1O2 Electricat Pouser System Analgsis

It is clear that as the angle 6 : (6p - 6n) gets smaller and smaller, the cosihe function appror1.0. The berter approximation of th6 sine of a small angle is the angle itself in radian becausevery small angle, ihe sine of an angle is approximately equal to the angle itself' Thus we now

*r: Llr,llu,lBe, (6p - 6q)'t=l

q, : ilu,llvnlc n,,rr!=l

2.U2.3 Approxlmate the Product of the Voltages to 1'OIn the per-unit system, the numerical values of voltage magnitudes l-V, l. and I V,l. are very ck1.0. Typical range under most operating conditions is 0.95 to 1.05. Fuithermore, its product ismuch closer to 1.0.

Given the discussed practical approximations, the power flow in the transmission systemapproximated using the following equation.

pr: IBeq (6e -6q)q =l

It is the called the DC power flow equation. It is commonly used in the optimal power(OpF) and economic dispatch problerns in the power systems. Since our point of interest is the fflow, we omit the power flow equation for reactive power'

SOLVED PROBLEMS

problem 2.1. The power system network is shown inFig. 2.11 bus I is considered as aof vohage 1.00 20" p.u. Thi line impedance are indicated in the network on lN MVAneglect the line shunt admittance-

(100 + j50) MVA (120 + j30) MVA2

0.02 + j0.04

g.s1 + j0.02 6.912 + j0.02

3

(30 + j50) MVA

Fig. 2.tI

l-oad Fla:u.r Studres 1O3

By using Gauss-Seidet metlnd at the end of first iteration, determine the magnitude of voltage andangles at buses 2 and i.

l1ln= fi : 002 +j004 : lo -Zoj:20

- i4Or t5 o.ol + jo.o2

ht= ffi:22'0588 - j36'7647Yrr : ln * ln: 10 - 2Oj + 20 - 40j : (30 -j60)y;;: y), + yi = l0 - zoj + 223s88 - j36.7647 :32.0585 -is6.7647Yr; = yt, + ytz:2O- j40 + 22.0588 - j36.7647 :42.0588 -i76.7647Yr2: -!0: - 10 +.120

I tro-ioo) (-lo+i2o) (-20+ i4o) .lybu. : I t- ro + izo) (32.0588 - is6:647) (- 22.0s88 + i3616a7)l

Lt- ,o + i40) (- 22.0s88 + i36.7647) (42.0588 - i76:647\ )Assuming initial bus vltages, Vro : Vro : I p.u.Yoltage at bus p is

A2=(-02

- j0.2)

43:

32Os88 - i56.7647

(- 0.3 + j0.s)42.0588

- j76.7647

vj*' : *-$r*vf *'- frroul(v;) o;,, q=p+tor=*,rorp:2,3

4. : Pz -,lQz : (Pcz - Prr) l-r (Qoz - Q,z)- -z Yrr. Y,

= (0.001162 - j0.004179)

= (- 0.006656 - j 0.000261)

,* : *,

for p : 2, 3and q : 1,2, 3; p * q.

Y.,B.,:3=Y,,

- l0 +j20

32.0588 -

js6.76n = (- 0.3425 + j0.0173)

104 Electrical Power System Analysis

_ -22.0588+j36i647

= (- 0.6574 - j 0.0173)

8,,

32.0s88 -

j56.7647

(- 20 +.i40\

832 :

V] :

Y,,

Y,, :Y,,

A2-.-----=

(vi )

42.0s88 -

j76.7647

- 22.0588 +36.7647

- (- 0.5105 +JO.01919)

: (- 0.4894 -j0.019193)

42.0s88 -

j76.7647

-BI Vl -B23 Vro

0.001162 -

j0.004t79-

(- o.342s +j 0.0173) (1.04) - (- 0.6s74 - j0.0173)lr.014762

-

j0.0048711.01477 z- 0.275" p.u.

+- - B,, vl - 8,, vj(vi)(- 0.006656

-

j0.000261)-

(- 0.5105 + J0.01919) (1.04) - (- 0.4894 -70.I

: 1.02096 -7O.0032186

: 1.02096 Z- 0.18' p.u.

(r.01476 -

j0.

. Problem 2.2.The power system network shown inFig. 2.12. Each line has a series0.A2 + j0.08) p.u. and a total slrunt admittance of j0.02 p.u. The specified quantities ofare given in tlrc.ligure. A controllable reactive power source is available at bus 3, with the0 S Qc: S 1.5 p.u. Using the Gauss-Seidel method, ftnd the voltage at bus '2' and '3' afieriteration.

= (0.6 +p.3) p.u.

Y,r= 1.0210 P.u.

I 56 = (1.5 +p.6)p.u.lVrl=1.Mp.u.Qos

Brg. 2.L2

.,1.J

Lmd Flou Studies lOsSol. Let bus (1) is blackbus

bus (2) is load busbus (3) is PV bus

Using the nominal r model for the transmission lines, the elements of Yru. are computed as :Series admittance of each line

o.o2+iooT : Q'g4t -i:t'764)P'u'

The elements of Ysu* are given by self admittances.

Yrr = Yzz : Y33 : ZIZ.S+1- j1.764. ryz.lL - 2)= 5.8E2 - j23.508 p.u.

