω, 250Ω and 300Ω respectively. The clearance

between the tower side and the phase conductor is 2 meter. Assume that the

breakdown of air is at 30kV per cm and the string insulator flashover at 1000kV.

Consider the coupling factor between ground wire and phase wire is 0.25. The

lighting strike when the A.C voltage is at [ 550√2√3 ] sin 110°kV.

Figure Q3(a) Transmission system with the top of tower was struck by lighting

(b) Figure Q3(b) shows a schematic diagram of a tilted transmission line

tower with two parallel ground wire and an impulse current wave-shape, i(t).

Consider the tower top is struck by the lighting current i(t) and voltage rises

to u(t).

Figure Q3(b)

(c) By referring to Fig Q3(b), discuss the validity of equation (write the

detailed derivation in order to prove it):

UTT (t) = u(t) - α T (1+β)[u(t -2TT) – (αTβ)u(t -4TT) + (αTβ)2u(t -6TT)….]

Where:

UTT (t) = potential distribution on the top of tower

α (Alpha) = coefficient of reflection

β (Beta) = coefficient of reflection on the tower top side

Solution:

(a) The ground potential rise at the point where lighting strikes the tower

Vsurge = 40 x 103 x [Z g∨¿ Zg∨¿ Z t ]

= 40 x 103 x [ Zg

2 x Z t

Zg

2+Z t

]= [ 250

2x 200250

2+200] 40kV

= [ 125 x 200325 ] 40kV = 3.08MV

(b) Zeq = Zg|| Zg|| Zg|| Zg||ZT

1Zeq

= 1Zg+ 1

Zg+ 1

Zg+ 1

Zg+ 1

ZT

= ZT+ZT+ZT+ZT+Z g

Zg ZT

=4 ZT+Zg

Z g ZT

Zeq = Z g ZT

4 ZT+Zg

So u(t) = i(t) . Zeq = Z g ZT

4 ZT+Zg

. i(t)

(c)

V1 = u(t) t = 0

V2 = (α + βα) u(t) t = 2TT

V3 = (α2β + α2β2) u(t) t = 4TT

V4 = (α3β2 + 32β3) u(t) t = 6TT

Β =UT Tα=α T

TT

Zg

Β α u(t)

V3

V2

TT

6TT

5TT

4TT

3TT

2TT

α u(t)

u(t)

Β α2 u(t)

Β2 α2 u(t)

Β2 α3 u(t)

Β3 α3 u(t)

V4

V1

VTT (t)=v1+v2+v3+v 4

= u(t) + (α + βα) u (t- 2TT ) + (α2β + α2β2) u (t- 4TT ) + (α3β2 + 32β3) u (t- 6TT )

= u(t)+α(1+β)[u(t -2TT) – (αTβ)u(t -4TT) + (αTβ)2u(t -6TT)….]

So the expression for VTT as given in the question is right

Q4 (a) Discuss in detail about the lighting phenomenon starting with the

formation of thunderclouds.

(b) A peak lighting current of 40kA has struck a ground wire at mid-span (at

the middle of two transmission towers). If the ground wire surge impedance is

given as Z=500Ω, calculate the generated voltage at the point of strike. State

the assumptions you made to answer this question.

Solution:

(a) During storms, charges are accumulated in clouds and equal charges of opposite

polarity are formed in earth. As these charges increases, the voltage gradient in

the air adjacent to the charge centre in the cloud increases.

When the gradient exceeds the insulation strength of air, a low current streamer

starts downward from the cloud and continues to grow. When the streamer

makes contact with the earth, it is like closing a switch between the two charges

of opposite polarity, one is the earth and the other in the streamer channel and in

the clouds . Thus large current flows.

In the middle stage of formation of a cloud, strong wind turbulence takes place

causing the separation of charges into several layers which are:

Intra-cloud (within the clouds)

Cloud to cloud

Cloud to ground

Cloud to air

Positive stroke or “Blue sky lightning”

(b)

I = 40kA

ZL =500ΩZL =500Ω

Assume:

1) Zs = infinity

2) No return voltage at the Earth line Zeq=Z /2

V at strike point, V=IZ /2=40 K x500 /2=10uV

ZS

Z Z

A

Q5 (a) Discuss two(2) of the followings

i. Lightning phenomena

ii. Direct strike and indirect strike

ii. Switching overvoltage

(5 marks)

(b) Lightning strike at mid-span of a transmission line ground wire at point a as

shown in Figure Q5(b). This wave travels in both direction of the

transmission line. Determine the transmission and reflection coefficients, β

and α at points b and d.

(10 marks)

Figure Q5(b)

(c) A lightning surge of magnitude 10 kA with the voltage waveshape of 1.2/50

µs strike a ground conductor at midspan of a transmission line. If the

channel surge impedance is 1500 Ω and the ground wire surge impedance

is 600 Ω, determine at the point of strike;

i. The equivalent circuit and equivalent impedance,

ii. The peak current, and

iii. The peak voltage.

R1

R2

ed

ZT1 ZT2

c Zg3Zg2bZg1

(10 marks)

Solution

(a)

i. Lightning Phenomena

Benjamin Franklin has proved lightning is an electrical phenomena. An Eletrical

phenomena carries the concept of charges involvement. So two types of charges are the

reasons for the cloud to be considered as a ‘cell’. The charge are a) positive type and

b)negative type. Fig. shows the typical thundercloud structure.

Not all clouds are lightning cloud generator. It is only the cumulonimbus cloud type that

can generate lightning. The Ice Splinter can be used to explain on the electrification of

the cloud. The moistures and precipitation particles being is suspension in air and due to

upwards action of updraft, causing supercooling to take place and resulting moistures to

become ice. The ionic migration of OH- and H- in the moistures built, leaving the OH- in

the front and II+ being lighter are pulled out to settle in the outer layer. The resultant two-

layer ice structure split due to different rate of ice expansion (the inner and outer layer).

The splinters are basically of positive-charged and negative-charged. The lighter

splinters are pulled upwards while the lighter negatively charged splinters settle at the

lower point of the cloud. If the electric field between the cloud and ground exceed the

dielectric strength of air, streamers will appear and propagate toward the ground. The

last jump of these streamers to the ground result upward streamers to move to attach

itself to the downwards-moving streamers. The attachment results in the process of

20km --

15km --

10km --

3km --

+ + + + + + + + + ++ + + + + + + + ++ + + + + + + +- - - - - - - - - --- - - - - - - - - --- - - - - - -

charges neutralisation of the positive and negative charges. This is known as return

strike. It causes large currents to flow to the ground.

[ 2.5 Marks ]

ii. (a) Direct Strike

Ground flash activities of lightning involve direct and indirect strike. If the

intend facilities or building or even structure for protection are struck by

lightning, which could end with structure damage and others, this is direct

strike.

(b) Indirect Strike

If the damage to building and equipment due to surge propagation is because

of inductive and capacitive affect. This is termed as indirect strike.

[ 2.5 Marks ]

OR

iii. Switching Voltage

With the steady increase in transmission voltages needed to fulfil the required

increase in transmitted power, switching surges have become the governing factor in

the design of insulation for EHV and UHV systems. In the meantime, lightning

overvoltage come as a secondary factor in these networks. There is a great variety of

events that would initiate a switching operations of greatest tolerance to insulation

design can be classified as follows:

a. Energization of a line

b. Load rejection

c. Switching on and off a equipment

d. Fault initiation and clearing [ 2.5 Marks ]

(b)

Point b is to be considered:

∝=Zg 2Zg 2

¿¿¿¿

Zt 1−Zg 1Zt 1+Zg 1

β=1+α

α=( Zg2 xZT 1

Zg 2+ZT 1−Zg 1)

Zg 2 x ZT 1Zg2+ZT 1

+Zg1=

Zg2 xZT 1−Zg 1(Zg 2+ZT 1)Zg2 xZT 1−Zg 1(Zg 2+ZT 1)

Point d:

α=R 1−ZT 1R 1+ZT 1

β doesnot exit

(c)

i. Based on the equivalent circuit:

Zg 1× Zg1

2Zg 1=600

2=300

ii Based on the circuit:

Zg1 Zg1

I peak

10kA10kA

I peak

1500 300

I peak= 15001500+300

×10=0.833× 10=8.33kA

iii . The peak voltage:

V surge=Zeq× Ipeak=300 ×8.33 kA=2499 kV=2.5MV

Q6. (a) A transformer has an impulse insulation level of 1175 kV and is to be

operated with an insulation margin of 15 % under lightning impulse conditions.

The transformer has a surge impedance of 400 Ω. A short length of overhead

earth wire is to be used for shielding the line near the transformer from direct

strikes. Beyond the shielded length, direct strokes on the phase conductor can

give rise to voltage waves of the from 1000e-0.05t kV (where t is expressed in µs).

If the corona distortion in the line is represented by the expression,

¿∆x

= 1β

(1−VoV ) μs /m

Where B = 100 m/µs and V₀ = 200 kV, determine the minimum length of

shielding wire necessary in order that the transformer insulation will not fail

due to lightning surges.

(10 marks)

Solution:

(a)

β= 2× 1600

1600+400¿1.6

For B.I.L of 1175 Kv, and an insulation margin of 15%, the maximum permissible

ZT=1600Ωβ

α1000e-0.05t

Zo =400Ω

voltage

¿1175×85

100

= 998.75kV

Since the voltage is increased by b= 1.6 times at the terminal equipment (transformer), the

maximum permissible incident voltage must be decreased by this factor, hence the

maximum permissible incident voltage is

=998.75/1.6

=624.22 kV

Therefore the shielding wire must reduce the surge to 624.22 kV(by virtue of the corona

distortion), that is

1000e-0.05t= 624.22

t = 9.425 µs

Hence ΔT = 9.425 µs

The corona distortion is given by

∆ tx= 1

B [1− 200624.22 ] μs/m

x=1525.53m

Therefore the minimum length of shielding wire required = 1.526 km.

Q7 (a) Discuss the following:

i. Backflashover of a lightning strike [2 Marks]

ii. Front time and tail time of lightning impulse voltage [2 Marks]

iii. The causes of switching surges [4 Marks]

(b) Prove that the reflection coefficient, α = (Z2 – Z1)/(Z2 + Z1) and the

transmission coefficient, β = 2Z2/(Z2 + Z1) for an incident lightning surge

on a transmission line where Z1 is the surge impedance of line 1 and Z2

is the surge impedance of line 2.

[4 Marks]

(c) A lightning current with the rate of rise of 25 kA/µs reaches a peak walue

in 1.6 µs has struck a ground wire at mid-span (at the middle of two

transmission towers). If the ground wire surge impedance is given as Zg

= 250 Ω, calculate the generated voltage at the point of strike. State and

justify all assumptions made.

If the striking point is changed to the top of one tower (not the end

tower) with a surge impedance of 100 Ω, calculate the generated voltage

at the point of strike.

[8 Marks]

Solution

(a)

i. Backflashover of a lightning strike

When a direct lightning stroke occurs on a tower, the tower has to carry

huge impulse currents. If the tower footing resistance is considerable , the

potential of the tower rises to a large value, steeply with respect to the line

and consequently a flashover may take place along the insulator strings.

This is known as “back flashover”.

(2 Marks)

ii. Front time and tail time of lightning impulse voltage

Front – time taken for the lightning impulse waveshape to rise from start

to peak value

Tail time – time taken for the lightning impulse waveshape to decay to

50% of it’s peak value

(3 Marks)

iii. The origin of switching surges in power system;

De-energizing of transmission lines, cables, shunt capacitor,

banks, etc.

Disconnection of unloaded transformers, reactors, etc.

Energization or reclosing of lines and reactive loads

Sudden switching off to loads

Shorcircuits and fault clearances

Resonance phenomenon like ferro-resonance, arcing grounds,

etc.

1.0

0.3

T2

T

T1

T1 =1.67TT’=0.3T1 =0.5T

T’

0.5

0.9

(3 Marks)

(b)

where, E & I – incident voltage and current

ET & IT – transmitted voltage and current

ER & IR – reflected voltage and current

No discontinuity of potential and current at junction J,

: E + ER = ET and I + IR = . . . (1)

E = Z1I, ET = Z2IT, ER = -Z1IR . . . . (2)

Substitute (2) into (1)

E/Z1-ER/ZR=(E+ER)/Z2.........(3)

By solving equition (3) gives;

Reflected coef., α , ER/E = (Z2 – Z1)/(Z2 – Z1)

Transmitted coef.,β, ET/E = 2Z2/(Z1 + Z2) (4 marks)

(c) i.

I = 25 x 1.6 = 40 kA,

Vsurge = IZeqv By assuming Zs=α;

= I(Zg//Zg)

= I(Zg/2)

Incident Wave

ER IR

ET IT

JE I

Reflected Wave

Transmitted Wave

I ZS zg Zg

Equivalent Circuit

VSURGE

I1

= 40 kA x 125

= 5 MV (4 Marks)

ii) By assuming ZS =α; Zg=250Ω; Zt =100Ω

Zeqv = (Zg/2) //ZT = Zt(Zg/2) =55.56Ω

(Zt + Zg/2)

: Vsurge = IZeqv

= 40 kA x 55.56

= 2.22 MV (4 marks)

Q8. (a) Fig. Q8 (a) shows a schematic diagram of a tilted transmission line

tower and an impulse current waveshape, i(t). Consider the tower top is struck

by the lightning current i(t) and voltage rises to u(t).

TT ZR

ZT u(t)

UR

Rt

Fig. Q8(a) Inverted tower for analysis

Also Zg is the surge impedance of the ground wire

Zt is the surge impedance of the tower

u(t) is the impulse surge function

i(t) is the current wave function

TT is the time of surge propagation from tower top to the tower footing

Rt is tower footing resistance

UTT(t) is the potential distribution on the top of tower

α is the coefficient of reflection on the tower bottom side

β is the coefficient of reflection on the tower top side

UTT(t)

i(t)βα

i(t)

Time (microsec)

Current (kA)

(i) Show that

u(t)= Zg Zt / (Zg + 2Zt ). i(t) (3 marks)

(ii) Determine whether the following equation is right or wrong (write

the detailed derivation in order to prove it)

UTT(t)=u(t)+αT(1+β)[u(t-2TT)+(αTβ)u(t-4TT)+(αTβ)2u(t-6TT)+……..]

V1=u(t)

V2=(α+ βα)

V3=(α2β+α2β2) u(t)

V4=(α3β2+α3β3) u(t)

t=0

t=2TT

t=4TT

t=6TT

TT

2TT

3TT

4TT

UTT

UtUt

=αUt

=αβ Ut

=α2β Ut

=α2β 2Ut

=α3β 2Ut

=α3β 3Ut

5TT

(b) A lightning current surge with the wave shape as shown in Fig Q8 (b), strike a

tower, which has a single ground wire in both directions. The characteristic are

as follows :

Surge impedance of lightning channel, Zl = infinity

Surge impedance of the tower, Zt = 150 Ω

Surge impedance of ground wire, Zg = 340 Ω

Velocity of wave propagation on lines = 298 m/ µs

Velocity of wave propagation on tower = 240 m/ µs

Height of tower = 30m

Effective tower footing resistance = 40 Ω

Lightning current peak magnitude = 40 kA

Based on Fig.8 (b), determine the maximum tower top potential for a duration 5

times the time of surge propagation from the tower top to the tower base after

the lightning strike the tower.

(15 marks)

Solution:

(a) i. u(t)= [Zg // Zt // Zg ] x i(t)

= Zg2

// Zt x i(t)

¿

Zg2

xZt

Zg2−Zt

x i (t )

¿ Zg x ZtZg−2Zt

x i ( t )

IDOrientation of propagation

20 µs 1 µs

Fig.Q8 (b) The simplified lightning current wave shape

ii. UTT=V1+V2+V3+V4

=u(t)- (α+βα ) u(t-2TT) - (α2β+α2β2) u(t-4TT) – (α3β2+α3β3) u(t-6TT)

=u(t)- α(1+β) [u(t-2TT) - αβu(t-4TT) – α2β2u(t-6TT)+…..]

Replace α = - αt

=u(t)- αt (1+β) [u(t-2TT) - αtβu(t-4TT) – αt2β2u(t-6TT)+…..]

The equation proves that UTT is right.

UTT=Ut(t) + β1α1ut (t-2 T) + β1α12 α2ut (t-4 T) + β1α1

3α22ut (t-6 T) +…..

=Ut(t) + (1 - β) α1ut (t-2 T) + β (1 - β) α12 α2ut (t-4 T) + (1 - β) α1

3α22ut (t-6 T) +…..

=Ut(t) + α1 (1 - β) [ut (t-2 T) - α1α2ut (t-4 T) + α12α2

2ut (t-6 T) +…..]

=Ut(t) + αT (1 - β) [ut (t-2 T) – (αTβ) ut (t-4 T) + (utβ)2 ut(t-6 T) +…..]

=Ut(t) + 0.579 (1 – 0.0625) [ut (t-2 T) – (0.579 x 0.0625) ut (t-4 T)]

=Ut(t) + 1.036 [ut (t-2 T) – (0.036)ut (t-4 T)]

=Ut(t) + 1.036ut (t-2 T) – 0.036ut (t-4 T)]

(b) The lightning current waveshape 1/20 µs

Comparison:

1 µs = 1/0.125 = 8 T

20 µs = 20/0.125 = 160

The Vsurge=

8 T

3.3 MV=1 pu

16 T

=Ut(t)

The peak voltage is 0.91 MV

Q9. (a) Discuss two (2) of the followings:

i. Lightning Phenomena (3 marks)

ii. Direct Strike and Indirect Strike (3 marks)

iii. Switching Overvoltage (3 marks)

(b) Lightning strike at mid-span of a transmission line ground wire at point a as

shown in figure Q9(b). This wave travels in both direction of the transmission

line. Determine the transmission and reflection coefficients, α and β at points b

and d.

UTT=Ut(t) + β1α1ut (t-2 T) + β1α12 α2ut (t-4 T) + β1α1

3α22ut (t-6 T) +…..

=Ut(t) + (1 - β) α1ut (t-2 T) + β (1 - β) α12 α2ut (t-4 T) + (1 - β) α1

3α22ut (t-6 T) +…..

=Ut(t) + α1 (1 - β) [ut (t-2 T) - α1α2ut (t-4 T) + α12α2

2ut (t-6 T) +…..]

=Ut(t) + αT (1 - β) [ut (t-2 T) – (αTβ) ut (t-4 T) + (utβ)2 ut(t-6 T) +…..]

=Ut(t) + 0.579 (1 – 0.0625) [ut (t-2 T) – (0.579 x 0.0625) ut (t-4 T)]

=Ut(t) + 1.036 [ut (t-2 T) – (0.036)ut (t-4 T)]

=Ut(t) + 1.036ut (t-2 T) – 0.036ut (t-4 T)]

(b) The lightning current waveshape 1/20 µs

Comparison:

1 µs = 1/0.125 = 8 T

20 µs = 20/0.125 = 160

The Vsurge=

R1 R2

ed

ZT1 ZT2

c Zg3Zg2bZg1

Figure Q9 (b)

(6 marks)

(c) A lightning surge of magnitude 10kA with the voltage waveshape of 1.2/50 µs strike a ground conductor at midspan of a transmission line. If the channel surge impedance is 1500Ω and the ground wire surge impedance is 600Ω, determine at the point of strike :

i. The equivalent circuit and equivalent impedance (2 marks)

i) The peak current (3 marks)

ii) The peak voltage (3 marks)

(d) A lightning current surge with the wave shape of figure 7 strikes a tower which

has a single ground wire in both directions. The characteristics are as follows;

Surge impedance of lightning channel = infinity

Surge impedance of tower = 150Ω

Surge impedance of ground wire = 340Ω

Velocity of the wave propagation on lines = 298 m/µs

Velocity of the wave propagation on tower = 240 m/µs

Coupling factor of phase conductors = 0.25

Height of tower = 30m

Effective tower footing resistance = 40Ω

Determine the maximum tower top potential, after 0.4 µs the tower has been

struck by the lightning. Please show clearly all the calculations involving

coefficients of the reflection and refraction. Show the surge progressions in

the form of the Bewley Lattice Diagram.

a) What will happen if the tower footing resistance increases in the value?

b) Provide one reason for the tower footing resistance increase in the value.

c) Why the speed of surge is higher in the conductor than in the tower

structure?

(17 marks)

25 kA

20 µs t 1 µs

Figure 9(d) the wave shape of the lightning current

Solution:

(a) i. Benjamin Franklin has prove lightning is an electrical phenomena. An

electrical phenomenon carries the concept of charges involvement. Two

types of charges are the reasons for the cloud to be considered as a ‘cell’.

The charges are:

a) Positive Type

b) Negative Type

Figure above show the typical thundercloud structure. Not all clouds are

lightning cloud generator. It is only the cumulonimbus cloud type that can

generate lightning. The Ice Splinter can be used to explain on the electrification

of the cloud. The moistures and precipitation particles being is suspension in air

and due to upwards action of updraft, causing super cooling to take place and

resulting moistures to become ice.

