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C O N T E N T S
CONTENTS
40. D.C. Transmission and Distribution ...1569
Transmission and Distribution of D.C. Power—Two- wire and Three-wire System—Voltage Drop and Transmission Efficiency—Methods of Feeding Distributor— D.C.Distributor Fed at One End— Uniformly Loaded Distributor— Distributor Fed at Both Ends with Equal Voltages — Distributor Fed at Both Ends with Unequal Voltages—Uniform Loading with Distributor Fed at Both Ends— Concentrated and Uniform Loading with Distributor Fed at One End— Ring Distributor— Current Loading and Load-point Voltages in a 3-wire System—Three-wire System— Balancers—Boosters—Comparison of 2-wire and 3-wire Distribution Systems—Objective Tests
41. A.C. Transmission and Distribution ... 1603 -
General Layout of the System— Power Systems and System Networks — Systems of A.C. Distribution-Single-phase, 2- wire System-Single-phase, 3-wire System-Two-phase, 3-wire System-Two-phase, 4-wire System—Three-phase, 3-wire System-Three-phase, 4-wire System—Distribution—Effect of Voltage on Transmission Efficiency —Comparison of Conductor Materials Required for Various Overhead Systems—Constants of a Transmission Line—Reactance of an Isolated Single-phase Transmission Line—Reactance of 3-phase Transmission Line—Capacitance of a Single-phase Transmission Line—Capacitance of a Three-phase Transmission Line-Short Single-phase Line Calculations— Short Three-phase Transmission Line Constants — Effects of Capacitance—Nominal T-method- “Nominal” π- method—Ferranti Effect-Charging Current and Line Loss of an Unloaded Transmission Line—Generalised Circuit Constants of a Transmission Line—Corona-Visual Critical Voltage—Corona Power —Disadvantages of Corona—
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Underground Cables—Insulation Resistance of a Single-core Cable—Capacitance and Dielectric Stress— Capacitance of 3-core Belted Cables—Tests for Three-phase Cable Capacitance—A.C. Distribution Calculations—Load Division Between Parallel Lines — Suspension Insulators— Calculation of Voltage Distribution along Different Units— Interconnectors—Voltage Drop Over the Interconnector— Sag and Stress Analysis— Sag and Tension with Supports at Equal Levels —Sag and Tension with Supports at Unequal Levels-Effect of Wind and Ice — Objective Tests.
42. Distribution Automation ... 1683 - 1698
Introduction—Need Based Energy Management (NBEM) — Advantages of NBEM—Conventional Distribution Network—Automated System — Sectionalizing Switches — Remote Terminal Units (RTU’s) — Data Acquisition System (DAS) — Communication Interface — Power line carrier communication (PLCC) — Fibre optics data communication — Radio communication — Public telephone communication — Satellite communication — Polling scheme — Distribution SCADA — Man - Machine Interface — A Typical SCADA System — Distribution Automation — Load Management in DMS Automated Distribution System — Data acquisition unit — Remote terminal unit (RTU) — Communication unit — Substation Automation — Requirements — Functioning — Control system — Protective System — Feeder Automation — Distribution equipment — Interface equipment — Automation equipment — Consumer Side Automation — Energy Auditing—Advantages of Distribution Automation — Reduced line loss —Power quality — Deferred capital expenses – Energy cost reduction
- Optimal energy use — Economic benefits — Improved reliability — Compatibility — Objective Tests.
43. Electric Traction ... 1699 - 1766
General—Traction Systems—Direct Steam Engine Drive — Diesel-electric Drive-Battery-electric Drive- Advantages of Electric Traction—Disadvantages of Electric Traction — Systems of Railway Electrification—Direct Current System—Single- phase Low frequency A.C. System—Three- phase Low frequency A.C. System—Composite System —
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of DC Motors—Rheostatic Braking Torque—Rheostatic Braking of Induction Motors — Regenerative Braking— Energy Saving in Regenerative Braking - Objective Tests.
45. Rating and Service Capacity ... 1795 - 1822
Size and Rating — Estimation of Motor Rating — Different Types of Industrial Loads—Heating of Motor or Temperature Rise—Equation for Heating of Motor — Heating Time Constant — Equation for Cooling of Motor or Temperature Fall — Cooling Time Constant — Heating and Cooling Curves — Load Equalization — Use of Flywheels — Flywheel Calculations — Load Removed (Flywheel Accelerating) — Choice of Flywheel — Objective Tests.