Mutual admittancesYr-2 = Y2r= Y3r = Yr3 = Yzt=Y3z: GZ.g4l + jll.764)p.u.

Assume Yz: 1.0 Z:0"= 1.0 +,/0

md 6r = 0o.Bus voltages are calculated asAt Bus (2)

l-tvl,,: *l'r ,r#- i,",, u,j,,i- Yr, | {u;y,', : z't , j

T

=#3508tm_{(z.slt*jll.76a)xl.a2+(-2.g4l+jll.7tA)x,*}]: (1.0368

- j0.0272)p.u.

For Bui (3)

Q!') :- rm [{vior}{Y,, v, +Y,2 vl',) *v,, v{"}]:

- Im tt .04 {(- 2.941 + jtt.764\ x 1.02 Zo"

+ (-2.94t + jtt.764) (1.0368 _ jo.0272) + (5.E82 _i23.508) (1.04 z|")ll: 0.179 p.u.

"' Qcr :0'179 + ati

,

: 0.179 + 0.6 : 0.779 p.u.

Qf) = Qc., Qr. :0.179P.u.

lOG Electrtcal Power Sgstem Analysis

or Qc, 0.179 + QL3.'. Qo, = 0.179 + 0.6 : 0.779 P.u.

i.e., within specified limits. Therefore, bus (3) rs acting as a generator busand Q!') satisfies the constraint

Hence,

l- ^ -.^.) Ivl,, :

=l I !;{a#' - {",, v, + Y,, ,,**}l, y,,L[ui?]..]- r,, J-

: ---l- [ - t,'; r9=l' - {(- z.o+r + jr r.7 64') xt.oz + (- z.s+r + in. 764X1 .0368 - j0.5.882-i23.508 L t.O4Z0o t\

: 1.0179 Z- 4.04 p.u..'. vj" : v3rr". .26!)

: 1.04 l- 4.04" p.u.Problem 2.3. Repeat the problem 2.2, if the reactive power constraint on generator 3 is 0 S

< 0"5.

Sol. Since Qc, calculated in problem 2.2 corresponding to initial specified voltage Vr : l.Mis 0.779 p.u., which is greater than the maximum specified limit. Hence Qc, is set equal toi.r., Qc, : 0.5 p.u.

Bus 3, therefore, becomes a PQ bus from a PV bus. Therefore, I V3 [ can no longer remainat 1-04 p.u. The value of V3 at the end of the first iteration is calculated as follows :

Note. V-,0 : I + 70 by virtue of a flat start

uj : + t"*-[v,, v,.\, v,]]: =-- L -_.f -t'l-rqtzs -{lz.s+r+ jfi.764)xl.oz+(-z.s4t+ jt].7A)x(1.0368-5.882-i23.508'L l-i6 t\ -'- ' '" '''--'-- ' \ -'-

: l.02136 -

j0.0756: l.O2416 Z- 4.24" p.u.

Load Flou: Studres fiVProblem 2.4. The power system shown in Fig. 2.13 each line has sgries impedance of (0.03 +

j0.07) p.u. emd shunt admittance of j0.01 p.u. The specified quantities at buses are also shown.Determine the element of Jacobian mntrix by rectangular co-ordinate formulation of N-R method.

Vr = 1.02 Z 0" P.u. S., = (0.6 +p.3) p.u.

Fig. 2.13

Sol. lVr I :1.04P.u.Bus (1) : Swing bus. Vr : 1'02 Z0 P.u.Bus (2) : PQ bus, Pz -/Qz : - 0.6 + i0.3Bus (3) : PV bus,

I v: I : 1'04 P'u'P::-1'5P'u'

Step 1. Formation of Ysu,

Self admittance, Yr, :, [-=*= + i -o'oll-'Lo03*J'007-t z ): (10.34s

-

j24.128) p.a.

And mutual admittances , Y r, = - 0.03 + 70.07

--5.172+it2.O69-

5.172 + j12.0691034s

-

j24.r28-

5.172 + j12.069

-5.172 + itz.M9f-

s172 + izalel10.345

- i24:28 )I to.:+s - i24.128

Y8,,, : G -jB =l- 5.5172 + i12.069

l- s.sttz + ir2.o6eGrr : Gzz: G33 : 10.345 P,u.

108 Electrical Pouer System AnalgsisBrr : Bzz : B33 : 24.128p.u.Gt2: G2r = Gtr: G6: G32: Gzl = 5.172p.u.Brz = Bzr : B3l : Brr = Brz: B23: - 12.069p.u.

Step 2. Assume that. V, : 1.0 + j0.0 = ez + jfz and 62 - 0'Vr : 1.02 + j0.0 p.u. (specified)

Now, V::1.04 +70.0p.u.: et+ jfiStep 3. For the power system the performance equation in matrix form is

where J1 , Jz, lt, Jo, J5 and J6 are Jacobian matricesOff-diagonal elements of J,,

?P. dP,de,

.0u,JP. )P.6". AA.)Qr aQ:

Ide. d", I

EV, aV.14 1;

I#]

= pGpq-frGrn 4*p

: 2Gtt-ItGzt

: 1 x -

5.172 -g = -5.172p.u.

