circle diagram

The General Circle Diagram of ElectricalMachinery

BY FREDERICK EMMONS TERMAN,* CECIL LOUIS FREEDMAN,tAssociate, A. I. E. E. Non-member

THEODORE LOUIS LENZEN,t and KENNETH ALFRED ROGERStAssociate, A. I. E. E. Non-member

Synopsis.-The well known circle diagram of a transmission net- The transmission network circle diagram can be applied to thework is applied to electrical machinery, giving circle diagrams of induction motor, yielding in the approximate representation thealternators, synchronous motors, synchronous condensers, and Heyland diagram, which is merely a special case of the more generaltransformers. These diagrams give a graphical representation of diagram. The transmission network method of attack gives athe machine performance under all possible conditions. Such straight-forward solution of many induction motor problems thatquantities as power loss, power input, power output, field current, would otherwise be difficult to handle, such as a motor equipped withetc., for any operating condition can be obtained by inspection. a phase advancer.These diagrams have the same field of usefulness as the circle diagram Methods of obtaining network constants by measurements ratherof the induction motor. than computations are described.

INTRODUCTION tive power is used when the sending end voltage isANY electrical network connecting two pairs of constant, while received power-reactive power co-

terminals can have its electrical characteristic ordinates go with a constant receiver voltage. Irnexpressed in terms of four constants A, B, C, and either case, the diagram can be readily constructed by

D through the equations laying out the proper system of power coordinates,Es= A Er + B Ir computing the centers and radii of the desired circlesIs = C E, + D Ir (1) from the network constants according to formulas thatIrtheequivalentCquations E+have been derived,2 and doing the rest with a compass.Or the equivalent equations After this has been done the results can be gone overEr = D Es-B Is for errors by applying a series of simple graphicalIr = AI,s-CEs (2) checks.

The subscripts s and r denote sending and receiving and The circle diagram as described was developed for thequantities, respectively. The coefficients A, B, C, and purpose of representing the properties of power lines andD, which are known as network constants, take into transmission networks. In a highly specialized formaccount the composition of the transmitting network this general circle diagram becomes the well knownand the frequency. They can be readily obtained for induction motor circle diagram. In both applicationsany particular case by methods and formulas that have this graphical method has been of inestimable valuebeen worked out by Evans and Sels,1 or by the measure- because it shows on one drawing all possible conditions.ments described in Appendix I. Only three of the four After the circles have once been laid out no further com-constants are independent, since the relationship putations are necessary, and a complete visualizationA D = I + B C must always exist between them. of the network performance is easily obtained.For a symmetrical network A = D. The circle diagram method of graphical representa-When either the sending end or the receiving end tion can be used to show the performance of alternators,

voltage is constant, the electrical properties of the net- synchronous motors, transformers, and synchronouswork can be represented graphically by a circle diagram condensers with the same advantages that are alreadyconsisting of families of circular loci drawn on a power- well known in the cases of the transmission line and thereactive power coordinate system. This circle diagram induction motor. The transmission network circleis a graphical representation of Equations (1) and (2), diagram is applied to such electrical equipment byand when once drawn, an inspection of the one diagram substituting for the actual machine an equivalentwill give such quantities as power, current, admittance, electrical network for which a circle diagram is thenand power factor at both sending and receiving ends of drawn. The details of the transformation from ma-the network, also efficiency of transmission, power lost chine to electrical network, and the special problemsin transmission, and so on almost without limit. involved are taken up in the following sections.A coordinate system of sending end power and reac-*Assistant Professor, Stanford University, Calif. CIRCLE DIAGRAM OF THE TRANSFORMERtGraduate Student, Stanford Uniiversity, Calif. The action that takes place in a transformer is1. Evans and Sels, Power Limitations of Transmission Systems,

A. I. E. E. TRANS., Vol. 43, 1926, p. 26. Also a series of articles 2. F. E. Terman, The Circle Diagram of a Transmissionby Evans and Sels in Electric Journal, 1921. Network, A. I. E. E. TRANS., VOl. 45, 1926, P. 1081.

