27190563 Regulation Losses of Transmission Lines

8/4/2019 27190563 Regulation Losses of Transmission Lines

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REGULATION AND LOSSE SG. D. McCann

IS chapter deals with problems relating to theperformance of transmission lines under normaloperating conditions. The analytical expressions for

and voltages and the equivalent circuits for trans-lines are first developed for short lines and foreffects of distributed line capaci-

taken into account). A simplification is pre-in the treatment of long lines that greatly clarifiesand reduces the amount of work necessary

calculations. Problems relating to the regulation andof lines and their operation under conditions of fixedvoltages are then considered. The circle diagramsdeveloped for short lines, long lines, the general equiv-circuit, and for the general circuit using ABCDThe circle diagrams are revised from the previ-editions of the book to conform with the conventionreactive power which is now accepted by the Americanof Electrical Engineers, so that lagging reactivepositive and leading reactive power is negative.

When determining the relations between voltages and cur-a three-phase system it is customary to treat thema per phase basis. The voltages are given from lineneutral, the currents for one phase, the impedances forconductor, and the equations written for one phase.

system is thus reduced to an equivalentHowever, vector relationships be-and currents developed on this basis areto line-to-line voltages and line currents if thedrops are multiplied by t/3 for three-phase

and by 2 for single-phase two-wire systems.Most equations developed will relate the terminal con-at the two ends of the line since they are of primaryThese terminals will be called the sending endiving end with reference to the direction of normalof power, and the corresponding quantities designated

the subscripts S and R.

I. EQUIVALENT CIRCUITS FORTRANSMISSION LINES

Short Transmission LinesFor all types of problems it is usually safe to apply thetransmission line analysis to lines up to 30 miles inor all lines of voltages less than about 40 kv. The

of distributed capacitance and its chargingvaries not only with the characteristics of the linealso with the different types of problems. For thisno definite length can be stipulated as the dividinglong and short lines.

CHAPTER 9OF TRANSMISSION LINES

Revised by :R. F. Lawrence

Neglecting the capacitance a transmission line can betreated as a simple, lumped, constant impedance,

Z= R+jX=xs=rs+jxsWhere

z = series impedance of one conductor in ohms per miler* = resistance of one conductor in ohms per milex* = inductive reactance of one conductor in ohms permiles = length of line in miles

The corresponding per phase or equi valent single-phasecurcuit is shown in Fig. 1 together with the vector diagram

EOUIVALENT TRANSMISSIONCIRCUIT TO NEUTRALFig. lEquivalent circuit and vector diagrammission lines. for short trans-

relating the line current and the line-to-neutral voltages atthe two ends of the line.The analytical expression for this relationship is givenby the equation:

Es=ER+ZI (1)

2. Long Transmission LinesThe relative importance of the charging current of theline for all types of problems varies directly with the volt-

age of the line and inversely with the load current. Toappreciate this fully it is necessary to consider the analysisof long lines.A long transmission line can be considered as an in-finite number of series impedances and shunt capacitancesconnected as shown in Fig. 2. The current IR is unequal to19 in both magnitude and phase position because somecurrent is shunted through the capacitance between phase

*These quantities can be obtained from t he tables of condu ctorchara cteristics of Chap. 3.

265


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266 Regulation and Losses of Transmission Lines Chapter 9

Fig. 2Diagram representing long transmission lines.

and neutral. The relationship between ES and ER for along line is different from the case of the short line be-cause of the progressive change in the line current due tothe shunt capacitance. If Es and ER are considered asphase-to-neutral voltages and Is and Ia are the phase cur-rents, the classical equations relating the sending-end volt-ages and currents to the receiving-end quantities are:--Es= ER cash (sdzy)+IR z sinh (~4) (2)Y

Is=% sinh (sG)+In cash (sd&) (3)lzIIThe susceptance, y, heretofore has been used most fre-

quently in these expressions. However, with the advent ofthe new form of tables giving characteristics of conductors,the shunt-capacitive reactance is obtained as a function ofthe conductor size and equivalent spacing. The reciprocalof y, which is x is therefore a more convenient quantity touse. For this reason the concept of shunt-capacitive re-actance is used through t/ ut this chapter. Eqs. (2) and (3)then become :

(4

(5)where z is the series impedance of one conductor in ohmsper mile, z is the shunt impedance of the line in ohms permile, s is the distance in miles.

z= -jx(10)6X * = capacitive reactance in megohms per mile.

Equations (4) and (5) can be written conveniently interms of the conventional ABCD constants.4 For the caseof a transmission line the circuit is symmetrical and D isequal to A. (Refer to Chapter 10, Section 21 for definitionof ABCD constants.)

Es=AER+BIR (6)Is=CER+DIR=CER+AIR (7).ER=AEs-BIS (8)IR= -CEs+DIS= -CEs+AIs (9)

whereA=cosh (s&)=cosh $

*This quantity can be obtained from the tables of conductor char-acteristics in Chap. 3. It is given in megohms in tables a s it is thenof the same order of magnitude as the inductive reactance.

in whichZ=zs and Z=c. S

(12)

The values of the hyperbolic functions can be obtainedfrom tables2 or charts3 or from evaluation of their equiv-alent series expressionscash (s$)=cosh 8=(l+;+;+;+.-) (13)

Fig. 3Variation of the real and imaginary components of A,B, and C for a 795 000 circular mils ACSR, 25-foot equivalentspacing, transmission line.T =0.117 ohm per mile.2 =0.7836 ohm per mile.x = 0.1859 m egohm per m ile.


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Regulation and Losses of Transmission Lines 267

sinh (s$)=sinh B(B+E+E+E+. . . .) (14)Expressed in terms of their equivalent series expansions,

ABC constants become

The series are carried out far enough so that the ABCcan be determined to a high degree of ac-However, for lines approaching one quarter wavethe series do not converge rapidly enough. In suchcase it is better to determine the ABC constants for thein two sections and combine them as described in10, Table 9.The ABC constants can be determined easily for anyof line by an evaluation of the cash and sinh func-

using the hyperbolic and trigonometric functions.procedure is outlined briefly here.J

-e=s i=cx+jpcx and p are in radians.

cash 8 = cash Q! cos @+j sinh a! sin /3sinh 0 = sinh (I! cos p+ j cash CY in pE=+ cacash cy=- 2c-e --Qsinh a=- 2

Figure 3 shows the variation of the ABC constants as aunction of line length for the line of Fig. 18. The real andimaginary parts of A, B, and C are shown for a completewave length.3. The Equivalent ?r of a Transmission Line

There are several equivalent circuits that represent theabove transmission line equations and thus can be usedfor the representation of transmission lines. One suchcircuit is the equivalent 7r shown in Fig. 4.Referring to this figure the equations relating theterminal conditions for this circuit are

Fig. 4Equivalent a circuit for representing long transmia-don lines.By equating like coefficients of the equivalent Eqs. (18)and (6) Z 0q=B (20)1+$%4 (21)Giving for the equivalent impedance Z&

(22)Expressed in terms of the corresponding hyperbolicfunctions and their equivalent series the equations for the

impedances areZ, = l/ZZ sinh

dZZ

+ 23 Z*30 240Z13- 1207 6OOZ*

+. . .

z2.-720Z12

(23)

4. Equivalent T of a Transmission LineAnother equivalent circuit for a transmission line isshown in Fig. 5. The equations for the impedances of thiscircuit areZ -L+2Lp- 17z3T=:-=-- 122 120Z2 20 16OZ

+ 3124 3 628 8OOZ*- - ->z+;=zt l-$+&2-l;;2;z,3(

+ 127Z*604 80024- l *

(25)

(26)

Fig. 5Equivalent T circuit for representing long transmis-sion lines.


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268 Regulation and Losses of Transmission Lines Chapter 95. Comparison of the Equivalent r vs. ABCD Con-stants

The choice of the use of the equivalent 7r vs. ABCD con-stants in calculating transmission-line constants is largely amatter of personal preference. However, each offers cer-tain advantages over the other. When the network calcu-lator is to be used, it is necessary to set up an actual circuitin the form of the equivalent 7r. The equivalent 7r affords abetter physical picture of transmission-line performanceand makes the comparison between long and short linesand the effect of charging current easier to visualize.

On the other hand, when a problem is to be solvedanalytically, the use of ABCD constants has a definite ad-vantage over the equivalent 7r because of the availabilityof the independent check: AD-BC= 1. This is partic-ularly desirable when other circuits are to be combinedwith the transmission line circuit.

The equivalent 7r or ABCD constants can be used to rep-resent any line, section of line, or combination of lines andconnected equipment. Either one represents accurately allconditions at the two terminals of the system. The equiv-alent circuit or ABCD constants being considered here per-tains only to a single line or line section. The generalequivalent circuit and general ABCD constants, if sodesired, can be determined by the combination of theequivalent circuits for the rest of the system as discussed inChapter 10.6. Expressions for Transmission Line Constants byFirst Two Terms of Their Series

When considering the accuracy with which transmissionline circuit constants need be determined, it should berealized that the resistance, inductance, and capacitance ofa line can rarely be known to within 3 or 4 percent andprobably never within one per cent. This is due to con-ductor sag, its variation with different spans, and the varia-tion that exists in conductor spacing together with theeffects of temperature upon conductor resistivity and sag.For this reason equations for the above circuit constantsthat are accurate to within 0.5 percent should be satis-factory.The effect of neglecting all but the first two terms of theseries in the above expressions can best be shown by con-sidering an actual line. For a 300-mile line with 250 000circular mil stranded copper and a 35-foot spacing thethird term in all of the above series expressions is largerthan normal.