The ionic migration of OH- and H- in the moisture built, leaving the OH- in the front

and H- being lighter is pulled out to settle in the outer layer. The resultant two

layer ice structure split due to different rate of ice expansion (the inner and outer

layer). The splinters are basically of positive charged and negative charged. The

lighter splinters are pulled upward while the lighter negatively charged splinters

settle at the lower point of the cloud. If the electric field between the cloud and

ground exceed the dielectric strength of air, streamers will appear and propagate

towards the ground. The last jumps of these streamers to the ground result in

upward streamers to move to attach itself to the downwards moving streamers.

The attachment results in the process of charges neutralization of the positive

and negative charges. This is known as return strike. It causes large currents to

flow to the ground.

(a) ii.

Direct Strike

Ground Flash activities of lightning involve direct and indirect strike. If the

intend facilities or building or even structure for protection are struck by

lightning, which could end with structured damage and others, this is

direct strike

Indirect Strike

If the damage to building and equipment due to surge propagation is

because of inductive and capacitive affects. This termed as indirect

strike.

(a) iii.

With the steady increase in transmission voltages needed to fulfill the required

increased in transmitted power, switching surges have become the governing

factor in the design of insulation for EHV and UHV systems. In the meantime,

lightning overvoltage come as a secondary factor in these network. There is a

great variety of events that would initiate a switching surge in a power network.

The switching operations of greatest tolerance to insulation design can be

classified as follows:

a. Energization of a line

b. Load rejection

c. Switching on and off of equipment

d. Fault initiation and clearing

(b) Point b is to be considered:

α=Zg 2∨¿Zt 1−Zg 1Zg2∨¿Zt 1+Zg 1

β=1−α

α=

Zg2x ZT 1Zg2x ZT 1

−Zg1

Zg2 x ZT 1Zg2 x ZT 1

+Zg1

α=Zg 2 x Zt 1−Zg 1(Zg2+ZT 1)Zg2 x Zt 1+Zg1(Zg 2+ZT 1)

Point d:

α= R 1−ZT 1R 1+ZT 1

; β does not exist

(c) i.

Zeq= Zg1 x Zg12Zg 1

Zeq=6002

Zeq=300

(c) ii.

10kA

VTITi

ZgZgZs

Zg1Zg1

Ipeak= 15001500−300

x 10

Ipeak=0.833 x10

Ipeak=8.33kA

(c) iii The peak voltage

Vsurge=Zeq x Ipeak=300 x 8.33kA=2499kV=2.5 MV

(d)

Lightning Current Waveshape

25 kA

20 µsec t 1 µsec

Surge impedance of lightning channel = infinity

Surge impedance of tower = 150Ω

Surge impedance of ground wire = 340Ω

Velocity of the wave propagation on lines = 298 m/µs

Velocity of the wave propagation on tower = 240 m/µs

Coupling factor of phase conductors = 0.25

Height of tower = 30m

Effective tower footing resistance = 40Ω

Zeq = Zg//Zg//Zb

a

= 340//340//150

= 170x 150170+150

= 79.7

Step 1: To determine the potential at the top of tower at the point of lighting strike at the top

of the tower point of lighting strike

Equivalent circuit representation:

Vsurge = Î x Zc /¿ Zeq

= 25 x103 x Zeq Zc = Open Circuit

= 25 x103 x79.7

= 1.992 MV …………………………………………Assume this as 1 p.u

Step 2: To determine time for surge propagating from point a to foot of tower

t tt=30

240m /µs

= 0.125 µsec equivalent to T

Step 3: Calculation of coefficient

(i) Reflection

(ii) Refraction

VSurgeZeqZcÎ

Bewley Lattice Diagram

For a duration of a Tstrike = 0.4

0.125

= 3.2 T

α1 = R−Zb

R+Zb

= 40−15040+150

= -0.58

α2 = Z g/¿Zg−Z t

Z g/¿Zg+Z t

= 170−150170+150

= 0.0625

β = 2Z g/¿Z g

Z g/¿Zg+Z t

= 340320

= 1.063

Potential at tower top a

ua=ut (t )+ βα 1ut (t−2T )

Using graphical method to determine solution

PT 0

121+4+12

1

T µsec

The maximum tower top potential

(0.2 – 0.6p)

= 0.23 x 1.992 MV

= 0.456 MV

(i) The ground potential rise at the tower increases with the increase in tower

resistance.

(ii) Loss of moisture due to draught reason and the soil cannot retain moisture

due to bad soil condition.

(iii) The travelling wave is given by the relation V = 1

√LC . Since the tower has

higher conductance value as well as the capacitance, therefore

V t=1

LT CT

< −1LCCC

i.e; LT CT>LCCC

So, V c > V T

The speed of surge is higher in the phase conductor than in the tower structure.

Q10. (a) Figure Q10 (a) shows a schematic diagram of a tilted transmission line

tower and an impulse current waveshape, i(t). Consider the tower top is struck

by the lightning current i(t) and voltage rises to u(t).

Figure Q10(a)

Show that

u (t )=Zg Zt−(Zg+2Zt ) . i(t)

Where Zg is the surge impedance of the ground wire

Zt is the surge impedance of the tower

U(t) is the impulse surge function

T is time of surge propagation from tower top to the cross-arm

i(t) is the current wave function

TT is the time of surge propagation from tower top to the tower footing

TA is the time of surge propagation from tower cross-arm to the tower

footing

Rt is tower footing resistance

(2 marks)

(b) Show whether the following equation is right or wrong (write the

detailed derivation in order to prove it)

U TT(t)= u(t) – αT(1+ β) [ u(t-2TT)-( αT β)u(t-4TT) + ( αT β)2 u(t-6TT) ]

Where UTT(t) is the potential distribution on the top of tower

UTA(t) is the potential distribution on the tower cross-arm

Alpha (α) is the coefficient of reflection

Beta (β) is the coefficient of reflection on the top side

(5 marks)

c.

Figure Q11 (c) shows a partly distribution system of electrical power network

where an overhead line is connected to a set of three underground cables

connected each other point A, B and C respectively. At those respective points

a resistor R is connected to the ground. The line is struck by lightning at point

0H, 100m away from the underground cable UG1. The form of the lightning

current is:

I(t) = 3.0x1010t , 0 < t < 2.0 μs

= 6.0x1010 - 3.0x108t , 2 < t < 101 μs

= 0 , t > 101 μs

Each underground cable is 300m length where Zoh, Za, Zb and Zc is equal to

450Ω, R=80Ω. Assume Z of lighthning channel impedance is infinity. Traveling

wave have the following velocities:

a) On the overhead conductors, 2.98x108 m/s

b) On the underground cable 2.68x 108 m/s

Calculate the first peak of voltage at point B after 3 microseconds the lightning

strike the overhead conductor and the time when it occurs.

(13 marks)

Solution:

(a) Zeq = Zg // Zg // Zt = Zg/2 // Zt

¿Z g /2x Z tZ g/2+Z t

x (t )=i ( t ) x Zeq

¿ i ( t ) x Z g /2x Z tZ g/2+Z t

=i (t ) x Zgx Z tZ g+2 Z t

U TT(t)= ut(t) + αUt(1+ β) (t-2TT) + α2 βu(t-4TT) + α2 β2 u(t-4TT) + α3 β2 u(t-6TT) + α3 β3

u(t-6TT)

= ut(t) + α (1+ β) ut (t-2TT) + α2 β3 (1+ β) u(t-4TT) + α3 β2 (1+ β) u(t-6TT)

= ut(t) + α (1+ β)[ ut (t-2TT)] + α β u(t-4TT) + (α + β)2 u(t-6TT)

(b)

(c)

Î = 3x1010 x 2x10-6

= 6x104 = 60KA 0<t<2us

Î =( 6x104 ) – (3x108 x 2x10-6 )

= ( 6x104 ) – ( 6x102 )

= 60000 – 600

= 59400 = 59.4KA

I = 0 Vsurge = 59.4 x103 x (450/2)

= 1.34mV

Time of travelling

At overhead line conductor

ΔOH – UG1 = 100 / (2.98 x108)

= 33.6 x10 -8

= 0.34μs = ΔT

ΔUG1 = Δ UG2 = Δ UG3

= 300 / (2.98 x108)

= 111.9 x10 -8

= 1.12 μs

ΔT = 1.12 μs

ΔUG1 = 0.34/1.12= 3.294

To simply the drawing

ΔUG1 = 3.00 AT

ΔOH – UG1 = (0.34/1.12) x 3.00 = 0.91 AT

Time frame = (3/0.34) x 0.91 AT = 8.02

α1 = (450-450)/(450 + 450)

= 0

β1 = 1 + α1 = 1

α2 = (6.79-450)/(67.9+ 450)

= - 0.738

β 2 = 1 + α2 = 0.262

α3 = -0.738, β3= 0.262

α4 = (80 - 450)/ (80 + 450)

= -0.698

β 2 = does not exist

Ut at B after 3 microseconds the lightning strike the overhead line = β 1 β 2 β 3 u(t-7ΔT)

VB = β 1 β 2 β 3 ut(t-7ΔT)

= 1 x 0.262 x 0.262 x ut(t-7ΔT)

= 0.069 ut(t-7ΔT)

The Zeq at lightning point of strike

= 450/2 = 225 Ω

Î =60KA

Vsurge = 225 x (59.4 x103 )

= 13.34MV

= 1 p.u

So that first peak voltage at point B after 3 microseconds the lightning strike the overhead

conductor is

= 0.069 x 13.34 MV

= 0.92 MV

The time it happen is 7ΔT + 2.0

= 7 x 0.34 + 2.0

= 4.38 μ

Q11 a. Discuss any two (2) of the followings:

i. Mechanism of lightning strike.

(4 marks)

ii. The difference between Simpson’s Theory and Reynold & Mason Theory

in explaining the phenomena oh charge formation in the clouds.

(4 marks)

iii. How does lightning strike can induce over voltages in power

transmission line?

(4 marks)

b. Protection of power transmission system from lightning strike is an important

aspect in the design of the system. A good design is to minimize the

disturbances due to lightning. Describe three (3) types of lightning protection.

It’s function and operation as a protection to the power system.

(6 marks)

c. Switching surge is an overvoltage that causes disturbances to power

transmission lines. With an increase in transmission voltage to over 400 kV,

the switching surface have caused problems as same as the lightning

overvoltage. Explain what is meant by switching overvoltage. It’s characteristic

and the methods of controlling the surge to prevent damage to the power

equipment. (6

marks)

Solution:

(a) i. Mechanism of lightning strike.

Stepped leaders propagating towards the ground

Strong attraction from the downwards stepped leaders cause oppositely charge streamers

appearing on various corners of the structure of the ground

Attachment of –ve charge downwards streamer with +ve charge upwards streamer. The final

jump is to the transmission power tower.

(a) ii. The difference between Simpson’s Theory and Reynold & Mason Theory in

explaining the phenomena oh charge formation in the clouds.

Teori Simpson

Menurut teori simpson ada tiga ruang penting dalam awan bagi pembentukan cas sepetri

didalam rajah diatas. Dibawah ruang A, kelajuan arus udara adalah melebihi 800cm/s dan

tidak terdapat titisan hujan yang jatuh. Dalam ruang A, kelajuan udara adalah tinggi untuk

memechkan titisan hujan menyebabkan percikan cas positif terbentuk pada awan dan cas

negative dalam udara. Percikan cas positif tersebut ditiup keatas tetapi dengan

pengurangan kelajuan udara, cas positif titisan air tersebut akan bercampur dengan titisan

yang lebih banyak dan jatuh kebawah. Ini menyebabkan ruang A akan bercas positif dan

ruang B keatas akan bercas negative. Pada ruang di bahagian atas awan suhu adalah

rendah dan hanya terdapat hablur ais. Kesan udara pada hablur ais tersebut menjadikannya

bercas negative.

Teori Reynold dan Manson

Menurut reynold dan manson, petir terhasil pada ketinggian antara 1 ke 2 km sehingga 12

ke 14 km dari paras bumi. Bagi pembentukan awan petir dan cas perlu ada arus udara,

lembapan dan julat suhu yang tepat. Arus udara yang dikawal oleh kecerunan suhu

bergerak keatas membawa lembapan dan titisan air. Titisan air didalam awan petir ditiup

keatas oleh arus udara dan mejadi beku sebagai hablur ais.pada suhu beku, hablur akan

bertambah dan bergerak ke bawah. Hablur tersebut membawa bersama-sama cas negative

ke ruang bawah awan sementara titisan air yang ditiup ke ruang atas awan membawa cas

positif. Proses pergerakan titisan air dan hablur ais ini yang menyebabkan berlakunya kilat.

(a) iii. How does lightning strike can induce over voltages in power transmission

line?

Sambaran Langsung

Kilat akan menyambar pada objek yang paling tinggi pada satu-satu kawasan. Talian

penghantaran adalah terdedah kepada sambaran kilat secara langsung. Sambaran terus

merupakan proses discas secara terus antara awan dan menara/dawai pengalir. Bila kilat

menjalankan cas negative pada hujung bumi, objek bumi (talian dan menara) akan

mengaruh cas positif. Disebabkan oleh penebat ke pengalir talian penghantaran. Talian

penghantaran akan berlagak sebagai kapasitor bercas positif. Cas yang teraruh ini akan

menyebabkan voltan lampau dan akan menghasilkan sambaran kilat. Voltan teraruh akan

merambar sepanjang talian, separuh akan dipantul (reflect) kebahagian lain talia. Pantulan

voltan teraruh ini boleh menyebabkan lampau kilat pada menara.

Sambaran tak langsung

Sambaran tak langsung atau sambaran teraruh merupakan discas awan pada sekitaran

menyebabkan keupayaan pengalir meningkat. Cas negative pada bahagian dasar awan

menyebabkan cas positif pada pengalir talian penghantaran yang berhampiran. Gelombang

begerak (travelling wave) voltan dan arus pada arah depan dan belakang sepanjang talian

penghantaran.

(b).

Peranti-peranti perlindungan kilat bagi system kuasa:

i. Sela Pembuangan (Expulsion Gap)

satu peranti yang mempunyai sela bunga api dengan pelindap kejutan arka (arc

quenching) yang memadamkan arka arus bila selaa tersebut pecahtebat

disebabkan oleh voltan lampau. Pada keadaan voltan lampau, kedua-dua sela

akan pecahtebat seretak dimana arka yang terhasil didalam tiub disebabkan oleh

arus kilat akan mengewap sebahagian bahan gentian. Gas yang terhasil dari

campuran wap air dan gentian akan memadamkan arka dan laluan menjadi litar

buka.

ii. Tiub Perlindungan (Protector Tubes)

Disambung dibawah pengalir pada menara talian penghantaran. Prinsip dan

kendalian sama seperti sela pembuangan. Bila voltan lampau berlaku, sela akan

pecahtebat dan arus akan dihadkan oleh rintangannya dan rintangan kaki

menara. Voltan lampau pada talian dikurangkan ke nilai kejatuhan voltan

melintangi tiub perlindungan.

iii. Penangkap Pusuan (surge arrester)

Peranti yang digunakan pada pencawang dan penghujung talian untuk mendiscas

voltan lampau kilat dan luruan penyuisan. Disambung secara selari dengan

peralatan yang dilindungi pada hujung talian atau tempat paling hampir dengan

pencawang. Ia akan mengurangkan voltan lampau ubahtika ke tahap bersamaan

dengan had penebatan peralatan. Elemen penghad arus adalah blok logam

oksida (metal oxide) yang mempunyai cirri-ciri rintangan yang tidak linear. Bila

arrester dipicu, blok logam oksida akan berada pada nilai rintangan yang rendah

untuk meghadkan arus. Bila arus luruan berkurangan, blok akan kembali ke nilai

rintangan tinggi.

(c).

Voltan lampau pensuisan merupakan voltan lampau yang terjana dalam system talian itu

sendiri. Voltan lampau tersebut boleh meningkat magnitudnya sehingga 6 kali ganda dari

voltan normal. Gelombang dedenyut pensuisan mempunyai tempoh masa yang lebih lama

dibandingkan dengan dedenyut kilat dan ia member kesan yang lebi teruk dari kilat.

Dedenyut pensuisan boleh terjadi disebabkan oleh penutupan dan pemutusan litar elektrik

menggunakan perkakasan suis (switchgear) dalam system kuasa yang mempunyai

kapasitan dan induktan yang tinggi. Dalam operasi pemutusan litar pusuan pensuisan

dengan kadar kenaikan voltan yang tinggi akan menyebabkan restriking berulangan dan

boleh merosakkan alat pemutus litar. Dedenyut pensuisan mengandungi frekuensi yang

tinggi serta pantulan gelombang dalam system.

Ciri-ciri pusuan pensuisan adalah berbeza-beza dan bergantung kepada punca asal

berlakunya dedenyut tersebut. Punca-punca berlakunya pusuan adalah :

‘De-energizing’ pada talian penghantaran, kabel shunt kapasitor

Pemutusan penyambungan transformer tanpa beban dan reactor

‘Energization’ atau penutupan(reclosing) talian dan beban reaktif.

Penutupan beban secara tiba-tiba.

‘Clearence’ kerosakan dan litar pintas

Fenomena salun seperti ferro-resonance, arching grounds

Rupabentuk gelombang dedenyut pensuisan adalah seperti dalam rajah dibawah.

Kawalan voltan lampau disebabkan oleh pensuisan boleh dilakukan dengan;

‘Energization’ bagi talian penghantaran dalam satu atau lebih langkah dengan

menyambungkan perintang dan mengeluarkannya semula selepas selesai proses

tersebut.

Kawalan fasa semasa penutupan pemutus litar

Pembuangan cas yang terperangkap sebelum pembukaan pemutus litar

Penggunaan reactor pirau (shunt reactor)

Pengurangan pusuan pensuisan menggunakan penangkap pusuan yang

bersesuaian.

Q12. (a) There are several different types of discharges in and around the

thundercloud. 90 percent of the discharges are cloud to ground lightning

discharge. A lightning strike basically involves attachment of downward leader

and the upward streamer originating from the ground or grounded structures.

Sometimes depending on the height of the building structures, lightning can

also triggered by an upward leader being attached to a downward stepped

leader coming from the cloud base. Based on these brief description of

mechanism of lightning flash, describe the basic of the Franklin Rod method of

lightning protection.

(3

marks)

(b) Fig. 12 (b) shows the damage of a building related to lightning strike. It also

shows the building is protected with a Franklin rod which is used as the basis for

lightning protection standard like IEC 62305 and others. Provide two reasons for

the lightning protection rod failure to provide protection to the building against

lightning strike.

Fig 12 (b) Lightning struck the sharp edge of a building top where the Franklin Rod is overlooking

(c) Figure 12 (c) shows the simulation of lightning strike a tower.

Figure 12 (c) Tilted Transmission Tower

(5

marks)

Based on Fig. 12 (c), shows that:

u (t )= ZgZt(Zg+2 Zt )

x i( t)

When Zg is the equivalent surge impedance of the ground wire

Zt is the surge impedance of the tower

u(t) is the impulse surge function

∆T is time of surge propagation from tower top to the cross-arm

i(t) is the current wave function

TT is time of surge propagation from tower top to the tower footing

TA is time of surge propagation from tower cross-arm to the tower

footing

Rt is tower footing resistance

Alpha(α) is the co efficient of reflection

Beta(β) is the co efficient of reflection on the tower top side

(2

marks)

(d) Fig. 12 (d) shows a tower with a 3 circuit ground wire system of Zg, 200

ohm(single circuit).The tower 30 meters high with surge impedance of 300 ohm.

The lightning current rises linearly to a peak of 60 kA in 2µs before commencing to

decline. Compute the potential UTT at the top of tower 0.21 microsecond after the

lightning strikes the tower top. Assume the lightning channel impedance is

i(t)

Time (microsec)

-αT β

i(t)

∆T

TT

UTT

ZG

RGUTA

UR

TA

infinity. The speed of lightning surge in the tower is 2.98x 108 meter per second.

The tower footing resistance is 10 ohm.

Fig.12 (d) A typical transmission line tower (10 marks)

Solution:

(a) Describe the basis of Franklin Rod Method of Lightning Protection

Based on research, electrode with sharp edges will experience high electrical stress

on its tip due to electric field intensification when high voltage is applied. Before

downward streamer is within striking distance from a facility, an upward opposite

streamer will be launched from the facility as well as the Franklin Rod. A strike will

happen when both of these streamers attach together and cause lightning return

stroke. In another word, the Franklin Rod is the sacrificial point of lightning strike. The

current is the discharged to the ground terminal via the down conductor.

(b) Failure of Franklin Rod to function

i) Several upward streamers are produced at the approach of highly charged

downward streamers from the cloud base. The competitive upward streamers

attachment with cloud downward streamers can result in the streamers

Tower Footing Resistance R is 10 Ohm

Tower Footing

Tower Cross Arm

Ground Wires

emerging from the building edges to overcome the streamers emerging from

the Franklin Rod.

ii) The lightning leader is of low current magnitude causes it to strike the building

from the side of the building. So the edge of the building is seen on the most

probable point of lightning strike.

