HIGH VOLTAGE
TECHNOLOGY
QUESTION AND ANSWER
PREPARED BY
ASSOC. PROF. DR. ZOLKAFLE BIN
BUNTAT
CHAPTER 1
ELECTRIC FIELDS
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.
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
1st iteration 2nd iteration
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
to 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
Where the average is field in the gap and is equal to the applied potential/gap separation between the electrodes
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
Dependent on material purity, T and E
Due to polarization effects, σ also depends upon time of application of the stress
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
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 0kV 1 2 3 6 5 4 7 8 9 50kV 50kV
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 photo-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)
Substituting two different set of values:
and =>
Therefore, α =5.76 ionizing collisions/cm The breakdown criterion is given by:
Therefore the Townsend‟s secondary ionization for the breakdown to be occurred at gap distance 0.9 cm is:
c. In gas, the effective ionization coefficient is given by: = 27.7[ ] – 2460
Where α is the effective ionization coefficient in , 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 ∫ , 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 at 4 bar and 60 °C. ANSWER: Given Therefore, – ;
Multiply both sides with , gives
Given; ∫ The normalized pressure of 4 bar at 60°C is;
From (1) and (2);
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:
Since ( )
Then at breakdown, ( )
» (
) ,( ( )- +
The ionization of gases is related to
( )
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: Iteration 1:
Iteration 2: Therefore:
QUESTION 5
a. State the most useful equation which can be used directly to solve electric field problem using Finite Difference Method?
ANSWER:
]
Where , , , 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:
Iteration 2: Iteration 1:
Iteration 2:
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 changedforce unionof 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
Iteration 1: Iteration 2:
So: Potential at 1 = 100 V Potential at 2 = 98.90 V Potential at 3 = 111.76 V Potential at 9 = 52.4 V Potential 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. ii.
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:
For node 2:
Not connected Not connected 50V V = 0 1 2 3 4 5 6 7For node 3:
For node 4:
For node 5:
For node 6:
For node 7:
Iteration 2
Iteration 3:
Summary of node potential
CHAPTER 1
INTRODUCTION TO
HIGH VOLTAGE TECHNOLOGY
(QUESTION AND ANSWER)
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;
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 1 2 3 4 5 6 Figure Q1 ANSWER: 1st iteration 2nd iteration L1 L2
Ground
The potential at point 2, 4 and 6 after 2 iterations are; V2 = 61.97 kV, V4 = 21.63 kV, V6 = 47.73 kV
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: ⁄ { ⁄ ] }
where,
p – pressure
Vs – sparkover voltage
F & g – different functions
ANSWER:
By neglecting attachment, breakdown criterion ( ) ………….. (1)
Since (Paschen‟s Law), ⁄ ⁄ and ⁄ , where significant different function.
At breakdown, ⁄
⁄ ………. (2)
And ⁄ ⁄ ……… (3)
Substitute (2) & (3) into (1) gives;
⁄ { ⁄ ] }
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
ANSWER:
Using equation,
For d = 1.3 cm, For d = 2 cm, (This suggests that for this gap γ starts to be active).
The value of can be found from the equation;
] 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
(for homogeneous dielectric medium)
Where v is the average field in the gap and is equal to the applied potential/gap
Emax = f.V
d
Field utilization factor µf = (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
dependent on material purity, and –
due to polarization effects, σ also depends upon time of application of the stress
ii. Dielectric permittivity
Dependent on , 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
is dependent on , Power loss =
⁄
iv. Loss angle see (iii)
v. Dissipation factor Similar to
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 6 5 4 7 8 9 0kV 50kV 50kV
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:
( ) ( )
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:
( ) ( ) (i)
The Townsend‟s breakdown criterion;
] (ii) Substituting (i) Into (ii);
( )
(iii)
Taking in on both sides of (iii)
* + ] (vi)
For uniform field;
b
Therefore,
Or
b ] ; b
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 P schen’s L w.
