GUDLAVALLERU ENGINEERING COLLEGE (An Autonomous Institute with Permanent Affiliation to JNTUK, Kakinada)
Seshadri Rao Knowledge Village, Gudlavalleru – 521 521 356.
Department of Electrical and Electronics Engineering
HANDOUT ON POWER SYSTEMS-I
UNIT-4 DISTRIBUTION SYSTEMS
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UNIT-IV
DISTRIBUTION SYSTEMS
Learning Objectives:
To familiarize the concept of distribution systems.
To classify the distribution systems.
To calculate voltage drops in DC distributors.
To introduce various distributors.
Syllabus:
DC & AC Distribution Systems
Definitions of Distribution System, various components of Distribution System: feeder, distributor, service mains; Classification of Distribution Systems: Nature of current, Type of construction, Scheme of connection; D.C. Distribution: 2wire DC system, 3wire DC system; Design features of Distribution Systems. Radial distribution, Ring main distribution, Voltage Drop Calculations in D.C Distributors for Radial D.C Distributor fed at one end and at the both the ends (equal/unequal Voltages) – Numerical problems, Ring Main distributor and stepped distributor.AC distribution, Comparison of DC and AC distribution.
Course outcomes:
Students will be able to 4.1 Identify different components of a distribution system. 4.2 Classify the distribution system. 4.3 Describe various design features of a distribution system. 4.4 Determine the voltage drop for various types of distributors. 4.5 Compare DC and AC distribution systems.
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UNIT-4 DISTRIBUTION SYSTEMS 4.1
Distribution System
The part of power system which distributes electric power for local use is known as distribution system. In general, the distribution system is the electrical system between the sub-station fed by the transmission system and the consumer’s meters. It generally consists of feeders, distributors and the service mains. Fig. 4.1 shows the single line diagram of a typical low tension distribution system.
Fig.4.1 single line diagram of a typical low tension distribution system. (i) Feeders. A feeder is a conductor which connects the sub-station (or localized generating station) to the area where power is to be distributed. Generally, no tappings are taken from the feeder so that current in it remains the same throughout. The main consideration in the design of a feeder is the current carrying capacity. (ii) Distributor. A distributor is a conductor from which tappings are taken for supply to the consumers. In Fig. 4.1, A B, BC, CD and DA are the distributors. The current through a distributor is not constant because tappings are taken at various places along its length. While designing a distributor, voltage drop along its length is the main consideration since the statutory limit of voltage variations is ± 6% of rated value at the consumers’ terminals. (iii) Service mains. A service main is generally a small cable which connects the distributor to the consumers’ terminals.
4.2
Classification of Distribution Systems
A distribution system may be classified according to ; (i) Nature of current. According to nature of current, distribution system may be classified as (a) D.C. distribution system (b) A.C. distribution system. 4
(ii) Type of construction. According to type of construction, distribution system may be classified as (a) Overhead system. (b) Underground system. The overhead system is generally employed for distribution as it is 5 to 10 times cheaper than the equivalent underground system. In general, the underground system is used at places where overhead construction is impracticable or prohibited by the local laws. (iii) Scheme of connection. According to scheme of connection, the distribution system may be classified as (a) Radial system (b) Ring main system (c) Inter-connected system.
4.3
D.C. Distribution
It is a common knowledge that electric power is almost exclusively generated, transmitted and distributed as a.c. However, for certain applications, d.c. supply is absolutely necessary. For instance, d.c. supply is required for the operation of variable speed machinery (i.e., d.c. motors), for electro-chemical work and for congested areas where storage battery reserves are necessary. For this purpose, a.c. power is converted into d.c. power at the substation by using converting machinery e.g., mercury arc rectifiers, rotary converters and motor-generator sets. The d.c. supply from the substation may be obtained in the form of (i) 2-wire or (ii) 3-wire for distribution. (i) 2-wire d.c. system. As the name implies, this system of distribution consists of two wires. One is the outgoing or positive wire and the other is the return or negative wire. The loads such as lamps, motors etc. are connected in parallel between the two wires as shown below. This system is never used for transmission purposes due to low efficiency but may be employed for distribution of d.c. power.
Fig.4.2 2-wire and 3-wire d.c distribution. (ii) 3-wire d.c. system. It consists of two outers and a middle or neutral wire which is earthed at the substation. The voltage between the outers is twice the voltage between either outer and neutral wire as shown in above fig. The principal advantage of this system is that it makes available two voltages at the consumer terminals viz., V between any outer and the neutral and 2V between the outers. Loads requiring high voltage (e.g., motors) are connected across the outers, whereas lamps and heating circuits requiring less voltage are connected
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between either eit her outer and a nd the neutral. The methods of obtaining 3-wire 3-wir e system are discussed in the following article.
