Question

Figure 3, shows the one-line diagram of a simple three-bus power system with generation at buses 1 and 3 . The voltage at bus 1 is \(V_{1}=1.025 \angle 0^{\circ}\) per unit. Voltage magnitude at bus 3 is fixed at \(1.03\) pu with a real power generation of \(300 \mathrm{MW}\). A load consisting of \(400 \mathrm{MW}\) and \(200 \mathrm{Mvar}\) is taken from bus 2. Line impedances are marked in per unit on a 100-MVA base. For the purpose of hand calculations, line resistances and line charging susceptances are neglected.

(a) Using Gauss-Seidel method and initial estimates of \(V_{2}^{(0)}=1.0+j 0\) and \(V_{3}^{(0)}=1.03+j 0\) and keeping \(\left|V_{3}\right|=1.03 \mathrm{pu}\), determine the phasor values of \(V_{2}\) and \(V_{3}\). Perform two iterations.

(b) If after several iterations the bus voltages converge to

$$ \begin{array}{l} V_{2}=1.001243 \angle-2.1^{\circ}=1.000571-j 0.0366898 \mathrm{pu} \\ V_{3}=1.03 \angle 1.36851^{\circ}=1.029706+j 0.0246 \mathrm{pu} \end{array} $$

determine the line flows and line losses and the slack bus real and reactive power.

Answer 1

M-1.025 L6 pu Pa=300MW j0.05 Ivs1= 1-03pu fo.025 jo 025 460 MW 200 MVAY 400+200) pu- (4+j2) p CRytj@)=400+ 200) MVA 100 3= 0:0 pu PLg=0.0pu 300 300 MW I00 Psch =Paen-) 0-4) pu40 pu Psch s'on-Ps)= (3.0-00)pu 3.0pu. sch =Cagon-Bu2)--2pu-- 20pu Rsangan-a) = 0-0)p-opu. at busk, cuunent injection, = 2 Pruhp-schn TK

Petny che (1) YKK jz1 erforming paaes ftna using hoed for Gars-Seidel method Gauabian (1) will be were iteratich no is deusted by i A (K-) NA YKK jAK for the givan Aystews Y-bus mato iven by jo.025 O,05 0.025 50.05 Y= 0.025 O025 50.025 a025 0+0g5 25 O 0 0 025 0.0s 0400 520.07 j60.0 j80.0 540.0 j20. 0 540.0 -j60.0

a leck bus CrS Bus 2 Pa bus is Aus 1025L0 pu. this value is Gmott a for alk the ituatin iwtial estinates, -0jo.pu itelatian C Pschaschz Y21 Vi Y22 C4.Otj20) 40)(02526)-04)030) 80.0 0025-j0.05 )1.0037-2-8553 Pscns-JOschs Y31VY (3.0-jo0)(j20) (1015 /0) - G40)( 0037255 -j60.0) L0.8629 P 2 .01008 ottaga -Controlled bus So, V| =1-03 nadlr Bu 3 a We anyune that @ the V03/0.86 24 M exetive supply infiredte auy amout & Reep the Voltage V at bus 3 Can at Tequired Iteration (2) (Peycc) Fri1 C4.0tj2-0) 1I-0037-2.9 553)* -j) (10256) J.0 0091 /-2. 8337pu (2) 21 Y22 j80.0) 4(030.829

(Psas-jAschs) 2 ale 132 ~(5-0-jo.0) 2 G 60030.86 29 ye G20)(1-025L0) -G) .00012.333 1-0079 21:2148 u medity he take Vs 1.03 and S 12148pu vat 'nesult V3se the we - :03 2=1.00243 -2.pu 1:03 /1.3685I pu

Line flors from -2 S2 1 V Z12 025 0-1. D01243/-2 60.025 (1.50425-j1.0016)p line flas from 2-1 10012437-2.1 - 1:02520 boa025 (1001243 21 E1.50425+jo92381) pu Pi2= 1.50425 pu Pioas, 2- 2tP) O pu s P21= -150425 pu. aich is quite nakural as the line ts rfecty redive -0.92389 p u 8912= 1.0016 Pu c0.07マ71u. une flors fromn 2-3 F -(-2. 4957+07615u S28 24957-.26035) pu 2-4957; Pns25Past2)-0pu ine flow fram 3-2 V3 32 Z32 22 P23=24957 pu. Par -1.07615 pu; Q32- 1.26035, Qesst 2) o. 1867 p.u. 1

Line floa from 1-3: .504 28jo.09648) pu 212 line flors fom 3-1: D-VI Z31 0.504284-j0-00.l0902) p-u pu is0.50428 bu ; Pa=0,50428pu, Prist f 0.10902 pu + Quris01254pu Q13=-0.09648pu Slack bus power APpareut paser Anijected ble ck bus 1.0-30.90513) pu slack bus neat porser injectian, = 1000 pu slack bus reactive pacoer gectia, Q0.0513 u