ELEC4611-15-Lec2 - Electric Stress in HV Equipment

ELEC4611: Electric Stress in High Voltage Equipment p. 1

ELEC4611 Power System Equipment

ELECTRIC STRESS

IN HIGH VOLTAGE EQUIPMENT 1. DESIGN CONSTRAINTS All electrical equipment requires insulation to prevent the possibility of dielectric breakdown and unwanted fault current flow. Insulation is also required to provide for the safety of personnel who may be in the vicinity of the equipment. The insulation is needed whether the equipment is a 500kV high voltage transmission line, a 230V appliance or an integrated circuit with voltage of only a few volts DC applied. It is not the total voltage over the insulation that is important in the equipment design, but the electric field (or stress) applied to the insulation at particular parts where the stress may be concentrated. When the electric field is considered the field magnitudes may be similar in equipment of greatly varying voltages. For example, modern integrated circuits, with ever-decreasing gaps between conducting tracks, are now having problems with dielectric breakdown across the substrate surface. If the gaps are of micron level, even a few volts can give an electric field of a few kV per mm, similar to those that occur in high voltage equipment insulation. In designing the insulation system for an item of electrical equipment or a number of items making up an electrical


system, a number of factors must be taken into account to achieve a viable design of the insulation system that will prevent breakdown under normal conditions and under typical anticipated abnormal overvoltages and operating conditions. The factors that must be considered include the electrical, mechanical and environmental requirements: these requirements are listed and discussed below. 1.1 Electrical requirements These include both operational aspects and also the asset management aspects of maintenance, testing and condition monitoring requirements. The various considerations that must be taken into account are:

Operational: (1) Whether AC or DC voltage (2) Frequency and magnitude of impulse voltages (3) EMC and EMI requirements (4) Operating temperature and ambient conditions (5) Environment: e.g. indoor or outdoor (6) Insulation coordination with other equipment (7) Expected lifetime of the insulation (8) Load duty cycles

Testing: (1) Continuous over-voltage: AC or DC (2) Insulation resistance (3) Temperature rise (4) Impulse tests: lightning and switching impulses


(5) Dielectric dissipation factor (DDF, tan) (6) Partial discharge testing (7) Continuous on-line monitoring requirements 1.2 Mechanical requirements These requirements vary considerably from one item of equipment to another. The range of possible considerations includes any of the following: Compression and tension capacity of insulators for

overhead lines, transformer winding spacers etc. Tensile strength of electrical conductor insulation. Vibration response of insulation in transformer coils

and circuit breakers. Bending stresses on underground cable insulation. Electro-dynamic forces on insulation of rotating

machine end windings, transformer windings, cables and insulated busbars.

Expansion of insulated electrical contacts and similar joints.

Pressure rise effects on insulation of sealed items with internal arcing.


1.3 Environmental aspects The insulation must be compatible with the prospective environment in which it is to operate. This will include (possibly) the following aspects: Pollution of the air: humidity, salt spray, industrial

pollution The ambient temperature (this will affect the thermal

ratings) Possible presence of moisture on insulators, cables etc Effect of radiation (UV): this can cause polymeric

insulation degradation. Wind conditions and impact on overhead lines Lightning strike frequency (keraunic level) Flora and Fauna interaction: e.g. trees and power lines

and birds etc Effect of extreme temperatures: e.g. in sub-arctic

regions or in cryogenic equipment 1.4 Operational Fields When the insulation material type and general configuration has been determined within the above constraints, it is then necessary to know the operational impact of the electric stresses (fields) that will be imposed on the insulation during normal and abnormal operation. There will also be a