For this line, from the conductor tables of Chap. 3r=O.237 ohms per mile~=x,+.rd=O.487+0.431=0.918 ohms per mile

~=~,+.~~=0.111+0.106=0.217 megohms per mileZ = rs+j.cs = (77.1 +j275.4) ohms

Z'= q'106-= -j723.3 ohmsZ 77.1 Ij275.4rl44 - j723.3 - -0.3807+jO.1066

z2- = 0.0056 - jO.003424Z2This term is thus about 0.6 percent of one (the first term).For the third term in the expression for ZT.

Z2- = 0.0011 - jO.00067120Zf2which is about 0.1 percent of one (the first term).

For all the rest of the constants the term is less than0.1 percent.Since these terms vary with the fourth power of thelength of the line, they decrease rapidly for lines less than300 miles in length and can be neglected. For instance fora 150-mile line the terms are one-sixteenth as large as fora 300-mile line.

Thus the above transmission line constants can be ex-pressed sufficiently accurately by the following equationswhich were derived from Eqs. (15), (16), (17), (23), (24),(25), and (26) by neglecting all but the first two terms ofeach series expression.

B=Z,,=lOOrS l-( &>+jlOOxS l-zs2+Tzsz

( GOOx 600~~C=$[(l-s)+j$]10-4

Z&= -j$[(l-&,)+j&]lO*

zr=5o~s( 1 +S)

+j5OxS ( T2S21 t s-m )Z~=-~$[(l+&)-j$]10(

(27)

w9(29)(30)

(31)

(32)In these equations:

S=length of line in hundreds of miles.x and r are in ohms per mile, and x in megohms per mile.

7. Simplified Method of Determining the I mped-ances of the Equivalent r Circuit for Transmis-sion LinesThe following method greatly simplifies the determina-tion of the impedances of the equivalent 7r circuit and still

enables them to be determined to within 0.5 percent forall practical power transmission lines.Equations (28) and (30) can be expressed in the follow-ing form: Z,, = lOOrSK,+ jlOOxS& (33)

z;,= -j2$(k.+jk,)lO* (34)gz=0.1335-jO.08117L

For the third term in the series expression for A.where


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Regulation and Losses of ransmission Lines 269

K,=l-

k,=l-

k 7S2==izG2

r2sz (36)

(37)

Examination of the above equations shows that for afactors K,, K,, and k, differ from 1 by athat is proportional to the square of the length of theHowever, a study of the characteristics of lines whichis economical to build and that have been built in theStates reveals that for a given length the variancethese correction factors from a mean is very slight. Init is only the lines with smaller conductor sizesequivalent spacings for which the correction factors

TABLE 1 MINIMUM CONDUCTOR SIZES AND SEPARATIONSFORWHICH THE MEAN VALUES OF THE CORRECTION FACTORSARE APPLI CABLE TO AN ACCURACY OF WITHIN ONE-HALF OF ONE PERCENT

Fig. 6Correction factors for, the equivalent ?r transmissionline impedances and AB C constants at 60 cycle s.

Table 1 gives minimum conductor sizes and spacings forvarious lengths of line for which the use of mean correctionfactors will give sufficient accuracy. For lines up to 300miles in length with conductor sizes and spacings equal toor greater than given by this table, the use of mean valuesfor I(,, K,, and k, gives an accuracy of within 0.5 percent.The correction factor k, is never greater than about 0.005and can be neglected. Thus, the shunt impedance Z&can be considered as a pure capacitor.In Fig. 6 are plotted the curves for K,, K,, and k, as afunction of line-length. The values on these curves con-form to those of the most common type of line construc-tion that is used for a given line length. Thus, in most

S =length of line in hun dreds of miles.T- =condu ctor resist ance in ohms per mile.z =inductive reactance in ohms per mile.x = capacitive rea ctance in megohms per mile.

be expressed to sufficient accuracy as parabolic equationsof the type 1 -KS2. In Table 2 are tabulated the cor-rection factors expressed in this form. The curves con-structed from these equations conform closely to the curvesof Fig. 6. Table 2 shows that K, can be considered as 1 upto 50 miles, K, as 1 up to 75 miles, and k, as 1 up to 100miles. Since in practically all cases the individual sectionsof line to be considered are not over 100 miles long, thecorrection factors can be neglected entirely if an accuracyof better than lyz percent is not desired. The largestdeviation from unity is in K, which at 100 miles is only1.4 percent.

- -cases the use of these values will give an accuracy con-siderably better than 0.5 percent. The factors can also

Example I-As an example of the use of this methodin determining the equivalent 7r of a transmission line, con-sider a three-phase, GO-cycle, 230-mile line of 500 000circular mil stranded copper conductors at an equivalentspacing of 22 feet.

From the Tables of Chap. 3Table 2EXPRESSIONS FOR THE CORRECTION FACTORS FORTHE EQUIVALENT r IMPEDANCES

T = 0.130 ohms per milex =0.818 ohms per milex = 0.1917 megohms per mileFrom the curves of Fig. 6 for a 230 mile line

K, = 0.93 1K, = 0.9G4k, = 0.982S is the length of the line expressed in hundr eds of miles. From Eqs. (33) and (34) or Fig. 6


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270 Regulation and Losses of Transmission Lines2, = (0.130) (230) (0.931) +j(O.SlS) (230) (0.964)

= (27.8+j181.4) ohmsZ&= -2 ;3gd7(0.982) ( 104)2w = -j1635 ohms

The equivalent circuit for this line is shown in Fig. 13.8. Adaptation of Simplified Method of DeterminingEquivalent 7r to Determining ABC Constants

The foregoing method can be adapted with an acceptabledegree of accuracy to determining the ABC constants of atransmission line. The ABC constants of the line should bedetermined by a more accurate method if the line is to becombined with other circuit elements. Eq. (27) can bewritten as follows:

A =K,fji(l-K,), (39)where XS2K,=l--200s.Since K, is the same form of correction factor as K, (Eq.(35)), a new curve for the correction factor can be plottedas shown in Fig. 6. The constant A is readily obtainedfrom the correction factor K, and Eq. (39). The constantB is equal to 2, and is determined through the use of thecorrection factors K, and K, of Fig. 6.From Eqs. (16) and (17) it can be seen that

c- B - Bxm6ZZ xx- jrx (40)where

r= conductor resistance in ohms per mile.x=inductive reactance in ohms per mile.X' = capacitive reactance in megohms per mile.

Example I(a)-Determine the ABC constants of thetransmission line of example 1.From the curves of Fig. 6, for a 230-mile line

K* = 0.897From the curve for K* of Fig. 6 and from Eqs. (39) and(40)

A=0.897+~~~*130( l-0.897)=0.897+jO.O164

B=Z, =27.8+j181.4 ohms (from example 1)(27.8+j181.4)(10e6)

= (0.818) (0.1917) - j(O.130) (0.1917)I=I 0.00000639+j0.001156

II. REGULATION AND LOSSES9. Analytical Solution for Voltage Regulation ofShort Lines from Known Receiver Conditions

The commonest type of regulation problem is one inwhich it is desired to determine the voltage drop for knownreceiving-end conditions. For the solution of this problemit is more convenient to make ER the reference vector asshown in Fig. 7(a). Unless denoted by the subscript L allvoltages will be taken as line-to-neutral voltages. If line-

Chapter 9

(b) FOR KNOWN SENDING END CONDITIONSFig. I-Vector diagrams for determining voltage regulation ofshort lines.

to-line voltages are applied to the following voltage equa-tions the impedance drop must be multiplied by fi forthree-phase lines or by 2 for single-phase lines.In the follow ing equations, (41) through (61), the sign ofthe pow er factor angle c$, depends upon w hether the currentis lagging or leading. For a lagging pow er facto r, 4 an d sin d ,are negativ e; for a leading pow er fact or, C#J nd sin 4 arepositiv e. The cos of t$ is posit ive for either lagging or leadingcurrent.

ER = & = referen_ceI =I cos +R+jI sin c$aZ=R+jX=rs+jxsEs=ER+IZ (41)orEs= (J!&+~R cos &-IX sin &)+j(fX cos $n

+IR sin $a). (4%Jn magnitude

(43)If the 7R a,nd IX drops are not over 10 percent of J%,Es can be determined for normal power-factors to within ahalf percent by neglecting its quadrature component.ThenEs = ER+I;R cos ~#JR TX sin $a (44)

The voltage regulation of a line is usually consideredas the percent drop with reference to ER.