4.4
A.C. Distribution
Now-a-days electrical electri cal energy is generated, transmitted tr ansmitted and distributed distri buted in the form of alternating cur-rent. One important reason for the widespread use of alternating current in preference to direct current curre nt is the fact that alternating alt ernating voltage can be conveniently changed in magnitude by means of a transformer. Transformer has made it possible to transmit a.c. power at high voltage and utilize it at a safe potential. High transmission and distribution voltages have greatly reduced the current in the co nductors and the resulting line losses. There is no definite line between transmission and distribution according to voltage or bulk capacity. However, in general, the a.c. distribution system is the electrical system between the step-down step-down substation fed by the transmission system and the consumers’ meters. The a.c. distribution system is classified into (i) primary distribution system and (ii) secondary distribution system. (i) Primary distribution system. It is that part of a.c. distribution system which operates at voltages somewhat higher than general utilisation and handles large blocks of electrical energy than the average low-voltage consumer uses. The voltage used for primary distribu tion depends upon the amount of power to be conveyed and the distance of the substation required to be fed. The most commonly used primary distribution voltages are 11 kV, 6·6 kV and 3·3 kV. Due to economic considerations, primary distribution is carried out by 3-phase, 3-wire system.
Fig.4.3 primary distribution system
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Fig. 4.3 shows a typical primary distribution system. Electric power from the generating station is transmitted at high voltage to the substation located in or near the city. At this substation, voltage is stepped down to 11 kV with the help of step-down transformer. Power is supplied to various substations for distribution or to big consumers at this voltage. This forms the high voltage distribution or primary distribution. (ii) Secondary distribution system. It is that part of a.c. distribution system which includes the range of voltages at which the ultimate consumer utilises the electrical energy delivered to him. The secondary distribution employs 400/230 V, 3-phase, 4-wire system.
Fig. 4.4 shows a typical secondary distribution system. The primary distribution circuit delivers power to various substations, called distribution sub-stations. The substations are situated near the consumers’ localities and contain step-down step-down transformers. At each distribution substation, the voltage is stepped down to 400 V and power is delivered by 3 phase, 4-wire 4- wire a.c. system. The voltage volta ge between betw een any two phases is 400 V and between bet ween any phase and neutral is 230 V. The single phase domestic loads are connected between any one phase and the neutral, whereas 3-phase 400 V motor loads are connected across 3-phase lines directly.
Fig. 4.4 shows a typical secondary distribution system.
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4.5
Overhead versus Underground System
The distribution system can be overhead or underground. Overhead lines are generally mounted on wooden, concrete or steel poles which are arranged to carry distribution transformers in addition to the conductors. The underground system uses conduits, cables and manholes under the surface of streets and sidewalks. The choice between overhead and underground system depends upon a number of widely differing factors. Therefore, it is desirable to make a comparison between the two. (i) Public safety. The underground system is more safe than overhead system because all distribution wiring is placed underground and there are little chances of any hazard. (ii) Initial cost. The underground system is more expensive due to the high cost of trenching, conduits, cables, manholes and other special equipment. The initial cost of an underground system may be five to ten times than that of an overhead system. (iii) Flexibility. The overhead system is much more flexible than the underground system. In the latter case, manholes, duct lines etc., are permanently placed once installed and the load expansion can only be met by laying new lines. However, on an overhead system, poles, wires, transformers etc., can be easily shifted to meet the changes in load conditions. (iv) Faults. The chances of faults in underground system are very rare as the cables are laid underground and are generally provided with better insulation. (v) Appearance. The general appearance of an underground system is better as all the distribution lines are invisible. This factor is exerting considerable public pressure on electric supply companies to switch over to underground system. (vi) Fault location and repairs. In general, there are little chances of faults in an underground system. However, if a fault does occur, it is difficult to locate and repair on this system. On an overhead system, the conductors are visible and easily accessible so that fault locations and repairs can be easily made. (vii) Current carrying capacity and voltage drop. An overhead distribution conductor has a considerably higher current carrying capacity than an underground cable conductor of the same material and cross-section. On the other hand, underground cable conductor has much lower inductive reactance than that of an overhead conductor because of closer spacing of conductors. (viii) Useful life. The useful life of underground system is much longer than that of an overhead system. An overhead system may have a useful life of 25 years, whereas an underground system may have a useful life of more than 50 years.
