III/IV B.Tech (Regular/Supplementary) DEGREE EXAMINATION

Transcription

1 Hall Ticket Number: 14EE503 October, 2018 Fifth Semester Time: Three Hours Answer Question No.1 compulsorily. Answer ONE question from each unit. III/IV B.Tech (Regular/Supplementary) DEGREE EXAMINATION Electrical & Electronics Engineering Transmission and Distribution Maximum: 60 Marks (1X12 = 12 Marks) (4X12=48 Marks) 1. Answer all questions (1X12=12 Marks) a) Why all overhead lines use ACSR conductors? b) On what factors the Skin effect depend? c) For what purpose bundled conductors are primarily used in transmission lines? d) Mention the factors that affect sag in the transmission line. e) What is meant by Ferranti effect? f) Define corona. g) State the advantages of interconnected system. h) What is meant by distributor? i) Define Feeder? j) What is meant by string efficiency? k) What is meant by grading of cables? l) Define thermal resistance of cables. UNIT I 2. a) Derive expression for loop inductance of a single phase transmission line? b) Find the inductive reactance of a 3- phase bundled conductor line with 2 conductors per phase with spacing of 40 cm. phase to phase separation is 7 cm in horizontal configuration. All are ACSR with diameter of 3.5 cm. (OR) 3. a) Derive the expression for capacitance of a single circuit 3-phase transmission line with unsymmetrical spacing. b) A 3- phase, 50 Hz, 132 KV overhead line conductors placed in horizontal plane 4m apart. Conductor diameter is 2 cm if the line length is 100 Km. Calculate the charging current per phase assuming complete transposition. UNIT II 4. a) A 3-phase, 50-Hz overhead transmission line 100 km long has the following constants : Resistance/km/phase = 0.1 Ω Inductivereactance/km/phase = 0 2 Ω Capacitive susceptance/km/phase= siemen Determine (i) the sending end current (ii) sending end voltage (iii) sending end power factor and (iv) transmission efficiency when supplying a balanced load of 10,000 kw at 66 kv, p.f. 0 8 lagging. Use nominal T method. b) Develop ABCD constants for Medium transmission line.(nominal-t & Nominal- π) (OR) 5. a) Develop an expressions for regulation and efficiency of a short transmission line and draw required circuit and phasor diagram. b) Develop ABCD constants for Long transmission line. 14EE503

2 UNIT III 6. a) Discuss and compare Radial and Ring main distribution system. What is the role of interconnectors in distribution system? b) A 2-wire d.c. distributor ABCDEA in the form of a ring main is fed at point A at 220 V and is loaded as under : 10A at B ; 20A at C ; 30A at D and 10 A at E. The resistances of various sections (go and return) are : AB = 0.1 Ω ; BC = 0.05Ω ; CD = 0.01Ω; DE = 0.025Ω and EA = 0.075Ω. Determine : (i) the point of minimum potential (ii) current in each section of distributor. (OR) 7. a) Explain briefly the classification of substations? b) With a neat sketch explain double bus with double breaker and double bus with single breaker and state their advantages and disadvantages. UNIT IV 8. a) In a 33 kv overhead line, there are three units in the string of insulators. If the capacitance between each insulator pin and earth is 11% of self-capacitance of each insulator, find (i) the distribution of voltage over 3 insulators and (ii) string efficiency. b) Discuss the methods of improve the string efficiency of overhead line insulators. (OR) 9. a) Describe the general construction of underground cable with neat sketch. Explain each part? b) Deduce an expression for insulation resistance of a single core cable in terms of specific resistance of dielectric its core and sheath diameter.

