Dutta, S N

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43D. C. THREE WIRE ELECTRICAL POWER TRANS­MISSION NETWORK AS SOLVED BY RELAXATION METHODS. N. DUTTADepabtmei t^ of Applied P hysics, Calcutta Uuivkhsity.(Received September 26, 1966)ABSTRACT. This paper reveals how the rolaxatioii iuotlit>d can be stiitably applied to detomiine some iirportant quantities such as line currents in D. C. Three Wire Transmis­sion System. The equivalent circuit of this network system is used and the problem is solved easily by applying the principle of relaxational solution of D.C. network, considering the heat­ing effects of steady currents flowing in it. The values of the required quantities thus ob­tained are compared with those found out by the normal method of network aTialysia.I N T l i O D U O T I O ND.C. Three Wire Transmission system (Starr, 1946) is used for having consider­able economy in feeders and distributors, when the electrical energy to be supplied is fairly large. Although there are different methods for solving this problem the relaxation method proves to be of advantage for yielding many useful infor­mations simultaneously.In this problem an unbalanced D.C. Three Wire Transmission system, that means the outer lines carrying unequal currents resulting a flow of current in the neutral line, is considered as shown in Fig. 1. The equivalent circuit diagram can be conveniently drawn indicated in Fig. 2, which is then solved by the relaxa­tion method.First of all Southwell and Black (1938), and later on Dutta (1966), applied the relaxation method in the problem of D.C. networks and showed its useful­ness in solving the network problem represented by Fig. 2 .AFig. 1. Diagram for D.C. three wire Transmission-Network.Fig. 2, Equivalent Circuit diagram for D.C. three wire Transmission Net-work.362THE METHODD. O. Three Wire Electrical Power Transmiseon, etc. 353In the method described in this paper the heating effects of steady currents are considered. An electrical theorem in connection with the heating effects is used  (Southwell and Black, 1938) and it can be enunciated as follows :*‘In a network of conductors to which specified currents are supplied at two or more nodal points the actual distribution of currents is such that the total generation of heat less twice the energy e x p e a d e d  in supplying the specified cur­rents from a source at datum potential has its minimum value consistent with the satisfaction of Kirohoff’s second law.”Two nodal points A and 0  of the network shown in Fig. 2, are considered and they are joined by a conductor of resistance UjiG* By Ohm's law a current ofR ag will flow from A  to G, (F^ and Vq the potentials at A  and 0). IfI^  and I g be the currents flowing towards A and O respectively then ia^ 9 ao{Va ^ ^ g)> where Qj^ q =  IIRag* When all the conductors connected to A are considered, it can be put as,S gAoiVa ’-^Vq)+ Ij o^ == 0A (1)denoting the current supplied to A  from outside. Then the heat generated in AO is gAG(VA'^VQ)^ the total heat generated in the network con be writtenas,s  gAGiVA^VG)^ ... (2)2H mS meaning the summation extending to every conductor. Also the rate of ex-mpondituro of energy is measured by lA o{yA ^^o )i if the current I^o  is supplied to A from an outside source a t the datum jjotential Vo- Then the total expendi­ture of energy is given by,W ao(Va - V o) ) ^ - E (3 )S denoting the summation extending to every nodal point. Hence the equationn(1 ) IB typical of the conditions for a minimum value of the quantity,Q =  H ^ E  == \^{g ^a (V A - Vg)^ }+ W ao(Va - V o)} ... (4)lit nbecause it is equivalent to.Due to the presence of the souroo of E.M.F. (generators) the problem has to bo modified as indicated below and then can be solved easily by relaxation method using the above theorem. I t  can be modified by assuming tha t the whole E.M.F. of each source is utilized to pass currents to earth through its own resistance 6354 8. N. DvMaand the datum distribution of known currents to enter and to leave the network at nodal points is obtained Then the effects of neutralising currents are simply to be calculated and superposed a t those points.Considering only the source of E.M.F. ‘E j’ let A  and F  be supposed to be joined by a wire of zero resistance as shown by dotted line. Then the current passing through that source from A to F  would return by that wire and hence inEthe datum distribution a current of enters the system at A  and leaves at-«(?iF, Bai being the internal resistance of the source (generator). Next the eurrent distribution is to bo calculated and superposed which will result when the neu-E Etralising currents +  and — are supplied a t F  and A after the source of Mgi ^ giE.M.F* is removed.Considering both the generators and all the branches of the network shown in Fig. 2, the expression for Q and the residuals can be written with the help of equation (4) as given below :2C»  + 2  | l {Fo- F , - ( K „ - V J )Jtc^ Q a Mgx^  { V g - V j ^  ^  + 2  ^ . { V o - V d - ( V o - V f ) }  ... (6)J i f C  G ■ tijX} -»Oa -“02Hence,dQ _ V a -V ^ Vg- V odVa B f s 'g R p G  G BoadQ Vg- V ^ V ^ - V p • Z i  = _ - ^ = F a^AO