: 3 G3z- fzGzz

= 1.04x-5.172 -Q = -5.379p.u.

APzAP:

4a,AVr

IaPr llot, II

--_

I -lao,l -

L^v;j

aez

Let

I*Lfz

dP, ?P,dfz dt?P, DP.EE EE

-ra;--;0;aE6-?\ -- iLdlz dft

[r, l, ILr. i:l

or

OP,dro

DP,Ae,

0P.d",

Diagonal elements of f,aP_

i; : erBno+frGrr+ d,

#:ezB2z+f2Gzz+dz

Ioad, Flotu Studres 109

: I x24.128 + 0 + 0.73414 = 24.862

# : erBrr+f3G33+ dj: 1.04 x24.128 + 0 - 0.71374 :24.379

Oft'-diagonal elements of J,

aP_#: : erBp, * foBoo, Q*Po J.r

+? = erB,rt|zBztdtt:1.0x(- 12.069) * 0:

-12.069

#: qBrr*hBzz:1.04x(- 12.069) + 0:

-12.55176Diagonal elements of J,

P : enB* * foGrr- doop

*

: zBzz * f'G22: d2

: 1.0 x24.128+ 0 - 0.734 L4 :23.374From equation (2.49)

cp : pG* t frB* + fro o oo + fq Bpn,l;,

Cz: zGzz * fzB,2 * elGzr * etGzs * fiBzr *.fzBzs(': P :1 is a slack bus)

: 1 x 10.345 + 0 + 1.02x5.172x 1.04 x-5.172 + 0 + 0:

- 0.309

C3: tGlr +/3 By * e1G1 * ezGtz +/r B:r * fzBy: 1.04x 10.345 + 0 + l.O2x- 5.172 * 1 x - 5.172 + 0 + 0

d from equation (2.50) ,t

dp:frGw-eoBr, + I (foG*-roBrn)tg=l*p

110 Electricol Power Analyslsdz : fz Gzz - ezBzz : (fi Gx * fi Gzt - erB21 - qBp)

: 0 -

1 x24.128 + 0 + 0 -

1.02 x -

t2.06g- 1.04 x -

:0.73414dt : ft Gn - es Bll +./r Gy + fzGsz - erB31 - e2 e,

:0- 1.04x24.128 + 0 + 0- 1.02x- 12.069- 1x- I:

- 0.71374.

Now Diagonal elements of J, are

aP..

u;:nGr,'-frBoo+co

+: etGzz-f.B2z+Czdt

Off-diagonal elements of J,

DQpdt,

EQz0r,

Diagonal elements of Jo

: I x 10.35 -0 + (-0.309) : 10.041

: e., Gr, + C.,

: 1.04 x 10.345 + 0.31136: 11.07016

: pBprt + fnGon' Q*P

: 2823 +.fzGzt

: 1 x -

12.069 : -

12.069p.u.

dP,d",

Depdfp

a!adfz

: fpBpp- eoGo, + C,

: IzBzz- e2G22 * C2

:0- 1.0x 10.128 + (-0.309):

- 10.437

Off-diagonal elements of Jo

*: -ep.Gw+fecpq,q*p

lLoad" Flow Studlles 111

Diagonal elements of Jt{udr,

- -2Gx+I2G23

:-1x-5.t72:5.172p.a.

- 2ro

3Qzdft

alv, I =2xr.o4=2.08dr,

Off-diagonal elemEnts of J5

av_J:Odro

Diagonal elements of J6

DY,:odf

,,

Off-diagonal elements of Ju

Dvp:o

Jacobian matrix Efq

l- 10.041 - 5.172 i 24562 - 12069'l-

| - s^tzs 11.07016 i -rzsstt6 2$7s Ir -

I --l-------------------- I

'

: l?1II _'_ :_rl_:yl2_ j_ __:l9fr7-

- - - -

s ttz I[- z.os -- ---0 ---T---- 0

---------b--- J

problem 2.5. For the problem 2.4, determine the Jacobian matrix and the unknown values at thecnd of first iteration using polar co-ordinate method'

Sol. Bus (1) is a swing bus Vr : 1.02 ZO p'u'Bus (2) P-Q, bus Pz - iQz: (- 0.6 + 70'3) p'u'Bus (3) P-V bus with P3 : - l '5 P'u'

I v, I : 1'04 P'u'The elements of Ysu. are

Yl : Yzz =Ytt:26'252 l'66'8" P'u'Ytz : Yzt : Yt3 = Y3l : Y23 = Y3z: L3'13 Zll32o p'u'0l'P : 66'8o and0" : - ll3'2o

:7i2, eirit*at power Sgstem AnalgsisLet us assuming,

Vz : 1.0 Z0' p.u.i.e.. I Vz I = 1.0;62 = 0", 63 = 0"

From equations (2.75) and [2'80(a)] the real and reactive powers at the buses are given byP: : G:z I u: l: + I Vz I t I yzr I I vr I cos (0,, * 6z - 6r) + I yzr I I v, I cos (+ 6r _ 63)l

: 10.345x 12 + l.o t13.13xr.a}eos (- rr3.2 +0-0) + 13.13 x 1.04xcos(- 1t3.2 + 0_0)l

= _ 0.31 p.u.

