Question

= A 50 Hz synchronous generator with an inertia constant of 3 MWS/MVA and Xd' 0.3 per unit delivers power to a large metropolitan system considered as an infinite bus of V = 1.05200 per unit through a transmission line of X = 0.1 per unit. The infinite bus absorbs P = 0.8 per unit at 0.85 power factor lagging. All per unit values are given on common base. A fault occurs near the generator terminals reducing its power output to zero. When the fault is cleared after 0.1 s, the original network conditions again exist. а

a) Calculate the generator internal voltage E’ and the initial generator power angle δ prior to the fault.

b) Write the swing equation for the machine. What will be the generator angle δ' at the instant when the fault is cleared after 0.1 s?

c)Determine whether the synchronous generator will remain stable after the fault is cleared.

d) If the fault clearing is delayed, determine the critical clearing angle and critical clearing time.

Answer 1

TL mm mm x=0.1 PU ~) Infinite Bus N=1.05 Lo f-502 9 xd'=0.3pu P = 0.8 pu at 0.85 power factor P = V 1 cos o 0.8 = 6.05) (1) (0.85) =) oid e-cor(0.8) 1:09) (0.65) 1 = 0.89632-39-78° po 9 m om xd! X V= 1.05200 s E = Vt vt 1 (jxd + } ] 1:0520 + (0.8963 <-3178°) (0.34001 2909 E 1.05 t 0.3585 L 58.220 " E = 1:05+. 013589 Com (58.223 tj 0.3585 Sim (58-22) 1.23 8 t О, ПОЧЕ 31.2 382 + 0.80442 <tant (0.30471 1.238 1.245 213.820 -> |EJ= 1.2750 8 =13.82 Pleion to the fault

swing Equation, 2h des Wo (Pm- Pe) pu Pe=0 fault . when fault occurred , Pm = Pe before Pe = E.V. sing = 0.8 pu Уl+X 228 2x3-14x50 de 2 =(0.8-0) pu = das 0.88 2x3.14850 41.86 elec-read 52 2x3 s ds 41.86t + に t=0 41.86 2² + so 8= - 0 0 EV Sims =0.8 Xatx 1-295x105 (Sims) = 0.8 o'y 18= 13.820 $ = 41.86x0'1127 13.820 y 2 + 13.82 0.209BX 180 3.14 12+13.82 8 = 25.82

remains &table (") If 8<8c ý system Pmax PMA EV = 6.275) (ros) Xarx O'Y PM = 0.epu Semo Ilmax = 3.34 SC Aso pu so Using equal ana aiterions Siem- o) do s (Pmax sims - Pm) de 2 Sc > > Pm C 8c-so) Pmax ( cossct cosso - Pm (4-so-sc) Pm se - lm so Pmax (cosse + cosso) - Pm (7-86) +Pm8c Pm (A-80-so) = Pmax (cossct cosso) 08 (180– 13.820) 3:34 ( cos &ct cos 13.82° 3.829) 0.8 x 166:18° X3.14 3.34 (Cosset con 13.820) 180° 2.319 = cossct 0.971 3.34 cosoca -0.276 106.020 SC sa sc 3 stable system 2 106.02-13.82 = 20.987² Se = 41.66 +2 + So ite = 0.277 sec