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(ix) Maintenance cost. The maintenance cost of underground system is very low as compared with that of overhead system because of less chances of faults and service interruptions from wind, ice, lightning as well as from traffic hazards. (x) Interference with communication circuits. An overhead system causes electromagnetic interference with the telephone lines. The power line currents are superimposed on speech currents, resulting in the potential of the communication channel being raised to an undesirable level. However, there is no such interference with the underground system.
4.6
Connection Schemes of Distribution System
All distribution of electrical energy is done by constant voltage system. In practice, the following distribution circuits are generally used: (i) Radial System. In this system, separate feeders radiate from a single substation and feed the distributors at one end only. Fig. 4.5 (i) shows a single line diagram of a radial system for d.c. distribution where a feeder OC supplies a distributor A B at point A. Obviously, the distributor is fed at one end only i.e., point A is this case. Fig. 4.5 (ii) shows a single line diagram of radial system for a.c. distribution. The radial system is employed only when power is generated at low voltage and the substation is located at the centre of the load.
Fig.4.5 single line diagram of a radial system for d.c. distribution This is the simplest distribution circuit and has the lowest initial cost. However, it suffers from the following drawbacks: (a) The end of the distributor nearest to the feeding point will be heavily loaded. (b) The consumers are dependent on a single feeder and single distributor. Therefore, 9
any fault on the feeder or distributor cuts off supply to the consumers who are on the side of the fault away from the substation. (c) The consumers at the distant end of the distributor would be subjected to serious voltage fluctuations when the load on the distributor changes.
Due to these limitations, this system is used for short distances only. (ii) Ring main system. In this system, the primaries of distribution transformers form a loop. The loop circuit starts from the substation bus-bars, makes a loop through the area to be served, and returns to the substation. Fig. 4.6 shows the single line diagram of ring main system for a.c. distribution where substation supplies to the closed feeder LMNOPQRS. The distributors are tapped from different points M, O and Q of the feeder through distribution transformers. The ring main system has the following advantages.
Fig. 4.6 shows the single line diagram of ring main s ystem (a) There are less voltage fluctuations at consumer’s terminals. (b) The system is very reliable as each distributor is fed via two feeders. In the event of fault on any section of the feeder, the continuity of supply is maintained. For example, suppose that fault occurs at any point F of section SLM of the feeder. Then section SLM of the feeder can be isolated for repairs and at the same time continuity of supply is maintained to all the consumers via the t he feeder SRQPONM.
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(iii) Interconnected system. When the feeder ring is Energized by two or more than two generating stations or substations, it is called inter-connected system. Fig. 4.7 shows the single line diagram of interconnected system where the closed feeder ring ABCD is supplied by two substations S1 and S 2 at points D and C respectively. Distributors are connected to points O, P, Q and R of the feeder ring through distribution transformers. The interconnected system has the following advantages :
Fig. 4.7 shows the single line diagram of interconnected system (a) It increases the service reliability. (b) Any area fed from one generating station during peak load hours can be fed from the other generating station. This reduces reserve power capacity and increases efficiency of the system.
4.7
Requirements of a Distribution System
A considerable amount of effort is necessary to maintain an electric power supply within the requirements of various types of consumers. Some of the requirements of a good distribution system are: proper voltage, availability of power on demand and reliabilit y.
4.7.1 Design Considerations in Distribution System Good voltage regulation of a distribution network is probably the most important factor responsible for delivering good service to the consumers. For this purpose, design of feeders and distributors requires careful consideration. (i) Feeders. A feeder is designed from the point of view of its current carrying capacity while the voltage drop consideration is relatively unimportant. It is because voltage drop in a feeder can be compensated by means of voltage regulating equipment equipment at the substation. 11
(ii) Distributors. A distributor is designed from the point of view of the voltage drop in it. It is because a distributor supplies power to the consumers and there is a statutory limit of volt-age variations at the consumer’s terminals (± 6% of rated value). The size and length of the distributor should be such that voltage at the consumer’s terminals is within the permissible limits.
4.8
Voltage drop calculations in D.C Distributors
4.8.1 D.C. Distributor Fed at one End — Concentrated Concentrated Loading Fig. 4.8 shows the single line diagram of a 2-wire d.c. distributor A B fed at one end A A and having concentrated loads I loads I 1, I 2, I 3 and I and I 4 tapped off at points C , D, D, E and F and F respectively. respectively.