need to know how the insulation will deteriorate under typical operational conditions. This will then define the appropriate tests and monitoring procedures necessary to achieve a required insulation life. It is thus necessary to calculate the electric fields under such normal and abnormal conditions. In early equipment designs the electric field determination in complex equipment such as transformers was very difficult and it was necessary to use electrolytic tanks to plot fields in scale models of the equipment. Even this was not ideal because of problems of achieving 3-D results in an essentially 2-D test tank configuration. As a result a large degree of redundancy was generally incorporated in the insulation design. However modern high speed computer capabilities, the development of finite element analysis methods of field calculation for 3-D configurations and better knowledge of material properties allow much more accurate modeling of the insulation operation and this has allowed much better design of insulation and more economic use of materials. However the trend to operation of insulation at or close to its limits with the minimum material requirements means that insulation life may be reduced and this will require better methods of condition monitoring and assessment to be developed and applied. 2. ASPECTS OF THE ELECTRIC FIELD IN HV

INSULATION


In many cases the electric field distribution in the insulation of electrical equipment is quite non-linear and modern design methods require the application of finite element analysis packages to solve Laplaces equation to calculate electric field levels and distributions. However there are also many applications where the electric fields are reasonably well-defined and amenable to simple calculation using standard formulae. The following section discusses some of the simpler field structures in equipment and uses them to illustrate typical insulation requirements and constraints. 2.1 Bare parallel conductors

We consider two bare electrical conductors of radius a, separated by distance d, with a voltage V between them, typical of a single phase overhead line (the three phase line is a simple extension of the following). The electric field in the air between the lines is:


V/m2 ln

VE rd ar

a

The maximum field occurs at the conductor surface r = a and is:

V/m2 ln 2 ln

V VE ad a da a

a a

Note that E is independent of the dielectric permittivity ( = ro) and depends only on the voltage on the insulation and the geometry. The above equation holds for both AC and DC voltage. This independence of permittivity is the case whenever the insulation is homogeneous. However when the insulation has different materials, the permittivity will be involved for AC fields only. For DC fields with mixed insulation the insulation resistivity will be involved. The insulation of such lines is (almost) always air and if the maximum electric field level is high enough it may cause ionization of air molecules at the conductor surface, with the result that corona discharge is caused. This will generally occur when the electric field level in the air at normal pressure is 30kV/cm or higher. The insulation design must thus determine the peak operating electric field and keep it below that level during normal operation. Typical operating peak field values may be 15-20 kV/cm at the conductor. Either the radius a or the separation d can be increased to reduce the electric field. For overhead lines,


typical values of d and a are (for 11 kV for example) about 100 cm and about 1 cm respectively. 2.2 Coaxial Cables

For a coaxial cable with inner (HV) conductor radius a, outer metal sheath inner radius b and with a voltage V across the cable insulation, the electric field distribution over the insulation is:

( ) V/mln

VE rbra

and the peak field occurs at the inner conductor:


( ) V/mln

VE abaa

Thus to reduce E(a) in the insulation we must increase either b or a. Increase of a is obviously the most effective method. It should be noted that the insulation must be used to withstand the peak field at the inner conductor, but this means that the outer insulation is then very under stressed and thus inefficient in terms of its economic utilization. Note that the field E is independent of the permittivity ( of the insulation if the voltage is AC. The field E depends only on the voltage V and the geometry, providing that the insulation is homogenous. For DC voltage the field E is determined by the electrical resistivity () but is given by the same equation and so is independent of resistivity if the material is homogeneous with constant . Typical maximum operating values of E in modern cross-linked polyethylene (XLPE) cables are about 200 kV/cm (20 kV/mm). This is about ten times the maximum operating electric field level able to be used in air. 2.3 Parallel plate capacitor For a voltage V across the two plates of a parallel plate arrangement, with plate separation d, the electric field in the


insulation between the plates is uniform and constant and is given by:

V/mVEd

This equation is applicable to power capacitors where, although they are made up of very large areas and lengths of flexible dielectric with metal foil layers on either side, they are nevertheless simple parallel plate geometries and the above equation can be used for electric field determination. Such power capacitors (used extensively in power systems for power factor correction applications) have arguably the highest operating electric field levels of any item of HV power equipment. Typical electric fields in such capacitors are up to 50 kV/mm, with plate (foil) separations of about 10 microns (10 x 10-6 metres). The same parallel plate geometry is also used for heating of non-conducting dielectric materials with high voltage application at high frequency to generate dielectric heating in the insulating materials. The power density p and total power P generated by an AC field is:

2 3 2 W/m cf Wr op E tan P CV tan

where p is power density, P is total power, = 2f (where f is the application frequency), r is the relative permittivity of the material, and o is the vacuum permittivity.


