Lood Flow Studies

2.7. INTRODUCT]ONThe various operation aspects of an electrical power system, the symmetrical steady state operatiorithe most important mode of operation. A knowledge of this mode of operation is essential tosupply of real and reactive powers demanded by various loads, with the fiequency and the variousvoltages maintained rvithin specified tolerances and with optimum economy. Study of this modetrperation is carried out to arrive at the most satisfactory layout at the planning stage and to maiqualityand economy of power supply while the systems is in operation.

The rnathematical formulation ol the kjad flow study results in a system of non-linear eq'fhese quantities can be written interms of either the bus admittance matrix or bus impedance'l'he former is more suitable to digital computer analysis. because of the cases with which theadmittance matrix could be formed and modified for network changes in subsequent changes.this approactr (adrnittance matrix) is the most economical from the point of view of computerand memory requirement , i.e., we can store n

(n-+ l) terms are to be stored in yru. for an

2system.

The symmetrical steady state operation can be classified as follows :l. System modeling and load flow analysis2. Optimum eenerating strategy and3. System control.In this chapter. we present only systen"r modeling and load flow analysis.

2.2 NECESSITY OF LOAD FLOW STUDIESLoad flow solution is a solution of the network under steady state operation subjected toinequality constraints under which the system operates. These constraints can be in the tbrm ofnode voltages, reactive power generation of the generators, the tap settings of the taptransformers under load. etc.

Load tlow studies are conducted at the stage ofplanning, operation and control. They are useddetermine the magnitude and phase angle of load buses and active and reactive power injected atand also active and reactive power flows over transmission lines. This information is necessary forfollowing functions ,

66

Load. flow Studres 57

(i) To keep the voltage level of certain buses within closed tolerances by proper reactive powerscheduling.

(ii) The total active power generation must be equal to the load demand plus losses. This should bedivided amongst the generators in a unique ratio for optimum economic operation. Load flowstudies are used to maintain the above ratio.

(iii) The effects of disturbances, which may result in systern failures, can be minimized by properprefault load flow strategies.

(rv) To analyze the effectiveness of alternative plans for future system expansion to meet the in-creased.load demand or for designing a new system.

(v) To determine the best location for capacitors or voltage regulator for improvemenf Ofroltageregulation.

(vi) Load flow studies are required at various stages of transient or dynamic stability analysis.Definition: The study, which gives steady state solution of power system network, for a specified

bus conditions, with certain unequal contraints, is known as load flow study or power flow study andthe solution obtained is known as load flow solution.

2.3 DATA FOR LOAD FLOW STUDIES2.3.L Data at the BusesIn general, a bus in an electrical power system is fed from generating units which inject active andreactive powers into it and loads receive active and reactive pewers from it. In the load flow studies,the generator and load (complex) powers are lumped into a net (complex) power. This net (complex)power is called the bus injected power.

The net power injected in the bus is given byS : P *7Q : (Pc +/Qo)

- (Pn +.lQn)

S: (pc_pD) +/(ec_eo).where Po, Qc : Generation real and reactive powers

Pp, Qn : Load real and reactive PowersP, Q : Injected real and reactive powers

ln addition to the above quantities, ryragnitude and phase angle of the voltage are also associatedwith each bus of the four quantities at a bus, viz. , active bus power, reactive bus power, bus voltagemagnitude and bus voltage phase angle, two quantities are specified, the remaining two quantities tobe obtained through the load flow solution. When all the four quantities at every bus in the powersystem are known. active and reactive power flows in all the transmission lines can be calculated.

Depending upon which quantities have been specified, the buses are classified in the followingthree categories:

(i) Load Bus : Load bus is one at which the active and reactive powers are specified. It is desiredto determine the magnitude and phase angle of voltage through load flow study. It is requiredto specify only Pp and Qp at such a bus voltage can be allowed to vary with in permissiblevalues i.e.,5%. Also bus voltage phase angle is not very important for the load.

(ii) Generator Bus or Voltage Controlled Bus : Generator bus or voltage controlled bus is one atwhich the voltage magnitude corresponding to the generation voltage and active power (P6)corresponding to its ratings are specified. It is required to determine the reactive power genera-tion (Qc) and bus voltage phase angle.

il . eU*rical Power Sgstem A(iii) Slack or Swing or Reference Bus : This type of bus is distinguished from the other two

by the fact that active and reactive powers at this bus are not specified but voltageand phase angle are specified. Normally there is only one bus of this type in a givensystem.The need of such a bus is as tbllows :

In a load flow study active and reactive powers can't be fixed a priori at all the buses as thecomplex power flow into the network is not known in advance, the system power loss beingtill the load flow solution is completed. It is therefore necessary to have one bus (i.e., slack bus)which complex power is unspecified so that it supplies the difference in the total system loadlosses and the sum of the complex powers specified at the remaining buses. By the same reasoningslack bus must be a generator bus.

The Table 2.1 summarized the above discussion.Table 2.1

Bus rype Specified quantities Can be determinedquantities

Load busGenerator busSlack bus

P,QP, lvllv 1,5

I vl,5Q,5P,Q

2.3.2 Representatlon of Transmlssion LlnesSince the load flow study is an aspect of the symmetrical steady state operation. the three phaseis solved on per phase basis. Also, only positive sequence equivalent circuits of the systemare considered.

The network model of a power system, it is sufficiently accurate to represent a short line by aimpedance and a long line by a nominal n model.