Percent Reg. = lOO(E, -E,)E R (45)


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Regulation and Losses of Transmission Lines 271For exact calculations formula (43) can be used withUsing the approximate formula (44) Eq. 45 can be

1OOs;i:Percent Reg. = _E (r cos C#Q-x sin (by)R (46)The load in kva delivered to the receiving end of ae line is given by the equation--KVA _ 3&J _ u~EJ

1000 1000 (47)where EL is the line voltage at the receiving end.expressed in terms of the load and the line-voltage can be written

Percent Reg. = 100 OOO(kva)(s)-2EL (r cos $R-x sin 4~) (48)These equations show that the amount of load that cantransmitted over a given line at a fixed regulation varieswith its length. Using the regulation calculated

these equations to determine the receiver-end voltagegive this quantity to y2 percent if neither the resist-reactive drops exceed more than 10 percent of the

voltage. The percentage variance of the regula-own correct value, however, may be great,upon its actual magnitude and for this reasonequations are not accurate for determining load limits

Example 2The use of these equations can be illus-by calculating the regulation on a three-phase linemiles long having 300 000 circular mil stranded copperat an equivalent spacing of four feet and carry-load of 10 000 kva at 0.8 power-factor lag and aline voltage of 22 000 volts.

r=O.215 oh ms per mi and x = 0.644 ohms per mi.Applying Eq. (48)Reg. =(100 000)~10 000) (5)

(22 000)2 (0.215) (0.8) - (0.644) ( -0.6) 1Reg. = 5.8yoVoltage Regulation of Short Lines from KnownSending-End Conditions

To calculate the receiving-end voltage from knownd conditions it is more convenient to use Es asreference vector as shown in Fig. 7(b). For this case

Es = J!?S= referenceER=Es-IZ (49)(50)(51)e quadrature component of ER:

ER=E~--IRcosc#Q+TX sin& (52)Problems Containing Mixed Terminal Condi-tions

Sometimes problems are encountered in which mixedconditions are given, such as load power factor

and sending-end voltage, or sending-end power factor andreceiver-end voltage, and it is desired to determine theunknown voltage for given load currents. Such problemscan not readily be solved by analytical methods. Forinstance, if it were desired to determine the receiver voltagefrom known load power factor, sending end voltage, andcurrent, it would be necessary to solve for ER in Eq. (43)by squaring both sides of the equation and obtaining aquadratic equation for ER. This is somewhat cumber-some. Trial and error methods assuming successive valuesof one of the two unknown quantities, are often more con-venient. Also, it is sometimes found easier to solve suchproblems by graphical means. The more important prob-lems of this type can be solved by use of the Regulationand Loss Chart as shown in Sec. 28(d) of this chapter.12. Taps Taken Off Circuit

Quite frequently the main transmission circuit is tappedand power taken off at more than one point along the cir-cuit. For such problems it is necessary to solve each in-dividual section in succession in the same manner as dis-cussed above, starting from a point at which sufficientterminal conditions are known.13. Resistance Losses of Short Transmission Lines

The total RI2 loss of a three-phase line is three times theproduct of the total resistance of one conductor and thesquare of its current.Loss = 3Rr2 in watts. (53)In percent of the delivered kw. loadPercent Loss= - 173rs1EL ~0s ~JR

It is sometimes desired to determine the amount ofpower that can be delivered without exceeding a givenpercent loss. This is given by

KW= '% cos2 +R (yo Loss)100 OOOrs (55)

This equation shows that the amount of power that canbe transmitted for a given percent loss varies inverselywith the length of the line and directly with the loss.14. Regulation of Long Lines from Known ReceiverConditions

The effect of charging current on the regulation oftransmission lines can be determined from the equivalent?r circuit. In Fig. 8(a) are shown the vector diagrams forthe case of known load conditions. The voltage drop inthe series impedance Z, is produced by the load current

ER1~ plus the charging current z flowing through the shuntimpedance at the receiver endif the line. For a given linethis latter current is dependent only upon the receivervoltage ER.

There are two methods of taking this charging currentinto account. One of these is to determine first the net> that flows through Zm togetherangle &. Using the equivalent


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Fig. E&-Vector diagrams for determining voltage regulation oflong lines.series impedance Z,, an d this current instead of the loadcurrent all of the analytical expressions developed forshort lines are applicable. The equivalent, terminal con-ditions to use are shown in Fig. 8 (a).Example 3As an example of the use of this methodconsider the line of example 1, operating at a line voltageat the receiver end of 110 kv delivering a load current IRof 50 amperes at 0.9 power-factor lagging.

En = (1 lO,OOO+jO)/& = 63,5OO+jOIR = 50e-j25.80 = 5O[cos ( - 25.8) +j sin(

= 45 - j21.8 amps- 25.8)]

I,,=I,+I~=45-j21.8+j38.8=45+j17=48.1~20~70Zeq= 27.8+j181.4 = 183.5?.280Es = 63,500+ (48.1~~~~-~)183.5~~~.~~)

= 61,7OO+j8640*Sine of negat ive angle is (-), of positive an gle is (+).

15. Regulation of Long Lines from Known S ending-End ConditionsFor this case the equivalent current flowing through

Z& can be determined as the difference between IS andI; the current in the shunt reactance at, the sending endof the equivalent circuit. The vector diagram and equa-tions for this case are shown in Fig. 8 (b).16. Effect of Line Capacitance on Regulation Ex-pressed in Terms of a Correction Factor

As an alternative method the voltage relations can bedetermined in a form equivalent to adding a correctionfactor to the terminal voltage instead of to the current.This method has an advantage in that an average valuecan be taken for this correction factor which is a functiononly of the length of the line.Referring to the vector diagram of Fig. 8(a) for knownreceiving-end conditions and lagging power-factor, it isseen that the vector equation for the sending-end voltageEs can be written in the following form in terms of theload current IR and receiving-end voltage ER if the cur-rent IRf is expressed in terms of ER:

cos +R+Repf~ sin &t (56)When the quadrature component of Es is neglected, itsmagnitude can be expressed asEs= l-xe( ) ER+IL~IR cos +R--XJR sin +R (57)eqFrom the same considerations that enabled average

values to be taken for the correction factors of the equiv-alent 7r impedance discussed in Sec. 7 an average valuecan be assumed for 2 in Eq. (57).eq

x-4ry- = 0.0201s2 (58)09where S is the length of the line in hundreds of miles. Anapproximate expression can thus be obtained for the regu-lation of long lines similar to that of Eq. (46).

1OOIRApp. y0 Reg. = ~(~.,~os~~-X~~sin~~)-2.01~~(59)R

Similar analysis can be applied to problems involvingknown sending end conditions. A comparison of Cqs.(59) and (46) shows that when Z,, is used for long lines,the equations are of the same form with the exception ofthe correction factor ( -2.01S2). For lines up to 100 milesin length short line formulas can usually be applied to agood degree of accuracy by merely adding this term to theresult. This, of course, neglects the correction factors K,and k, for Z,,.17. Determination of Voltage at IntermediatePoints on a Line

The voltage at intermediate points on a line may becalculated from known conditions at either terminal bysimply setting up the equivalent circuit for the line be-


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Chapter 9 Regulation and Losses of Transmission Linestween the terminal and the intermediate point. For theline thus set up any of the methods given above may beused.18. Resistance Losses of Long Lines

The effect of charging current on line losses can betreated as in Sec. 14 for regulation. Referring to Fig. 8the loss can be considered to be due to the currentIe,=IR+IR=IS-19 flowing through the equivalentresistance (n,,).Thus in terms of the load current

Los~=312,,(1~+1~)~ watts (60)-- -2ERsin 9n+z2 1atts (61)eaIII . CIRCLE AND LOSS DIAGRAMS

Equations for line currents, power, and resistance lossescan be expressed as functions of the terminal voltages andsystem constants. Such equations and graphical repre-sentations of them are found convenient not only for themore common types of performance problems but also inconnection with system stability. The graphic form ofthe power and current equations are very similar and areknown as circle diagrams. Of these the power circlediagram is the most important. In the past this diagramhas been primarily limited in its use to transmission sys-tems. However, it is thought that if its simplicity and theclarity with which it depicts system performance arebetter understood, it will be applied more frequently toboth transmission and distribution problems.19. Vector Equations for Power

In previous editions of this book, lagging reactive powerwas considered as negative and leading reactive powerpositive. This conformed to the standard adopted by theAmerican Institute of Electrical Engineers at that time.The convention has now been adopted as standard by theInstitute that lagging reactive power be considered aspositive and leading reactive power negative. Using thisnotation the vector expression for power can be written asthe product of the voltage and the conjugate of thecurrent. P+jQ=Ef (6%

This can be shown with reference to Fig. 9.E = E cos O ,+ j,@ sin 8,I =I COS Oi+jT sin 0iI=1 cos ei-j1 sin 0i

Ef=E (cos O,+j sin &)I (cos ei---j sin 0i)--= EI [(cos 8, cos Bi+sin 8, sin 0,) +j(sin 8, cos 8i- cos 8, sin ei)]

since, cos (&---0,) =cos 8, cos Bi+sin 8, sin 0iand sin (0,---0J = sin 0, cos 0i-cos 0, sin 8i-- --Ef=EI cos (O,--Oi)+jEI sin (0,-e,)

Let + be 8,--8i; then for lagging or inductive powerfactors 4 is positive and

273

Fig. 9Diagram for determining the vector equation forpower.