2.4 HV Bushings When high voltage conductors have to be taken into earthed chambers such as the tank of a transformer, an insulating bushing is needed to insulate the HV conductor from the earthed metal of the tank to prevent breakdown. The bushing consists simply of a layer of insulation between the conductor and the flanges of the tank wall opening. It is usually the case that high voltage bushings are coaxial in their structural geometry and thus the electric field will be determined from the same equation as given for the electric field in the coaxial cable above.

( ) V/mln

VE rbra


( ) V/mln

VE abaa

E(a) is the peak value of the electric field. As noted previously for cables the 1/r variation of the electric field with radius means that the outer regions of the insulation are under-stressed and are not utilized effectively in terms of material economy. While it is difficult to change the field distribution in cables because of their long length and consequent manufacturing difficulties, it is possible to design and manufacture HV bushings so as to make the electric field level more uniform and thus to reduce the material dimensions required. Bushings, particularly those used for very high voltages of 66 kV and above are often capacitively graded by subdividing them into a number of insulated coaxial layers each separated by a very thin layer of metal foil. The result is that the bushing is effectively a number of series capacitors connected between the HV conductor and earth (they are called capacitor or condenser type bushings). By varying the axial length of the individual foil layers it is possible to specify (grade) the separate layer capacitances in such a way that the voltage across each layer is controlled and thus the electric field in each layer can be controlled to make better overall use of the insulation thickness. The capacitance of each isolated layer will determine the voltage across, and hence the electric field in,


each layer. This design method will be outlined in detail later. 2.5 Electric Field Distribution in AC and DC High

Voltage Equipment When a single homogenous insulation material is used for HV equipment, the manner in which the voltage is distributed over the insulation volume is determined by the capacitance distribution for AC voltage application and by the electrical resistance variation in the case of DC voltage application. As is well-known for any dielectric material which is subject to an electric field, the resistance and capacitance of the insulation configuration are related by the equation:

RC =

or R 1/C and, because we have V = Q/C for a capacitor and V = IR for a resistor, we have:

dV d(1/C) for a capacitor and

dV dR for a resistance


Thus, because of the constancy of the RC product and the above proportionalities, it would be expected that AC and DC electric fields would have identical distributions for AC and DC excitation for the same insulation material and dimensions. This equality is, in fact, the case but only under a certain condition. In practice, this required condition is not met by the insulation in operating equipment and thus the AC and DC field distributions can vary significantly. The required condition is that the values of and are constant over the whole insulation and do not vary during operation. While this is true of the permittivty of an insulator, it is certainly not true for the resistivity of an insulation material. In particular, the resistivity of an insulator material is very dependent on material temperature and exhibits an exponential variation with temperature because of the effect of temperature on electrical conduction in dielectrics. Increased temperature will generate higher energies on the structure and thus free more electrons and allow an increase in leakage current flow and thus decrease in resistance of the material.

Ileakage exp(aT)

Rleakage exp(aT) The end result is that while the AC field distribution is constant and invariable in operating insulation subject to AC voltage, when insulation is subject to DC voltage, any non-uniform temperature gradient produced by operation will cause changes in the resistivity distribution and thus