2.3.3 Representatlon of TransformelsA power transformer without tap-changing facility is represented by a lumped series positiveThe transformers with tap changing facility and the phase shifting transformers are discussed below :2.3.3.1 fixed top:elting lronrlormers: A transformer with a fixed tap setting and connectedbuses 7' and'q' is represented by its positive sequence series impedance /admittance in series withideal auto transfbrmer having a turns ratio of d : 1 as shown in Fig. 2.1.

bus

a:tFig. 2.1 Transformer with a rtxed. tap se*itg

Load FIow Studies 69

From the Fig. 2.1,v, l,oT-

and lrr: (Vr -Vq)Ypq

From equations (2.1) and (2.2)

...(2.r\

...(2.2)

...(2.3)

...(2.4\

...(2.s',)

adSubstituting V, from equation (2.4) ln equation (2.3)

'r:4:W"r,'aaY,:+

,,:(\-%l L' \4 ',) a

I,r : (Vs -Y)Yrn

: (Nq-U)+

"o: *, YB :

:(: ,)"-

".: (' -i)'-

Similarly

' " '(2'6',)The above transformer connected between the buses 'p' anid '4' is represented by an equivalent n

ndel as shown in Fig. 2.2.bus P,

lo+

ftom Fig. 2.2, Eig' 2'2 7

ln: (Yo-Vr)Yn + VpYBI lo= (Yr-Vr)Ye + V4YcSolving equations (2.5) to (2.8), we ger

...(2.7)

...(2.8)

...(2.e)!b mathematical model given in equation (2.9) is used to represent a transformer with fixed tap

in load flow sordies.Iop thonging under lood tronslormer : In the case of a tap changing under load (TCUL)

, the tapping is changed i.e., the value of 'c'is varied to maintain the voltage magnitudehr the specified tolerances. The load flow equationi are solved by numerical methods involvingrErlrin number of iterations. The value of 'a' is changed normally once in two iterations, with thislq: of transformer also represented by an equivalent n model.

6 busq

7A Elrctrical Power2.3.3.3 Phose:tiltirq lronelorners: It is used to advance the phase angle of the bus,voltage. Itrepresented by an ideal transformer as in previous cases but wittr a turns ratio which is anumber. Let the turns ratio be (a, + 7br). The transformer is represented as shown in Fig. 2.3.

Fictitiouslro

Jtr(a. +j bJ, ,r,!. '*

Fig. 2.3 Phose sbifting *ansfonnnNeglecting losses, for the transformer

and lrn: (V, -Y,)Yp,tFrom equation (2.10)

,: l'o -

l,-

'P (a, + ib')" a' - ib'Substituting I., from equation (2.11) in equation (2.L2)

.

-

(v,-vr)Y,'n--f,j4

In a similar way

For the convenience of nodal analysis, the phase shifting transformer is representedlbllowing model

lr:YoYro*YoY*lr=YrY*+YoY*

:[#_",] #rv.: (Vp

-(a, * ibr)Vq) dhIo = (Y, - V) Ypq:[',

.ut ]t: [(a, + jb)Vq-Vi&

: u,,, from equationas + Jos

and

Load. Flow Studies 71

t3.4 Gharacterlstics,of Load Models

Ynn(a,

-

jb,):Y

>Ib response of nearly all loads to voltage changes can be represented by some combination ofEsant impedance, constant current and constant power (or MVA). Actually, the constant currentd is unnecessary as it is nearly equivalent to 50Vo constant impedance load combined with 50%iE lrnt power load. It has been found convenient to retain the constant current model as it is easilyoqrehended and is frequently used in the absence of more complete data. Figs. 2.4 and2.5 showth relationships of load current aqd power with voitage for three simple load types.

Tb constant power type load representation is the most severe representations from the systemffiity point of view because of the affect in amplifying voltage oscillations. For example, a drop in*te will cause an increase in load current resulting a further voltage drop.

f.mversely, constant impedance load have a decided damping effect on voltage oscillations.

Loaa cunent f 0.8

1'o ----rl.'*n" 1'4

Eig. 2.4 Retatiowhip between load curyent and nodc wbage for siwple had. types

Solving equations (2.13) to (2.15), we getYro:

1o*ur1,Ypq: -

Y --

Y*

It can be observed *^rJ;:r-9'DrydYw

1.0 1.2 1.4-->Voltage

Relotionship between load. ond. node wltage for situple toad typesFig. 2.5

Load Flotu Studres 77

(ii) The other elements are called otT-diagonal elements. These are weaker in thbir magnitudes.(iifi As the number of buses are increased, most of the off-diagonal elements will be zero.(iv) If an inducting shunt element is added to the bus, it witl strengthen the magnitude of total

self admittance of that bus and does not change the magnitude of mutual elements. This will bevice-versa if the added shunt element to the bus is capacitive.

The net injected complex power into bus 1Sr : Pr +7Q, : V, Il

Frrom equation (2.17)Sr : Pr *"/Qr : V1 (Y11 Vl + Yt2 V2 t Y13 V3)*

: lvrlei6,,ly,,l e.r3r, I v1 | d6, + I Yrz I fen lY2l d6, + I Y,r I eis" lYzl ru')-: lVr2y11 lg-10,, + lvr Vzy,z lel(6,-62-6,,) + lVrV:y,: lel(6,-6:-E':) ...(z.zl)

g{arating the real and imaginary parts of equation (2.21)fb real part is,

Pr : lV,'Y,,1 cos 0,, + lVr Vz Yrzl cos (6, - 6z - 6,i + lVr V: Y,rl cos (5, - 6:- 0r:)(2.22)

t imaginary part isQr: [l V,'Y,, lsin0,r + lvrv:Yrz lsin(61-62-0rz) + lvrvrYr: lsin(6,-6r-0r:)]

]hilarlyrealandreactivepowerinjectedintobus2andbus3.hgrneral. tbr 3 bus system

3

Pr: I Yo'Yo, I cos0r, + IN, YrY*lcos(5r-Dq-ee,)Q=l

*p

3

: I lV, v, Yr, I cos (6, - 6, - opq), for p : 1,2,3 ...(2.24)q=lIhilerly,

Qp : IV, YnYorlsin (6r - 6q- 0/,q). for p : 1,2, 3q=l

GAUSS SEIDEL METHODiterative algorithm for solving a set of non-linear algebraic equations. To start with reasonable

ralues, assumed tbr all the unknown quantities. The assumption is based on experience. Forhrses, the magnitude and phase of (bus) voltages are unknown. Some reasonable values are

forthese. If we choose the same values forall the P-Q bus voltages, usually 1.0 Z0o p.u.,*srmption is called a flat voltage start, for P - V buses, some reasonable value (usually 0') is

for the voltage phase angle. Then, using these assumed values as well as specified values, ag of bus voltage magnitudes and phase angles are computed as explained below. The voltageriude and phase angle are known or specified for a swing bus and hence remain fixed, i.e.,

...(2.25)

throughout the computational process.