For leading or capacitive power factor, 4 is negative andthe imaginary component will be negative. A completediscussion of the direction of the flow of reactive power isgiven in Chap. 10, Sec. 2.20. Current and Power Equations and Circle Dia-grams for Short Lines

Using the above notation the per phase power at eitherend of a line is given by the product of the line-to-neutralvoltage and the conjugate of the current at the particularend in question. If Is is chosen as positive for currentflowing into the line, positive sending-end power indicatespower delivered to the line; and if In is taken as positivefor current flowing out of the line, positive receiving-endpower indicates power flowing out of the line.Referring to Fig. 1: Is=IR=I

Ps+jQs=&f&+jQR=ERf

The current canvoltage as follows :be expressed in terms of the terminal

I=~s-E~---z---

also f=H2

(63)Thus

&&-J%BRPs+jQs= 2-ERER+ER&

PR+j&R=- 2

If ER be taken as the reference, then Es = Escjeacd--E&!?s=E~; ESBR =ESER E~~; EnBR =E&; and ERES=ERESE .je The expressions for sending- and receiving-end powerbecome --Ei ESERE~Ps+jQs=2--~

--P+jQ=Ef=EI cos $+jEI sin +. the vector to the center of the sending end circle diagram.

(64)

(65)The sending and receiving end real and reactive power isthe sum of two vector quantities. Furthermore, if thevoltages Es and ER are held constant, there is only one re-

maining variable, 8. The interpretation of Eqs. (64) and(65) in the form of power circle diagrams is an importantconcept. Tts simplicity is self evident by referring to Eq.(64) and Fig. 10.

The first term 2 is plotted as shown on Fig. 10 and isL2


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274 Regulation and Losses of Transmission Lines Chapter

Fig.10Power circle diagram f or short lines.--The second term EsERej--, which is the radius of the circle2is added to this first term so that the resultant is the send-ing end real and reactive power. A complete sending endcircle diagram is obtained by first determining the center--of the circle from I?; EsE&-, and second, the radius- --)2 2letting 0 = 0.The receiving-end circle diagram is obtained in the samemanner.Equations (64) and (65) can be reduced in general termsto Cartesian coordinate form in which the real and reactiveparts are separated. However, it is simpler to insert theproper numerical values in the vector and conjugate formand solve by polar and Cartesian coordinates, from whichthe circle diagrams can then be plotted.

If Eq. (65) is reduced to Cartesian-coordinate form itcan be shown that the maximum power that can be re-ceived over the line is obtained when 6 is equal toy = tan-l 4, the angle of the line impedance.r The expres-sion for the maximum receiving power is--

PR max= -++E+s 036)It can also be seen from Fig. 10 that PR is maximum when8=7.

When the line-to-neutral voltages are expressed in volts,

the coordinates of the diagram are per-phase real volamperes and per-phase reactive volt-amperes. When epressed in kilovolts, the coordinates become thousands kilowatts and thousands of reactive kilovolt-amperes.Total three-phase power is three times the per-phasepower. All of the expressions for power written contain--products of Ei, Ei, or EJ3R. When given in terms line-to-line voltages, they are all three times as great when line-to-neutral voltages are used and thus thequations then represent total three-phase power.Referring to Fig. 10 for the operating condition indcated by the given angle 8 the point A of the power circdiagram shows the value of Ps and Qs being delivered the line at the sending end and the point B the valuof PR and Qn drawn from the line at the receiver end. Thdifference between Ps and PR is the RI2 loss of the linitself for this operating condition.The value of Q at each end is the reactive power whicmust be supplied to the line in the case of the sending enor drawn from the line in the case of the receiving end order to maintain the chosen terminal voltages. At threceiving end the reactive power drawn Gy the load itseat the particular load power factor may not be equal that required to maintain the desired voltage. If a sychronous condenser is used at the receiving end, the dference must be supplied by the condenser to maintain thvoltage.It will be noted that for a given network and givevoltages at both ends there is a definite limit to the amounof power which may be transmitted. If the angle 8 increased beyond this point, the amount of power transmitted is reduced. The critical value of 8 for this conditionwas shown by Eq. 66 to be 8 = y. The only way the powlimit may be increased for a given network is by increasingthe voltage at either or both ends. Increasing the voltagat one end increases the radius of both circles in direcproportion and moves the center at that end only awafrom the origin, along a line connecting the original centeto the origin, proportional to the square of the voltage that end. Where the network is subject to change, changein network constants will also change the power limiReferring to Fig. 10 and Eq. (66)) it is evident that decrease in the magnitude of 2 will result in an increase the power which may be transmitted. Thus any changwhich decreases the series impedance such as the additionof parallel circuits will increase the power limit.

Since the conjugate of the phase current, in amperes, the per-phase power in volt-amperes at either end divideby the phase voltage at the same end, either the sendingend or receiving-end power circles, when placed in thproper quadrants, can be used to represent the locus of thcurrent with a proper change in scale of the coordinates.Referring to the sending end circle diagram of Fig. 1Ps+jQs=Ef and for the point A, Qs is positive lagginreactive power. Therefore the imaginary component of thconjugate of the current is positive; the imaginary component of the current is negative. If the power circle digrams are rotated about the real power axis so that thcenter of the sending-end circle is in the fourth quadrantI?;

-z- will then be the vector to center), and the center


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Chapter 9 Regulation and Losses of Transmission Lines 275the receiving end circle is in the second quadrant, then the gether with the quantities for laying out the diagram.power circle diagrams properly represent the current circle Since the coordinate of the center of the power circlesdiagrams if the appropriate change in scale of the coor- depends only on ER which is fixed, all the circles have thedinates is made. Lagging reactive current is negative and same center but different radii corresponding to the differ-leading reactive current is positive. ent values of sending end voltages.If the sending-end circle is used the current is referredto the sending end voltage as the reference vector and thecoordinates should be divided by the sending end voltage.For instance, if the sending-end power diagram were con-structed using line-to-line voltages in kilovolts resulting inpower coordinates given in thousands of total three-phasekilovolt-amperes, the power coordinates should be dividedby 43 times the line-to-line sending end voltage in kilo-volts giving current coordinates in thousands of amperes.If the receiving end circle is used, the current is referred tothe receiving end voltage as reference. For the currentcircle diagrams the angle 0 still, of course, refers to theangle between the two terminal voltages.

Examination of this figure shows, for example, that themaximum load at 0.9 power factor lag which can be carriedby the line at 5 percent regulation without reactive powercorrection is that indicated by point A or about 2600 kw.If it is desired to transmit a load of 5000 kw indicated bypoint B, the regulation would be about 11 percent withoutrkva correction. To reduce the regulation for this load to5 percent would require that the receiver and load condi-tions be that indicated by the point C, and it is evidentthat about 2400 lagging reactive kilovolt-amperes must besupplied to the receiver end of the line to attain this condi-tion by having capacitors or a synchronous condensersupply that amount of lagging reactive kilovolt-amperes.For a study of the performance of a system it is some-

times found convenient to plot on the power circle diagrama family of circles corresponding to various operating volt-ages. The most common case is one in which the line is tooperate at a fixed receiver voltage and it is desired to de-termine the line performance for various sending-end volt-ages. For such a case the receiver diagram is usually allthat is needed.

21. Current and Power Equations and Circle Dia-grams for Long LinesRepresenting long lines by their equivalent r circuit asshown in Fig. 6 results in modifying the form of the simpleshort line equivalent circuit by the addition of the shunt

capacitive reactances at each endExample 4-An example of this type of problem isshown in Fig. 11. There the line constants are given to- Thus the equations for the terminal currents have anadditional term as shown in Fig. 8.

Fig 11Family of receiver power circle s for a 15-mile line withNo. 0000-19 strand-copper conductors and I-foot equivalentspacing.

Receiver voltage E, = 22-kv line-to-line.T =0.303 ohm per m ile.z =0.665 ohm per mile.Z = zs = 10.94 ohms.

I s- Eg+g; r,=7 2,.s-&i; 8seq eq -4 eq (67)I Es-& ER &-3, $RRZ------

z eq -z'; fR=------;-.eq s e q z d q (68)The sending- and receiving-end power is determined inthe same manner as for the short line.

Ps+.iQs = EsfsL eq L -4 &?2eq

Rewriting Eq. (69) in a slightly different formPs+jQs=(~+~)-~.

Similarly for receiving end power:(70)

A comparison of Eq. (70) with (64), and (71) with (65)shows them to be of the same form consisting of a fixedvector with a second vector constant in magnitude butvariable in phase, added to it. The power circle diagramcan be plotted as shown in Fig. 12. The circle diagram ismost easily obtained by the numerical and vector substitu-tion for the voltages and impedances. The center and theradius of the circle can then be calculated by reductionusing a combination of polar and Cartesian coordinates.Example 5 illustrates the method and shows the powercircle diagrams which are obtained in Fig. 13.


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276 Regulation and Losses of Transmission Lines Chapter

Fig. 12Power circle diagram for long lines.In Eqs. (70) and (71) the terms

are not a function of the angle 8 and therefore add directlyto the short line fixed vector so that the effect is to sh iftthe center of the power circles in the direction of volt-amperes only. The presence of the shunt reactancesdecreases the amount of positive reactive volt-amperes putinto the sending end of the line for a given amount of realpower and increases the positive volt-amperes delivered atthe receiving end. This decreases the amount of leadingreactive volt-amperes which would have to be absorbed bysynchronous condensers or capacitors for a given loadcondition. It does not affect the real power conditions fora given operating angle or the load limit of the line. Thesefactors are determined entirely by the series impedance ofthe line.Referring to Fig. 12, if the radius of the receiving-endcircle for 8=0 were plotted with the origin as the center,the vector would be at an angle y with the real power axis.The angle indicated on Fig. 12 is therefore equal to y,the angle of the equivalent series impedance. The max-imum real power that can be delivered over the line occurswhen 8 = y.

