Electric power can be transmitted or dis

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Principles of Power System

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C H A P T E R

Underground Cables

11.1 Underground Cables 11.2 Construction of Cables

11.3 Insulating Materials for Cables

11.4 Classification of Cables

11.5 Cables for 3-Phase Service

11.6 Laying of Underground Cables

11.7 Insulation Resistance of a Single-Core Cable

11.8 Capacitance of a Single-Core Cable

11.9 Dielectric Stress in a Single-Core Cable

11.10 Most Economical Conductor Size in a Cable

11.11 Grading of Cables

11.12 Capacitance Grading

11.13 Intersheath Grading

11.14 Capacitance of 3-Core Cables

11.15 Measurements of Ce and Cc

11.16 Current-Carrying Capacity of Under-ground Cables

11.17 Thermal Resistance

11.18 Thermal Resistance of Dielectric of a Single-Core Cable

11.19 Permissible Current Loading

11.20 Types of Cable Faults

11.21 Loop Tests for Location of Faults in Underground Cables

11.22 Murray Loop Test

11.23 Varley Loop Test

Intr IntrIntr Intr

Introductionoductionoductionoductionoduction

E

lectric power can be transmitted or dis tributed either by overhead system or by underground cables. The underground cables have serveral advantages such as less liable to damage through storms or lightning, low maintenance cost, less chances of faults, smaller voltage drop and better general appearance. However, their major drawback is that they have greater installation cost and introduce insulation problems at high voltages compared with the equivalent overhead system. For this reason, underground cables are employed where it is impracticable to use overhead lines. Such locations may be thickly populated areas where municipal authorities prohibit overhead lines for reasons of safety, or around plants and substations or where maintenance conditions do not permit the use of overhead construction.

The chief use of underground cables for many years has been for distribution of electric power in congested urban areas at comparatively low or moderate voltages. However, recent improve-ments in the design and manufacture have led to the development of cables suitable for use at high voltages. This has made it possible to employ underground cables for transmission of electric

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power for short or moderate distances. In this chapter, we shall focus our attention on the various aspects of underground cables and their increasing use in power system.

11.1 11.1 11.1 11.1

11.1 Under Under Under Undergr Undergrgrgrground Cablesound Cablesound Cablesound Cablesound Cables

An underground cable essentially consists of one or more conductors covered with suitable insula-tion and surrounded by a protecting cover.

Although several types of cables are available, the type of cable to be used will depend upon the working voltage and service requirements. In general, a cable must fulfil the following necessary requirements :

(i) The conductor used in cables should be tinned stranded copper or aluminium of high con-ductivity. Stranding is done so that conductor may become flexible and carry more current. (ii) The conductor size should be such that the cable carries the desired load current without

overheating and causes voltage drop within permissible limits.

(iii) The cable must have proper thickness of insulation in order to give high degree of safety and reliability at the voltage for which it is designed.

(iv) The cable must be provided with suitable mechanical protection so that it may withstand the rough use in laying it.

(v) The materials used in the manufacture of cables should be such that there is complete chemical and physical stability throughout.

11.2 11.2 11.2 11.2

11.2 Construction of Cables Construction of Cables Construction of Cables Construction of Cables Construction of Cables

Fig. 11.1 shows the general construction of a 3-conductor cable. The various parts are :

(i) Cores or Conductors. A cable may have one or more than one core (conductor) depending upon the type of service for which it is intended. For instance, the 3-conductor cable shown in Fig. 11.1 is used for 3-phase service. The conductors are made of tinned copper or alu-minium and are usually stranded in order to provide flexibility to the cable.

(ii) Insulatian. Each core or conductor is provided with a suitable thickness of insulation, the thickness of layer depending upon the voltage to be withstood by the cable. The commonly used materials for insulation are impregnated paper, varnished cambric or rubber mineral compound.

(iii) Metallic sheath. In order to pro-tect the cable from moisture, gases or other damaging liquids (acids or alkalies) in the soil and atmosphere, a metallic sheath of lead or aluminium is provided over the insulation as shown in Fig. 11.1

(iv) Bedding. Over the metallic

sheath is applied a layer of bedding which consists of a fibrous material like jute or hessian tape. The purpose of bedding is to protect the metallic sheath against corrosion and from mechanical injury due to armouring.

(v) Armouring. Over the bedding, armouring is provided which consists of one or two layers of galvanised steel wire or steel tape. Its purpose is to protect the cable from mechanical injury while laying it and during the course of handling. Armouring may not be done in the case of some cables.

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material (like jute) similar to bedding is provided over the armouring. This is known as serving.

It may not be out of place to mention here that bedding, armouring and serving are only applied to the cables for the protection of conductor insulation and to protect the metallic sheath from mechanical injury.

11.3 11.3 11.3 11.3

11.3 Insulating Materials for Cables Insulating Materials for Cables Insulating Materials for Cables Insulating Materials for Cables Insulating Materials for Cables

The satisfactory operation of a cable depends to a great extent upon the charac-teristics of insulation used. Therefore, the proper choice of insulating material for cables is of considerable importance. In general, the insulating materials used in cables should have the following properties :

(i) High insulation resistance to avoid leakage current.

(ii) High dielectric strength to avoid electrical breakdown of the cable. (iii) High mechanical strength to withstand the mechanical handling of cables. (iv) Non-hygroscopic i.e., it should not absorb moisture from air or soil. The moisture tends to decrease the insulation resistance and hastens the breakdown of the cable. In case the insulating material is hygroscopic, it must be enclosed in a waterproof covering like lead sheath.

(v) Non-inflammable.

(vi) Low cost so as to make the underground system a viable proposition. (vii) Unaffected by acids and alkalies to avoid any chemical action.

No one insulating material possesses all the above mentioned properties. Therefore, the type of insulating material to be used depends upon the purpose for which the cable is required and the quality of insulation to be aimed at. The principal insulating materials used in cables are rubber, vulcanised India rubber, impregnated paper, varnished cambric and polyvinyl chloride.

1. Rubber. Rubber may be obtained from milky sap of tropical trees or it may be produced from oil products. It has relative permittivity varying between 2 and 3, dielectric strength is about 30 kV/mm and resistivity of insulation is 1017Ω cm. Although pure rubber has reasonably high insu-lating properties, it suffers form some major drawbacks viz., readily absorbs moisture, maximum safe temperature is low (about 38ºC), soft and liable to damage due to rough handling and ages when exposed to light. Therefore, pure rubber cannot be used as an insulating material.

2. Vulcanised India Rubber (V.I.R.). It is prepared by mixing pure rubber with mineral mat-ter such as zine oxide, red lead etc., and 3 to 5% of sulphur. The compound so formed is rolled into thin sheets and cut into strips. The rubber compound is then applied to the conductor and is heated to a temperature of about 150ºC. The whole process is called vulcanisation and the product obtained is known as vulcanised India rubber.

Vulcanised India rubber has greater mechanical strength, durability and wear resistant property than pure rubber. Its main drawback is that sulphur reacts very quickly with copper and for this reason, cables using VIR insulation have tinned copper conductor. The VIR insulation is generally used for low and moderate voltage cables.