3 Bapatla Engineering College: Bapatla (AUTONOMOUS) EEE Department Class: 3/4 EEE SCHEME OF VALUATION Date: 18/10/2018 Sub. Code: 14EE503 TRANSMISSION AND DISTRIBUTION Max.Marks:60M a) Aluminum conductor steel-reinforced cable (ACSR) is a type of high-capacity, high-strength stranded conductor typically used in overhead power lines. The outer strands are high-purity aluminum, chosen for its good conductivity, low weight and low cost. b) Factors affecting skin effect. Frequency Skin effect increases with the increase in frequency. The shape of the conductor Skin effect is more in the solid conductor and less in the stranded conductor because the surface area of the solid conductor is more. c) Bundled conductors are primarily employed to reduce the corona loss and radio interference. d) Factors affecting the sag. Conductor weight Sag of the conductor is directly proportional to its weight. The weight of the conductors is increased due to ice loading. Span Sag is directly proportional to the square of the span length. e) In electrical engineering, the Ferranti effect is an increase in voltage occurring at the receiving end of a long transmission line, above the voltage at the sending end. This occurs when the line is energized, but there is a very light load or the load is disconnected. f) A corona discharge is an electrical discharge brought on by the ionization of a fluid such as air surrounding a conductor that is electrically charged. Spontaneous corona discharges occur naturally in high-voltage systems unless care is taken to limit the electric field strength. g) The function of generating station is to provide power to a large number of consumers. But consumers power requirement will vary as per their activities. On the other hand, to get maximum efficiency the alternators in the power station have to run at rated capacity. h) It generally consists of feeders, distributors and the service mains in general, distribution system is electrical between substation fed while designing a distributor, voltage drop along its length main through use split phase power, can have both 120 volt receptacles 240 those taping which extracted for supply. i) In electric power distribution, Feeder is voltage power line transferring power from a distribution substation to the distribution transformers In an electrical wiring circuit in a building which Feeder is a wire/line that carries power from a transformer or switch gear to a distribution panel. j) String efficiency is an important consideration since it decides the potential distribution along the string. The greater the string efficiency, the more uniform is the voltage distribution. Thus 100% string efficiency is an ideal case for which the voltage across each disc will be exactly the same. k) Grading Of Cables. Grading of cable is the process of achieving uniform distribution of dielectric stress or voltage gradient in a dielectric of cable. Voltage gradient or dielectric stress is maximum at the surface of the conductor and minimum at the inner surface of a sheath. l) Thermal resistance is a heat property and a measurement of a temperature difference by which an object or material resists a heat flow. Thermal resistance is the reciprocal of thermal conductance. It is the thermal resistance of unit area of a material.

4 UNIT I 2 a) Consider two solid round conductors with radii of r 1 and r 2 as shown in Fig. One conductor is the return circuit for the other. This implies that if the current in conductor 1 is I then the current in conductor 2 is -I. First let us consider conductor 1. The current flowing in the conductor will set up flux lines. However, the flux beyond a distance D + r 2 from the center of the conductor links a net current of zero and therefore does not contribute to the flux linkage of the circuit. Also at a distance less than D - r 2 from the center of conductor 1 the current flowing through this conductor links the flux. Moreover since D >> r 2 we can make the following approximations Inductance due to internal and external flux written as We can rearrange L 1 as follows Substituting r 1 = r 1 e -1/4 in the above expression we get The radius r 1 can be assumed to be that of a fictitious conductor that has no internal flux but with the same inductance as that of a conductor with radius r 1. In a similar way the inductance due current in the conductor 2 is given by herefore the inductance of the complete circuit is If we assume r 1 1 = r 2 1 = r, then the total inductance becomes 2 b)

5 (OR) 3 a) Fig. shows a 3-phase transposed line having unsymmetrical spacing. Let us assume balanced conditions i.e. Q A + Q B + Q C = 0. Considering all the three sections of the transposed line for phase A, Potential of 1st position, Potential of 2nd position, Potential of 3rd position, Average voltage on condutor A is

6 Capacitance from conductor to neutral is 3 b) 4 a) UNIT II

7 4 b) i. Nominal Π Representation of a Medium Transmission Line: Now applying KCL, at node P, we get. Similarly applying KCL, to node Q. Now substituting equation (2) to equation (1) Now by applying KVL to the circuit, Comparing equation (4) and (5) with the standard ABCD parameter equations