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
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 3.04 3.26 3.47 3.69 3.81 4.38 4.66 5.02 5.54 6.06
Taking In on both sides of (1);
,
i. Plot graph versus ;
From the graph, interception © at I , gives;
(12-4=8) (5-3.5=1.5) , 2.94 3.04 3.26 3.47 3.69 3.81 4.38 4.66 5.02 5.54 6.06 0 1 2 3 4 5 6 7 In I
ii) Townsend‟s primary ionization coefficient, α ; Gradient of the graph (m) shows the value of α.
Townsend‟s secondary ionization coefficient,
]
Substituting for higher value of;
– – ] 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.
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 is kept constant)
Substituting two different sets of values;
(1) and (2) (1) ÷ (2) α= 5.76 ionizing collisions/cm
The breakdown criterion is given by;
Therefore the Townsend‟s secondary ionization for the breakdown to be occurred at gap distance 0.9cm is; QUESTION 7
In SF6 gas, the effective ionization coefficient is given by;
[ ]
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, ∫ , 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: [ ] ; ; (1) ∫ ; (2)
The normalized pressure of 4 bar at 60 °C is;
From (1) and (2) 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:
P1 P2 P3 P4 P5 P6 P7 P8
Iteration 1 Iteration 2:
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
[ ] [ ]
Then at breakdown, * + * +
[ ] . [ ] /
The ionization of gases is related to
[ ]
i.e. 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.
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
Nonoptimize
Optimized
2
4
6
8
10
15
20
25
30
0
Distance Z (cm)
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: Vaa Vbb
Iteration 1 Iteration 2:
QUESTION 12
Coefficient, α is given by; [ ] where, d- gap distance - total current - initial current ANSWER:
Total no. of electron at anode, . At steady state, average current in gap at distance x,
and
Total number of current,
* +
[ ]
b. The following data in table Q12b are given for two parallel plates while the electric field, E is kept constant.
Gap distance, d(cm) Ratio of Current and Initial Current, 0.5 1.2 1.3 1.6 2.0 2.3 ANSWER: By using equation, For d = 0.5 cm, or For d = 1.3 cm, or For d = 2 cm, or
(This suggest that for this gap start to be active) The value of can be found from the equation;
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 is doubled. The values for the ratio of electric field and pressure, and ratio of first ionization coefficient and pressure, are given in Table Q2c.
Table Q2c (v/cm mm Hg) (ion pairs/cm mm Hg) 41 0.0196 42 0.0222 ANSWER: d = 22.8mm = 2.28 cm, p = 200 mm Hg = 19.15KV = 42 v/cm mm Hg, = 0.0222 = 41 v/cm mm Hg, = 0.0196
Find when is doubled?
From secondary Townsend Breakdown Process,
and E = Breakdown criteria: ( ) or E = = 8.40 KV/cm = 8400V/cm = 42 v/cm mm Hg From table; = 0.0222 = 0.0222(200) = 4.44
From breakdown criteria (is doubled, ), - ) = ln 2 - = By Interpolation; v/cm
41
42
2
0.0196
0.02068
0.0222
2
𝛼 𝑝0
QUESTION 13
a. State the most useful equation which can be used directly to solve electric field problem using Finite Difference Method?
ANSWER:
]
Where , , , 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:
Iteration 2: Iteration 1:
Iteration 2: QUESTION 14
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
⁄ { ⁄ ] }
Where,
- gap distance at sparkover voltage - pressure
- sparkover voltage - different function
ANSWER:
By neglecting attachment, breakdown criterion ( ) ………….. (1)
Since (Paschen‟s Law), ⁄ ⁄ and ⁄ , where significant different function.