The current circle diagrams for the sending- and re-ceiving-end currents can be obtained as discussed in Sec.20. The sending-end current diagram is obtained from thesending-end power circle and is referred to the sending-end voltage vector as reference. The receiving-end currentdiagram is obtained from the receiving-end power circle andis referred to the receiving-end voltage.

Fig.13Equivalent circuit and power circle diagram for a 2mile l ine with 500 000 circular mll. stranded copper conductoand an equivalent spacing of 22 feet.Opera tin g volta ges; ES = 230-kv, El< = 200-kv, line-to-line.For t his line T =0.130 ohms per mile.

z =0.818 ohms per mile.x=O.1917 megohms p er mile.

From curves of Fig. 6 for 230 milesK, =0.931K, = 0.964k, = 0.982

Z,, = (27.8+j181.4) ohms ; ZIeq = -j1635 ohms .Example 5 Fig. 13 shows th e power circle diagra

constructed for an actual line.The power circle diagrams are obtained from Eqs. (and (71). If line-to-neutral voltages in kv are used, results must be multiplied by three to obtain real and active power in mw and mvar. If the line-to-line voltagin kv are used, the results are three-phase power in mand mvar. .a E;Vector to center = z+z

(230)2= 27.8 --j181.4

+ (23OY+j1635

(230)2 + (230)2= 183.4~-~~~~~~ 1635~~~~


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Chapter 9 Regulation and Losses of Transmission Lines 277= 288d81.28+32.4E-fgo= 43.6+j284+32.4 = 43.6+j251.6--EsE&Radius of the sending end circle = - v for 8=0.z w

230 x 200=- 27.8-j181.4 = _ 25 1p.28 = - 38.0 -j248Ps+jQs (for 0=0) =43.6+j251.6-38.0-j248=5.6+j3.6Similarly for t,he receiving circle:

E,z ER2Vector to center = -If -rsil- (200)2 (200)2-- = -33.0 -jr90.5= 27.8~$81.4 +j1635.--E&tfor 8 = 0, Radius = Tzw = 38.0 +j248

and PR+~QR = -33.0-j190.5+38.O+j248 = 5.O+j57.5Figure 13 shows the power circle diagrams plotted fromthe calculated results given above. Suppose it is desired todeliver a load of 100 mw at 0.9 power factor lagging;i.e., P+jQ= lOO+j48. From the curves of Fig. 13, for a

delivered power of 100 mw the angle 0 is 23.5. The fol-lowing values from the circle diagrams are Ps+jQs = 108+jll and PR+jQR = lOO+j20. These values arendicatedon the diagram of Fig. 14. The arrow indicates the direc-

Fig. 14Reco rded values of power flow as obtained fromFig. 13 and Example 5.tion of positive real power flow. Inductive lagging re-active power in the same direction is positive and is thevalue in parenthesis. These designations and nomencla-ture follow present-day network calculator practice.At the receiving end there is a deficit of lagging reactivepower. A synchronous condenser operating overexcitedwould be required to supply 28 mvar. If the condenser isconsidered as a load the direction of the arrow can bereversed with a minus sign in front of the value for thereactive power. The synchronous condenser is then takingnegative, or leading reactive power.22. Current and Power Equations and Circle Dia-grams for the General Equivalent T Circuit

The circle diagrams are applicable to the study of theperformance of an overall system. Such a system can berepresented by an equivalent 7r circuit of the form shownin Fig. 15. For such a case the shunt impedances usuallyare not equal and have resistance components introducedby the presence of other equipment containing resistance.

If the shunt impedances take the completely generalform of 2s and ZR, the equations for sending- and re-

Fig. 15Power circle diagram for the general equivalent ?rcircuit.ceiving-end power can be written directly from equations(70) and (71).

fs+jQs=(~+$)-E~ (72)

and p,+jQ.=( -E?i)+sF (73)

The construction of the power circle diagrams is thesame as for the long lines as shown in Fig. 12. In the caseof the general equivalent a, Zs replaces Z,, at the sendingend and ZR replaces ZCq at the receiving end. The effectof resistance and reactance in the shunt branch at thesending or the receiving end can be visualized better if theimpedance is expressed in Cartesian coordinate form.Referring to Eq. (72), the second quantity in the first termbecomes

(74)This quantity is added to the short line vector to center,

-2ESz!,

This point as applied to the sending end circle diagramis illustrated in Fig. 15. The complete vector to center is-2Es E:shown as -+-;-, as the sum of the two individual vector

geq 2 , Equantities, and as the sum of the vector 2 and the Car-L@Rstesian coordinates - @Xs eqZA2 and jp zs2 -


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278 Regulation and Losses of Transmission Lines Chapter Referring to Fig. 15 and Eq. (74) the effect of resistanceis to shift the center of the circle in the direction of in-creased positive real power. A positive reactance shifts the

center in the direction of increased positive reactive power;a negative reactance shifts the center in the direction ofdecreased positive reactive power.In the case of the receiving-end circle diagram, the effectof resistance is to shift the center of the circle in thedirection of increased negative real power. A positivereactance shifts the center in the direction of increasednegative reactive power; a negative reactance shifts thecenter in the direction of decreased negative reactivepower.

The current circle diagrams for this case can be de-termined as discussed in Sets. 20 and 21.23. Loss Diagram

Although the resistance loss can be taken from the powercircle diagram, it can be obtained more accurately andconveniently from the Loss Diagram.

Loss = Ps - PRFor the case where the transmission line alone is beingconsidered

E; EsERLossFFR+R cos 8-X sin 0)--E2 EsER+FfR- r (R cos 8+X sin 0)--

= (@+&)!&2E+R COS e (75)The graphical representation of this equation is given inFig. 16. I

Fig. 16The transmission line loss diagram (when solving forgeneral equivalent A loss, substitute R ,, for R and z,, for z).

For the general equivalent 7r circuit, the equation forloss isLos s= @+,!?; 2+&i;[ 1

E2

eq c l )2--

+ R 2EsERR cos 8(2;) lx- z;, eq (76)

As shown by Fig. 16 this is equivalent to the formula fothe loss on the transmission line alone except for the terms

A!?; E;-23; and mRL which represent the losses in th6%) 2 Rresistance components of the shunt impedances 2; an2;.As was the case for the previous power equations, line-to-neutral voltages are used, the loss is on a pephase basis; and if line-to-line voltages are used the totalthree-phase loss is represented.An equation for the load which can be delivered at given percent line loss on lines regulated by synchronouscapacity is important in determining their performance.Upon the assumption of equal sending- and receiving-endvoltages a very simple approximate equation can be derived which gives an accuracy of a fraction of a percent ovethe practical operating range of loss and regulation. Whenloss is expressed as a percentage of PRY this equation is:

PR= y$$ Loss E:X;q[ 1loo+% Loss) R,qzZqA corresponding equation for &R is

(77

PR in Eq. (77) is, of course, independent of the loadpower factor and from Eq. (78) the required amount osynchronous capacity to maintain equal sending- andreceiving-end voltages for the delivered load PR can bobtained by subtracting the reactive kva of the loadfrom &R.24. Current and Power Relations in Terms of theABCD Constants

In many cases it is desirable to use ABCD* constantsbecause of the desirability of the check AD -BC = 1. Thisis particularly true where there are several combinations ocircuits including transmission lines, series impedances andshunt impedances. Expressions for sending and receivingend power can be obtained readily and the circle diagramscan be drawn. E&!=AER+BIR (7%

Is = CER+DIR (80ER=DEs-BIS (81IR= -CEs+AIS (82

Solution of the above equations for IS and IR gives:fj ZRIs=;Es-9; f,=B&-B.

IR=Ps+jQs=Esfs

=E 2 c ES*R-ssB B=E2g E&de

sB--z---*For definition of ABCD const ant s see Chap. 10 Sec. 21.

(83

(84

(85


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Chapter 9 Regulation and Losses of Transmission Lines 279

036)where

A=Al+jA2=A+; A=A1-jAz=Ae-jB= Bl+jB2=Zi@ fi = B1 - jB2 = BE-jfiD=Dl+jDz=DP fj=D,--jD,=~~-j8

The sending- and receiving-end power can be obtainedreadily from solution of Eqs. (85) and (86) by numericalsubstitution using polar and Cartesian coordinates. Eqs.(85) and (86) take the familiar form (see Sec. 20) of a fixedvector plus a vector of constant magnitude but variable inphase position. The circle diagram construction is shownin Fig. 17. The maximum real power that can be de-

Fig. 17Power ci rcle diagram in terms of ABCD constants.livered occurs when 0= p, which is the angle of the con-stant B. The angle p is indicated on Fig. 17.A breakdown of Eqs. (85) and (86) into their Cartesiancoordinate form gives the equation for loss in the form

Loss=Ps-PR=E2S E2R~(u,D~+B,D,)+B2(B~A1+B,A2) -T+037)

Further discussion of the use of ABCD constants andpower angle diagrams is given in Chapter 10, Sec. 21.

IV. TYPICAL TRANSMIS SIONLINE CHARACTERISTICS

In any detailed analysis of power flow, voltage regulation,and losses involving a transmission line circuit, each lineshould be considered individually with regard to itsspecific characteristics. However, for rough approxima-tions there are certain rules of thumb that apply to anaverage line and that can be used for orientationreasons.

