Question

2. nBase MVA 100p LI Base KV G1 20kV Base KV Lines 220kv G13 phase 20 kv, 100 MVA Xd-1, Xd-0.3, Xd"-0.2, Xo-0.8 H-1s T1 3 phase 100 MVA20kv -220kv, DY,X-0.1 L1, L2 Ro X1-0.2 pu on given base Y ignored L2 The infinite bus voltage is 1 PU The generator supplies 0.6 pu real power at 0.8 pf lagging to the infinite bus a. Sketch an impedance diagram appropriate for transient stability analysis b. Calculate the internal transient voltage E5 for the generator G1 C.Calculate the power-angle curve for the intact system

Answer 1

infiny Bus bar vi- o pu 20kv 220V # Base Values 100 MVA, 20kv for G)210kv ener alor or Genera tor Tans formcr me t For transient Stabiliy Anal enerator is a (trans ent reactana ) .30 # Irn pedane diachon) Or transient Hnal ven power Systen Aletook

Dlerna ran Sient Voltaqe (Sj LS) er nalor uShor at bar . Pf angle at bus ba 0 8 nte o 1PU :310.1t 어

03)+ (1-225) E;二 1.2612. Substitut iCO 2 Loe Ge SinS: o-2318 >- Sin (o-2318)13.518 C power -Angle Curve. 2522 P 25- SnS P- S Curv

ran) com ro -lc[ive e [evevc coCr i n (I 261) C1) . and Ma n tude D2.lO2 Sin only o Coad power a ttor 2.521 System oih *ppin 2lo 2 we 2

lengtn from G, O-2 0.3 3 Ytlo Ej Ls ot p.dan 3 (0-1)(0-/K) , 00334 O-24 0-1+0 ヒ o o334 o 066 3 (022) (%4 ) 3 - 6-622-3 o-o223 ヒ o 4339 b661 0 0223

0.1334 о-0667. 23 3 (0 4 339)(ooh67 lt (0.0223)(0.0 667 ) t(o·q339)(6.02 23) .627,3 6.0 1-8 PU O O223 (6.6223) (6.0669)1 (0433๑)(6 (093341 (6.6667) ㅓ 6.123 pU o223) Co o 66) C 431 (o o223) o 0 66 stem nletuaDb LS USba cr ewur oalor rom o·4334 P- 2I8

3· 0261 Sin hen ysd dlanrrans former bus bor ts ine 2 has a bolted 3-ф phane ault at Sing ern Iven tripped ● P-on C.) → When@i lhe one of tte ltveA PS Curves -for. different Cord,Tr