Fig. 4.8 Let r 1, r 2, r 3 and r 4 be the resistances of both wires (go and return) of the sections A C , CD, CD, DE and EF and EF of the distributor respectively. = I 1 + I 2 + I 3 + I + I 4 Current fed from point point A A = I 1 + I 2 + I 3 + I + I 4 Current in section A section A C = I 2 + I 3 + I 4 Current in section CD Current in section DE section DE = I 3 + I 4 Current in section EF section EF = I 4 Voltage drop in section A section A C = r 1 ( I 1 + I 2 + I 3 + I 4) Voltage drop in section CD = r 2 ( I 2 + I 3 + I 4) Voltage drop in section DE section DE = r 3 ( I 3 + I 4) Voltage drop in section EF section EF = r 4 I 4 Total voltage drop in the distributor = r 1 ( I I 1 + I + I 2 + I + I 3 + I + I 4) + r 2 ( I I 2 + I + I 3 + I + I 4) + r 3 ( I I 3 + I + I 4) + r 4 I 4 It is easy to see that the minimum potential will occur at point F point F which which is farthest from the feeding point A point A
∴
.
4.9.2 Uniformly Loaded Distributor Fed at One End Fig 4.9 shows the single line diagram of a 2-wire d.c. distributor A B fed at one end A A and loaded uniformly with i amperes per metre length. It means that at every 1 m length of the distributor, the load tapped is i amperes. Let l metres metres be the length of the distributor and r ohm ohm be the resistance per metre run.
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Fig. 4.9 Consider a point C on the distributor at a distance x x metres from the feeding point A as shown in Fig. 4.9. Then current at point C is is = i l − l − i x amperes x amperes = i (l (l − − x) x) ampere Now, consider a small length dx near dx near point C . Its resistance is r dx and dx and the voltage drop over length dx is dx is dv = i (l − x) r dx = i r (l − x) dx Total voltage drop in the distributor up to point C is is
− − = − 2 0
The voltage drop upto point B ( B (i.e i.e.. over the whole distributor) can be obtained by putting x putting x = = l in in the above expression.
∴
Voltage drop over the distributor AB distributor AB
= irl ∗ l −
)
= = = 12
Where i l = I , the total current entering at point A point A r l = R, the total resistance of the distributor Thus, in a uniformly loaded distributor fed at one end, the total voltage drop is equal to that produced by the whole of the load assumed to be concentrated at the middle point.
4.9.3 Distributor Fed at Both Ends — Concentrated Concentrated Loading Whenever possible, it is desirable that a long distributor should be fed at both ends instead of at one end only, since total voltage drop can be considerably reduced without increasing the cross-section of the conductor. The two ends of the distributor may be supplied with (i ) equal voltages (i i ) unequal voltages. 1.
Two ends fed with equal voltages.
Fig. 4.10 All the currents tapped off between points A A and E (minimum p.d. point) will be supplied from the feeding point A while A while those tapped off between B and B and E E will will be supplied from the feeding point B. B. The 13
current tapped off at point E point E itself itself will be partly supplied from A from A and and partly from B from B.. If these currents are x are x and y and y respectively, respectively, then, I 3 = x + y Therefore, we arrive at a very important conclusion that at the point of minimum potential, current comes from both ends of the distributor. Point of minimum potential. It is generally desired to locate the point of minimum potential. There is a simple method for it. Consider a distributor A distributor A B having B having three concentrated loads I loads I 1, I 2 and I and I 3 at points C points C , D and E and E respectively. Suppose that current supplied by feeding end A end A is I is I A . Then current distribution in the various sections of the distributor can be worked out as shown in Fig. 4.11 (i ( i). Thus
IAC = IA ICD = IA − I IDE = IA − I − I IEB = IA − I − I − I3
Fig. 4.11
Voltage drop between A between A and and B B = = Voltage drop over A over A B Or V − V = I A R AC + ( I I A − I 1) RCD + ( I I A − I 1 − I 2) R DE + ( I I A − I 1 − I 2 − I 3) R EB From this equation, the unknown I A can be calculated as the values of other quantities are generally given. Suppose actual directions of currents in the various sections of the distributor are indicated as shown in Fig. 4.11 (ii ( ii). ). The load point where the currents are coming from both sides of the distributor is the point of minimum potential i.e. i.e. point E point E in in this case. 2. Two ends fed with unequal voltages. Fig shows the distributor AB fed with unequal voltages one end A being fed at V1 volts and end B at V2 volts. Thus Voltage drop between A and B =Voltage drop over AB Or V1-V2=Voltage drop over AB
Fig. 4.12 14
4.9.4 Uniformly Loaded Distributor Fed at Both Ends Now determine the voltage drop in a uniformly loaded distributor fed at both ends. There can be two cases viz . the distributor fed at both ends with (i ( i) equal voltages (ii ( ii)) unequal voltages. The two cases shall be discussed separately. (i ) Distributor fed at both ends with equal voltages. Consider a distributor A distributor A B of length l metres, having resistance r ohms ohms per metre run and with uniform loading of i amperes per metre run as shown in Fig. 13.24. Let the distributor be fed at the feeding points A and A and B B at at equal voltages, say V volts. volts. The total current supplied to the distributor is i l . As the two end voltages are equal, therefore, current supplied from
each feeding point is
i.e. i.e.