3. Impregnated paper. It consists of chemically pulped paper made from wood chippings and impregnated with some compound such as paraffinic or napthenic material. This type of insula-tion has almost superseded the rubber insulainsula-tion. It is because it has the advantages of low cost, low capacitance, high dielectric strength and high insulation resistance. The only disadvantage is that paper is hygroscopic and even if it is impregnated with suitable compound, it absorbs moisture and thus lowers the insulation resistance of the cable. For this reason, paper insulated cables are always

Underground cable

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* Special precautions have to be taken to preclude moisture at joints. If the number of joints is more, the installation cost increases rapidly and prohibits the use of paper insulated cables.

provided with some protective covering and are never left unsealed. If it is required to be left unused on the site during laying, its ends are temporarily covered with wax or tar.

Since the paper insulated cables have the tendency to absorb moisture, they are used where the cable route has a *few joints. For instance, they can be profitably used for distribution at low voltages in congested areas where the joints are generally provided only at the terminal apparatus. However, for smaller installations, where the lenghts are small and joints are required at a number of places, VIR cables will be cheaper and durable than paper insulated cables.

4. Varnished cambric. It is a cotton cloth impregnated and coated with varnish. This type of insulation is also known as empire tape. The cambric is lapped on to the conductor in the form of a tape and its surfaces are coated with petroleum jelly compound to allow for the sliding of one turn over another as the cable is bent. As the varnished cambric is hygroscopic, therefore, such cables are always provided with metallic sheath. Its dielectric strength is about 4 kV/mm and permittivity is 2.5 to 3.8.

5. Polyvinyl chloride (PVC). This insulating material is a synthetic compound. It is obtained from the polymerisation of acetylene and is in the form of white powder. For obtaining this material as a cable insulation, it is compounded with certain materials known as plasticizers which are liquids with high boiling point. The plasticizer forms a gell and renders the material plastic over the desired range of temperature.

Polyvinyl chloride has high insulation resistance, good dielectric strength and mechanical tough-ness over a wide range of temperatures. It is inert to oxygen and almost inert to many alkalies and acids. Therefore, this type of insulation is preferred over VIR in extreme enviormental conditions such as in cement factory or chemical factory. As the mechanical properties (i.e., elasticity etc.) of PVC are not so good as those of rubber, therefore, PVC insulated cables are generally used for low and medium domestic lights and power installations.

11.4 11.4 11.4 11.4

11.4 Classification of Cables Classification of Cables Classification of Cables Classification of Cables Classification of Cables

Cables for underground service may be classified in two ways according to (i) the type of insulating material used in their manufacture (ii) the voltage for which they are manufactured. However, the latter method of classification is generally preferred, according to which cables can be divided into the following groups :

(i) Low-tension (L.T.) cables — upto 1000 V (ii) High-tension (H.T.) cables — upto 11,000 V (iii) Super-tension (S.T.) cables — from 22 kV to 33 kV (iv) Extra high-tension (E.H.T.) cables — from 33 kV to 66 kV

(v) Extra super voltage cables — beyond 132 kV A cable may have one or more than one core depending upon the type of service for which it is intended. It may be (i) single-core (ii) two-core (iii) three-core (iv) four-core etc. For a 3-phase service, either 3-single-core cables or three-core cable can be used depending upon the operating voltage and load demand.

Fig. 11.2 shows the constructional details of a single-core low tension cable. The cable has ordinary construction be-cause the stresses developed in the cable for low voltages (upto 6600 V) are generally small. It consists of one circular core of tinned stranded copper (or aluminium) insulated by layers of

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impregnated paper. The insulation is surrounded by a lead sheath which prevents the entry of mois-ture into the inner parts. In order to protect the lead sheath from corrosion, an overall serving of compounded fibrous material (jute etc.) is provided. Single-core cables are not usually armoured in order to avoid excessive sheath losses. The principal advantages of single-core cables are simple construction and availability of larger copper section.

11.5 11.5 11.5 11.5

11.5 Cables for 3-Phase Service Cables for 3-Phase Service Cables for 3-Phase Service Cables for 3-Phase Service Cables for 3-Phase Service

In practice, underground cables are generally required to deliver 3-phase power. For the purpose, either three-core cable or *three single core cables may be used. For voltages upto 66 kV, 3-core cable (i.e., multi-core construction) is preferred due to economic reasons. However, for voltages beyond 66 kV, 3-core-cables become too large and unwieldy and, therefore, single-core cables are used. The following types of cables are generally used for 3-phase service :

1. Belted cables — upto 11 kV

2. Screened cables — from 22 kV to 66 kV 3. Pressure cables — beyond 66 kV.

1. Belted cables. These cables are used for voltages upto 11kV but in extraordinary cases, their use may be extended upto 22kV. Fig. 11.3 shows the constructional details of a 3-core belted cable. The cores are insulated from each other by

lay-ers of impregnated paper. Another layer of impreg-nated paper tape, called paper belt is wound round the grouped insulated cores. The gap between the insu-lated cores is filled with fibrous insulating material (jute etc.) so as to give circular cross-section to the cable. The cores are generally stranded and may be of non-circular shape to make better use of available space. The belt is covered with lead sheath to protect the cable against ingress of moisture and mechanical injury. The lead sheath is covered with one or more layers of armouring with an outer serving (not shown in the fig-ure).

The belted type construction is suitable only for low and medium voltages as the electrostatic stresses developed in the cables for these voltages are more or less radial i.e., across the insulation. However, for high voltages (beyond 22 kV), the tangential stresses also become important. These stresses act along the layers of paper insulation. As the insulation resistance of paper is quite small along the layers, therefore, tangential stresses set up **leakage current along the layers of paper insulation. The leakage current causes local heating, resulting in the risk of breakdown of insulation at any moment. In order to overcome this difficulty, screened cables are used where leakage currents are conducted to earth through metallic screens.

2. Screened cables. These cables are meant for use upto 33 kV, but in particular cases their use may be extended to operating voltages upto 66 kV. Two principal types of screened cables are H-type cables and S.L. H-type cables.

(i) H-type cables. This type of cable was first designed by H. Hochstadter and hence the name. Fig. 11.4 shows the constructional details of a typical 3-core, H-type cable. Each core is insulated by layers of impregnated paper. The insulation on each core is covered with a metallic screen which usually consists of a perforated aluminium foil. The cores are laid in such a way that metallic screens * Separate single-core cable for each phase.

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make contact with one another. An additional

conducting belt (copper woven fabric tape) is wrapped round the three cores. The cable has no insulating belt but lead sheath, bedding, armouring and serving follow as usual. It is easy to see that each core screen is in electrical con-tact with the conducting belt and the lead sheath. As all the four screens (3 core screens and one conducting belt) and the lead sheath are at †earth potential, therefore, the electrical stresses are purely radial and consequently dielectric losses are reduced.

Two principal advantages are claimed for H-type cables. Firstly, the perforations in the metallic screens assist in the complete impregnation of the cable with the compound and thus the possibility of air pockets or voids (vacuous spaces) in the dielectric is eliminated. The voids if present tend to reduce the breakdown strength of the cable and may cause considerable damage to the paper insula-tion. Secondly, the metallic screens increase the heat dissipating power of the cable.