At breakdown, ⁄
⁄ ………. (2) And ⁄ ⁄ ……… (3)
Substitute (2) & (3) into (1) gives;
⁄ { ⁄ ] }
QUESTION 16
Breakdown voltage measurement of a uniform field gap in air at 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
Determine the breakdown voltage of a 20 mm gap at a pressure of 3 bars and temperature of . ANSWER: ⁄ ⁄ ⁄ ⁄ ⁄ ⁄ ⁄ From the data given,
⁄ ⁄
…………. (1) 26.0 ⁄ ⁄
………. (2) From (1) and (2); A=23.55 and B =6.42 For the case of atmospheric air;
⁄ ⁄ ⁄
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
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.
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:
Iteration 2:
Therefore:
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.
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 .
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 is doubled. Data‟s for the ratio of electric field and pressure, and ration of first ionization coefficient, are given in the Table Q19.
Table Q19 (V/cm mm Hg) (ion pairs/cm mm Hg) 41 0.0196 42 0.0222 ANSWER:
Find Vswhen
is double?From secondary Townsend Breakdown Process,
] Breakdown criteria: ] From Table:
From breakdown criteria ( is double, ) By interpolation 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:-
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.
ii) Using Poisson‟s equation if the medium has a space charge density.
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: Iteration 2:
Iteration 3 :
Therefore:
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 = ]
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 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 , i – ions do not have enough energy to ionize gas molecules directly but may release electrons on colliding with the cathode surface.
ii) Photon, 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, 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
= i + p + 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 electrons are produced at the cathode per ionizing collision in the gap, then n0 „ = n0” ] Thus , no “ = no + n0” ] no “ = ] Therefore n(d) = no “ exp(αd) = ]
Under steady state conditions , I = ]
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.
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, ( ), determine;
i. The relative air density ρ referred to standard atmospheric conditions of 1013 25 mbar and 20°C. ANSWER:
0.90kV
Vs
C,
30
,
1034
,
25
.
0
cm
mbar
t
d
ii. The value of constants A and B
ANSWER: 9 9 3 g/m V V Vll ∆
V
V9(1 ∆V
)kV
91
.
0
)]
2
(
002
.
0
1
[
90
.
0
(1 /m3 V12 g Vll ∆V
)kV
38
.
88
)]
1
(
002
.
0
1
[
56
.
88
30 273 20 273 1013 1034 20
99 . 0 303 293 1018 1034 88.56kV Vs C, 15 , 1180 , 7 . 2 cm mbar t d
288 298 1013 1180 20
21
.
1
________ 2 2 2 2 2 1 1 1 1 1 d B d A V d B d A V s s 5 . 0 25 . 0 81 . 1 27 . 3 91 . 0 25 . 0 38 . 88 27 . 3 Biii. The breakdown voltage of a 3 cm gap spacing at a pressure of 3000 mbar and a temperature of 20°C. ANSWER: 5 . 0 25 . 0 81 . 1 27 . 3 5 . 0 91 . 0 81 . 1 38 . 88 5 . 0 25 . 0 81 . 1 27 . 3 91 . 0 38 . 88 A B A
2
______
)
81
.
1
(
)
27
.
3
(
38
.
88
1
________
)
5
.
0
(
)
25
.
0
(
91
.
0
_______
81
.
1
27
.
3
)
7
.
2
(
21
.
1
5
.
0
25
.
0
)
25
.
0
(
99
.
0
2 2 2 2 1 1 1 1B
A
B
A
Dari
d
d
d
d
75
.
35
19
.
1
54
.
42
45
.
0
64
.
1
65
.
1
19
.
44
)
25
.
0
(
81
.
1
)
5
.
0
(
27
.
3
)
91
.
0
(
81
.
1
)
5
.
0
(
38
.
88
07 . 16 19 . 1 10 . 22 98 . 2 19 . 1 ) 25 . 0 ( 38 . 88 ) 91 . 0 ( 27 . 3 C
20
t
3000mbar,
3cm,
d
98
.
2
d
) 98 . 2 ( 07 . 16 ) 88 . 8 ( 75 . 35 Vs QUESTION 23a. 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
88 . 8 ) 3 ( 96 . 2 96 . 2 293 293 1013 3000 20 d
kV
57
.
269
89
.
47
46
.
317
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.