Current supplied from each feeding point
= 2 Fig. 4.12
Consider a point C at at a distance x distance x metres metres from the feeding point A point A.. Then current at point C is is
= − = −
Now, consider a small length dx near dx near point C . Its resistance is r dx and dx and the voltage drop over length dx is dx is
Voltage drop upto point C
= ( (2 − ) ) = (2 −) = ∫0 − = −
Obviously, the point of minimum potential will be the mid-point. Therefore, maximum voltage drop will occur at mid-point i.e. i.e. where x where x = = l /2. /2. Max voltage drop =
Where
− −
= ∗ − 2 2 4 = 18 = 18 = 18 15
(i i ) Distributor fed at both ends with unequal voltages. Consider a distributor AB AB of length l metres having resistance r resistance r ohms per metre run and with a uniform loading of i i amperes per metre per metre run as shown in Fig. 13.25. Let the distributor be fed from feeding points A and A and B B at at voltages V A and V B respectively.
Suppose that the point of minimum potential C is situated at a distance x x metres from the feeding point A point A.. Then current supplied by the feeding point A point A will will be i x. x.
Fig. 4.13 As the distance of C from feeding point B is (l-x) ,therefore, current fed from B is (l-x). Voltage drop in section BC =
2 − 2 v
Voltage at point C, Vc= V A-drop over AC
= − 2 Voltage at point c, Vc=VB-drop over BC
− = − 2 From equations (i) and (ii), we get,
− = − 2 = − 2 Solving the equation for x, we get,
= − 2 As all the quantities on the right hand side of the equation are known, therefore, the point on the distributor where minimum potential occurs can be calculated.
4.9.5
Ring Distributor
A distributor arranged to form a closed loop and fed at one or more points is called a ring distributor . Such a distributor starts from one point, makes a l oop through the area to be served, and returns 16
to the original point. For the purpose of calculating voltage distribution, the distributor can be considered as consisting of a series of o f open distributors fed at both ends. The principal advantage of ring distributor is that by proper choice in the number of feeding points, great economy in copper can be affected. The simplest case of a ring r ing distributor is the one having only one feeding point as shown in above Fig. Here A is the feeding point and tapings are taken from points B and C. For the purpose of calculations, it is equivalent to a straight distributor fed at both ends with with equal voltages.
4.9.6 Ring Main Distributor with Interconnector Sometimes a ring distributor has to serve a large area. In such a case, voltage drops in the various sections of the distributor may become excessive. In order to reduce voltage drops i n various sections, distant points of the distributor are joined through a conductor called interconnector . Fig. shows the ring distributor ABCDEA distributor ABCDEA.. The points B points B and and D D of of the ring distributor are through an interconnector BD interconnector BD.. There are several methods for solving s olving such a network. However, the solution of such a network can be readily obtained by applying Thevenin’s theorem. The steps of Procedure are:
Fig. 4.14
Current in interconnector BD =
+
Therefore, current distribution in each section and the voltage of load points can be calculated.
5.
AC DISTRIBUTION:
Now-a-days, electrical energy is i s generated, transmitted and distributed dist ributed in the form of alternating current as an economical proposition. The electrical energy produced at the power station is transmitted at very high voltages by 3-phase, 3-wire system to step-down sub-stations for distribution. The distribution system consists of two parts viz . primary distribution and secondary distribution. The primary distribution ci rcuit is 3 phase, 3-wire and operates at voltages (3·3 (3· 3 or 6·6 or 11kV) somewhat higher than general utilisation levels. It delivers power to the secondary distribution circuit through distribution transformers situated near consumers’ localities. Each distribution transformer steps down the voltage to 400 V and power is distributed to ultimate consumers’ by 400/230 V, 3-phase, 3 -phase, 4-wire system.. 5.1. A.C. Distribution Calculations
A.C. distribution calculations differ from those of d.c. distribution in t he following respects: 17
(i ) In case of d.c. system, the voltage drop is due to resistance alone. However, in a.c. system, the voltage drops are due to the combined effects of resistance, inductance and capacitance. (ii ) In a d.c. system, additions and subtractions of currents or voltages are done arithmetically but in case of a.c. system, these operations are done vectorially. (ii i ) In an a.c. system, power factor (p.f.) has to be taken into account. Loads tapped off form the distributor are generally at different power factors. There are two ways of referring power factor viz (a) It may be referred to supply or receiving end voltage which is regarded as the reference vector. (b) It may be referred to the voltage at the load point itself.