(c)

At the point of strike,

Zequivalent (Zeq )=Zg/¿ Zt /¿Zg

¿

Zg2

x Zt

Zg2

x Zt

¿ ZgZtZg+2 Zt

The voltagesurge at the point of strike=I ( t ) x Zeq

¿ i (t ) x ZgZtZg+2Zt

¿ZgZt

Zg+2 Ztx i(t)

(d)

i(t)I(t)

Zg

Zg

Zt

δ

UTT60kA

2µs

Zt=300Ω

i(t)

-αT

β

Zg=200Ω

30

10Ω

Zg = 200Ω , Zt = 300Ω, R = 10Ω, Zl = infiniti

Zeq¿ZgZt

Zg+6 Zt

¿ 200 x300200+1800

¿30Ω

V surge ¿30Ω x60kV

¿1800kV

¿1.8 MV

Transit time from the tower top to the ground

∆t¿30

2.98x 108=10.07 x10−8

= 0.1 µsec

αT¿10−30010+300

=−0.935

-αT¿0.935=δ

β¿Zg /¿ Zg/¿ Zg/¿Zg /¿Zg /¿ Zg−Zt

ZTT+Zt

¿

Zg6−Zt

Zg6+Zt

¿

2006−300

2006+300

¿ 33.3−30033.3+300

=−0.80

ZTT

Zg

δβUt

βUt

δUt

Ut

Utt

3∆t

2∆t

∆t

UTT = Ut + αβUt

= U(t) + Ut(t-2∆t)(δ + β)

= Ut(t) + Ut(t - 2∆t)(δ + β)

= Ut(t) – 1.74 Ut(t-2∆t)

Resultant

-2p

-1p

1p

1.74 Ut(t-2∆t)

2µs

Ut(t)

Q13. (a) State the following :

i) Lightning strike mechanism

ii) Lightning strike parameter and characteristics

iii) Direct strike

iv) Indirect strike

v) Surge arrester ( 15 marks )

b) A 300m length of overhead earth conductor which is laid between two towers

is strike by a lightning. The lightning struck to a point of a distance of one third of

the length from one the tower. The surge earth conductor and tower impedances

were 300ohm and 100ohm respectively. The lightning strike carries 25 kA current

within 1.5µs and the surge impedance is about 1200ohm.Determine the surge

voltage at the tower top after the strike.

( 5 marks )

Solution:

i) Lightning strike mechanism

Apabila keamatan medan elektrik pada awan melebihi kekuatan dielektrik udara

yang terion yang lembap(~10kV/cm).Satu penjurus elektrik bergerak ke arah bumi

dengan kelajuan 1/10 dari kelajuan cahaya.Walaubagaimana pun, pergerakan

penjurus itu akan berhenti(selepas 50m) dan memancarkan kilat atau cahaya yang

terang.

Selepas lebih kurang 100µs, penjurus muncul kembali dan mengulangi proses tadi

untuk beberapa kali.Oleh kerana ia memerlukan beberapa penjurus pandu untuk

sampai ke bumi, maka ia dinamakan penjurus pandu bertangga(stepped

leader).Jumlah masa yang diperlukan bagi pandu bertangga untuk sampai ke bumi

adalah lebih kurang 20ms.

-2p

Sambaran kilat dan discas elektrik yang disebabkan oleh kilat diterangkan melalui

teori penjurus untuk discas bunga api di antara dua elektrod pada jarak yang panjang

dan medan tak sekata.Kilat mengandungi beberapa discas bermula dengan discas

pandu dan berakhir dengan discas kembali atau discas utama.Penjurus kembali

terbentuk apabila penjurus pandu telah mencecah bumi.Semasa pergerakan

penjurus pandu, cas positif bertumpuk pada hujung penjurus.Apabila mencecah atau

menghampiri bumi, keamatan medan elektrik pada bumi cukup besar untuk

menghasilkan penjurus kembali.Oleh itu cas positif tadi berpatah balik ke awan untuk

meneutralkan cas negative awan dan oleh itu arus yang besar mengalir dalam laluan

tersebut.Magnitud arus adalah dari 1000 ke 250 000 A.Kelajuan sambaran pandu

pemula ialah 1.5x107 cm/s.Sambaran seterusnya 108cm/s dan penjurus kembali

1.5x109 ke 1.5x1010 cm/s(~0.05 ke 0.5 kelajuan cahaya).

Tempoh sambaran utama atau sambaran kembali ialah 100µs atau lebih.Setelah

sambaran kembali selesai, arus dalam julat yang lebih kecil (100 ke 1000 A) mungkin

terus mengalir untuk selama lebih kurang 20ms.Pengaliran arus ini akan

menyebabkan wujudnya sambaran-sambaran ulangan.Walau bagaimana pun,

sambaran ulangan ini bergerak pada halaju yang lebih kecil(~1% kelajuan cahaya)

dan tidak bercabang.Sambaran balik atau sambaran utama untuk sambaran ulangan

ini juga mempunyai arus yang lebih kecil.Jarak masa diantara sambaran ulangan

ialah 0.6 hingga 500ms dengan purata 30ms.Jumlah masa kesemua sambaran

berbilang ini mungkin mencecah lebih dari 1 saat.

Beberapa graf penting kilat ditunjukkan seperti dalam Rajah 1 hingga Rajah 8.

ii) Lightning strike parameter and characteristics

Diantara parameter penting kilat ialah

Amplitude arus dan agihan kebarangkaliannya.

Kadar naik arus dan agihan kebarangkaliannya.

Bentuk gelombang voltan.

Gelombang arus mempunyai bahagian yang mempunyai yang mempunyai

nilai yang kecil(beberapa ampere) tetapi untuk tempoh yang lama iaitu

beberapa milisaat.Ini boleh mengakibatkan kerosakan haba berlaku pada

peralatan kuasa.

iii) Direct strike

Masa ke puncak dan kadar naik juga merupakan cirri yang

penting.Kebanyakan arus sambaran kilat mempunyai kadar naik melebihi

7.5kA/µs manakala sebahagian (10%) yang lain pula melebihi 25 kA/µs.

Daripada pengukuran yang dibuat, sambaran kilat boleh mengakibatkan nilai

voltan pusuan setinggi 5000kV wujud pada talian penghantaran.Secara

purata, kebanyakan sambaran kilat menghasilkan voltan pusuan kurang dari

1000 Kv.Masa depan(front time) gelombang pusuan voltan biasanya di antara

2 hingga 10µs manakala masa ekor(tail time) pula dari 20 ke 100µs.Kadar

voltan pula ~1MV/µs.

iv) Indirect strike

Sambaran terus berlaku apabila proses discas secara terus berlaku diantara

awan dan menara atau dawai pengalir talian penghantaran.Manakala

sambaran teraruh berlaku disebabkan oleh proses discas diantara dawai

pengalir(yang mempunyai cas teraruh dari awan) dan bumi.

v) Surge arrester

Medan elektrik yang terhasil diantara awan dan bumi biasanya tidak

mendatangkan apa-apa kesan kepada dawai pengalir talian penghantaran

kerana ianya ditebat melalui penebat gantungan.Walau bagaimana pun jika

kecerunan medan adalah tinggi, kebocoran cas positif boleh berlaku dimana

cas tersebut mengalir dari menara ke dawai pengalir melalui permukaan

penebat gantungan.Proses ini mengambil masa beberapa ratus saat.Apabila

proses discas diantara awan dan sesuatu objek pada bumi berlaku, cas-cas

positif yang banyak masih lagi tertinggal pada dawai talian dan ini

mengakibatkan satu pemuat yang besar wujud diantara dawai talian dan

bumi.Oleh itu voltan lampau akan terhasil didalam dawai dan akhirnya satu

sambaran berlaku ke bumi.Oleh itu sambaran ini disebut sebagai sambaran

kilat teraruh(induced lightning stroke).

(b)

Litar setara pada pengalir bumi

Ia=IL xZg

Zg+Zg/¿ ZL

¿25k x300

(300+300 x 1500300+1500 )

= 13.636 kA

Litar setara diatas menara (TT)

n=30m

1/3 2/3

IaZg Zg

IL

ZL

Zg = 300Ω

Zt =100Ω

ZL = 1500Ω

IL = 25kA

IaZg

Zt

TTTTZgZg Ia

Zt

Z setaradiatasmenara=Zg /¿ Zt

¿ 300x 100300+100

¿75

Vpusuan diatas menara = Ia x 75

= 13.636 x 75 k

= 1022.7 kV

= 1.0227 MV

Q14. (a) Figure Q14(a) shows a schematic diagram of a titled transmission line

tower and an impulse current wave shape, i(t). Consider the tower top is struck

by the lightning current i(t).

Picture of Q14 (a)

Show that,

u(t )=Z gZ t /(Z g+2Z t) .i(t )

where,

Z g is the surge impedance of the ground wire.

Z t is the surge impedance of the tower.

u(t ) is the impulse surge function.

ΔT is time of surge propagation from tower top to the cross-arm.

i(t) is the current wave function.

T Tis time of surge propagation from tower top to the tower footing.

T A is time of surge propagation tower cross-arm to the tower footing.

RT is tower footing resistance.

(2 marks)

(a) Show whether the following equation is right or wrong (write the detailed

derivation in order to prove it).

UTT (t) ¿ u(t )– α T (1+β)¿ – (αT β)u(t – 4 TT ) +(α T β )2u¿

where

UTT ( t ) is the potential distribution on the top of tower.

UT A (t) is the potential distribution on the tower cross-arm.

Alpha (α ) is the coefficient of reflection.

Beta (β ) is the coefficient of reflection on the tower top side.

(4 marks)

(b) Figure 14 (b) shows two towers (1 and 2) of a transmission line, which are

joined by overhead ground wires. The line is struck by lightning at point y,

100m away from point 2. The form of the lightning current is,

I (t)=3.0 x 1010 t ,0<t<2.0µs

¿6.0 x10 4−3.0x 10 8 t ,2<t<101 µs

¿0 , t>101 µs

Each tower is 30 m tall and stands on a square base with a 8m side.

Zg=450Ω,R=80Ω, assume Z of lightning channel ¿1500Ω.

Travelling waves have the following velocities;

i. On the conductors, 2.98 x10 8m / s .

ii. On the towers, 2.68 x10 8m / s .

Calculate the first peak of voltage at tower 1 and the time when it occurs.

(13 marks)

Solution:

(a) u(t )=Zeq i(t )

Zeq=Z g /¿Z g /¿Z T

1Zeq

= 1Zg

+ 1Zg

+ 1ZT

1Zeq

=ZT+ZT+Z g

Z g ZT

Zeq=Zg ZT

2ZT+Zg

u(t )=Zg ZT

2ZT+Zg

. i(t)

Proved.

(b)

UT A(with time consideration).

¿M 1+M 2+M 3+M 4+M 5+M 6

¿u(t )+αu(t)+αβu(t)+α 2βu (t)+α 2 β2u(t )+α 3 β2u(t )

UT A (with time consideration).

¿u(t – ΔT )+αu [ t – (2T – ΔT )]+αβu [t – (2T +ΔT )]+α 2 βu[ t – (4T T – ΔT )]+¿

α 2 β 2u [t – (4T T +ΔT )]

Replace αwith −α T

UT A (t )=u(t – ΔT )– α T u[ t – (2T – ΔT )] – α T βu [t – (2T+ΔT )]+α T 2 βu [t – (4T T – ΔT )]+α T 2β 2u [ t – (4T T+ΔT )]

Without time consideration.

UT T=V 1+V 2+V 3+V 4

¿u(t )+αut (1+β)+α 2βu (t)(1+β )+α 3 β2u(t )(1+ β)

With time.

UT T=u(t )+α(1+β )u (t –2T T )+α 2 β(1+ β)u (t)(t – 4 T T )+α 3 β 2(1+β)u(t )(t –6T T )

¿u(t )+α(1+ β)[u(t – 2T T )+αβu (t – 4T T )+α 2 β2ut (t –6T T )]

Replace α by −α T;

¿u(t )– αT (1+β )[ut (t – 2T T )– α T βu(t – 4T T )+(αTβ)2u(t – 6T T )]

(c)

ZT=30 ln2(h¿¿2+r 2)

r2 ¿

ZT=30 ln2(30¿¿2+42)

42 ¿

ZT=30 ln [114.5 ]

¿142.2Ω

Zg/¿ZT=480 X 142.2480+142.2

Zg/¿ZT=480 X 142.2

622.2

ZT=109.7

α 1=Zg /¿ ZT−Zg

ZT+Zg

α 1=109.7−450142.2+450

α 1=−0.6

β1=1+α=0.4

αR=R−ZT

R+ZT

αR=80−142.280+142.2

αR=−0.28

α T=Zg/¿ Zg−ZT

Zg/¿ Zg+ZT

α T=225−142.2225+142.2

α T=82.8

367.2

α T=0.225

βT=1+αT=1+0.225=1.225

Therefore, the equation UTA as provided by the equation is wrong and UTT equation is right.

Bewley Lattice Presentation.

Transit Time: From the point of strike on ground wire to the tower 1.

¿ 250−100

2.98 X 108

¿ 150

2.98 X 108

¿50.3 X 10−8

¿0.5 µsec

Transit Time from tower top to tower footing

¿ 30

2.68 X 108

¿11.19 X 10−8

¿0.11 µsec

If ΔT = 0.11 µsec, so 0.5 µsec ¿0.50.11

= 4.5 ΔT

Voltage at the top of tower 1

¿V 1+V 2+V 3+…

¿ β1u( t)'+βT αR β1u(t )' '+αR2 βT β1 αT u (t) ' ' '

¿0.4u (t) '+(1.225)(−0.28)(0.4)u(t) ' '+(−0.280)2(0.4)(1.225)(−0.28)u(t )' ' '

u(t )=I (t)Z eq=30 X 225

¿6.75 MV=1 p .u .

(without Z lightning channel)

u(t )’=I (t)Z eq /¿Zlightningchannel

¿30 X 195.7

¿5.87 MV .

Graphical Representation.

CHAPTER 4

Question 1:

a) What is the basic difference between self-restoring and non self-restoring insulation?

b) Briefly explain, with the aid of suitable diagrams, the statistical method of insulation

coordination.

c) In a laboratory, switching impulses are applied to a post insulator in order to

determine the basic switching impulse insulation level (BSL) of the post insulator.

The results of the test are shown in Table Q1. These test results are then plotted on

a linear graph paper as in Figure Q1.

i) Determine the critical flashover voltage (CFO)

ii) Calculate the BSL using statistical equation

iii) Show in the graph of Figure Q1 the value of BSL, and what is the percentage of

flashover at this condition?

Table Q1

Applied crest voltage, kV

No. of "shots"

No. of "flashovers

900 100 21000 40 201050 40 331075 100 93960 40 7980 40 16960 40 10

850 900 950 1000 1050 11000

20

40

60

80

100

Applied crest voltage, kV

No.

of "

flash

over

s

Figure Q1

Answer Q1:

a) Self-Restoring (SR) Insulation

- Insulation that completely recovers insulating properties after a disruptive

discharge (flashover) caused by the application of a voltage is called self-

restoring insulation.

- This type of insulation is generally external insulation.

Non Self-Restoring (NSR) Insulation

- This is the opposite of self-restoring insulators, insulation that loses insulating

properties or does not recover completely after a disruptive discharge caused

by the application of a voltage.

- This type of insulation is generally internal insulation.

b) The aim for statistical method is to quantify the risk of failure of insulation through

numerical analysis of the statistical nature of the overvoltage magnitudes and of

electrical withstand strength of insulation.

In a statistical study, what has to be known is not the highest overvoltage possible,

but the statistical distribution of overvoltage.

The risk of failure of the insulation is dependent on the integral of the product of the

overvoltage density function fo(V) and the probability of insulation failure P (V). Thus

the risk of flashover (R) per switching operation is equal to the area under the curve.

R=∫0

∞

f o (V ) .P .dV

Since a suitable insulation cannot be found such that the withstand distribution does

not overlap with the overvoltage distribution, in the statistical method of analysis, the

insulation is selected such that the 2 % overvoltage probability coincides with the 90

% withstand probability as shown.

c)

i) CFO is defined as a voltage level at the condition of the insulation results in a

50 % probability of flashover, i.e, half the impulse flashover. Therefore, from

figure Q1 we can determine the CFO = 1000 kV.

ii) BSL=CFO(1−1.28σ

CFO )=1000(1−1.2850

1000 ) = 936

Therefore, BSL = 936 kV

Where σ is the voltage difference between the 16 % and 50 % points or

between the 50 % and 84 % points. From the Figure Q1, σ = 50 kV.

iii)

850 900 950 1000 1050 11000

20

40

60

80

100

Applied crest voltage, kV

No.

of "

flash

over

s

Therefore, the percentage of flashover at BSL is 10 %

10%

936

Question 2:

a) Fig. Q4 (a) shown an insulation coordination practice using gaps/arcing horns. Briefly

describe what you understand from the diagram.

Fig. Q2(a) Coordination using gaps/arcing horns

b) Fig. Q2(b) shown an evaluation of risk factor in an insulation coordination practice using

the statistical technique. Briefly describe what you understand from the diagram as Fo

(V) and P(V) move to Either direction (left and right).

c) A 500 kV steep fronted wave (rate of rise 1000kV/µs) reaches a transformer of surge

impedance 1500 Ὠ through a line surge impedance 500 Ὠ and protected by a lightning

Transformer insulation

1.0m line gap/arcing horns

0.66 mm co-ord gap

arrester with a protective level of 700 kV, 60 m from the transformer. sketch the voltage

waveforms at the arrester location. Determine the time at which the arrester operates.

Assume all waves travel at 3.0 x 108 m/s

Fig.Q4 (b) Evaluation of risk factor

Answer Q2:

a) Consider the typical co-ordination of a HV transmission line between the transformer

insulation. A line gap ( across an insulator string) and a co- coordinating gap ( across the

transformer bushing), in a rural distribution transformer, a lightning arrester may not be

used on account of the high cost and a coordinating gap mounted on the transformer

bushing maya be the main surge limiting device. In co-coordinating the system under

consideration, we have to ensure that the equipment used is protected, and that

inadvertent interruptions are kept to a minimum. The co-coordinating gap must be chosen

so as to provide protection of the transformer under all conditions (slow as well as fast

front waves). However, the line gaps protecting the line insulation can be set to a higher

characteristic to reduce unnecessary interruptions. A typical set of characteristics for

Lower overvoltage

Overvoltage Distribution

fo (V)

P(V)Insulation withstand distribution

Higher insulation

fo (V) .P (V)

Risk of failure

90% withstand probability

2 % overvoltage probability

Voltage

Probability (%)

insulation coordination by conventional methods, in which lightning impulse voltages are

the main source of insulation failure, is shown in the above figure. For the higher system

voltages, the simple approach used above is inadequate. Also, economic considerations

dictate that insulation co-ordination be placed on a more scientific basis.

b) The aim of statistical methods is to quantify the risk of failure of insulation through

numerical analysis of the statistical nature of the overvoltage magnitudes and of

electrical withstand strength of insulation. The risk of failure of the insulation is

dependent on the integral of the product of the overvoltage density function fo(V) and

the probability of insulation failure P(V).Thus the risk of flashover per switching

operation is equal to the area under the curve ∫ fo (V) *P(V)*dV, as shown in the

diagram. Since we cannot find suitable insulation such that the withstand distribution not

overlap with the overvoltage distribution, in the statistical method of analysis, the

insulation selected such that the 2 % overvoltage probability coincides with the 90 %

withstand probability shown.

c)

500kVpeak α

Z0=500ὨD=60m β ZT=1500Ὠ

Arrester

If the separation is 50 m, the travel time in the line ( Between arrester and transformer)

τ =60

100

=0.2µs

Transimision and reflection coefficients

β =2 X 1500

1500+500

=1.5

α =1500−5001500+500

=0.5

Question 3:

a) A transformer has an impulse insulation level of 1175 kV and is to be operated with an

insulation margin of 15% under lightning impulse conditions. The transformer has a

surge impedance of 1600 and is connected to a transmission line having a surge

impedance of 400 . A short length of overhead earth wire is to be used for shielding

the line near the transformer from direct strikes. Beyond the shielded length, direct

strokes on the phase conductor can give rise to voltage waves of the form 1000e−0.05 tkV

(where t is expressed in µs).

If the corona distortion in the line is represented by the expression,

Δtx= 1

B [1−VoV ]

Where B = 110m/µs and Vo = 200kV, determine the minimum length of shielding wire

necessary in order that the transformer insulation will not fail due to lightning surges.

b) A 500kV steep fronted wave (rate of rise 1667kV/µs) reaches a transformer of surge

impedance 1600 through a line of surge impedance 400 and protected by a

lightning arrestor with a protective spark-over level 700 kV, 75 m from the transformer.

Sketch the voltage waveforms at the arrestor location. Determine the time at which the

arrestor operates. Sketch also the voltage waveforms at the transformer location.

Determine the maximum voltage at the transformer and the duration it appears across

the transformer. Assume the surge travels at 300 m/us in the line and no reflection is

considered at the arrester.