46. Electronic Control of AC Motors ... 1823 - 1832
Classes of Electronic AC Drives — Variable-Frequency Speed Control of a SCIM—Variable Voltage Speed Control of a SCIM—Speed Control of a SCIM with Rectifier Inverter System—Chopper Speed Control of a WRIM—Electronic Speed Control of Synchronous Motors—Speed Control by Current fed D.C. Link—Synchronous Motor and Cycloconverter— Digital Control of Electric Motors — Application of Digital Control—Objective Tests.
47. Electric Heating ... 1833 - 1860
Introduction—Advantages of Electric Heating—Different Methods of Heat Transfer — Methods of Electric Heating— Resistance Heating—Requirement of a Good Heating Element—Resistance Furnaces or Ovens—Temperature Control of Resistance Furnaces—Design of Heating Element —Arc Furnaces—Direct Arc Furnace—Indirect Arc Furnace — Induction Heating—Core-type Induction Furnace— Vertical Core-Type Induction Furnace—Indirect Core-Type Induction Furnace—Coreless Induction Furnace—High Frequency Eddy-current Heating—Dielectric Heating — Dielectric Loss-Advantages of Dielectric Heating— Applications of Dielectric Heating — Choice of Frequency — Infrared Heating — Objective Tests.
48. Electric Welding ... 1861 - 1892
Definition of Welding—Welding Processes—Use of
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Electricity in Welding—Formation and Characteristics of Electric Arc—Effect of Arc Length — Arc Blow—Polarity in DC Welding — Four Positions of Arc Welding— Electrodes for Metal Arc Welding—Advantages of Coated Electrodes—Types of Joints and Types of Applicable Welds—Arc Welding Machines—V-I Characteristics of Arc Welding D.C. Machines — D.C. Welding Machines with Motor Generator Set—AC Rectified Welding Unit — AC Welding Machines—Duty Cycle of a Welder — Carbon Arc Welding — Submerged Arc Welding —Twin Submerged Arc Welding — Gas Shield Arc Welding — TIG Welding — MIG Welding — MAG Welding—Atomic Hydrogen Welding—Resistance Welding—Spot Welding—Seam Welding—Projection Welding —Butt Welding-Flash Butt Welding—Upset Welding—Stud Welding—Plasma Arc Welding—Electroslag Welding—Electrogas Welding — Electron Beam Welding—Laser Welding—Objective Tests.
49. Illumination ... 1893 - 1942
Radiations from a Hot Body—Solid Angle—Definitions — Calculation of Luminance (L) of a Diffuse Reflecting Surface — Laws of Illumination or Illuminance — Laws Governing Illumination of Different Sources — Polar Curves of C.P. Distribution—Uses of Polar Curves - Determination of M.S.C.P and M.H.C.P. from Polar Diagrams—Integrating Sphere or Photometer — Diffusing and Reflecting Surfaces: Globes and Reflectors—Lighting Schemes —Illumination Required for Different Purposes — Space / Height Ratio— Design of Lighting Schemes and Lay-outs—Utilisation Factor or Coefficient of Utilization [η] — Depreciation Factor (p) —Floodlighting —Artificial Sources of Light — Incandescent Lamp—Filament Dimensions —Incandescent Lamp Characteristics—Clear and Inside—frosted Gas-filled Lamps—Discharge Lamps—Sodium Vapour Lamp—High- Pressure Mercury Vapour Lamp — Fluorescent Mercury— Vapour Lamps— Fluorescent Lamp— Circuit with Thermal Switch —Startless Fluorescent Lamp Circuit — Stroboscopic Effect of Fluorescent Lamps — Comparison of Different Light Sources — Objective Tests.
50. Tariffs and Economic Considerations ... 1943 - 2016
Economic Motive—Depreciation—Indian Currency —
C H A P T E R
D.C.