8 We derive the parameters of a medium transmission line as: ii. Nominal T Representation of a Medium Transmission Line: Applying KVL to the above network we get, Now the sending end current is, Substituting the value of V M to equation (9) we get, Again comparing equation (8) and (10) with the standard ABCD parameter equations, The parameters of the T network of a medium transmission line are

9 5 a) (OR) 5 b) Here a line of length l > 250km is supplied with a sending end voltage and current of V S and I S respectively, where as the V R and I R are the values of voltage and current obtained from the receiving end. Lets us now consider an element of infinitely small length Δx at a distance x from the receiving end as shown in the figure where. V = value of voltage just before entering the element Δx. I = value of current just before entering the element Δx. V+ΔV = voltage leaving the element Δx. I+ΔI = current leaving the element Δx. ΔV = voltage drop across element Δx. zδx = series impedence of element Δx yδx = shunt admittance of element Δx Where, Z = z l and Y = y l are the values of total impedance and admittance of the long transmission line.

10 Therefore, the voltage drop across the nfinitely small element Δx is given by Now to determine the current ΔI, we apply KCL to node A Since the term ΔV yδx is the product of 2 infinitely small values, we can ignore it for the sake of easier calculation. Therefore, we can write Now derivating both sides of eq (1) w.r.t x, Now substituting from equation (2) The solution of the above second order differential equation is given by. Derivating equation (4) w.r.to x. Now comparing equation (1) with equation (5) Now to go further let us define the characteristic impedance Z c and propagation constant δ of a long ransmission line as Then the voltage and current equation can be expressed in terms of characteristic impedance and propagation constant as Now at x=0, V= V R and I= I r. Substituting these conditions to equation (7) and (8) respectively. Solving equation (9) and (10), We get values of A 1 and A 2 as,

11 Now applying another extreme condition at x = l, we have V = V S and I = I S. Now to determine V S and I S we substitute x by l and put the values of A 1 and A 2 in equation (7) and (8) we get By trigonometric and exponential operators we know Therefore, equation (11) and (12) can be re-written as Thus comparing with the general circuit parameters equation, we get the ABCD parameters of a long transmission line as, UNIT III 6 a) Electrical Power Distribution System Radial In early days of electrical power distribution system, different feeders radially came out from the substation and connected to the primary of distribution transformer. Radial electrical power distribution system has one major drawback that in case of any feeder failure, the associated consumers would not get any power as there was no alternative path to feed the transformer. In case of transformer failure also, the power supply is interrupted. In other words the consumer in the radial electrical distribution system would be in darkness until the feeder or transformer was rectified. Ring Main Electrical Power Distribution System The drawback of radial electrical power distribution system can be overcome by introducing a ring main electrical power distribution system. Here one ring network of distributors is fed by more than one feeder. In this case if one feeder is under fault or maintenance, the ring distributor is still energized by other feeders connected to it. In this way the supply to the consumers is not affected even when any feeder becomes out of service. In addition to that the ring main system is also provided with different section isolates at different suitable points. If any fault occurs on any section, of the ring, this section can easily be isolated by opening the associated section isolators on both sides of the faulty zone transformer directly. In this way, supply to the consumers connected to the healthy zone of the ring, can easily be maintained even when one section of the ring is under shutdown. The number of feeders connected to the ring main electrical power distribution system depends upon the following factors. 1. Maximum Demand of the System: If it is more, then more numbers of feeders feed the ring. 2. Total Length of the Ring Main Distributors: It length is more, to compensate the voltage drop in the line, more feeders to be connected to the ring system. 3. Required Voltage Regulation: The number of feeders connected to the ring also depends upon the permissible allowable, voltage drop of the line. The sub distributors and service mains are taken off may be via distribution transformer at different suitable points on the ring depending upon the location of the consumers. Sometimes, instead of connecting service main directly to the ring, sub distributors are also used to feed a group of service mains where direct access of ring distributor is not possible.