the electric field distribution will change substantially with temperature gradient and will be different to the AC distribution. Thus, in the case of a coaxial bushing with a radial temperature gradient, under AC excitation the field will have its standard 1/r variation as above, but for DC excitation with a temperature gradient (high T at the conductor, low T at the earth flange) the electric field distribution may be inverted, with almost a linear variation with r. 3. MULTI-DIELECTRIC STRUCTURES In many high voltage insulation systems multi-dielectric structures are used with dielectric materials with different relative permittivities in order to provide better use of insulation by reducing the overall insulation thickness and making the field more uniform within the overall layer structure. In other cases the use of multi-dielectric structures is unavoidable, such as in the case of a parallel plate gap with a slab of insulation material inserted for dielectric heating purposes as described previously. An air gap must be left to allow access for insertion and removal of the object to be heated and the electric field in the air gap and the dielectric breakdown properties of the air at the application frequency will determine, to great extent, the maximum allowable


electric field in the material to be heated and thus the power density of the heating in the material In both of these cases the relative permittivities (r) (and hence the total permittivities, ) are important because the voltage across the different dielectric layers is determined by the capacitances of the different dielectric layers and the capacitance is dependent on the total permittivity of the layer material. The electric field is thus determined by the permittivity also. [Note that this will only be true for AC voltage application: when DC voltage is used across multi-dielectric structures, the voltage across each layer will be determined by the electrical resistance of the various dielectric layers and the voltage distribution may be different to that when AC is used as the resistivity and permittivity variation will be different for different materials]. 3.1 Parallel plates with two dielectrics

In the above, two different dielectrics are used with different total permittivities 1 and 2.


We have V = V1 + V2 and d = d1 + d2 and also 1 2 1 1 2 2V V V E d E d The field properties of dielectrics require that the electric displacement D (the total charge in this case) is constant in each material: i.e.

1 2 1 1 2 2 = constantD D D E E wherer1andr2 Using these relationships we get:

1 11 1 2

2

11 2 1

2

EV E d d

d d E

and

11 1

1 11 2 1 2

2 2

V dE V Vd d d d


22 2

2 22 1 2 1

1 1

V dE V Vd d d d

[Using the method of capacitance voltage division to determine the voltages, we get, for example:

2

2 21

1 2 1 2

1 2

1

11 2

2

=

=

AC dV V V

C CA

d d

d Vd d

That is, the same result as was obtained above.] The result can be generalized for any number of dielectric layers. The electric field and voltage in the kth layer is given by:


1 2 31 2 3

1 2 31 2 3

-------

. -------

kk k k

kk k k

k k k

VEd d d

dV E d Vd d d

For a flat gas-filled void in a slab of solid dielectric, the electric field in the void can be elevated to a level which is higher than the breakdown field strength of the gas. Consider the simple case of the gas void shown below:

The electric displacement (D) is not a medium dependent property and is thus the same value in both the void and the solid insulation, so that we have:

( )

( )

constant =

void void solid solid

r solidsolidvoid solid solid

void r void

D E E

E E E


Because the void is gas-filled, with relative permittivity r = 1.0, we have

( )void r solid solidE E and since r(solid) is typically about 2-4, the electric field in the gas-filled void is typically 2-4 times the electric field in the surrounding solid dielectric material. The designed operating field Esolid in solid dielectrics is normally up to 10 times that possible in air, because of the much higher breakdown strength of solids as opposed to air. Thus the result is that the electric field in the void will almost certainly be high enough to initiate a gas discharge (a partial discharge) in the air void. The temperature of the ionized gas in the discharge is very high and this will cause some chemical change (e.g. carbonization) in the insulation material and degrade it, and thus start deterioration of the solid insulation material at the walls of the void. For example, in the most commonly used power cable insulation material, cross linked polyethylene (XLPE), the relative permittivity is 2.3 and the breakdown field strength is 50 kV/mm. The normal maximum designed operating electric field in XLPE is typically about 20 kV/mm. Thus if an XLPE slab operating at 20 kV/mm has an air void contained within it, the electric field in the air in the void will be 2.3 x 20 = 46 kV/mm or 460 kV/cm. The breakdown field strength in air at normal pressure is 3 kV/mm or 30 kV/cm. Thus breakdown (or partial discharge


breakdown) will certainly occur in the air in the void under these conditions. Eventually the damage will increase the void size until full breakdown of the whole insulation thickness may occur with catastrophic results. 3.2 Coaxial geometry In this case of two coaxial layers with different permittivities 1 and 2, the peak electric field in each layer (at the inner radii of the respective layers) is given by:

2 3

1 1 2 2

1

ln lnx xVE r r rr

r r

max(1)32 1

11 2 2

max(2)32 2

21 1 2

ln ln

ln ln

VErrr

r r

VErrr

r r

V = V1 + V2


Example If r1 = 20mm, r2 = 25 mm and r3 = 31 mm, and the two layers are composed of: (1) Paper (r1 r2): with r = 3.2 and electrical resistivity = 1014 m (2) Oil (r2 r3): with r = 2.2 and electrical resistivity = 1013 m Determine the peak electric fields in each layer of insulation if the total applied voltage is 210 kV(peak). For AC application Application of the previous formulae for coaxial geometry gives: E1m = 19.6 kV/mm [V1 = 87.4 kV] E2m = 22.8 kV/mm [V2 = 122.6 kV] For DC application For the same voltage level, but DC, where the electrical resistivity controls the voltage distribution the two field levels are, E1m = 42.9 kV/mm [V1 = 192 kV] E2m = 3.4 kV/mm [V2 = 18 kV]


If there is a temperature variation in the materials, the resistivity will change and hence the electric field distributions will change from the above. 3.3 Improvement of the insulation use The example sketched below shows the effect of using different permittivity layers for the insulation. The different permittivities of the layers change the capacitances and this changes the voltage distribution and thus the electric field distribution in the overall insulation structure. The use of different dielectrics with different permittivities to make better use of the insulation allows either of two results to be achieved: (i) An increase in the allowable total voltage across the

insulation [e.g. in the example below, the 200 kV AC is able to be increased to 264 kV]

(ii) A decrease in the insulation overall thickness [e.g. in the example below, the outer radius b=40mm decreases to b=35.2 mm]

[Note that for DC voltage where voltage is distributed according to resistance, the resistivities are temperature dependent so that the distribution of the electric field distribution is load dependent. It is possible for the maximum DC electric field to occur near the outer edge of the coaxial layers because of the temperature effects.]


. lnT mbV E aa

For VT = 200 kV, a = 20 mm, b = 40 mm: Em = 14.4 kV/mm


V1 = Em a ln(b1/a) (b a)/4 = 5mm b1 = 25 mm For Em = 14.4 kV/mm V1 = 64.3 kV etc VT = 64.3 + 65.6 + 66.6 + 67.3 = 264 kV [ or for VT = 200 kV, V1 = 50 kV, gives b1 = 23.8 mm and new b = 20 + 4x3.8 = 35.2 mm ]


Multiple dielectric structures are difficult to manufacture so that they are used only in special applications such as insulating bushings. Layered bushings, however, are relatively easy to make and are commonly used to grade voltage distribution. They do not use different insulation materials but use capacitive grading of the insulation layers. The capacitance is graded by using different lengths of the foil layer sections to alter capacitance. When the voltage distribution is changed, the electric field is also changed (graded). 3.4 Capacitively-graded (condenser) bushings The general construction of a capacitance-graded bushing is shown below. The layers are normally of the same thickness and are separated by thin metal foil layers to act as the capacitance electrodes. The axial length of the foil (l) determines the capacitance and hence the voltage and electric field. The electric field and capacitance in each layer are given by:

V/mln

VE rbra

and

2 Fln

lCba


Capacitor bushing in a porcelain housing and the electric field distribution.

Comparison of equipotentials of (a) a non-graded bushing and (b) a capacitor bushing. Note the relative distribution of the voltage in the

insulation within the flange area.