78 Etectrical Power System Analysis2.5.L P-V Buses are AbsentLet there be a total 'n ' number of bpses.

WhenthererisnoP-Vbusesarepresent,thatmeans,wehaven-lloadbtrsesorP-Qbusespresent. the remaining one bus is swing bus.

Unknown variables are

Yp: I Vr 1 ei6' p :2, ..., n* 1, is slack bus

which are (n -

l) complex unknowns Vz, Vr , ... ,Y oand complex power Sr : Pr + jQ1, at thebuscanonlybedetermined.if theunknown I Vp land6ratthe(n - l)loadbusesaredeterminedThe current entering intopth bus of an'n 'bus'system is given by

Ir: iv, Yr, -vtYpp+ Luouo*p:2,3,...,nq=l n;;The complex power injected into pd'bus is

Sp : P, * jQo: Yol,Si : P, -iQr: v*rt,to: (pr_ierl tV)

Note. The load flow solution can be solved more easily if I, rather than I|.From equation (2.26). we get

Substituting I, from equation (2.27) in equation (2.28\

(or)

u,: *[,, arr,J

For Gauss iterative method / \

v"**r- I In, -J?, -frr,,u;lo Y,, I t"; t" ,,ri,' * ' )The bus voltage equation (2.30) can also be solved by the Gauss-Seidel method. For this

the new calculated voltage Vr* * t immediately replaces Vrt and is used in the solution of sutcalculations. For Gauss Seidel method the equation (2.30) can be modified as

"r"rlv-,+,- I tt+ -ir,,v;*,- t'P -'-[

tr,, ,;,0 q=p+t

I-oad. Ftow Shrdres 79

The second term on the R.H.S. of the equation (2.31) is clear beeause the voltage prior to bus p'should correspond to the value as calculated during the current iteration.

Computing time can be reduced to considerable extent by performing as many arithmeticoperations as possible before starting the repetitive iterative computation' This means that thearithmetic operations which do not change during the iterative process should performed once and forall at the staiting. Since P, Q and Y at a bus don't changO-with iterations'

Let 6. : P'-

"lQn , forp = 2, 3, ..., tt

" Y" * l, slackbus

,*:?*,

...(2.32)

...(2.33')d for p : 2; 3, ..., n

4:1,2, "' n, q*PThe equation (2.31) can be written as

, ,

v^r * r : ol, - ir, vk +r - Lrrrul ..,(2.34\'P (vj). h q=p+lL5.l.t Acrelerolion lodor: The convergence Ii-. ,i*". be speeded up*by the

.use .of the acceleration

;; r"r tfr. p;us, the accelerated value of voltage at the (/c + l)th iEration is given byvj,l"'.,:vo*+cr(vpft*t-vr*) "'(2'35)

rhere cr is a real number called the acceleration factor. Use of acceleration factor reduces the totalnmber of iterations required considerably. A generally recommended range is 1'3 < cr S 1'6' Arrong choice of cr may indeed slower convergence or some times even result in divergence from theolution. The Gauss-ieidel method with the use of acceleration factor is known as the method ofsccessive over relaxation.

15.1.2 Algorithm lorGours -seidel method when PV bus is obsentl. Read system data and formulate the Y"r.2. AssumeinitialbusvoltagesVro : 1 +i0' for p :2,3, "', n

+ 1, slackbus3. Set iterationcountft : Oand I AV.r* | : e4. Compute A, and Br., from equations (2.32) and (2'33)5. Set bus count P : I6. If bus is PQ bus, then

(i) Comput" Vr* * I from equation (2.34)(il) Compute aVpk - voo*' -vro

If lAvr*l)lAV,o,*l(lii) Assign new voltage to old..r.,yrr

_ vro *

Otherwise (slack bus) go to next bus7. Increment the bus count, P : P t IE. Check all buses are taken into account. IfP < n, then go to step 6 and repeat.9. Checkconvergence

If I 4V,,,,* | 2 e, then increment iterationcount k : k * 1, go to step 5 repeat'10. Calculate line power flows and slack bus power-

80 Electricat Pouer System Ancrlysls2.5.1 .3 Ilow, thort lor Gouss.Seidel ,method when PV bus is obsent

2.5.2 Gaus+Seldel Method when PV Buses are PresentSome of the buses in a 'n' bus power system are voltage controlled buses. where P and I V I specifiedbut Q and 6 are unknowns. Let the buses be numbered as

p: l, slackbusp :2,3, ..., i, PQ busesp : i + l, i + 2,..., n, PVbuses

system data and formulate the Yor"

Assume initial bus voltages Voo = 1 + j0, P = 1, 2, 3, ...,n*1

Set iteration count k = 0, and lA V-,, | = e

Set bus count P = 1

Compute V^ fronr eqn. (2.34)

kk+1klaV, l=Vo -Vp

BuscountP=P+1

into account

aVr", l

I-oad FIow Studres 81

At the voltage controlled buses the rnaximum and minimum reactive powers are specified and therelue Q must lie between these limits and also voltages at these buses are also specified.i-e-. Qp rnin < Qp ( Qp ,nu^d lvrl:lvpl,p"".

For PQ bus the voltage calculation as usual in previous case.For the pth PV bus, we have to maintain the magnitude of voltage at a specified value I Vp lrp"..

Tb voltage at the PV bus is controlled by controlling Qr. Therefore, the values of Q, and 6, are toh rpdated in every iteration, this is as follows :

fnlQr, : - m I vrZrorv, l, for P : i + l, i * 2, ..., nlq=r I

The revised value of Q, is obtained from the abore equation by substituting most updated values ofnteqes on the right hanh side. Thus for (k + l)th iteration

Qk+l:-Im

The revised value of 6, is obtained frorn equation (2.34) as

[ef Z",,vi*' * (uii,i,", rr]for P:i+l,i+2,.'.,n

...(2.36)

...(2.37)d{i!,t--

6r**'-zYr**'

I ao*t p-l6,t'* t : Angle l "' , - 2rrr vf *' -

L (vr) o*=;for P:i+I,i+2,

^ k+r Pr_jQI,-'

t ,,, u;lq=p+t l.... n

*!c ...(2.38)

Th limits of reactive power can be checked and fixed as given below :

[Qr.,, , ir Ql*r ( Qr *i,I

Q,,o* t : lQrmax, ifQ;.' >Qp,"* (p : i + l,i + 2, ..., n) --.(2.39)

loi.' , if Qp.in . Qf .' . Q^no*ffiany limit (either maximum or minimum) is violated, then that bus is treated as the PQ bus. But

ffir subsequent computation iteration, Qro * ' cornes within the limits then the bus is convertedE to the PV bus.