There are several ways of solving a.c. distribution problems. However, symbolic notation method has been found to be most convenient for this purpose. In this method, voltages, currents and impedances are expressed in complex notation and the calculations are made exactly as in d.c. distribution. 5.2. Methods of Solving A.C. Distribution Problems
In a.c. distribution calculations, power factors of various load currents have to be considered since currents in different sections of the distributor will be the vector sum of load currents and not the arithmetic sum. The power factors of load currents may be given (i) w.r.t . receiving or sending end voltage or ( ii) ii) w.r.t . to load voltage itself.
(i )
Power factors referred to receiving end voltage. Consider an a.c. distributor AB distributor AB with concentrated loads of I of I 1 and I and I 2 tapped off at points C and B as shown in Fig. Taking the receiving end voltage VB as the reference vector, let lagging power factors at C and B and B be be cos φ1 and cos φ2 w.r.t. VB. VB. Let R Let R1, 1, X X 1 and R and R2, 2, X X 2 be the resistance and reactance of sections AC sections AC and CB of the distributor.
Fig. 4.15
Impedance of section AC section AC , Z AC = R1 R1 + j + j X 1 Impedance of section CB, CB, Z CB R2 + j + j X 2 CB = R2 Load current at point C , I 1 = I = I 1 (cos φ1 − j − j sin φ1) Load current at point B point B,, I 2 = I = I 2 (cos φ2 − j − j sin φ2) 18
Current in section CB, CB, I CB = I 2 (cos φ2 − j − j sin φ2) CB = I 2 = I Current in section AC section AC , I AC = I 1 + I + I 2 = I 1 (cos φ1 − j − j sin φ1) + I + I 2 (cos φ2 − j − j sin φ2) Voltage drop in section CB, CB, V CB − j sin φ2) ( R2 R2 + j + j X 2) 2) CB = I CB CB Z CB CB = I 2 (cos φ2 − j Voltage drop in section AC section AC , V AC = I AC Z AC = ( I I 1 + I + I 2 ) Z ) Z AC = [ I I 1(cos 1(cos φ1 − j − j sin φ1) + I + I 2 (cos φ2 − j − j sin φ2)] [ R1 R1 + jX + jX 1] 1] Sending end voltage, VA = VB +VCB +VAC Sending end current, IA current, IA = I 1+ I2 The vector diagram of the a.c. distributor under t hese conditions is shown in Fig. 2. Here, the receiving end voltage VB is taken as the referenc r eferencee vector. As power factors of loads are given w.r.t. VB, VB, therefore, I therefore, I 1 and I and I 2 lag behind VB by VB by φ1 and φ2 respectively.
(ii)
Power factors referred to respective load voltages. Suppose the power factors of loads in the previous Fig. 14.1 are referred to their respective load voltages. Then φ1 is the phase angle between VC and I and I 1 and φ2 is the phase angle between VB and I 2. 2. The vector diagram under these conditions is shown in Fig.3.
Fig. 4.16 Voltage drop in section CB = I2 ZCB = I2 I2 (cos φ2 − j − j sin φ2) (R2 (R2 + j X2) Voltage at point C = VB + Drop in section CB = VC Now I1 = I1
∴
I1 = I1
∠
∠
∠
α (say)
− φ1 w.r.t. voltage VC
− (φ1 − α) w.r.t. voltage VB
i.e. I1 = I1 I1 [cos (φ1 − α) − j − j sin (φ1 − α)] Now IAC = I 1+ I2 = I1 I1 [cos (φ1 − α) − j − j sin (φ1 − α)] + I2 (cos φ2 − j − j sin φ2) Voltage drop in section AC = IAC ZAC
∴
Voltage at point A = VB + Drop in CB + Drop in AC
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6. STEPPED DISTRIBUTOR
It is also known as tapered distributor. distr ibutor. It is necessary to design each distributor with the minimum volume of conductor material by satisfying the voltage limits.
