Basu, R N

Dutta, S N

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ON THE PERFORMANCE CALCULATIONS OF THREE PHASE TRANSMISSION LINES BY RELAXATION METHODR. N. BASU* AND S. N. DUTTADepartment op Applied  P dysics, Calcutta University  {Received October 10, 1966)ABSTRACT. Tho application of relaxation technique in the uominal-T method of solution of traitsmission lino problem is discussed with an illustration. Tho principle under­lying the relaxational solution of o,c. nolworks having comidox circuit constants (i.e., oonsi- doring tho roaotunco^ along witli tho is tan cos in diffeiunt bramthos js utilised and tho resultsthus obtained arc cfiraparod with those calculated by normal inoihod of solution. It also discuss(3s about the applica-tion of tho method to tho solution oflrmg transmission lines wlion' tho oquivalent-T method is used instead of (he nominal-T method nomially used for medium linos.48I N T 11 O D U C T T O NTIio performance calculations of three phase transmission lines under different given conditions at the receiving end and sending end are quite familiar in Elec­trical Engineering. Of tho different methods available for such calculations, the nominal-^ method is one. In the nominal-T method tho line capacitance is assumed to be localised at the mid-point of the line as shovm in Fig. 1. Dependingy zB-rmnn— aaaa-© . Ltnc./ / / / / / / / 'D'CcapcioIIanceLoad/7 7 7 T 7 'DFig. 1Diagram for nominal T.method.7 7 7 7 7 7 7 7D♦ Present address ! Rosoarch and Developir ont Department, Northern Eleotrio Co., Ottawa, Canada.382On the Performance Galcidations of Three Phase, etc. 383on tho available informations regarding the sending end and receiving end condi­tions, three distinct cases can be considered separately, viz.,(i) when receiving end voltage and load are known,(ii) when sending end voltage and receiving end impedance are known, and(iii) when sending end voltage and receiving end load are known.When the receiving end terminals of the 2’-network, shown in Fig. 1, arc terminated by the load-impedance, the same network can* be rcprt^senled in a convenient form as shown in Fig. 2a. The network thus fomicd is solved here with the help of relaxation method, and it is shown that such a relaxaiional solution yields a number of useful iiiformaticms simultanciously.2aEquivalent impodauce network diagram nominal T-method,Fig. 2bEquivalent admittanee network diagram for nominal T-mothod.In using the technique of relaxation to the solution of a.c. networks, additional complications (in comparison with d.c. networks) aris(  ^due to tho complex nature of ihc branch-impedances, which include resistances as well as reaetanees, Southwell and Black (1938) have applied relaxation method to the solution of an a.c. network having complex circuit constants. The principle underlying such a solution reveals tho usefulness of tlio application of relaxation technique for solving the network shown in Fig. 2a, i.e., for the performance calculations of transmission lines.P R I N  C I r  E E O F T H E  M E T H O 1)In  Fig. 2b, F ^c, • • • etc. arc equal to the reciprocals of the impodanoes^ABt ••• respectively, and tlie voltage between A  and D  stands for the sending end voltage measured between phase and neutral. Now writing the ad- mittemoe of any branch, say BC, as—F^c =  9BC-\-jit>BC ••• (1 )and assuming the potentials a t  the points B  and C to bevb =  ■ )\ ... (2)and VC — v,^C i+ jv„^o^ J384 jR. N. Basu and S. N. DtiUathe current flowing from B to C  through the branch BO may be represented by I b c =  Y  bc{vb—vc)— [?nc{Ve(B)—Ve(C)}—6£c{%n>—»V(C)}]—» » ( ( ? ) } + —»«!