A study was made of recently constructed transmissionlines in the United States in the voltage range from 69 to230 kv and Table 3 shows the results. This table is a goodrepresentative cross section of existing lines and gives im-portant characteristics of typical lines. The conductorsizes, spacings, and type of tower construction representthe most common usage. For the middle value of spacing,the characteristics of the aluminum conductor and itscopper equivalent are given to illustrate the differencebetween types of conductors. In previous years, copperconductors were used more frequently although the presenttrend seems to be toward the use of ACSR conductors.The spacings given were modified slightly in some in-stances so as to follow a smooth curve of spacing vs.voltage for the different types of construction. Regardingthe type of construction, it appears that the particularlocale dictates the material used. As a matter of fact, incertain sections of the world reinforced concrete poles areused because of the unavailability and high cost of eithersteel or wood.The 60-cycle series reactance in ohms per mile is givenfor each line in the table. The average of these values is0.7941 ohm per mile, which indicates that the rule ofapproximately 0.8 ohm per mile for a transmission line isapplicable. Frequently, it is desired to know the per-cent reactance per mile of a line and for convenience thisvalue is also given. The percent reactance varies directlywith the kva base so that for some base other than 100mva, the percent reactance can also be determined con-veniently.

As previously mentioned, the use of susceptance is lessat present because of the manner in which tables of con-ductor characteristics are given. The shunt-capacitivereactance in megohms per mile is therefore included in thistable. The susceptance can be determined by taking thereciprocal of the shunt-capacitive reactance. The suscept-ance is in micromhos per mile. Shunt-capacitive react-ance varies inversely with the distance in miles.The average value of the shunt capacitive reactances inTable 3 is 0.1878 megohm per mile. A good rule is that 0.2megohm per mile may be used for the shunt-capacitivereactance. It is significant to note that regardless of thevoltage, conductor size, or spacing of a line, the seriesreactance and shunt-capacitive reactance are respectively,approximately 0.8 ohm and 0.2 megohm per mile.

The charging kva per mile of line is a convenient valuefor reference and is given in column 9 of the table. Thisvalue varies with the voltage of the line. Some convenient


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280 Regulation and Losses of Transmission Lines Chapter 9TABLE 3. TYPICAL TRANSMISSIONLINE CHARACTERISTICS T 60 CYCLES

rules are given for estimating charging kva in the followingdiscussion.The surge impedance of a transmission line is numeri-

cally equal to 4zC It is a function of the line inductance

and capacitance as shown and independent of line length.A convenient average value of surge impedance is 400 ohms.As shown in the table, this value is more representative ofthe larger stranded copper conductors than it is for theACSR conductors. Compared to the average value of 356ohms from the table, 400 ohms is a good approximation.

resistance; i.e., Es =ER+fR. Surge-impedance loading initself is not a measure of maximum power that can bedelivered over a line. Maximum delivered power must takeinto consideration the length of line involved, the impe-dance of sending- and receiving-end equipment, and ingeneral all of the major factors that must be consideredwith regard to stability. The relation of surge-impedanceloading to line length, taking into account the stabilityconsideration, is covered in Chap. 13, Part IX.

Following is a summary of approximations that may beapplied to transmission lines for estimating purposes:

Surge-impedance loading in mw is equal to(kv L--d2Surge Impedance

and can be defined as the unit power factor load that canbe delivered over a resistanceless line such that the q2Xis equal to the charging kva of the line. Under this condi-tion the sending-end and receiving-end voltages and cur-rents are equal in magnitude but different in phase posi-tion. In the practical case of a line having resistance, themagnitude of the sending-end voltage is approximatelyequal to the magnitude of the receiving-end voltage plusthe product of the magnitude of the current and the line

1. Series reactance of a line=0.8 ohm per mile.2. Shunt-capacitive reactance of a line=0.2 megohm per mile3. Surge impedance of a line = 400 ohms. (km-d24. Surge-impedance loading, (SIL) in mw = 400 orinkw=

2.5(kvLpL)?5. (a) Charging kva for a hundred miles of line is 20.5 percentof the SIL.

(b) Charging kva of a line is also=5000 100L)(!F%$)*,where L= line length in miles,kvn-L = line- to-line voltage in kilovolts.


17/25


Fig .18Distribution of voltage and current along a 300-miletransmission line , 795 000 circular mils, ACSR conductor, 25-foot equivalent spacing.- Voltage._ _ _ Current

r =0.117 ohm per milex =0.7836 ohm per milez= O.1859 megohm per mile

The effect of the distributed capacitance of a transmis-sion line on the voltage and current distribution along theline is illustrated in Fig. 18. The calculated results arebased on a transmission line 300 miles in length, 230 kv,795 000 circular mils, and 25-foot equivalent spacing. Theloo-percent surge-impedance loading of the line is 139 000kilowatts. The current corresponding to this load at 100percent voltage is 348 amperes, The voltage and currentare shown as a function of the line length for 100 percent,50 percent surge-impedance loading at the middle of theline and for zero delivered load. The voltage at the middleof the line was maintained at 230 kv and Es and ER wereallowed to vary depending upon the load condition.At loo-percent surge-impedance loading, the voltagesEls = 240 kv and ER= 219 kv. The current is a constantvalue of 348 amperes. If the surge-impedance loading isassumed at the receiving end of the line, the magnitude ofthe current is slightly different at the sending end becauseof line resistance. The amount of this difference dependsupon the ratio of line reactance to resistance and the lengthof the line. Based on the calculated voltages of Es and ER,the regulation of the line is 9.5 percent. The value ofregulation as determined from the product of the magni-tude of the current and the resistance is also 9.5 percent.For 50-percent surge-impedance loading the current is aminimum value at the middle of the line. If the surge-impedance loading is taken at the receiving end, the cur-rent decreases to a minimum at the receiving end. In Fig. 18surge-impedance loading is taken in the middle of the line

for purposes of exposition. Generally the surge-impedanceloading should be considered at the receiving end becausethe delivered load is usually the quantity of most interest.V. 60-CYCLE TRANSMIS SION LINEREGULATION AND LOSS CHARTS

The voltage regulation and efficiency of a transmissionline or distribution feeder are fundamental properties ofits performance. In determining these quantities for ex-isting systems or in designing new systems to meet givenload requirements, it is thought that the charts presentedhere will save a great deal of time and labor that wouldin many cases be necessary if analytical methods were used.For low voltage lines without synchronous or staticcapacitors, voltage regulation is usually the more impor-tant consideration. For instance, in the design of a lineto carry a certain load one wishes to determine the propertransmission voltage and conductor size. Based on an assum-ed allowable regulation several voltages and conductor sizeswill be found to transmit the load, the final choice beingbased upon economics for which the line efficiency is de-sired. The performance of higher voltage regulated lines,however, is determined primarily by the line loss.The charts presented here were developed with thesetwo points of view in mind. Quite frequently it is desiredto obtain quickly an approximate solution. The QuickEstimating Charts afford a simple method for such cases.For more accurate calculations the Regulation and LossChart is provided. It is important to be able to considermore than just the line itself. The transformers are oftenthe determining factor in the choice of the proper line volt-age. The Regulation and Loss Chart is constructed so thatfrom the knowledge of the equivalent impedance of asystem its performance can be determined.25. Quick Estimating Charts

In Figs. 19 and 20 are plotted curves showing the powerwhich can be transmitted at five percent regulation to-gether with the corresponding percent line loss for variousvoltages and conductor sizes. These curves afford therapid estimation of such problems as the regulation for aknown load, the load limit of a line for a given regulationand the determination of voltage and conductor size for thetransmission of a given load at a given regulation. Fig. 21is an aid for interpolation between the values of powerfactor given on the curves.

The curves of Fig. 22 give the power which can be trans-mitted for various conductors and voltages at a line loss offive percent. These curves are most useful in determiningthe performance of lines regulated by synchronous or staticcapacitors,Charts Ba sed Upon Regulation- Fig. 19 appliesspecifically to stranded copper conductors, but it can beused for copperweld-copper conductors with an accuracyof two to three percent. Fig. 20 applies to ACSR con-ductors. The load which can be transmitted over a lineat a fixed regulation varies inversely with its length so thatfor a given line the actual load is the value read from thecurves divided by the line length. For 220 to 440-volt linesthe values on the curves are given in kilowatts times hun-


18/25


Fig. 19Quick Estimati ng Charts Based Upon 5 Percent Reg-ulation-stranded Copper Conductors.

The cur ves give load in kilowatt s X miles or kilowatts X hundredsof feet which can be received at 5 percent regulation togetherwith corr esponding line loss.

For a given lengt h of line, power is equal to value r ead from curvesdivided by length of line.

Power for other r egulations is approximately equal to values% Regrea d from curves mult iplied by ___5 *For power factors other than given in charts, multiply values read

from curves for un ity power factor by fractions given in Fig. 21.Percent loss for other regulations and power factors than found on

charts are given by equation(Percent Loss ) 2= Percent Loss)I X ~-- X(Kw Load) I (Power Fa ctor);For single phase lines divide power read from charts by 2 and

percent loss by d3.


19/25

Chapter 9 Regulation and Losses of Transmission Lines

Fig. 20Quick Estimating Charts Based Upon 5 Percent Reg-ulation-A.C.S.R. Conductors. dreds of feet. For higher voltages they are in kilowattstimes miles.

The cur ves give load in kilowatt X miles which can be receivedat 5 percent r egulation tog&her with corresponding line loss.