Answer Q3:

a)

1000e−0.05 t α

Z0 =400 Ω β ZT=1600Ω

β = 2 X 1600

1600+400

= 1.6For BIL of 1175 kV, and an insulation margin of 15%, the maximum permissible voltage

= 1175 X85

100

= 998.75 kVSince the voltage is increased by β = 1.6 times at the terminal equipment (transformer),

the maximum permissible incident voltage must be decreased by this factor, hence the

maximum permissible incident voltage is

= 998.75

1.6= 624.22 kVTherefore the shielding wire must reduce the surge to 624.22 kV (by virtue of the corona

distortion), that is

1000e−0.05 t=624.22

t = 9.425 µs

Hence Δt = 9.424 µs

The corona distortion is given by

Δtx= 1

B [1−VoV ] µs/m

Substituting

9.425x= 1

110 [1− 2006.24 .22 ] µs/m

X = 1525.53 m

Therefore the minimum length of shielding wire required = 1.526 km.

b)

500kV peak α

Z0 =400Ὠ ZL = 75m β ZT=1600Ὠ

Arrester

If the separation is 75m, the travel time in the line

τ = 75

300

= 0.25 µs

Transmission and reflection coefficients

β = 2 X 1600

1600+400

= 1.6

α = 16500−4001600+400

= 0.6

From the diagram, arrester operates at t = 0.7 µs

Then, the maximum voltage across the transformer = 800kV

So, the duration of this voltage across the transformer = 0.4 µs

0

Reflected surge

Time (µs)

(0.6 x 1667 = 1000 kv/µs)

Resultance voltage across arrestor

Max votage of TransformerArrestor operates at 700kv (spark over)

Arrestor voltage (kv)

200

400

600

800

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1ך ך

0

Transmitted surge

Time (µs)

(1.6 x 1667 = 2667 kv/us)

Transfomer operating time 0.95 µs

Arrestor operates at 0.7 us + 0.25 µs ( at tx )

Arrestor voltage (kv)

200

400

600

800

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1

Question 4:

a) What is the basic difference between self-restoring and non-self restoring insulation?

Describe a method for the implementation of insulation coordination to a power system

with self-restoring and non-self restoring insulation. In each case emphasis should to be

given to the description of Risk Failure and Critical flashover Voltage V50.

b) Describe the differences between Basic Lightning Impulse Insulation Level (BIL) and

Basic Switching Impulse Insulation Level (BSL) and explain how they affect in the design

of power system networks?

The following results are obtained for the determination of volt-time curve obtained for a

non-self restoring and self-restoring insulation in a multi-level test.

Table 4c (i) The volt-time curve characteristic of string insulator

Voltage level

(kV) 1300 1600 2280 2600 2800 2900 3000

No. of impulses

applied to the

insulator10 10 10 10 10 10 10

Time to flash

(μsec) 35 30 20 15 10 5 2

Table 4c(ii) The volt-time curve characteristic of a pair of arc-horn

Voltage level

(kV) 1200 1880 2800 3100 3300 3400 3500

No. of impulses

applied to the

insulator10 10 10 10 10 10 10

Time to flash

(μsec) 35 30 20 15 10 5 2

Answer the following questions:

(i) Based on the above volt-time curve of the Table 4c(i) and 4c(ii)

(ii) Can the arc-horn I protects the string insulator I against destructive

discharges?

(iii) If the flashover voltage of the arc-horn I is reduced by 20% while maintaining

same time to flashover, is the volt-time curve of arc-horn I suitable for insulation

coordination with the string insulator I? Give reasons.

If the volt-time curve from Table 4c(i) and 4c(ii) respectively is used for the line

insulation coordination, will there be a destructive discharge if a lightning surge with a

peak of 2000 kV at 1.5μsec and afterwards decrease linearly to zero at rate of 20 kV per

μsec, arrived at the string insulator point of connection with the conductor? If yes, where

will be the discharge takes place?

Answer Q4:

a) The basic difference between self restoring and non-self restoring insulation is:

Self-Restoring Insulation: Insulation that completely recovers insulating properties after

a disruptive discharge (flashover) caused by the application of a voltage is called self-

restoring insulation. This type of insulation is generally external insulation.

Non-Restoring Insulation: Insulation that losses insulating properties after a disruptive

discharge (flashover) caused by the application of a voltage. This type of insulation is

generally internal insulation.

Critical Flashover Voltage (CFO)

CFO is defined as a voltage level at the condition of the insulation results in a 50%

probability of flashover, i.e., half the impulses flashover.

Method of describing the risk of failure.

1. Over voltage distribution–Gaussian function.

2. Insulation breakdown probability–cumulative distribution

R=∫0

∞

Pb (Vk ) p ο(Vk )δu

Insulation strength characteristic for non

Self-restoring insulation

Insulation strength characteristic for

self restoring Insulation

BIL=CFO¿)

BSL=CFO(1−1.28σf

CFO )

b) For equipment rated at less than 300 kV, it is a statement of the Basic

Lightning Impulse Insulation Level (BIL) and the short duration power frequency

withstand voltage. For equipment rated at greater than 300 kV, it is a statement of the

Basic Switching impulse Insulation Level (BSL) and the power frequency withstand

voltage.

Spark gaps for surge protection

The selected gap spacing should no only be capable of withstanding the highest normal

power frequency voltage but should flash-over when over voltages occur, protecting the

equipment. However, this is not always possible due to the voltage-time characteristics

gaps and equipment having different shapes. Rod gaps are generally used as a form of

back up protection rather than the main form of protection.

Surge Diverters

Surge diverters (or lightning arrestors) generally consist of one or more spark gaps in

series, together with one or more non-linear resistors in series. Silicon Carbide (SiC) was

the material most often used in these nonlinear resistor surge diverters. However, Zinc

Oxide (ZnO) is being used in most modern day surge diverters on account of its superior

volt-ampere characteristic. In fact the ZnO arrestor is often used gapless, as its normal

follow current is negligibly small.

(i) No.

(ii) The new volt-time curve of the arc-horn I is suitable for the protection of the

string insulator I. This is true when the volt-time curve of arc-horn I lies lower

than the string insulator I.

(iii) Destructive discharge will take place. It will happen on the string insulator I.

Question 5:

a) Briefly discuss the implementation of insulation coordination for high voltage

transmission line system using statistical method?

b) Explain the differences between the insulation of a high voltage system and low voltage

system. Confine your explanation to a transmission line entering a high Voltage

substation substation whereas for the low voltage system confine to the AC mains point

entry in a house to the electrical and electronic equipment.

c) A test was conducted on a pair of arc-horn that was used for transmission line for the

Protection of string glass insulator against the lightning inducing effect on phase

conductor. This test is important because Malaysia is in the region of high isoceraunic

level where in every second there is not less that 100 flash to the ground. Before the

installation of the string insulator, lightning impulse tests are carried out in a high voltage

laboratory to determine the BIL and BSL. The results of BIL tests are shown in Table 5c

and the test set up is shown in Fig 6. Determine the BIL of the string insulator.

Table 5c: Results of lightning impulse test of a 12-unit string insulator where NO means

NO FLASHOVER and YES mean FLASHOVER

Kv\N

(injection)

1 2 3 4 5 6 7

700 NO NO NO NO NO NO NO

750 NO NO NO NO NO NO NO

800 NO YES NO NO NO NO NO

820 YES YES YES NO NO NO YES

840 NO YES YES YES YES YES NO

850 YES YES YES YES YES YES NO

860 YES YES YES YES YES NO YES

870 YES YES NO YES YES YES YES

880 YES YES YES YES YES YES YES

890 YES YES YES YES YES YES YES

S Rf

ClRtCi

Fig 5 The schematic diagram of an impulse generator where Rf is the front resistor, Rt is

the tail resistance, C1 is the load capacitance and Ci is generator capacitor.

Answers Q5:

a) Statistical Method

This is based on knowledge of the statistics of both overvoltage occurrence and of

flashover probability. The design is based on an acceptable risk of flashover. If at a

voltage level v, the probability of failure is P(v) and the frequency of occurrence if surge

of that level is f(v) then the risk of function is defined as,

r (v )=P (v ) f (v)

The risk function is shown below,

For a given insulation design, the total risk of failure will be,

R∫0

∞

P (v ) f (v )dv

The risk is R is therefore determined by the area under the r(v) curve, as the insulation

is strengthened by using a large gap or layer insulation.

Figure 5

For a given insulation design the total risk of failure will be

R∫0

∞

P (v ) f (v )dv

The risk R is therefore determined by the meas under the r(v) curve as the insulation is

strengthened (e.g by using a layer gap or longer insulation string), so this risk at failure

shall reaches.

The effect of increasing, Vg on the risk of failures. It is therefore possible to press the

P(v) curve, we can then choose a reference probability and quote the voltage as a

statistical withstand voltage Vw.

The peak value of a switching or lightning impulse test voltage at which insulation

exhibits, under the specified conditions, a 90% probability of withstand has a 10%

probability of breakdown.

b) The system design engineer will therefore specify that the insulation must have a

withstand voltage of Vw, based on the acceptable risk criterion, and the insulation

design engineer will develop insulation suitable for what withstand level.

Line insulation coordination.

1. The tower strike distance or clearance between the phase conductive and the

grounded tower sides and upper tress.

2. The insulator string length.

3. The number and type of insulation.

4. The need for and type of supplemental lower grounding.

5. The phase to ground midiper clearance.

6. The phase strike distance or clearance.

7. The need for rating and location of line surge arrestable.

c) Insulation coordination in line voltage, installation and equipment. The classification of

expected overvoltages in 230/400V consume installation resulted in the following fault

categories. The worst case is considered is the transient overvoltages occurs at the

peak of the surge voltage.

Category J: Sites that are protected a transient overvoltage’s.

Category II: Sites at which external overvoltages are not expected.

Category III: sites of category II at which particular demands are made for reliability

and availability .

Category IV: sites at which external overvoltages as especial.

CFO = 826.0 kV

σ=27.5kV

Applied Crest Voltage, kV

Kv\ N( injection) Proability of breakdown (%)

700 0

750 0

800 1/7 ×100 = 143%

820 4/7 ×100 = 67.14%

840 5/7 ×100 = 71.4%

850 6/7 ×100 = 85.7%

860 6/7 ×100 = 85.7%

870 6/7 ×100 = 85.7%

880 7/7 ×100 = 100%

890 7/7 ×100 = 100%

BIL=CFO (1−1.28σf

CFO ) ¿826 (1−1.28

27.5826 )

¿826 (1−0.0426 )

¿826× 0.957

¿790.5kV

The BIL of the string insulator is 790.5 kV

650 700 750 800 850 900 9500.00%

20.00%

40.00%

60.00%

80.00%

100.00%

120.00%

0.00% 0.00%

14.30%

67.14%71.40%

85.70%85.70%

85.70%

100.00%100.00%

Proability of breakdown (%)

Kv\ N( injection)

Proa

bilit

y of

bre

akdo

wn

(%)

Question 6:

a) What is the basic difference between self-restoring and non-self-restoring insulation?

b) Describe the differences between Basic Lightning Impulse Insulation Level (BIL) and

Basic Switching Impulse Insulation Level (BSL) and explain how they affect in the

design of power system network?

Following results were obtained for a self-restoring insulation in a multi-level test.

Voltage level (kV)

200 210 220 230 240 250

No. of impulse applied

10 10 10 10 10 10

No. of Flashovers

0 1 5 9 10 10

c) Determine the following:

(i) 50% flashover

(ii) P U standard deviation

(iii) Statistical withstand voltage

(iv) Statistical flashover voltage

(v) BIL of this insulation.

Answer Q6:

a) Self-restoring insulation regains its dielectric integrity following the occurrence of a

breakdown. The examples are gas or liquid insulations. Non-self-restoring insulation

losses its insulating property after a breakdown or disruptive discharge and must be

replaced. Solid insulation, as in cables, transformers, and machines, belongs to this

category of insulation.

b) BIL: This is the reference insulation level which is expressed as peak impulse voltage

having a standard lightning impulse waveform of 1.2/50 μsec. It is determined by tests

made using impulses of the standard lightning impulse wave shape.

BSL: Similar to BIL but is defined for a standard switching impulse voltage of 250/2500

μsec waveform. It is generally specified for equipment with rated voltage at >= 300

kVrms.

The level of voltage supply from customer side to the generation side can be

categorized as

i) Low Voltage (LV)

ii) Medium Voltage (MV)

iii) High Voltage (HV)

iv) Extra High Voltage (EHV)

v) Ultra High Voltage (UHV)

These systems are affected by transient overvoltage generally due to external source

such as lightning phenomenon and internally generated is the switching activities. In

order system components can function as they should, they have to be design to

withstand against lightning related surges as well as switching activities. Related to

withstand on lightning surge BIL is to be referred and BSL for switching. System working

more than 300kV BSL is more significant than BIL.

(c)

i) 220kV

ii) P = 16%, V = 211.2kV

P = 84%, V = 228.1kV

Standard Deviation = V50 – V16 = 220 – 211.8 = 8.2kV

Standard Deviation = V84 – V50 = 228.1 – 220 = 8.1kV

Coefficient of variation = 8.1/220 = 0.037 pu

iii) Statistical withstand voltage

= (1 – 3 Coeff)V50

= (1 – 3*0.037)*220

= 195.6kV

iv) Statistical flashover voltage

= (1 + 3 Coeff)V50

= (1 + 3*0.037)*220

= 244.4Kv

200 210 220 230 240 250 2600

10

20

30

40

50

60

70

80

90

100

Breakdown Voltage (kV)

Perc

enta

ge (%

)

84%

16%

Question 7:

a) Discuss three (3) mechanisms of solid insulation breakdown.

Answer Q7:

1. Intrinsic Breakdown

When the high voltage applied for a long time.

Electrical power is determined by the intrinsic strength.

Depend on the presence of free electrons that move through the lattice of

solid.

These electrons produce a current flow.

2. Electromechanical Breakdown

Less rigid insulation material (rubber, PVC).

Electrostatic forces that act exceed the strength of solid mechanics.

Mechanical damage will occur.

3. Thermal Breakdown

Leakage current flow when the electric stress applied.

Solid temperature increased by the heating process.

Heat transferred to the insulation surroundings by conduction and radiation

process.

Limiting the maximum thickness of a solid.

4. Chemical And Electrochemical Breakdown

Chemical changes if the electrical stress is continuously react with air and

gas.

Reactions that occur - oxidation, hydrolysis, chemical reaction.

Can be reduced by checking the material carefully.

5. Treeing And Tracking Breakdown

Two effects when electric stress is long:

i. the presence of conductive path across the surface.

ii. effects of leakage current due to sparks through the path that

produces sparks.

During the process oftracking,the spread ofsparksin the form

ofbranchescalledtreeing.

6. Internal Discharge Breakdown

Cavitycontains airinthe solidinsulation.

Fieldin the cavity larger than the field in theinsulation.

Breakdown exists in the cavity.

Question 8 :

a) Explain what is meant by the B.I.L of a high voltage equipment.

Answer Q8:

a) For equipment rated at less than 300 kV, it is a statement of the Basic Lightning Impulse

Insulation Level (BIL) and the short duration power frequency withstand voltage.

Question 9 :

a) Describe in details with the appropriate diagrams, the methods of insulation coordination

for over voltages.

b) In transmission system, components whose failure would have very severe

consequences are often protected by protective devices. Describe the coordination of

protected insulation with regards to a high voltage transformer.

Answer Q9:

a) There are two methods of insulation coordination for overvoltages:

i) Conventional Method.

This is the method used when the probability of failure for a given overvoltage is

unknown, ie. With non-self restoring insulation. The known probability distribution of

overvoltage amplitudes is used only to determine the maximum surge which is liable

to occur. To this is added a safety margin to allow for the unknown flashover

probability distribution. The resultant voltage level is specified as a ‘withstand’

voltage, V w which the insulation must be able to hold off.

ii) Statistical Method

This is based on knowledge of the statistics of flashover probability. The design is

then based on an acceptable risk of flashover. If at a voltage level V, the probability

of failure is P(V) and the frequencies of occurrence or surges of that level is f(V) then

the risk function is defined as, r(V) = P(V).f(V).

The risk function is shown below,

For a given insulation design the total risk of failure will be:

∫0

∞

P (V ) .F (V )dV

The risk is therefore determined by the area under the r(V) curve. As the insulation is

strengthened by using larger gap or longer insulation strength, the risk of failure

diminishes as in the diagram below.

For a chosen risk of failure it is therefore possible to position the curve. We can then

choose a reference probability and quote the corresponding voltage as a statistical

withstands voltageV w. In practice the reference probability is normally taken as P(V)

= 0.1, ie. V w is the voltage at which the insulation has a 90% probability of withstand.

The system design engineer will therefore specify that the insulation must have a

withstand of V w based on the acceptable risk criterion, and the insulation design

engineer will develop insulation suitable for that withstand level.

b) In transmission systems, component whose failure would have very severe

consequences are often protected by devices such as surge diverters or spark gaps.

These devices are designed to break down in preference to the insulation of the

equipment to be protected such as high-voltage transformer. The coordination of the

protective devices insulation with regards to the protected equipment can be made by

using the conventional and statistical methods. In applying the conventional method, the

maximum allowable overvoltage is coordinated with the protective level of the device, ie.

The protective device is set to flashover if V w for the more expensive equipments is

exceeded.

In applying the statistical method the flashover probabilities Pp(V) and P(V) of the

protective device and the equipment which it protects are used to calculate a probability

P*(V) that the main insulation will flash over in spite of the protective device. The risk is

then found form R*=∫0

∞

P∗(V ) .F (V )dV

If the protective device is a spark gap, it itself will be dimensioned by normal methods to

give a risk of flashover which although greater than other devices, is still limited. If surge

diverters are used there is no reason to limit the risk of flashover as surge diverters

automatically restore the system voltage due to their non-liner resistance characteristics

without damage. However if a surge diverter is triggered by a high-voltage switching

surge or lightning transient.

Question 10:

(a) Describe the secondary processes which can lead to an electron avalanche and how

these processes may be identified. Show that the discharge current in a multi avalanche

Townsend process in a non-attaching gas is given by

I=I 0 eαd

1−γ [e¿¿(αd−1)]¿

Where I 0 – initial voltage

Α – first Townsend ionization coefficient

γ - Second Townsend ionization coefficient

d – Gap distance in cm

(b) Measurement of breakdown voltages in a uniform field spark gap in air gave the result as

shown in the Table Q10

Table Q10

Gap spacing (mm) Pressure (Bar) Temperature (C ˚)Breakdown voltage,

V s(kV )

2.5 1.03 30 0.91

27 1.18 15 88.38

Use the expression derived from Paschen’s law V s=A (pd )+B (√ pd ) determine

i) The relative air densityρ, referred to standard atmospheric conditions of 1013.25

mbar and 20˚C.

ii) The value of constant A and B.

iii) The breakdown voltage of a 3cm gap spacing at pressure of 3000 mbar and a

temperature of 20˚C.

Answer Q10:

a) The electrical breakdown of a gas is brought about by various processes of ionization.

These gas processes involving the oscillation of electrons, ions and photons with gas

molecules and electrode processes which take place at or near the electrode surface.

When a pair of electrodes is immersed in a gas and a voltage applied across them the

current voltage characteristics of figure below is observed

At low voltage the observe current is due to collection of free carriers in the gap and as

the voltage is increased a level is reached at which the free electrons gain enough

energy to ionize. Electrons produced may cause further ionization so that an electron

avalanche is generated. Ionization is the process by which an electron is removed from

an atom, leaving the atom with a net positive charge. The probability of ionization due to

the electron will depend on the number of collision made per unit distance with

coefficient referred as the primary ionization coefficient which is the number of ionization

collisions per unit electron per cm travel. With the primary ionization alone the

discharged is not self sustaining. If the source of initial electrons is removed the current,

the current I falls to zero. This suggests that processes other than simple α - process are

occurring. This additional current is produced by secondary emission processes. A

secondary ionization coefficient γ is defined as the number of secondary electrons

produced at the cathode produced in the gap

These processes for secondary electrons liberation can be identified by:

i) Positive ion γ i- ions do not have enough energy to ionize gas molecules directly but

may release electrons on colliding with the cathode surface

I

Breakdown

Non-Self-sustain

Self-sustain

VVs

Ionization

Charge collection

Io

INon-Self-sustain S

elf-sustain

Vs

Ioniz

Charge

Io

I

ii) Photons γ p- a proportion of the collision in the gap cause excitation of neutral gas

molecules which in return to the ground state may be emit photons which release

electrons by photoemission.

iii) Metastablesγ p- metastables molecules may diffuse to the cathode and release

electrons

One or more secondary mechanism may exist giving a total secondary effect

described

γ=γ p+γ p+γ p

Let n0 = number of initials electrons at cathode

n0' = The number of secondaries

n0' '= the total emission including secondaries

i.e n0' '=n0+n0 '

At x ,n ( x )n=nο e ^ (αx)

The total number of new electros produced, n (d )=nο ( e ^ (αd-1)

dx

n0

Volts

Distanceo d

- +

If γ electrons are produced at the cathode per ionizing collision in the gap, then

nο '=γnο ( e ^ ( αd - 1)

Thus, nο = nο + γnο (e(αd−1) )¿

nο = nο over 1 - γ ( e ^ ( αd - 1) ^ <?> )¿

∴n (d )=nο e ^ ( αd ) = nο e ^ ( αd ) over 1 - γ ( e ^ left (αd - 1 right ) )

Under steady state conditions, I=Iοe(αd )

1−γ (e (αd−1))

Gap spacing (mm) Pressure (Bar) Temperature (C ˚)Breakdown voltage,

V s(kV )

2.5 1.034 30 0.91

27 1.180 15 88.38

d1=0.25cm=10mBar , t=30 ˚C ,V s=0.90kW

f 20=P

1013−273+20

273+ t

¿ 10341013

.293288

=1.19

Question 11:

a) What is the basic different between self-restoring and non-self-restoring insulation?

b) Describe the differences between Basic Lightning Impulse Insulation Level (BIL) and

Basic Switching Impulse Insulation Level (BSL)

c) Name the types of BIL that are mentioned in the standards?