(C)}]Lot —L I bc be the total current flowing into B  from allnthe branches attached to it, so that,and —  icslBn == — «a( C) }— %C) } ]B } ... (3)I f  =  *®(S)2+j»v(B»8 stands for the current supplied to B  from outside, then by Kirchoff’s law—and *g{B) =  »e(B)l+*’«(J)8 =  0  iff(B) =  VB)l+»»r{B»a = } (4)Supposing that the vector potential of 2? is unity and those of the points A, B ,C  to be zero, the currents passing from Z) along the branches B B  and BC  will be given byand I d b  —  Y b d  —I dc =  Y  CD — fifcn+j^cD } (5)and no current will flow in any other branch of the circuit. In order that the as­sumed potential may be correct a current —//ja — I db+ ^dc =  ixiD)2+Pv(V)2 will have to be supplied to D from outside. Under such a condition the currents I db ^dc will leave the network a t B  and G respectively. But in fact no current passes to or leaves the network a t B  and C. Therefore it is required to superpose on the assumed potentials those which would result if currents I b2{= I db) aud /pj( == I bc) were supplied at B  and C and allowed to leave the network a t D  and A the latter points being held a t zero potential.It is thus obtained initially*«lB) =  ^c(B)8 ~  9BD  » *«(C) =  *e(C)8 =  9CD >»V(B) =  V B )2  =  &BI) 5 »»(C) — »v(0)2 =  ^OD ;=  »e(Z»8 =  —iSBD+gCD) \»B<0) =  VP)* =  —(6b® +6ob)»aud =  0.For liquidating these initial values (viz, ix{C)y a-nd ip(C)) byimposing suitable vector potentials a t B  and C only, a table of standard opera­tions can be prepared by the use of the following ;On the Performance Calculations of Three PhasCj etc. 385^  ^h(C)Sv,'x(B) d v p i B )=  bBC =9Vy^ B)and di^^x(B)dVy^ B)dVy(B) - s  (gBc) ;diyiB) -S (bjBc) ; B. . .  (6 )which are easily obtained from relations (3).On liquidation of the residuals (i,e,, the initial values of ia,{B)y iy(B)y x^(C) nnd iyio) the vector values of the currents at A  and 2), and also of the potentials at B and O will be obtained These values will naturally correspond to the case when unit potential difference is applied between the terminals D and A.The procedure involved in and the information available from such a method of solution will be apparent from the illustration given below :I L L U S T R A T I O NFor illustrating the method the following example which is worked out by Starr (1963), is considered.Example. A three-phase transmission line 100 miles long has the following constants :Resistance per mile =  0.26a; Reactance per mile =  0.8Q; Susceptance per mile == 14 X 10~® mho; Receiving end voltage =  66kv. I t  is required to determine (using nominal-T method),(i) the sending end voltage and (ii) the sending end current when the line is delivering 16000 kw at 0.8 power factor lagging. The vector admittances of the different branches of the network (Fig. 2b), as calculated for the problem under consideration, are :Y ab == 10-8(7.1 ~j22.79),Y bc =  10-8(7.1 ~j22.79),Yen =  10-8(3.446-j2.68),Ybi> «  10-8(0+;1.4).10386 B, N. Basu and S. N. DuttaFirst of all it may be assumed for convcnionco that the potential a t i)  is 1 kv (one kilovolt), and tlie potentials at J5, and C are all zero. Then the currents in the branches DC and DB will bo== 3.446-j2.58,and I db =  O+jl.4.Therefore, the current —//>2 to bo fed from outside will be given by :=  3.446-~jl.l8.Thus it is obtained initially,== 5 x^{C) =  3.446; =  —3.446;iy{B) 1-4; iy(C) =  —2.58; iy^ D) =  1.18; and .^r(^ ) “  '^ v(A)Using the relations (6) and keeping in mind that all the '{/' and ‘6’ values mentioned there are to bo multi}>lied by 10^  (since hen^ one kilo volt is considered in place of one volt mentioned previously) the unit operation table (Table I) is readily obtained for liquidating the initial values.TABLE IUnit operation table5%(B) 5 C) 6^ X{7))7.1 - I t . 2 7.1 0 -2 2 .7 9 44.18 - 2 2 . Id 1.40 7.1 -10 .546 3.446 0 - 2 2 . Id 25.37 -2 .5 8V y { B ) ^ l 22.79 -4 4 .1 8 22.79 - 1 . 4 7.1 -1 4 .2 7.1 0M 0 )= 1 0 22.79 -2 5 .3 7 2.58 0 7.1 -10 .546 3.446With the help of the procedure suggested previously by Basu (1958), the final operation table (Table II) is prepared which enables complete liquidation of the initial values in throe steps only. I t  may, however, be mentioned that this procedure is not an essential one for the method under consideration. Tlui relaxation table is then worked out as shown in Table III.Thus when vpA (i e., sending end voltage) is 1 kv,(i) the receiving end voltage,VDc =  VD-vc =  (0.788-j0.143)X10  ^ volts =  0.8/Z.