27190724 Steady State Performance of Electrical Systems

7/27/2019 27190724 Steady State Performance of Electrical Systems

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CHAPTER 10STEADY-STATE PERFORMANCE OF SYSTEMS INCLUDINGMETHODS OF NETWORK SOLUTION

Orig inal Author:E. L. Harder

POWER system must generate, transmit, and thenA istribute electric power to the desired points, reli-ably and in good condition. The electrical perform-ance of the system as dealt with in this chapter is themeasure of how well it performs this task and is expressedby such quantities as voltage regulation, loading of linesand equipment, efficiency and losses, and real and reactivepower flow. Stability, of vital importance also, is dealtwith in Chap. 13.

The key to the determination of such system quantitiesis the netw ork solution, or determination of currents andvoltages throughout the system for any prescribed con-ditions. From the network solution can be determined allof the essential electrical characteristics that are dependentupon the fundamental-frequency currents and voltages.

Network solution is based on Kirchoffs two laws:First, that the vector sum of all the voltages acting

around any closed loop is zero.And second, that the vector sum of all the currents

flowing to any point is zero.In the course of applying these elementary principles

to the solution of thousands of linear networks for manyyears, various investigators have found several powerfultheorems that follow directly therefrom, such as the super-position theorem, the reciprocal theorem, and Theveninstheorem. These theorems not only assist in visualizing thephenomena taking place in the circuits, but also greatlysimplify and systematize the work of solution for the speciesof networks to which they apply.

The method of symmetrical components, given in Chap.2 is a highly developed special application of the super-position theorem, taking advantage of the symmetry ofthe several phases of the usual polyphase power system.

The direct use of Kirchoffs Laws can be designated asSolution by Equations, to distinguish it from Solutionby Reduction in which portions of a system are progres-sively replaced by simpler equivalents until a single branchremains. This latter makes use of the superposition theoremin treating one emf at a time. Also, it utilizes equivalentcircuits, many of which are now available.Thevenins theorem and the superposition theorem haveprovided direct methods for obtaining solutions in net-works of several fixed emfs, with enormous simplification.

Solutions of networks can be expressed in many forms,each one being particularly adaptable to certain types ofnetworks or certain problems. Thus, the expression of so-lutions as Self and Mutual Drops and Current Division

Revi sed by E. L. Harder

is particularly well suited to regulation and apparatusloading studies. The method of Driving Point and Trans-fer Admittances or Impedances is well suited to powerflow or stability studies on multiple-entrance systems, andthe General Circuit Constants, ABCD, or the equivalentPi and T are similarly advantageous for the transmission-type network having two significant terminals.

These methods of network representation and solutionconstitute a highly developed science with extensive pres-ent literature. However, as they constitute the heart ofthe problem of steady-state performance of systems as wellas of many other system problems, a large part of thischapter will be devoted to them. In general, the mostcommonly used methods will be outlined and illustratedby examples. For further information a bibliography ofselected references is included.

Network solution, once accomplished largely by analyt-ical methods, is now performed to an increasing extent bya-c and d-c network calculators. However, many problemsare still solved analytically and also a thorough knowledgeof methods of network representation and solution is asessential as ever to the system designer. Fortunately, how-ever, the calculator has removed the enormous burden ofroutine calculation and has made it economically possibleto solve complicated systems. Analytic methods are stilllargely used for the simpler studies or where networkcalculators are not available.

I. NETWORK REPRESENTATION1. Single-Line Diagram. Fig. 1

In dealing with power systems of any complexity, oneof the first essentials is a single-line diagram, in which eachpolyphase circuit is represented by a single line. Strippedof the complexity of several phase wires, the main powerchannels then stand out clearly, and the general plan ofthe system is evident. Most power companies maintainup-to-date single-line diagrams of their systems.

This diagram is a short-hand or symbolic representationof the principal connections, showing the equipment in itscorrect electrical relationship and usually having indicatedon it, or in supplementary tabulations, data essential forthe determination of the impedance diagram. The recom-mended symbols for apparatus are given in Table l(a).In addition, auxiliary symbols, Table l(b), are inscribednear the devices in question, to indicate the winding con-nections and the grounding arrangement, if any, at the

290


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Chapter 10 St ead@-Stat e Performance of Syst ems Includi ng M et hods of Netw ork Sol ut ion 291TABLE 1(a)-GRAPHICAL SYMBOLS FOR DIAGRAMS-EQUIPMENT SYMBOLS

neutral. The use of these auxiliary symbols is illustrated and its implications are discussed in detail here.in Fig. 1.Similar diagrams showing circuit breakers and dis-

connecting switches are used as power-system operatingdiagrams. Or they can be marked with suitable symbols toshow the relay (See Chap. 11) or lightning protection.2. The Sign of Reactive Power

The + sign used with the reactive-power terms in theloads of Fig. 1 designate lagging-reactive power in accord-ance with the standard notation approved by the AIEEStandards Committee on Jan. 14, 1948 and recommendedfor adoption to the American Standards Assn. and theIEC. Since this is a change from the convention used ineditions 1 to 3 of this book the history of this standard

The complete specification of real- and reactive-powerflow in a circuit requires:

First, an indication of the direction spoken of, i.e., areference-positive direction.Second, numerical values and associated signs. Thenumerical values give the magnitude of the real- andreactive-power components respectively. The associatedsigns show whether they flow in the reference-positivedirection or not.

Third, there must be a convention as to whether it islagging-reactive power or leading-reactive power, the di-rection and magnitude of which is being specified.

Lagging-reactive power is that which is generated orsupplied by an over-excited synchronous machine or by a


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292 SteadpStat e Perfo rmance of Syst ems In cludi ng M et hods of Netw ork Solut i on Chapter 10TABLE l(b) .GRAPHICAL SymBOLS FOR DIAGRAMS- WINDINGCONNECTION SYMBOLS

static capacitor and used by inductive loads such as induc-tion motors, reactors, and under-excited synchronousmachines.

According to the convention recommended by AIEE in1948 and used throughout this book the positive sign forreactive power indicates that lagging-reactive power isflowing in the reference-positive direction. The vectorrelationship for power is therefore:

P+jQ = Ef, the symbol A designating conjugate.For example if E is taken as reference, E =E and if

I =I- jl is a lagging current, 1 and 1 being positivequantities, the real power is P =Efl and the lagging-reactive power is Q = El.The expression, - -P+jQ=Ef=E(I +j i)=Ep+ji?f lresults in the proper sign for the P and Q terms, whereasl ?I would give the right values but the wrong sign for theQ term. With this new convention, and taking E as refer-ence, the power vector P+jQ lies along the conjugate ofthe current vector. Consequently current and power circlediagrams lie in conjugate quadrants.Historical SummaryOriginally there was one schoolof thought, typified by Evans, Sels6, and others, that usedthe positive sign for lagging-reactive power for the samereasons that it has now finally been adopted. The principalreasons were these. Like real power, lagging-reactive poweris generally used in the load and must be supplied at someexpense in the supply system. It is thus the commoditydealt with by the practical power-system designer, anddispatched by the operators. This concept is consistentmathematically with the following forms:

Power associated with voltage E and current I is:P+jQ=Ef

and power in an impedance 2 to a current I is:

Fig. lSingle-line diagram of a power system.

P+jQ =I22The form P+ jQ = E2Y is then erroneous and gives thewrong sign for Q.

For the conventional transmission line, with this concept(lagging-reactive power positive) the center of the sending-end power-circle diagram lies in the first quadrant and thecenter of the receiving circle in the third quadrant.

The other school of thought used leading-reactive poweras positive, lagging-reactive power as negative. This hadthe theoretical advantage of throwing current and powercircle diagrams into the same quadrant, but the disadvan-tage that lagging-reactive power, the reactive commodityusually dispatched by power-system operators, was thena minus quantity. This concept is consistent with themathematical forms:

Power associated with a voltage E and a current I is:-P+jQ=l?I

Power flowing into an admittance Y due to a voltageE is:

P+jQ=E2Y


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Chapter 10 Steady-Stat e Performance of Systems Includi ng M ethods of Netw ork Solut ion 293The form P+jQ = r22 is then erroneous and gives the

wrong sign for Q.The latter school, (leading-reactive power positive) won

out, for the time being, on the basis of the theoreticalconsiderations, and on August 12, 1941 the AmericanStandards Association approved this convention as anindustry standard, C42-1941, Section 05.21.050. The firstthree editions of this book followed this standard conven-tion. However, the convention was never followed bysystem-planning and operating people to any extent. Theycontinued to dispatch lagging-reactive power which theycalled simply reactive, and to mark on their flow chartsthe direction in which lagging-reactive power flowed. Theycould not be converted to selling a negative amount ofleading-reactive power for positive money, but preferredto sell a positive amount of lagging-reactive power.

A majority of engineers have now come to considerlagging-reactive power as the commodity being dealt with.The AIEE Standards Committee recognizing this faitaccompli recommended to ASA in 1948 adoption of theconvention making lagging-reactive power positive. Thisreference book has, starting with the fourth edition, 1950,been changed to conform with what will undoubtedly bethe standard from now on, namely, lagging-reactive powerpositive.

Teachers and writers can materially aid in eliminatingconfusion by discontinuing all use of the term leading-reactive power which after all is simply an unnecessaryname for the negative of lagging-reactive power. Such aterm is no more necessary than a name for the negativeof real power. Eventually if this is followed the adjectivelagging can be dropped, as reactive power will alwaysmean lagging-reactive power.3. Impedance Diagram. Fig. 2

The second essential in analytic study of a power systemis the impedance diagram, on which are indicated on acommon basis, the impedances of all lines and pieces ofequipment related to the problem. Because of the sym-metry of phases it is usually sufficient to represent onlyone phase-called the reference phase, or a phase. Underbalanced conditions of operation, the currents and volt-ages in the other two phases are exactly equal to those in aphase and merely lag behind the a phase quantities by 120and 240 electrical degrees. Hence, when the a phase quan-tities have been determined, the others follow directly.