374

30-4

Jan. 1930 TERMAN, FREEDMAN, LENZEN, AND ROGERS 375

accurately given by the well known network shown in posed to operate with substantially constant voltageFig. 1, which has the network constants at one terminal.

A = D = 1 + Z Y/2 The transformer is a very efficient piece of equipment,B = Z (1 + Z Y/4) with small voltage drops and low losses. As a conse-C = Y (3) quence some of the circles have large radii and appear to

Ratios of voltage transformation other than the unity be almost straight lines on the diagram. This is thevalue to which Equations (3) apply can be taken into case with the secondary voltage and load power factoraccount in the network constants. Calling the ratio loci in Fig. 2, and is ordinarily to be expected. Circlesof primary to secondary voltage N (N more than one too large for a beam compass can be readily drawn with

the aid of a piano wire.Z12 Z/? In drawing circles giving power dissipated in the

2 > transformer, great care must be taken in computing thecircle radii, and it is necessary that the network con-stants A and D be known to a number of significantfigures sufficient to incorporate to a fair accuracy the

FIG. 1-EQUIVALENT CIRCUIT OF TRANSFORMER effect of the Z Y term which they contain. LogarithmZ-Series resistance and reactance representing primary and secondary tables of at least five places are usually advisable in

copper losses and leakage reactanceY-Shunt admittance representing core loss and magnetizing current computations leading to the loss circles. Instead of

computing the loss circles in the usual manner, it is morefor a step-up ratio and N less than one for a step-down satisfactory under ordinary circumstances to obtaintransformer), then when the transformer leakage re- them by the much simpler method described in Appen-actance Z and admittance Y are referred to the secon- dix IIdary side, the network constants have the values It is of course possible to use equivalent transformer

A = N (I + Z Y/2) circuits other than that shown in Fig. 1. For example,A = N Z(1 + Z Y/2) the impedance Z can be considered as lumped in oneB = NZ (1 + Z Y/4) piece, and placed either on the sending or receiving sideC = Y/ND = (1 + Z Y/2)/N (4a)

When Z and Y are referred to the primary side thenetwork constants become

A = N(1+ZY/2) (B (1+Z Y/4)/IN CZ nar Vltgeii2 Kv.

C =YN 21D =(1+Z Y12)/IN (4b) >_

Equations (3) and (4) consider the leakage reactance Secondary Voltage 108 Kyequally distributed between primary and secondary, > 200which is the usual assumption. Network constants mz 6 1 06Kv.for other divisions are readily obtained by using a_/methods outlined by Evans and Sels. ?

The circle diagram of the transformer is a trans-mission line type of circle diagram drawn for the net-work constants given in Equations (4). When appliedto an actual example the result is as shown in Fig. 2, 200 400 600 800 1000 1200tothe result is as shown in Fig. 2, ~~~~~~~PRIMARY POWER KILOWATTSwhich has been drawn for a 1000 kv-a. transformer FIG. 2-CIRCLE DIAGRAM OF 1000-KV-A. TRANSFORMER WITHoperating with constant primary voltage. To avoid PRIMARY VOLTAGE KEPT CONSTANT WITH 110 KV.confusion, only power loss, power factor, and voltagecircles are shown, although of course many other types of the admittance Y. Such an equivalent circuit hasof loci can be drawn when desired. slightly simpler expressions for network constants thanThe transformer circle diagram gives a complete and does Fig. 1, and involves only a slight approximation.

exact graphical representation of the properties of the In making approximations it is very necessary that thenetwork shown in Fig. 1, and therefore shows the network constants be exactly the constants of the net-transformer performance to the same degree of pre- work in question. Simplifying the formulas for net-cision as does the usual equivalent circuit. In con- work constants by dropping apparently insignificantstructing a circle diagram it is necessary that either the terms will sometimes yield amazing results. Thussending end (primary) voltage or the receiving end dropping the Z Y terms of A,B, andDin Equations (3)(secondary) voltage be kept constant. This restriction introduces an error in these constants of only about oneordinarily involves no limitation to the usefulness of per cent, but this omission is equivalent to neglectingthe diagram because transformers are normally sup- all of the iron loss, and affects practically nothing else.