82 Electricat Pouser Sgstem AnalUsis2.5.2.1 Algorithrn lor Gouss-Seidelmelhod when PV bus is presenl

l. Read system data and formulate the Yso.2. Assume initial bus voltages Vro : 1 + j0, for p : 2,3, ..., n

* 1 slack bus

3. Setiterationcountk : 0and I AV.r* | : e4. Compute A, and B, fiom equations (2.32) and (2.33)

' 5. set bus count p : 26. If bus is PQ bus, then

(i) Comput" Vr* * 1 from equation (2.34)(ii) Compute AVpft: Yoo *'-Yro

If lAVrol>lAv.,*l(iii) Assign new voltage to old

i.e., Y ,k - Vro

*

7. If bus is PV bus, then(i) Comput" go* I for the PV bus using equation (2.36)(ii) Check the limits ol Qro * I and set according to equation (2.39)

If no limit is violated, set Q1;* : 0If any limit is violated, set Q1;* : I

(iii) lt Qri, : l, compute the voltage using equation (2.34)(iv) If Qri,n : 0, then

vro * ' : Vp.p"" Z6; * twhere 5ro *

' is calculated from equation (2.37)

(v) Assign new voltage to oldi.e.,Yrk

- Vro -

8. Increment the bus count, p : p * I9. Checkall buses are taken into account

lf p < n, then go to step 5 and repeat.10. Checkconvergence

IflAVmaxl2e,thenincrement iteration count & : k + 1, go to step 5 and repeat.

.1. Calculate line power flows and slack bus power.I,1

:

Load. Flott, Studies 83

,J'J..Z How chort lorGousr'seidelmethod wten PV bus is present

Read system data and formulate the Y

Assume initial bus voltages 3 = 1 +j0, p=2,3,...,i, PQbus' *1.slack bus

Set iteration count k = 0, and IAV* | = e

Compute Ap for P = 2,3, ..., i and Bpq lot p = 2, 3, .'., nfrom eqns. (2.32) and (2.33) q = 1'2, ..-, n' p *

and Vohl= lVe l"*" z6ro'

buscount,p=p+

lncrement iteration count k= k +

ilf .flea*.ieel' Power Analysls2-6 NEWTON-RAPHSOf{'S METHOD

Consider

and

The number of nurnerical solution methods are available for load flow solution, a.mong those NrLlX,l""#X31^::.,::::::'i1:l'-1]'^PJ *' ;;!; r,"n"'*i'i,o powe,rur technique rtquadratic rate of convergence, whereas the Gauss-s"iu.i rra"ttoffi ffi:;'J:#"#t:JConsequently, for this method does not necessitate the use or

"c".i.r"tion factor and alsoto the selection of slack bus.N-R method can be appried to the load flow sorution in a number of ways, the mostused (l) rectangular coordinates and (il) polar coordinates.

2.6.1, N-R Method UsinE Rectangular CoordlnatesIn this formulation the load flow equations are expressed in rectangular form. consider an ,n,power system.

We know the power at bus p is

Pp-iQp: V;I/, : vofv*v,y): ro-tf, q=lYq:eqtjfq

Y ro : Grr

- iBn,Substituting rhe above quantities'in

.q*iio, (2.40)Pp

-iQp : @p-tfp) i(o, - .Ero)(ro * j.fn)Q=l

The real part of equarion (2.41) is

and the imaginary part of equation e.a\ is

The real and reactive power at each bus are the function of ,e, and,f ,ThusPp : 8t (e, J)

ll;';lr,T*l1Y,T*::::8,1i'f* given ngJver.sv-stem consisringn buses, ,,,,',,,; ;:1'1a slack bus and all remainins buses are toaaLuses. The diife;;f;#il:"riiilXlil[ fiiin real and reactive po*"rrio change in .e, and ,f, takethe form as^Pe: i-*o,.f,* o,^s:E**,.2**

,r: 2Vr("n Gon * frBnr)* fn(fo G* - rnBr)l

o, : I[, (ro G,o * f,B*) - ,o(fo c* - e,B*)f

and

Load Ftou: Studies '85

rhere AP, and AQ, represent the differences between the specified and the calculated values of P, andQ, respectively using equations (2.42) and (2.43).In short the equations (2.45) md (2.46\ for all(a

- 1) buses can be written in matrix form as

l-pl I-1, I r"l l-ael...(2.47)

loal :

L:;i:;lL^;lrhere Jl, J2, J3 and Ja are the elements of the Jacobian and are first order partial derivatives.

The elements of the Jacobian matrix can be derived from the equations (2.42) and (2.43'1.The diagonal elements of J1 are

aP- nfr : 2e, G* + frBrr'- fpB* * Ek, G rn + fn B *) .Q.48)

*J'From equatior. (2.26) net current injection is

Y: fv,nv,q=l: Yppr, * fronu,

,;irp: co + idp: (cpp-iBpp)(eo+if) * i(o* - iB*\(rn* ifn)

,=*,

Separating the real and imaginary parts

The reat part, cp: pGpp * frBo, * LP, Gnn + fnB*) ...(2.4g)o =*'o

rrd imaginary part, dp: foG*- eoBo, * fio G oo - rrB*) ...(2.50)*p

The equation (2.48) can be written asaP..

fr: eoGw--frBrr+ c, ...(2.51)The off-diagonal elements of J1 are

aP.-

fr : ,pG*-f,,Boo, e*p ...(2.52)The diagonal elements of J2 are

aP- nd : 2frGo, * 7,(t, G,, - e,Bo,)