Even when unbalances, such as a line-to-ground fault,or one-wire-open, occur at one or two points of an otherwisebalanced polyphase system, the impedance diagrams forthe reference phase are sufficient, if use is made of themethod of symmetrical components as outlined in Chap. 2.The impedance diagram, corresponding to the systemshown in Fig. 1, is given in Fig. 2. Generator impedancesare not shown as they do not enter into the particularproblem. All impedances on this diagram have been ex-pressed in ohms, and admittances in mhos, on a 110-kvbase. Actually there are several choices, such as percentor per unit on various kva bases or ohms on voltage basesother than 110-kv. The relations between these severalmethods, and factors affecting the choice are discussedsubsequently.

Fig. 2-Impedance diagram of a system shown in Fig. 1.

4. Determination of ImpedancesIn general, accurate cable and overhead-line impedances

can be obtained from the tabulations of Chaps. 4 and 3, interms of the wire size and spacing. Average apparatusconstants are given in the tabulations of Chap. 6, MachineCharacteristics, and Chap. 5, Power Transformers andReactors. More accurate figures, if required, can be ob-tained from name-plates or direct from the manufacturerof the equipment. The impedances of lines and trans-formers in Fig. 2 have been obtained, for example, from thetables of Chaps. 3 and 5 based on the data given in Fig. 1.Representation of Loads-The necessary representa-tion of loads in the impedance diagrams depends uponthe use intended. In short-circuit studies, loads are mostfrequently neglected. In stability studies, they must gen-erally be considered. Several methods of representing loadsare as follows:

a. Shunt impedance that draws the same kilowatts andreactive kva at normal voltage as is drawn by theactual load.

b. An impedance circuit, which, for any conditions ofvoltage is adjusted to draw the desired amount ofreal and reactive power.


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Steady-Stat e Performance of Systems Includi ng M ethods of Netw ork Solut ion Chapter 10

c.

d.

In network calculator studies, the use of a sourcetadjusted, in phase angle and magnitude, to draw thedesired real and reactive power from the system.Given any characteristics of variation of real andreactive power with voltage, the load can be con-verted to impedance at the expected voltage, thisimpedance used in determining the system voltages,and then the load impedance corrected to the newvoltage if such correction is warranted.

Conversion of Load Kw and Reactive Kva toOhms or MhosLoads given in kilowatts and reactivekva can be converted to impedance or admittance form bythe following equations:Let P = kw (three-phase)Q = reactive kva lagging $(three-phase)

EL--L =line-to-line voltage in kv at which the conver-sion is to be made.

2 = vector impedance value, ohms line-to-neutral.Y = vector admittance value, mhos, line-to-neutral.

z lOOOE2L-L lOOOE2L-L= P-j& = p2+&2 V'+jQ)lOOO(kv) 2

= kw - j reactive kva (lagging) ohms, line-to-neutral (1)Y= P-j& = kw - j reactive kva (lagging)1000E2L-L 1000(kv)2mhos, line-to-neutral (2)

For example, at 13.8 kv a load of 10 000 kva at 80percent power factor lagging may be expressed as:

P=8ooo kw Q = +6000 reactive kvaThe impedance required to represent it is:

1000(138)2 = (6000)2+ (8000)2 (SOOO+j6000) = 15.2+j11.4 ohms,line-to-neutral

and the admittance is:y SOOO-j6000

= lOOO(13.8)2 = 0.0420 -jO.O315 mhos, line-to-neutral,Y and 2 as given above are the admittance or impedance

values to be used in the single-phase impedance diagram inwhich only the reference phase and neutral are represented.Shunt Capacitors are built to a tolerance of -0 to+10 percent of their rated kva, +5 percent being theaverage. It is generally sufficiently accurate to considerthe reactance to be 100 percent based on 105 percent ofthe rated kva base.Series Capacitors-The determination of reactanceof a series capacitor can best be explained by example.Suppose ten standard 15-kva single-phase, 440-volt, shunt-capacitor units have been used in parallel in each phase,or a total of 150 kva per phase. The capacitive reactancepresented in series in each phase is then:

The "Sources are voltage regulator-phase shifter circuits from amain power bus and can be readily adjusted to either draw or feedthe desired quantities of real and reactive power.

X 1000(kv)2 1000(0.44)2s=1.05*(kva)= 1.05X150 = 1.22 ohms (5)This ohmic value can be converted to percent by Eq. (12).Shunt Reactors have 100-percent voltage drop acrossthem when connected to normal voltage, or have 100-per-cent impedance based on the kva drawn from the systemat normal voltage.Series ReactorsThe reactance of a series reactor isfrequently expressed in percent, but the kva of its partsis given. Thus, if a 6-percent reactor is desired in a circuithaving a rating of 10 000 kva, three-phase, the reactorrating will be 600 kva, three-phase (6 percent of 10 000kva). Three 200-kva single-phase reactors might be used.These would ordinarily be referred to as three 200-kva,6-percent reactors, whereas actually they constitute athree-phase reactance in the circuit having 6-percent re-actance on a 10000-kva base. Care must be taken, there-fore, to determine the reactance value on the through ortransmitted kva base, or 10000-kva base in the examplecited. The relation between reactor kva, and through kvaare as follows:

Reactor three-phase kva rating = aThrough or transmitted kva rating = APercent reactance on the transmitted kva base=XThen (3)

Given the reactor three-phase kva rating, a, the throughkva rating isA=Ta (4)

The reactor has a reactance of X percent on the kva base A.In the case cited above of a 600-kva, 6-percent reactor,

Eq. (4) gives A = yX600 = 10 000. Whence, the reactorhas a reactance of 6 percent on a 10 OOO-kva base.

The standard reactance tolerance of current-limitingreactors is -3 percent to +7 percent for single-phase and-3 percent to + 10 percent for three phase. The ratedreactance is generally used in system calculations unlesstest figures are available.5. Conversions. Percent to Ohms and Ohms toPercent Method IIf a base kva (three-phase) and kv (line-

to-line) are selected, the corresponding normal or basecurrent, line-to-neutral voltage, and impedance values canbe immediately determined.They are: kvaNormal Current I, = -v%W amperes (6)

Normal Voltage E _ lOOO(kv)(line-to-neutral) n - d3 voltsE, ohms per phase,Normal Impedance 2, = Tn line-to-neutral. (8)

*I f the ratio ofplace of 1.05.

actual to rated kva is known, i t should be used inQ is positive for lagging reactive kva. (Note: Per Unit is percent divided by 100).


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Chapter 10 Steady-Stat e Performance of Systems Includi ng M ethods of Netw ork Solut ion 295

From these relationsverted to ohms. any

percent impedance can be con-

Ohms = (normal impedance) percent impedance100(9)

Conversely any ohmic figure can be converted to percent.Percent = 100 OhmsNormal Impedance )=lOO(qf) (10)Method 2-The magnitude of 2, from (8), (7), and

(6) can be substituted in (9) and (10) and gives directconversions :

(11)

For example, a 15000-kva, 13.8-kv to 66-kv transformerbank has a reactance of 8 percent on the 15000-kva base.Let it be required to determine its impedance in ohms ona 66-kv base.Normal current:

I 15 000- w = 131 amperes.-66d3Normal voltage :

E 66 000- - = 38 100 volts, line-to-neutral.ll- &iNormal impedance:

z 38 100Il- 131- ~ = 291 ohms per phase, line-to-neutral.

Transformer impedance = 8 percent of 291=23.3 ohms per phase at 66 kv.The direct determination from (11) is,Transformer impedance

= w33)2(10)15 000 =23.3 ohms per phase.

The first method is longer, but gives other informationgenerally required in the problem, and has some advantagein visualizing the procedures.6. Conversions to a Different Kva Base

From (12) it is apparent that for a given ohmic imped-ance the percent impedance varies directly with the kvabase selected. Thus lo-percent impedance on a 10 OOO-kvabase becomes 100-percent impedance on a 100 OOO-kvabase. When using percent impedances, all percentagesshould be expressed on the same kva base.7. Conversions to a Different Voltage Base

In system studies if impedances are expressed in ohmsit is desirable to convert them all to a common voltage baseso that transformer turns ratios need not be considered inthe subsequent calculations. The terms voltage ratio

and turns ratio are often used loosely as synonymousterms, until more precise or important calculations arebeing made for which it is desired to be quite accurate.Then the question sometimes arises as to whether imped-ances should be transferred to the voltage base on theother side of a transformer on the basis of its voltage ratioor its turns ratio. It is actually the turns ratio that countsand should be used as will be shown later in this section.The turns ratio is the same as the nameplate voltage ratiobut differs from the terminal voltage ratio under load.

Also in approximate calculations it is frequently assumedthat for all parts of the system of the same nominal voltagethe same transformation ratio can be used to the desiredvoltage base. This is a rough approximation and becomesexact only if the transformer turns ratios between parts ofthe system at the same nominal voltage are all unity.Barring this, one correct procedure is to select some onepoint of the system as a base and transform all otherimpedances to this base by multiplying by the square ofthe intervening turns ratios. Once all impedances are ona common base they can all be transformed by a singlemultiplier to any other voltage base.When impedances are in percent on a given kva base thepercent refers to a given normal voltage. Thus strictly twoconditions must be fulfilled in sequence for percent im-pedances to be used in network solutions. First, the normalvoltages to which the percentages refer must be in the sameratios as transformer turns ratios throughout the system.Second, the normal voltage used in converting the answersfrom percent to amperes and volts must be the same as thenormal voltages on which the percent impedances arebased. Otherwise approximations are involved. Theseapproximations can be eliminated by suitable transforma-tions beyond the scope of this chapter except for thefollowing general method.Where doubt exists as to the correct direct transforma-tion of percent impedance, the impedance of each elementcan be converted to ohms. The ohmic values can be con-verted to a common base as described above and combined.The result can be reconverted to percent on any desiredkva and voltage base. This is the general procedure bywhich rules for direct percent-impedance transformationsare derived.