TRANSMISSION
AND
DISTRIBUTION
Learning Objectives
➣➣ ➣➣➣ Transmission and Distribu- tion of D.C. Power ➣➣ ➣➣➣ Two-wire and Three-wire System ➣➣ ➣➣➣ Voltage Drop and Transmission Efficiency ➣➣ ➣➣➣ Methods of Feeding Distributor ➣➣ ➣➣➣ D.C.Distributor Fed at One End ➣➣ ➣➣➣ Uniformaly Loaded Distributor ➣➣ ➣➣➣ Distributor Fed at Both Ends with Equal Voltage ➣➣ ➣➣➣ Distributor Fed at Both ends with Unequal Voltage ➣➣ ➣➣➣ Uniform Loading with Dis- tributor Fed at Both Ends ➣➣ ➣➣➣^ Concentrated and Uniform Loading with Distributor Fed at One End ➣➣ ➣➣➣ Ring Distributor ➣➣ ➣➣➣^ Current^ Loading^ and Load-point Voltage in a 3-wire System ➣➣ ➣➣➣ Three-wire System ➣➣ ➣➣➣^ Balancers ➣➣ ➣➣➣ Boosters ➣➣ ➣➣➣ Comparison of 2-wire and 3-wire Distribution System
(1) Electricity leaves the power plant, (2) Its voltage is increased at a step-up transformer, (3) The electricity travels along a transmission line to the area where power is needed, (4) There, in the substation, voltage is decreased with the help of step-down transformer, (5) Again the transmis- sion lines carry the electricity, (6) Electricity reaches the final consumption points
Ç
1570 Electrical Technology 40.1. Transmission and Distribution of D.C. Power By transmission and distribution of electric power is meant its conveyance from the central station where it is generated to places, where it is demanded by the consumers like mills, factories, residential and commercial buildings, pumping stations etc. Electric power may be transmitted by two methods. ( i ) By overhead system or ( ii ) By underground system—this being especially suited for densely- populated areas though it is somewhat costlier than the first method. In over-head system, power is conveyed by bare conductors of copper or aluminium which are strung between wooden or steel poles erected at convenient distances along a route. The bare copper or aluminium wire is fixed to an insulator which is itself fixed onto a cross-arm on the pole. The number of cross-arms carried by a pole depends on the number of wires it has to carry. Line supports consist of ( i ) pole structures and ( ii ) tower. Poles which are made of wood, reinforced concrete or steel are used up to 66 kV whereas steel towers are used for higher voltages. The underground system employs insulated cables which may be single, double or triple-core etc. A good system whether overhead or underground should fulfil the following requirements :
1. The voltage at the consumer’s premises must be maintained within ± 4 or ± 6% of the declared voltage, the actual value depending on the type of load. 2. The loss of power in the system itself should be a small percentage (about 10%) of the power transmitted. 3. The transmission cost should not be unduly excessive. 4. The maximum current passing through the conductor should be limited to such a value as not to overheat the conductor or damage its insulation. 5. The insulation resistance of the whole system should be very high so that there is no undue leakage or danger to human life. It may, however, be mentioned here that these days all production of power is as a.c. power and nearly all d.c. power is obtained from large a.c. power systems by using converting machinery like synchronous or rotary converters, solid-state converters and motor-generator sets etc. There are many sound reasons for producing power in the form of alternating current rather than direct current. ( i ) It is possible, in practice, to construct large high-speed a.c. generators of capacities up to 500 MW. Such generators are economical both in the matter of cost per kWh of electric energy produced as well as in operation. Unfortunately, d.c. generators cannot be built of ratings higher than 5 MW because of commutation trouble. Moreover, since they must operate at low speeds, it necessi- tates large and heavy machines. ( ii ) A.C. voltage can be efficiently and conveniently raised or lowered for economic transmis- sion and distribution of electric power respectively. On the other hand, d.c. power has to be generated at comparatively low voltages by units of relatively low power ratings. As yet, there is no economical method of raising the d.c. voltage for transmission and lowering it for distribution. Fig. 40.1 shows a typical power system for obtaining d.c. power from a.c. power. Other details such as instruments, switches and circuit breakers etc. have been omitted. Two 13.8 kV alternators run in parallel and supply power to the station bus-bars. The voltage is stepped up by 3-phase transformers to 66 kV for transmission purposes* and is again stepped down to 13.8 kV at the sub-station for distribution purposes. Fig. 40.1 shows only three methods com- monly used for converting a.c. power to d.c. power at the sub-station.
- According to Indian Electricity Rules, voltage fluctuations should not exceed ± 5% of normal voltage for L.T. supply and ± 12 12 % for H.T. supply. ** Transmission voltages of upto 400 kV are also used.
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Fig. 40.
40.2. Two-wire and Three-wire Systems
In d.c. systems, power may be fed and distributed either by ( i ) 2-wire system or ( ii ) 3-wire system.
