CHAPTER – 40 D.C. TRANSMISSION AND DISTRIBUTION

CHAPTER – 40 D.C. TRANSMISSION AND DISTRIBUTION

LEARNING OBJECTIVE 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 Uniformaly Loaded Distributor Distributor Fed at Both Ends with Equal Voltage Distributor Fed at Both ends with Unequal Voltage Uniform Loading with Distributor Fed at Both Ends

LEARNING OBJECTIVE 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 Voltage in a 3-wire System Three-wire System Balancers Boosters Comparison of 2-wire and 3-wire Distribution System

TRANSMISSION AND DISTRIBUTION OF D.C. POWER

EXPLANATION OF DIAGRAM Electricity leaves the power plant Its voltage is increased at a step-up transformer The electricity travels along a transmission line to the area where power is needed There, in the substation, voltage is decreased with the help of step-down transformer Again the transmission lines carry the electricity Electricity reaches the final consumption points

Electric power may be transmitted by two methods. By overhead 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 pole structures and tower.

By underground system This being especially suited for denselypopulated areas though it is somewhat costlier than the first method. The underground system employs insulated cables which may be single, double or triple-core etc. A good system whether overhead or underground should fulfill the following requirements : 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.

The loss of power in the system itself should be a small percentage (about 10%) of the power transmitted. The transmission cost should not be unduly excessive. The maximum current passing through the conductor should be limited to such a value as not to overheat the conductor or damage its insulation. The insulation resistance of the whole system should be very high so that there is no undue leakage or danger to human life

PRODUCING POWER IN THE FORM OF ALTERNATING CURRENT RATHER THAN DIRECT CURRENT ? It is possible 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. A.C. voltage can be efficiently and conveniently raised or lowered for economic transmission 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.

A TYPICAL POWER SYSTEM FOR OBTAINING D.C. POWER FROM A.C. POWER

Three methods commonly used for converting a.c. power to d.c. power at the sub-station. A 6-phase mercury-arc rectifier gives 600 V d.c. power after the voltage has been stepped down to a proper value by the transformers. This 600-V d.c. power is generally used by electric railways and for electrolytic processes. A rotary converter gives 230 V d.c. power. A motor-generator set converts a.c. power to 500/250 d.c. power for 3-wire distribution systems.

A SCHEMATI DIAGRAM OF LOW TENSION DISTRIBUTION SYSTEM FOR D.C. POWER

The whole system consists of a network of cables or conductors which convey power from central station to the consumer’s premises. The station bus-bars are fed by a number of generators (only two shown in the figure) running in parallel. From the bus-bars, the power is carried by many feeders which radiate to various parts of a city or locality. These feeders deliver power at certain points to a distributor which runs along the various streets. The points FF are known as feeding points. Power connections to the various consumers are given from this distributor and not directly from the feeder. The wires which convey power from the distributor to the consumer’s premises are known as service mains (S).

TWO-WIRE AND THREE-WIRE SYSTEMS
In d.c. systems, power may be fed and distributed either by 2-wire system 3-wire system 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. The potential difference and current have their maximum values at feeding points and The standard voltage between the conductors is 220 V.

3-wire has not only a higher efficiency of transmission but when used for distribution purposes, it makes available two voltages at the consumer’s end. 3-WIRE SYSTEM

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.

VOLTAGE DROP AND TRANSMISSION EFFICIENCY
Consider the case of a 2- wire feeder . AB is the sending end and CD the receiving end. Obviously, the p.d. at AB is higher than at CD. The difference in potential at the two ends is the potential drop or ‘drop’ in the cable. 2-WIRE FEEDER

Suppose the transmitting voltage is 250 V , current in AC 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 AB = 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 = R = 2 × 102 × 0.5 = 100 W

efficiency of transmission = In the present case, power delivered by the feeder is = 2500 W and power received by it as 2400 W. η= 2400 × 100/2500 = 96%

For calculating the total voltage drop along the entire length of a distributor, following information is necessary. value of current tapped at each load point the resistance of each section of the distributor between tapped points Example : A DC 2-wire feeder supplies a constant load with a sending-end voltage of 220 V. Calculate the saving in copper if this voltage is doubled with power transmitted remaining the same.