The pitfall of ignoring near-unity turns ratios extends tovoltage also. Suppose a 13.8-kv generator feeds throughstep-up and step-down transformers to a 13.8-kv distribu-tion system and that impedances have been expressed onthe distribution system voltage base. Suppose further thatthere is a resultant 1:l.l step-up turns ratio between thegenerator and the distribution system. Then a generatoroperating at 13.8 kv would be at 13.8X l.l= 15.18 kv onthe 13.8-kv voltage base of the distribution system, andmust be so treated in the calculations. Similarly, for cal-culations in percent, the same machine must be treatedas operating at 110 percent voltage. The theoretical basisupon which all such transformations rest, and examples oftheir correct use is given in the following paragraphs.From an energy or power standpoint, no change is madeif all voltages are multiplied by a constant, N, all currentsdivided by N, all impedances multiplied by N2, and alladmittances divided by N2. When two circuits are sepa-


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296 Steady-Stat e Performance of Systems Includi ng M ethods of Netw ork Solut ion Chapter 10

Fig. 3Power invariant transformation.

rated by an ideal transformer* of turns ratio N, such anoperation performed on the quantities on one side of thetransformer with a corresponding change in the trans-former ratio, has the advantage of bringing the currentsand voltages on the two sides to an equality. A directconnection can be made and the ideal transformer can beomitted from the diagram (see Fig. 3). Solutions can bemade with the quantities on the fictitious or transformedvoltage base, and they can be reconverted to actualquantities whenever desired.

An actual transformer differs from an ideal transformerin two respects only. It has primary and secondary resist-ances and leakage reactances, which are no different thanthe same impedance connected externally. Its primaryand secondary ampere-turns differ by a small quantityof exciting ampere-turns that excite the core. A shuntbranch can be connected which draws the requisite excitingcurrent if important in the particular problem.Example-As an example consider the circuit of Fig. 4,a generator, transformer and high-voltage line with a three-

phase short circuit at the end. Suppose the short-circuitcurrents are to be determined. This problem will also illus-trate that calculations can be made interchangeably withimpedances in ohms on any voltage base or in percent onany kva base.

The generator reactance (assumed 15%) in ohms isfrom (11):*A transformer having zero exciting current and zero leakage

impedance.

(15)(10>(13.8>250 000 =0.571 ohms at 13.8 kv

The transformer reactance in ohms is:(9)(10>(13.8>2

50 000 =0.343 ohms at 13.8 kv.The line impedance is, from Chap. 3.

9.75+j57.9 ohms at llO-kv. [See Fig. 4(c)]

Fig. 4Problem illustrating the expression of ohms on vari-ous voltage bases and the relation to percent on a kva base.

The shunt impedances of this line are high (line CD Fig.2) and will be neglected for simplicity in this problem.Use of Generator Voltage BaseIf the current inthe generator is desired, it will be most convenient toexpress all impedances on the generator voltage base. Thegenerator and transformer impedances are already on thisbase. The line impedance is converted to it by multiplyingby the square of the turns ratio, usually taken as thenameplate voltage ratio corresponding to the taps in use.Thus the line impedance is:

(9.75+j57.9) ( >&f 2=0.154+j0.910 ohms at 13.8 kv.The impedance diagram of Fig. 4(b) results, in which

all impedances are expressed in ohms on a 13.8-kv base.The fault current in the generator is then:I= 13 8002/3( 1.830) = 4350 amperes.

The current flowing in the line is:13.8I=4350 110( > = 545 amperes.

Use of L ine Voltage Base-A similar result would beobtained if the generator and transformer reactances had


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Chapter 10 Steady-Stat e Performance of Systems Includi ng M ethods of Netw ork Solut ion 297been converted directly from percent to ohms at llO-kv.The impedance diagram, Fig. 4(c) would then result, thefault current being calculated directly for the line andrequiring a conversion 110multiplication by 13 to deter-.mine the current in the generator.

Use of Percent on Kva BaseA third method ofapproach is to convert the line impedance to percent ona kva base, and work in percent. A convenient basewill be 50 000 kva since two of the impedances are alreadyknown on this base. The line impedance is, from Eq. (12) :(g*75+57*g)(50O)4 O+j23.9)% on 50 OOO-kva base.(10)(110)2 *The impedance diagram Fig. 4(d) results, the percentagesbeing shown as decimal fractions or per unit to facilitatecomputation.

In this case the current is:1=& = 2.08 per unit or 208 percent of the normal. current, corresponding to the selected kva

base.This normal current is:I 50 ooo=&13 8 or 110) =2090 amp. at 13.8 kv. or 262 amp. at 110 kv

The generator and line currents are, therefore, 208 percentof 2090 and 262 or 4350 and 545 amperes respectively,which agree with the preceding calculations.

The base selected obviously is immaterial. Had a100 000-kva base been used, the impedances in Fig. 4(d)would all be doubled and the resulting percent currentshalved. But the normal currents to which these percent-ages refer would be twice as great, and thus the samenumber of amperes would be obtained.8. Phase Shifts in Transformer Banks

In addition to magnitude transformation, the voltage ofthe reference phase in general undergoes a shift in angularposition. For balanced conditions, that is, consideringpositive-sequence quantities only, this is generally of nosignificance. For example, in the problem just worked out,the current in the reference phase of the line may or maynot have been in phase with the reference or a phasecurrent in the generator. If the transformer were delta-delta, the currents would have been in-phase; if delta-starthey would have been 30 degrees out-of-phase, using theusual conventions.However, it should be recognized that an angular trans-formation has been made whenever the single-phase circuitor impedance diagram is used for the calculation of currentsand voltages in a circuit including a star-delta connectedtransformer bank. The following statements should aid indetermining the treatment required in any particular case.Radial SystemsIn radial systems, the angle trans-formation is not usually significant as few phenomenainvolve comparisons of the phase angles of line currentson opposite sides of a transformation. Since currents andvoltages are shifted alike, power or impedance determina-tion at any one point in the circuit is unaffected by theangle transformation.

Transformer Differential ProtectionA typical ex-ception is the differential protection of a transformer bank.Here the currents on opposite sides of the transformationare purposely compared and measures must be taken tocorrect for the shift if the devices used are sensitive tophase angle.Sequence Voltages and CurrentsPositive-sequencevoltages and currents are shifted the same as the referenceor a phase in progressing through a symmetrical trans-formation. Negative-sequence voltages and currents, ifpresent, are shifted the same amount as the reference phasebut in the reverse direction. Zero-sequence voltages andcurrents are not shifted in progressing through a trans-formation.Ideal TransformationThe shifts referred to have todo with the ideal transformer only, deleted of all leakageimpedance and exciting current. That is, they depend onlyon how many turns of primary and secondary are usedon each core and how these are grouped to form the phaseson the primary and secondary sides. Symmetry with respectto a, b, and c phases is assumed.Regulating TransformersA symmetr ical threephase bank of regulating transformers may involve bothratio and phase-angle transformation. Suppose that inprogressing through a particular bank of this type, aphase-angle advance of 10 degrees exists in the referencephase. Then, in progressing through the transformer inthe same direction, positive-sequence quantities (currentsand voltages) are advanced 10 degrees, negative-sequencequantities retarded 10 degrees, and zero-sequence quantitiesnot shifted at all.Standard Angular ShiftsThe angular shifts ofreference phase for various transformer connections aregiven in Chap. 5, Sec. 13. The American Standard* is a30-degree advance in phase in progressing through eithera star-delta or a delta-star connected transformer from alower to a higher voltage. When carried out consistently,this will permit interconnections at various system volt-ages without difficulty in phasing. However, at presentpractically all possible connections are in use throughoutthe industry.9. Loop Systems That Close

Transformations of magnitude or angle in a system in-volving one or more loops can be treated similarly to aradial system provided that:

a. The product of the magnitude transformation ratiosfor the reference phase, taken in a common directionaround each closed loop is unity.b. The sum of the reference phase angular shifts takenin a common direction around each closed loop is zero.

If each transformation ratio is expressed vectorially asNP, including angular significance in the term vectortransformation ratio, then a and b above can be combinedinto the single requirement:

c. The product of the vector transformation ratiosaround each closed loop is lcj.If the requirements a and b, or c are fulfilled, then the

circuits of the system can be divided into zones separated*ASA Standards C-57.


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Fig. 5Ratio and angular transformations.

from each other by transformations. One zone, usuallythe one of greatest interest in the particular problem, canbe taken as the reference zone.

Example-For example, in Fig. 5 currents in variousparts of the system are to be determined for a balancedthree-phase fault on the 132-kv line.

There is one closed loop in which:

= 1 EJO (13)Therefore, the reference-phase impedance diagram can

be prepared from the single-line diagram without showingany transformations.

Let Zone 3 be taken as the reference zone and all im-pedances expressed in ohms on 132-kv base. The faultcurrent, If, and the distribution of currents I ,, and I , arenow readily determined. So also are the voltages through-out the network. It is recognized that in Zone 3 these arethe actual reference phase currents and voltages. In Zones1 and 2 they are the actual quantities transformed to theZone 3 base, and hence, must be transformed to their ownrespective bases to obtain the actual quantities. Since theyare all positive-sequence currents and voltages, that is,normal balanced three-phase quantities, the actual cur-rents and voltages of the reference phase, which have beenindicated on the single line diagram, are as follows:

In Zone 3

In Zone 2I [ = I t (14)E,=E, (15)

I : = 3$ = I ,N /

In Zone 1

The Zone 1 quantities may also be expressed as follows,illustrating the general method to be followed when thezone in question is separated from the reference zone byseveral transformations.

NW2I+----- ejel p - I fN 1N2e-j+efiE tx ejel=~~2e -m+rw

The power at any point s, for example, can be cal-culated without transforming. For:

(f,N,~+j@) = EJ, (20)and P,+jQB=E,f, (21)

These are the same. In other words the transformationsdescribed thus far and ordinarily used in analytical workare power invariant. They differ from transformations toa model scale for setting on a network calculator, in whichpower must obviously be scaled down.10. Loop Systems That Do Not Close

If the product of vector transformation ratios around aclosed loop is not unity, special consideration needs to begiven. This case will be sub-divided into three parts, viz-(a) product of ratios not unity, (b) sum of angular shiftsnot zero, and (c) product of ratios not unity and sum ofangles not zero.Product of Ratios Not UnityMany transformersare provided with taps in one or more windings. With staror delta connected windings, use of these taps changes theratio only, without affecting the angular shift through thetransformer. Thus, by far the largest number of cases ofnon-unity vector transformation ratio around closed loopsfalls in this category of ratio discrepancy only.Example-An example is shown in Fig. 6, in which twocircuits A and B differ in capacity, the taps having beenincreased on the B circuit to make it carry more of the load.The power factor of the portion of load that can be thusshifted from B to A depends on the impedance phase anglesof the A and B circuits being nearly pure wattless for purereactive circuits, and pure watts for pure resistive circuits.Thus, for 60 degrees impedance angle circuits the shifted*See Section 2.


