Question

Question 1: A single line diagram of a three-bus power system is shown in Fig 1. Bus 1 is the slack bus with a voltage of 1.020 per unit, bus 2 is a voltage-controlled bus (PV-bus) with a voltage magnitude of 1.05 pu and real generated power of 1 00 MWand the reactive power in the range Q.(20MVAR) < Q<Q-60M¥AR .BUS 3 is PQ bus with P 300 MW and Q= 200 Mvar. Take 100 MVÅ susceptance are neglected as a base and assume the line resistance and a) [5 points] Using Gauss-Seidel method, determine Qa, 6 and Va. Use an initial estimate of V 1.05.40 and V=1.040 and perform TWO iterations only. b) [5 points] Find the real and reactive power S: of the last iteration of the voltage iteration obtained in (Part a). 2 12=-j40 P=100 M 13=j20 923=j20 Slack Bus V21= 1.05 1.020 300 MW 200 Mvar -juo -j20 J60 j60 -j20 -juo 1300 juo Spch a 0-100)t(0-0) ja0 -j20 sch0-300) H0-200) Oo! 100

Answer 1

100 MW Mi340 (a) Slack hu 200 MVAr MW S13 Yn Ypus 1GK Y33 Edha-y) Y12 12 -(tat to EE j20 360 -340 -320 S40 360 346 j20 j20 O0 TOO 100 MW = .0pu 『2- Pecn2=Caz-fら)-(+-0-0.0) - 200 200 MNAT -TOD 069 Pao 3.0 pu 200 2.0pu 300 MW 100 ColaQh2.0pu - Oschy Qう Pchs Poz- Co-3.0u3.0pu 22

jected apparent prser fom hus s Si Vi NB Vi "N NB (Rjai) C-Jai NB

iteration, vottrge the it hn Frr kt NB K-I) 4 J (KI) (k-I) NB 1 (K-) > ja1 inutial quen VI0L0pu

0sC-j40 X0jo xiow-jsoxi)] L3 3.15) Tm - -3.15 P.u C20p MVAR) 0.2 pu GO MVAY Qnax Quin= 0.2 pu Loas CP8) bus bu 2 ile be treated an 1st iteration: a sleck bus. IOL0 bu buo YI V 23 /2 (1.O-70.2) 560 1.91692-014554 pu rojag Ya, VY32 133 3tj20)po)x0o) - Ej2o(1ousi 540 1.54992.4" p'u (1) ) o) 0)+0) x -ImD1169 Lo455 97.564 pu. Qui'n - O.2pu bus 2 aiet be treated ao P-Q(LOa& bus

tnd iteration (1J Y22 - 40) C1-Dlo) i20C1541/2-48) IO-j0.2) 560) L-9969/-o4554) 11812/0.6802 pu P (2) Y31 V- Yas 3-0Fj2-0) G40 .5491/2.48) 20 )x.0L') -20)(-1812/0.4802) L2.8996p4 1.1221 IV31= 14221u 2.8990 &2=0.6802 -Im( , I 1812206802(40) x6o)+j6)1g12b4) 8岛990 -11.782 bu ag- Quin 0.2pu Qnin

(b) Real & rea ctive poaer injectian at buo 1, S V V22+Yig V3) r2 P-ja=(1-0L0)*C60)x C-0Lo) E0)(-1812/0.6802) -(1.6959-39.658o pe + (320)C1221/2.&790')7 ral poses irjecbian, 1-695 pu-169.59 MW = 965.80MMAY Au neactive paser injectien, 9.6580bu