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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
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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 =
Pl (^) = Pl σ σ ( ii ) 440 V Supply V 2 = (^220)
Pl σ
%age saving in Cu = 1 2 1
Pl Pl V V V Pl
− σ σ × = × σ
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^ =^
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
l irl^ i^ l^ r^ l ir 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 =
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 −
irx^2 shows that the diagram of drop of a uniformly-loaded distributor fed at one end is a parabola.
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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
=
×
= 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
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
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.
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Solution. The distributor along with currents in its various sections is shown in Fig. 40.21. Resistance/metre single = 0.4/1000 = 4 × 10 −^4 Ω Resistance/metre double = 8 × 10 −^4 Ω Voltage drop on both conductors of 200-metre long distributor is = 8 × 10 −^4 [50 i + 26 ( i − 20) + 25( i − 60) + 50( i − 85) + 50( i − 115)] = 8 × 10 −^4 (200 i − 12,000) volt This drop must be equal to the potential difference between A and B.
∴ 8 × 10 −^4 (200 i − 12,000) = 236 − 237 = − 1 ∴ i = 53.75 A Current in section AC = 53.75 A Current in section CD = 53.75 − 20 = 33.75 A Current in section DE = 53.75 − 60 = −−−−− 6.25 A Current in section EF = 53.75 − 85 = −−−−− 31.25 A Current in section FB = 53.75 − 115 = −−−−− 61.25 A The actual current distribution is as shown in Fig. 40.22. Obviously, minimum voltage occurs at point D i.e. 75 metre from point A (or 125 m from B ) Voltage drop across both conductors of the distributor over the length A D is = 8 × 10 −^4 (50 × 53.75 + 25 × 33.75) = 2.82 V ∴ potential of point D = 236 − 2. = 233.18 V
Fig. 40. Example 40.11. A distributor cable AB is fed at its ends A and B. Loads of 12, 24, 72 and 48 A are taken from the cable at points C, D, E and F. The resistances of sections AC, CD, DE, EF and FB of the cable are 8, 6, 4, 10 and 5 milliohm respectively (for the go and return conductors together).
The p.d. at point A is 240 V, the p.d. at the load F is also to be 240 V. Calculate the voltage at the feeding point B, the current supplied by each feeder and the p.d.s. at the loads C,D and E.
(Electrical Technology ; Utkal Univ.) Solution. Let the current fed at the feeding point A be i. The current distribution in various sections becomes as shown in Fig. 40.23. Voltage drop on both sides of the distributor over the section AF is = [8 i + 6( i − 12) + 4( i − 36) + 10( i − 108)] × 10 −^3 = (28 i − 1296) × 10 −^3 V
Fig. 40.
D.C. Transmission and Distribution 15851585
Fig. 40. Since points A and F are at the same potential, the p.d. between A and F is zero. ∴ (28 i − 1296) × 10 −^3 = 0 or i = 46.29 A Current in section AC = 46.29 A Current in section CD = 46.29 − 12 = 34.29 A Current in section DE = 46.29 − 36 = 10.29 A Current in section EF = 46.29 − 108 = − 61.71 A Current in section FB = 46.29 − 156 = −109.71 A The actual current distribution is as shown in Fig. 40.24.
Fig. 40. Current applied by feeder at point A is 46.29 A and that supplied at point B is 109.71 A. Voltage at feeding point B = 240 − drop over FB = 240 − (− 5 × 10 −^3 × 109.71) = 240.55 V. Voltage at point C = 240 − drop over AC = 240 − (8 × 10 −^3 × 46.29) = 239.63 V Voltage at point D = 239.63 − drop over CD = 239.63 − (6 × 10 −^3 × 34.29) = 239.42 V Voltage at point E = 239.42 − drop over DE = 239.42 − (4 × 10 −^3 × 10.29) = 239.38 V Example 40.12. A two-wire, d.c. distributor PQ, 800 metre long is loaded as under : Distance from P (metres) : 100 250 500 600 700 Loads in amperes : 20 80 50 70 40 The feeding point at P is maintained at 248 V and that at Q at 245 V. The total resistance of the distributor (lead and return) is 0.1 Ω. Find (a) the current supplied at P and Q and (b) the power dissipated in the distributor.
Solution. As shown in Fig. 40.25 ( a ), let x be the current supplied from end P. The other currents in the various sections of the distributor are as shown in the figure.
D.C. Transmission and Distribution 15871587
It should be noted that location of point of minimum potential is not affected by the uniformly- spread load of 0.5 A/m. In fact, let us, for the time being, imagine that it is just not there. Then, assuming i to be the input current at A, the different currents in various sections are as shown. Since points A and B are fed at equal voltages, total drop over the distributor is zero. Distributor resistance per metre length (go and return) is 0.1/1000 = 10−^4 Ω.
∴ 10 −^4 [200 i + 200( i − 120) + 300( i − 180) + 200( i − 280) + 100( i − 320)] = 0 or 1000 i = 166,000 ∴ i = 166 A Actual current distribution is shown in Fig. 40.26 ( b ) from where it is seen that point D is the point of minimum potential. The uniform load of 0.5 A/m upto point D will be supplied from A and the rest from point B.