12 6 b) (OR) 7 a) Classifications of Substations Step-up or Primary Substations Such types of substations generate low voltage like 3.3, 6.6, 11, or 33kV. This voltage is stepped up by the help of a step-up transformer for transmitting the power over large distances. It is located near the generating substation Primary Grid Substations This substation lowered the value of primary stepped up voltages. The output of the primary grid substation acts as the input of the secondary substations. The secondary substation is used for stepping down the input voltage to more lowered for further transmission. Step-down or Distribution Substations This substation is placed near the load centre where the primary distribution is stepped down for sub-transmission. The secondary distribution transformer feeds the consumer through the service line Transformer substations In such type of substation transformers are installed for transforming the power from one voltage level to another level as per need. Switching Substations The substations use for switching the power line without disturbing the voltage is known as the switching substations. This type of substations is placed between the transmission line.

13 Converting Substations In such types of substations, AC power converting into DC power or vice versa or it can convert high frequency to lower frequency or vice versa. High Voltage Substations (HV Substations) Involving voltages between 11 KV and 66 KV. Extra High Voltage Substations Involving voltages between 132 kv and 400 KV. Ultra High Voltage Operating voltage above 400 KV. Grid Substations This substation is used for transferring the bulk power from one point to another. If any fault occurs on the substation, then the continuity of whole of the supply is affected by it. Town Substations These substations step down the voltage at 33/11 kv for more distribution in the towns. If there is any fault occurs in this substation, then the supply of the whole town is blocked. Indoor Type Substations In such type of substations, the apparatus is installed within the substation building. Such type of substations is usually for the voltage up to 11 KV but can be raised for the 33 KV or 66 KV when the surrounding air is polluted by dust, fumes or gasses, etc. Outdoor Substations These substations are further subdivided into two categories Pole Mounted Substations Such Substations are erected for distributions of power in the localities. Single stout pole or H-pole and 4-pole structures with relevant platforms are operating for transformers of capacity up to 25 KVA, 125 KVA, and above 125KVA. Foundation Mounted Substations Such types of substations are used for mounting the transformers having capacity 33,000 volts or above. 7 b) Double Bus Double Breaker: In this scheme there are two buses and two circuit breakers per circuit are used (See Fig-D). In normal state both the buses are energised. Any circuit breaker can be removed for maintenance without interruption of the corresponding circuit. Also the failure of one of the two buses does not interrupt any circuit as all the circuits can be fed from the remaining bus and isolating the failed bus. By shifting circuit from one bus to other the loading on the buses can be balanced. The substation with this configuration requires twice as much equipments as single bus scheme. This scheme has high reliability. But due to more equipments this scheme is costly and requires more space. This scheme is usually used at EHV transmission substation or generating station where high reliability is required. Double Bus Single Breaker: This scheme is shown in Fig-E. This scheme has two buses. Each circuit has one breaker and connected to both buses by isolators as shown. There is one tie breaker between two buses. The tie breaker is normally closed. For the tie breaker in closed position the circuit can be connected to either of the buses by closing the corresponding switch. It is clear that fault on one bus requires isolation of the bus and the circuits are fed from the other bus. From the figure you can guess that the configuration has some

14 bus scheme. This scheme is costlier and requires more space than the single bus scheme. Many EHV transmission substations use this schemee with an additional transfer bus. 8 a) UNIT IV