Fig. 40.
D.C. Transmission and Distribution (^15731573) In the 2-wire system, one is the outgoing or positive wire and the other is the return or negative wire. In the case of a 2-wire distributor, lamps, motors and other electrical apparatus are connected in parallel between the two wires as shown in Fig. 40.3. As seen, the potential difference and current have their maximum values at feeding points F 1 and F 2. The standard voltage between the conductors is 220 V. The 2-wire system when used for transmission purposes, has much lower efficiency and economy as compared to the 3-wire system as shown later. A 3-wire has not only a higher efficiency of transmission (Fig. 40.4) but when used for distribu- tion purposes, it makes available two voltages at the consumer’s end (Fig. 40.5). This 3-wire system consists of two ‘ outers ’ and a middle or neutral wire which is earthed at the generator end. Its potential is midway between that of the outers i.e. if the p.d. between the outers is 460 V, then the p.d. of positive outer is 230 V above the neutral and that of negative outer is 230 V below the neutral. Motors requiring higher voltage are connected across the outers whereas lighting and heating circuits requiring less voltage are connected between any one of the outers and the neutral.
Fig. 40.4 Fig. 40. The addition of the middle wire is made possible by the connection diagram shown in Fig. 40.4. G is the main generator which supplies power to the whole system. M 1 and M 2 are two identical shunt machines coupled mechanically with their armatures and shunt field winding joined in series across the outers. The junction of their armatures is earthed and the neutral wire is taken out from there.
40.3 Voltage Drop and Transmission Efficiency Consider the case of a 2- wire feeder (Fig. 40.6). A B is the sending end and CD the receiving end. Obviously, the p.d. at A B is higher than at CD. The difference in potential at the two ends is the potential drop or ‘drop’ in the cable. Suppose the transmitting volt- age is 250 V, current in A C is 10 amperes, and resistance of each feeder conductor is 0.5 Ω, then drop in each feeder conductor is 10 × 0.5 = 5 volt and drop in both feeder conductor is 5 × 2 = 10 V. ∴ P.d. at Receiving end CD is = 250 − 10 = 240 V Input power at A B = 250 × 10 = 2,500 W Output power at CD = 240 × 10 = 2,400 W ∴ power lost in two feeders = 2,500 − 2,400 = 100W The above power loss could also be found by using the formula Power loss = 2 I^2 R = 2 × 10 2 × 0.5 = 100 W The efficiency of transmission is defined, like any other efficiency, as the ratio of the output to input
Fig. 40.
D.C. Transmission and Distribution (^15751575) ( i ) 220 V Supply Current per feeder conductor I 1 = P / Area of conductor required A (^) 1 = I 1 /σ = P/220 σ Volume of Cu required for both conductors is V 1 = 2 A (^) 1 l = 2202^ Pl σ^ = 110 Pl σ ( ii ) 440 V Supply V 2 = (^220) Pl σ
%age saving in Cu = 1 2 1
100 110 220 100 110
Pl Pl V V V Pl
− σ − σ × = × σ
= 50%
40.4. Methods of Feeding a Distributor Different methods of feeding a distributor are given below :
1. feeding at one end 2. feeding at both ends with equal voltages 3. feeding at both ends with unequal voltages 4. feeding at some intermediate point In adding, the nature of loading also varies such as ( a ) concentrated loading ( b ) uniform loading ( c ) combination of ( a ) and ( b ). Now, we will discuss some of the important cases separately.
40.5. D.C. Distributor Fed at One End In Fig. 40.7 is shown one conductor A B of a distributor with concentrated loads and fed at one end. Let i 1 , i 2 , i 3 etc. be the currents tapped off at points C , D , E and F and I 1 , I 2 , I 3 etc. the currents in the various sections of the distributor. Let r 1 , r 2 , r 3 etc. be the ohmic resistances of these various sections and R 1 , R 2 , R 3 etc. the total resistance from the feeding end A to the successive tapping points****. Then, total drop in the distributor is = r 1 I 1 + r 2 I 2 + r 3 I 3 + .......... Now I 1 = i 1 + i 2 + i 3 + i 4 + ......; I 2 = i 2 + i 3 + i 4 + .......; I 3 = i 3 + i 4 + ...... ∴ = r 1 ( i 1 + i 2 + i 3 +.......) + r 2 ( i 2 + i 3 + i 4 + ........) + r 3 ( i 3 + i 4 + ......) = i 1 r 1 + i 2 ( r 1 + r 2 ) + i 3 ( r 1 + r 2 + r 3 ) + ...... = i 1 R 1 + i 2 R 2 + i 3 R 3 + ....... = sum of the moments of each load current about feeding point A****.