(How many type of Basic Lightning Impulse Insulation Level and name them?)

d) Name the types of BSL that are mentioned in the standards?

(How many type of Basic Switching Impulse Insulation Level and name them?)

e) In relation to BIL and BSL, standards provide two types of tests to determine the best

method to obtain BIL and BSL.

(i) Explain those methods as provided by the standards document.

(ii) Discuss which method is the best by not neglecting to discuss points for instance

probability of passing test, probability of flashover per-impulse, manufacturer’s risk

and ideal test and other relevant factors.

Answer Q11:

a) Self-Restoring Insulation : Insulation that completely recovers insulating properties after

a disruptive discharge (flashover) caused by the application of a voltage.

Non Self -Restoring Insulation : Insulation that losses insulating properties after a

disruptive discharge (flashover) caused by the application of a voltage.

b) For equipment rated at less than 300 kV, it is a statement of the Basic Lightning Impulse

Insulation Level (BIL) and the short duration power frequency withstand voltage. For

equipment rated at greater than 300 kV, it is a statement of the Basic Switching impulse

Insulation Level (BSL) and the power frequency withstand voltage.

c) 2 type, statistical and conventional

d) 2 type, statistical and conventional

e)

i) 1) The n/m test : m impulses are applied. The test is passed if no more than n result

in flashover.

2) The n + m test : n impulses are applied. If none result in flashover, the test is

passed. If there are two or more flashover, the test is failed. If only one flashover

occurs, m additional impulses are applied and the test is passed if none of these

result in a flashover.

ii) These alternate tests can be analysed statistically to determine their characteristic.

That is a plot is constructed of the probability of passing the test as a function of the

actual but unknown probability of flashover per application of a single impulse. The

characteristics for the above three tests are shown in Fig 1.1

Question 12:

a) Discuss three (3) mechanisms of solid insulation breakdown

b) Show that the breakdown criterion in gas according to Paschen’s Law is given

by :

g¿¿¿

where, d s- gap distance at sparkover voltage

p-pressure

V s - sparkover voltage

f&g-different functions

c) In nitrogen gas, the static breakdown voltage V s of a uniform field gap may be

expressed as,

V s=Apd+B √ pd

Where A and B are constants, p is the gas pressure in torr referred to a temperature of

20 ° and d is the gap legth in cm.

A 1 cm uniform field gap in nitrogen at 760 torr and 25 ° is found to breakdown of 33.3

kV. The pressure is then reduced and after a period of stabilization, the temperature and

pressure are measured as 30 °C and 500 torr respectively. The breakdown voltage is

found to be reduce to 21.9 kV. If the pressure is further reduced to 350 torr while the

temperature of the closed vessel is raised to 60 ° and the gap distance is increased to 2

cm, determine the breakdown voltage.

Answer Q12:

a) Three (3) mechanisms of solid insulation breakdown :-

i) Intrinsic/Ionic breakdown

ii) Electromechanical breakdown

iii) Thermal breakdown

Intrinsic/Ionic breakdown.

Occurs at a very short duration of HV applied (10-8 s).

Depends on the presence of free electron, which capable of migration thru the lattice

of the dielectric.

G CREATIVE AND INNOVATIVE MIND

Electromechanical breakdown.

Due to electrostatic compressive forces that exceed mechanical compressive

strength.

The highest apparent electric stress before breakdown, if the thickness of specimen

do

is compressed to d under applied HV;

Mechanical instability occurs when d/do = 0.6, and Y : Young’s modulus and

depends on mechanical stress.

Thermal breakdown

Conduction current flows – heats up the specimen and the temperature rises.

Heat generated transfers to the surrounding medium by conduction and radiation.

Breakdown occurs when heat generated > heat dissipated.

Heat generated is proportional to the frequency – thermal breakdown is more serious

at high frequency.

Thermal breakdown stresses are lower under a.c. condition then d.c.

b) By neglectingattachment, breakdown criterion;

γ (ead−1 )=1…………… (1)

Since (Paschen’s Law) ,αp=f ( E

p) and γ=g (E

p) where f &g signify different function.

At breakdown, α ds=pd f (E s

p)………………. (2)

= pds f (V s

pds

)…….……….... (3)

Substitute (2) &(3) into (1) gives

g( V s

pds)exp[ pd s . f ( V s

pds)]−1=1proved

c) V s1=Ap1 d1+B √ ( p1 d1) p1=760 torr ,t 1=25 ˚ C ,d1=1cm ,V s1=33.3kV

Corrected pressure to standard temperature of 20 ˚ C,

p1=760(273+20)

273+25=747.25 torr √ p1 d1=27.3

33.3=A (747.25 ) (1 )+B (27.43 )(1)

¿747.25 A+27.34 B…….(1)

V s 2=Ap2 d2+B √( p2 d2) p2=500 torr ,t 2=30 ˚ C ,d2=1cm ,V s 2=21.9kV ,

Corrected pressure to standard temperature of 20˚C

p1=500(273+20)

273+30=483.50 torr ,√ p2d2=21.99

21.9=A (483.50 ) (1 )+B (21.99 )(1)

¿483.50 A+21.99B……………..(2)

(1 )×21.99 :732.27=16432.03 A+601.21B………(3)

(2 )×27.34 :598.75=13218.89 A+601.21B…...….(4)

(3)-(4):13352=3213.14 A A=0.042∧B=0.072

V s 3=Ap3 d3+B √( p3d3) p3=350 torr , t 3=60 ˚C ,d3=2cm ,V s 3=?

Corrected pressure to standard temperature of 20 ˚ C,

P3=350(273+20)

273+60=307.96 torr ,√ p3 d3=24.82

V s3=(307.96 ) (2 ) (0.042 )+ (24.82 )(0.072)

=27.66kV

CHAPTER 5

INSULATION DIAGNOSTIC & PARTIAL DISCHARGES

(QUESTION AND ANSWER)

Question 1

a) The dissipation factor or loss tangent is an indication to determine the performance of

insulating property of insulators. The most common method used to determine the lost

tangent is by using a.c bridges.

i) Sketch the circuit diagram of a high voltage Schering bridge for the

measurement of loss tangent (tan δ).

ii) Derive the expression for tan δ of the unknown series model of tge tested

sample when the high voltage standard capacitor used in the Schering Bridge is

a loss-free type.

Answer

i)

C : Capacitance of the sample

r : Resistance of the sample

C2 : Standard capacitor

R3 , R4 and C4 : Variable components of Schering bridge

ii)

Z1=r− jωC

; Z2=1

jωC2

; Z3=R3 ;Z4=R4

1+ jωC4 R4

AC Supply

Sample

C4R4R3

r

C2

C

At balance condition ; Z1 Z 4=Z2 Z3 ;

Z1=Z2 Z3

Z4

=[ R3

jωC2] [ 1+ jω C4 R4

R4]= − j R3

ωC2 R4(1+ jωC4 R4 )

Z1=r− jωC=

C4 R3

C2

− jR3

ωC2R4

By equating the real and imaginary parts of both side;

Therefore;

r=C4 R3

C2

;∧C=C2 R4

R3

For series model, tan δ = 𝝎Cr,

tan δ=ω(C2 R4

R3)(C4 R3

C2)=ωC4 R4

b) Breakdown in solid or liquid dielectrics arise from the action of the electrical discharges

in internal gaseous cavitties. Draw an equivalent circuit for such an arrangement and

derive the expression for energy dissipated in the cavity in one discharge.

Answer

Void in solid dielectric material Equivalent circuit of void in the structure

a

b2

b1

c tVaT

void

Cc

Cb

CaVa

Cb=ε0 εr A1

T−tC c=

ϵ 0 A1

t

Energy dissipated in the cavity in one discharge;

¿12 (C c+

CaCb

Ca+Cb )V c ²

Where

V c=Cb

Cb+Cc

∙V a

Therefore

¿12 (C c+

CaCb

Ca+Cb )( Cb

Cb+C c ) ² V a ²

¿12 (Cb²

Cc )V a²

c) A solid dielectric specimen of 5cm diameter and 10mm thickness has dielectric

constant of 3. It has an embedded air filled void of 2.5mm diameter and 1mm

thickness. It is subjected to a voltage of 100k Vrms. Find the voltage at which an

internal discharge in the void can occur. The breakdown strength of air is 3kV/mm. Also

determine the charge value each time there is discharge inside the void and what will

be the value of charge as measured on the detector?

Answer

The inception voltage of the stressed void Vi is given by:

V i=Ecb t(1+ 1εr(d /t−1))

Where

Ecb - breakdown stress of the void = 3 kV/mm

d - thickness of the specimen = 10mm

t - thickness of the void = 1mm

ε r - dielectric constant of the specimen = 3

when t = 1mm

V i=3∗1(1+ 13(10/1−1)) = 12kV peak

Capacitance of void

C c=ε0 εr A

t

When t = 1mm, area of void = π (2.5 x 10ˉ³)² / 4 = 491 x 10ˉ⁶ m²

C c=(8.854 x 10ˉ ¹²)(1)(491 x10 ˉ ⁶)

1x 10 ˉ ³=4.34 x10 ˉ ¹⁴ F

Capacitance of slab

Ca=ε0 εr A

d

Ca=(8.854 x10 ˉ ¹²)(3)(0.025) ² (3.14)

0.01=5.21x 10 ˉ ¹² F

Cb=ε0 εr A

d−t=¿ 5.79 x 10ˉ¹² F

The measured change

Qa=V i ∙Ca=(12 x103 ) ∙ (5.21x 10ˉ 12 )=65.52nC

Therefore,

Qc=Cb+Cc

Cb

× Qa=5.79x 10 ˉ ¹²+4.34 x 10ˉ ¹⁴

5.79 x10 ˉ ¹²∙62.52nC ≈ 63nC

Question 2

a) Using the circuit for series and parallel models of an insulating sample, derive the loss

tangent (tan δ ) equation in terms of the capacitance (C) and resistance (R) of the

sample.

Answer

Solid Insulator Model Series Model

Rs = Rp

1+ω2 C2 p R2 p

V

Vr

V

Vc

I

θ

δ

I

Vc

Vr Rs

Cs

Cs = Cp+ 1

ω2 C pR2 p

Tan δ = ¿Vr∨ ¿¿Vc∨¿¿

¿ = IRs

I1ωs

= ωCsRs

Solid Insulator Model Parallel Model

tan δ=¿¿ Ir∨ ¿

¿ Ic∨¿=(

VRp)

(V ωCp)=

1ωCpRp

¿

¿¿

b) What are partial discharges? Describe some of the typical partial discharges.

Answer

Some of the typical partial discharges are:

I

Ir Ic

VRp Cp

Ic

I

Ir

θ

δ

i) Corona or gas discharge. These occur due to non-uniform field in sharp edges of

the conductor subjected to high voltage especially when the insulation provided is

air or gas or liquid. [Fig. (a)]

ii) Surface discharges and discharges in laminated materials on the interfaces of

different dielectric material such as gas/solid interface. [Fig. (a) and (b)]

iii) Cavity discharges: When cavities are formed in solid or liquid insulating materials

the gas in the cavity is over stressed and discharges are formed. [Fig. (d)]

iv) Treeing Channels: High intensity fields are produced in an insulating material at

its sharp edges and this deteriorates the insulating material. The continuous

partial discharges so produced are known as treeing channels. [Fig. (e)]

c) A sample of impregnated paper (Ԑᵣ=3.5) of thickness 2mm is placed between large

parallel-plate electrodes. A cubical air-filled cavity with 2x2 mm length of its edges and

0.05mm deep, with its axis at right angles to the electrodes is located in the centre of

the dielectric. A sinusoidal alternating voltage is applied between the electrodes. The

breakdown voltages of air at different pd level is shown in figure Q5

(i) Determine the discharge inception stress in kV/mm in the dielectric for a pressure

of 550mm Hg.

(ii) Calculate the apparent discharge magnitude.

(iii) If the pressure in the void is doubled, determine the new inception voltage.

Answer

i) Discharge inception voltage, Vᵢ,

V ᵢ=C b+C cC b

.U=(1+C cC b ) .U

where U : breakdown voltage of the void

V i=[1+( ԐoA id1

ԐoԐrAid−di

)] .U=[1+( d−d1Ԑrd 1 )] .U

[1+( 2−0.053.5 x0.05 )] .U=12.14U

U is taken from the graph

At pd = (500)(0.05) = 25; therefore U≈0.46kV

Vi = 12.14 (0.46) = 5.58kV

Inception stress in kV/mm Ei = Vi/d = 5.58/2 = 2.79kV/mm

ii) Apparent discharge qₐ;

qₐ = δVₐ[C ₐ+ C bC cC b+C c

¿ ;∧δ V ₐ=δ Vc ( C bC ₐ C b

)

since Cb << Cc << Ca ; therefore δVa = δVc (CbC a

)

And qₐ = δVc (CbC a

) [ Ca + Cb ] ≈ δVcCb = U.Cb

where Cb =Ԑ oԐrAid−d 1

¿(8 .85x 10 ˉ 12) (3 .5 )(4 x10 ˉ 6)

1 .95 x10 ˉ ³ = 0.064pF

qa = (0.46x10 ˉ ³)(0.064 x10 ˉ 12)=29.44 pF

iii) If p = 2x500 = 100 mmHg

pd = 1000(0.05) = 50

U = 0.74 kV (from the graph)

Vi = 12.14 (U) = (12.14)(0.74) = 8.98kV

Question 3

a) Explain the purpose of insulation diagnostic tests on electrical power equipment. What

are the parameters or properties normally measured when investigating the insulation

performance?

Answer

The purpose of insulation diagnostic testing on the electrical equipment is to estimate

the important of any deterioration that has taken place. The tests are done on

equipment that has been in service for some time or on new equipment. For the in-

service equipment, the results of the test indicate if the equipment can safely be

returned to service or needs a major overhaul.

The parameters or properties that normally measured when investigating the insulation

performance are:

i) Breakdown strength

ii) Conductivity / resistivity

iii) Loss tangent (tan d) / dissipation factor

iv) Permittivity

v) Partial discharge

vi) Non-electrical properties – optical, acoustic, chemical, mechanical etc.

b) The circuit diagram for Schering bridge is shown in Fig. Q5. Both ends of the sample

and the standard capacitor are connected to the high voltage side of the bridge. The

standard capacitor used in the circuit has losses and can be represented as a

capacitance (C2) and resistance (r2) in series.

Show that at balanced condition, the capacitance and the resistance of the sample are:

C=C2

R4

R3

; r=(r 2+R4 C4

C2) R4

R3

Fig. Q5 Schering Bridge

Answer

At balance condition

AC Supply

Standard capacitor with losses

Sample

C4

R4R3

r r2

C2C

Z1 Z 4=Z2 Z3 - equation (i)

Where

Z1=r+ 1jωC

Z2=r+ 1jωC2

Z3=R3

Z4=R4/¿1

jωC4

=R4

1+ jω C4 R4

From equation (i)

Z1=Z2 Z3

Z4

r− jωC=(r2+

1jωC2

)(R3 )

( R4

1+ jω C4 R4)=(r2 R3+

R3

jω C2)(1+ jωC4 R4

R4)

r− jωC=

r2 R3

R4

+R3 C4

C2

− j( R3

ωC2 R4

−ωC4 r2 R3)Equal both side, gives;

r=r2 R3

R4

+R3 C4

C2

; → r=R3

R4(r2+

R4C4

C2)

and

1jωC

=R3

ωC2 R4

−ωC4 r2 R3=R3−ω2C2C4 r2 R3 R4

ωC2 R4

Therefore

C=C2 R4

R3−ω2C2C4 r2 R3 R4

=R4

R3 ( C2

1−(ωr 2C2) (ωR4 C4 ) )

Normally, ωr 2C2≪1 and ω R4C4 ≪1

Hence C=C2

R4

R3

PROVEN

c) Show that for a solid insulating material of relative permittivity εr, containing a cylindrical

air-filled cavity of depth t, which is small in relation to the thickness T of the dielectric,

the voltage across the sample (Vo) is given by the expression;

V a=Tεr t

× V c

Where Vc is the voltage across the cavity.

From the above equation, explain why the partial discharge can occur in the cavity

even though only the normal service voltage is applied across the insulating material.

Answer

Void in solid dielectric material Equivalent circuit of void in the structure

Void capacitance,

C c=ε0 A1

t

Capacitance series with void,

a

b2

b1

c tVaT

void

Cc

Cb

CaVa

Cb=ε0 εr A1

T−t

Remaining capacitance,

Ca=ε0 εr A1

T

Voltage across cavity,

V c=Cb

Cb+CC

× V a

Therefore

V a=Cb+CC

CC

×V c=( ε0 εr A1

T−t+

ε 0 A1

t )( ε0 A1

t )×V c

Since T >>t;

V a=( εr A1

T+

A1

t )( ε r A1

T )×V c=(1+ T

εr t )× V c

Yields V a=V c+Tεr t

V c proven

In term of electric field strength

V a=Ea T and V c=Ec t

Hence EaT=(1+ Tεr t )E c t

And Tεr t

≫1

EaT=( Tεr t )Ec t ⟹ Ec ≈ εr Ea and ε r>1

Electric stress in the void is greater than the dielectric stress across the sample.

Partial discharge occurs due to the very small gap of the void even at the normal

service voltage.

Question 4

a) The dissipation factor or loss tangent is an indication to determine the performance of

insulating property of insulators. The most common method used to determine the loss

tangent is by using a.c. bridges.

i) Sketch the circuit diagram of a high voltage Schering bridge for the measurement

of loss tangent (tan δ )

ii) Derive the expression for tan δ of the unknown series model of the tested sample

when the high voltage standard capacitor used in the Schering bridge is a loss-

free type.

Answer

C: capacitance of the sample

r: resistance of the sample

C2: standard capacitor

R3, R4 and C4: variable components of Schering Bridge

ii) Given;

AC Supply

Sample

C4R4R3

r

C2

C

Z1=r− 1jωC

Z3=R3

Z4=R4

1+ jω C4 R4

At balance condition

Z1 Z 4=Z2 Z3 - equation (i)

Z1=Z2 Z3

Z4

=( R3

jω C2)( 1+ jωC4 R4

R4)

Z1=r− jωC=

R3C4

C2

− j( R3

ωC2 R4)

By equating the real and imaginary parts of both side;

Therefore: r=C4 R3

C2

; and C=C2 R4

R3

For series model, tan δ=ωCr,

tan δ=ω(C2 R4

R3)(C4 R3

C2)=ωC4 R4

b) A solid insulating material of relative permittivity, ε r comprises a cylindrical air-filled

cavity of depth t. The material has thickness T, where its value is much greater

compared to the depth of the cavity. Show that the voltage across the sample

(material), V ais given by the expression.

V a=Tεr t

× V c

Where V c is the voltage across the cavity.

From the above equation, explain why partial discharge can occur in the cavity even

though only normal service voltage is applied across the insulating material.

Answer

Void in solid dielectric material Equivalent circuit of void in the structure

a

b2

b1

c tVaT

void

Cc

Cb

CaVa

Void capacitance, C c=ε0 A1

t (i)

Capacitance series with void, Cb=ε0 εr A1

T−t (ii)

Remaining capacitance, Ca=ε0 εr A1

T (iii)

Voltage across cavity, V c=Cb

Cb+CC

× V a (iv)

Therefore

V a=Cb+CC

CC

×V c=( ε0 εr A1

T−t+

ε 0 A1

t )( ε0 ε r A1

T−t )×V c

Since T >> t;

V a=( εr A1

T+

A1

t )( ε r A1

T )×V c=(1+ T

εr t )×V c

Yields

V a=V c+Tεr t

V c

proven

In term of electric field strength

V a=Ea T and V c=Ec t

Hence EaT=(1+ Tεr t )E c t

And Tεr t

≫1

EaT=( Tεr t )Ec t ⟹ Ec ≈ εr Ea

Since ε r>1, the electrical stress in the void is greater than the electrical stress across

the sample. Thus, partial discharge occurs due to the very small gap of the void even

at the normal service voltage.

Question 5

a) Partial discharge detection and measurements were limited to the laboratory due to

high levels of electrical noise at the switchyards. With the aid of proper diagrams

describe the main sources of interferences or noise which hampered the partial

discharge detection process and the techniques to suppress the interferences.