376 GENERAL CIRCLE DIAGRAM OF ELECTRICAL MACHINERY Transactions A. I. E. E.

CIRCLE DIAGRAM OF THE SYNCHRONOUS MOTOR from it apply only to a single phase, it is possible toThe performance of a synchronous motor can be label this single-phase diagram with the corresponding

shown by a circle diagram based on the equivalent three-phase quantities. The practical way of doingcircuit of Fig. 3, which represents one phase of the this is to compute the circle centers and radii on themachine. In this figure Er is the actual induced phase single-phase basis, then lay out a coordinate systemvoltage, as determined by air-gap flux, E, is the sending calibrated directly in three-phase power quantities andor terminal phase voltage, which is constant in the case draw the circles using radii and center coordinates threeof a motor, Xa and Ra are the armature leakage reac- times the calculated single-phase values. The circlestance and effective a-c. resistance (including armature themselves can also be marked with the correspondingcopper and stray load losses), respectively, per phase, three-phase values. Thus a circle representing a single-

phase loss of 10 kw. can be marked 30 kw. on the three-Ra Xa phase diagram, and will then represent total three-

phase loss.ES G Er The circle diagram as derived from Fig. 3 does not

include windage and friction losses, but can be made todo so byasmle expedient. Since these losses repre-FIG. 3-EQUIVALENT CIRCUIT OF SYNCHRONOUS MOTOR AND by a simp

SYNCHRONOUS CONDENSER sent mechanical power developed in the machine, butRa - Series resistance representing armature copper and load losses not available at the shaft, they can be taken into ac-Xa - Series reactance representing armaturereactance count by suitably labeling the diagram. That is, a lossG - Shunt conductance representing ironlosses circle would be drawn to represent a certain network lossEs - Terminal voltageEr - Induced voltage but would be labeled with this loss plus the windage and

friction loss, and an output power (receiver power)while the iron losses are accounted for by the conduc- circle drawn representing a certain mechanical powertance G. The motor is thus reduced to a transmission output would be labeled with this power minus windagenetwork through which power is transmitted with a and friction, to give actual net shaft power.sending end voltage of E, and a receiving voltage of Er The circle diagram of a 100-hp. synchronous motor isThe power which in the equivalent circuit of Fig. 3 is shown in Figs. 4 and 5, in which the labeling is in termsdelivered to the receiving voltage is the mechanical of three-phase quantities, and concludes windage andpower produced by the motor and so is the shaft outputplus the windage and friction. The motor losses ex-clusive of field loss are the losses in the equivalent circuit 6__ ___|___wi 80plus the windage and friction.The network constants corresponding to Fig. 3 are ,, __

A + (Ra+Xa) d4

B =Ra+Xa 40_

C GD=1 _ __owe

In a given machine the usual stray power test will give E 4j

Ra, G, windage, and friction. The iron losses can be - - -represented by a conductance placed as shown in Fig. 3because these losses are very nearly proportional to the r Loss Center o _ ___square of induced voltage (and hence of air-gap flux). !5 1Since Fig. 3 applies to only one phase of the machine, 40 -in a polyphase motor the total iron losses are divided .

60 80 01 1 Iequally among the phases. An iron loss of Pi watts at KILOWATT INPUTan induced voltage of Er accordingly leads to a conduc- FIG. 4-CIRCLE DIAGRAM OF 100-HP. SYNCHRONOUS MOTORtance G = (Pi/N)/Er'2 in a machine with N phases.The leakage reactance Xa can be computed, or can be friction. The motor diagram has been divided intoobtained approximately by measurement. The fidelity two parts to avoid confusion, but by the use of coloredof the circle diagram is fortunately not appreciably inks all necessary loci could satisfactorily go on oneaffected by reasonable uncertainties in the value of Xe. figure. It is of course understood that Figs. 4 and 5 doThe circle diagram of the synchronous motor is based not show all the circular loci that could be drawn.