(ii) S.L. type cables. Fig. 11.5 shows the constructional details of a 3-core *S.L. (separate lead) type cable. It is basically H-type cable but the screen round

each core insulation is covered by its own lead sheath. There is no overall lead sheath but only armouring and serving are provided. The S.L. type cables have two main advan-tages over H-type cables. Firstly, the separate sheaths minimise the possibility of core-to-core breakdown. Sec-ondly, bending of cables becomes easy due to the elimina-tion of overall lead sheath. However, the disadvantage is that the three lead sheaths of S.L. cable are much thinner than the single sheath of H-cable and, therefore, call for greater care in manufacture.

Limitations of solid type cables. All the cables of

above construction are referred to as solid type cables because solid insulation is used and no gas or oil circulates in the cable sheath. The voltage limit for solid type cables is 66 kV due to the following reasons :

(a) As a solid cable carries the load, its conductor temperature increases and the cable com-† The four screens and lead sheath are in electrical contact and lead sheath is at earth potential.

* In this arrangement, each core is separately lead sheathed and hence the name S.L. cable. H-Type Cables

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pound (i.e., insulating compound over paper) expands. This action stretches the lead sheath which may be damaged.

(b) When the load on the cable decreases, the conductor cools and a partial vacuum is formed within the cable sheath. If the pinholes are present in the lead sheath, moist air may be drawn into the cable. The moisture reduces the dielectric strength of insulation and may eventually cause the break-down of the cable.

(c) In practice, †voids are always present in the insulation of a cable. Modern techniques of manufacturing have resulted in void free cables. However, under operating conditions, the voids are formed as a result of the differential expansion and contraction of the sheath and impregnated com-pound. The breakdown strength of voids is considerably less than that of the insulation. If the void is small enough, the electrostatic stress across it may cause its breakdown. The voids nearest to the conductor are the first to break down, the chemical and thermal effects of ionisation causing perma-nent damage to the paper insulation.

3. Pressure cables For voltages beyond 66 kV, solid type cables are unreliable because there is a danger of breakdown of insulation due to the presence of voids. When the operating voltages are greater than 66 kV, pressure cables are used. In such cables, voids are eliminated by increasing the pressure of compound and for this reason they are called pressure cables. Two types of pressure cables viz oil-filled cables and gas pressure cables are commonly used.

(i) Oil-filled cables. In such types of cables, channels or ducts are provided in the cable for oil circulation. The oil under pressure (it is the same oil used for impregnation) is kept constantly supplied to the channel by means of external reservoirs placed at suitable distances (say 500 m) along the route of the cable. Oil under pressure compresses the layers of paper insulation and is forced into any voids that may have formed between the layers. Due to the elimination of voids, oil-filled cables can be used for higher voltages, the range being from 66 kV upto 230 kV. Oil-filled cables are of three types viz., single-core conductor channel, single-core sheath channel and three-core filler-space channels.

Fig. 11.6 shows the constructional details of a single-core conductor channel, oil filled cable. The oil channel is formed at the centre by stranding the conductor wire around a hollow cylindrical steel spiral tape. The oil under pressure is supplied to the channel by means of external reservoir. As the channel is made of spiral steel tape, it allows the oil to percolate between copper strands to the wrapped insulation. The oil pressure compresses the layers of paper insulation and prevents the possibility of void formation. The system is so designed

that when the oil gets expanded due to increase in cable temperature, the extra oil collects in the reservoir. How-ever, when the cable temperature falls during light load con-ditions, the oil from the reservoir flows to the channel. The disadvantage of this type of cable is that the channel is at the middle of the cable and is at full voltage w.r.t. earth, so that a very complicated system of joints is necessary.

Fig. 11.7 shows the constructional details of a single-core sheath channel oil-filled cable. In this type of cable, the conductor is solid similar to that of solid cable and is paper insulated. However, oil ducts are provided in the metallic sheath as shown. In the 3-core oil-filler cable shown

in Fig. 11.8, the oil ducts are located in the filler spaces. These channels are composed of perforated metal-ribbon tubing and are at earth potential.

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The oil-filled cables have three principal advantages. Firstly, formation of voids and ionisation are avoided. Secondly, allowable temperature range and dielectric strength are increased. Thirdly, if there is leakage, the defect in the lead sheath is at once indicated and the possibility of earth faults is decreased. However, their major disadvantages are the high initial cost and complicated system of laying.

(ii) Gas pressure cables. The voltage required to set up ionisation inside a void increases as the pressure is increased. Therefore, if ordinary cable is subjected to a sufficiently high pressure, the ionisation can be altogether eliminated. At the same time, the increased pressure produces radial compression which tends to close any voids. This is the underlying principle of gas pressure cables. Fig. 11.9 shows the section of external pressure cable designed by Hochstadter, Vogal and Bowden. The construction of the cable is similar to that of an ordinary solid type

except that it is of triangular shape and thickness of lead sheath is 75% that of solid cable. The triangular section reduces the weight and gives low thermal resistance but the main reason for triangular shape is that the lead sheath acts as a pressure membrane. The sheath is protected by a thin metal tape. The cable is laid in a gas-tight steel pipe. The pipe is filled with dry nitrogen gas at 12 to 15 atmospheres. The gas pressure produces radial compression and closes the voids that may have formed between the layers of paper insulation. Such cables can carry more load current and operate at higher voltages than a normal cable. Moreover, maintenance cost is small and the nitrogen gas helps in quenching any flame. However, it has the disadvantage that the overall cost is very high. 11.6

11.6 11.6 11.6

11.6 Laying of Under Laying of Under Laying of Under Laying of Undergr Laying of Undergrgrgrground Cablesound Cablesound Cablesound Cablesound Cables

The reliability of underground cable network depends to a considerable extent upon the proper laying and attachment of fittings i.e., cable end boxes, joints, branch

con-nectors etc. There are three main methods of laying underground cables viz., direct laying, draw-in system and the solid system.

1. Direct laying. This method of laying underground cables is simple and cheap and is much favoured in modern practice. In this method, a trench of about 1·5 metres deep and 45 cm wide is dug. The trench is covered with a layer of fine sand (of about 10 cm thickness) and the cable is laid over this sand bed. The sand prevents the entry of moisture from the ground and thus protects the cable from decay. After the cable has been laid in the trench, it is covered with another layer of sand of about 10 cm thickness.

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The trench is then covered with bricks and other materials in order to protect the cable from mechani-cal injury. When more than one cable is to be laid in the same trench, a horizontal or vertimechani-cal inter-axial spacing of atleast 30 cm is provided in order to reduce the effect of mutual heating and also to ensure that a fault occurring on one cable does not damage the adjacent cable. Cables to be laid in this way must have serving of bituminised paper and hessian tape so as to provide protection against corrosion and electorlysis.

Advantages

(i) It is a simple and less costly method.

(ii) It gives the best conditions for dissipating the heat generated in the cables.

(iii) It is a clean and safe method as the cable is invisible and free from external disturbances. Disadvantages

(i) The extension of load is possible only by a completely new excavation which may cost as much as the original work.

(ii) The alterations in the cable netwok cannot be made easily. (iii) The maintenance cost is very high.

(iv) Localisation of fault is difficult.

(v) It cannot be used in congested areas where excavation is expensive and inconvenient. This method of laying cables is used in open areas where excavation can be done conveniently and at low cost.