10/52

Steady-Stat e Performance of Systems Includi ng M ethods of Netw ork Solut ion 299

Fig. 6Product of ratios around a closed loop not unity.at about 50 percent power factor. The amountis nearly constant, and not a percentage of the

load. Thus, at no load, there is a circulation overSuppose that the network of Fig. 6 is to be solved, andis desired to work on the Zone 2 basis. Zone 3 can be

transformed to this basis as explained in a preced-section. However, no transformation can be found for

that will result in both transformer ratios beingThe best that can be done is to make one of them,

for example A, unity by transforming the voltages of Zone1 in the ratio E813.8 and currents and impedances bythe corresponding factors. This leaves an uncompensatedor remnant ratio to be accounted for in B, which may berepresented as an autotransformer, Fig. 6(b).

In a-c network calculator studies, small auto-transform-ers of the remnant ratio are used and no further consider-ation need be given. For analytic studies the simplestmethod is to neglect the remnant transformation ratio,provided great accuracy is not required. The order ofmagnitude of the circulating current can be estimated bydividing the inserted voltage by the loop impedance to seewhether it can be neglected in the problem at hand. Forexample, if the remnant ratio is 1.05 the inserted voltage isof the order of 0.05 per unit under normal load conditions.If the loop impedance is 0.50 per unit the order of magni-tude of circulating current is 6 =O. 10 per unit.

If this cannot be neglected the following approximationis suggested in cases where the remnant ratio is close tounity. The accuracy of the method is indicated later byan example.

a. Treat as though the ratio were unity and determinethe resulting shunt voltage at the location of theauto-transformer.

b. Determine the resulting series voltage introduced,in this case 138-132 -6132 = 1324.5 percent of the shunt

C.voltage, and in phase with it.Determine the current circulated in the network bythe action of this series voltage alone, setting thegenerator emf, E, equal to zero. Determine the volt-ages for this condition also.

d.

e.

Superpose this set of circulating currents on the cur-rents previously calculated. Superpose the voltagessimilarly. The resulting solution is in error only by acorrection factor of the second order which usuallycan be ignored, as will be shown subsequently.Where several such auto-transformers are requiredto close the impedance diagram, the circulatingcurrents can be calculated separately, treating theratios of the others as unity at the time. All of theresulting circulating currents can then be superposed.The resulting voltages can likewise be superposed.

This approximation is based on the concept that theauto-transformer could be replaced by a shunt load thatdraws the same current from the system as the main sectionof the auto-transformer and a series emf that impressesin series the same voltage as the short extension of theauto-transformer. With this substitution, the solution bysuperposition is exact. If the auto-transformer introducesfive-percent voltage in series and the impedance to theresulting circulating current is 50 percent, then ten-percentcurrent will flow. With five-percent voltage this amountsto 0.5-percent load, which is supplied from the system tothe shunt winding of the auto-transformer, thence, to theseries winding and back into the system, as 12X and 12Rlosses of circulation. As this load drawn by the shunt


11/52

300 Steady-Stat e Performance of Systems Including M ethods of Netw ork Solut ion Chapter 10winding is quite small, 0.5-percent in the case just cited,it is most frequently ignored.

In general, introduction of the series voltage raises thevoltage on one side of the auto-transformer and lowers iton the other side, as compared with the voltage that wouldbe present if the auto-transformer were not there. Thus,if the series voltage is five percent, the shunt voltageapplied to the auto-transformer will differ by not overfive percent from that calculated with the auto-transformerremoved. A correction of five-percent in the shunt voltagewould change the series voltage from five-percent to 4.75percent. This small correction usually is not required.Thus, the steps as outlined from a to d above will usuallybe sufficiently accurate.ExampleA comparison of the exact solution (byequations-see Sec. 13) and the approximation in thecase of Fig. 6 will illustrate the procedure and indicatethe degree of accuracy to be expected. Assume a set ofconstants as follows, in per unit on the generator kva base.Assume the voltage to be maintained constant on thegenerator bus.

A =jo. 10 E=l+jOB =jO.lOc=0.10+j0.173D=O.10+~0.173F=jO.lOG=0.90+j0.50

Following the steps suggested above, the series voltageis first set equal to zero and the solution using the generatorvoltage alone is obtained, Fig. 6(c). The series voltage is4.5 percent of 0.9745 -jO.O329 or equal to 0.0443 -jO.OOlSand is directed to the right. Then, setting the generatorvoltage equal to zero and applying the series voltage alone,the solution of Fig. 6(d) is obtained. Adding vectoriallythe corresponding quantities in these two solutions, thesuperposed solution, Fig. 7(a), is obtained for the simul-taneous application of the generator voltage, E, and the

Fig. 7-Comparison of exact and approximate solutions ofFig. 6.

series voltage e, and is a good approximation to the exactsolution of the circuit of Fig. 6(a) and (b) as will be shownby comparison with Fig. 7(b).

Th e Exact Solut ion for the currents and voltages in Fig.6(b) can be obtained by writing Kirchoffs Law for the

TABLE 2 COMPARISON OF RESULTS BY APPROXIMATEANDEXACT METHODS OF SOLUTION WHEN PRODUCT OFVECTOR TRANSFORMATIONRATIOS IS NOT UNITY.(REFER ALSO TO FIG. 7).

drops around each of the three loops and setting up a fourthequation stating that the total ampere-turns on the perfecttransformer are zero.

The solution of these simultaneous equations with thenumerical values of the impedances A to G substituted,is given in Fig. 7(b). Table 2 shows the error in variousquantities by the approximate method. The voltages arewithin 0.2 percent. The largest current error is 3.35 percentin I ,. The sum of errors in I , and I , are about 4.5 percent.This is necessary since these two currents are taken thesame in the approximate solution and differ by 4.5 percentin the exact solution.Sum of Angular Shifts Not ZeroRegulating trans-formers or regulators as well as special connections of trans-formers can introduce angular shift. If the net shift arounda closed loop is not zero but is small, the treatment issimilar to that for ratio discrepancies except that the seriesvoltage is introduced at right angles to the shunt voltage.On the a-c network calculator, transformers cannot beused to obtain a shift since the circuits are single phase.Power sources must be used to introduce the necessaryseries voltages.Sum of Angular Shi fts Not Zero and Sum ofRatios Not UnityThe series voltage can be introducedat any desired angle, corresponding to the net vector trans-formation ratio, and the currents superposed as outlinedabove, with appropriate phase relations.


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SteadpState Performance of Systems Includi ng M ethods of Netw ork Solut ionII. NETWORK SOLUTION

Network Theorems**The Superposit ion Theorem states that each emf

s currents in a linear network* independently ofby any other emf. It follows that the

currents of a given frequency can be treatedof those of any other frequency, and ofThe superposition theorem is a direct result offact that the fundamental simultaneous differentialof the network are linear. (See any standard

on Differential Equations.) (See Sec. 13.)The Compensation Theorem states that if the im-of a branch of a network be changed by an

AZ, the change in current in any branch is thewould be produced by a compensating emf -AZ1,

in series with the modified branch, I being the origi-current in that branch. By compensating emf is meantwhich, if it were inserted, would neutralize the drop

AZ. This theorem follows directly from theThe Reciprocal Theorem states that when a sourceelectromotive force is connected across one pair of

of a passive linear network and an ammeter isacross a second pair of terminals, then the source

electromotive force and the ammeter can be inter-without altering the reading of the ammeterneither the source nor receiver has internal

). This follows from the fact, that, in Eq. (56),all emfs are zero except E2 and E3 for example, then if

EP=O, I,,E*, while if ES=O, 13=E+,that if ES= Ez, I3 = I 2 for the conditions of the theorem.

that A23 =&,.Reference Current and Voltage Directions

To specify uniquely a vector current or voltage in asome system must be adopted to label the pointswhich the voltage is being described or the branch

which the current, flows. This system must also indicatereference or positive direction. Two common methodsuse of reference or positive direction arrows and

double subscript notation.ReferenceDirection Arrows-Fig.8(a)and(b)-Whennetwork is to be solved to determine, for example, the

flow for a given set, of impressed emfs, the networkmarked with arrows to indicate the reference

direction of each current and voltage involved.be drawn arbitrarily, although if the predom-directions are known, their use as reference-direction

r interpretation.The use of open voltage-arrowheads and closed current,-

eads will avoid confusion in numerical work, whereE and I symbols are not used.

It must be decided at this point whether the voltage**See also Thevenins Theorem, Sec. 18.*A linear network is one in which each impedance is linear; that is,

a straight line relation between current and voltage drop.tA passive network is one having no internal emfs as distinguished

0 p---rn 6) nFig. 8Methods of notation used in network solution.

arrow is to represent a rise or a drop. In system calcula-tions it is generally used as the rise in voltage. Whilethis decision is arbitrary, once made, it must be adheredto consistently.Finally a vector value must be assigned to the voltageor current. It can be expressed as a complex number, orin polar form or graphically and gives the magnitude andrelative phase of the quantity with respect to some ref-erence. A symbol, such as I ,, can be used to designatethis vector quantity.

The known vectorial voltages or currents must be asso-ciated with the reference arrows in a manner consistentwith the conditions of the problem. For example, supposethe problem in Fig. 8(a), is to determine the currents thatwould flow with the two voltage sources 180 degrees outof phase, and 100 volts each, rms, 60-cycle. Then if E,is taken as l00+jO, E, must be taken as -lOO+jO. Hadthe arbitrary reference-direction arrows been taken as inFig. 8(b), then for the same problem a consistent, set ofvoltages would be :

E, = l00+jO, E, = l00+jOOrdinarily the reference-direction arrows for shunt voltagesare directed from the neutral to the phase conductor asin Fig. 8(a).