+p

(r *:pBpp+fpcpp+dp

...(2.53)

86 Electrica.L Power

i.e.,

and

i.e.,

Off-diagonal elements of J2 areaP_

t:roB*+fpce(t,q*p iSimilarly,Diagonal elements of J, are

V : eoB* * frGoo- d,Off-diagonal elements of J3 are

*

: erB* * f oGoq, q* pDiagonal elements

" :b:*ff:f'Bon-enG*tc'

and off-diagonal elements of Jn are

}: -rpG*+fpBpq,q*pSymmetry property in Jacobian matrix : By careful examination of the off-diagonal

of the submatrices asfJipq: -UqjpqdP,

_ _19"dr, a fq

Uzlpq : IlzlpqaPe : aQeof,t dro

This relationship reduces the efforts considerably as it is enough only to determine the offelements of any two sub-matrices.2.6.1,2 When PV buses ore presenl: For a PV bus, the reactive power Q, is not specified but themagnitude I V, I is specified.

Thus lVul': ro'+fiTherefore, the following equation replaces equation (2.46) for each pv bus

l^vpl,:u{;f a,e,*ry**,The total number of equations considered in this case are fixed and equal to 2 (n

- l\.

When PV bus is present the equation (2.47) can be modified as

lil l:fir i:l;;rL

The elements of J1, Jr, J3 and Ja are given by equations (2.51) to (2.58). The elements of J5can be determined from equation Q.6l).

laad Flow Studies 87

a!.,

The diagonal elements of J, are

ry*:zpThe diagonal elements of Ju are

ulu,1, :.,,{f,, _ .r,The off-diagonal elements of J, and Ju submatrices are zaro

ryr : ?lv,r' :oduo trfe

...(2.64\

...(2.6s)

...(2.66)lrlS Algodthrn for N'R reetonglor oordinore method whin pv bus ir obsent

l. Read system data and formulate the yn*2. Assume initial bus voltages Vrl : t * j0, for p : 2,3, ..., n

* 1, slack bus3. Set iteration count, t : 0 and eonvergence criteria = tl. Set bus count, p : I5. Check type of bus. If bus is slack bus go to step 6.

Otherwise. if bus is PQ bus, then(i) Evaluate the active and reactive powers Prt and Qr& using equations (2.42) and, (2.43)(ii) Compute Lppk : pp rp." - ppt.r1and 4Qro : Qp rp.. - erk

"ur6. Increment the bus count, p : p + I7- check all buses are taken into account .lf p e(i) compute rhe elements of Jacobian matrix using equations (2.51) to (2.5g)(ii) Cgmpute voltage increments A, eok andA/rt using equation (2.47)and new bus voltages

o k + 1 _o k-p -o *liernand frr*t=1rk+Lfok(iir) go*Oute I V,,r I, cos 6, and sin 6, for all buses(iD Check the limits of voltage and ser

l,ntn': lV, l*,, sin 6ol*.*: I vp l*"* sin 6,

I vrol : I vp lnm,if I vr* I < I vpCompute ,oo *': I Vp lr,ncos IoanAff " i

(or) | vrk | : lv, Lo,*, if I vro I > I v,Compute ,,,r *

' :

I ur, l*"*cos 6, and$t + 1

otherwise go to step 11.Increment iteration counr, k :lk + I and go to step 5.Compute line power fldws ani{ slack bus power.srpp,;', \

,i

D.!t-12.

88 Electrtcctl Power System Analysis2.b.1.4 [low rhort lor l{-R reclongulor toorrlinote mefiod when PV bus is obssnt

Assume initial bus voltages Vro= 1 + P, p = 2,3, ..., n

Set iteration count. k = 0. and converqence criteria =

using eqns. (2.421and (2.43)

Compute A Pok = P"ogq - Ppk 1@D ?nd

a Q^k= Q-*- Q^k

Compute th6 largest absolute of the residue

using eqns. (2.51 ) to (2.58)

Compute a e"kanO A fot using eqn. (2.47)

Calculate erk'1= erk+ 6 grkgn6

fok*1= frk+6;rk

o*l';vk+1ap = lvol*cos 6,hlfp = lVrl*sin 6,

Lmd Ftow Shrdies 89ttl.5 Algrirhm tor x'x rffigrfur oodh*enetudur$y bu, htrree.ilr

l. Read system data and formulate the yru,2. Assume initial bus voltages Vro : I +7O, for p : 2,3, ..., n3. Set ireration count, ft = 0 and converg"o."

"ri "rif ! ;'"""uut4. Set bus count, p : I5. Check ifp is a slack bus. If yes go to step 7.

Otherwise(i) Compute Pre qnd ert using equilrions (2.42) and (2.43)(ii) Compute A ppft : pp.p".

-

pp*6 and6. Check type ofbus. IfpV bus

(i) Check the limits ol ert and serQro : Q, min, if Qr* S 9.,n: Qp.,u*, 6 Qf > Qp

^u*

rf no rimit is violated *,;,*q: 'Jo"t a Q'o ( Qp*"*

If any limit is violated ser e1;. : I(rr) If Qri,,, = 0, then,

Cornfur. I A Vr* l, : I v, lr,*" - lyrr l,l r,-^-^9Tar:e ifbus is PQ bus, compure6 er* : ep.u* - ept,r or A ert : epmin - ept*rI .- rncrement the bus count, p : p + II t check all buses are taken into accsunt.ltp eI ,i, compute the elements of Jacobian marrix using equations (2.51) ro (2.5g)I t"l Cornpute voltage increments L erk andA/rt using equation (2.47)and new bus voltages1 ,nft+l- eok+a,erkutd

I t , compute , {} ,-,"; g rL r* t otherwise so ro srep 12.!L lf bus is PV bus, rhenI comPute ,,,0 * t : I Vp lsp"q cos 6, andI tk+t - |vp 1.p".sin6,I (Xherwise if bus is pe bus. thenI (l) Check the limits of voltage and setI lvrol : I v, l*rn, if lvrols I V, l_inI Compute ,;*.'= | V, l^,ncos Lrandfrk *, : I Vp l.insin6,I (or) lvr*l:lv, l.u*,iflvrol>lv2 l*u*I Comoure ,ur* l, : I v, lru* cos Lrandff * i r I V, l*"*sin6,I C-oprre lrne power tiows and slack bus power.E sroo.It-