General structure of the layers for the calculation of required foil lengths ln

For the various layers in the above structure, we can write:

1

2 . Fln

nn

n

n

lCaa

1

V/mln

nn

nn

n

VE rara

1

V/mln

nn

nn

n

VE aaaa


The total capacitance is:

11total

n n

C

C

and the layer voltages are:

totaln

n

CV VC

The other condition that holds for the layers is that the charge (also the electric displacement D in this case) on each capacitance much be the same: i.e

Q = constant = C1V1 = C2V2 = C3V3 = = Cn Vn 3.5 Bushing design There are two approaches to implementing the design of the capacitively graded bushing insulation to improve efficiency of insulation use: (i) Designing to have Vn = constant for each layer

i.e. totalnVV

n


(ii) Designing to have En(max) (= En(a)) = a constant for each layer.

Having decided which method is to be used, it is then necessary to find the various values of foil lengths ln needed to achieve the result for either condition (i) or (ii). It is normal to have the thickness of each insulation layer identical. Example: (ii) To have the same peak electric field in each layer. We have Q = Cn Vn = constant. Thus we have:

1

2 constantln

nn n n

n

n

lC V Vaa

(1)

Also we require:

max1

constantln

nnn

nn

n

VE E aaaa

(2)


We divide equation (1) by equation (2) to get:

1

max 1

ln2 . . .ln

nn

nn n

nn n

n

aaal VQ

E Vaa

2 . . a constant

n nl a

Thus we have lnan = a constant as the requirement for this design. If we define the length l1 of the first (foil) layer and also specify the radii an then this will define the other l values to achieve the required result. Note that it is the length of the foil conductor (capacitor electrode) that determines the capacitance, NOT the length of the dielectric layer. In some cases the outer foil layer on a section may be the same length as the (inner) dielectric layer, but in other cases the outer foil layer of a section may be the same length of the outer dielectric layer that it is part of.


4. EFFECTS OF LEAKAGE CAPACITANCE ON

VOLTAGE DISTRIBUTION In graded bushings the leakage capacitance to earth is broken up into several different series capacitances which are adjusted in their values to achieve better and more efficient voltage distribution over the insulation of the bushing. However in many cases the leakage capacitance to earth of any metal components at elevated voltages will also affect, in an undesirable manner, the voltage distribution in some particular insulation design configurations that involve the metal components. This effect of earth-leakage capacitance must be taken account of in the design and if necessary some counteractive changes may be necessary to restore a more uniform voltage distribution. The two most commonly-encountered insulation configurations that are affected by leakage capacitance are:

(i) the insulator string used to support overhead line conductors

(ii) the windings in transformers or the stator windings in rotating machines

In both of these configurations the primary insulation between the live line and the neutral has an equivalent circuit that contains a number of separate capacitance sections and the problem that arises is one of a non-linear


voltage distribution, caused by leakage capacitance to earth, over these capacitance sections. The non-linear distribution means that some insulation may be more highly stressed by voltage: this may need to be corrected by design. In the two examples above the non-uniformity effect occurs at different conditions. In the insulator string the effect is present at power frequency voltage at all times. In the windings, the non-linearity only occurs for high frequency transient overvoltages. 4.1 String insulators on overhead lines Insulator strings on overhead lines are connected to the steel towers or to the wood or concrete poles, all of which are effectively at earth potential. The strings support the HV conductors and thus have the full phase voltage applied over them. The strings are actually composed of a number of series connected cap and pin insulator units. Each individual cap and pin unit is an insulator with a rated voltage of about 10-12 kV. For higher voltage lines, the number of cap and pin units in the string is increased accordingly. An 11 kV line may have just one unit while a 330 kV line (197 kV phase voltage) may have 20 or more units in a string. Each cap and pin unit represents a single identical capacitor (of about 30-40 picofarads) and thus each string is a system of series capacitances of equal value. The string should ideally have a uniform voltage distribution along its length