2. Draw-in system. In this method, conduit or duct of glazed stone or cast iron or concrete are laid in the ground with manholes at suitable positions along

the cable route. The cables are then pulled into position from manholes. Fig. 11.11 shows section through four-way underground duct line. Three of the ducts carry transmis-sion cables and the fourth duct carries relay protection con-nection, pilot wires. Care must be taken that where the duct line changes direction ; depths, dips and offsets be made with a very long radius or it will be difficult to pull a large cable between the manholes. The distance between the manholes should not be too long so as to simplify the

pull-ing in of the cables. The cables to be laid in this way need not be armoured but must be provided with serving of hessian and jute in order to protect them when being pulled into the ducts.

Advantages

(i) Repairs, alterations or additions to the cable network can be made without opening the ground.

(ii) As the cables are not armoured, therefore, joints become simpler and maintenance cost is reduced considerably.

(iii) There are very less chances of fault occurrence due to strong mechanical protection pro-vided by the system.

Disadvantages

(i) The initial cost is very high.

(ii) The current carrying capacity of the cables is reduced due to the close grouping of cables and unfavourable conditions for dissipation of heat.

This method of cable laying is suitable for congested areas where excavation is expensive and inconvenient, for once the conduits have been laid, repairs or alterations can be made without

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ing the ground. This method is generally used for short length cable routes such as in workshops, road crossings where frequent digging is costlier or impossible.

3. Solid system. In this method of laying, the cable is laid in open pipes or troughs dug out in earth along the cable route. The troughing is of cast iron, stoneware, asphalt or treated wood. After the cable is laid in position, the troughing is filled with a bituminous or asphaltic compound and covered over. Cables laid in this manner are usually plain lead covered because troughing affords good mechanical protection.

Disadvantages

(i) It is more expensive than direct laid system.

(ii) It requires skilled labour and favourable weather conditions.

(iii) Due to poor heat dissipation facilities, the current carrying capacity of the cable is reduced. In view of these disadvantages, this method of laying underground cables is rarely used now-a-days.

11.7 11.7 11.7 11.7

11.7 Insulation Resistance of a Single-Cor Insulation Resistance of a Single-Cor Insulation Resistance of a Single-Cor Insulation Resistance of a Single-Core Cable Insulation Resistance of a Single-Core Cablee Cablee Cablee Cable

The cable conductor is provided with a suitable thickness of insulating material in order to prevent leakage current. The path for leakage current is radial through the

insula-tion. The opposition offered by insulation to leakage current is known as insulation resistance of the cable. For satisfactory operation, the insula-tion resistance of the cable should be very high.

Consider a single-core cable of conductor radius r1 and internal sheath radius r2 as shown in Fig. 11.12. Let l be the length of the cable and ρ be the resistivity of the insulation.

Consider a very small layer of insulation of thickness dx at a radius x. The length through which leakage current tends to flow is dx and the area of X-section offered to this flow is 2π x l.

∴ Insulation resistance of considered layer = ρ

π

dx x l 2 Insulation resistance of the whole cable is

R = 2 2 1 1 1 2 2 r r r r dx dx x l l x ρ ρ = π π

∫

∴ R = ρ π 2 2 1 l r r e log

This shows that insulation resistance of a cable is inversely proportional to its length. In other words, if the cable length increases, its insulation resistance decreases and vice-versa.

Example 11.1. A single-core cable has a conductor diameter of 1cm and insulation thickness of 0·4 cm. If the specific resistance of insulation is 5 × 1014Ω-cm, calculate the insulation resistance for a 2 km length of the cable.

Solution

Conductor radius, r₁ = 1/2 = 0·5 cm Length of cable, l = 2 km = 2000 m

Resistivity of insulation, ρ = 5 × 1014Ω-cm = 5 × 1012Ω-m Internal sheath radius, r₂ = 0·5 + 0·4 = 0·9 cm

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∴ Insulation resistance of cable is

R = ρ_π _π 2 5 10 2 2000 0 9 0 5 2 1 12 l r r e e log = × log × ⋅⋅ = 0·234 × 109Ω = 234 M ΩΩΩΩΩ

Example 11.2. The insulation resistance of a single-core cable is 495 MΩ per km. If the core diameter is 2·5 cm and resistivity of insulation is 4·5 × 1014Ω-cm, find the insulation thickness.

Solution.

Length of cable, l = 1 km = 1000 m Cable insulation resistance, R = 495 MΩ = 495 × 106Ω Conductor radius, r₁ = 2·5/2 = 1·25 cm

Resistivity of insulation, ρ = 4·5 × 1014 Ω-cm = 4·5 × 1012Ωm Let r₂cm be the internal sheath radius.

Now, R = ρ π 2 2 1 l r r e log or log_er r 2 1 = 2 2 1000 495 10 4 5 10 6 12 π ρ π l R ₌ × × × ⋅ × = 0·69 or 2·3 log₁₀ r₂/r₁ = 0·69 or r₂/r₁ = Antilog 0·69/2·3 = 2 or r₂ = 2 r₁= 2 × 1·25 = 2·5 cm ∴ Insulation thickness = r₂− r₁ = 2·5 − 1·25 = 1·25 cm

Example 11.3. A single core cable 5 km long has an insulation resistance of 0·4 MΩ. The core diameter is 20 mm and the diameter of the cable over the insulation is 50 mm. Calculate the resistiv-ity of the insulating material.

Solution.

Length of cable, l = 5 km = 5000 m Cable insulation resistance, R = 0·4 MΩ = 0·4 × 106Ω Conductor radius, r₁ = 20/2 = 10 mm Internal sheath radius, r₂ = 50/2 = 25 mm

∴ Insulation resistance of the cables is R = ρ π 2 2 1 l r r e log or 0·4 × 106 = _π ρ 2 5000 25 10 × ×loge ∴ ρ = 13.72 ××××× 109 ΩΩΩΩΩm TUTORIAL PROBLEMS TUTORIAL PROBLEMSTUTORIAL PROBLEMS TUTORIAL PROBLEMSTUTORIAL PROBLEMS

1. A single-core cable has a conductor diameter of 2.5 cm and insulation thickness of 1.2 cm. If the specific resistance of insulation is 4·5 × 1014_Ω_{cm, calculate the insulation resistance per kilometre length of the}

cable. [305·5 MΩΩΩΩΩ]

2. A single core cable 3 km long has an insulation resistance of 1820 MΩ. If the conductor diameter is 1.5 cm and sheath diameter is 5 cm, calculate the resistivity of the dielectric in the cable.

[28.57 ××××× 1012 ΩΩΩΩΩm] 3. Determine the insulation resistance of a single-core cable of length 3 km and having conductor radius 12·5 mm, insulation thickness 10 mm and specific resistance of insulation of 5 × 1012_Ω_m. _{[156 M}_Ω_Ω_Ω_Ω_Ω_]

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11.8

11.8 11.8 11.8

11.8 Ca Ca Capacitance of a Single-Cor Ca Capacitance of a Single-Corpacitance of a Single-Core Capacitance of a Single-Corpacitance of a Single-Core Cae Cae Cae Cablebleblebleble

A single-core cable can be considered to be equivalent to two long co-axial cylinders. The conductor (or core) of the cable is the inner cylinder while the outer cylinder is represented by lead sheath which is at earth potential. Consider a single core cable with conductor diameter d and inner sheath diameter D (Fig. 11.13). Let the charge per metre axial length of the cable be Q coulombs and ε be the permittivity of the insulation material between core and lead sheath. Obviously *ε = ε₀ε_rwhere ε_ris the relative permit-tivity of the insulation.