90 Electricat Power Sgstem Analysls2.6.1.6 tlow thort Ior N'R rectongulor toordinote method when PY bus il prerent

Assume initial bus voltage Vpu =* 1 slack bus

of the.absolute of the residue

I

I

tro

Compute aej ano arj using qn. (2.47)k+1kk

Calculate ep = ep + Aeo andktlkkfp =fp +Afp

Set iteration count k = 0 and convergence criteria =

Set bus count P = 1

P" and Q,, unsing eqns (2.421and (2.43)

Compute APj = P, "o*-

Pko*,

compute lavo l'='

t2 '

kr2lVDl spec- IVD I

buscountP=P+ 1

Load. Ftow Shrdies 91

N-R Method Using Polar Coordinatesh this formulation' the load flow equations are expressed in polar form. The total number ofqltions in rectangular coordinate version are2. (n- lf and in polai coordinate version are (2n

- Z

- S)F.n bus system having one slack bus and 'g' voltage ront.oll"d (pV) buses. Thus the use of polarlh results in lesser number of equations and a smaller size of Jacobian as compared with themcangular form' This is a definite advantage of polar form over the rectangular forrnand therefore,rbgrneral polar form is used.

fle know fiom equation (2.40)

Pp-iQp: ir; Yn,Y,A=l

Ler Ypr:Gro-iBrr: lypql Z-0*: lypql e-i|*V;: IV, I e-i6, andV,, :1y,,1ei6,

Snbstituting the above quairtities ir"iequation (2.67;

Q=l

P, -iep : t lYryonyole-iP*+5,, -6,)

ftparating the real ano imagina'lrt r.,, of equation (2.6s)

*, : t lv, Yo, v,lcos (or, * 5o - gr)q=l !

_n

Qr, : I lYrYrrv,lsin(or, *6, - 6r)The real and reactive powers

"lalro bus are the function of magnitude and phase angle of bus*rge. ThusPo= gr (6, I v 11Q,: gr (6, l v l)

Tbe real part,

dthe imaginary part,

...(2.67)

...(2.68)

...(2.6e)

...(2.7O)

lV"'I, cos 6p and sin 6p

= lVpl min cos 6r

= lVpl .;n sin 6r= lVol"*" cos 6p= lVol"r"" sin 6p

ej.l = lvol .", cos 6ol*1 = lvel .u* sin Ep

lncrement iteration count, k = k +

Wl Electrical Pouer2.6.2.1 Wtcn PV hscs arc nol FiGnft For a given power system consisting 'n' number of buassuming bus 1 is a slack bus and all remaining buses are taken as load buses.

The differential equations which relate the change in real and reactive power to changemagnitude and phase angle of bus voltage take the form as

Ap, = E*AE, +E #r^vq I ', . ...(2Aep = EHA6, +,4 #,^vq I ...(z

where AP, and AQ, represent the differenees between the specified and the calculatedvalue of P,Q, respectively using equations (2.69) and (2.70).

In shorr, the equations (2.71) and (2.72) for all (n - l\ buses can,be represented in matrix form[APl [], i lrl I aa IL^el = Li;'i:;lLovrl'

The elements of the Jacobian matrix can be derived from the bus power equations.The real power equation (2.69) can be written as

P, = | Yt,l2 lYr, lcos0r, * t lvt,Ywvnlcos(0,, *6, -5r),,=*,

The diagonal elements of J, are

\.. aP. *,- -

*; : AlurYrrYn lsin(orr+6r-6r)and the off-diagonal elements of 1,tri,

aP_

d : lYrY*V, lsin(0rot6r-60),e*pThe diagonal elements of J2 are

I = z I ypy* I cos sro* flyorYrlcos(0r, +6, - 6r)alVpl t t' P't' o=*r,

and the off-diagonal elernents of l, areaP-

-1"r1 : I V, Yr, I cos (o,o * 6,- 6)' 4* P

The reactive power equation (2.7q can be written asn

' Qp: lv, l'lY* | sinor, + ) FtrYrrv, lsin (o*+ 6p-6q)q-l*p

II The diagonal elements of J, are

I B = E ,, Y*y,l cos (0,, + !r - 6/ ...t2.80(r)I| *a rn, off-diagonat elements oylj.aLJur*- ,:- - -,, t

: - lv' YooYol cos (0,0 + 6o ' 6i)it' q*p "'(2'sl)

I *ffi ":'rrrr*l

sin(or, * 6r- 6),4*p ...(2.83)

! ', may be noted that, do not:T-rh" symmetry jn,the Jacobian, if polar coordinates are used.I f,o*.ur., if replace A I v I * +/ in equation (2.74),the equarion (2.74)can be modified asII [fi] :[? Ifl+l e84,I Ih" equations for the off-diagonal terms ai.e, -I From equation (2.77)I - aP^I Hpq,:5d : lYrYorv, lsin (lrq+ 6e-6q) ...(2.8s)I From equation (2.83) \I t': DQo ,] Lrr: ffi lvo | : lYoYrov, lsin(oro+ 6,e-6q) ...(2.86)I From equarion (2.79')

] *r, = #, I v, | : lYpywvo I cos (opr*6e-64) ...(2.s7)J From equation (2.81)I ,--: EQ' -I uoobservedthat

t'o: at = - lv'Y'oY' lcos(0" + 6'-6') "'(2'88)[ :t

tt 'Dserveo

tnat Hpq: Lpq

I - ..