1. Hence, the drop at the far end of a distributor fed at one end is given by the sum of the moments of various tapped currents about the feeding point i.e. ν = Σ i R. 2. It follows from this that the total voltage drop is the same as that produced by a single load equal to the sum of the various concentrated loads, acting at the centre of gravity of the load system. 3. Let us find the drop at any intermediate point like E. The value of this drop is = i 1 R 1 + i 2 R 2 + i 3 R 3 + R 3 ( i 4 + i 5 + i 6 + .....)*
Fig. 40.
- The reader is advised to derive this expression himself.
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=
sum of moments moment of load beyond upto assumed acting at
(^) + (^) E (^) E E In general, the drop at any intermediate point is equal to the sum of moments of various tapped currents upto that point plus the moment of all the load currents beyond that point assumed to be acting at that point. The total drop over both conductors would, obviously, be twice the value calculated above.
40.6. Uniformly Loaded Distributor In Fig. 40.8 is shown one conductor A B of a distributor fed at one end A and uniformly loaded with i amperes per unit length. Any convenient unit of length may be chosen i.e. a metre or 10 metres but at every such unit length, the load tapped is the same. Hence, let i = current tapped off per unit length l = total length of the distributor r = resistance per unit length of the distributor Now, let us find the voltage drop at a point C (Fig. 40.9) which is at a distance of x units from feeding end A. The current at point C is ( il − ix ) = i ( l − x ).
Fig. 40.8 Fig. 40. Consider a small section of length dx near point C. Its resistance is ( rdx ). Hence, drop over length dx is dv = i ( l − x ) ( rdx ) = ( ilr − ixr ) dx The total drop up to point x is given by integrating the above quantity between proper limits. ∴ 0 dx
ν
∫ =^0 (^ )
x
∫ ilx^ − ixr^ dx ∴^ ν^ =^
1 2 2 2 2 ilrx − irx = ir^ ^ lx −^ x The drop at point B can be obtained by putting x = l in the above expression. ∴ drop at point B = 2 2 2 (^ )^ (^ )^1 2 2 2 2 2 ir^ ^ l − l^ ^ = irl =^ i^ ×^ l^ ×^ r^ × l = IR = I × R where i × l = I – total current entering at point A ; r × l = R –the total resistance of distributor A B. Total drop in the distributor A B = 12 IR ( i ) It follows that in a uniformly loaded distributor, total drop is equal to that produced by the whole of the load assumed concentrated at the middle point. ( ii ) Suppose that such a distributor is fed at both ends A and B with equal voltages. In that case, the point of minimum potential is obviously the middle point. We can thus imagine as if the distribu- tor were cut into two at the middle point, giving us two uniformly-loaded distributors each fed at one end with equal voltages. The resistance of each is R /2 and total current fed into each distributor is I /2. Hence, drop at the middle point is = (^1) ( ) ( )^1 2 2 2 8 ×^ I^ × R^ = IR It is 1/4th of that of a distributor fed at one end only. The advantage of feeding a distributor at both ends, instead of at one end, is obvious. The equation of drop at point C distant x units from feeding point A = irlx − 12 irx^2 shows that the diagram of drop of a uniformly-loaded distributor fed at one end is a parabola.