Similarly.pr: G$ lV: l, + iVs lly:r Vrcos(013 * 6l_61) + y23V2cos(ozl +5:_e)l

: 10.345x(1.04)2 + r.04x [r3.l3x r.02 cos (- rr3.z +0-0) + 13.13 x lcos(-113.2+0-0)l

= 10.345 x 1.0816 + 1.04 t- l0.448l: 0.323 p.u.

Qz = Bzr lv, l'+ vrly2rv,sin(ozr *62-6r) r yzrvrsin(ozl +E_Ul: 24.128x 12 + 1.0 [l3.13x 1.02 sin(- I 13.2 +0-0) + t3.13x 1.04sin(_ 1l+0_0)j= 24.128 + (- 24.8605) :

- 0.733 p.u.

Similarly,e: : B$ I Vl 12 + Vr [y3r V, sin(0l3 * 0z _6r) * yrzV,sin(02: * St_L)]

:24.128x 1.042 + 1.04t13.13x 1.02sin(- 113.2) + 13.13x l xsin(_ 113= 26.0968 - 25.353= 0.743g p.u.

Therefore. the change in real and reactive powers are

laprl [-o.o-(-o3l)l f-o.2etso loo, l:l-rj-0323'l:l-1.8231Loo,j L-o.r-(-0.733)l L o.orrl

The elements of the Jacobean are calculated as follows :aP"

# : - Q2 + Bzz I Yz 12 : o.733 + 24.128x t2:24-861

?P,E6,

: -Q3 + B, I Yrl' : -0.7438 + 24.t2gxt.O42:25.353

Load Flow Shrdres 113

k : Yzr vz V3 sin (ozr + 6'? - s,): 13.13 x 1.0 x 1.04 sin (- 113.2) : - 12.551

# : Yu vr v, sin (0,, + 6s - 6z)= 13.13 x 1.04 x 1 sin (- 113.2\: - 12.551

ap- P, =10.34sx1*f_g'l =10.035.titl =Gzz I vz I + u, ( 1"0 )- Lv'vJJ'

#;l = vs Yrz cos (032 + 6s - L) : 1'o+ x 13.13 cos (113.2)

: -

5.3794

H : -Gzzl vi I + P2: -10.345x 12 + (-0'31) = - 10.655'

k : - | vz I I Yx I I vr I cos(023 +L-6r):

- 1.0 x 13.13 x 1.04 cos (113.2) : 5.3794

a;ffi = 4z lvz I + 3 : 24'128x 1 + !-ffi : 23'3e5The corrections vector is given by

[aa!'l f z4.s6t -tz.sst 10.03s-I'f-oee'lI ao!" | : I - 12.ssr zs.3s3 - s.37s4l l- r.szl ILoui"l [- ro.oss s.37s4 23.3es] | o.+lll

[- o.oos+l: l-o.tosal r"o

[0.0,,,0]i. The new values at the end of first iteration are

6t) = 610) - 661tr = O.0684 rad6!) : 610) - 46trt = 0.1034rad

I V, l(,) : Vjo) - A Vjt) : 0.98t86 p.u.

Problem 2.6. solve the probrem 2.4 by decoupled andfast decoupted methods.SoI. Decoupled MethodThe performance equation by using decoupled method is

|--o-o-l : [Lj-9] [-16rL^aj : Ltil,;j L^vlAP = Jl A6andAQ: JaAV

From the solution of problem 2.5, the J1 and Jo elements are

I z+.aa -n.sst I o I: l-rz.ssr 2s.3s3 i o IL-- o--------o----i zrio5l

lorrl : [ 24.861 _l2.s5ll laarlLap,l l_- rz.ssr zs.rss.j Loa,.J

[i3:] : [-?i lll j3ii]l ' [--,:;:][- o.oorql: l-o.,rrr_J

t"o

AQz : [23.3951 LVz

AV1 : $= :0.0185' 23.395

The new values are

and

6!" : 0 -

(- 0.0639) : 0.0639 rad6!" : 0

- (- 0.1035) :

- 0.1035 rad

vj" | : l.o- 0.0185 : 0.9815 p.u.Fast Decoupled Method :The performance equation by using fast decoupled method is

tiil: fx:; x:: ll[;: tLoo,J lo o 6llav,j

Inad, tuw Shrdres 115From the equarions (z.tOZ\ and (2.103), Hzz,Hzt, Hrr, ilr, andL22are

Hzz: hz : Bzz I V, l? : 24.l2}x t2 :24.12EH33 : Bl: I Vt 12 : 24.128x I .042 : 26.097H2t=H32=BztlVzllV:l

= I2.M9x I x 1.04 - - tZ.S52laprl | 24.128

-tz.ss2f l-as.lLoo,l = l-o.ssz 26.w7) L^u,Jlaorl _ [ 24:28 -t2.ssz1-t [_o2elLoa,J - l-n.ssz z6.os7) l_r.szll

I-- 0.064s1: L -o.rorl 'ud

And AV2 : ff"| u, I: mx l'0 : o'018 P'u'

The new estimated values are

5f) = 0 + (-0.0645) : -0.0645 rad

st') = o + (- o.lol) - - o.lot radlrj I : 1.0 + 0.018 : l.0l8p.u.