on the network constants of Equations (5), and is Loci giving input current, induced voltage, angle be-constructed in the usual way using a constant terminal tween terminal and induced voltage, etc., could havevoltage and a system of input power-reactive power been included.coordinates. There is, however, considerable flexibility The usefulness of the synchronous motor circle di-possible in the labeling of the diagram. Thus, although agram is greatly increased by superimposing constantthe equivalent circuit and hence the diagram obtained field current lines upon the power-reactive power


coordinate system, as has been done in Fig. 5. The the approximate and simple method of drawing losslocation of these lines can be obtained by either mea- circles, given in Appendix II, is recommended as beingsuring or computing the combinations of reactive and the most satisfactory.real power that with the field current in question will CIRCLE DIAGRAM OF THE SYNCHRONOUS CONDENSERgive the terminal voltage for which the diagram is SInCE the On cds iS CONouSdrawn. The field current loci are approximately Snet esnhooscnesri ycrnudirawn.r Thes, fid ourrbeenactlociarclheapo mat motor operated without a shaft load, the circle in Figs. 4cuarrearcs,o anuldwouldbetr rexactlies if thearma-ue and 5 for zero output represents synchronous condenserture reaction could be truly replaced by an armaturereactance of constant value, action. The entire discussion on the motor appliesThe circle diagram as described does not take into here without change, and so need not be repeated.

account field copper loss. It is possible, however, to CIRCLE DIAGRAM OF THE ALTERNATORmark each field current line with the corresponding Since an alternator is merely a synchronous motor

operated backwards, that is with power supplied to theshaft rather than taken from it, a circle diagram similar

tuy i_ rOtZ to that of the motor can be drawn for the alternator.E 80 The equivalent circuit of the alternator is given in

Fig. 6, and the corresponding network constants are60 A 1

B Ra±+Xa

,E20 t 4a < S Thenotation is thesameasexlie enetowta- -qutonG()

1Si O\|\g\ .I The alternator circle diagram is drawn from these= 1l K~o% , ,i 91 network constants inR a manner similar to that followedg:20Ct a with the synchronlous motor. As in the case of the

_____ _____ ____ ________ motor the final diagram c:an be labeled to represent40 20 40 60 806 10 three-phase quantities although the actual computa-

KILOWATT INPUT tions are on the basis of one-phase. Windage andFIG. 5-CIRCLE DIAGRAM OF 100-HP. SYNCHRONOUS MOTOR friction losses, although not taken care of in the equiva-

(FIG. 4 CONTINUED) lent circuit, can be included by proper labeling. Thusthe shaft driving power equals the windage and friction

field loss, and in this way to obtain from the circle loss plus the three-phase sending end power of thediagram the total power loss of the alternator for a given equivalent circuit, and the total machine power loss isload power and power factor. The procedure is tolocate the point on the coordinate system corresponding XAto the desired load conditions. The field current line _f .passing through this point shows the field power and the s Ri Erloss circle at the point gives the other losses (i. e., iu by' iwindage, friction, armature copper, load, and iron) sothat the total loss is the sum of these components. :FIG. 6 EQUIVALENT CIRCUIT OF ALTERNATORIt is also possible to draw total loss loci, several of Ra - Seriesresistancerepresenting armature opperand load losseswhich are shown in Fig. 5.These lines are computed Xa - Seriesoreactance rresenting armature reactance

locatethe pont on he cordinat systecorrepondig -Shncodtae reResntn irnose

point by point from the loss circles and field currentE-nInducedvoltagelines, and are almost but not exactly circular. Er - Terminal voltageThe accuracy of the motor circle diagram is approxi-

mately that of the usual stray power test method. The the windage and friction plus the three-phase loss ndi-fundamental assumptions are: (1) iron losses are con- cated onthelosscircle.sidered proportional to the square of the induced volt- As in the case of the motor, lines of constant fieldagei and pendento armature current; (2) load currentccanbepdrawnton the coordinate system. Theselosses are assumed proportional to the square of the lines must be corputed point by point, and are approxi-armature current; and (3) the armature is assumed to mately but not exactly circular arcs. They can behave a constant leakage reactance. None of these marked with the field copper loss theyrepresent, and inassumptions introducesappreciableerror. this way the diagram readily gives total loss exactly