Consider a cylinder of radius x metres and axial length 1 metre. The surface area of this cylinder is = 2 π x × 1 = 2 π x m2

∴ Electric flux density at any point P on the considered cylinder is D_x = Q

x 2π C/m

2

Electric intensity at point P, Ex =

Dx ε = Q x 2π ε = Q x _r 2π ε ε₀ volts/m

The work done in moving a unit positive charge from point P through a distance dx in the direc-tion of electric field is E_xdx. Hence, the work done in moving a unit positive charge from conductor to sheath, which is the potential difference V between conductor and sheath, is given by :

V = E dx Q x dx Q D d x d D r d D r e / / / / log 2 2 0 2 2 0 2 2

z

=

z

_{π ε ε} = _{πε ε}

Capacitance of the cable is

C = Q V Q Q D d r e = 2π ε ε₀ log F m = 2π ε εo r e D d log ( / ) F/m = 2 8 854 10 2 303 10 π× ⋅ × ×ε ⋅ –12 log ( / ) r D d F m = εr D d 41 4 ₁₀ 10 9 ⋅ log ( / )× − F m If the cable has a length of l metres, then capacitance of the cable is

C = εr l D d 41 4 10 10 9 ⋅ × − log F

Example 11.4. A single core cable has a conductor diameter of 1 cm and internal sheath diameter of 1·8 cm. If impregnated paper of relative permittivity 4 is used as the insulation, calculate the capacitance for 1 km length of the cable.

Solution. Capacitance of cable, C = εr l D d 41 4 ₁₀ 10 9 ⋅ log ( / )× − F

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276

Here ε_r = 4 ; l = 1000 m

D = 1·8 cm ; d = 1 cm Substituting these values in the above expression, we get,

C = 4 1000 41 4 ₁₀1 8 1 10 9 × ⋅ log ( ⋅ / )× − F = 0·378 × 10 −6 F = 0·378 µF Example 11.5. Calculate the capacitance and charging current of a single core cable used on a 3-phase, 66 kV system. The cable is 1 km long having a core diameter of 10 cm and an impreg-nated paper insulation of thickness 7 cm. The relative permittivity of the insulation may be taken as 4 and the supply at 50 Hz.

Solution. Capacitance of cable, C = εr l D d 41 4 ₁₀ 10 9 ⋅ log ( / )× − F Here, εr = 4 ; l = 1000 m d = 10 cm; D = 10 + 2 × 7 = 24 cm Substituting these values in the above expression,

C = 4 1000 41 4 ₁₀24 10 10 9 × ⋅ ×log ( / )× − F= 0·254 × 10 –6 F = 0·254 µF Voltage between core and sheath is

V_ph = 66 3 = 38·1 kV = 38·1 × 103 V Charging current = V_ph/ X_C = 2π f C V_ph

= 2π × 50 × 0·254 × 10–6 × 38·1 × 103A = 3.04 A

Example 11.6. A 33 kV, 50 Hz, 3-phase underground cable, 4 km long uses three single core cables. Each of the conductor has a diameter of 2·5 cm and the radial thickness of insulation is 0·5 cm. Determine (i) capacitance of the cable/phase (ii) charging current/phase (iii) total charging kVAR. The relative permittivity of insulation is 3.

Solution.

(i) Capacitance of cable/phase, C = εr l D d 41 4 ₁₀ 10 9 ⋅ log ( / )× − F Here εr = 3 ; l = 4 km = 4000 m d = 2·5 cm ; D = 2.5 + 2 × 0.5 = 3.5 cm Putting these values in the above expression, we get,

C = 3 4000 10 41 4 ₁₀3 5 2 5 × × ⋅ × ⋅ ⋅ –9 log ( / ) = 1984 ××××× 10 –9 F (ii) Voltage/phase, V_ph = 33 10 3 3 × = 19·05 × 103 V Charging current/phase, I_C = V X ph C = 2π f C V_ph = 2π × 50 × 1984 × 10–9 × 19·05 × 103 = 11·87 A (iii) Total charging kVAR = 3Vph IC = 3 × 19·05 × 10

3

× 11·87 = 678·5 ××××× 103 kVAR TUTORIAL PROBLEMS

TUTORIAL PROBLEMSTUTORIAL PROBLEMS TUTORIAL PROBLEMSTUTORIAL PROBLEMS

1. A single core cable has a conductor diameter of 1 cm and internal sheath diameter of 1.8 cm. If the impregnated paper of relative permittivity 3 is used as insulation, calculate the capacitance for 1 km

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277

* It may be recalled that potential gradient at any point is equal to the electric intensity at that point.

2. Calculate the capacitance and charging current of a single core cable used on 3-phase, 66 kV system. The cable is 1 km long having a core diameter of 15 cm and impregnated paper insulation of thickness 22.5 cm. The relative permittivity of the insulation may be taken as 3.5 and supply at 50 Hz.

[0.144 µµµµµF ; 1.74 A] 3. An 11 kV, 50 Hz, single phase cable 2.5 km long, has a diameter of 20 mm and internal sheath radius of 15 mm. If the dielectric has a relative permittivity of 2.4, determine (i) capacitance (ii) charging current (iii) total charging kVAR. [(i) 0.822 µF (ii) 2.84 A (iii) 31.24 kVAR]

11.9 11.9 11.9 11.9

11.9 Dielectr Dielectr Dielectr Dielectric Str Dielectric Stric Stric Stress in a Single-Coric Stress in a Single-Core Caess in a Single-Coress in a Single-Coress in a Single-Core Cae Cae Cablee Cableblebleble Under operating conditions, the insulation of a cable is subjected to elec-trostatic forces. This is known as dielectric stress. The dielectric stress at any point in a cable is infact the potential gradient (or *electric inten-sity) at that point.

Consider a single core cable with core diameter d and internal sheath diameter D. As proved in Art 11.8, the electric intensity at a point x metres from the centre of the cable is

Ex =

Q x

o r

2π ε ε volts/m

By definition, electric intensity is equal to potential gradient. There-fore, potential gradient g at a point x metres from the centre of cable is

g = Ex

or g = Q

x

o r

2π ε ε volts/m ...(i)

As proved in Art. 11.8, potential difference V between conductor and sheath is V = Q o r 2π ε ε loge D d volts or Q = 2π ε εo r e V D d log ...(ii)

Substituting the value of Q from exp. (ii) in exp. (i), we get, g = 2 2 π ε ε π ε ε o r e o r V D d x log / = V x D d e log volts/m ...(iii)

It is clear from exp. (iii) that potential gradient varies inversely as the distance x. Therefore, potential gradient will be maximum when x is minimum i.e., when x = d/2 or at the surface of the conductor. On the other hand, potential gradient will be minimum at x = D/2 or at sheath surface.