Nr, :-'-lo, ...(z.Bg)

I n. equarions for the diagohal elelrients are

[ ,.requation (2.76), Hm:B: lvolrBor-eo .p.so)I

I r-* equation (z.lsl, Npp : .#, I V, | : I v,l, cpp + pp ...(z.gt)

From equation (2.80), rr, : H - pn-Gpe lvp 12

Fromequarion(2.82), t,: #l I V, I : e, + | ypf Bpe

appear on the left hand side of equation (2.84)*O lllil does not appear on the right hand' lvrl

Thus with slight modification in equation (2J4), we can get the symmetry propertyJacobian, which is observed in the case ofexpressing Jacobian in rectangular coordinates.2.6.2.2 When PV buses ore presenl: Now consider when PV buses are included in 'n' bus powerFor a PV bus, the reactive power Q, is not specified and I v, I is fixed i.a., specified; A e,

94 Electricat Potuer System A

equation (2.s4) since A I v, | : 0. considerpe and qth buses are PQ buses and f and sthPV buses. Then the Jacobian matrix can be written as

4ftbur sthbu,

pth PQ bus p'h bus A6

Alvqllv, I

A6.,

qth Pq bus

where. aP,.: 55;'NP,:-dQr r -- aS., 'uP'l -

rtl' PV busrth bus

Hrn

Ipq

sth PV bus

aP_

a 1-v,1 I v, I

ffir,,r2.6.2.3 Algorifim lor N-R polor roordinote method when pv buses ore obsent

l. Read system data and formulate the Y"r.2. Assume initial bus voltage magnirudes I vr, I ano phase angles 610 forp : 2,3, ..., k

3. setiterationcountk:0andconvergencecriteria:e *l'slackbus4. Set bus countp : I5. Check rype of bus, if bus is slack bus go to step 6. Otherrvise, if bus is PQ bus, then(i) Evaluate the active and reactive powers prft and ert using equations (2.69) and, (2.70'11(ii) Computg A ppt : pp.p...

- Fp.ur

and AQrt : Qprp"" -Qpk*r6. Increment the bus countp : p + I7. check all buses are taken into account.lt p < n, then go to step 5 and repeat.8. Compute the largest of the absolute of the residue.9. Check convergence. If the residue > e(i) Compute the elements of Jacobian matrix using equations (2.76) ro (2.83) using the

mated I V,, I and 6, from step 2.

H,, N,, H,,lu, L,,, J,,,

H,, N,, H,,

I-ood Prous Studres 95

(ii) Compute A 6rk and A I Vrt I using equation (2.74)-(iii) Compute new bus voltage magnitude and phase angle of all load buses6ro*t:6r*+a6r*t.e.,

and Ivro* t I: Ivet|+AIvroIOtherwise go to steP 9.

10. lncrementiterationcountk : k + l withthesemodified lV, land6randgotostep5'11. Compute line power flows and slack bus power.12. Stop.

u.2.4 flow rhort lor ll-R polor oordinole method rrlten PV busel orc obsenl

Assuming initial bus voltage magnihtde lVr"l and

phase angle 16o0; for p = 2, 3, ...n,

Set iteration count k = O, and convergence criteria =

Set bus count P = 1

P"^ and Qo^ using eqns. (2.69) and (2.

lncrement bus count P = P + 1

th largest of the absolute of the

Compute the element of Jacobiusing eqns. (2.75\lo (2.42)

a6okand alvokl using e9n. (2.

av^k'1 l= v^kl + a lv^kl

lncremnt interation count f = k +

96 Ebctncal power Sgrstem2.6.2.5 Algorithm lor l{-I poftr oordhote morlnrd when pytums oro presant

1. Read system data and formulate ysu"2. Assume initiar bus voltage magnitude I v} | and phase angle 6! for all pe buses and

angle 610 at all pV buses (except slack bus)3. Set iteration count & : 0 and convergence criteria : eo

Slffr*t Prt and Qrt fo, all PQ buses and prt for alt pv buses using equations ,(2.69)

5' compute A prt and A erk for a[ pe buses and Apr* for a[ pv buses using thbA Prt : Pprp".

- Ppk.r1 rlld

6 Qr* : Qrr*" - Qrt"ur6. Compute the largest of the absolute of the residue7. Check convergence

If the residue ) e(i) Compute the elements of Jacobian marix(ii) Compute A 6, and

+iusing equation (2.84)!'(iii) Update the riew vaJue-s of bus voltage magnitude and phase angle of all pe buses andvalue ofph4$e angle for all pV UusJs using expressions - --.6r**r-6r*+a6pr

and lyoo* t_l : lvrol *A lVprlOtherwise go to step 9.

t *ir::-9fo {or ail-pV buses and check e,oin ( ep* ( ep r.*. If yes return to step 4mcrement the iteration count /< : ,t * 1.If not, set Qrt : Qp,rtn(or).set Qpk : Qo.", as the case may be treat this bus as a pe bus, return to step 4 and ithe iteration codnt /c : k + 1Coqpute line power flows and slack bus power.Stop.

9.

10.

Laad Ftou; Studies W

2.6.2.b tlow drort lor ll'R polor toord[mta method when PV burr ora present

nssume 6f for p = 2, 3, ..., n

and V^ for p = 2,3, ... m i.e., PQ buses

Set iteration count k = 0 and convergence criteria = e

for p - 2, 3, ...,

for P = 2, 3, ..., n

for p = 2, 3, ...,

Compute Hro , and Nooforg = 2' 3. "''nJoo,forq=2,3,...

lncrement buscountP=P+1

Compute A16o I, for p = 2, ...,n

and#, forp - 2,3 ...,m

Compute 6o for p = 2, 3, ...,nk+1Vo tor p = 2,2, ...,^

lncrement iteration count k = k + 1

PVbusp>m

9! pkrrrlvr, fuuser SustemL7 I'EGI'I'PLED LOAD FIOW METHODAn important characteristic of.an electric power transmission system operating in steady state,changes which occur in bus active power due to small changes inius voltage magnitude is very sras compared to their changes due to smail changes in bus vortage phase angre (... , = T,, ulT:::ki:1t""-:,^.:T::""f .1?Irtrix of J, of the racobian Matrix can be neglected and to be rak

( o = *

,t - v)) ' so, all the elements of J, of the Jacobian Matrix can also be negtected and totaken as zero.