I hoblem 2.7. For the system shown in Fig. 2.14 with bus I as slack bus, obtain the power flowi Jurion using G-s methoi after the end rffTrtiirr;;;;'."* ' ^

Fig.2.r4

116 Elecf'rcal Pouter System Analgsris

Bus CodeP o

Line Impedance in p.u. Half line charging admittance

I2

3

2

3

3

i0.1ja.2j0.2

000

Bus loading dataBus code Generation

PAload

PAlvl Reactive power limil

Qnin Q^*

Typeof bus

I2

3

5 03.5 0.s

1.01.1 0 J

Slack busP-V busPQ- bus

Assume,

Vr : 1.0 10" p.u.Vz = 1.1 10" p.u.

Since bus (2) is P-v 0".,lJ";"'r;0,"'3rl[,,ef) : - rm [vjolt", u,+y, vr(o) *"rr r,]]

- Im [.] UlO x 1.0 * (-ils) x 1.1 +75 x 1.0]l

= - [m [l.] x (/10 -j16.5 + J5)]Qf) = 1.65 p.u. and this is within specified limits6f) : Angle or (vj'))

: Ansre"r[+{V,"-Yz, v, -",, ul',}]

: Ansre,t [* {# - jro x r -i5 x 1}]: Ansle

"r [* {4.s4s - jl6i}]: 15.4o

sor. YBu.: I ll; ji: flL ,t js -j101

Lood. Ftaus Studres 117 iThe voltage of bus (2) is

vj') : 1.1 /-15.4" p.u.The voltage of bus (3) is

r- 'n -l1r(r)

-

I l P, -iQ, _ E, v, - v, vj', I'r Y, L vj'' -r't '52' '2 )= 7o.1 [- a.s - j0.5 - 7 5 x I - / 5 x l.l zl5.4f=70.1 [- 3.5 -70.5 - j5 + 1.4606 -y5.3025):70.1 (- 2.0394

-

j10.8025): 1.08025 - j0.2A394: 1.09933 2-1O.69" p.u.

Problem 2.8. Solve the problem 2.7 after first iteration in polar coordinate form of N-R method.Sol.

ftsz-w" tozgo" 5zw" 1ybu,:|rcaso" rsz-go" srgo l

IL5l9o" 5z9o".toz-90'lVr : 1.0 ZOo P.tt.Vz = 1.1 10" P.n.Vr: 1.0 Z0o P.u.

Since bus (2) is P-V bus, P2, P, and Q3 are to be calculated from the following expressions

*r: Ll uo ,o Yr, lcos (or, * 6, - sr)tl=l

a, : il v, v, Yr, lsin (oon *sl sr)q=|Pz : Vz Vr Yz, cos (02, * 6z - 6,) + V, YzYzzcos 022 + V2 V3 Yzg cos (}zt + 6z- L)"'Pr : Vl Vr Y:, cos(0r, t 6:-6,) + V, YzYtzcos(0r, + 6r- 6, + V3 V: Ysrcos 0r3Qr : Vr Vr Yrr sin(031 t 6r- 6,) + V, V, sin(0lz t 6t- 6) + V3 V3Y33 sin033

From above formulae

Pjo) :0Pju) :0Q!o) : 5 sin (- 90') + 5.5 sin (- 90') + 10 : - 0.5 p'u.

118 Elrct'fral Pouer SgstemThe power mismatches are

AP2 : P21.pe", -

Pjo) : 5 -0 : 5 p.u.AP3 : P31rp..;

- Pjo) :

- 3.5

-0 - - 3.5p.u.AQ: : Q:(,p..) - Q!o) : - 0.5 - (- 0.5) : 0p.u.

The perfbrmance equation in matrix form is

I ap, ap, ap. Ilap,l I a6-; a6l a6l [aa,llror l lffi ffi ffill*;l

L 06, ?6r av3 I -

= - [vz vr yzr sin (02, f 6z - El) + v2 vsyzt sin (04 + 6' -

63)]

- - (- l1

- 5.5) : 16.5

aP.a6;

: v' v, sin (ozr + 6z - 6r) : - 5'5

AP^ I

ffi : V: Yzlcos(0'3 * 0z-6r) : 0aP.# : V3 V" Y.2 sin (032 t 0t - 6J : - 5.5ddzaP.

fr : -[V:Vr Y3, sin(gir + 5r-6,) + VrV,Yrrsin(0lz + L-6)]:

- [- 5.0 - 5.5] : 10.5

# = Vr yrr cos (01r + 6:

- 6r) + Vr yrrcos (0rz + 6l- 6z) + 2[3y3rcos0r,

-0

H : - Vr Vz Yucos (orz + 6s - 0r):Q

dQr06.