with the same voltage across each unit. However this does not happen because each cap and pin unit will also have some leakage capacitance from the cap to earth (represented by the steel tower or pole) and this will distort the string voltage distribution and make it non-uniform. As a result the HV line end units have a higher voltage across them than the tower (earth) end units. The cap and pin insulators have a steel cap and pin which are designed to slot together and attach to each other. They also represent the two electrodes of the equivalent capacitor unit. The insulator material is either porcelain or glass and this is cemented to the cap and pin sections. There is a long surface path with the various skirt sections underneath to prevent surface breakdown due to any surface contamination, such as water and/or dust and salt. There is a trend now to use polymeric insulation to replace these porcelain units. The polymeric or composite insulators comprise a fibre-glass core embedded within a polymer outer cover. In contrast to the cap and pin units, composite insulators are formed in one single extended section. The polymer is generally either a silicone material or EPDM (ethylene propylene diene monomer). Silicone has the better characteristics.


Cap and Pin insulator detail.

Two strings of 28 cap and pin units for a 400kV quad-conductor transmission line.


Equivalent capacitance circuit for a line support with four cap and pin units.

C is the self-capacitance of each cap and pin insulator unit. Typically, for porcelain units, C is about 30 40 pF and Co is leakage capacitance to the tower or pole earth. In the diagram, Co is the leakage capacitance to earth for each unit. It is taken as constant because the distance to the main frame of the tower or the pole is similar for each unit in the string. Typically, Co is about 0.1 0.2 C or about 3 6 pF A ratio of C/Co = 10 is about the highest possible for an overhead line: more typically the ratio might be about 5.


The higher the ratio of self-capacitance to leakage capacitance the smaller is the effect on the uniformity of voltage distribution along the string. A higher ratio to reduce non-uniformity requires a lower value of leakage capacitance Co and this can only be achieved by increasing the separation of the string units from the earthed structure of the tower or pole. This can be done by increasing the cross-arm length for example, but such measures obviously increase the tower structure costs. 4.2 Voltage distribution calculation While it is possible to develop a rigorous mathematical analysis to determine the voltage distribution for any number of units in a string, for a small number of units it is easier and more instructive to determine the voltage distribution from a simple circuit analysis method. For example, we derive below the voltage distribution for the five insulator string with the equivalent capacitance circuit shown in the diagram below, where we take Co to be equal to 0.1C. We assume that there is a voltage V across the tower end unit (#1) and we assume that there is capacitive current I1 through the self-capacitance of unit #1 flowing directly to the tower. We then proceed to work through the various units in sequence by determining capacitive current in each unit to determine the voltage across each of them in terms of the tower end unit voltage V.


We use the fact that Ic = CVc for a capacitance and from this we then use the following dependencies:

Ic C and

Vc Ic for any capacitance. The calculation then proceeds to determine the various In values and hence the individual Vnm of each unit and the various Vn levels at each module.


The final result is the total voltage T nV V in terms of the voltage V across the tower end unit. Then, when the total voltage VT is specified from the rated voltage, the value of V is then determined and the other voltages are then determined from V. For unit #1: Voltage = V = V1 I11 = 0.1 I1 I2 = 1.1 I1

V12 = 1.1 V This is the Unit #2 voltage

and V2 = V1 + V12 = 2.1 V I21 = 0.21 I1 and I3 = 1.1 I1 + 0.21 I1 = 1.31 I1 thus V23 = 1.31 V This is Unit #3 voltage and V3 = V2 + V23 = 3.41 V I31 = 0.341 I1 and I4 = 1.31 I1 + 0.341 I1 = 1.651 I1 thus V34 = 1.651 V This is Unit #4 voltage


and V4 = V3 + V34 = 5.061 V I41 = 0.5061 I1 I5 = 1.651 I1 + 0.5061 I1 = 2.157 I1 V45 = 2.157 V This is Unit #5 voltage and V5 = V4 + V45 = 7.218 V

Thus, the relative voltages for the five units are: #1 1.0 V #2 1.1 V #3 1.31 V #4 1.65 V #5 2.16 V

or a total of 7.218 V for the string. The string efficiency is defined as:

Total string voltageString Efficiency = x100no. of units x voltage on line end unit

For example, if the string is for use on a 66 kV overhead line, then the total voltage over the string is Vtotal = 66/3 = 38.1 kV.