∴ Maximum potential gradient is g_max = 2V

d D

d

e

log

volts/m [Putting x = d/2 in exp. (iii)] Minimum potential gradient is

gmin =

2V

D D

d

e

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278

Principles of Power System ∴ g g max min = 2 2 V d D d V D D d e e log / log / = D d

The variation of stress in the dielectric is shown in Fig. 11.14. It is clear that dielectric stress is maximum at the conductor surface and its value goes on decreasing as we move away from the conductor. It may be noted that maximum stress is an important consideration in the design of a cable. For instance, if a cable is to be operated at such a voltage that *maximum stress is 5 kV/mm, then the insulation used must have a dielectric strength of atleast 5 kV/mm, otherwise breakdown of the cable will become inevitable.

Example 11.7. A 33 kV single core cable has a conductor diameter of 1 cm and a sheath of inside diameter 4 cm. Find the maximum and minimum stress in the insulation.

Solution.

The maximum stress occurs at the conductor surface and its value is given by; gmax = 2V d D d e log Here, V = 33 kV (r.m.s) ; d = 1 cm ; D = 4 cm Substituting the values in the above expression, we get,

g_max = 2 33

1 4

×

×log_e kV †/cm = 47·61 kV/cm r.ms. The minimum stress occurs at the sheath and its value is give by ;

g_min = 2V D D d e log = 2 33 4 4 × ×log_e kV/cm = 11·9 kV/cm r.m.s Alternatively ; gmin = gmax× d D = 47·61 × 1/4 = 11·9 kV/cm r.m.s.

Example 11.8. The maximum and minimum stresses in the dielectric of a single core cable are 40 kV/cm (r.m.s.) and 10 kV/cm (r.m.s.) respectively. If the conductor diameter is 2 cm, find :

(i) thickness of insulation (ii) operating voltage Solution.

Here, g_max = 40 kV/cm ; g_min = 10 kV/cm ; d = 2 cm ; D = ? (i) As proved in Art. 11.9,

g g max min = D d or D = g g d max min × = 40× 10 2 = 8 cm ∴ Insulation thickness = D–d 2 = 8 2 2 – = 3 cm (ii) g_max = 2V d D d e log * Of course, it will occur at the conductor surface.

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∴ V = g d D d max e e log _log 2 40 2 4 2 = × kV = 55.45 kV r.m.s.

Example 11.9. A single core cable for use on 11 kV, 50 Hz system has conductor area of 0·645 cm2 and internal diameter of sheath is 2·18 cm. The permittivity of the dielectric used in the cable is 3·5. Find (i) the maximum electrostatic stress in the cable (ii) minimum electrostatic stress in the cable (iii) capacitance of the cable per km length (iv) charging current.

Solution.

Area of cross-section of conductor, a = 0.645 cm2 Diameter of the conductor, d = 4a_π = 4× ⋅0 645

π = 0·906 cm

Internal diameter of sheath, D = 2·18 cm (i) Maximum electrostatic stress in the cable is

gmax = 2 2 11 0 906 2 18 0 906 V d D d e e log log = × ⋅ _⋅⋅ kV cm = 27·65 kV/cm r.m.s.

(ii) Minimum electrostatic stress in the cable is

gmin = 2 2 11 2 18 2 18 0 906 V kV cm D D d e e log = log × ⋅ _⋅⋅ = 11·5 kV/cm r.m.s.

(iii) Capacitance of cable, C = εr l D d 41 4 10 10 9 ⋅ × − log F Here ε_r= 3.5 ; l = 1 km = 1000 m ∴ C = 3 5 1000 41 4 2 18 0 906 10 10 9 . log × ⋅ ⋅ ⋅ × − = 0·22 ××××× 10–6 F

(iv) Charging current, I_C = V XC =2π f C V = 2π× 50 × 0·22 × 10−6× 11000 = 0·76 A 11.10 11.10 11.10 11.10

11.10 Most Economical Conductor Size in a Cable Most Economical Conductor Size in a Cable Most Economical Conductor Size in a Cable Most Economical Conductor Size in a Cable Most Economical Conductor Size in a Cable

It has already been shown that maximum stress in a cable occurs at the surface of the conductor. For safe working of the cable, dielectric strength of the insulation should be more than the maximum stress. Rewriting the expression for maximum stress, we get,

g_max = 2V d D d e log volts/m ...(i)

The values of working voltage V and internal sheath diameter D have to be kept fixed at certain values due to design considerations. This leaves conductor diameter d to be the only variable in exp. (i). For given values of V and D, the most economical conductor diameter will be one for which gmax

has a minimum value. The value of gmax will be minimum when d loge D/d is maximum i.e.

d dd d D d e log

L

NM

O

QP

= 0 or log_e D . .– d d d D D d + ₂ = 0 or log_e (D/d) − 1 = 0 or log_e (D/d) = 1 or (D/d) = e = 2·718

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280

∴ Most economical conductor diameter is

d = D

2 718. and the value of gmax under this condition is

g_max = 2V

d volts/m [Putting loge D/d = 1 in exp. (i)]

For low and medium voltage cables, the value of conductor diameter arrived at by this method (i.e., d = 2V/g_max) is often too small from the point of view of current density. Therefore, the conduc-tor diameter of such cables is determined from the consideration of safe current density. For high voltage cables, designs based on this theory give a very high value of d, much too large from the point of view of current carrying capacity and it is, therefore, advantageous to increase the conductor diameter to this value. There are three ways of doing this without using excessive copper :

(i) Using aluminium instead of copper because for the same current, diameter of aluminium will be more than that of copper.

(ii) Using copper wires stranded round a central core of hemp. (iii) Using a central lead tube instead of hemp.

Example 11.10. Find the most economical value of diameter of a single-core cable to be used on 50 kV, single-phase system. The maximum permissible stress in the dielectric is not to exceed 40 kV/cm.

Solution.

Peak value of cable voltage, V = 50 × 2 = 70.7 kV

Maximum permissible stress, gmax = 40 kV/cm (assumed peak)

∴ Most economical conductor diameter is d = 2 2 70 7

40 V

g_max = ×

⋅ _{= 3·53 cm}

Example 11.11 Find the most economical size of a single-core cable working on a 132 kV, 3-phase system, if a dielectric stress of 60 kV/cm can be allowed.

Solution

Phase voltage of cable = 132/ 3 = 76·21 kV Peak value of phase voltage, V = 76·21 × 2 = 107·78 kV Max. permissible stress, gmax = 60 kV/cm

∴ Most economical conductor diameter is

d = 2 2 107 78 60 V

g_max = ×

⋅ _{= 3·6 cm}

Internal diameter of sheath, D = 2·718 d = 2·718 × 3·6 = 9·78 cm

Therefore, the cable should have a conductor diameter of 3.6 cm and internal sheath diameter of 9·78 cm.

11.11 11.11 11.11 11.11

11.11 Grading of Cables Grading of Cables Grading of Cables Grading of Cables Grading of Cables

The process of achieving uniform electrostatic stress in the dielectric of cables is known as grading of cables.

It has already been shown that electrostatic stress in a single core cable has a maximum value (gmax) at the conductor surface and goes on decreasing as we move towards the sheath. The

maxi-mum voltage that can be safely applied to a cable depends upon gmax i.e., electrostatic stress at the

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281

electric must be more than g_max. If a dielectric of high strength is used for a cable, it is useful only near the conductor where stress is maximum. But as we move away from the conductor, the electro-static stress decreases, so the dielectric will be unnecessarily overstrong.