: Vl Vr Y, cos (0tr * 6: -

6r) + V3 V2 y, cos (0, + 6, _ 5r)-Q

Ao-# : vr Yrrsin(031 ri:-6r) + v2y32sin(ore + 6s_L) +2y3y33sin033ov3

=-5-5.5 +20=9.5

DP,

45,

Load. Ftoto Studies 119

[-i] :ll::i q,l[^t]Lol Lolffil l':: 'ij ,ll [{]

I o.ott+ 0.03846 o I l- 5l:lo.ora+o o.ns4 o ll-l.slI o o o.roszo]l oJ

laorl l-o.ztzsz1loo, l=l-o.ztrs+l.uaLou,J L o l

6!') = 6!u' * AD2 : o + o.23zs2 : 0.23252rud : 13.32"6!' : E!u) + A6, = 0-0.21154 : -O.2n54rad : - t2.l2o

v3(r) -

vj') + AV3: t.o + o: l.op.u.hoblenr Z.9.-singte line diagram of a simple power systemwith generators at buses I and 3 aswn in Fig- 2.15. The magnitude of voltage at bus t is 1.05 p.u. Vottage magnitude at bus 3 isI at 1.04, p.u. with active power generation of 200 MW. A load coisisting of a00 MW and

MVAR is takenfrom bus 2.Line impedance are markcd itt p.u. on a 100 MVA base and the line charging susceptances are

Determine the fottowing using Giuss-seidel method at the end of first iteration.fr) Voltage at buses 2 and 3fr) Slack bus power

Direction of line flows and line losses

0.0125 + j0.025

@ lvrl=t'o+Ps = 200 MW

o

Fig. 2.r5

12O Electricat Power SystemSol. The line admittances are

Yrr: ! : l0-j2oLtzYrr : 10 -730 and Yy : 16 - i32

I zo-7so -to+jlo -to+j3olYBu* : l-to+7zo 36- j52 -16+ fi21

f- ro+7ro -16+ j32 26- j6z )Sz = Pz -"/Q2 : -

(400+ j2s0)100

= - 4 - j2.5 p.u. (since load bus)Pr : 2'0 P'u'

Bus I istakenastheslackbus. Startingfromaninitialvoltageof V! ! t +j.0, V! = 1.04T1

vi*, = + 16j." -fr,,v,.,- ,>;-r,7ul :+[h1- Y''vi

-""-u']l-+- jz.s _,_ l

32 _ is2l-5 - (- 10 + i20)x 1.05 - (- 16 + jaz) x t.o+l: (0.97462

- jO.Ot423O7) p.u.

Bus 3 is a regulated bus where voltage magnitude and active power are specified. For acontrol bus, first reactive power is computed as

e! : -,o, [{r,'). (v, vi +\, vj *v,, r,')]:

_!mtl_._Qa(.10.+ j30)x 1.05 + (- t6 + j32)x(0.e7462- jo.o4naT +(26- j62) x l.0a)l: 1.16 p.u.

The value of Q! is used for the computation of voltage at-bus 3.

uj :+[t--Y''vi -"""j]

: *+t%* - (- r0 +i30) x 1.05 - (- ro + i3z) (0.s7462 - io.o4

: 1.03783 -

j0.00517 : 1.03783 2-0.2854p.u.Since I V: lrp.. is held constant at 1.04 p.u. and 63 can obtain only

V:' : t.04 Z- 0.2854 p.u.: 1.039987

-7O.00517 p.u.

load Flow Studies 121Slack bus power can be calculated

Sr, : P, -JQp : v; fvoorvnq=l

Sr = Pr -,lQr : Vi [Yrr Vr + Yrz V2 + Yr3 V3 ]= 1.05 t(20-i50)x 1.05 + (- 10 + j20)x[0.97462-jo.042307) + (- l0 +7'30)

(1.039987 - 70.00s17)l

: 2.1842 + jI.4085 p.u.Th complex power flows can be calculated as

Spq: Pp, f ./Qp,/ = YrIh = V, (Vj -Vi\ y.r,

s12 : v1(vi - vllyl' : Q67 .97 + jl13.88) MVA

s21 : v2(Vi -

vi)y; : (- 160.51 -j98.94) MVAS13 : (26.8 + j26.ll) MVASrr = (- 26.67 - j25.73) MVASz:: (- 224.06

-i136.51) MVAs32:233.13 + j154.58) MVA

Ib line losses areSlorrp4 : Plorrp, *JQ,o.rrs = Spn * S*

: (7.47 + j14.94) MVAhilarly, Sloos 13 : (0.13 + 7O.38) MVA

Sross23 : 9.06 + jl8.08) MVAlb power flow diagram is shown in Fig. 2.16, where the active power direction is indicated by

thb reactive power direction indicated by ..>

II

V {zta.nt

('.' Charging admittance is zero)

.852

1-140

167.97------+

160.51

-->

,*t'3

116III

Fig. 2.16

122 Elbctrical Pouer System AnalysisProblem 2.10. The transmission line is a 230 kV, 200 km long having the following data, find Y1

rnatrix Jbr the two bus sltstem. Express all values in p.u. on 230 kV and 300 MVA base.R : 0.074 ohm/km, cttL : 0.457 ohm/km

I

ac

Sol. Base kV :230Base MVA : 300

Total resistance, R : 0.074 x 200 : 14.8 f)X. : joll- : j4.457 x 200 : j91.4 Ox.:4:-jo.zllxto6x2oo(DU

: -

jSS.4 x 106e

7zBase -

Rp.u. :

XLp.u. :

Xc p.u. =

Line charging admittance,

Mutual admittance,

:

Yu:Y!: :

I'l&ts -

r^ _ Base kVA . _ 300 x 103 : 753.06 ArBase - {3 ,-(B^. kv) ,!-3 ** - t 't

V.".. 230 x 103= r05.4QI"r." 753.06

+-* = *+ : 0.04845 p.u.zr^r, 305.4j9t

'4 = i0.299 n.u.305.4

,## : -ir81.4 x ro3 p.u.