In constructing the motor circle diagram some dif- as in the case of themotor.ficulty will be experienced in determining the radii of The circle diagram of a 25,000-ky-a, alternator isloss circles unless at least five place tables are used in shown in Fig. 7. This diagram is drawn for a constantcomputing radii and network constants. In most cases terminal voltage and so utilizes a coordinate system of


load (or receiver) power and reactive power. The the equivalent induction motor circuit, a circle drawnlabeling gives three-phase quantities and takes into for a unity power factor load is the operating circleaccount windage and friction losses. Thus, as the of the usual motor diagram.windage and friction is 122 kw., a loss circle computed The equivalent circuit of the induction motor asfor a single-phase loss of 76 kw. is marked as used by different investigators varies somewhat, ac-3 X 76 + 122 = 350 kw. Only power loss and in- cording to the assumptions made. The basis of theduced voltage circles are shown on Fig. 7, although Heyland diagram is the equivalent circuit of Fig. 8A,many other types of circles could be added if desired. which is only approximately correct. The more cor-The entire discussion given in connection with the rect equivalent circuit of Fig. 8B also leads to a circle

motor circle diagram applies to the alternator diagram diagram, but one more difficult to draw.The network constants corresponding to Fig. 8A are

A=1

cc Loss Cswhilethoseapplyng tig 8 are0- ~~~~~~~~B~Z Z(1Z Y

:: _Ac

UM ~~~~D1+ZrYZ (7a)

____8 ___ A transmission line type of cirleb diagram may be°ni Iv 1 X a ~~~~~~~drawnfor either of these equivalent circuits, or for

F 12> I_ ___ / |_ V;oltaged any modified arrangement, using a constant terminal

C I___ ___ ____ _wht In any case, the circle representing a unity power factoro 4 5 126O 20 24 2 oad is the operating circle of the motor, and the inter-

J LOD THUAD OF KIOWTT

FIG. 7-CSIRCLE DIAGRAM OF 25,000-KV-A. ALTERNATOR

with only minor and obvious modifications. Inpar- s Lle1 I'Qz Aaticular, the method given in Appendix II will generally 0 ||||| | / | {|be ofconsiderable assistance in drawing losscircles.E0 1 1 |ff0pret ,/

0 4 oll~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~Ef 0prcn

CIRCLE DIAGRAM OF THE INDUCTOR MOTOR F - _ %>The classical induction motor circle diagram is d

merely a special application of the more general circle

w ed~~~~~~~~~~~~

2,400 Ef 70percenti

>~~~~~~~~~~~~ LIEPWRPRaHS-Al'00e 10 voltage,~~~~and_input __ Si=poe- ercentepo erceodnats

o ,} FIG. 9 CIRCLE DIAGRAMOF A SMALL INDUTCTION- MOTOR

i y~~ ~ ~~ ~I ansecioseof this circle riheffiientigy,iitpowerloss, towr

_OADT_ loutheoet., circles o the motor, performance.

FIG.-CIRLE DAGto saOr Zr00-VA fLTRicTiOnR ;;9cRlLEiAGAh S0 oe

(b) Also, since each load resistance in the equivalent circuitFIG. 8-EQUIVALENT INDUCTION MOTOR NETWORKS corresponds to a certain value of slip, load resistance