Solution. Let l = length of each conductor in metre σ = current density in A/m2 P = power supplied in watts

METHODS OF FEEDING A DISTRIBUTOR
Different methods of feeding a distributor are given below : feeding at one end feeding at both ends with equal voltages feeding at both ends with unequal voltages feeding at some intermediate point In adding, the nature of loading also varies such as concentrated loading uniform loading combination of concentrated loading and uniform loading .

D.C. DISTRIBUTOR FED AT ONE END
As shown Fig 40.1 one conductor AB of a distributor with concentrated loads and fed at one end Fig 40.1

Let , , etc. be the currents tapped off at points C, D, E and F and , , etc. the currents in the various sections of the distributor. Let , , etc. be the ohmic resistances of these various sections and etc. , , the total resistance from the feeding end A to the successive tapping points. Then, total drop in the distributor is

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. V = 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. Let us find the drop at any intermediate point like E. The value of this drop is

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.

UNIFORMLY LOADED DISTRIBUTOR
As shown in Fig 40.2 one conductor AB 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 Fig 40.2

Now, let us find the voltage drop at a point C which is at a distance of x units from feeding end A. The current at point C is (il − ix) = i (l − x). (Fig 40.3) 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 Fig 40.3

The drop at point B can be obtained by putting x = l in the above expression where i × l = I – total current entering at point A; r × l = R–the total resistance of distributor AB Total drop in the distributor AB =

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. 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 distributor 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

Example : A uniform 2-wire d.c. distributor 200 metres long is loaded with 2 amperes/ metre. Resistance of single wire is 0.3 ohm/kilometre. Calculate the maximum voltage drop if the distributor is fed (a) from one end (b) from both ends with equal voltages.

DISTRIBUTOR FED AT BOTH ENDS WITH EQUAL VOLTAGES
As shown in Fig (a) a distributor fed at two points and with equal voltages. The potential of the conductor will gradually fall from onwards, reach a minimum value at one of the tapping, say, A and then rise again as the other feeding point is approached. Fig (a)

All the currents tapped off between points and A will be supplied from while those tapped off between and A will be supplied from . The current tapped at point A itself will, in general, be partly supplied by and partly by 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 (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 Fig 40.4

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. Suppose, there is a difference of ν volts between ends and with being at higher potential. Convert ν volts into ampere-metres with the help of known value of resistance/metre. Since is at a lower potential, these ampere-metres appear in the column for as initial drop.

DISTRIBUTOR FED AT BOTH ENDS WITH UNEQUAL VOLTAGES
If, for example, ν is 4 volts, then since resistance/metre is 5.04 × Ω, initial ampere-metres for are The table of respective moments will be as follows -4

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 = volts

UNIFORM LOADING WITH DISTRIBUTOR FED AT BOTH ENDS
Since the distance of M from Q is (l − x), the drop from Q to M is = volts Since potential of P is greater than that of Q by ν volts

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. Example : A 2-wire distributor 500 metres long is fed at P at 250 V and loads of 40A, 20A, 60A, 30A are tapped off from points A, B, C and D which are at distances of 100 metres, 150 metres, 300 metres and 400 metres from P respectively. The distributor is also uniformly loaded at the rate of 0.1 A/m. If the resistance of the distributor per metre (go and return) is Ω, calculate the voltage at (i) point Q and (ii) point B.