Answer

Typical noise source :

i) Power supply

ii) Voltage regulator

iii) High voltage source

iv) Filtering of the HV source

v) Feeder line and electrodes

vi) Coupling capacitor

vii) Loose conductive objects in the vicinity of the last location

viii) Pulse shaped interferences

ix) Electromagnetic waves by radio transmitter (harmonic interferences)

x) Interference currents in ground system of PD measuring instrument

The interferences can be reduced by;

i) Filter grounding

ii) Shielded room

iii) Separate source

iv) Filter AC source (Harmonics)

b) A solid dielectric has a small cavity. Draw an equivalent circuit for such an arrangement

and define all symbols. Derive the expression for electric field strength across the

cavity.

Answer

Void in solid dielectric material Equivalent circuit of void in the structure

In the absence of the void, the electric field within the insulation would everywhere be

Eₐ = Vₐ/d. Neglecting conductance, the insulation, which its void, can be represented

by the three capacitances a, b, and c. The void capacitance is represented by c. If the

void has length d1 and the cross sectional area A1 (perpendicular to d1) then;

a

b2

b1

c tVaT

void

Cc

Cb

CaVa

C = Ԑ₀ A ₁

d₁

The void is connected to the conducting plates through two capacitance b1 and b2 .

Their series combination is represented by the single capacitance b. Clearly,

b = Ԑ₀Ԑᵣ A ₁(d−d₁ )

Where Ԑᵣ is the relative permittivity of the insulating material. The remaining capacitor

a has the value,

a=Ԑ₀ԐᵣA(d )

Where A is the cross sectional area of the insulation, minus the (usually small) cross

sectional area of the void.

The voltage across the void is;

V c=C b

C b+Cc.Vₐ

And substituting for the capacitances gives,

V c=VₐԐᵣd ₁

Ԑᵣd ₁+(d−d₁ )

In the term of electric field strength we have,

Vₐ =Eₐ.T and Vc = EC.t

Where Ec is the dielectric field strength in the void. Substituting in equation gives,

EC=EₐԐᵣd ₁

Ԑᵣd ₁+(d−d ₁ )

Usually the void will be small, so that d₁<< d and Ԑᵣd₁<< d. Consequently,

EC ≈ ԐᵣEₐ

This shows that the field strength I n the void is greater than the field strength in the

insulating material. Because the gas in the void is likely to have a lower breakdown

strength that the insulating material, partial discharge are very probable.

c) While doing studies on partial discharges in cavities of cylindrical disc of 1.0cm

diameter and 1cm thickness, a cylindrical cavity of 1mm diameter and 1mm thickness

is made its center. The discharge voltage measured across the specimen is 0.2 V with

sensitivity of 1pC/volt. Determine the magnitude of charge transferred from the cavity

by taking Ԑᵣ of the disc to be equal to 2.5.

Answer

1pC = 1volt, since the discharge magnitude measured = 0.2pC.

Capacitance of the cavity, Cₐ = Ԑ0 xarea

Thickness = Ԑ₀( π

4 ) (1 x10¯ ³ )²(1x 10¯ ³ )

= πԐ0 x10¯ ³

4F

Capacitance, Cb = Ԑ0 xarea

Thickness =

Ԑ0 X 2 X 5 X ( π4 )X (1x 10¯ ³ ) ²

9 x10¯ ³

= (Ԑ ₀ Xπ4 )X ( 2.5

9 )X 10¯ 3

Thus qc = C ₐ+C b

C b X 0.2 = 0.92pC

Question 6

a) Breakdown in solid or liquid dielectrics arises from the action of electrical discharges in

internal gaseous cavities. Draw an equivalent circuit for such an arrangement and

device the expression for energy dissipated in the cavity in one discharge.

Answer

Void in solid dielectric material Equivalent circuit of void in the structure

a

b2

b1

c tVaT

void

Cc

Cb

CaVa

Cc= εₒA ₁d ₁

Cb= εₒ εᵣ A ₁(d−d ₁)

Ca= εₒ εᵣ Ad

Energy dissipated in the cavity in one discharge;

= ½ [Cc+(CaCb )(Ca+Cb)

]Vc ²

= ½ [Cc+(CaCb )(Ca+Cb)

][ Cb(Cb+Cc)

] ² Va ²

= ½Cb ²Cc

Va ²

b) A Schering Bridge as shown in figure 5 is used to measure the dielectric loss and

capacitance of the insulation of an electrical power equipment. Prove that the

capacitance, C and tangent loss are given by;

C=C ₂R ₄R ₃

and tan δ=ωC ₄R ₄

Answer

The circuit of the Schering Bridge is shown in Figure. The unknown is represented by C

and r in series. ThusZ1=r+ 1j ωC

, and the unknown has loss tangent, tan δ=ωCr .

C₂ is standard capacitor. Often an air capacitor is used in this position, being loss free.

To balance the bridge it is necessary to have two variable components. Normally one

of these is C₄ and the other one is either R₃ or R₄. Balance occurs when,

Z₁ Z₄ = Z₂ Z₃

AC Supply

Sample

C4R4R3

r

C2

C

That is when,

(r –j

ωC )R 4

1j ωc 2

R 4+1

j ωc 4

=R ₃

j ωC ₂

Re-arranging and separating into real and imaginary parts gives the balance

condition,

r = R ₃C ₄C ₂

and C = C ₂R ₄R ₃

Substituting in tan δ=ωCr

Gives tan δ=ωC ₄R ₄ proved

c) A solid dielectric specimen of 5cm diameter and 10mm thickness has dielectric

constant of 3. It has an embedded air filled void of 2.5mm diameter and 1mm

thickness. It is subjected to a voltage of 100kVrms. Find the voltage at which an

internal discharge in the void can occur. The breakdown strength of air is 3kV/mm. Also

determine the charge the value each time there is discharge inside the void and what

will be the value of charge as measured on the detector?

Answer

The inception voltage of the stressed void Vi is given by;

Vi=Ecb t 1+ 1ε r ( dt −1)

Where

Ecb - breakdown stress of the void = 3 kV/mm

d - thickness of the specimen = 10mm

t - thickness of the void = 1mm

εr - dielectric constant of the specimen = 3

when t = 1mm

Vi=3∗1 1+13 ( 10

1−1)=12kV peak

when t = 0.1mm

Vi=3∗0.11+ 13 ( 10

0.1−1)=10.2kV peak

This shows that the smaller is the void then the lower becomes the discharge inception

voltage

Capacitance of void

Cc=εo εr A/ t

When t = 1mm, area of void = π (2.5 x10 ˉ ³) ² /4=491 x10 ˉ ⁶m ²

Cc=(8.854 x 10ˉ ¹²)(1)(491 x10 ˉ ⁶)

1x 10 ˉ ³

¿4.34 x10 ˉ 14

Capacitance of slab

Ca=εoεr A /d

Ca=(8.854 x 10 ˉ 12) (3 ) (0.025 )2 (3.14 )

0.01

¿5.21 x10 ˉ 12 F

Cb=ε o ε rA

[d−t ]

¿5.79 x10 ˉ 12 F

The measured change

Qa=Vi .Ca

Qa=(12x 10 ³)(5.21x 10 ˉ ¹²)

¿62.52nC

Qa= CbCb+Cc

× Qc

Therefore,

Qc=Cb+CcCb

×Qa

¿( 5.79 x10 ˉ ¹²+4.34 x10 ˉ ¹⁴5.79x 10 ˉ ¹² )×62.52nC

Qc=63nC

Therefore measured change is around 1.0% less than the actual charge

Question 7

a) A solid dielectric has a small cavity. Draw an equivalent circuit for such an arrangement

and define all symbols. Under what conditions will this system have a partial

discharge?

Answer

Fig 5.1 A cavity in a dielectric and its “abc” equivalent circuit.

Cc = Capacitance of the cavity

Cb = Capacitance of the dielectric which in series with Cc

Ca = Capacitance of the rest at the dielectric

Va = Voltage apply to the dielectric

Vc = Voltage across the cavity

b) From a survey conducted on faulty 22kV polyethylene cables that have been in service

from at least one year to tens of years were found to experience fault due to several

factors. The factor were due to unknown cause, treeing, insulation failure and others.

Cables which were just within of one year of installation faulty due to unknown reasons

while the one being installed within three years were found to have traces of treeing

initiated from the cable sheath to the cable conductor. Explain briefly the initialization of

treeing in cable insulation and how it can cause insulation failure.

Answer

Electrical treeing initiate and propagate in a dry dielectric due to high and divergent

electric stress at metallic or semiconducting contaminant and/or void due to partial

discharge such trees consist of hollow channel, resulting from decomposition of

dielectric materials by the PDs. The tree shows up clearly in translucent solid dielectric

when examined with an optical microscope and transmitted light. Electrical tree

channels are permanently visible and there is a great variety of the visual appearance

of stems and branches of such trees as well as the circumstances in which initiation

and growth of such trees occur

a) Branch type b) Bush Type

c) A solid dielectric specimen of 5cm diameter and 10mm thickness has dielectric

constant of 3. It has embedded air filled void of 2.5mm diameter and 1mm thickness.

It is subjected to a voltage of 100kVrms. Find the voltage at which an internal

discharge in the void can occur. The breakdown strength of air is 3kV/mm. Also

determine the charge value each time there is discharge inside the void and what will

be the value of charge as measured on the detector? If the cavity thickness is

reduced to 10µm, what will be its effect on the discharge inception voltage?

Answer

The inception voltage of the stressed void Vi is given by as:

Vi=Ecb t 1+ 1εr ( d

tc−1)

where,

Ecb = breakdown stress of the void = 3kV/n

t = thickness at the specimen = 10mm

tc = thickness of the void

= 1mm or 0.1 mm for the last case

εr = dielectric constant of the specimen = 3

When t = 1mm then

Vi=3 x1 1+ 13 (10

1−1)

¿12kV peak

When t = 0.1mm then

Vi=3 x0.1 1+ 13 ( 10

0.1−1)

¿10.2kV peak

This shows that smaller is the void them lower becomes the discharge inception

voltage.

Capacitance of void

Cc=εo εrAt

When tc = 1mm, area of void

¿ π (2.5 x10 ˉ ³) ²/ 4=491 x10 ˉ ⁶m ²

Cc=(8.854 x 10ˉ ¹²)(1)(491 x10 ˉ 6)

1 x10 ˉ ³

¿4.34 x10 ˉ 14

Capacitance of slab

Ca=εoεr A / t

Ca=(8.854 x10 ˉ ¹²)(3)(0.025)² (3.14)

0.01

¿5.21 x10 ˉ ¹² F

Cb=εo εr A /[ t−tc ]

¿5.79 x10 ˉ ¹² F

The measured change

Qa=Vi .Ca

Qa=(120x 10 ²)(5.21x 10 ˉ ¹²)

¿62.52nC

Qa= CbCb+Cc

× Qc

Therefore,

Qc=Cb+CcCb

xQa

¿ 5.79x 10ˉ 12+4.34 x10 ˉ 14

5.79 x10 ˉ 12 x 62.52nC

Qc=63.0nC

Therefore measured change is around 1.0% less than the actual charge.

Question 8

a) A solid dielectric has a small cavity. Draw an equivalent circuit for such an arrangement

and define all symbols. Under what conditions will this system have a partial

discharge?

Answer

Fig 5.1 a cavity in a dielectric and its “abc” equivalent circuit.

Cc = Capacitance of the cavity

Cb = Capacitance of the dielectric which in series with Cc

Ca = Capacitance of the rest at the dielectric

Va = Voltage apply to the dielectric

Vc = Voltage across the cavity

b) From a survey conducted on faulty 22kV polyethylene cables that have been in service

from at least one year to tens of years were found to experience fault due to several

factors. The factor were due to unknown cause, treeing, insulation failure and others.

Cables which were just within of one year of installation faulty due to unknown reasons

while the one being installed within three years were found to have traces of treeing

initiated from the cable sheath to the cable conductor. Explain briefly the initialization of

treeing in cable insulation and how it can cause insulation failure.

Answer

Electrical treeing initiate and propagate in a dry dielectric due to high and divergent

electric stress at metallic or semiconducting contaminant and/or void due to partial

discharge such trees consist of hollow channel, resulting from decomposition of

dielectric materials by the PDs. The tree shows up clearly in translucent solid dielectric

when examined with an optical microscope and transmitted light. Electrical tree

channels are permanently visible and there is a great variety of the visual appearance

of stems and branches of such trees as well as the circumstances in which initiation

and growth of such trees occur.

a) Branch type b) Bush Type

c) A solid dielectric specimen of 5cm diameter and 10mm thickness has dielectric

constant of 3. It has embedded air filled void of 2.5mm diameter and 1mm thickness. It

is subjected to a voltage of 100kVrms. Find the voltage at which an internal discharge

in the void can occur. The breakdown strength of air is 3kV/mm. Also determine the

charge value each time there is discharge inside the void and what will be the value of

charge as measured on the detector? If the cavity thickness is reduced to 10µm, what

will be its effect on the discharge inception voltage?

Answer

The inception voltage of the stressed void Vi is given by as:

Vi=Ecb t 1+ 1εr ( d

tc−1)

where,

Ecb = breakdown stress of the void = 3kV/n

t = thickness at the specimen = 10mm

tc = thickness of the void

= 1mm or 0.1 mm for the last case

εr = dielectric constant of the specimen = 3

i) When t = 1mm then

Vi=3 x1 1+ 13 (10

1−1)

¿12kV peak

ii) When t = 0.1mm then

Vi=3 x0.1 1+ 13 ( 10

0.1−1)

¿10.2kV peak

This shows that smaller is the void them lower becomes the discharge inception

voltage.

iii) Capacitance of void

Cc=εo εrAt

When tc = 1mm, area of void

¿ π (2.5 x10 ˉ ³) ²/ 4=491 x10 ˉ ⁶m ²

Cc=(8.854 x 10ˉ ¹²)(1)(491 x10 ˉ 6)

1 x10 ˉ ³

¿4.34 x10 ˉ 14

Capacitance of slab

Ca=εoεr A / t

Ca=(8.854 x10 ˉ ¹²)(3)(0.025)² (3.14)

0.01

¿5.21 x10 ˉ ¹² F

Cb=εo εr A /[ t−tc ]

¿5.79 x10 ˉ ¹² F

The measured change

Qa=Vi .Ca

Qa=(120x 10 ²)(5.21x 10 ˉ ¹²)

¿62.52nC

Qa= CbCb+Cc

× Qc

Therefore,

Qc=Cb+CcCb

xQa

¿ 5.79x 10ˉ 12+4.34 x10 ˉ 14

5.79 x10 ˉ 12 x 62.52nC

Qc=63.0nC

Therefore measured change is around 1.0% less than the actual charge in the

void.

Question 9

A solid can be modeled either as parallel or series model as shown in below. Derive the

equation for tangent loss, tan δ for both models.

Solid Insulator Model Series Model

Cs

RsVr

Vc

I

δ

θ

I

Vc

V

Vr

V

Solid Insulator Model Parallel Model

Answer

δ

θ

Ir

I

Ic

CpRpV

IcIr

I

V

Vr

V

Vc

I

θ

δ

I

Vc

Vr Rs

Cs

Solid Insulator Model Series Model

Rs = Rp

1+ω2 C2 p R2 p

Cs = Cp+ 1

ω2 C pR2 p

Tan δ = ¿Vr∨ ¿¿Vc∨¿¿

¿ = IRs

I1ωs

= ωCsRs

Solid Insulator Model Parallel Model

tan δ=¿¿ Ir∨ ¿

¿ Ic∨¿=(

VRp)

(V ωCp)=

1ωCpRp

¿

¿¿

I

Ir Ic

VRp Cp

Ic

I

Ir

θ

δ

Question 10

a) How does the “internal discharge” phenomena lead to breakdown in solid dielectrics?

Answer

Cause gradual deterioration of the insulating materials, sometimes over a period of several

years, leading perhaps to eventual failure.

b) A solid dielectric has a small cavity. Draw an equivalent circuit for such an arrangement

and define all symbols. Under What condition will this system have a partial discharge?

Answer

Void in solid dielectric material Equivalent circuit of void in the structure

Usually the void will be small, so that d₁<< d and Ԑᵣd₁<< d. Consequently,

EC ≈ ԐᵣEₐ

a

b2

b1

c tVaT

void

Cc

Cb

CaVa

This shows that the field strength In the void is greater than the field strength in the

insulating material. Because the gas in the void is likely to have a lower breakdown

strength that the insulating material, partial discharge are very probable.

c) A solid dielectric specimen of 5cm diameter and 10mm thickness has dielectric

constant of 3. It has an embedded air filled void of 2.5mm diameter and 1mm

thickness. It is subjected to a voltage of 100kVrms. Find the voltage at which an

internal discharge in the void can occur. The breakdown strength of air is 3kV/mm. Also

determine the charge the value each time there is discharge inside the void and what

will be the value of charge as measured on the detector? If the cavity thickness is

reduce to 100 μm, What will be its effect on the discharge inception voltage?

Answer

The inception voltage of the stressed void Vi is given by;

Vi = Ecb t 1+1/εr (d/t-1)

Where

Ecb - breakdown stress of the void = 3 kV/mm

d - thickness of the specimen = 10mm

t - thickness of the void = 1mm

εr - dielectric constant of the specimen = 3

when t = 1mm

Vi = 3 * 1 1 + 1/3 (10/1 - 1) = 12 kV peak

when t = 0.1mm

Vi = 3 * 0.1 1 + 1/3 (10/0.1 - 1) = 10.2 kV peak

This shows that the smaller is the void then the lower becomes the discharge inception

voltage

Capacitance of void

Cc = εo εr A/t

When t = 1mm, area of void = π (2.5 x 10ˉ³)² / 4 = 491 x 10ˉ⁶ m²

Cc = (8.854 x 10ˉ¹²) (1) (491 x 10ˉ⁶)

1 x 10ˉ³

= 4.34 x 10ˉ¹⁴

Capacitance of slab

Ca = εo εr A/d

Ca = (8.854 x 10ˉ¹²) (3) (0.025)² (3.14)

0.01

= 5.21 x 10ˉ¹²F

Cb = εo εr A/[d-t]

= 5.79 x 10ˉ¹² F

The measured change

Qa = Vi . Ca

Qa = (12 x 10³) (5.21 x 10ˉ¹²)

= 62.52nC

Qa = ___Cb____. Qc

Cb + Cc

Therefore,

Qc = _Cb_+ Cc_. Qa

Cb

5.79 x 10ˉ¹² + 4.34 x 10ˉ¹⁴ . 62.52nC

5.79 x 10ˉ¹²

Qc = 63nC

Therefore measured change is around 1.0% less than the actual charge.

Question 11

a) A Schering Bridge as shown in Figure Q7b is used to measure the dielectric loss and

capacitance of the insulation of an electrical power equipment. Prove that the

capacitance, C and tangent loss are given by;

C=C2 R4

R3

and tan δ=ωC4 R4

ANSWER

A Schering Bridge as shown in Figure Q7b is used to measure the dielectric loss and

capacitance of the insulation of an electrical power equipment. Prove that the

capacitance, C and tangent loss are given by;

C=C2 R4

R3

and tan δ=ωC4 R4

Z1 Z 4=Z2 Z3

That is when,

(r− j /ωC )R4

1jωC2

R4+1

jωC4

=R3

jω C2

Re-arranging and separating into real and imaginary parts gives the balance condition,

AC Supply

Sample

C4R4R3

r

C2

C

r=R3

C4

C2

and C=C2

R4

R3

Substituting in tan ω=ωCr gives,

tan δ=ωC4 R4 proved

Question 12

a) In a strongly inhomogeneous field, external partial discharges occur at electrodes of

small radius of curvature when a definite voltage is exceeded. These are referred to as

corona discharges and depending upon the voltage amplitude, they result in a large

number of charge pulses of very short duration. With the aid of diagram where

appropriate;

i) Define the terms corona.

ii) Types of corona and how it occurs.

iii) The problems which are created by corona discharges on high voltage

transmission lines.

Answer

i) The term corona is used to describe the discharge phenomena which occur at

highly stressed electrodes prior to the complete breakdown of the gap between

the electrodes. It is a partial discharge in air around a sharp point or thin wire in a

strong, non-uniform field. It is characterized by a visible glow, an audible noise,

radio interference, chemical effects such as production of ozone and loss of

electrical power. It occurs whenever the local voltage gradient exceeds the

ionization value of the air and depends on the air density, humidity and in outdoor

situations whether it is fair weather or raining and also on the roughness of the

conductor surface.

ii) Corona can be classified into:

DC Corona.

Positive corona (anode)

When the highly stressed electrode is the anode, the following corona

modes are observed as the voltage is increased.

- Onset streamers

Also known as ‘burst’ pulses, these are intermittent, filamentary discharges

which propagates only a short distance from the highly stressed electrodes.

- Hermstein glow

As the voltage is increased, the intermittent streamer discharges give way

to a steady glow discharges. This transition occurs when a large enough

negative ion space is generated near the anode to give a quasi-uniform

field in that region.

- Breakdown streamer

Eventually, the shielding effect of the glow discharge is not able to prevent

the formation of large streamers which propagates well into the gap.