For a given length of line, power is equal t o value read from curvesdivided by length of line.

Power for other r egulations is approximately equal to values read7% Regfrom curves multiplied by 5,

For power factors other tha n given in charts, mu ltiply values readfrom cur ves for unity p ower factor by fra ctions given in Fig. 21.

Percent loss for other regulations and power factors than found oncharts are given by equat ion

(Percent Loss) 2 = (Percent Loss)I X (Kw Load)2 (Power FactorI?(Kw Load)1 (Power Fa ctor):For single phase lines divide power r ead from charts by 2 and

percent loss by 4%

For each voltage a common equivalent conductor spac-ing is assumed and the curves are drawn so that it is pos-sible to interpolate to a good degree of accuracy for othervoltages than those given. In addition the relationshipthat the power is proportional to the square of the volt-age may be used. Since the percent loss does not varymore than about a tenth of one percent for each conductorsize in each set, of curves, mean values are given asshown.For the same line voltage, conductor, equivalent spac-ing, and regulation half as much load can be transmittedon a single-phase two-wire line as for a three-phase line.For this reason the curves can be used to good accuracyfor this kind of line by simply dividing by two the loadread from them. For this single-phase load the percent


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Charts Based Upon Loss-In Fig. 22 (a) are plottedcurves for short lines which show the power in kilowattstimes miles which can be transmitted under two condi-tions. The solid curves are based on five percent loss andequal receiving- and sending-end voltages. These are use-ful for lines where little regulation can be allowed such ason interconnected systems. The dotted curves are for themaximum power which can be transmitted at the givenload voltage and five percent loss. For this condition theregulation varies but in no case does it exceed about fivepercent.Fig. 22 (b) is for higher voltage lines long enough thatdistributed capacitance of the line need be considered.Only the condition of equal sending and receiving endvoltages is considered here since regulation does not greatlyeffect the power for the conductors and spacings practicalto use. For all of these curves an arbitrary coordinate sys-tem has been used for the abscissa beneath which isplotted the correct sizes for the various conductors. Thecurves here are based on 10 percent loss.

Equation (77) was used for determining the curves forequal voltages at both ends of the line and its examinationshows that, for the practical range of losses, power forother values of percent loss are very nearly that read from70 Lossthe curves multiplied by ~5 or 10 If greater accuracy isdesired the factor 70 Loss100 + 70 Loss of Eq. (77) can be used.

Fig. 21Effect of power factor on load that can be carried at Eq. (55) was used for the curves based on the maximuma fixed regulation. power at five percent loss. For this case power is directlyCurves apply specifically for three foot equivalent spacing and five porportional to loss. For both sets of curves it is propor-

percent regulation, but can be used with good accuracy for normal tional to the square of the receiving-end voltage.spacing and regulation range. The power which can be transmitted over a single-phaseline is one half that of a three-phase line of the same equiv-loss will be that read from the charts divided by [fi alent spacing and line-to-line voltage. Thus Fig. 22(a) can(or 1.732)]. be used to good accuracy for single-phase lines by dividingCurves are presented for three common power factors: the values read from the curves by two.unity, 0.9 lag, and 0.8 lag. It is difficult to interpolate forother power factors, however, especially between unity

and 0.9. To facilitate this the curves of Fig. 21 are pro-26. Examples of the Use of the Quick Estimating

vided showing the effect of power factor on the load that Example 6(a)Determine the maximum load atcan be transmitted at a fixed regulation in terms of that unity power factor and five percent regulation which canat unity power factor. The curves apply specifically to be transmitted over a three-phase five-mile line havingstranded copper conductors at a three foot equivalent 300 000 cir mil stranded copper conductors and operatingspacing and for five percent regulation, but they will give at a load line voltage of 22 kv.an accuracy within 10 percent for conductor spacings up to From the unity power factor curves of Fig. 19 for this20 feet and for the same copper equivalent in other com- conductor size and voltage, 100 000 kw times miles is ob-mon conductors. The error however may be as high as tained. The load is then 100 00025 percent for spacings as small as 8 inches. - = 20 000 kilowatts. The5The Quick Estimating Curves can also be used for other percent loss read from the curves is 4.2.values of regulation if the approximation is made that the Example 6(b)What is the load for this line at thisload which can be transmitted varies directly with the regulation but 0.95 power factor lag? Referring to Fig.regulation. 21 it is seen that for this conductor size 0.58 as much loadAfter having determined the load for other power fac- can be transmitted at 0.95 power factor as at unity.tors or regulations than those for which the curves are Thus the load is 20 000X -58 = 11 600 kilowatts. Thedrawn, the percent loss can be determined from the re- percent loss as determined from Eq. (88) islation

(Percent Loss)2 = (Percent Loss)1 Percent Loss = (4.2) 11 600(1)220 OOO( 95)2 = 2.7%X (Kw Load)2 X (Power Factor):(Kw Load)1 (Power Factor)! (88) Example 6(c)-What load can be transmitted over thisI- . ,a line at unity power factor but 15 percent regulation7



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Fig. 22Quick Estimating Charts Based Upon Percent Loss.The solid curves are based on percent loss and equal receiving-and sending-end voltages. The dotted curves are for the maximumpower which can be received at a given receiving-end voltage andpercent loss.For the curves of Fig. 22 (a) line capacitance has been neglectedand power for a given length of line is value read from curves dividedby line length in miles. Loss base is 5 percent.In Fig. 22 (b) line capacitance has been taken into account and thedata is thus a function of line length. Loss base is 10 percent.For all curves:For other values of percent loss multiply power read from curves- - -

by % Loss- for (a) and-* for (b).5 10For single-phase lines divide power read from charts by 2.



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Fig. 23Regulati on and Loss Chart for transmission lines.


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The answer is (20 OOO)? = 60 000 kw560 000The percent loss is (4.2)%x= 12.6%

Example 7Determine the conductor size and voltagenecessary to transmit 10 000 kw at 0.9 power factor lag fora distance of ten miles.This corresponds to 100 000 (kw times miles). Referringto the 0.9 power factor curves for both copper and ACSRconductors for this load, it is seen that the following linescan be used:

Stranded Copper ACSRVolta ge Cond. Size y& Loss Cond. Size 70 Loss33 000 300 000 cir mil 2.5 636 000 cir mil 1.944 000 x0. 0 4.0 No. 0000 3.766 000 h-0. 4 4.5 Ko. 2 5.0

If it were desired to allow a ten percent regulation in-stead of five percent, the value of kilowatt miles to referto on the curves would then be 50 000 instead of 100 000.The use of the Quick Estimating Charts based uponline loss is quite similar. For instance, if the line of ex-ample 6 were equipped with capacitors so that regulationwould not be excessive, examination of Fig. 22 shows thatit could deliver a maximum of 116 000_____ = 23 200 kw at5 >five percent loss.27. Regulation and Loss Chart

Several valuable voltage regulation charts have beendeveloped. Perhaps the best known of these are theDwight7 and Mershon8 charts. The chart shown in Fig.23 provides a means of solving not only regulation butloss problems to a high degree of accuracy. It is just assimple in its use as any of the previous ones, but has thedistinct advantage that it is based upon an exact solutionof the vector diagram for any circuit which can be repre-sented by a single lumped impedance. For this reasonproblems involving the determination of the load whichcan be transmitted for a given regulation can be solvedmuch more accurately than from charts based upon ap-proximations.

The chart is developed on the principle that for a givendifference in magnitude between the sending-end and re-ceiving-end voltages, the impedance drop (ZI) is fixedentirely by the angle p = y ++ where (,=tan-lf) is theimpedance angle of the line and (b is the power factor angle.For lagging power factors $ is negative and for leadingpower factors + is positive. Thus, corresponding to vari-ous values of percent regulation, the corresponding percentZI can be plotted as a function of the angle p. These arethe set of curves on the chart for voltage drops from 0 to15 percent and voltage rises from 0 to 5 percent. Thevalue of the percent (XI) is the same whether p is positiveor negative. It depends only upon its magnitude.

Since the use of the chart requires a knowledge of y and4, additional curves are provided to facilitate their deter-mination. One of these is a cosine curve for determining 4from the power factor. For obtaining y from a knowledgeof the resistance and reactance of the line, tangent andcotangent curves are plotted so that y can be obtainedfrom the ratio z/r or r/x. However, a simpler means isprovided for standard conductors, by the set of curves atthe top and bottom of the main portion of the chart. Thesecurves give y for various conductors as a function ofequivalent spacing. The resistance of the conductor permile is necessary, and it is given for each conductor. Thevalues on the chart are for a conductor temperature of50C.Although the chart is developed primarily for problemsinvolving known receiver voltage and power factor, it canalso be used for problems where the sending-end voltageand receiving-end power factor and either load current orsending end kva are known. This is the commonest typeof problem involving mixed terminal conditions.28. Use of the Regulation and Loss Chart f or Short

Lines(a) Regulation from Know n Load Conditions-t ocalculate regulation when receiving-end (or load) voltage,power factor, and current or kva are known:(1) Determine p = r++ where the sign of 4 is dependentupon jvhether the current is leading or lagging.