Summarizing then, the complete specification of aquantity in the reference-direction-arrow system involvesthree elements:

a. The reference-direction arrow, drawn arbitrarily.b. An agreement, consistently followed as to what the

reference-direction arrow means; particularly whetherthe voltage arrow means the voltage of the pointabove the tail or the drop from tail to point.

c. A vector to represent the magnitude and relativephase of the quantity with respect to a reference.

Suggested convention: For voltages, the vector quantityshall indicate the voltage of the point of the arrow abovethe tail, that is, the rise in the direction of the arrow. Itthen is also the drop from point to tail.


13/52


Mesh Currents and Voltages-Refer t o Fig. 8(c)-The mesh current system involves a somewhat differentpoint of view. Here each current is continuous around amesh and several currents may flow in the same branch.(I, and I, flow in 2.) The branch current1 is the vector sumof all the mesh currents in the branch, taken in the refer-ence direction for the branch current. If such a networkcan be laid out flat, it is most convenient to take thereference direction for mesh currents as simply clockwisefor example. Or circular arrows can be used as shown inFig. 8(c). The example of solution by equations in Sec. 9illustrates the use of mesh currents.

The same reference directions can be conveniently usedfor mesh emfs, which are the vector sum of all emfs actingaround a particular mesh, taken in the reference direction.Double-Subscript NotationFig. 8(d)-A double-subscript notation is sometimes used and is of courseequivalent to the drawing of reference arrows. Here againan arbitrary decision must be made as to what is intended.Suggested Convention. Refer to Fig. 8(d).

I ,t means the current from s to t.E,, means the voltage of s above nIt is apparent then that E,, = --E,,; Iet = -Its, etc.Setting Up EquationsI f the work is analytical, bythe method of equations, the equations must be set upconsistent with the reference direction arrows, regardlessof th e val ues of any k now n currents or vol tages. Consistentequations for Figs. 8(a), (b), (c), (d), are as follows, usingKirchoffs Laws (See Sec. 1) : The voltage equations arewritten on the basis of adding all of the voltage rises in aclockwise direction around each mesh. The total must ofcourse be zero. The current equations are written on thebasis that the total of all the currents flowing up to a pointmust equal zero. Arrows and double subscripts have themeanings given in the suggestions above.

Referring to Fig. 8(a).E,-- I ,& -IZ=O (26)- I ,& -E,+IZ=O (27)

&-I-I,=0 (28)Referring to Fig. 8(b).

E,--I& --IZ=O (29)- I ,Z,+E,+IZ=O (30)

& I - I ,=0 (31)

Referring to Fig. 8(c).E,-I,(Z,+Z)+I,Z=O (32)I ,& I,(Z+Z,) -E,=O (33)

Referring to Fig. 8(d) or 8(e).E,,- I ,& -I & = 0 (34)I t ,Z- I& -E,,=0 (35)

I,t-It,-It,=0 (36)In the double-subscript system, the voltages and cur-

rents could of course be indicated on the figure. They havebeen purposely omitted in Fig. 8(d), however, to emphasizethat the specification of these quantities in Eqs. (34) and(35) is perfectly definite from the subscripts alone.

Also, the inclusion of reference-direction arrows on thediagram, even when the double-subscript system is used,may aid in writing equations, although they are notstrictly required. If used, they must be consistent withthe double-subscript system. That is, each arrow mustbe directed from the second subscript toward the first forvoltages, and from the first subscript toward the secondfor currents. Fig. 8(e) illustrates such a diagram con-sistently labeled.13. Solution by EquationsRepresentation-A network of n meshes can be

represented as having n independent currents, I I to I ,,as shown in Fig. 9. The branch currents are combinationsof these. See Brunch Currents below.Mesh Impedances are defined generally as: ZPp=voltage drop in the reference direction in mesh q per unitof current in reference direction in mesh p. The curvedarrows indicate reference directions in each mesh. Ingeneral

z PQ=GlJThe impedances ZPP and Z,, are called self and mutualimpedances.

Specifically in Fig. 9.(First subscripts i and c indicate inductive and capacitivereactances respectively.)

zll=R,+Rb+j(Xia+Xib-Xorr-Xcb) (37)z 12~ -Rb-j(Xib-Xob) (38)213=0 (39)

Fig. 9General flat network.


14/52

Chapter 10 Steady-Stat e Performance of Systems Includi ng M ethods of Netw ork Solut ion 303214=0 (40)Etc.Zn=Rb+Ro+Rd+R,+j(Xlb+XI,+Xld-X,b-x~~)

(41)z 23 = - jxide (42)[The polarity marks signify that the mutual flux links

the two windings in a manner to produce maximum volt-ages at the same instant at the marked ends of the wind-ings. See Fig. 9(b) .]

z 24 = -L (43)Zzs= 0, etc. (44)212 = 221 (45)Z13 = &I etc. (46)

Mesh Em&--Reference-positive directions for branchemfs, E,, &, etc., are shown by arrows associated there-

The cofactor of a term in a determinant is the minordeterminant obtained by eliminating the row and columncontaining that term, this minor being prefixed by a +- sign depending on whether the sum of column num-zr and row number is even or odd respectively. In (57),the first subscripts define columns, the second rows. Thusthe cofactor of a term has a + sign if the sum of sub-scripts on the term is even. Since Z,, = Z,,, it can be shownthat A,, = A,,.

In (56), the term involving El is the current that flowsin mesh p if all emfs are set equal to zero except El . Simi-

&A p2larly, the term D is the current that flows in mesh pif Ez alone acts and El, ES, etc., are equal to zero.Specifically :

with.A mesh emf is the sum of the branch emfs acting around

that particular mesh in the reference direction.where D is

The same reference direction will be used for mesh emfsAnd:

as for mesh currents.Specifically in Fig. 9.

El=E,--EtE2= -ECE3= -Et,Ec=& +Er+.Eo

(47)(48)(49)(50)Etc.

E,=E,EquationsKirchoffs Law

drop around each closed meshpressed in that mesh.

L.%1+~2~21+~3&1+ ul2+~2~22+u32+ *Ml3+~2%3+&&,+ *

. . .

. . .

. . .

(51)states that the voltagemust equal the emf im-

~~+I,Znl=Er (52)-+I,Zn2=E2 (53)l l +I&,, = Et (54). .

. .

. .

Mesh Currents*-Equations (52) to (55) can besolved for the mesh currents 11 to I,. The solution forcurrent in any particular mesh, p, is:

I E&I E&z &A,,P =D+D+D+. . .+yLT (56)

where D is the determinant of coefficients~d21Z,1* * * zll&2222&2 * * - zl2D = 2& 23233~ . . Zn3 (57). . . .

. . . .

. . . .ZnZ2n.L * - * zlrl

and A,, is the cofactor of Z,, in the determinant (57).*The method of determinants is used to state the solution here.

However, any method of solving the simul taneous equations (52)to (55) for the unknown currents 11 to I, may be used.

given by (57).Z12&2* * 42Azl = - Z& f& . . Z,a (59). . .. . .. . .

&l&n - l - zm

Currents can now be obtained by combina-vector sum of all mesh currents flowing in aion. Inc.branch is the branch current.

Specifically with reference to Fig. 9.

Etc.

I,==II&,=&-I2I,= I2Im=14--la

(62)(63)(64)(65)

Brunch Voltages-The branch voltages, E,, EYI,etc., or the voltages between any two conductively con-nected points in the network, as E,,, can be obtained byvectorial addition of all voltages, both emfs and dropsthrough any path connecting the two points.

The voltage drop from x to y, Dxy, and the voltage ofpoint x above point y, Exy, are the same.(Note that drop is measured from first subscript to sec-ond. The voltage of the first subscript is measured abovethe second.)


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304 Steady-Stat e Performance of Systems I ncludi ng M ethods of Netw ork Solut ion Chapter 10

.Dxy=Exy=ICRC+jl ,(Xi,-XA (66)Dys=&z = IdRd + jIdXid -jIaidtt (67)D wa E,,= -Eo+LRm 033)D =E,, =I,R,+jl,(Xi,-X,)+IdRd

=+jIdXid - jISide. (69)Note that

DXZ = Dx,+D,, (70)E xz=Eys+Exy. (71)

Example of Solution by Equations(a) Given theimpedances and emfs of a network, Fig. 10, required to

Fig. 1Example of a solution by equations.

find the currents. Note: The headings (b), (c), etc., ref erto the corresponding paragraphs b, c, etc., in which themethod and equations are given.(b) Mesh Impedances

Z~~=O+j(7.55+136) =O+j143.55212 = 221 - j113.0222=5+j(150.5-2.65) 15+jl47.85

(c) Mesh EmfsEl= 2.9 +j56.6E 2= -0.29 - j5.66

(d) Equations. It is unnecessary to write these outcompletely since only the solutions are desired. Howeverfor completeness they are:

& (O+j143.55) +12( -4 13.0) = 2.9 +j56.611( -j113.O )+12(5+j147.85) = -0.29-j5.66

(e) Mesh CurrentsD = O+j143.6 -j113.0

-j113.0 5+j147.9 = -84OO+j718. A 11= 5+j147.9A 12 =A 21= +j113AZ = O+j143.6

I E& &An1- D+D

L (2.9+j56.6)(5+j147.9)+(-0.29-j5.66)(jl13)-84OO+j718 -84OO+j718=0.921+j0.007I &&I &h.

2- 0+0

F (2.9+j56.6)(jll3)+(-0.29-j5.66)(+j143.6)-84OO+j718 -84OO+j718= 0.662+jO.O34

& API .Note that the term D=z is the transfer admit-tance between meshes 1 and 2, or is the current in either

one of these meshes per unit of emf impressed in the other.Thus the voltage E2 is -0.1 of El and likewise the currentin mesh 1 resulting from E2 s -0.1 of the current in mesh2 due to El .14. Solution by Reduction

General-The currents flowing in a network of knownimpedances, caused by a given set of applied emfs, can bedetermined by the method of superposition (See Sec. 11).First the solution (currents in all branches of interest) isobtained considering one emf acting with all others setequal to zero. Following the same procedure for each emfin turn, a number of current solutions are obtained. Bythe principle of superposition, the current in any branch,when all emfs are acting at once, is the sum of currents inthat branch caused by each emf acting independently withthe others set equal to zero. The principle of superpositionpresupposes a linear network. The same reference direc-tions must be adhered to for all solutions if the super-position is to be a simple vector addition of the severalcurrent solutions.