Onset streamers Hermstein glow Breakdown streamer

Negative corona (cathode)

When the highly stressed electrode is negative, three modes are again

observed.

- Trichel pulses

These differ from the burst pulses in that their magnitude and repetition

frequency are both very regular.

- Cathode glow

As the voltage is raised a critical Trichel pulse repetition frequency is

reached and the repetitive discharge is replaced by a steady cathode glow.

- Negative streamer

These discharges are usually known as negative feathers to avoid

confusion with positive streamer discharges. They develop out of the glow

mode and a long rise time compared with other pulsating coronas.

Trichel pulses Cathode glow Negative streamer

AC Corona

With an alternating voltage applied, the same basic corona types will

appear although their characteristics may be altered to an extent which

depends on the gap length

- Small gaps (d ≤ 100cm)

Here, the ions generated in any of the corona modes above are able to

cross the gap during one half cycle to the next and the corona modes will

therefore be similar to those for direct voltages, although all three modes

may be observed in one half cycle.

- Large gaps (d > 1m)

For those gaps, space charge can persist from one half cycles to the next

and can have an effect on the corona modes observed. The usual effect is

to enhance the positive glow phase. Further, the negative streamer is never

observed in ac stressed gaps, since its onset potential is higher than the

positive polarity breakdown voltage. Breakdown always occurs on the

positive half cycle.

ii) Transmission line corona

The above description of the types of corona discharge referred particularly to the

point plane gap where there is a single site for discharges to occur. On a

transmission line, corona may occur anywhere on the line and the average

corona currents will be much higher.

iii) The problems which are created by corona discharges on high voltage

transmission lines are;

Power losses

The power losses depend upon the maximum gradient for which the line is

designed. For the single conductor, this occurs at the conductor surface.

For the given cross section of conductor required for current carrying

capacity, the maximum stress may be reduced by using bundled

conductors in which 2,4 or 6 wire assembly is used.

Radio interference

Radio interference is caused only by the pulse corona modes and only

Trichel pulses and positive streamers are of interest. The positive

streamers usually have shorter rise time than the Trichel pulses and greater

amplitude, so that the rate of change of current is greater and their RI effect

is therefore greater. These corona discharges cause radiation of

electromagnetic waves.

Audible noise

Recent studies of EHV and UHV lines indicate that audible noise may be a

problem where such lines pass near inhabit areas. Difficulties arise in

monitoring such noise levels as the ‘apparent noise’ is a non-linear function

of frequency. Measuring instruments have thus been developed which have

a similar response to the human ear and levels have been se based limits

at which most people find the noise objectionable.

b) Partial discharge detection and measurements were limited to the laboratory due to

high levels of electrical noise at the switchyards. With the aid of proper diagram;

i) Describe in details the main sources of interference or noise which hampered the

partial discharge detection and measurement process.

ii) The technique which are widely used to suppress the noise in order to measure

the partial discharges.

Answer

The source of the noise is listed as below:

i) The power system's noise through the apparatus outlets, which may excited by

the internal discharge of other equipments in power system, such as discharge of

the bus bar, switching of the breaker and so on.

ii) The high frequency noise such as coupling by capacitor and inductor form the

generator's rotator DC excitation. These noise may originated by the thyristor of

excitation system.

iii) The external noise from the environment outside, such as broadcasting

interference of AM radio and high frequency signals from mobile phone.

iv) The noise in diagnosis system itself, such as noise of circuit or switch power

supply

v) The technique which are widely used to suppress the noise in order to measure

the partial discharges. Band pass Filter, FFT Filter, Wavelet Filter, Neural

Network Filter to eliminate the noise.

Question 13

a) (i) Define the breakdown of electrical insulation.

(ii) List all mechanism of breakdown in solid and liquid insulation.

Answer

i) Electrical breakdown in insulation the maximum voltage that can be covered by

an insulating material

ii) List breakdown mechanisms in insulating

Electrodynamic

Heat

Chemical and electrochemical

Internal discharge

Tracking

b) Explain the Townsend mechanism of breakdown in the gas medium and prove that

creation of breakdown is given by (all symbols have their usual meanings):

ɤeᵉᵈ=1

Answer

Townsend breakdown mechanism used to reduce the breakdown in the gas business.

This mechanism did in low pressure conditions and the small distance between the

electrodes, the voltage supplied to the electrodes until breakdown occurs. Breakdown

process begins with the creation of the first and subsequent calculation of the

coefficient of voltage increase will occur and thus the distribution of secondary

breakdown occurs. Currents equation taking into account the secondary generator.

I= Io x eᵉᵈ1−γ (eᵉᵈ−1)

Take the breakdown criteria by taking the equation below:-

1−γ (eᵉᵈ−1)=0

γ (eᵉᵈ−1)=1

Because of; γ eᵉᵈ ≫1

Their for; γ eᵉᵈ=1

c) Breakdown voltage measurement of a uniform field pressurized gas is shown in table

below.

Distance (mm) 20.3 28.8

Pressure (mbar) 500 700

Temperature (°C) 20 25

Breakdown voltage (kV) 31.6 53

Answer

At a distance; d1 = 20.3 mm, P1 = 500mb, t1 = 20 °C, V1 = 31.6 kV

P= P11013 ( 273+20

273+t )

P= 5001013 ( 273+20

273+t )

P=0.4936

31.6=A x0.4936 x 20.3+B √0.4936 x 20.3

31.06=10.02 A+B (3.165) ………. E1

At a distance; d2 = 28.8 mm, P2 = 700mb, t2 = 29 °C, V2 = 53 kV

P= P21013 ( 273+20

273+25 )

P= 5001013 ( 293

298 )

P=0.6794

53=A x 0.6794 x 28.8+B√0.6794 x 28.8

53=19.57 A+B (4.424 ) ……. E2

When E1 x 19.57

618.412=196.09 A+61.94B

When E2 x 10.02

531.06=196.09 A+44.33 B

E1 to E2

87.352=17.61B

B=4.96

From E1

A=( 31.6−15.710.02 )

A=1.587

At a distance; d3 = 15.5 mm, P3 = 250mb, t3 = 25 °C, V3 =? kV

P= P21013 ( 273+20

273+25 )

P= 5001013 ( 293

298 )P=0.2427

V 3=1.587 x 0.2427 x15+4.96√0.2427 x1

V 3=15.24 kV

d) (i) Explain what is meant by partial discharge in high voltage engineering.

(ii) What are the effects to high voltage apparatus if the partial discharge level is

greater than the maximum allowed.

Answer

i) Partial discharge is an electrical discharge that only partially bridge the insulating

medium.

ii) The cable will damage because of the high partial discharge level is greater from

the limit of insulation limit.

Question 14

a) Write short discussions on the followings:

i) Paschens Law and its significance in gas breakdown phenomena;

ii) Thermal breakdown of solid insulators

Answer

a) i) Paschen’s Law is found to be valid over a wide range of partial discharge values.

At higher partial discharge values, the breakdown voltage in some gases is found

to e slightly higher than the values at smaller gaps for the same values of partial

discharge.

ii) Thermal breakdown is based on the experimental observation of extremely large

current just before breakdown. These high current pulses are believed to

originate from the tips of the microscopic projections on the cathode surface with

densities of the order of 1 A/cmᵌ.

CHAPTER 6

HIGH VOLTAGE GENERATION

(QUESTION AND ANSWER)

QUESTION 1

a) Describe and give an example, the significance of routine tests, type tests and

maintenance tests on high voltage equipment.

ANSWER

Routine tests are done on equipment for the purpose of eliminating equipment with

manufacturing defects by non-destructive tests. These are generally easily verifiable. Made

by manufacturer on every finished piece of product to fulfills the specifications. Example:

Resistance measurement on power transformer or partial discharge measurement in high

voltage cables.

Type tests are done on equipment to establish that the particular design is suitable for a

particular purpose. They are normally done once on new designs and when specifically

requested by consumers purchasing in bulk quantities. Performed on each type of equipment

before their supply on a general commercial scale – demonstrate performance

characteristics.

Example: One minute rain test on porcelain insulator where the insulator is sprayed

throughout the tests with artificial rain or temperature rise test and lightning impulse test on

power transformers.

Maintenance tests are usually carried out after maintenance/repair of the equipment and

conducted according the schedule provided. Purpose of the test is to ensure the equipment

lifetime is achieved.

Example: Partial discharge measurement on cables and oil breakdown test on transformer.

b) Describe briefly, with the aid of suitable diagrams the cascade arrangement of

transformers to obtain high alternating voltage for testing purposes.

ANSWER

Figure 1 shows a typical cascade arrangement of transformers used to obtain up to 300 kV

from three units each rated at 100 kV insulation. The low voltage winding is connected to the

primary of the first transformer, and this is connected to the transformer tank which is

earthed. One end of the high voltage winding is also earthed through the tank. The high

voltage end and a tapping near this end is taken out at the top of the transformer through a

bushing, and forms the primary of the second transformer. One end of this winding is

connected to the tank of the second transformer to maintain the tank at high voltage. The

secondary of this transformer too has one end connected to the tank at high voltage.

Figure 1

The secondary of this transformer too has one end connected to the tank and at the other

end the next cascaded transformer is fed. This cascade arrangement can be continued

further if a still higher voltage is required. In the cascade arrangement shown, each

transformer needs only to be insulated for 100 kV, and hence the transformer can be

relatively small. If a 300kV. Transformer had to be used instead, the size would be massive.

High voltage transformers for testing purposes are designed purposely to have a poor

regulation. This is to ensure that when the secondary of the transformer is short circuited (as

will commonly happen in flash-over tests of insulation), the current would not increase to too

high a value and to reduce the cost. In practice, an additional series resistance (commonly a

water resistance) is also used in such cases to limit the current and prevent possible damage

to the transformer.

c) A six-stage impulse generator designed to generate the standard waveform

(1.2/50µs) has a per stage capacitance of 0.06µF to be used to test transformers with

an equivalent winding to earth capacitance of 1nF. A peak output voltage of 550kV is

required for testing the transformer.

ANSWER

i) With the aid of appropriate calculations select the values of resistive elements in the circuit to produce the required waveform. State any assumptions made.

c1=6 of 0.06 μF=0.06/6=0.01μF (Capacitor effectively∈series )c2=1nFStandard waveform1.2 /50 μs ¿ time ,t 1=3 R1¿

1.2×10−6=3R1

(0.01×10−6× 110−9)(0.01× 10−6+1×10−9)

¿3 R1

(1×10−17)(1.1×10−8)

¿2.73×10−9 R1

∴R1=440Ω.

Tail time , t 2=0.7 (R1+R2) (C1+C2 )

50×10−6=0.7 (440+R2 ) (0.01× 10−6 )

50×10−6=0.7¿

7.7×10−9=(440+R2 )

∴R2=6494kΩ

¿1.01kΩ per stage

ii) Draw the basic circuit diagram of the multi-stage impulse generator indicating

all relevant values on it. Indicate also on the diagram the wavefront and

wavetail control resistors and the charging resistors.

ANSWER

QUESTION 2

a) Describe the types of test conducted on high voltage equipment .

ANSWER

i) Routine Tests

Routine tests are made by the manufacturer on every finished piece of product to

make such that it fulfills the specifications. Acceptance and commissioning tests

are made by the purchaser and self-explanatory. Routine testing such as a

power-frequency overvoltage tests is performed on every unit at the

manufacturer’s factory and possibly after receipt of the unit by the purchaser.

ii) Type tests

Type tests are performed on each type of equipment before their supply on a

general commercial scale so as to demonstrate performance characteristics

Wavefront control resistor

Wavetail control resistors

Wavetail control resistors

meeting the intended application. These tests are of such a nature that they need

not be repeated unless changes are made in the design of the product. Type

testing, such as an impulse voltage test, is done to prove the specifications of a

new design and is probably restricted to one or two units of each design.

iii) Maintenance Tests

Maintenance tests are usually carried out after maintenance or repair of the

equipment. Normally this tests is conducted according to the schedule provided.

The purpose of the maintenance test is to ensure the lifetime of the equipment is

achieved.

iv) Special TestSpecial tests are tests other than the above.

b) With the aid of suitable diagrams discuss the generations of high voltage direct

current (HVDC) using the full wave rectifier circuit.

ANSWER

Full Wave Rectifier for HVDC generation

Full wave rectifiers produce dc voltage less than ac maximum voltage. Ripple or voltage

fluctuation will be present and this has to be kept within a reasonable limit by means of filters.

During the +ve half-cycle, rectifier A conducts and charged up capacitor, C (smoothing

capacitor). During the –ve half-cycle, rectifier B conducts and charged up C. The source

transformer requires a centre tapped secondary with rating of 2V. The output waveform is

shown in the figure below:

c) Lightning impulse voltage is simulated in the laboratory using an impulse

generator and is used to conduct lightning impulse tests on high voltage

equipment based on standard test procedures. An impulse generator

(unmodified Marx) has five (5) stages. The following circuit elements are

available: 100kV – rated 0.2µF capacitors, a 300Ω front resistor, and a 2500Ω tail

resistor. The load capacitor is given as 100pF.

ANSWER

i) Determine the output impulse waveshape of generator and give comments on the

waveshape as compared to the standard test procedure.

Charging capacitor ,C1=0.2/5=0.04 μF

Load capacitor ,C2=0.001 μF

Front timet 1=3.0 R 1C1C2

C1+C2

¿3 (300¿ 0.04 ×10−6 × 0.001×10−6

0.04×10−6+0.001×10−6 =0.88 μs

Tail time , t 2=0.7 (R1+R2 ) (C1+C2¿

¿0.7 (300+2500 ) (0.041× 10−6 )=80.36 μs

The front time of the wave is in the tolerance of standard waveshape which is 1 .2μs± 30 % while the tail time is out of the tolerance of standard waveshape which is 50 μs± 20 %.

ii) What is the maximum output voltage of the generator if the charging capacitor is

charged up to the maximum rated voltage?

Hint:

V out max=V ¿

βCL R f

β ≈CS+CL

R f CS CL

DC charging voltage for 5 stages ,V C=5×100 kV=500kV

Maximunoutput voltage ,V O /p=V c

R1 C2 (α−β¿ [exp (−α t 1¿–exp (−β t 1 )…………… (1 )

α=1/R1 C2=3.33× 106

β=1/R2 C1=1×104

substitute the values into equation (1 )gives ,

V o /p=471.88kV

QUESTION 3

a) Explain with diagram, different types of rectifier circuit for producing high dc

voltages.

ANSWER

Rectifier circuits for producing high dc voltages from ac sources may be half-wave, full-

wave or voltage doubler-type rectifier. The rectifier may be an electron tube or solid-

state device. The schematic diagram of the rectifiers can be seen in the figures below.

b) Explain the different schemes for cascade connection of transformer for

producing very high ac voltages.

ANSWER

Schematic diagram of two types of cascade transformer can be shown in the figure below:

Figure (a) shows the cascade transformer units in which the first transformer is at the

ground potential along with is tank. The second transformer is kept on insulators and

maintained at a potential of V 2, the output voltage of the first unit above the ground.

ii. Half wave rectifier i. Full wave rectifier

iii. Simple voltage double iv. Cascade voltage double

Figure (a)

Figure (b) below for providing the excitation to the second and the third stages is shown.

Isolating transformer Is1 , Is2 ,∧Is3 are 1:1 ratio transformers insulated to their respective tank

potentials and are meant for supplying excitation for the second and the third stages at their

tank potentials. Power supply to the isolating transformers is also fed from the same ac input.

This scheme is expensive and requires more space. The advantage is the natural cooling is

sufficient and the transformers are light and compact.

Figure (b)

c) An impulse generator has eight stages with each condenser rated at 0.16µF and

125kV. The load capacitor available is 1000 pF. Find the series resistance and

the damping resistance needed to produce 1.2/50 µs impulse wave. What is the

maximum output voltage of the generator, if the charging voltage is 120kV?

ANSWER

Equivalent circuit for impulse generator:

C1, the generator capacitance: 0.16

Ɛ = 0.02µF

C2, the load capacitance = 0.001µF

t 1 = 1.2µs (time to front)

t 1 = 3R1 C1C2

C1+¿C2¿ ⟹ R1 =

13

t 1 C1+C2

C1C2

= 13

(1.2×10−6) (0.02×10¿¿−6)+(0.001×10−6)(0.02×10−6 )×(0.001×10−6)

¿

⟹ R1= 13

×0.0252× 10−12

2× 10−5 ×10−12 = 420Ω

t 2 = 0.7 (R1+R2¿ (C1+C2¿⟶ time total

50 × 10−6 = 0.7 (420+R2 ¿ ( (0.02×10−6 ¿+(0.001×10−6)¿

50×10−6 = (294+0.7R2 ¿ (0.021×10−6¿

294+0.7R2 = 50× 10−6

0.021× 10−6 = 2380.95

0.7R2 = 2380.95-294 = 2086.95

R2 = 2086.95

0.7 = 2981Ω

The DC charging voltage for 8 stages:

V = 8×120kV = 960kV

The maximum output voltage:

V o = V

R1C2(α−β) ( e−α t1−e−β t 1¿

α = 1

R1C2 =

1

420×0.001×10−6 = 2.38×106

β = 1

R2C1 =

1

2981× 0.02× 10−6 = 0.017×106

V o = V (e¿¿−2.38×106 ×1.2×10−6−e−0.017 ×106 ×1.2× 10−6

)420× 0.001×10−6 (2.38−0.017)×10−6 ¿

V o = 960kV0.9925

(0.0575-0.9798) = 967.25(-0.9223) = -892.1kV ≈ 892kV

QUESTION 4

a) With the aid of suitability labeled diagram, discuss the generation of high voltage

alternating current (HVAC) using series resonant transformer and state the output voltage

equation across the test object.

ANSWER

HVAC generation using resonant transformer:

Series Resonant Transformer

The equivalent circuit of the high voltage testing transformer consists of windings

leakage reactance & resistance, magnetizing reactance and shunt capacitance

across the output terminal due to bushing of hv terminal and test object. Utilised in

testing at very high voltage and on occasion requiring large current such as cable

test, dielectric loss and pd measurement. A voltage regulator of either auto-

transformer or induction regulator is connected to the main supply. The secondary

winding of the exciter transformer is connected across the voltage reactor, L and the

capacitive load, C. The inductance of the reactor is varied by varying its air gap and

operating range is set in the ration of 10:1. Capacitance C comprises the capacitance

of the test object, capacitance of the measuring voltage divider, capacitance of the

bushing etc.

The output voltage equation;

V c = − jV Xc /¿))

= V Xc

R

= V/ωCR

b) High voltage testing can be classified into two types which are destructive and

non- destructive tests. Discuss in details destructive and non-destructive tests

and give example of an application of the tests towards high voltage

equipment.

ANSWER

Destructive Test

The test is the deliberate application to equipment of a voltage higher than its normal

working voltage, for a specific period of time, to discover if the insulation withstands or

breaks down under that voltage. By definition, a breakdown test is a destructive

technique to measure the dielectric strength the insulation and it is usually made on a

sample piece of the material. The term “destructive” relates to phenomena associated

with the failure of the insulation under test. Normally the equipment or apparatus

underwent destructive cannot be used in the service.

Non-destructive Test

Because of their cost, the testing of components or complete equipment is usually by a

non-destructive technique designed to ensure that the level of insulation is adequate for

service conditions. Unless breakdown is intended, the test voltage is not raised high

enough to cause failure in good equipment.

The test is mainly done to assess the electrical properties, such as resistivity, dielectric

constant and loss factor. This test is done to detect any deterioration or faults in the

internal insulation of the apparatus. The apparatus is not destroyed during the non-

destructive test and can be used again.

c) The high voltage laboratory of the Institute of High Voltage and High Current

(IVAT) is to conduct a lightning impulse test on high voltage equipment. A

competent test engineer has to apply a series of reduced lightning impulse

wave, full impulse wave and chopped impulse wave on the tail. Based on the

statement given, discuss in detail the test procedures by considering the

equipment under test, standards referred, test connections and sequence.

ANSWER

The test carried out are in accordance with the BS 171; Part 3. Generally, it is carried

out at room temperature with the transformer not energized. In preliminary test, the

front and tail resistors of the generator are adjusted to give a wave as near to the

standard 1.2/50 µs as the load will permit. These tests are done at a voltage between

50 and 75% of the full test value, after which no change is made in the circuit

parameters. There are cases, however, where this standard impulse shape cannot

reasonably be obtained, because of low winding inductance or high capacitance to

earth. The resulting impulse shape is then often oscillatory. Wider tolerances may in

such cases be permitted by agreement between the customer and manufacturer.

Test Connections

The impulse test-sequence is applied to each of the line terminals of the tested

winding in succession. In the case of a three-phase transformer, the other line

terminals of the winding shall be earthed directly or through a low-impedance, such

as a current measuring shunt. If the winding has a neutral terminal, the neutral shall

be earthed directly or through a low-impedance such as a current measuring shunt.

The tank shall be earthed.