1578 Electrical Technology
= 2 1.78 10 8 32, 700 A
× × − × ∴ A = 3.56 × 327 × 10 −^7 m^2 = 1,163 cm 2
40.7. Distributor Fed at Both Ends with Equal Voltages It should be noted that in such cases ( i ) the maximum voltage drop must always occur at one of the load points and ( ii ) if both feeding ends are at the same potential, then the voltage drop between each end and this point must be the same, which in other words, means that the sum of the moments about ends must be equal. In Fig. 40.11 ( a ) is shown a distributor fed at two points F 1 and F 2 with equal voltages. The potential of the conduc- tor will gradually fall from F 1 onwards, reach a minimum value at one of the tapping, say, A and then rise again as the other feeding point F 2 is approached. All the currents tapped off between points F 1 and A will be supplied from F 1 while those tapped off between F 2 and A will be supplied from F 2. The current tapped at point A itself will, in general, be partly supplied by F 1 and partly by F 2. Let the values of these currents be x and y respectively. If the distributor were actually cut off into two at A —the point of minimum voltage, with x amperes tapped off from the left and y amperes tapped off from the right, then potential distribution would remain unchanged, showing that we can regard the distributor as consisting of two separate distributors each fed from one end only, as shown in Fig. 40.11 ( b ). The drop can be calculated by locating point A and then values of x and y can be calculated. This can be done with the help of the following pair of equations : x + y = i 4 and drop from F 1 to A (^) 1 = drop from F 2 to A (^) 2. Example 40.4. A 2-wire d.c. distributor F 1 F 2 1000 metres long is loaded as under : Distance from F 1 (in metres) : 100 250 500 600 700 800 850 920 Load in amperes : 20 80 50 70 40 30 10 15 The feeding points F 1 and F 2 are maintained at the same potential. Find which point will have the minimum potential and what will be the drop at this point? Take the cross-section of the conduc- tors as 0.35 cm^2 and specific resistance of copper as (1.764 × 10^ −^^6 ) Ω -cm. Solution. The distributor along with its tapped currents is shown in Fig. 40.12. The numbers along the distributor indicate length in metres.
Fig. 40. The easiest method of locating the point of minimum potential is to take the moments about the two ends and then by comparing the two sums make a guess at the possible point. The way it is done is as follows : It is found that 4th point from F 1 is the required point i.e. point of minimum potential. Using the previous two equations, we have
Fig. 40.
D.C. Transmission and Distribution (^15791579) x + y = 70 and 47,000 + 600 x = 20,700 + 400 y Solving the two equations, we get x = 1.7 A, y = 70 − 1.7 = 68.3 A Drop at A per conductor = 47,000 + 600 × 1.7 = 48,020 ampere-metre.
Resistance/metre =
8 4
ñ 1.764 10 1 0.35 10
l A
− − = ×^ × ×
= 50.4 × 10 −^5 Ω/m
Hence, drop per conductor = 48,020 × 50.4 × 10 −^5 = 24.2 V Reckoning both conductors, the drop at A is = 48.4 V Moments about F 1 Sum Moments about F 2 Sum in ampere-metres in ampere-metres 20 × 100 = 2,000 2,000 15 × 80 = 1,200 1, 80 × 250 = 20,000 22,000 10 × 150 = 1,500 2, 50 × 500 = 25,000 47,000 30 × 200 = 6,000 8, 40 × 300 = 12,000 20, 70 × 400 = 28,000 48, Alternative Solution The alternative method is to take the total current fed at one end, say, F 1 as x and then to find the current distribution as shown in Fig. 40.13 ( a ). The drop over the whole distributor is equal to the sum of the products of currents in the various sections and their resistances. For a distributor fed at both ends with equal voltages , this drop equals zero. In this way, value of x can be found. Resistance per metre single = 5.04 × 10 −^4 Ω Resistance per metre double = 10.08 × 10 −^4 Ω ∴ 10.08 × 10 −^4 [100 x + 150( x − 20) + 250 ( x − 100) + 100 ( x − 150) + 100 ( x − 200) + 100( x − 260) + 50( x − 290) + 70( x − 300) + 80 ( x − 315)] = 0 or 1000 x = 151, ∴ x = 151.7 A This gives a current distribution as shown in Fig. 40.13 ( b ). Obviously, point A is the point of minimum potential.
Fig. 40. Drop at A (considering both conductors) is = 10.08 × 10 −^4 (100 × 151.7 + 150 × 131.7 + 250 × 51.7 + 100 × 1.7) = 10.08 × 10 −^4 × 48,020 = 48.4 V — as before. Example 40.5. The resistance of a cable is 0.1 Ω per 1000 metre for both conductors. It is loaded as shown in Fig. 40.14 (a). Find the current supplied at A and at B. If both the ends are supplied at 200 V. (Electrical Technology-II, Gwalior Univ.)