The solving of a network involving several emfs is thusreduced to the more fundamental problem of solving anetwork with one impressed emf. This can be accomplishedby the method of reduction.

Fig. 11Solution by reduction. Bridge network current distribu tion.


16/52

Chapter 10 Steady-State Performance of Systems Including Methods of Network Solution 305Solution by Reduction consists of replacing portionsof network, such as Fig. 11 a), by simpler equivalent sec-

tions, Fig. Il(b, c, d), until a simple series circuit results,Fig. 11 (e), which includes the impressed emf and one im-pedance branch. The current is readily calculated. Then,using current distribution factors obtained in the courseof reduction, a reverse process is carried out, expanding thenetwork to its original form and determining the divisionof currents in the process. The methods and equivalentcircuits for carrying out this procedure in general are givenin the subsequent paragraphs.The Network Equi valents will first be given. Networkconstants can be expressed either in admittance or im-pedance form. Some transformations are more readilyperformed in impedance form, such as adding impedancesin series, or delta-to-star transformations. Others are moreconveniently performed in admittance form, such as ad-ding admittances in parallel, or star-to-delta transforma-tions. For more complicated transformations, it is best toconvert constants to the most convenient form for theparticular transformation. For simpler ones, it is usuallypreferred to use one form or the other throughout theproblem.

The common transformations are presented in bothforms. The more complicated and unusual ones only in theform best suited. Impedances (symbol 2) are reciprocalsof admittances (symbol Y) and vice versa.That is

z :1=- 1

y ;1=- 1

(72)

Fig. 12Solution by reduction. Bridge network current dis-tribution.

In all cases, the equivalent circuits are equivalent onlyinsofar as the labeled terminals are concerned. For ex-ample, when a star with mutuals is reduced to a starwithout mutuals, the potential of the center point is notthe same in the equivalent.

Delta and star forms used in general networks are iden-tical with Pi and T forms used in specialized transmissionforms of networks. See Fig. 15. The difference is simplyin the manner of drawing the circuit. Thus the star-deltaand delta-star transformations are at once, T to Pi andPi to T transformations. The arrow between parts of thefigures indicates that the figure on the left is being trans-formed to the figure on the right. It is assumed then thatthe currents are determined for the figure on the rightand the equations under the figure on the left are fordetermining the resulting currents (or voltages) in it.15. Transformations in Impedance Forma. Impedances in Ser ies (Fi g. 13)

Fig. 13Impedances in series.21 21

el=Z1+Zz=Z (75) z=z,+z, (74)e2=Z1+Z2=Z (76)b. Impedances in Parall el (F ig. 14)~The parallel

of two impedances is the product divided by the sum.

. 2221=z1+zz

Fig. 14-Impedances in

. 21zp=zl+zz

parallel.

(78) z=$$ (77)1 2(79

Fig. 15Pi and delta; t and stars or Y are the same.


17/52

306 Steady-State Performance of Systems Including Methods of Network Solution Chapter Suggested order of calculation (80)

D=Zl+Zz22il = -D = current in Z1 per unit current in 2. (81)

i2 = 1 -iI = current in Z2 per unit current in 2. (859(il and i2 are current distribution factors.)z = il& (83)c. Delta to Star Transformation or Pi to T (F ig.16)The star impedances are the product of adjacent

delta impedances divided by the sum of all delta imped-ances.

Fig. 16Delta to star-impedance form.

I sb = - %I.+$Ib (88a) (84)@W

(85)

I,,= -%I+5 zabIa (goa) Zc=!!i$.f= - ibcIc + iabIa (gob) D = Zab+Zbc+Zca (87)

Suggested order of calculation*D = &b+Zbc+Zca (87)

ziab LbD (91)za = Zcaiab

zb = &bib0&ai,, = -D (95)& = Zbcica (96)(i&, ibc, i,, are current distribution factors.)d. Star to Delta Transformation or T to Pi(F ig. 17)b

Fig. 17Star to delta-impedance form.*Then after ,, b, nd I, ave been ound, ,,,ab, nd Ih can be

determined sing qs. (8Sb), 89b), nd (90b).

I, = La IabIb = Iab - IbcI, = Ibo - ha

(101)(102)(103)

zab Dz,&,&,c= Dzbz,Z,, = DZ,Z,D=$+++j- (1a b 0

Alternative forms of the transformation formulas lows :

Num. =Z,Zb+Z,Z,+ZbZ, (1Num.z =-jj--- zazbab 0

=za+zb+z 0Num.

zbo= -jj- a=zb+zc+y a

zbzo

(1(1

Z Z ZaNzm ,+Za++a=----= (1e. Star with Mutual s to(Fig. 18) Star without Mutua

Fig. 18Star with mutuals to star without mutuals-impedance form.U

ZB=Zb+Zca-Zbc-Zab (1ZC=Zc+Zab-Zca-Zbo u

Polarity marks require that with all reference directiofrom center outward as shown, all self and mutual dare from center outward. That is, it is understood with the polarity marks as shown, the voltage drop the center to a will be written:

Dna = I aza+ I bZab + I cZcaand the numerical (vector) values and signs assignedZ& and Z,, must be such as to make this true. It folthat Z& is defined as the voltage drop from n to a divby the current from n to b that causes the drop.

Special case: Star w ith one mu tual between two branto star with out mu tual . (Fig. 19.)

Fig. 19Star with one mutual to star without mutuimpedance form.ZA=Za+ZbczB=zb-zb,,i&C=& --ii&,


18/52

Chapter 10 Steady-State Performance of Systems Including Methods of Network Solution 307

f . Two Self Impedances and a Mutual Trans-formed to an Equi valent Star or T. Or the Equi v-alent Circuit of a Two-winding Transformer (Fi g.20)

Fig. 20Two self impedances and a mutual transformed toan equivalent star or T. Or the equivalent circuit of a twowinding transformer.

z1= 11- 12 (1 4z2 222zn (115)NOTE: This transformation involves bringing b and d to the

same potential and is permissible only when these potentials arenot otherwise fixed. Strictly, the form on the right is equivalentto that on the left with switch S closed. However, if the closureof S would not alter the current division, it can be consideredclosed and the equivalent circuit used. The resulting potentialsEab, and IL will be correct but the potentials E,, and Edb, whichare definite in the equivalent, are actually indeterminate in theoriginal circuit and must not be construed as applying there.See note under e for meaning of polarity marks, considering band d as point n.

This is the familiar equivalent circuit of a two-windingtransformer, provided all impedances have first beenplaced on a common turns basis. In this case .Z12 s theexciting impedance and 2 = Z1+Zz the leakage impedance.16. Transformations in Admittance Forma. Admi ttances in Seri es (F ig. 21 )

Fig. 21Admittances in series.

Y2el=yl+yz (117) KY2y=a 016)Yl--e2- Yl+Y, 018)

Suggested order of calculation.D= Yl+Yz (119)

y2el=- D (120)e2=1-erY = Ylel, or Y2e2

(121)(122)

b. Admi ttances in Parall el (F ig. 22)i~=*~=$ (124) Y= Yl+Yz (123)

V V

Fig. 22Admittances in parallel.

c. General Star to Mesh Transformation, orElimination of a Junction (Fi g. 23)

Fig. 23General star to mesh transformation, or eliminationof a junction-admittance form.

NOTE: A network can be solved by eliminating one junctionpoint after another until a single-branch mesh remains.

RULE : A mesh branch is the product of adjacent starbranches divided by the sum of all star branches.The mesh contains n/2 (n - 1) branches, where n is the numberof star branches.Ir=Izl+&L+ l 9 $-In1 (130) KY2&2=-g- (126)12=112+132+ - - * +1,2 (131) 1 3Y13 D- (127)

etc. etc.I,= L,+Iz,+ - . . +Lp (13% Y,,=Y (128)

D=Y,+Y2+&+-+Yn (129)In which the positive reference direction for any meshcurrent I,, is toward terminal q.Suggested order of calculation.D= yl+y2+y3+ l * * +Yn (133) +c2 (137)

Yl--I=D k1 (134) Y23= D2-2 = k2Y3 (138)KY2

D12=-----=k y1 2 (135) Y2~=~=hYI (139)

KY3Y -,=klY33 -etc.

(136) etc.

d. Star to Delta or T to Pi(Specia1 case of c) Fig. 24)I a = ,, - Lb (144) y.,=y (140)Ib = Iab - Ibo


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308 Steady-State Performance of Systems Including Methods of Network Solution Chapter

Fig. 24Star to delta-admittance form.IO = Ibc - I,, (146) Y,,= y (142)

D= Y,+Yb+Yc (143)Suggested order of calculation same as for general

transformation. (c)e. Delta to Star Tramformation, or Pi to T (F ig. 25)

to star-admittance form.I b = -i,,I,+ibcIb (151)I c = -ia&,+icaIo (152)I ca -ib&+ia& (153)

ya = D yea yabYb=DYabYbcyo = D ybo yea

(147)(148)(149)

1where &b = -Dyab (154) D=++&++ (150)ab 081

ibo = F (155)ux bo1i,, = -D Yea (156)

17. Examples of Solution by ReductionA power-distribution system network is solved by the

method of reduction in Sec. 20.The following example is also given showing the method

when two sources of emf are present.Solve for the currents in the network of Fig. 11(a) bythe method of reduction for applied voltages as follows:

Case 1. E,= O+jlOO voltsEd = 50 +jo voltsCase 2. E, = 3O+j40 voltsEd = 10 +jSO volts

Obtain unit current division and total current divisionas indicated in Tables 4(a) and 5.The method generally is to resolve the network for one

applied voltage at a time, with the other one set equalto zero.

After solutions have been obtained for each appliedvoltage acting independently, these solutions are super-posed to obtain the current flow with both applied volt-ages acting simultaneouslv.