Test Sequence and Test Criteria

The peak voltage applied for the full wave test depends on the system highest

voltage and on the BIL (breakdown insulation level) for the transformer as specified

by the international standards. To ensure that any winding not under test is not

damaged by a transferred surge, its terminal are either earthed directly or through

resistors of such a value that the transferred voltage is limited to less than 75% of its

test value.

A sequence of impulses is applied to each line terminal in the following order;

1. One reduced full wave (normally 60%)

2. One 100% full wave

3. Two reduced chopped wave (60%)

4. Two 100% chopped wave

5. Two 100% full wave

During these tests, each impulse voltage and current are recorded.

Failure Detection and Location

Insulation failure in a transformer arising from an impulse test may be detected by

any, or all, of the following methods:

i) A change in the wave shape of the voltage and current oscillograms both before

and after the chopped waves has been applied.

ii) Acoustic noise within the transformer.

iii) Visual signs of flashover under oil such as the presence of carbon, bubbles and

fibre bridges in the oil.

The general opinion is that records of the current waves provide better evidence

than the voltage waves for the detection of failure in transformers.

Whilst it is relatively easy to detect failure, the possibility of locating a point of

failure from either voltage or current oscillogram is rather limited. According to

Hickling, 1968, the surge behaviour of a transformer is similar to that of a

transmission line. The propagation velocity for an impulse wave through oil-

immersed windings is about 130m of conductor length per µs. A voltage

oscillogram may provide some clue to the position of a breakdown as the

disturbance on the trace must occur at twice the wave travel time from the line to

the point of failure.

QUESTION 5

a)The type of tests conducted to high voltage equipment are as follows;

ANSWER

i) Routine Tests

Routine tests are made by the manufacturer on every finished piece of product to

make such that it fulfills the specifications. Acceptance and commissioning tests

are made by the purchaser and self-explanatory. Routine testing such as a

power-frequency overvoltage tests is performed on every unit at the

manufacturer’s factory and possibly after receipt of the unit by the purchaser.

ii) Type tests

Type tests are performed on each type of equipment before their supply on a

general commercial scale so as to demonstrate performance characteristics

meeting the intended application. These tests are of such a nature that they need

not be repeated unless changes are made in the design of the product. Type

testing, such as an impulse voltage test, is done to prove the specifications of a

new design and is probably restricted to one or two units of each design.

iii) Maintenance Tests

Maintenance tests are usually carried out after maintenance or repair of the

equipment. Normally this tests is conducted according to the schedule provided.

The purpose of the maintenance test is to ensure the lifetime of the equipment is

achieved.

iv) Special Test

Special tests are tests other than the above.

b) Discuss with the aid of suitable diagrams the generation of high voltage

direct current (HVDC) using full wave rectifier circuit.

ANSWER

Full Wave Rectifier for HVDC generation

Full wave rectifiers produce dc voltage less than ac maximum voltage. Ripple or

voltage fluctuation will be present and this has to be kept within a reasonable limit by

means of filters. During the +ve half-cycle, rectifier A conducts and charged up

capacitor, C (smoothing capacitor). During the –ve half-cycle, rectifier B conducts and

charged up C. The source transformer requires a centre tapped secondary with rating

of 2V. The output waveform is shown in the figure below;

c) Lightning impulse voltage (LIV) is simulated in the laboratory using impulse

generator to conduct lightning impulse test on high voltage equipment based

on standard testing procedures. An impulse generator has 5 stages with the

charging capacitor value of 0.2microfarad rated at 100kV, load capacitor of

1000pF, front resistor of 300ohm and tail resistor 2500ohm.

ANSWER

i) Determine the output impulse waveshape of the generator and give comments on

theswaveshape as compared to the standard testing procedure.

Charging capacitor ,C1=0.2/5=0.04μF

Load capacitor ,C2=0.001 μF

Front timet 1=3.0 R 1C1C2

C1+C2

¿3 (300¿ 0.04 ×10−6 × 0.001×10−6

0.04×10−6+0.001×10−6 =0.88 μs

Tail time , t 2=0.7 (R1+R2 ) (C1+C2¿

¿0.7 (300+2500 ) (0.041× 10−6 )=80.36 μs

The front time of the wave is in the tolerance of standard waveshape which is

1 .2μs± 30 % while the tail time is out of the tolerance of standard waveshape which

is 50 μs± 20 %.

ii) What is the maximum output voltage of the generator if the charging capacitor is

charged up to the maximum rated voltage.

DC charging voltage for 5 stages ,V C=5×100 kV=500kV

Maximunoutput voltage ,V O /p=V c

R1 C2 (α−β¿ [exp (−α t 1¿–exp (−β t 1 )…………… (1 )

α=1/R1 C2=3.33× 106

β=1/R2 C1=1×104

substitute the values into equation (1 )gives ,

V o /p=471.88kV

iii) Sketch the 5 stage impulse generator.

ANSWER

Multi stage Impulse Generator

QUESTION 6

a) What is the basic difference between self – restoring and non-self- restoring insulation?

ANSWER

Self-Restoring (SR) Insulation Non-Self-Restoring (NSR) Insulation

1)Insulation that completely recovers insulating properties after a disruptive discharge (flashover) caused by the application of a voltage is called self-restoring insulation

1)This is the opposite of self-restoring insulators, insulation that loses insulating properties or does not recover completely after a disruptive discharge caused by the application of a voltage

2)This type of insulation is generally external insulation.

2)This type of insulation is generally internal insulation

b) Describe the differences between Basic Lightning Impulse Insulation Level (BIL)

and Basic Switching Impulse Insulation Level (BSL) and explain how they affect in

the design of power system network?

ANSWER

Basic Lightning Impulse Insulation Level (BIL) Basic Switching Impulse Insulation Level (BSL)

1)The reference insulation level which is expressed as peak impulse voltage having a standard lightning impulse waveform of 1.2/50ᵤs

1)BSL is the electrical strength of insulation expressed in terms of the crest value of the “standard switching impulse”

2) It is determine by test mode using impulse of the standard lightning impulse wave shape.

2)The BSL may be either a statistical BSL or a conventional BSL.BSLs are universally for wet conditions.

c) Following result were obtained or a self-restoring insulation in a multi-level test

Voltage level (KV) 200 210 220 230 240 250

No of impulse applied 10 10 10 10 10 10

No of Flashover 0 1 5 9 10 10

P = m/n (%) 0 10 50 90 100 100

Determine:

i) 50% Flashover

ii) P.u standard deviation

iii) Statistical withstand voltage

iv) Statistical flashover voltage

v) BIL of this insulation

ANSWER

i) 50% Flashover

refer figure,50% Flashover is 220kV

ii) P.u standard deviation

P = 16% V= 211.8kV

P = 84% V= 228.1kV

Standard deviation

V50 – V16

= 220 – 211.8

=8.2kV

Standard deviation

V84– V50

= 228.1 - 220

=8.1kV

Standard deviation(coefficient of variation)

= 8.1/220

=0.037 pu

iii) Statistical withstand voltage

=(1- 3 (0.037 pu))V50

=(1- 3 (0.037 pu)) 220

= 195.6Kv

iv) Statistical flashover voltage

= (1 + 3 (0.037 pu) ) V50

=(1+ 3 (0.037 pu)) 220

= 244.4Kv

v) BIL of this insulation

Question 7

a) Discuss the suitable diagram the generation of high voltage direct current (HVDC) using full wave rectifier.

ANSWER

Full Wave Rectifier for HVDC generation

Full Wave Rectifier produce dc voltage less than the ac maximum voltage.

Ripple or voltage fluctuation will be present and this has to be kept within a

reasonable limit by means of filters.

During the + half cycle, rectifier A conduct and charge up capacitor,

C(smoothing capacitor)

During the –half cycle , rectifier B conduct and charge up C

The source transformer requires a centre tapped secondary with rating of 2V

The output waveform is shown in the figure below;

b) With the aid of suitable labelled diagram, describe the measurement of high

voltage alternating current (HVAC) using sphere gap and discuss factor of

affecting the sparkover voltage of the gap.

ANSWER

High voltage measurement using sphere gap is based on the BS 359 Standard. In

this standard tables of breakdown voltage are given for various sphere gap

configuration. The sphere gap arrangement is show in figure below.

Sphere Gap For Voltage Measurement

Uniform field spark gap have a sparkover voltage within a known tolerance and can

measure peak value of voltage if gap distance is known. Eg. Vs = 30kV peak at 1 cm

spacing in air at 200C and 760 torr . Arrangement : vertically with lower gap grounded

or unwanted oscillation in the source voltage when b/d occurs.

Factors affecting sparkover voltage of gap:

i) Nearby earthed object

ii) Atmospheric condition & humidity

iii) Irradiation

iv) Polarity & rise time of voltage waveform.

c) Lightning impulse voltage (LIV) is simulated in the laboratory using impulse

generator to conduct lightning impulse test on high voltage equipment. An

impulse generator has 10(ten) stages with the following parameter:

Charging capacitor of 0.2uF rated at 120kV

Load capacitor of 1000pF

Impulse wave of 1.2/50us

i) Determine the front and tail resistors,

ANSWER

Charging capacitor, C1 =0.2/10 =0.2uF

Load Capacitor, C2 =0.001uF

Front time, t1 = 1.2us = 3.0 R1 C1C2/C1 + C2

Then,

R1 = 1.2x10 -6 (C1C2) 3 C1 + C2

= 420 Ω

Tail Time, t2 = 50 us = 0.7(R1 + R2)(0.021x10-6) =0.7(420 + R2)(0.021x10-6)

Then,

R2 = 2981 ohm

ii) What is the maximum ouput voltage of the generator if the charging voltage is

120Kv,

DC charging voltage for 10 stages, Vc = 10 x 120kV=1.2MV

Maximum output voltage, Vo/p = V c ____ [exp (-αt1)-exp(-Betat1)--------------------(1) R1C2(alpha –Beta)

α= 1/R1C2= 2.38x106

β = 1/R2C1 =0.0168x106

Substitute the values into equation (1) gives,

Vo/p = 1.12MV

iii) Multi-stage Impulse Generator

QUESTION 8

a) i)Sketch a single stage equivalent circuit of a lighting impulse generator.

ANSWER

Single stage equivalent circuit:

ii) For the given circuit, give equation for the front and tail times (for 1.2/50 µs wave),

and the efficiency of the generator.

ANSWER

Tf=3 R2C2C1

C1+C2

Tt=0.7 R1(C1+C2)

ɳ=C1

C1+C2

×100 %

iii) If a 1.5 µF capacitor is available, determine the values of others elements of

the circuit assuming a load capacitance of 15pF.

C1=1.5µF

C2=15 pF

1.2μ=3R2(1.5μ)(15 p)

1.5μ+15 p

∴R2=1.2 μ(1.5 μ+15 p)

3(1.5 μ)(15 p)

¿26 ,666.9Ω

50 μ=0.7 R1(1.5μ+15 p)

∴R1=50 μ

0.7 (1.5 μ+15 p)

¿47.62 Ω

iii) A voltage divider with high voltage arm capacitance of 1000pF is now connected

to the output terminal. If a maximum output voltage is 100kV, determine the

minimum value of the low voltage arm capacitance required if the low voltage

reading is not exceed 100V peak.

ANSWER

Vout = 100kVp

CH+CL

CH

=100kV100V

∴CL=100kV CH−100CH

100V

¿1k CH−CH

¿1k (1000 p)−1000 p

¿9.99×10−7

¿0.999 μF

b) With the help of suitable labeled diagrams, describe how a sphere gap can be

used for high voltage measurements. Discuss the advantages and

disadvantages of this method of high measurement voltages.

ANSWER

A sphere gap can be used for high voltage measurements based on the BS358

standard. In the standard, tables of breakdown voltage were for variance sphere gap

configuration.

By following the series sphere gap configure (D and d) one can determine the applied

voltage which causes the break gap by referring K the relevant tables.

Advantages

i) Ease of use anywhere around the world

ii) Standard high voltage values one guaranteed within the series tolerance.

Disadvantages.

i) ±3% or ±5% (for DC) can be too large for today’s HV measurement.

ii) Not suitable for waveform view and continuous readings.

c) Sketch the following circuits:

i) A voltage doublers

ii) A 3-stage voltage multiplier (Cockcroft-Walton)

iii) A 3-stage impulse generator.

ANSWER

i. Voltage doublers

ii. A 3 stage voltage multiplier (Cockcroft-Walton)

iii. A 3-stage impulse generator

QUESTION 9

a) Explain the advantages of generating ac high voltage using cascade transformer as

compared to the normal step-up transformer.

ANSWER

Advantages

- Cheap

- Use less coil.

- Lack of insulation problem

- Small size

b) Explain the method to measure ac high voltage using capacitor voltage divider

& sphere gap.

ANSWER

i) Capacitor voltage divider

The use of capacitive voltage dividers with an electrostatic voltmeter is to eliminate

the errors due to harmonics.

The applied voltage V1 is given by,

ii) sphere gap

-Sphere gaps technique are reliable only for certain gap configurations.

-Normally, only sphere gaps are used. In certain casesuniform field gaps and rod gaps are also used, but theiraccuracy is less.

-The actual breakdown voltage Udat air density d may be found from the tabulated value, Udoby the following formula;

c) Explain the operation of Cockroft-Walton three stage multiplier for direct

current (dc) high voltage generation.

i) Used for higher voltages.

ii) Generate very high dc voltage from single supply transformer by extending

the simple voltage doubler circuit

ANSWER

d) With the aid of suitable diagrams, explain the principle of operation three

stages Mark impulse generator.

ANSWER

The 1st stage being replaced by a pulse transformer. For the trigger process, using

this transformer a voltage pulse with a polarity opposite to the charging voltage of

capacitor C1 is included resulting in an over voltage across the first stage spark gap

FS1. This over voltage causes the spark gap to break down and the Marx-generator

is erected in the usual manner.

As the spark gaps operate in the same way as in the free running mode, i.e. without

an additional trigger electrode, a longer lifetime of the spark gaps is to be expected.

QUESTION 10

a) Discuss in detail about the phenomenon starting with the information of thunderclouds

ANSWER

Terdapat banyak faktor-faktor yang menyumbang kepada pemebentukan atau

menumpuan cas di dalam awan. Secara ringkasnya semasa rebut petir, cas-cas

positif dan negative terpisah disebabkan oleh pengerakkan arus udara yang besar

berserta hablur ais pada bahagian atas awan dan kejadian pada bahagian bawah

awan. Pemisahan ini bergantung kepada ketinggian awan iatu di antara 200 ke 10

000m. suhu di dalam awan boleh mencapai nilai 1 ke 100 ˚C. Manakala voltan pula di

antara 107 ke 108 V dengan medan elektrik di antara 100kV/cm. Tenaga yang wujud

mungkin mencapai nilai 250kWh.

Adalah dipercayai bahawa kawasan arus awan bercas positif manakala bahagian

bawah arus bercas negative. Medan electric pada permukaan bumi boleh mencapai

nilai 300V/cm dibandingkan pada keadaan cuaca biasa (tanpa rebut petir) nilainya 1V/cm.

Disebabkan oleh kecerunan suhu, arus udara bergerak keatas sambil membwa

kelembapan dan titisan air. Suhu ketinggian 4kn ialah 0˚C dan pada ketinggian

12km pula -50˚C. titisan ait tidak akan membeku apabila mecapai suhu 0˚C

sebaliknya mebeku pada -40˚C dalam bentuk hablur ais yang membesar. Suhu

beku efektif adalah di antara -33 ke -40˚C.

Disebabkan beratnya, hablur ais yang tebentuk mula bergrek ke bawah. Pasa masa

yang sama titisan air yang ditiup ke atas oleh arus udara dan seterusnya titisan air

menjadi dingin melampau bergantung pada ketinggian. Oleh itu satu awan petir yang

dingin lampau bergerak ke atas dan satu hujan bergerak ke bawah.

Dalam masa penegrakan tersebut titisan air akan menjadi beku, mula-mula pada

bahagian luar titisan. Satu kelompang yang mengandungi air di tengahnya pun

terbentuk. Apabila air di dalam kelompang itu membeku, kelompang akan pecah

disebabkan oleh pengembangan yang berlaku seterusnya menghasilkan biji-

biji kecil ais yang membawa cas positif. Biji-biji ini kan di bawa oleh arus udara

kebahadian atas awan.

Hujan batu yang bergerak pula membawa cas negative yang sama bnyaknya. Oleh

itu, terbentuklah cas positif pada bahagian atas awan dan cas negative pada bahagian

bawah cas.

b) Briefly describe the followings:

i) Back flashover

ii) Ground flash density

iii) Insulation co-ordination

iv) Temporary overvoltage’s

c) A peak lighting current of 40kA has struck a ground wire a mid span (at the

middle of the transmission towers). If the wire surge impedance is given as

Z=500Ω, calculate the generated voltages at the point of strike. State the

assumption you made from this question

ANSWER

Anggap

i) Zs infinitiii) Tiada balikan voltan pada talian bumi

Zsetup: Z/2

V pada titi ukuran, V=IZ2

= 40 k×500

2

= 10 MV10.

Zl=500ΩZc=500Ω

I=40kA

QUESTION 11

a) Explain what is meant by the terms “T1/ T2 Impulse Wave” and outline the

method of lightning impulse voltage production in the laboratory.

ANSWER

An impulse voltage wave is a unidirectional voltage which rises rapidly to a maximum

value and then decays rather more slowly to zero as shown in the diagram below:

The wave shape is generally defined in terms of the time T1 and T2 in microsecond.

T1 is the time taken by the voltage wave to reach its peak value. i.e from 10% to 90% of

the voltage wave. T2 is the total time from the start of the wave to the instant when it has

declined to one half of its peak, i.e from start of the wave to 50% of the peak during decay.

T2T1

t

V

100%

90%

50%

10%

b) It is required to impulse test a 200pF, 500kV capacitor at twice its rated voltage

using a 50/250 µs impulse. A number of 0.01 µF, 100kV capacitors are available

and high voltage resistors can be constructed as required.

i) Design and sketch a suitable Marx multistage generator utilizing the maximum number of capacitor.

ANSWER

ii) Calculate the maximum peak voltage possible when each capacitor is

charged to its maximum value of 100kV.

Given

C1= 0.001µF, 100kV

C2 = 200pF, 500kV

T1 = 50µs

T2 = 250µs

ANSWER

Twice = 2(500kV)

Choose stages = 10

C1 = the Generator Capacitance

= 0.01µ

10

= 0.001µF

T1 = 50µs

50µ = 3 R1.[C 1.C 2C 1+C 2

]

50µ = 3 R1. [(0.001µ ) (200 p )(0.001 µ )+(200 p )

]

R1= 100kΩ

T2 = 0.7(R1+R2) (C1+C2)

250µ = 0.7(100k+R2) (0.001µ+200p)

R2 = 197.6kΩ

The DC Charging Voltage of 10 stages

V = 10 x 100k

V = 1MV

The Maximum Output Voltage

Vo = V

R 1.C 2 (α−β ).(e−αt 1−e−βt 1)

α=1

R 1.C 2β =

1R 2.C 1

= 1

(100k )(200 P) =

1(197.6k )(0.001 µ)

=50 x103 = 5.061 × 103

Vo = 1 M

(100k× 200 p )(50k−5.061k).(e (−50 k )(50 µ)−e(−5.061 k)(50 µ))

= 767.7kV

QUESTION 12

a) Surge diverters (or lightning arrestors) generally consist of one or more spark gaps in series, together with one or more non-linear resistors in series.

ANSWERINSPIRING CREATIVE AND INNOVATIVE MINDS

Silicon Carbide (SiC) was the material most often used inthese nonlinear

resistor surge diverters.

Zinc Oxide (ZnO) is being used in most modern day. Surge diverters on

account of its superior volt-ampere characteristic. In fact the ZnO arrestor is

often used gap less, as its normal follow current is negligibly small.

The volt-ampere characteristics of SiC and of ZnO non-linearelements are

shown below for comparison with that of a linearresistor.

b) Discuss in detail about the lightning phenomenon starting with the formation of

thunderclouds.

An electrical phenomenon carries the concept of charges involvement. So two

types of charges are the reasons for the cloud to be considered as a cell. The

charges are positive type & negative type.

ANSWER

Figure shows the typical thundercloud structure.

Not all clouds are lightning cloud generator. It is only the cumulonimbus cloud type

that can generate lightning. The ice Splinter can be used to explain the electrification

of the cloud. The moistures and precipitation particles being is suspension in air and

due to upwards action of updraft, causing super cooling to take place and resulting

moisture to become ice.

The ionic migration of OH- & H- in the moisture built, leaving the OH-in the front and

H- being lighter are pulled out to settle in the outer layer. The resultant two-layer ice

structure split due to different rate of ice expansion ( the inner & the outer layer).

The splinters are basically of positive-charged & the negative-charged. The lighter

splinters are pulled upwards while the lighter negatively charged splinters settle at the

lower point of the cloud. If the electric field between the cloud & ground exceed the

dielectric strength of air streamers will appear and propogate towards the ground.

The last jump of this streamers to the ground result in upward streamers to move to

attach itself to the downwards-moving streamers. The attachment results in the

process of charges neutralization of the positive & negative charges. This is known as

return strike. It causes large currents to flow to the ground.