4, the power factor angle, can be obtained from thecosine curve.y, the impedance angle, can be obtained by reading itfrom the conductor curves or by calculating r/x or X/Twhichever is less than one and reading from the cor-responding curve. r and x are the conductor resistanceand reactance in ohms per mile.(2) Calculate percent ZI where

(43 rs I) 100Percent 21 = pp = 100 000 rs (kva)EL cos y E2L cos yfor three-phase lines (89)

= (2 rs I) 100 = 200 000 rs (kva)EL cos y lY2L cos yfor 2-wire single-phase lines 039a)

EL is the line voltage in volts. s is the length of theline in miles.(3) For the calculated values of p and percent ZI readpercent regulation from curves of constant regulation.

(b) Load Limit ationf or Fixed Regulati on-To de-termine load limit for a given value of regulation:(1) IMermine p as in above and from chart for givenvalue of regulation and p read the correspondingpercent %I.

Load in kva= (70 ZI)Ef, cm! y100 000 1sfor three-phase lines

= (% zr>ls;, cos y200 000 1S

for single-phase 2-wire lines

w-u

NOa)


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288 Regulation and Losses of Transmission Lines Chapter 9(c) Line Deficiency-The line loss in percent of the

load kva is given by the equationPercent Loss = y0 RI = 70 ZI cos y (91)

where cos y can be read off its cosine curve from the knownvalue of y. The loss can be determined in percent of theload in kilowatts by dividing the value obtained fromEq. (91) by the power factor. If it is desired to determinethe percent loss for a given regulation, the percent ZI canbe obtained without the use of Eq. (89). It is simplynecessary to determine p and for this angle and the givenregulation to read the (percent 21) from the chart.(d) Use of Chart for Know n Sending-End Volt ageand Receiving-End Power Factor-The chart can beused to as good accuracy as desired for problems of thisnature. As a first approximation the regulation, in percentof the sending-end voltage, can be obtained as outlined in(a) when the sending-end line voltage is used in Eq. (89).Either the line current or the load kva expressed in termsof the sending-end voltage can be used. The load (orreceiving-end) voltage can be calculated from this regula-tion and the sending-end voltage. This first approximationwill usually give the load voltage to an accuracy of aboutone percent, but the percent accuracy of the regulationmay be much worse depending upon its magnitude.A more accurate value can, however, be very easilyobtained by the following method of successive approxi-mations. Using this first determined value of load volt-age and then each successive value obtained, recalculatethe regulation. One or two such steps will usually givevery good accuracy. When calculating the percent ZI inthis process it is not necessary to solve Eq. 89 each time.The new value of percent ZI can be obtained by dividingthe first value calculated by the ratio of the load voltageto the sending-end voltage. This type of problem is illus-strated in Example 8(d).

It is, of course, obvious that the load limit for knownsending-end voltage, load power factor, and regulationcan be determined as in 28(b) after the load voltage iscalculated from the regulation and sending-end voltage.29. Examples of the Use of the Regulation and LossChart

Consider a three-phase line ten miles long with No. 0000stranded-copper conductors at an equivalent spacing ofsix feet and operating at a line voltage of 33 kv at the loadend.Example 8(a)-For rated voltage at the receiving endand a 9140 kva load at 0.9 power factor lag, determine theregulation.

Referring to the impedance angle curves for strandedcopper conductors at the bottom of the chart, the im-pedance angle for this conductor and spacing is y =67.2.Cos y is 0.390 and the conductor resistance is 0.303 ohmsper mile. Reading from the cosine curve the power factorangle for 0.9 power factor is 4=26, and the sign isminus p=y+$=67.2-260=41.20From Eq. (89) :

Percent ZI = (100 000) (0.303) (10) (9140)(33 ooop (0.390)

Reading from the chart for this percent ZI and p--41.2,the regulation is found to be 5.0 percent.Example 8(b)-Determine the maximum kva that canbe transmitted over this line at the same power factor fora regulation of no greater than 5 percent. Reading fromthe chart for 5 percent regulation and p of 41.2, the percentZI is found to be 6.54.Using Eq. (90) :

Load in kva= (6.52) (33 OOO)2(0.390)(100 000) (0.303) (10)= 9140

Load in kw = (9140) (0.9) = 8230.Example 8(c)-As an example of the calculation of

efficiency for the above case using Eq. (91) :Percent loss = (6.52) (0.390) = 2.55.

Example 8(d)--For this same line operating at asending-end line voltage (EsL) of 33 kv and a sending-endload of 9140 kva but a receiving-end lagging power factorof 0.9, determine the line voltage at the load end.As shown in Example 8(a) :

The value of percent ZI determined as a first approx-imation by using the sending-end voltage and kva inEq. (89) isand

Percent ZI = 6.52p =y++=41.2*Thus as a first approximation

Percent Reg. = 53==%=31.42 kv.

As a second approximationPercent ZI = (1.05) (6.52) = 6.85

reading from the chart for percent ZI =6.85 and p =41.2Percent Reg. = 5.20

EL= EsL-=31.35 kv.1.052As a third approximation

Percent ZI = (1.052) (6.52) = 6.87Percent Reg. = 5.25 (as closely as can be read from thechart)

BSLEL=-- 1.0525 -31.34 kv.30. Use of Regulation and Loss Chart for Long Lines

As shown in Sec. 16, methods of calculating regulationfor short lines can be applied to lines up to 100 miles inlength to a good degree of accuracy by simply adding thecorrection factor ( -2.01S2) to the percent regulationwhere S is the length of the line in hundreds of miles.

If greater accuracy is desired, the chart can be usedwith the equivalent load current and power factor ob-tained as described in Sec. 14. Using this method bothregulation and efficiency can be determined.31. Determination of Effect of Transformers onLine Performance

The chart can be used as described in Sec. 28 for de-termining regulation and efficiency of transformers al-6.52


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Regulation and Losse S of Transmission Lines 289the transformer charts in Chap. 5 are simpler. Inthe performance of a line and transformers

however, the chart can be used to advantage.impedance of the transformers can be combined withof the line into a single impedance. These impedancesexpressed either in ohms or in percent on somekva base. Transformer impedance is usuallyin percent. It can be expressed in ohms by the

z z (percent) ELk) (10)(ohms) = kva (92)The transmission line impedance in ohms can be t rans-to a percent basis by the equation

z z(percent) = (ohms) ckva)E2L(kv) (lo) (93)transmission line resistance can be read directly fromchart and the reactance obtained from the chartreading the line impedance angle y from the chart andof r/x: or x/r for this angle.

For problems of this type it is usually easier to use thein percent. After having obtained the totalR and percent X, the equivalent anglecan be read from the curves for the ratio of R/X or

The percent ZI can be calculated from the equationPercent ZI = (%RI = %R> (rated oad) (actual load) (g4)cos y (rated load)Example 9AS an example of the calculation of aof this type consider the 10 mile, 33 kv, 300 000ir mil stranded copper line found adequate for the(10 000 kw = 11 111 kva) load at 0.9 power factor lag of

xample 7.Assume that it has transformers at each end rated at12 000 kva with 0.7 percent resistance and 5 percent re-

actance, and let us calculate the total regulation and lossof the system.Reading from the chart

The line resistance is (0.215)(10) = 2.15 ohmsr/x for the line impedance angle of 71.6 is 0.330The line reactance is go = 6.51 ohmsThe percent impedance of the line on a 12 000 kvabase is from Eq. (93).

Percent 2~ = W~+j6.51)(12 000) =2 3,+j7 16(X3)2( 101 . .

The total impedance isPercent 2 = (2.37+j7.16) +2(0.7+j5)= 3.77+j17.16

?@=3.77=()21g%X 17.16Reading from the chart for this ratio

y = 77.7cos y = 0.219

For 0.9 power factor + = -26p=51.7From Eq. (94)

Percent ZI = (-.E) (g-g)= 15.94The regulation read from the chart for this percent ZIand the calculated value of p isRegulation = 10.5%

The loss in percent of the load in kw is from Eq. (91)Percent Loss= ~15.94)~0*219) =3 330.9 - *

5.

6.

7.8.

9.

REFERENCESPrinciples of Electric Power Tran sm ission, by L. F. Woodruff(a book), J ohn Wiley & Sons, Inc. Second E dition, p. 106.Tables of Complex H yperbolic and Circular Functions, by Ken-nelly (a book), Har vard University Press.Chart Atlas of Complex H yperbolic and Circular Fu nctions, byKennelly (a book), H arvard University Press.Transmission Line Circuit Constan ts, by R. D. Evans an dH. K. Sels, Th e Electric Journ al, J uly 1921, pp. 307-390 an dAugust 1921, pp. 356-359.Circle Diagra m for Tran smission Lines, by R. D. Evan s andH. K. Sels, Th e Electric Journ al, December 1921, pp. 530-536and Februar y 1922, pp. 53 and 59.Some Theoretical Considerat ions of Power Tran smission, byC. L. Fortes cue an d C. F. Wagner, A.I.E.E. Tran sactions, V. 43,1924, pp. 16-23.A Chart for the Rapid E stimating of Alternating Current PowerLines, by H. B. Dwight, Th e Electric Journ al, J uly 1915, p. 306.Electrical Characteristics of Tran sm ission Circuits, by WilliamNesbit (a book), Westinghouse Technical N ight School Press.Third Edition, pp. 43-45.Th e Tran sm ission of Electric Power, by W. A. Lewis (1948Lithoprin ted Edition of Book), Illinois Inst itut e of Technology.

27190563 Regulation Losses of Transmission Lines

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Transcript of 27190563 Regulation Losses of Transmission Lines