621.315.09:621.3.012.2 Monograph No. 89 SUPPLY SECTION TRANSMISSION-LINE ESTIMATIONS BY COMBINED POWER CIRCLE DIAGRAMS By F. DE LA C. CHARD, M.Sc, Member. (The paper was first received 24th June, and in revised form \2th October, 1953. // was published as an INSTITUTION MONOGRAPH, \5th January, 1954.) SUMMARY Power circle diagrams drawn for either sending or receiving ends of a transmission line do not have a common centre for the voltage circles, nor are the power axes within the same semicircle. There is also a possibility of confusion over the sign of reactive power when conditions at both ends of the line are estimated by means of separate diagrams. A quadrant diagram with current taken as reference, which complies with the convention of the American Institute of Electrical Engineers for reactive power, is shown to be a satisfactory base on which a combined send-receive power circle diagram may be constructed. If such a diagram is built as a calculating board, all sendingand receiving-end quantities may be simply read with sufficient accuracy for the estimation of system load conditions. Examples of the use of the calculating board are given. LIST OF PRINCIPAL SYMBOLS A,B,C,D— General network constants. Es, Is — Sending-end line voltage and current. Er, lr — Receiving-end line voltage and current. Zt/_P — Line impedance and angle, per phase. Y = Line admittance, phase to neutral. Zt,/_h = Transfer impedance and angle, per phase. Zs,/_Ps = Sending-end driving-point impedance and angle, per phase. Zr/Pr — Receiving-end driving-point impedance and angle, per phase. 8 — Transmission angle (angle between Es and Er). Ps,Pr ~ Sending- and receiving-end powers. Qs,Qr ~ Sending- and receiving-end reactive powers. WATTS IN, Pt — Power received or absorbed by the load at the receiving end of the line. WATTS OUT, Po — Power supplied or produced by the source at the sending end of the line. VARS IN, Qj -~ Reactive power absorbed at the sending or receiving ends of line, by source or load, i.e. that due to a capacitive source or an inductive load. VARS OUT, Qo — Reactive power produced at the sending or receiving ends of the line by the source or load, i.e. that due to an inductive source or a capacitive load. (1) INTRODUCTION The power circle diagrams for sending and receiving ends of a line are derived from the corresponding voltage phasor diagrams (Figs. 1 and 2). These phasor diagrams are usually drawn with the reference quantities in the same direction, although they refer in one case to a source and in the other to a load. The derived power circle diagrams have different centres for the voltage circles, with a common active- and reactive-power axis. They can, however, use the same set of Correspondence on Monographs is invited for consideration with a view to publication. Mr. Chard is in the Department of Electrical Engineering, University of Bristol. Fig. 1.—Receiving-end voltage phasor diagrams. DE, 0. r2 Fig. 2.—Sending-end voltage phasor diagram. circles, within the same semicircle, if the axes, in respect of the sending-end diagram, are rotated through 180°. This axis reversal might be objectionable were it not justified by considerations of usage. Power-system metering has long made use of the terms "WATTS IN," "WATTS OUT," "VARS IN" and "VARS OUT." Decrease Decrease \ Increase VArs, out Fig. 3.—Quadrant diagram, in accordance with the current American I.E.E. sign convention for reactive power. These terms imply a quadrant diagram (see Fig. 3) in which current sent or received is shown on the horizontal axis and sending- and receiving-end voltages appear in two diagonally [204] It follows that in calculating the in-phase and quadrature components of apparent power.1) Receiving-end Power Circle Diagram From the above equations. increase of excitation produces In this equation Z is the transfer impedance Zt. WATTS IN may be considered positive and WATTS OUT negative. while values of the receiving-end voltage Erl and the transmission increase in quadrature current moves them parallel to OQ. on a power diagram. OSQ the reactive-power axis. from the equations Es = AEr + BIr and Er = DES — BJS. the diagram of Fig.—Receiving-end power circle diagram.and receiving-end regions within the same (2. OrP. in order to obtain reactive power with the correct sign.CHARD: TRANSMISSION-LINE ESTIMATIONS BY COMBINED POWER CIRCLE DIAGRAMS opposite quadrants. Sufficient accuracy for a diagram is obtained by the use of a nominal-n. Taking A and D as or using the nominal-77. It is not the purpose of the paper to justify the quadrant Then the intercept ORF on the reactive-power axis is \YE} and diagram which is already in general use. Thus an inductive load demands WATTS IN 6=0 205 and VARS IN. ON ~ En the synchronous machine as a generator or motor is seen to be consistent. 