The unequal stress distribution in a cable is undesirable for two reasons. Firstly, insulation of greater thickness is required which increases the cable size. Secondly, it may lead to the breakdown of insulation. In order to overcome above disadvantages, it is necessary to have a uniform stress distribution in cables. This can be achieved by distributing the stress in such a way that its value is increased in the outer layers of dielectric. This is known as grading of cables. The following are the two main methods of grading of cables :

(i) Capacitance grading (ii) Intersheath grading 11.12

11.12 11.12 11.12

11.12 Capacitance Grading Capacitance Grading Capacitance Grading Capacitance Grading Capacitance Grading

The process of achieving uniformity in the dielectric stress by using layers of different dielectrics is known as capacitance grading.

In capacitance grading, the homogeneous dielectric is replaced by a composite dielectric. The composite dielectric consists of various layers of different dielectrics in such a manner that relative permittivity ε_r of any layer is inversely proportional to its distance from the centre. Under such conditions, the value of potential gra-dient at any point in the dieletric is *constant and is independent of its distance from the centre. In other words, the dielectric stress in the cable is same everywhere and the grading is ideal one. How ever, ideal grading requires the use of an infinite number of trics which is an impossible task. In practice, two or three dielec-trics are used in the decreasing order of permittivity ; the dielectric of highest permittivity being used near the core.

The capacitance grading can be explained beautifully by re-ferring to Fig. 11.15. There are three dielectrics of outer diameter d₁, d₂ and D and of relative permittivity ε₁, ε₂ and ε₃ respectively. If the permittivities are such that ε₁> ε₂> ε₃and the three dielec-trics are worked at the same maximum stress, then,

1 1 ε d = 1 2 1 ε d = 3 2 1 d ε or ε₁d = ε₂d₁=ε₃d₂ Potential difference across the inner layer is * As ε_r∝ 1

x ∴ εr = k/x where k is a constant.

Potential gradient at a distance x from the centre

= Q x Q k x x Q k r 2πε ε₀ =2πε₀( / ) =2πε₀ = Constant

This shows that if the condition ε_r∝ 1/x is fulfilled, potential gradient will be constant throughout the dielectric of the cable.

† g_1max = Q d π ε ε0 1 ; g_2max = Q d πε ε0 2 1 ; g_3max= Q d πε ε0 3 2

If g_1max = g_2max = g_3max = g_max (say), then,

1 1 ε d = 1 1 2 1 3 2 ε d =ε d †

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282

Principles of Power System V1 = g dx Q x d d d d / / / / 2 2 0 1 2 2 1 1 2

z

=

z

_{π ε ε} dx = Q d d g d d d e max e 2 ₀ ₁ 2 1 1 π ε ε log = log 2 0 1 2 max g Q d   =  _{π ε ε}  ∵ 

Similarly, potential across second layer (V₂) and third layer (V₃) is given by ;

V₂ = g d d d max e 2 1 2 1 log V3 = g d D d max e 2 2log ₂ Total p.d. between core and earthed sheath is

V = V₁+ V₂+ V₃ = g d d d d d d d D d max e e e 2 1 1 2 1 2 2 log + log + log

L

N

M

O

Q

P

If the cable had homogeneous dielectric, then, for the same values of d, D and g_max, the permis-sible potential difference between core and earthed sheath would have been

V′ = g d D

d

max e

2 log

Obviously, V > V′ i.e., for given dimensions of the cable, a graded cable can be worked at a greater potential than non-graded cable. Alternatively, for the same safe potential, the size of graded cable will be less than that of non-graded cable. The following points may be noted :

(i) As the permissible values of g_max are peak values, therefore, all the voltages in above ex-pressions should be taken as peak values and not the r.m.s. values.

(ii) If the maximum stress in the three dielectrics is not the same, then,

V = 1 1 2 max 1 2 3 2

1 2

log log log

2 2 2 max max e e e g d g d g D d d d d + d + d

The principal disadvantage of this method is that there are a few high grade dielectrics of reason-able cost whose permittivities vary over the required range.

Example 11.12. A single-core lead sheathed cable is graded by using three dielectrics of rela-tive permittivity 5, 4 and 3 respecrela-tively. The conductor diameter is 2 cm and overall diameter is 8 cm. If the three dielectrics are worked at the same maximum stress of 40 kV/cm, find the safe working voltage of the cable.

What will be the value of safe working voltage for an ungraded cable, assuming the same con-ductor and overall diameter and the maximum dielectric stress ?

Solution.

Here, d = 2 cm ; d₁ = ? ; d₂ = ? ; D = 8 cm

ε1 = 5 ; ε2 = 4 ; ε3 = 3 ; gmax = 40 kV/cm

Graded cable. As the maximum stress in the three dielectrics is the same,

∴ ε1d = ε2d1 = ε3d2 or 5 × 2 = 4 × d₁ = 3 × d₂ ∴ d₁ = 2·5 cm and d₂ = 3·34 cm * g_max = Q d π ε ε0 1 ∴ g_max d = Q π ε ε0 1 or gmax 2 d = Q 2πε ε_{0 1} *

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Permissible peak voltage for the cable

= g d d d d d d d D d max e e e 2 1 1 2 1 2 2 log + log + log

L

N

M

O

Q

P

= 40 2 2 2 5 2 2 5 3 34 2 5 3 34 8 3 34 loge ⋅ + ⋅ loge ⋅_⋅ + ⋅ loge _⋅

L

NM

O

QP

= 20 [0·4462 + 0·7242 + 2·92] kV = 20 × 4·0904 = 81·808 kV

∴ Safe working voltage (r.m.s.) for cable = 81 808

2

⋅ ₌_{57·84 kV}

Ungraded cable. Permissible peak voltage for the cable

= g d D d max e e 2 40 2 2 8 2 log = × log kV = 55·44 kV

∴ Safe working voltage (r.m.s.) for the cable = 55 44

2

⋅ ₌_{39·2 kV}

This example shows the utility of grading the cable. Thus for the same conductor diameter (d) and the same overall dimension (D), the graded cable can be operated at a voltage (57·84 − 39·20) = 18·64 kV (r.m.s.) higher than the homogeneous cable — an increase of about 47%.

Example 11.13. A single core lead sheathed cable has a conductor diameter of 3 cm; the diameter of the cable being 9 cm. The cable is graded by using two dielectrics of relative permittivity 5 and 4 respectively with corresponding safe working stresses of 30 kV/cm and 20 kV/cm. Calculate the radial thickness of each insulation and the safe working voltage of the cable.

Solution. Here, d = 3 cm ; d₁ = ? ; D = 9cm ε1 = 5 ; ε2 = 4 g1max = 30 kV/cm ; g2max = 20 kV/cm g_1max ∝ 1 1 εd ; g2max ∝ 1 2 1 ε d ∴ g g max max 1 2 = ε ε2 11 d d or d1 = g g d max max 1 2 1 2 30 20 5 3 4 ×ε_ε = × × = 5·625 cm

∴ Radial thickness of inner dielectric = d1 d 2 – = 5 625 3 2 ⋅ − = 1.312 cm Radial thickness of outer dielectric

= D d– 1 2 = 9 5 625 2 – ⋅ = 1.68 cm Permissible peak voltage for the cable

= g d d d g d D d max e max e 1 1 2 1 1 2 log + 2 log = 30 2 3 5 625 3 20 2 5 625 9 5 625 × log_e ⋅ + × ⋅ log_e _⋅ = 28·28 + 26·43 = 54·71 kV

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284

∴ Safe working voltage (r.m.s.) for the cable

= 54.71/ 2 = 38·68 kV

Example 11.14. A 66-kV single-core lead sheathed cable is graded by using two dielectrics of relative permittivity 5 and 3 respectively; thickness of each being 1 cm. The core diameter is 2 cm. Determine the maximum stress in the two dielectrics.