= Stator impedance ~~~~~(or conductance) circles can be labeled to give slip.Zr=RotorimeacrdcdottrieWe h ewr fFg 8A and the network con-Y Shunt mian v iron losses, windage and areused the resulting dg

ticular, exacty m d g n in AHeyland diagram, which isdiagram of a transmission network. This follows from accordingly only a special case of the general trans-the fact that the equivalent circuit of the induction mission network diagram, and the geometrical construc-motor is that of- a transformer operating with a resis- tions commonly used to get slip, efficiency, etc., on thetance load. Starting with the network constants of Heyland diagram are merely graphical means of


determining where different power loss, slip, and other by the phase angle of the impedance Rr + Z, where Rr iscircles intersect the unity power factor load arc. the rotor resistance. This circle is the operating circleA typical example of an induction motor circle dia- of the motor when equipped with a phase advancer of

gram drawn by the transmission network method is the type mentioned. Slip, shaft power of the inductionshown in Fig. 9, which is based on the equivalent circuit and commutator motors, etc., can be represented on theof Fig. 8B, and is similar to the Heyland diagram, but diagram by methods that are readily devised. Anslightly more accurate. By rotating Fig. 9 a quarter adequate discussion of this and similar cases wouldturn it will be readily recognized as an induction motor require a separate paper, however, and it is merely in-diagram. tended at this time to point out the possibilities of theAlthough any induction motor diagram can be drawn transmission network method of attacking induction

by following the transmission network mode of attack motor problems, particularly those out of the ordinary.that has been outlined, such a procedure is not advisable Induction generator action is represented by thewhere the usual Heyland diagram based on blocked negative power portion of the motor diagram so that therotor and no load tests will do. For this usual case, the discussion that has been entered into applies to theclssical procedure is so satisfactory and so well stand- induction generator with obvious modifications.ardized that there is no incentive to replace it by some- REPRESENTATION OF MACHINE CHARACTERISTICS ON Athing different, even if equally good. POWER COORDINATE SYSTEMThe transmission network method of approach does, In representing the relation between two variables

however, have a considerable field of usefulness in in- it is customary to use a coordinate system in which oneduction motor problems. For example, it gives a of these variables is plotted as a function of the other.straightforward way of obtaining the operating circle Thus in a synchronous motor, one might plot efficiencywhen equivalent motor circuits other than that Of as a function of shaft load, there being one such curveFig. 8A are used. Again, the usual construction for for each value of field current. Where a very compre-obtaining slip on the Heyland diagram is not accurate hensive picture of machine performance is desired,when the effective rotor resistance depends upon the however, it is preferable to use a power-reactiveslip frequency. This variation can be correctly taken power coordinate system and to superimpose the de-into account in the network constant method of attack. sired loci upon this as has been done to a limitedThe procedure is to assume a slip, and then draw a load extent in Figs. 2, 4, 5, and 7. Lines representing suchconductance circle for a load resistance computed from loci may be either circular or non-circular, dependingthe rotor resistance at the slip in question. The inter- upon the circumstances, but no matter how muchsection with the operating circle gives the operating saturation, etc., is present it is always possible to showpoint for the slip in question. correctly the complete performance of the apparatus onAn important use of the transmission network method a single diagram.

of viewing the induction motor is in the analysis of The use of a power coordinate system to show char-problems where electromotive forces are introduced acteristics has the advantage of permitting a readyinto the rotor circuit by commutator machines. The aceitc ha th adatg of pemttn aed*troiii.visualization of the entire behavior under all conditions.performance of induction motors equipped in this way Prom such a diagram it is possible to determine quanti-can ordinarily be represented by some sort of a circle tatively the operating characteristics of a piece ofdiagram, the exact nature of which depends upon the equipment for a given set of conditions by inspection.characteristics of the electromotive force generated by Thus in the case of a synchronous motor which is tothe commutator motor. e carry a certain load at a certain power factor, the pointAn example that illustrates the power of the network on the coordinate system that corresponds to this con-constant method of attack is the case ofinductionmotor dition can be readily found from the power factor andin which a series commutator motor driven at constant load power loci using Fig. 4. The input power, field

speed is inserted in the rotor circuit. Such a machine, current, total power lost, efficiency, etc., are then givencommonly called a phase advancer, is at its terminals by the various loci lines passing through the operatingequivalent to an impedance that is fixed both in magni- point, using diagrams such as Figs 4 and 5tude and phase angle by the machine construction,and is independent of frequency. To incorporate in the CONCLUSIONcircle diagram the effect of a phase advancer equivalent The principles presented in this paper have been putto a vector impedance Z one computes the network to laboratory test4 as far as the university facilities allowconstants of the motor circuit assuming the rotor re- and satisfactory agreement of theoretical and experi-sistance is zero. Using these constants a circle is mental results was found in all cases.