Particular attention is directed to certain relationships.The total current input resulting from a particular aplied emf is obtained by dividing it by the driving-pointimpedance* at that pair of terminals. Thus, when thdriving-point impedances, Z,, and Zdd, at the impressedvoltage terminals have been determined, and the uncurrent divisions developed, the network is solved. This illustrated by Cases 1 and 2, which differ only in magnitude of the applied voltages. The resulting currents these two cases are obtained from the same basic networksolution.Solutiona. Eliminate the mutual as per Eqs. (lll-113) and combine impedances in series forming &.

Zb =j20 -jlO =jlOZk =j30 -jlO =j20zg =j1o+j30 =j40

b. Let Ed = 0 and replace star Ze, Zd, Zh by its equivalent delta from Eqs. (97-100).

D&++?-320 j5 jl0 = -j(O.O5+0.20+0.10) = -jO.35Zeh = (- j0.35) (j20) (jl0) = j70Zde = ( - j0.35) (j20) (j 5) = j35Zdh= (-jO.35)(j 5)(jlO) =jl7.5

c. Parallel the branches Zde and Zr also, Zdh and 21, Fig. 11(c), obtaining Fig. 11 (d).

D=Zt+Zde=j45. Zde j35zf=z=j45 = 0.78

i&= l-0.78=0.22 (in Z&)Zm=0.22Xj35=j7.77(Parallel Zdh and Z,)

D=Zk+Zdh=j20+jl7.5=j37.5. Zdh j17.5zk=-=-D j37.5 = 0.468

idb= 1 -i&=0.532 (in Zdh)2, = 0.532 X j17.5 = j9.333d. Z,+Z, =Z,, = j17.10

Parallel Z,, with Zeh = 2,D=Z,,+Z,h= jl7.lO+j7O=j87.10ieh

Zz-.2.!?! jl7.10= ___ = 0.196 (through Zh)D j87.10i,, = 1 -&h = 0.804 (through Z,,)Z, = imnZmn = 0.804 X jl7.10 = j13.74

e. The impedance viewed from &a terminals is:Z sa=Z0+Z,=j13.74+j40=j53.74

f. Current Division for unit current in at (a).The symbol (i) has been used for current division

factors.Let prime symbols be used for the currents in terms one ampere total input to the network.*Driving-pointmpedance s that impedance easured look

into ny pair f terminalsf a passive etwork ith ll ther minals erminated n a specifiedanner. In this ase ll thterminalsre hort-circuited.


20/52

Chapter 10 Steady-State Performance of Systems Including Methods of Network Solution 309i*= l.O=&

ieh = ieh = 0.196Ii?mn 0.804i l k = fmn& = 0.804 X0.468 = 0.375 = ,,& h = mnid h 0.804 x0.532 = 0.429if = fmni f 0.804 X0.78 = 0.625

i~e=imni~e=0.804X0.22=0.179&=0.196+0.179=0.375id = 0.196+0.429 = 0.625 = i bid = 0.429 - 0.179 = 0.250

The six currents i,, i,, if, i,, id, ib are given in Table4(a) and constitute the current division corresponding tounit current entering the network at a. Figs. 12(a) to (c)illustrate the steps in dividing the current.

g. Transfer admittances. See Table 4(b):y $=---a=- j53.74 -jO.O1861asY,b = ibt Y,, =Y,, = ,Y,, =etc.

jO.01163jO.00698

Note: The transfer admittance, Ysb, is the current inbranch (b) in the reference direction per unitvoltage impressed in branch (a) in the referencedirection.

TABLE 3

In a similar manner, for voltage applied at (d), thedriving-point impedance Zdd and the current division andtransfer admittances can be obtained. These are given inTables 3 and 4. It is essential that the same referencedirections be maintained for all current divisions, in orderthat the solutions for applied voltages at different terminalscan be superposed.

The current divisions of Table 5 for the conditions in-dicated in the second and third columns are obtaineddirectly from the basic network solution Tables 3 and 4.For example with E, = O+ jlO0,

I =E,=+j100=1@31a3 Zaa j53.74 *(or aaEaYa,~ -jO.Ol86l(jlOO) = 1.861)

Multiplying by the unit current division correspond-ing to current in at a, the currents I , to I f , for 1.861 am-peres in at a, are determined.

This method is particularly advantageous when manydifferent combinations of applied voltages are to be appliedto the same network. It is also convenient to obtain thetransfer admittances, as shown in Table 4(b). These arethe currents in the various branches corresponding to unitvoltages applied at the respective driving points. It isnecessary only to multiply by any actual single appliedvoltage to obtain the corresponding current division. Thereis a check here, for the reciprocal theorem states that thecurrent at (d) for unit voltage applied at (a) must be thesame as the current at (a ) for unit voltage applied at (d).18. Solution by Thevenins Theorem

Thevenin's Theorem2 is useful in analyzing a net-work or part of a network when its reactions at a particularpair of terminals are of prime importance. Through itsuse, a complicated network consisting of several emfs andimpedances can be replaced by a simple series circuit of oneemf and one impedance supplying the pair of terminals ofinterest. The theorem can be stated as follows:

Wit h r espect to any single exter nal circui t conn ected t o anyTable 4(b)-TRANSFER ADMITTANCE (See ote under g)


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310 Steady-State Performance of Systems Including Methods of Network Solution Chapter 1given pai r of term in als of a network, the network can be the circuit of Fig. 26(a), so far as the terminals z, y arer epla ced by a single bran ch havi ng an i mpedance, Z, equalto the im pedan ce-measur ed at th ese termi nal s looki ng int o

concerned. E, is as measured with the terminals z, y opencircuited in Fig. 26(b). Z is as measured in Fig. 26(c) by

th e network (when all th e network emf s are mad e zer o) an dconta in in g a sin gle emf , Eo, equal t o the open-cir cuit voltage

applying any voltage E of the frequency under consider-of th e network acr oss the given-pair of termi nal s.

ation to 5, y and measuring the corresponding vector cur-rent I with the emfs El , Es and E3 short circuited. 2 i

The term emf as used here has a broader meaning than the vector quotient E / I .electromotive force. It is any voltage in the network that An Example of the use of this theorem is found in theremains constant while the impedance connected to theoutput terminals is varied. Thus, the voltage of a battery calculation of short-circuit current on a loaded system.The equivalent circuit of the system, up to the point oof negligible internal impedance is an emf, while the voltage fault, consists of an emf, E,, and an impedance, Z. Edrop in an impedance is not, unless the current is held is the voltage at the point of fault before the fault and iconstant. (See later paragraph). A generator having regu- usually a known system operating voltage. Z is the imlation is segregated into an emf and an internal impedance,back of which the voltage is constant for the particular

pedance looking into the system at the point of fault withall emfs set equal to zero. The short-circuit current is then:

problem and hence can be treated as an emf.The General Case is illustrated by Fig. 26. The emfs, I = $. Thus, it is unnecessary to determine the generatorEl , Ef , and E3 can be of any single -frequency. If morethan one frequency is present, the emfs of each frequency internal voltages. At a given operating voltage E,, andmust be treated separately, as the equivalent circuit will fixed generating capacity, increased load tends to increasenot usually be the same-for different frequencies. The short-circuit current by lowering Z, the driving-point imimpedances may be composed of resistances, inductances pedance at the fault with all system emfs set equal to zeroand capacitances, but must be linear within the accuracy

The method applies equally well to a network in whichnecessary for the problem at hand. A linear impedance is certain fixed currents are forced to flow, as by currentone that satisfies Ohms Law, E =ZZ, Z being a constant. transformers. Examination of the equations of a networkWith these considerations as a basis, Thevenins The- having fixed current input reveals its identity with a net-orem states that the circuit of Fig. 26(d) is equivalent to work of fixed emfs. For example, consider the circuit o

Fig. 27Application of Thevenins Theorem in a network ofixed input currents.

Fig. 27(a). The equations that involve the known inputcurrents are :-LA --&(Ac+D)+I,c+zmD=o (157

+zcB--I,(B+E+C)+I,C+Z,E=o (158The equations involving E, and E, in Fig. 27(b) are:

E,--I,(A+C+D)+Z,C+I,B=O (159Fig. 26-Determination of equivalent network by means of

+E,-L(B+E+C)+Z,C+I,E=O (160Equations for the remainder of the network are the same


22/52

Chapter 10 Steady-State Performance of Systems Including Methods of Network Solution 311for Fig. 27(a) or (b). It is apparent that (157) and (158)are identical with (159) and (160) respectively, if:

E *=-I&i (161)E o= -I&l (162)In other words, the terms - I,A and - IJ3 can be treatedas emfs in applying Thevenins Theorem, and the per-formance at terminals x, y treated through the use ofopen-circuit voltage and driving-point impedance. Thelatter is obtained with the input-current terminals, thatis, the a, b, and c leads from the current transformers,open circuited in Fig. 27(a), or the equivalent emfs, E,and Eb, of Fig. 27(b) set equal to zero.

A more complete discussion is given in ReferenceNumber 2.19. Solution by Circulating Currents

A ladder-type network common where transmission anddistribution circuits parallel each other as in a-c railwayelectrification3 is represented in Fig. 28. The example is,

Fig. 28A general ladder-type net work.however, for a three-phase system. Suppose it is desiredto determine the current division and regulation for theparticular loading condition shown, without making a gen-eral solution of the network. This problem lends itself tothe method of circulating currents.

The voltages at the load buses must first be assumedand the kva loads converted to currents. The sum of thethree load currents flow in the generator and constitutethe current I, in Fig. 29. These load currents and thegenerator current are assumed to be fixed for the balanceof the problem.

The division of I, between I,, and I, is next assumed.Now the voltage drops from 1 to 2, 2 to 5 and 2 to 3 can

Fig. 29Impedance diagram for the system of Fig.ing the method of circulating currents. 28. Show-

be calculated. The current Ih is obtained by subtractingthe load at 5 from I,, and then the drop from 5 to 6 canbe computed. Knowing the drops from 2 to 3 and 2 to 6,for the assumed current division, the drop from 3 to 6 isobtained by subtraction. This drop divided by the imped-ance j gives the current If. Now the balance of the currentscan be obtained, that is, Id and Ik. However unless aperfect guess has been made, the drop 34-7 will differfrom the drop 3-6-7, and an arbitrary voltage, AE, must beincluded to make the voltages around the loop add to Zero.So far an &act solution has been obtained for the case ofcertain load currents, a particular generator voltage anda voltage AE. However, the solution is desired withoutAE. To obtain it a solution is next obtained for -AEacting alone. According to the conditions of the problemthe load and generator currents are fixed so that thesebranches are considered open circuited when computingthe currents caused by -AE alone. This solution is there-fore quite simple and gives rise to a set of currents I,,I& . . . Ik, which are the circulating currents for whichthe method is named.