Solution. Fig. 11.16 shows the composite dielectric of a capacitance graded cable. The poten-tial difference V between conductor and earthed sheath is given by ;

V = g dx g dx d d d D 1 2 2 2 2 2 1 1 / / / /

z

+

z

= Q x dx Q x dx d d d D 2 ₀ ₁ 2 2 2 0 2 2 2 1 1 π ε ε π ε ε / / / /

z

+

z

= Q d d D d e e 2 1 1 0 1 1 2 1 π ε ε log +ε log

L

N

M

O

_Q

P

...(i) Now, g_1max = Q d π ε ε0 1 ...(ii) Putting the value of Q = g1maxπ ε0ε1 d from exp. (ii) in exp. (i), we get,

V = g d d d D d max e e 1 1 1 1 2 1 2 1 1 ε ε log +ε log

L

N

M

O

_Q

P

or g_1max = 2 1 1 2 1 V d d d D d e e log + log

L

NM

O

QP

ε ε Here, d = 2 cm, d₁ = 4 cm, D = 6 cm ; V = 66 3× 2 = 53.9 kV, ε1 = 5, ε2 = 3 Substituting the values, we get,

g1max = 2 53 9 2 4 2 5 3 6 4 × ⋅ + [log_e log_e ] kV/cm = 2 53 9 2 0 6931 0 6757 × ⋅ + ⋅. [ ] = 39·38 kV/cm Similarly, it can be *proved that :

g_2max= 2 1 2 1 1 1 V d d d D d e e ε ε log +log

L

NM

O

QP

...(iii) = 2 53 9 4 3 5 4 2 6 4 × + . [ loge loge ] kV/cm = 2 53 9 4 0 4158 0 4054 × ⋅ ⋅ + ⋅ [ ] = 32·81 kV/cm 11.13 11.13 11.13 11.13

11.13 Intersheath Grading Intersheath Grading Intersheath Grading Intersheath Grading Intersheath Grading

In this method of cable grading, a homogeneous dielectric is used, but it is divided into various layers by placing metallic intersheaths between the core and lead sheath. The intersheaths are held at suit-able potentials which are inbetween the core potential and earth potential. This arrangement im-* g_2max = Q

d

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proves voltage distribution in the dielectric of the cable and consequently more uniform potential gradient is obtained.

Consider a cable of core diameter d and outer lead sheath of diameter D. Suppose that two intersheaths of diameters d₁ and d₂are inserted into the homogeneous di-electric and maintained at some fixed potentials. Let V₁, V₂and V₃respectively be the voltage between core and intersheath 1, between intersheath 1 and 2 and between intersheath 2 and outer lead sheath. As there is a definite potential difference between the inner and outer layers of each intersheath, therefore, each sheath can be treated like a homogeneous single core cable. As proved in Art. 11.9,

Maximum stress between core and intersheath 1 is g_1max = V d d d e 1 1 2log Similarly, g_2max = V d d d e 2 1 2 1 2 log g_3max = V d _D d e 3 2 2 2 log

Since the dielectric is homogeneous, the maximum stress in each layer is the same i.e.,

g_1max = g_2max = g_3max=g_max (say)

∴ V d d d e 1 1 2log = V d d d e 2 1 2 1 2 log = V d D d e 3 2 2 2 log

As the cable behaves like three capacitors in series, therefore, all the potentials are in phase i.e. Voltage between conductor and earthed lead sheath is

V = V1 + V2 + V3

Intersheath grading has three principal disadvantages. Firstly, there are complications in fixing the sheath potentials. Secondly, the intersheaths are likely to be damaged during transportation and installation which might result in local concentrations of potential gradient. Thirdly, there are con-siderable losses in the intersheaths due to charging currents. For these reasons, intersheath grading is rarely used.

Example 11.15. A single core cable of conductor diameter 2 cm and lead sheath of diameter 5.3 cm is to be used on a 66 kV, 3-phase system. Two intersheaths of diameter 3·1 cm and 4·2 cm are introduced between the core and lead sheath. If the maximum stress in the layers is the same, find the voltages on the intersheaths.

Solution. Here, d = 2 cm ; d1 = 3·1 cm ; d2 = 4·2 cm D = 5.3 cm ; V = 66 2 3 × = 53·9 kV g_1max = V d d d V e e 1 1 1 2 1 3 1 2 log log = × ⋅ = 2·28 V1

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286

* This equation is obtained if we put the values of eq. (ii) in eq. (i). g2max = V d d d V e e 2 1 2 1 2 2 1 55 4 2 3 1 log log . . = ⋅ = 2·12 V2 g_3max = V d D d V e e 3 2 2 3 2 2 1 5 3 4 2 log log = ⋅ ⋅ ⋅ = 2·04 V₃

As the maximum stress in the layers is the same,

∴ g_1max = g_2max = g_3max or 2·28 V1 = 2·12 V2 = 2·04 V3 ∴ V2 = (2·28/2·12) V1 = 1·075 V1 and V3 = (2·28/2·04) V1 =1·117V1 Now V1 + V2 + V3 = V or V1 + 1·075 V1 + 1·117 V1 = 53·9 or V1 = 53·9/3·192 = 16·88 kV and V2 = 1·075 V1 = 1·075 × 16·88 = 18·14 kV

∴ Voltage on first intersheath (i.e., near to the core)

= V − V1 = 53·9 − 16·88 = 37·02 kV

Voltage on second intersheath = V − V1− V2 = 53·9 − 16·88 − 18·14 = 18·88 kV

Example 11.16. A single-core 66 kV cable working on 3-phase system has a conductor diam-eter of 2 cm and a sheath of inside diamdiam-eter 5.3 cm. If two intersheaths are introduced in such a way that the stress varies between the same maximum and minimum in the three layers, find :

(i) positions of intersheaths (ii) voltage on the intersheaths (iii) maximum and minimum stress

Solution.

Here, d = 2 cm ; D = 5.3 cm ; V = 66 2 3

×

= 53.9 kV

(i) Positions of intersheaths. Suppose that diameters of intersheaths are d₁ and d₂ cm respec-tively. Let V₁, V₂and V₃ respectively be the voltage between conductor and intersheath 1, between intersheath 1 and 2 and between intersheath 2 and outer lead sheath.

g_1max = V d d d e 1 1 2log ; g_2max = _d V _d d e 2 1 2 1 2log ; g_3max = V d D d e 3 2 2 2 log As the maximum stress in the three layers is the same,

∴ V d d d e 1 1 2log = V d d d e 2 1 2 1 2log = V d _D d e 3 2 2 2log ...(i) In order that stress may vary between the same maximum and minimum in the three layers, we have, d1/d = d2/d1 = D/d2 ...(ii) ∴ V d 1 = V d V d 2 1 3 2 = ...*(iii)

Electric power can be transmitted or dis

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