5 is drawn for particular of current moves the extremities of Es or Er along OP. Thus. The Es ( ~Er) circle then passes it. the current term must be conjugated as follows: P + jQ= vl An advantage inherent in the quadrant diagram and the use Fig.2) Sending-end Power Circle Diagram semicircle. Similarly. 4 is drawn for particular values of the sending-end voltage Esl and the transmission angle dt. The use of the terms VARS IN and VARS OUT avoids Watts in an explicit sign convention. E2 (1) From eqn. 4. It will be seen. A capacitive load absorbs active power and produces reactive power. This change results from conjugating Ir in order to comply with the convention that the reactive power due to an inductive load circuit (VARS IN) be considered positive. . it can be shown that the lines OP and OQ are at right-angles and that an increase in the power component From either of these equations.. (3) (2) THEORY The phasor diagrams of Figs. Thus. 1 and 2 are drawn for current in phase with the reference voltage. that the BIr phasor lies along the active-power axis but that the reactivepower axis is in the opposite direction to that which might have been expected from Fig. the components of apparent power are obtained in terms of voltages and impedances. In either case.arrangement which modifies eqn. which necessitates WATTS OUT and VARS OUT from the source.arrangement Ps+jQs = EsIs^El/P- (1 + j Z r / 9 0 + p). (2. but to show that a the diagram is normally drawn by first setting up this distance combined send-receive power circle diagram may be based upon from the power origin O R . (1) to • (2) Watts out Fig. 1 and 4. is that the behaviour of OF = Er. which necessitates the production of active power and the absorption of reactive power by the source. with its angle p. VARS OUT and decrease of excitation demands VARS IN. but the method described in the paper places both sending. 5. Previous "universal" power circle diagrams have had to through F. Fig. but the quadrant diagram shows an implicit convention by relating the American Institute of Electrical Engineers sign convention for reactive power to the Argand diagram.—Sending-end power circle diagram. OSP form the power axis and OrQ. and the line admittance Y is assumed to be a pure susceptance. VARS IN and VARS OUT. of the terms. 1. reactive power taken by an inductive load circuit (VARS IN) is in the +y direction and reactive power given by a capacitive load circuit (VARS OUT) is in the —j direction. use the entire circle. This compactness has made possible the conThe sending-end components of apparent power are struction of a line calculating board based on the combined diagram. by comparing Figs. (1). . Pr \in n^circle n circle O Fig.and receiving-end diagrams are now as shown in the quadrant diagram (Fig. Reactive / Power' [\ 11 ft \7 6 / 5 4 3 2 loutOl Power i i . are known. Similarly. perpendiculars are dropped to the active. and Er at an angle /Pt to the reference axis. whose radius is nEs where n = ErlEs. This circle gives the value of Er. 0=0. the sending-end reactive-power axis passes through G and the power axis intersects it at a distance \YE} vertically below G. Er. Measuring off the sending-end active and reactive power along their respective axes gives some point K on an intermediate circle. which is the same point as O in Figs. 6). Pr and Qri the diagram (Fig. Ps and Qs. although parallel. and the power-axis intersection is at a distance \YE} on the power scale. Alternatively.—Calculating-board settings for sending-end voltage determination. if the sending-end quantities. and a second circle of radius n2 times the unit value is chosen. 3). and a distance ^YE} is measured vertically below G to give the sending-end power origin Os. n and also the load angle 6{ are noted.1) Method of Using the Combined Diagram Given the receiving-end quantities. are displaced for the sending. To estimate the receiving-end power conditions. (3) THE COMBINED SEND-RECEIVE DIAGRAM The combined diagram starts from the origin O (Fig. if 0 = 0 This equality is the basis of a combined send-receive diagram having common circles and a common direction for zero transmission angle. and the transmission angle 62 is read from the direction of the line OK. which has been taken as UNIT volts. The axes for both sending. from which Ps and Qs are read off.206 CHARD: TRANSMISSION-LINE ESTIMATIONS BY COMBINED POWER CIRCLE DIAGRAMS The choice of a common origin for both sending. where Erl is a particular value of£"r. Fig. 6. 