Now let these two solutions be superposed; that is:I, = I,+ I,Id = Id+ Id etc.The resulting solution does not involve AE since the AE

of the first solution is canceled by the -AE of the secondsolution. It is therefore an exact solution for the loadcurrents assumed. The voltage drops from the generatorto he several load points can now be computed, since thecurrents are known. Also, from the new load voltages, andfrom the load currents that have been held fixed through-out the solution, new load kvas and power factors canbe computed.

The net result is an exact solution for a set of conditionsthat differs more or less from those originally assumed.While this can be used as a basis for a second approximationit is more generally considered the engineering answer.The loads are usually not known exactly; the solutionobtained provides an exact reference point in the regionof the loads assumed, and thereby provides a tangiblebasis for engineering judgment.

There is much to be said for this type of solution as asystem design tool, since it capitalizes experience and fore-knowledge of the order of magnitude of the answer. Asan example the network of Fig. 28 has been solved for theloads indicated thereon.Example of Method of Circulating Current-Thenetwork diagram, Fig. 29, is obtained from the single-linediagram as outlined in Secs. 2 and 3. The 15 OOO-kvatransformer impedance should be converted to ohms on a34.5-kv base and then multiplied by (13.8/33) to convertto the 13.8-kv base at the load. The resulting diagramis on the load-voltage base, a conversion being necessaryto change to or from the generator-voltage base, which isalso nominally 13.8 kv. Thus with the generator at 13.8kv, the corresponding voltage to be applied in Fig. 29 is13.8(34.5/13.8) (13.8/33) = 14.42 kv or 4.5 percent abovenormal. In an actual case transformer resistances shouldbe included as these are significant in regulation andloss calculations.


23/52

312 Steady-State Performance of Systems Including Methods of Network Solution Chapter As a first approximation, assume the regulation in step-

up and step-down transformers to total 10 percent with anadditional 10 percent in lines. Allowing for a 4.5 percentabove normal voltage at the generator, the loads should beconverted to currents based on approximately 85 percentvoltage or 11 700 volts. The load currents are given in thefollowing tabulation.

In the current distribution calculation which follows, thecurrent I , must be guessed, or taken arbitrarily. Latera circulating current is determined, which, added to thearbitrarily assumed value, gives the correct current. Itis advantageous to guess as close as possible so that thecorrecting circulating current is small. In fact if the guessis sufficiently close, the labor of calculating the distributionof circulating current can be saved.

Ia = 638I , - 300Ic=I,--I* = 338I s = 245Ih==& Ic , =I 55IO = 338z, - 0.53 +j1.176MO = 179 +j397I, = 300z3 c=l +jl.91LZtJ m + j573Ih = 55zh a 3.03+j6.37IhZh = 166.5 +j350LA c3 + j573D 266 = 166.5 +j923ICZ - 179 +j397IfZf = - 13.5 +j526Zf a +j1.91If 5=) 275 +j7.1IO CJ 338Id=& & = 63 - j7.1IhIf

= 55= 275 +j7.1

Ih+IfI6Ikzd

- 330 +j7.1= 295= 35 +j7.1D 0.53+j3.086

IdIdZdSC

~3 63 -j7.1- 55.3 +j190.7CI 3.03+j6.37,

IkIkzk

c=l 35 +j7.1= 60.8 +j244.5

IfZf = - 13.5 +j526D 67 = 47.3 +j770.5D 47=IdZd = 55.3 +j190.7AE = - 8.0 +j579.8

Solut i on for Circulat ing Cur rent .

Find the impedance as viewed from AE with load agenerator branches taken as constant current branchesi.e., open circuited for this calculation. Apply a voltagAE = negative of the AE required to close the mesh in tabove calculation.

Z* - 0.53 +j 1.176&A e3 +j 1.91zh - 3.03 +j 6.37Z 3266 = 3.56 +j 9.456Zf L3 +j 1.91D - 3.56 +j11.366l/D = 0.0251 -j 0.0801Zf m +j 1.91G/D = i3266 = O.l53O+j 0.0479if = 1 -is266 = 0.8470 -j 0.0479Zf El +j 1.91Z par = O.O915+j 1.618zd = 0.53 +j 3.086zk = 3.03 +j 6.37Z = 3.6515+j11.074l/Z =I 0.0269-j 0.0814-AE=AE= 8.0 - j579.8I : =AE/Z = -46.98 -j16.25Id= -Ik = 46.98 +j16.25&2S6 = O.l53O+j 0.0479IO = 6.41 +j 4.74I[=I,-Id= -40.57 -jll.51I i = -I , = - 6.41 -j 4.74Ihf = -I, = - 6.41 -j 4.74

The currents from the arbitrary distribution (requirinAE to close) and the circulating currents contributed AE = -AE, are now combined to get the actual current vision for the load currents assumed. The circulating crents are distinguished by prime symbols, the total divisiby double-primes. Fig. 30 illustrates this superposition.

I , =.I, = 638 +jOI, = 300 +jOI, = - 6.4 -j4.7I , ~=l 293.6 -j4.7I O 0 338 +jOIO c3 6.4 +j4:7I , = 344.4 +j4.7Ih = 55 +jOIh = - 6.4 -j4.7Ih If - 48.6 -j4.7If = 275 +j 7.1If = - 40.6 -jll.SIf II = 234.4 -j 4.4Id = 63 -j 7.1Id - 47.0 +j16.3Id II = 110.0 +j 9.2Ik - 35 +j 7.1Ik = - 47.0 -j16.3Ik II =--I2 -j 9.2

Check of Drops Around Loops. This solution can checked by checking voltage drops around each loop.


24/52

Chapter 10 Steady-State Per form ance of Systems In cludi ng M eth ods of N etw ork Solu ti on

Fig. 30-Solution of the network of Fig. 28 by the method ofcirculating currents.

Calcul ate th e dr op rom generator bu s to oad point 6.L = 638.0 +jOZ3 = 0.53+jU311L&z, = 338.1 +j1155.4I / Z , = 9.0 +j560.8D 26 = 347.1 +j1716.2

Regula ti on-Determi nat ion of L oad Volta ges. Themagnitude of the generator voltage is known, but not itsphase position. The phase position of the load voltageE6 is known but not its magnitude. It is at an 80-percentpower factor position with respect to 16. The drop fromgenerator to load point 5, D z6, is known vectorially with16 as reference. Thus the magnitude, E6, can be deter-mined by the solution of a quadratic equation as shownbelow, or graphically as indicated in Fig. 31.

E, = E,(0.8+j0.6)&+Dn,=E,E,2 (14427)~= ~ = (0.8E6+347.1)+ (0.6E6+1716.2)18E,+2(1307.4)E6-66.313x10~=0E 6 = - 1307.4 _+d( 1307.4)2+66.313 x lo6

Fig. 31Graphical determination of the load voltage & inFig. 28.

1. The load urrent a s long he eferenceine.2. Draw circlef adius 330 olts n which , must terminate.3. Draw constructionine t the load ower factor 80 ercent)along hich !j ust lie.4. Construct he oltage rop vector,&, and move it arallel

to itself,ith one end followinghe generator oltage ircle,until he ther nd fallsn the oad oltage onstructionine.

6. E, and E6 can then e drawn in nd their ector alues caledOff.

E, = - 1307.4 + 8248= 6941 volts L-N= 12 022 volts L-L= 87.12 percent of 13 800=83.33 percent of 14 427

i!& = 6941 x (0.8+j0.6)&= 5553+j4165

Check of Voltage Dr op from 1 to 6.E6= 5553+j4165

&= 347+j1716E, = 5900 + j5881= 8330.4 L-N

14428 L-LThe remaining load voltages are readily determined as

follows:if35 =IhZh =E8 =

==

Ee =IkWZk =E, ==

==

5553 +j4165177.2 +j295.35375.8 +j3869.7

6624 L-N11 473 L-L83.14a/, of 13 80079,52a/, of 14 4275375.8 +j3869.722.2 -j 104.35353.6 +j3974.0

6667 L-N11 547 L-L83.68% of 13 80080.04% of 14 427


25/52

314 Steady-State Performance of Systems Including Methods of Network Solution Chapter 10Load Power Calculations.

&3fa

= ( 5.553 -$.165&v= 735

P5+jQ,= (4081 +j3061)kva= 5101 kvaat 80.00 percent power factor lagging.

9,= (5.375 $869)kv= 885P6+jQ6 = (~7~~861+f$Wkva=

at 81.16 percent power factor lagging.E7 =( 5.354 +j3974)kv317 = 294 -jOP,+jQ,= (1574 +j1168)kva= 1960 kva

at 80.31 percent power factor lagging.Generator Output Power.

ffg=( 5.900 +j5.881)kv= 1914 -jO

P,+jQg= (11293 +jll256)kva= 15 945 kvaat 70.82 percent power factor lagging.

Loss Calculation.Pl+jQs= 4081 +j3061PB+jQ6 = 4757 +j3424P,+jQ,= 1574 +jll68

Total of Loads= 10 412+j7653P,+jQp= 11 293+jll 256Losses = 881 +j3603Kw Line Loss = 881, 1 I 293

=7.80 percent of generatoroutput.

In an actual case transformer resistances must be in-cluded in the diagram as these are significant in regulationand loss calculations. Transformer iron losses must beadded to the copper losses thus determined to obtain thetotal loss.

The solution given in Fig. 30 is exact for the conditionsshown on the figure, which differ slightly from the originalassumptions of Fig. 28. However, the total load is offonly 1.3 percent and the regulation values therefore applyclosely for the original conditions. In a practical problemit is not significant that the answer does not apply exactlyto the original load assumptions. If the work is done witha calculating machine so that s

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27190724 Steady State Performance of Electrical Systems