15 8 b) Methods of Improving String Efficiency The maximum voltage appears across the insulator nearest to the line conductor and decreases progressively as the crossarm is approached. If the insulation of the highest stressed insulator (i.e. nearest to conductor) breaks down or flash over takes place, the breakdown of other units will take place in succession. This necessitates to equalise the potential across the various units of the string i.e. to improve the string efficiency. The various methods for this purpose are : 1. By using longer cross-arms. The value of string efficiency depends upon the value of K i.e., ratio of shunt capacitance to mutual capacitance. The lesser the value of K, the greater is the string efficiency and more uniform is the voltage distribution. The value of K can be decreased by reducing the shunt capacitance. In order to reduce shunt capacitance, the distance of conductor from tower must be increased i.e., longer cross-arms should be used. However, limitations of cost and strength of tower do not allow the use of very long cross-arms. In practice, K = 0 1 is the limit that can be achieved by this method. 2. By grading the insulators. In this method, insulators of different dimensions are so chosen that each has a different capacitance. The insulators are capacitance graded i.e. they are assembled in the string in such a way that the top unit has the minimum capacitance, increasing progressively as the bottom unit (i.e., nearest to conductor) is reached. Since voltage is inversely proportional to capacitance, this method tends to equalise the potential distribution across the units in the string. This method has the disadvantage that a large number of different-sized insulators are required. However, good results can be obtained by using standard insulators for most of the string and larger units for that near to the line conductor. 3. By using a guard ring. The potential across each unit in a string can be equalised by using a guard ring which is a metal ring electrically connected to the conductor and surrounding the bottom insulator. The guard ring introduces capacitance between metal fittings and the line conductor. The guard ring is contoured in such a way that shunt capacitance currents i1, i2 etc. are equal to metal fitting line capacitance currents i 1, i 2 etc. The result is that same charging current I flows through each unit of string. Consequently, there will be uniform potential distribution across the units. (OR) 9 a) Construction of Cables : Figure shows the general construction of a 3-conductor cable. The various parts are : 1) 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 Figure is used for 3- phase service. The conductors are made of tinned copper or aluminium and are usually stranded in order to provide flexibility to the cable. 2) 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. 3) Metallic sheath:in order to protect 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 Figure.

16 4) 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. 5) Armouring:Over the bedding armouring is provided which consist of one or two layers of galvanised steel wire or steel tape. Its purpose is to protect the cable from mechanical injuries while laying it or handling it. Armouring may not be done in the case of some cables. 6) Serving: In order to protect armouring from atmospheric conditions, a layer of fibrous material like jute similar to bedding is provided over the armouring. This is known as serving. 7) Armouring and serving are only applied to the cables for the protection of conductor insulation and to protect metallic sheath from mechanical injury. 9 b) Insulation Resistance of a Single Core Cable: The cable conductor is provided with an insulation of suitable thickness so as to avoid leakage of current. The path for leakage current is radial, as shown in Fig. through the insulating material. The opposition offered by the insulation to the leakage current is called the insulation resistance of the cable. Consider a single core cable of conductor radius r 1, internal sheath radius r 2, length l and insulation material resistivity ρ. Let an elementary cylindrical section of the insulation of radius r and thickness dr be considered. Now the length through which the leaking current will flow is dr and area of x-section is 2πrl. The insulation resistance offered to the leakage current by elementary cylindrical section of the insulation under consideration is ρdr/2πrl. Insulation resistance of the cable, i.e. the insulation resistance of the cable varies inversely as the length of the cable (R INS α 1/l). If ρ is in Ω-m, r 2, r 1, and l are in metres then insulation resistance of the single core cable is in ohms. Average value of ρ for impregnated paper is about Ω-m at 15 C and decreases exponentially with temperature. Dielectric Stress in a Single Core Cable: Under operating conditions, the insulation of a single core cable is subjected to electrostatic stress, called the dielectric stress. Potential gradient at any point is defined as the rate of increase of potential at that point and is the same as the dielectric stress at that point.

17 Since single core cable is a form of cylindrical condenser, therefore, electric intensity at a distance x from the centre O of the cable is given by the expression: Since potential gradient = Electric intensity Hence, Substituting the value of Q from above expression in Eq. we have Since potential gradient g varies inversely as x (as obvious from the above expression), therefore, potential gradient will be maximum when x is minimum i.e., x = d/2 and potential gradient will be minimum when x is maximum i.e. x = D/2 Maximum and minimum values of potential gradient are given by: Hence the ratio between the maximum potential gradient and the minimum potential gradient i.e.,