Thus V = 38.1/ 7.218 = 5.28 kV and the actual voltages on the five units are: #1 5.28 kV #2 5.81 kV #3 6.92 kV #4 8.71 kV #5 11.40 kV

The string efficiency is: 38.1 100 66.8%5 11.4

If only four units are used for the same line voltage, the tower end unit voltage is V = 38.1/5.06 = 7.53 kV and the voltages are:

#1 7.53 kV #2 8.28 kV #3 9.86 kV #4 12.43 kV


If only three units are used for the same voltage, the tower end unit voltage is V = 38.1/3.41 = 11.2 kV and the voltages are: #1 11.2 kV #2 12.3 kV #3 14.6 kV



Grading rings are often attached to the line end insulator unit to give an increased effective capacitance of the line end unit and also the next unit along. This will decrease the voltages across them and increase the string efficiency. [See tutorial example]. A typical grading ring attached to an insulator string is shown below.


An analytical investigation of the voltage distribution can be performed by modeling the string as a distributed capacitance system. The analysis then gives the following expression for the voltage on the nth cap and pin unit in a string of no units:

0 0sinhsinhn

nV V

n

where: oCC

oV total voltage on the string 4.3 Impulse voltage distribution on windings At power frequencies (50 or 60 Hz), the voltage distribution along a transformer winding or a stator winding is determined by the distributed series inductance of the winding. The result of this is a uniform (linear) distribution of voltage along the length of the winding from the live end to the neutral end. However when there are high frequency (some 100s of kHz) transient voltage surges incident on such windings the voltage distribution of this surge along the winding is determined by the capacitance distribution of the winding. This capacitance distribution consists of both turn-to-turn


(series) capacitance and turn-to-earth (shunt) leakage capacitance as shown in the following equivalent circuit. The result of this leakage capacitance is to cause such surge voltages to have their voltage distributed non-linearly along the winding, with the line-end turns having a higher voltage across them than the earth-end turns of the winding.

In the diagram, Cg is the total leakage capacitance to earth and gC l is the leakage capacitance per unit length. Cs is the total series capacitance of the winding of length l and thus Cs.l is the turn-to-turn capacitance per unit length. Analysis of the initial voltage distribution over the winding when a steep fronted pulse of amplitude Vo arrives at the live end of the winding is similar to that for a standard transmission line. Solving the wave equation gives the following expression for the voltage distribution along the axial length of the winding, when the neutral end is earthed.


sinh

sinho

xlV x V

where: gs

CC

x distance from the neutral end For an isolated neutral, the voltage distribution is:

cosh

cosho

xlV x V

These (initial) voltage distributions are shown below for various values of and for an earthed neutral and unearthed neutral at the far end of the winding. The input turns can be seen to be highly stressed electrically. For example when = 10, 60% of the initial surge voltage appears across the first 10% of the winding turns. Reinforcement of the insulation at the top end of the winding makes the problem worse because the additional insulation will reduce the turn-to-turn capacitance and make the distribution more non-uniform because will then be increased.


There are two possible solutions to improving the voltage distribution: (i) Interleaving of the windings

The manner of winding the turns of the winding can be changed so as to interleave the turns of the winding, as shown below. The effect of this change in winding configuration is to change the distribution of turn-to turn capacitance and thus to increase the total series capacitance (Cs) while leaving the leakage capacitance Cg unchanged. The result is a decrease of and a more uniform voltage distribution. Typically, a reduction of from about 10-12 typically to a value of about 3-4 can be achieved by this method. It is, of


course, somewhat more difficult to manufacture but it is a commonly-used and very effective practice. (ii) Use of stress grading rings

In a generally similar approach to the use of capacitance grading rings on line insulators, such effective extensions of the windings can be used to change the winding capacitance distributions and hence the voltage distribution.


Examples of transformer windings and methods of grading voltage distribution.

Documents

ELEC4611-15-Lec2 - Electric Stress in HV Equipment