Solution. First, consider drop due to concentrated load only. Drop in PA = 150 × (100 × ) = 7.5 V Drop in AB = 110 × (50 × ) = 2.75 V Drop in BC = 90 × (150 × ) = 6.75 V Drop in CD = 30 × (100 × ) = 1.5 V ∴ total drop due to this load = 18.5 V Now, let us consider drop due to uniform load only. Fig 40.5

RING DISTRIBUTOR A ring distributor is a distributor which is arranged to form a closed circuit and which is fed at one or more than one points. For the purpose of calculating voltage distribution, it can be looked upon as consisting of a series of open distributors fed at both ends. By using a ring distributor fed properly, great economy in copper can be affected. If the ring distributor is fed at one point then, for the purposes of calculation, it is equivalent to a straight distributor fed at both ends with equal voltages

RING DISTRIBUTOR Example : A 400-metre ring distributor has loads as shown in Fig.40.6 (a) where distances are in metres. The resistance of each conductor is 0.2 Ω per 1,000 metres and the loads tapped off at points B, C and D are as shown. If the distributor is fed at A, find voltages at B,C and D. Fig.40.6

RING DISTRIBUTOR Solution. Let us assume a current of I in section AD [Fig. (a)] and then find the total drop which should be equated to zero. ∴ 70 I + 90 (I − 50) + 80 (I − 120) + 60 (I − 220) = 0 ∴ 300 I = 27,300 or I = 91 A. Fig 40.6

RING DISTRIBUTOR The current distribution becomes as shown in Fig. (b) from where it is seen that C is the point of minimum potential. Drop in AD = 2(91 × 70 × 0.2/1,000) = 2.55 V Drop in DC = 2(41 × 90 × 0.2/1,000) = 1.48 V Drop in CB = 2(29 × 80 × 0.2/1,000) = 0.93 V Drop in BA = 2(129 × 60 × 0.2/1,000) = 3.1 V Voltage at D = 240 − 2.55 = V Voltage at C = − 1.48 = V Voltage at B = 240 − 3.1 = V

CURRENT LOADING AND LOAD-POINT VOLTAGES IN A 3-WIRE SYSTEM
Consider a 3-wire, 500/250-V distributor shown in Fig The motor requiring 500 V is connected across the outers whereas other loads requiring lower voltage of 250 V are connected on both sides of the neutral. Fig 40.7

The current in positive outer AE flows from the left to right and in negative outer FB from right to left. The current through the various sections of the neutral wire NL may flow in either direction depending on the load currents on the positive side and negative side but is independent of the loads connected between outers. Since 150 A enters the +ve outer but 160 A comes out of the network, it means that a current of 10 A must flow into the neutral at point N. Once the direction and magnitude of current in NQ is known, the directions and magnitudes of currents in other sections of the neutral can be found very easily.

Since PH takes 40 A, currents meeting at P should add up to 40 A. As seen, 20 A of CQ and 10A of NQ flow towards P, the balance of 10 A flows from point M. Then, 30 A current of DM is divided into two parts, 10 A flowing along MP and the other 20 A flowing along ML to feed the load LG. Knowing the values of currents in the various conductors, voltage drops can be calculated provided resistances are known. Example : In a 3-wire distribution system, the supply voltage is 250 V on each side. The load on one side is a 3 ohm resistance and on the other, a 4 ohm resistance. The resistance of each of the 3 conductors is 0.05 Ω. Find the load voltages.

Solution. Let the assumed directions of unknown currents in the three conductors be as shown in Fig Applying KVL to closed circuit ABCDA, we have Fig 40.8

THREE-WIRE SYSTEM It consists of the ‘outer’ conductors (between which the voltage is twice the normal value for lighting) and the third wire which is called the middle or neutral wire. It is of half the cross-section as compared to any one of the two outers and is earthed at the generator end. The voltage of the neutral is thus approximately half way between that of the outers. If the total voltage between outers is 460 V, then the positive outer is 230 V above the neutral and the negative outer 230 V below the neutral. Motor loads requiring higher voltage are connected across the outers directly whereas lighting and heating loads requiring lesser voltage of 230 V are connected between any one of the outers and the neutral.