-10°18' kv.and(ii) the  sending end current,2 .468-jl.25== 2 .8 2 1 /^ -2 6 W ' amps.On the Performance GalcukUiona of Three Phaaey etc. 387TABLE IIFinal operation table6%(B) SVxiCi Svmri) 8«V,C) 3%(C).1 .2 0 0 .71 0 - 1 .4 ^ .689 -2.279 - .1 4 2.795 -.3 7 6-.188 -.377 .1 .194 .941 0 0 -.9 3 8 5.004 .22 -6 .6 1.376.276 .227 -.1 1 4 -.11 -.641 0 0 .636 -7 .068 7.689 0 - .6 2TABLE IIIRelaxation talkieSUa:(C) S%(C) X^(A) HiB) '^ x(6^  H{.D) '^ U(A) v^iB) h/iC).081Initial values 0 0 3.446 -3.44G 0 1.4 -2 .5 8  J.18.246 .492 -.122 - .2 4 5  -.043 -.0350.065.0180.126.017I1.7472.3592.4580000 -1.751 -5.606 1.055 4.296 .256 0 -2.361 -2.353 1.198 0 1.149 0 -2 .4 6  -1.257 0 0 1.242.212 .083 .143 {2.458 -2.46 -1 .267 0 0 1.242Therefore for a receiving end voltage of 66/\/3 (between phase and neutral) the following quantities will be obtained,Sending end current , I  a =  134.4 amps. [132.2 amps];Sending end voltage, ~  82.5 kv[82.3 kv]; and the sending end power factor =  cos 26^57'lagging [cos 26°58'].The corresponding values of I  a, Vad and the power factor as obtained by normal method of calculation (Starr, 1953) are given side by side within the brackets [ ], and it appears that they agree fairly well with each other.D I S C U S S I O NThe utility of the method described in this paper is appreciated when one consi­ders that a relaxational solution as explained, readily yields a number of informa­tions at a time viz., (i) vector currents a t the sending and receiving ends, (ii) phase angle between the sending end and receiving end voltages together with the ratio between the two voltages, the ratio being constant for a given load-impedance, etc. Moreover the fact that the number of initial values to be liquidated in such problems will never exceed four, always brings forth a limitation in the labour associated with the process of liquidation. I t  may, however, be borne in mind tha t this method always requires a prior knowledge of the load admittance or the load impedance (which may be given directly or may be obtained from the given data). But as a rule the value of the receiving end impedance is not known383 R. N. Baau and 8. N, DuUaunless the load, its power factor and the receiving end voltage are known. Under such conditions Case (ii), as mentioned earlier in the introduction, becomes identical with Case (i).So far as the application of relaxation technique is concerned, the solution of long transmission lines by equivalently method is in no way different from the nominal-y method of solution of medium transmission lines. In the former case the branch impedances (i.e., Zab> Zbc, and Z^j) of Fig. 1) only will have some changed values which can be calculated by using proper expressions given in standard text books.a c k n o w l e d g m e n t<1The authors take this opportunity to express their gratefulness to Prof. A. K. Sengupta, D.Sc., A.M.I.E.E. (London), Head of the Department of Applied Physics, Calcutta University, for his guidance throughout the progress of this work.Jl E F E R E N C E SBlack, A. N., and Southwell, H. V., 19.18, Proc. Roy. Soo. 164(/l), 447.BaiiHi, H. N., 1958, Journ, Aftsn. AjppUed Phyn» 5, 30.Sout?iw©ll, JR. V., 1951, Relaxation Methods in Engineering Science, Oxford University Press, London, Chapter VI,Starr, A, T., 1953, Oeneration, Transmission and Utilisation of Electrical Power, Chapter VI., Sir Isaac Pitman & Sons Ltd., London,

On the Performance Calculations of Three Phase Transmission Lines by Relaxation Method

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On the Performance Calculations of Three Phase Transmission Lines by Relaxation Method

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