Fig. 4.18
Loads when tapped from any distributor changes the current along the length of the distributor .for example ,a uniformly loaded distributor supplying voltage at one end carries a current that changes f rom maximum value at supply end to the minimum at the other end .If the uniform cross section conductor is used throughout the distributor, it affects savings. If the t he conductor cross section is determined based on the current t o be carried i.e, if the conductor is stepped, the required conductor material is minimum. When the conductor is tapered then the cross section of the conductor at any an y point is proportional to the square root of the distance from the far end off the distributor. Practically it is not possible .the stepped distributor is shown in fig.
Assignment-Cum-Tutorial Questions A. Questions testing the remembering / understanding level of students
I)
Objecti Objecti ve / Shor hor t answer nswer Questi Questi ons
1. Which of the following is not the distribution system normally used (A) 3 phase-4 wire
(B) 3 phase-3 wire (C) Single phase - 3 wire
(D) Single phase -4 wire.
2. Which distribution system is more reliable? `
(A) Ring main system
(B) Tree system (C) Radial system
(D) All are equally reliable.
3. Out of the following systems of distribution, which system offers the be st economy? (A) Direct current system
(B) AC single phase system
(C) AC 3 phase 3 wire system
(D) AC 3 phase 4 wire system.
4. What are the advantages of dc transmission system over ac transmission system? (A) DC system is economical (B) there is no skin effect in dc system (C) Corona limits are highest for dc circuits as compared to ac circuits
(D) All of the above.
5. The main advantage of ac. transmission system over d.c. transmission system is (A) Easy transformation
(B) Less losses in transmission over long distances 20
(C) ) Less insulation problems
(D) Less problem of instability.
6. Material generally used for bus bar is (A) copper
(B) aluminium
(C) steel
(D) tungsten.
7. Stability of a system is not affected by (A) Reactance of line
(B) Losses
(C) Reactance of generator
(D) Output torque.
8. The permissible voltage variable in voltage in distribution is (A) 0.1%
(B) 1%
(C) 10%
(D) 50%.
9. The service mains connect (A) Distributor and consumer terminals
(C) Distributor and relay system
(B) distributor and transformer
(D) transformer and earth.
10. Conductors which carry bulk loads from generating station or a substation are called: a) Busbars
b) distributors
c) feeders
d)service mains
11. The no. Of circuits through which a consumer is fed in a rin g main system is: a)1
b) 2
c)3
d)4
12.The voltage most commonly used for the primary distribution is a) 11KV
b) 230V
c)132KV
d) none
13. 13. Thevenin’s theorem Thevenin’s theorem is used to solve a) DC network
b)AC network
c)either DC or AC networks d)none
14. Which of the following power distribution system gives the better reliability? a) Ring main system b) radial system
c) neither a&b
d)none
a) greater security of service b) reduction in length of lines
c)none
15. An interconnected system provides
I I ) D escr scr i pti ve Quest Questii ons 1. Describe briefly the different types of D.C distributors? 2. Derive an expression for the voltage drop for a uniformly loaded distributor fed at one end? 3. What is the purpose of interconnector in a D.C ring main distributor? 4. Define terms a) feeder
b)distributor
c)service mains
5. Discuss the relative merits and demerits of dc distribution and ac distribution systems? 6. Explain about ring main distributor and advantages & disadvantages? 7. Explain the voltage drop calculation for the power factors refer red to receiving end voltage? 21
8. What are the important requirements for a good distribution system? 9. What is an inter-connector? Discuss the advantages of using it in distribution systems? 10. Compare radial and ring main distribution systems? B. Question testing the ability of students in applying the concepts
I ) M ultip ulti ple choi choi ce questions uestions 1. Cross section of the conductor is obtained, on the basis of current loading for a) Feeders
b) distributors
c) service mains
d) line
2. The point of minimum potential for a uniform distributor fed at one end is at a) The middle
b) the far end c)a point between the far end and the middle
3 .In solving AC networks and distributors the resistances have to be replaced by a) Admittance
b) impedance c) conductance
d) none
4. A distribution transformer usually is a a) star-star transformer
b) delta-delta transformer
c) star-delta transformer
d) none
5. Three phase, four wire AC system of distribution is used for a) Unbalanced load
b)balanced load
c)all types of load
d)none
6. A uniformly loaded distributor fed at one end only has power loss of “x” watts, if the distributor is now fed from both ends a) x/2
b)x/4 c)x/8
d)x/6
7. Advantages of radial system of distribution over other s ystems is its _______ initial cost a) Less
b) high
c) moderate
d) very high
8. The voltage drop at the far end of uniformly loaded distributor of 50m length of resistance 0.01ohm/m with a load of 2A/m length is a) 230V
b) 25V
c) 1V
d) 0.02V
9. In a distributor fed at both ends with concentrated loads, the point of minimum potential always occurs at a ______ point. a) Receiving end
b) mid
c) sending end
d) none
10. Which of the following distribution systems is the simplest and has lowest installation cost? a) Interconnected system b) radial system
c) ring main system 22
d) none
I I ) PROBLEMS PROBLEMS 1.