Uniform load from A to D = 400 × 0.5 = 200 A. Hence, I (^) A = 166 + 200 = 366 A. Similarly, IB = 154 + (600 × 0.5) = 454 A. Drop at D due to concentrated load is = 10−^4 (166 × 200 + 46 × 200) = 4.24 V. Drop due to uniform load can be found by imagining that the distributor is cut into two at point D so that A D can be looked upon as a distributor fed at one end and loaded uniformly. In that case, D becomes the other end of the distributor.
∴ drop at D due to uniform load (Art. 40.6) = irl^2 /2 = 0.5 × 10 −^4 × 4002 /2 = 4 V ∴ total drop at D = 4.24 + 4 = 8.24 V ∴ potential of D = 240 − 8.24 = 231.76 V Example 40.14. It is proposed to lay out a d.c. distribution system comprising three sections— the first section consists of a cable from the sub-station to a point distant 800 metres from which two cables are taken, one 350 metres long supplying a load of 22 kW and the other 1.5 kilometre long and supplying a load of 44 kW. Calculate the cross-sectional area of each cable so that the total weight of copper required shall be minimum if the maximum drop of voltage along the cable is not to exceed 5% of the normal voltage of 440 V at the consumer’s premises. Take specific resistance of copper at working temperature equal to 2 μ Ω -cm. Solution. Current taken from 350-m section is I 1 = 22,000/440 = 50 A Current taken from 1.5 km section, I 2 = 44,000/440 = 100 A ∴ Current in first section I = 100 + 50=150 A Let V = voltage drop across first section ; R = resistance of the first section A = cross-sectional area of the first section Then R = V / I = V /150 Ω
Now, A = l (^) I l R V
ρ = ρ = 80,000 × 2 × 10 −^6 × 150/V = 24/Vcm^2
Now, maximum allowable drop = 5% of 440 = 22 V ∴ voltage drop along other sections = (22 − V ) volt Hence, cross-sectional area of 350-m section is A 1 = 35,000 × 2 × 10 −^6 × 50/(22 − V ) = 3.5/(22 − V ) cm^2 Also, cross-sectional area of 1500-m section is A 2 = 150,000 × 2 × 10 −^6 × 100/(22 − V ) = 30/(22 – V ) cm 2 Now, total weight of copper required is proportional to the total volume. ∴ W = K [(800 × 24/ V ) + 350 × 3.5/(22 − V ) + 1500 × 30/(22 − V )] = K [1.92/ V + 4.62/(22 − V )] × 104 Weight of copper required would be minimum when d W / d V = 0
1588 Electrical Technology
dW dV
= 1.92 2 4.62^2104
K
V V
+^ ×
or 1.92 2 V
(22 − V )
or (22 − V ) 2 = 2.4 V^2
or V = 22/2.55 = 8.63 volt ∴ A = 24/8.63 = 2.781 cm 2 A 1 = 3.5/(22 − 8.63) = 0.2618 cm 2 A (^) 2 =30/(22 − 8.63) = 2.246 cm^2 Example 40.15. A d.c. two-wire distributor AB is 450 m long and is fed at both ends at 250 volts. It is loaded as follows : 20A at 60 m from A, 40A at 100m from A and a uniform loading of 1.5 A/m from 200 to 450 m from A. The resistance of each conductor is 0.05 Ω /km. Find the point of minimum potential and its potential. (Electrical Power-II, Bangalore Univ. 1993) Solution. In Fig. 40.27, let D be the point of minimum potential and let i be the current in distributor section CD. Then, current supplied to load D from end B is (40 − i ). If r is the resistance of the distributor/metre (both go and return), then
Fig. 40. Drop over A D = (20 + i ) × 60 r + i × 40 r Drop over BD = drop due to concentrated load + drop due to distributed load = (40 − i ) × 350 r + 1.5 × r × 2502 /2 — Art. 40. Since the two feeding points A and B are at the same potential of 250 V, the two drops must be equal. ∴ (20 + i ) × 60 r + i × 40 r = (40 − i ) × 350 r + 1.5 × r × 2502 /2 ∴ i = 132.6 Å Since (40 − i ) comes out to be negative, it means that D is not the point of minimum potential. The required point is somewhat nearer the other end B. Let it be F. Obviously, current in section DF = 132.6 − 40 = 92.6 A. Hence, distance of minimum potential point F from end A is = 60 + 40 + 100 + 92.6/1.5 = 261.7 m Total voltage drop over section AF is = 2 × 0.05 × 10 −^3 (152.6 × 60 + 132.6 × 40 + 92.6 × 100 + 92.6 + 61.7/4) = 2.65 V ∴ potential of point F = 250 − 2.65 = 247.35 V. Example 40.16. A two-wire d.c. distributor AB, 1000 metres long, is supplied from both ends, 240 V at A and 242 V at B. There is a concentrated load of 200 A at a distance of 400 metre from A and a uniformly distributed load of 1.0 A/m between the mid-point and end B. Determine (i) the currents fed at A and B (ii) the point of minimum potential and (iii) voltage at this point. Take cable resistance as 0.005 Ω per 100 metre each core. Solution. The resistance per 100 metres of both cores = 0.005 × 2 = 0.01 Ω. Let us take 100 m as the unit of length. Let current fed at end B be I (^) B as shown in Fig. 40.28.