Then equation (2.74) simplifies to

as zero' The changes which occur in bus reictive power due to;;ft;G:'t i|Ji,'|,"r"angle is quite small, when compared to their changes due to small changes in bus voltage mag

till :lt;ll^iiJ ,,,In terms of the sub-matrices, the equation (2.g5) yields the following two decoupled equations

AP:JrA6and AQ:JaAlvl

It can be seen that L*: Hrcfor q * p, q and p: 2,3, ..., n

...(2.The solution of these equations is less time consuming ind the computer memory requirenwould be less since the.-elements of J, u.ilJ; are zero and need not be stored. rurttreriimptincalcan be obtained by rewriting equatioi e.qil as follows :

=rlr[ffi]where Lr, from equation (2.g3) can be'rewritten as

L--: rv r ao"-pp ,'r'ffi:Qp*lvel2Bpp

andLrn from equation (2.g6)DepLnr: lu, | ffi = | v, yoryol cos(or, + 6p-64)Let J, be denoted by [H], then equation (2-96) gives

where H, from equation,rffi;jfle rlwrtten asH--: DP' : ,

andHorrromequation ,ri;;: #: lvP l'Bo'-Q'

Hpq:*: lYrYoqv, lsin (orq*6e-6q)

* l, slackbusand thus reduces the computation effort.

...(2.101)

Laad. Ftotu Studies 99

2.8 FAST DECOUPLED LOAD FLOW METHODFor routine solutions Newton-Raphson's method has more popular method. However, it is limiteduse for small core applications. Whereas weekly convergent Gauss-Seidel method is the mosteconomical and it is not fast converge. If some valid assumptions are considered for practical powerrransmission sysrem, the fast decoupled load flow algorithm (FDLF) is simple, faster and morereliable than N-R method and has lower storage requirements for entirely in core solutions.

The tbllowing valid assumptions can be made without introducing much error(D QP 11 Br, I V,l'(rr) I 6, - 6, I is very small and therefore, cos (60 - 6r) = 1'0(rrr) G, is verY small so, G, sin (6, - 6r) = 0'0In view of the assumption (i), the equations (2.98) and (2.100) becomes

Hpp:L*: lVpf Bep ..'(2.102)and in view of assumprions (ii) and (iii), the equation (2.99) (or) equation (2.101) can be written as

H*t:Lrr: lYoYrl {1Y,,, lsin0rrcos(6r-6r) + lY*l cosOrnsin(6'-6')l: I V, Y,, I {Bpq cos (6, - 6r) + G* sin (6p - 6q)

i:lvpv, l{Bp./+o}: lVpYolBr*forq*p "'(2'103)

Computation of the elements of [H] and [L] using the above approximate expressions would befaster. Hence. this method is called fast decoupled load flow method.

COMPARISON OF GAUSS.SEIDEL AND NEWTON-RAPHSON'S METHOD2.9

G-S Method

In this rnethod, tbr developing the programrectangular coordinates are used.It requires the fewest number of arithmeticoperations to complete an iteration andrequires less time per iteration.The rate of convergence of this method isslow, required more number of iteration toobtain the solution.The number of iterations of this methodincreases directly as the number of buses ofthe power system.In this method, convergence is affected bythe selection of a slack bus.For large system, it is less accurate andunreliable.It required less inemory space.

z.LO COMPARISON OF DECOUPLED METHOD AND FAST DECOUPLEDMETHOD WITH N.R METHOD

Decoupled Load FIow Method : It is simple and computationally efficient than the Newton-Raphson's method. The main advantage of this method compared to the N-R method is it reduces

4.

5.

6.

7.

N-R Method

In this method polar coordinates are preferredfor developing program.In this method, the elements of the Jacobianare to be computed in each iteration. So thetime per iteration is more.The rate of convergence of this method isfast and required less number of iteration toobtain the solution.It is independent of the number of buses ofthe system (vary 3 to 5 iteration).

It does not effect for the selection of slackbus on the convergence.It is more accurate and reliable.

It required more memory space.

memory requirement in storing the Jacobian elements. Time per iteration of this method is almost;ffi:: $",il lr[:tli#l*:t arwavs take more number orit"ration,'"** ;;;l"lir,," ,oru,Fast Decoupled Load FIow Method : It is superior than the N-R method from the point of rof speed and storage' For this method theconvergence is geometric. It required nonnally two toli:1,,?::lT.T:,::::::y:i f:,":*",",fr* **-r;;;r ilR method rhis is due to rhethat the valid assumptions are considerd. storage requirement,

"r";il'J:;;.illJ:fff"[*:i[ffit]:i::t-1t1fl]jY"|o.fl}h1 a"*upt"ito"a'n"* *",t"il inis me*roa is also usetul to sooptimization problems and conting"rr.y "n'au"tion for ry;r",,;;xi;X#ffi:il:"ff*ff:

ffi:"l[Xt;lh:::"'pute the l"'oti* matrix i, "r".v ii*Joo ", s matrix i, "orri*t "r",

l.fi, Electricat power

2,L7. CAtcutATtON OF L|NE Ftowsconsider the rine connecting between the busesp and 4 as shown in Fig. 2. r0.p

,ltpqlpso

zr lroflstnitsian l,ine connected belween bavs pt and qComplex power injected into the line from the ptr busCurrent injected by the ptr busVoltages at thepfr and 4fr buses

lpo = Line series adrnittancerp41, rq20 : Line charging admittance at busesp and 4 respectivery.The current flowing from bus p into the line is given by

Ipq=Ipqt*Ipqo

rhe comprex power ,or.","oi,,lu,* I3.kl J:J;T.bus 4 isSoq:Prq+iQpq=

yrlnrol, from equarion e.l}4) into the equation (2.105)

Fig.2.lO Ponspq :r_'pq -

V, and Vo :

PLpq: Sr, * s*losses can be determined by summing a, the line flows of the given

Sr, : v, (v"; _ v)y--

v'r) !'oo * v*ry*o,,lSimilarly, the complex oJ;r rri".r"oi"r, ,{"'rri1"ililr?; busp isSr, : Y, [(V*,

- V*,) y*n-Y")!*ro t y*n!*,

.llf I-"*.".loss in tt",p"fl, 1rr"'rtt*.'i*;ff.l lr:J"g:lthe power flows in the p _ q, lineth"p* bus to 4,h bqs

Substituting the

i-e.,Total transmission

system.

Power lroflstnissinn