D.C. Transmission and Distribution (^15811581)
Drop over DE = 10−^4 × 300 × − 61.4 = −1.84 V ∴ Voltage at C = 200 − 4.43 = 195.57 V Voltage at D = (195.57 − 2.7) = 192.87 V , Voltage at E = 192.87 − (−1.84) = 194.71 V Example 40.7. A 200 m long distributor is fed from both ends A and B at the same voltage of 250 V. The concentrated loads of 50, 40, 30 and 25 A are coming on the distributor at distances of 50, 75, 100 and 150 m respectively from end A. Determine the minimum potential and locate its position. Also, determine the current in each section of the distributor. It is given that the resistance of the distributor is 0.08 Ω per 100 metres for go and return. (Electric Power-I (Trans & Dist) Punjab Univ. 1993) Solution. As shown in Fig. 40.17, let current fed at point A be i. The currents in various sections are as shown. Resistance per metre of the distributor (go and return) is 0.08/100 = 0.0008 Ω. Since the distributor is fed at both ends with equal voltages, total drop over it is zero.
Hence, 0.0008 [ 50 + 25( – 50) + 25( – 90) + 50( – 120) + 50( – 145) i^ i^ i^ i^ i^ ]= 0
or 200 i = 16,750 or i = 83.75 A
Fig. 40.
Fig 40. The actual current distribution is shown in Fig. 40.18. Obviously, point D is the point of mini- mum potential. Drop at point D is = 0.0008(50 × 83.75 + 25 × 33.75) = 4.025 V ∴ minimum potential = 250 − 4.025 = 245.98 V
40.8. Distributor Fed at Both Ends with Unequal Voltages This case can be dealt with either by taking moments (in amp-m) about the two ends and then making a guess about the point of minimum potential or by assuming a current x fed at one end and then finding the actual current distribution. Consider the case already discussed in Ex. 40.4. Suppose, there is a difference of ν volts between ends F 1 and F 2 with F 1 being at higher potential. Convert ν volts into ampere-metres with the help of known value of resistance/metre. Since F 2 is at a lower potential, these ampere-metres appear in the coloumn for F 2 as initial drop. If, for example, ν is 4 volts, then since resistance/metre is 5.04 × 10 −^4 Ω, initial ampere-metres for F 2 are
= 4 104
× (^) = 7,938 amp-metres
The table of respective moments will be as follows :
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Moments about F 1 Sum Moments about F 2 Sum 20 × 100 = 2000 2,000 initial = 7, 80 × 250 = 20,000 22,000 15 × 80 = 1,200 9, 50 × 500 = 25,000 47,000 10 × 150 = 1,500 10, 30 × 200 = 6,000 16, 40 × 300 = 12,000 28, 70 × 400 = 28,000 56, As seen, the dividing point is the same as before. ∴ x + y = 70 and 47,000 + 600 x = 28,638 + 400 y Solving for x and y , we get x = 9.64 A and y = 60.36 A After knowing the value of x , the drop at A can be calculated as before. The alternative method of solution is illustrated in Ex. 40.12.
40.9. Uniform Loading with Distributor Fed at Both Ends
Consider a distributor PQ of length l units of length, having resistance per unit length of r ohms and with loading per unit length of i amperes. Let the difference in potentials of the two feeding points be ν volts with Q at the lower potential. The procedure for finding the point of minimum potential is as follows: Let us assume that point of minimum potential M is situated at a distance of x units from P. Then drop from P to M is = irx^2 /2 volts (Art. 40.6) Since the distance of M from Q is ( l − x ), the drop from Q to
M is = ( )^2 2
ir l − x volt.
Since potential of P is greater than that of Q by ν volts, ∴
2 2
irx (^) = ( )^2 2
ir l − x + ν or x = 2
l irl +^ ν
40.10. Concentrated and Uniform Loading with Distributor Fed at One End
Such cases are solved in two stages. First, the drop at any point due to concentrated loading is found. To this add the voltage drop due to uniform loading as calculated from the relation. 2 2 ir^ ^ lx −^ x As an illustration of this method, please look up Ex. 40.9 and 40.13. Example 40.8. Each conductor of a 2-core dis- tributor, 500 metres long has a cross-sectional area of 2 cm^2_. The feeding point A is supplied at 255 V and the feeding point B at 250 V and load currents of 120 A and 160 A are taken at points C and D which are 150 metres and 350 metres respectively from the feeding point A. Calculate the voltage at each load. Specific resistance of copper is 1.7_ × 10^ −6^ Ω -cm. (Elect. Technology-I, Bombay Univ.)
A circuit-board as shown above uses DC current
Fig. 40.