B gi B gi B gidQdVpV o - V p ,F ps’aVa - V f , B gi ^V o -V pB pc'gV p - V o _BaadQ Vg- V j> . Vp - V d , E f _ — FndVo Boa ^ Boi Bot Boa x» D0 =  F q(6)If  these residuals Fa, Fj^, Fp  and Fd obtained initially are liquidated, the poten­tials a t the points A, O, F  and D  are found out and the corresponding currents can be calculated from them knowing the required resistances of different branches. In  order to liquidate them the basic unit, group operation and the relaxation tables (Table I  and II) are prepared (Allen, 1954; Dutta, 1966). All these proce­dures are elaborately shown in the following illustration.The following example worked out by Christie (1962) using different method is taken for illustration.D. C. Three Wire Electrical Power Transmiaaont etc. 355In  the D.C. Three Wire Transmisaion System shown in Fig. 1 , the E.M.F. of the generators are E i =  ~  110  volts, the internal resistances of the generatorsare Boi =  B a  =  0.2 ohm, the resistance of the neutral wire is Bpo =  0.4 ohm, the outer line resistances are B ab =  •®oc — 0.1 ohm, the load resistances are Bgo =  8.0 ohms and B qc — 10.0 ohms. The currents flowing in the two outer and the neutral wires are to be found out.From the supplied data, the values of the resistances of different branches of the network shown in Fig. 2, are B aq =  B ab+ B bg == 8.1 ohms, Bdo — Boc-j- B og =  10.1 ohms, B p cc  =  Bpga  =  2 x  Bpg =  0.8 ohm.With the substitution of the numerical values in the relation (6) and (6), it can bo written as follows :2Q =  + 2  x 6 5 0 (F ^ -F |.)++  + 2 x 550(Ff -F ^ ,)or 2 0  = -i. i V A - V p r  , (Fo-Fn)* , (F^-Fn)*8.1 0.4 ^  0.2 10.1 " 0.2 “+2 x 560(F4-FF)+2 x 600(F/.-Fn) ... (5a)= -  I p - V a  _  V a - V pdVa 8.1 0.4 y o - F n _ n _10.1dQ _  Vo- V a _  Va - V fBVa '  8.1 0.2 -5500 =  Ea -5 5 0  =  F a... (6a). j? e  „  =  0 -dV r 0.4 ^  0.2 0.2■ m  = - r e x ” + - 0X -”  = = “ = J'oOn liquidating the residuals the potentials a t different nodal points are ob­tained from the relaxation table. Then the required currents can be easily cal­culated from those values of potentials and the resistances of the outer and neutral wires as given below :Fflf  ^=  106.4280 volts; where Voj is the potential of 0  with respect to A ,Vdo == 108.7914 ” ; ........  V m  ...........  D  .......  G,Vpg =  0.9432 ” ; ........ Vpo ...........  F  .......  O,Iga — 13.13 amps; where I qa is the current flowing in the wire joining Gfand^l,Inc *  10.78 ” ; ....... I dg .............  .......  tSt,IpG =« 2.46 .......  IpG356 8. N. DuttaI*-• E-i5  -2oOhO5S«oU«o«oa a>Od oo oo o o i 6ifi 1o XO oi t i 04 »o7'tHCO CO04fH o oo lO1iO■<**4C4 CO O i<M 04 O i1'- wo O041o 04 O1 1 1 -1J1 '1)1 1i~( I 1 11s o T £1 o  ^OiCOOD(MIoio0^a  1HP3wo04o o 04lOwo ofH 04 orM OCO1UO1^ l:^o COfH1o o6^olOlO1 o OoCb »oI>»o04fe. o 04 o04 04OCOCOoc1 wot«. o04«o ©4 COCO7o GOII COo 1 »oou pH 1eo 1 fH OiO*-Hu 1IIo00«oII 1 1 04CDO7G> 1 1 1«o 1 1 1r—1QO041—1 1—1 gopH WOt>04 PHX X X52, £CHa lIj'oWOQ>oo00(MThe values of the currents in the wires OA, DO and FG thus found by relaxa­tion method are compared with those calculated by conventional method of network analysis and they are found to be in good agreement as shown in the (Table III) below :TABLE in  Comparison of valuesD- G. Three Wire Electrical Power Transmisson, etc. 357UnknownquantitiesValues calculated by_____ __— ____IlolaxationMethod CoiivontiormlMethodlOA 13.13 amps 13.30 ampsUw 10.78 amps 10.90 amps2.46 amps 2.40 ampsD I S C U S S I O NThis method is seen to be a convenient one for having the values of all the unknown quantities obtained simultaneously, such as the potentials at the nodal points in the problem. With the increase of the number of branches of the network in the electrical problems, the conventional methods become laborious whereas this method can yield much quicker solution preferably with little practice of relaxation technique. In this method the internal resistances of the generators or the resistances of the paths A F  and F D  are to bo Imown in order that the cur­rent in the datum distribution may be calculated.A C K N ff W L E D G M E N 'fThe author is highly indebted to Prof. A . K. Sengupta, D.Sc., A.M.I.E.E. (London), Head of the Department of Applied Physics, Calcutta University, for liis help and guidance throughout the progress of this work.E E F E R E N C E SAllen, D. N. de G., 1934, R dam tw n Methods, (McGraw-ITill Book Co., Ine., Now York Choptor 1 and 2,Black, A. N. and Southwell, R. V., 19.98, Proc- Roy. Soo. A 164, 447.Christie, C. V., 1952, Electrieal Engineering, MoOniw-Ilill Book Co. Inc. Now York 91.Dutta, S. N., 1966, Indian J. Phya. 40, 7, 163.SoxMxwoMt'R.'V., t95\,Rdaxntion Methods in Engineering Seienee, Oxford University Press, London Chapter VI..Starr, A. T., 1946, Generation, Tranamiaaion and UtHiaation of Electrical Power, Sir Issao Pitm an & Sons Lid., London 33.

D. C. Three Wire Electrical Power Transmission Network as Solved by Relaxation Method

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D. C. Three Wire Electrical Power Transmission Network as Solved by Relaxation Method

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