Current Loading and Load-point Voltages in a 3-wire System

40.12. Current Loading and Load-point Voltages in a 3-wire System

Consider a 3-wire, 500/250-V distributor shown in Fig. 40.34. 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. 40.34

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.

1594 Electrical Technology

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. After that, voltages at different load points can be calculated as illustrated in Ex. 40.20.

Example 40.20. 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.

(Elements of Elect. Engg-I, Bangalore Univ.)

Solution. Let the assumed directions of unknown cur- rents in the three conductors be as shown in Fig. 40.35. Ap-

plying KVL (Art. 2.2) to closed circuit ABCDA, we have Fig. 40.35 − 3.05x − 0.05(x − y) + 250 = 0

or 310x − 5y =

− 25,000 ... ( i)

Similarly, circuit DCEFD yields

0.05 (x − y) − 4.05y + 250 = 0 or

5x − 410y = − 25,000 ... (ii) From ( i) and ( ii) , we get x = 81.64 A, y = 61.97 A. Since both currents come out to be positive,

it means that their assumed directions of flow are correct. ∴

Example 40.21. A 3-wire d.c. distributor PQ, 250 metres long, is supplied at end P at 500/250

V and is loaded as under : Positive side: 20 A 150 metres from P ; 30 A 250 metres from P. Negative side: 24 A 100 metres from P ; 36 A 220 metres from P. The resistance of each outer wire is 0.02 Ω per 100 metres and the cross-section of the middle

wire is one-half that of the outer. Find the voltage across each load point. Solution. The current loading is shown in Fig. 40.36. The current flowing into the positive side is 30 +

20 = 50 A. Since current flowing out of the negative side is 36 + 24 = 60 A, it means that 10 A must be flowing in the section NC of the neutral. To make up 24

A in load CD. 10 A are contributed by NC and 14 A by BC . The balance of 6 A flows through BF and adds up with 30 A of KF to make up 36A through load FE.

It is given that the resistance of outers is 0.02 Ω per 100 m. Since neutral is of half the cross-section, its resistance is 0.04 Ω per 100 m. Knowing them, volt- age at load points can be determined as under. Let us

see how we will calculate voltage across load A B. Fig. 40.36

D.C. Transmission and Distribution

Voltage at A B = 250 − drop in PA − drop in BC + drop in CN. It should be particularly noted that drop in CN has been added instead of being subtracted. The

reason is this : as we start from P and go round PABCN, we go along the current over section PA i.e .

we go ‘downstream’ hence drop is taken negative, but along CN we go ‘upstream’, hence the drop as taken as positive * . Proceeding in this way, we tabulate the currents, resistances and voltage drops in

various sections as given below :

Section

Resistance (W)

Current (A)

Drop (V)

P.D. across A B = 250 − 1.5 −

= 248.62 V

0.36 + 0.168 = 247.83 V P.D. across CD = 250 − 0.4 − 1.2 = 248.4 V P.D. across FE = 248.4 + 0.28 − 0.168 − 0.864 = 247.65 V

P.D. across QK = 248.62 − 0.6 −

Tutorial Problem No. 40.3

1. A 3-wire system supplies three loads (a) 10 (b) 20 and (c) 30 amperes situated at distances of 100, 150 and 300 metres respectively from the supply point on the positive side of the neutral wire. Connected between negative and neutral wires are two loads (d) 30 A and (e) 20A, situated 120 and 200 metres respectively from the supply point. Give a diagram showing the values and directions of the currents in various parts of the neutral wire. If the resistance of the outers is 0.05 Ω per 1000 metres and that of the neutral 0.1 Ω per 1000 metres,

calculate the potential difference at the load points (b) and (e); the pressure at the supply point being 100 V between outers and neutral.

[10A and 5A; Volts at ( b) = 99.385 V ; at (e) = 99.86 V]

2. A 3-wire d.c. distributor 400 metres long is fed at both ends at 235 volts between each outer and neutral. Two loads P and Q are connected between the positive and neutral and two loads R and S are connected between the negative outer and the neutral. The loads and their distances from one end (X) of the distributor are as follows : Load P, 50 A, 100 metres from X ; Load Q, 70 A, 300 metres from

X ; Load R, 60 A, 150 metres from X; Load S, 60 A,350 metres from X. Determine the p.d. at each load point and the current at each feeding point. The resistance of each outer is 0.25 Ω per 1000 metres and that of the neutral is 0.5 W per 1000 metres.

[Current into the +ve outer at X = 55 A ; other end Y = 65 A; Out from −−−−− ve outer at X = 45 A; other end Y = 75 A ; P.D. s at P = 233.125 V; Q = 232.375 V; R = 232.812 V; S = 233.812 V]