!rc: !zo: + : j5.5r3x lo-6 p.u.

1-l't::/zt:

-Lrz

I= (0.s28 - j3.26)p.u.

/p

0.5280.s28

.3253.2s

-0.=0.j0.3- j3.

iX,

):):-j+

Rp.u +,/

l-rz * -Yro

!!r * )zoI o.szs

-[-*rt

-.i3.2s-

j3.2s-0.528 + j3.2s10.s28

- josls)

Self admittance.

Inod FW[b Shrdtes 123

Problem 2.11,. The load Jtow data for the poruer system shoutn in Fig. 2. t7 is given in the fotlowingtables . The voltage magnitude at bus 2 is to be maintained at 1 .03 p.u. and reactive power 0 < Q2 < 35 .Obtain the voltage at bus 3 using G.S. method after first iteration.

Trble 2.2 Line dataBus codep-q

ImpedanceZ pq{nn.)

t-2t-32-3

0.8 + j0.240.a + j0.060.06 + .i0.18

Big.2.L7(November 2004)

I

Sol. Let base MVAThen p.u. values:

f{utual admittances

: 100

pc2 : 2i'!, =

g.2,ecz = o, Po: = 0.5, Qo, : 0.2vL 100

PG3 = 0, Qcl = 0, Por = 0.6, Qp3 : 0.25Pz : 0.2- 0.5 : - 0.3, 0< Q2 < 0.35P3:-0.6,Q::-0.25y-rr: yrr : +==-. !- =r.25- j3.7sr tL r Lt Zr, 0.08 + 70.24

Table 2.3 Bus dataBus codeand name

Assumedbus voltage

GenerationMW MVAR

LoadMW MVAR

I -

Slack2

-PV2_PQ

+j0+j0+j0

1.05t.01.0

020

0

000

0050 2060 25

124 Electrical Pouer System Ana.lysris

Ir.r:):r : +=5-,/1.5Ltllzt=ln:l-667-i5

Selfadmittances Yrr: Orz *)rJ = 1.25-i3.75 + 5-il.5:6.25-i5.25Similarly Y?2: 2.917 - i8.75

Yr3:6'667 -

j6'5I a.zs - js.zs *1.25 + j3.75 *5 + 71.5

yB,. = l-r.zs + j3.7s z.gr7 - js.75 -r.667 + i5[ -s *;r.s -1.667 + i5 6.GG7 - i6.s

: : ::; 1 :; ;; [(-o.g -_ {o_.ozzzs) _ (_r.zs + j3.7s)x 1.05 _ (_t.oot + i2.e17 - i8.7s L 1.03 \

: =-J- fz.aatz+- j9.o12sl2.917

-j8.75'= 1.01915 -.i0.0325 : 1.0196 Z-1.828" p.u.

[ -s * yr.s -1.667 + i5 6.GG7 - i6.s) IQ2

"ur = - r- {(vr',*" )- ["r, v,' + E, -'i.r* * Y* vr']] I= -rm{r.o:[(-r .25+13.75)x1.05 +(z.ott* j8.75)x1.03 +(-r.oot. jr){:

- IIn (0.0257 - j0.07725) |: 0.07725 p.u. I

Bus 2 acted as generated bus since Q2..1 is within specified limits Ini:*Ltr-Y,,vr -r,",] |

: 1sn 'rnlz'oatz+-

ie'ol2si I

vl : +[q-+-Y,, vi -",, uj-l, yr..L(q), r, , r. .)r [-0.6+j0.25

6.66:t - i.'{==: - (-s + 7l's) x 1'0s - (-1' 667 + i S\x(t'os) z

: (0.96627 -70.03696) p.u.

Problem 2.12. The loatl.flow data for the system shown in Fig. 2.18 given below in th:itbles.

Q2 car = - r- {(vrt,*" )- ["r, v,' + E, vl.o.. + "r, U i]= - rm {t.o: [(-t .25 + 13.75)x 1.05 + (z.ott * j8.75) x 1.03 + (-t.ooz + 7s):

- IIn (0.0257

-

j0.07725) r: 0.07725 p.u.

Bus 2 acted as generated bus since Q2..1 is within specified limits

vj : vl.*" z6\=t.o3l-1.828" =(1.02947 - j0.0329)p.u.

Load. Flotu Studres 125

Bus codep-q

knpedanceZ pq(po)

1-21-32-3

j0.0sj0.ri0.05

Table 2.4 Load data

Fig. 2.18

lVzl = 1.0 p.u., m&tcimum and minimum reactive power limits are given at bus 2 are 50 and- l0 MVAR's respectively. lJsing G.S. method, find the load flow solution upto first iteration.