6) is entered at O R . which is common for both sending and receiving ends.and receiving-end quantities. the corresponding receiving-end voltage is Er2 which shows VARS IN in the —j direction on Fig. n = Er\IEs E. is measured clockwise from the zero line for the receiving end and counter-clockwise for the sending end. and the intermediate circle is of radius (UNIT X Esl/Er). Efir\Zt. 2 and 5 shows that if the phasor diagram is superimposed on the power circle diagram. (3. 4 and 5. Alternatively. 4. This gives Esl in terms of Er. in which case the circle through F represents (UNIT X E}JEj) and the intermediate circle is (UNIT X Erl/Es). Eqns. 4) and will give WATTS OUT in the negative direction on the power axis. The voltage circle through G is then of radius (UNIT x EjJE}). below F. For Er — UNIT voltage. If a leading quadrature component of current Isq is added in Fig. Es is taken as UNIT volts and the sending-end power scale is positioned by setting the origin Os at a distance WEjx vertically below the point G. From the intersection of the n-circle and the line at an angle 6X to the load-angle zero in the counter-clockwise direction (point K). and Pr and Qr are set off along the appropriate axes. corresponding to the sending-end conditions. n — Ef\\Er For E. 5.—Combined send-receive diagram. Comparison of Figs. 7. gives the zero of angle on the transmission-angle scale. Es.and receivingend diagrams makes it inevitable that the active. (1) and (3) have a term of common magnitude.and reactive-power axes. The line OFG. the Bls phasor lies along the active-power axis. The receiving-end reactive-power axis is the vertical through F. = UNIT voltage. Rotation of the sending-end diagram through 180° also gives this common term the same angle to the reference (power) axis in both diagrams. since E s Fr . The intersection of this circle with the 6 = 0 line gives the point G. where Esl is the chosen or determined value of Es. (2. This is the opposite direction to that previously obtained for VARS IN in Fig.i and Er\ are particular values of E. the circle through G may be chosen to represent UNIT voltage. The load angle 9. 5) through 180° will make the reactive-power axis correspond with that of the receiving-end diagram (see Fig. 2. A number of circles whose radii are proportional to different values of Es and E} are drawn.3) The Combined Send-Receive Power Circle Diagram Rotating the sending-end diagram (see Fig.and reactivepower axes. the 771( angle 6{. The voltage circle through F may be taken as UNIT volts and is the Es (=Er) circle of the receivingend diagram. /_P scale.: "Electrical Transmission of Power and scale. M. their respective cursors (point N). tions" (McGraw-Hill. (2) THIELEMANS. Green circles are used for w:-circles in receiving-end power determinations. 1921. (9) American I. one division on the power scale = jj.1-04. Setting the transmission angle 6 to 6{ gives the point N from which the receiving-end active power Pr and reactive power Qr can be read. 1922. 610. p. Red circles are used for «2-circles in sending-end power determinations. The intersection of the 6— O line with this circle gives the point F. cursor and the UNIT black circle (point F). the particular circles used will therefore be distinguished by a colour and a number. (1) THIELEMANS. at a distance %YE}. pivoted at O. 1945. 41. and the reactive-power cursor can be offset by the required amount. D.2) Long-Line Calculating Board Having secured the advantages of common concentric circles for both sending. R. P. and so the VAr cursor must be set down a distance 26-6/29-5 = 0-9 division. (4) BIBLIOGRAPHY The following selected bibliography is confined to those references which deal with fundamental considerations relating to the solution of transmission-line problems by diagrammatic methods. 878 and 929. 1950). and Procedure: (g) Set the angle cursor with O SEND against 77-1° on the £P ibid. 1 02. W. New York. New York. O.1 The calculating board has three sets of circles. (12) MORTLOCK. and WANG. R. £ . A. The determination of Ps and Qs involves repositioning the (8) SCHWAGER. 1. tables of values can be prepared for these quantities.: "Principles of Electric Power Transmission" (John Wiley. diagrammes et regulation des lignes de transport d'energie a longue distance. a calculating board can be devised in which the active. is the active-reactive-power origin O R . scale. 1952). etc. up to 1-25. and HUMPHREY DAVIES. Qi) Find the intersection of the 1-23 red circle and the Signals" (John Wiley.: "New Transmission." Transactions of the American I. VI. 67. down to 0-8. 64. (d) Set the reactive-power cursor with the zero line 0-9 (4) HOLLADAY. C .. (3. Black circles are used for voltage determination and have radii from 10 to 12-5 in power-scale units.E. radius is 1 • 232 times the UNIT value. 