draw corespodingto aloadpowe facor rpresnte 4. The results are to be found in the following Stanford3. For example, see John I. Hull, Theory of Speed and Power University theses: T. L. Lenzen and K. A. Rogers, Application of

Factor Control of Large Induction Motors by Neutralized Polyphase the Transmission Line Circle Diagram to Transformers andA-C. Commutator Machines, A. I. E. E. TRANS., Vol. 39, 1920, Alternators; C. L. Freedman, Application of the Transmissionp. 1135. Line Circle Diagram to the Induction Motor.


The circle diagram of a transmission network, de- this is a transmission line. Since there are three inde-veloped to solve transmission line problems, seems to be pendent network constants any three independentthe fundamental circle diagram, applying to many measurements are sufficient. The measurements mostvarieties of electrical circuits and machinery. Its use in easily made are:connection with transformers and synchronous motors, (1) Sending end impedance Z1 with receiver open.generators, and condensers leads to circle diagrams fully (2) Sending end impedance Z2 with receiver shorted.as useful as the classical induction motor diagram. (3) Impedance Z3 at receiver with sending end open.The authors wish to express their appreciation for (4) Impedance Z4 at receiver with sending end

the assistance rendered by Mr. Monges of the General shorted.Electric Company in supplying data on commercial According to Equations (1) and (2) these impedancesequipment. are related to the network constants as follows:

Appendix I Zi = A/CDETERMINATION OF NETWORK CONSTANTS BY Z2 = B/D

MEASUREMENT Z3 = D/CDirect Measurement of A. From Equation (1) it is Z4 = B/A (12)

seen that with the receiver open: The simultaneous solution of any three of these fourA = Es/Er (8) equations together with the relation A D = 1 + B C

The vector ratio Es/Er can be measured in both magni- will enable the four network constants to be computed.tude and phase by the three-voltmeter method, or by The four equations of (12) cannot be simultaneouslysuccessive applications of the three-voltmeter method. solved themselves to give the four network constantsThe phase can also be measured by a wattmeter in because only three of these constants are independent.which the current in each of the coils is proportional to Appendix IIone of the two voltages. APPROXIMATE METHOD OF COMPUTING Loss RADII

Direct Measurement of B. From Equation (1) it is ReferE Toa previousLOSS Rcle

seen that with the receiver terminals short circuited: Reference to a previous article2 shows that loss circleB E, Ir (9) radii are given by an equation of the form

The phase of B can be determined by the use of a watt- Radius = m VL-meter in which E, is applied to the voltage coil and Ir where m and Lo are constants that can be computedflows through the current coil. from the network constants and the terminal voltage.Measurement of C. From Equation (1) it is seen In these computations, however, Lo is difficult to

that when the receiver terminals are open circuited: determine accurately as it is the small difference of twoC = Is/Er (10) nearly equal quantities.In the case of equipment such as the transformer,This vector ratio can be readily obtained by the use of alternator, etc., where the minimum possible loss with

ammeter, voltmeter, and wattmeter. the fixed terminal voltage is for all practical purposesMeasurement of D. With power supplied at the the no-load loss, it is permissible to substitute the no-

receiver terminals, and with the sending end terminal load losses for the quantity Lo. This is because theopen, Equation (2) shows: loss Lo is the minimum possible loss with the fixed

D = Er/Es (11) terminal voltage being used. With this simplificationMeasurement of Constants When Both Ends of Net- the loss radii may be easily computed with satisfactory

work are not Available at One Point. An example of accuracy.

circle diagram

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