(November 2007)Sol. Assume base MVA : 100 p.u. values

p, : 1# = 0.3 p.u., - o.l < Q, s o.z" 100

p, : \p=-0.2p.u., er =-0.2p.u., 100[-rlo izo i10 Iyr,. : | ;zo -j4o j2o IL ,,ro izo -i301

Q!",r = -rm{(vi,o*). ["r,vJ *vr,]vf.ro +",, q]:

-Im {r.o [;zo x 1.03 - 740 x 1.0 + 720 x l]]: * 0.6 p.u.

Table 2.5 Bus dataBus code Assumed

bus voltageGeneration Load

MW MVAR MW MVARI2?

1.03 + j01+ j0t + j0

0fl0

000

0Nn

010n

126 Elbctical Power

Q2.u1 violates the specified limits and it acts as load bus and fix Qz "ar

&s Q2 ,nin1.e., Qz:-0.1P.u.

y; : +L=t - i2ox1.o3 -izo xr]: (r.orzs + j7.5 x ro-3) R.u.

vl : +]ry-7r0xr.03- izo(r.orzs+ jT.srro*)]: (1.01 t667

- 1.6667 x 10-3) p.u.

Pr: Ilu, u, yr, l"or(er, *6, -6r)q=l

Problems 2.13. The power system networlcs shown in Fig. 2.19 obtain V3 using N - R methodfirst iteration.

Fig. 2.19

Sol. rBus -

3.9528 Zt0g.49.2233 t-71.65.27046 2108.4

Pp-jQp:The real power at bus 3 is

l5.8l14 tto8.4" 1s.27M6 ztOS.4 l

2l.o8le z-71.6)Itg.ioq z-71.6"I| 3.9528 2108.4Its.sr t4 ztor.4

Iui Yoyo,q=l

Bus code Asswnedbus voltages

Generation LMdMW MUAR MW MUAR

I)3

1.05 + j01.0 + j01.0 + j0

0n0

000

0ilfr

0n25

Load. fr1ow Studres 127

And reactive power is

Specified values,

Taking base

Power mismatches are

= Y,{ {Y t Y31 cos (0lr + 6: - 6r) + V2 Y32 cos (032 + 63 - 62)* Vr Yr3 cos 033)

= 1.0 [15.8114x 1.05 cos (108.4o)1G 5.27046x I xcos (108'4")+ 21.0819 x I x cos (- 71.6')

= - 0.25 p.u.

Qr : Vr tVr Yrr sin(031 t 6r-6r) + v2Y32sin(0rz + q-6, + v3Y3 sin0rr)

= 1.0 [15.8114x 1.05 sin (108.4') + 5.29046 x 1 x sin (108'4") +21.0819 x I x sin (- 71.6")l

= 0.75 p.u.

Pr=0:60=-60MWQl:0-25:-25MW

MVA = 100

-(n-;53inp.u. : J!ff. : -0.6 -i0.25AP: : - 0.6 - (-0.25) : - 0.35 P.u.AQr : - 0.25 - (0.75) : - 1.0 P.u.

Total Jaccobian matrix elements are

I Ep, aPr Il^P, l_lao, a%lLaq,l - laq, ae, I

LaE a",j

H = H - v3[Vr Y31sin(0, + 6r-&) +v2Y32srn(gsz + e2-Ul

: 1.0[1.05x 15 + I x5] = 20.75N: 9+ : [Vr Ytrcos(03, * 6r-63) + V2Y32cos(0p + E-S,

av3+ 2V3 Yr3 cos 0331

: 1.05 x (- 5) + I x (- l-667) + 2x 1x6.667 :6.417I : ?9' = V3 [Vr Y31 cos (0rr + 6r - 6l) + V2 Y32 cos (032 * 02

= 1.05 (- 5) + t (- 1.667) : - 6.917

L : * : - [Vr Y,sin(0rr + 6, -6r) +aV1

t, :-[1.05x15 + 1x5 +zxlx(-20)[-o.rsl _l zo.ts 6.4t71[oo,lL r ol l-e.en D.zs)Lav,lioa,l _ [-o.ozso+'lLav,l I o.o+rze

-]

AVr : O.O4l2g

YzYnsin (0r, * 6z -

+ 2Y, Y33 sin: 19.25

1.

fuis:,

Voltage at bus 3 after first iteration, V: = 1 - 0.04129 = 0.9587 p.u.

sHoRT QUESTTONS AND ANSWERSWhat are the diagonal elements of Ynr" known as?The diagonal elements of You, are known as the short circuited driving point admittances.At a particular bus in a power system, the load complex power aggregates to (100+ j50) MVA andgenerator complex power to (150

-

j 75) MVA. How is this power classified? What is the buspower?

Ans: Bus complex'power : (generator complex power) -

(load complex power)=

(150 -j75)- (100+js0)

: (50 _j25) MVA3. How long transmission line represents in load flow studies?

Ans: As positive sequence equivalent r model circuit.4. How is a fixed tap setting transformer represented for load flow studies ?

Ans: Represented as an equivalent n model circuit is shown in Fig.2.20

Fig.2.2O

The admittances given by

YA = Y.""=*(l:-r)r- and y6 =(,-*)".where Y, is the series positive sequence admittance of the transformer referred to bus '4'side andis the turns ratio.