599. For receiving-end determinations the inner black circle represents UNIT voltage.and reactive-power scales can be positioned by engraving them on Perspex cursors which can be moved horizontally and vertically respectively. Chap. determine Es and Angle 6 Line Data: ZT= \(A/11'\ ohm. . W. 1921. 512." Transactions of the (e) Scale Pr = 4-41 divisions and Qr ~ 1-45 divisions on American I. 70. A third cursor. 1948. 675.CHARD: TRANSMISSION-LINE ESTIMATIONS BY COMBINED POWER CIRCLE DIAGRAMS active-power axes must be repositioned according to the receivingend diagram.E.E. J." The only calculation involved is that of the vertical distances FO R = \YE2 and GOS = \YEl and the power scale in terms of unit voltage. The power scale remains as 1 division = 29-5 MW. p. 65. 475 and (a) Set the angle cursor with O RECEIVE against 77 1° on the 515. D. London. 0-96. 435. L. The outer green circle coincides with the outer black circle.: "Circle Diagrams for through F. 9. is set according to the line constants for the 6 ~ 0 position. II. G. Reset the watt cursor so that the zero Transactions of the American I. X. 205. C. The first red circle coincides with the inner black circle.and receiving-end diagrams..E.2) Given Er. up to 1 • 25. 7 on which the appropriate calculating-board settings are shown. (c) Set the active-power cursor with O IN line passing (3) EVANS.: "A Universal Power Circle Diagram. Pr = 1 3 0 M W I N . R. Since the radius of the UNIT black circle is 10 power-scale £2 divisions. Qr = 42-7MVAriN.: "Power System Interconnection" (Pitman. Chap. = 1-23 Er= 270 kV.." Comptes Rendus. p. 1940). and ibid." Electrical Engineering. In the estimations which follow..r." and in the clockwise direction for "receive. Examples of typical measurements are given in Section 5. Vol. 8. p. (6) WOODRUFF. 785. P.. 1938). Pr and Q r .18. Vertically below F. Subcommittee: "The Sign of Reactive Power. 1928). 1920.— 29-5 MW. and so the circles are labelled 1-0. 7): Generate de UElectricite. Y. which is the UNIT value times n2.." 6 = 0 line (point G)." Revue Procedure (see Fig.: "A Graphic Method for the Exact division below F (point O ). 1951.. The numerous references in which such methods are used have not been included. M. as are the black circles. 451. (11) GOODRICH. (5.E.1 02. 1170. 1 04. E. but the outer circle is now UNIT voltage and the remaining circles are labelled 0-992. This is done by selecting a circle corresponding to E}. and it reads transmission angles in the counter-clockwise direction for "send." Electric Journal. p.: "Calculs et diagrammes des lignes de (b) Find the intersection of the 6 — 0 line on the angle transport de force a longue distance. Set the VAr cursor with the zero line 1-36 divisions below G (point Os). For sending-end determinations the same set of black circles are used. 49. 1946. pp. 403. Yr=0 001083/ 90 mho.3) Determination of Ps and Qs under these Conditions (7) RISSIK.E.: "Power 0) \YEj = 40 = 40/29-5 = 1 . p. etc. and SELS.: "Calculs. 0-976.3 6 divisions on the power System Analysis" (Chapman and Hall. and the red circles are labelled 1 0. H. p. Transmission Systems. (5) DAHL. 2042. L. p. (5.1) Calculating-Board Estimations based on Section 3. and they are labelled 1-0 to 0-8 as are the black circles when used with a sending-end voltage as UNIT volts. pp. } = 26-6. 247. C : "Electric Circuits: Theory and Applicaif) Read results: 6 = 20°. H. etc. H.. line passes through G. F.E.. New York. K. pp. Receiving-end quantities: Er = 220 kV = UNIT voltage. 170. together with Fig. Chap.. 207 (5) APPENDIX (5. (10) KIMBARK. according to its own scale.watt and VAr cursors and the use of the red circle whose Line Diagrams. For a given transmission line.E. R Solution of Transmission Lines. 530. (m) Read results: Ps = 4-9 = 144-5 MW OUT. this is done in two stages by the method given in Section 3. The first stage.4) Given ESi P. Qr and 6 As in the previous examples. (5. Thus one division on the power scale becomes E}\\2-5Z.. determine Eri P f .1. . equal Qr = 41 MVAr IN Thesefigurescompare with the initial values given in Section 5. QS = i • 1 = 32-4 MVAr OUT. As an indication of the possible accuracy of reading. after four separate determinations involving repositioning of the various cursors.1.1. Repositioning the watt and VAr cursors and using the appropriate green circle gives Pr and Qr.208 CHARD: TRANSMISSION-LINE ESTIMATIONS BY COMBINED POWER CIRCLE DIAGRAMS to Es. and Q. gives ET and the angle 9. by using the outer black circle (radius 12-5 power scale divisions) as UNIT voltage. corresponding to Section 5. the receiving-end quantities were given as Er = 221 kV Pr= 131 M W I N (k) Find the intersection of the 9 = 20 line and 1-23 black circle (point k).