A 2-wire d.c. distributor 200 metres long is uniformly loaded with 2A/metre.Resistance of single wire is 0·3 Ω/km. If the distributor is fed at one end, calculate : (i) the voltage drop upto a distance of 150 m from the feeding point (ii) the maximum voltage drop
2. A 2-wire d.c. street mains AB, 600 m long is fed from both ends at 220 V. Loads of 20 A, 40 A, 50 A and 30 A are tapped at distances of 100m, 250m, 400m and 500 m from the end A respectively. If the area of X-section of distributor conductor conductor is 1cm2, find the minimum consumer voltage. Take ρ = 1·7 × 10− 10−6 Ω cm.
\ 3. A 2-wire, DC distributor 500 m long is loaded uniformly at the rate of 0.4A/m. If the voltage drop in the distributor is not to exceed 5V, Calculate the area of cross section of each conductor required when the distributor is fed at one end. Take resistivity of a conductor material as 1.7×10
-8
Ω-m.
4. A 300 m ring distributor has loads as shown below, where distances are in meters. The resistance of each conductor is 0.4 ohms per km and the loads are tapped off at points B, C and D as shown. If distributor is fed at A with 240 V supply, find voltages at B, C and D.
5. A two-wire d.c. distributor cable 1000 metres long is loaded with 0·5 A/metre. Resistance of each conductor is 0·05 Ω/km. Calculate the maximum voltage drop if the distributor is fed from both ends with equal voltages of 220 V. What is the minimum voltage and where it occurs ? 6.
A 2- wire DC distributor AB 300 m long and is fed from both ends and supplies a uniformly distributed load of 0.15amps per meter length together with the following concentrated loads: 50A at C; 60A at D and 40A at E, distances AC, CD and DE being 75, 100 and 50 meters respectively. If the supply voltages at A and B are 205 and 200 volts respectively and the resistance of each conductor is 0.00015 ohms per meter, calculate the current supplied at each end and the point of minimum potential.
7. A DC 2 wire distributor 600m long is fed at both ends A and B at 220V shown in figure. The load consists of 50A at 100m from A, 75A at 150m from A and a uniform loading of 0.5A per meter for the last 400m. The resistance of each conductor is 0.05Ω 0.05Ω/Km. Determine the location and magnitude of minimum voltage
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8. A 2- core distributor cable 400 metres long supplies a uniformly distributed lightning load of 1 amp per metre. There are concentrated concentrated loads of 120, 72, 48, and 120 amps at 40, 120, 200 and 320 meters respectively respectively from the end A. The cable has a resistance of 0.15 ohm per km run. Determine the position of the lowest – lowest – run run lamp and its voltage when the cable is fed at 250 volts f rom both ends A and B.
9. For the same voltage drop between the supply point A and the load point D in the network shown below, compare the weights of copper required with and without an interconnector between A and D. Conductors Conductors have the same cross sections in each scheme.
10. A single phase a.c. distributor AB 300 metres long is fed from end A and is loaded as under : (i) 100 A at 0·707 p.f. lagging l agging 200 m from point A (ii) 200 A at 0·8 p.f. lagging 300 m from point A The load resistance and reactance of the distributor is 0·2 Ω and 0·1 Ω per kilometre. Calculate Calculate the total voltage drop in the distributor. The load power factors refer to the voltage at the far end. 11. A 3-phase, 400V distributor AB is loaded as shown in Fig.14.8. The 3-phase load at point C takes 5A per phase at a p.f. of 0·8 lagging. At point B, a 3-phase, 3 -phase, 400 V induction motor is connected which has an output of 10 H.P. with an efficiency of 90% and p.f. 0·85 lagging. If voltage at point B is to t o be maintained at 400 V, what should be the voltage at point A ? The resistance and reactance reactance of the line are 1Ω and 0·5Ω per phase per kilometre respectively respectively..
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