THREE-WIRE SYSTEM In practice, although effort is made to distribute the various loads equally on the two sides of the neutral, yet it is difficult to achieve exact balance, with the result that some out-of-balance current does flow in the neutral as shown in Fig 40.9 This difficulty of unequal voltages with unbalanced loads is removed by using balancers Fig 40.9

BALANCERS In order to maintain p.d.s on the two sides of neutral equal to each other, a balancer set is used. The commonest form of balancer consists of two identical shunt-wound machines which are not only coupled mechanically but have their armatures and field circuits joined in series across the outers. The neutral is connected to the junction of the armatures as shown. When the system is unloaded or when the loads on the two sides are balanced, then both machines run as unloaded motors and since their speeds and field currents are equal, their back e.m.fs. are the same

BALANCERS When the two sides are unbalanced i.e. when the load supplied by + ve outer is different from that supplied by negative outer, then out-of-balance current through the mid-wire to the balancers where it divides into two halves each equal to /2 as shown in Fig.40.10 Fig.40.10

BALANCERS If is greater than , then +ve side is more heavily loaded (than −ve side) hence the p.d. on this side tends to fall below the e.m.f. of the balancer on this side so that runs as a generator. However, the p.d. on the lightly-loaded side rises above the e.m.f. of the balancers on that side, hence runs as a motor. In this way, energy is transferred through balancers from the lightly-loaded side to the heavily loaded side

BALANCERS Machine is running as a motor and driving machine
as a generator. Let

BALANCERS Hence, we find that the difference of voltages between the two sides of the system is proportional to — out of balance current and armature resistance of the balancer. For this reason, is kept very small and effort is made to arrange the loads on the two sides such that out-of-balance current is as small as possible.

BOOSTERS A booster is a generator whose function is to add to or inject into a circuit, a certain voltage that is sufficient to compensate for the drop in the feeders etc. In a d.c. system, it may sometimes happen that a certain feeder is much longer as compared to others and the power supplied by it is also larger. In that case, the voltage drop in this particular feeder will exceed the allowable drop of 6% from the declared voltage. This can be remedied in two ways by increasing the cross-section of the feeder, so that its resistance and hence drop is decreased or by increasing the voltage of the station bus bars.

BOOSTERS The second method is not practicable because it will disturb the voltage of other feeders, whereas the first method will involve a large initial investment towards the cost of increased conductor material. To avoid all these difficulties, the usual practice is to install a booster in series with this longer feeder as shown in Fig Fig

BOOSTERS Since it is used for compensating drop in a feeder, it is known as feeder booster. It is a (series) generator connected in series with the feeder and driven at a constant speed by a shunt-motor working from the bus-bars. The drop in a feeder is proportional to the load current, hence the voltage injected into the feeder by the booster must also be proportional to the load current, if exact compensation is required. In other words, the booster must work on the straight or linear portion of its voltage characteristic.

BOOSTERS Example : A 2-wire system has the voltage at the supply end maintained at 500. The line is 3 km long. If the full-load current is 120 A, what must be the booster voltage and output in order that the far end voltage may also be 500 V. Take the resistance of the cable at the working temperature as 0.5 ohm/kilometre. Solution. Total resistance of the line is = 0.5 × 3 = 1.5 Ω Full-load drop in the line is = 1.5 × 120 = 180 V Hence, the terminal potential difference of the booster is 180 V (i.e. 180/120 = 1.5 volt per ampere of line current). Booster-output = 120 × 180/1000 = 21.6 kW

COMPARISON OF 2 -WIRE AND 3 -WIRE DISTRIBUTION SYSTEM
For this comparison, it will be assumed that— The amount of power transmitted is the same in both cases. The distance of transmission is the same. The efficiency of transmission (and hence losses) is the same. Voltage at consumer’s terminals is the same. The 3-wire system is balanced and in the 3-wire system, the mid-wire is of half the cross-section of each outer

COMPARISON OF 2 -WIRE AND 3 -WIRE DISTRIBUTION SYSTEM

CHAPTER – 40 D.C. TRANSMISSION AND DISTRIBUTION

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CHAPTER – 40 D.C. TRANSMISSION AND DISTRIBUTION

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