Question

P2. (50p) A 100 kVA, 400 V synchronous machine has the following equivalent circuit parameters: X, = 2.0.2, R = 0.25 12, PART 1: When the machine is operating as a generator and connected to the infinite bus. a. (10p) What should be the power factor to have zero voltage regulation at 50% load (draw the phasor diagram)? b. (10p) Calculate the percent voltage regulation at full load with 0.9 leading power factor (draw the phasor diagram). PART 2: When the machine is operating as a motor and drawing 144 A current from the grid. c. (10p) Calculate the E, and 8 if the power factor is unity (draw the phasor). d. (10p) Calculate the Erand 8 if the power factor is lagging 0.8 (draw the phasor). e. (10p) Calculate the E, and if the power factor is leading 0.8 (draw the phasor).

Answer 1

Solution Given a Synchronous machine rated look VA, 4000 Sydhetsensus X (=2r, Re=0.25 Part 1) Given machine works as. Rs jxq Generator a - voltage regulation Ta is zero at 50% f =400 Lo pated of load full load= 100kva, --) 50%. =) 50KVA V3 Upts = 50x10² 3 Isa 50x10 400 x 3 Is = 72.174 Condition of Zero voltage regulations cos Coto) - Substitute -Igts E phase zut values. Z Rstjes, Izdavant xa 2²+0.25² Es 2.015 m

Cat - - Itan ( Clan ) Clan 2 0.25 82.87° cos(Osto -Ist z Vt =-72.17X 2.015 2x 400 √3 coslost 8) = -0.314 cost [-0.314) 82.87+0 - Ost0 = 108.351 &= 25.989 (p. 25.490 0.902740 -) Coco = Power factor= cos(25.49) - 0.9027 -> o is positive Is leads Ut factor power is 0.9027 leading

calculating 6: - cal EJ = Vt Lo + Is L&* [Rstjxz] 400 at 72.17L25.481 x2.0154 (82.87° √3 4004 145.42 (108-711 √3 23094 + (46641 +1 137.93) 184.291 +3137:33 230.07 L36.772 G 6 = 36.77 phasor diagram Taks 8= 36.7720 Tats, Te Reo 2514910 t

14 load = full verload with 0.9 leading Power Lector S= look va 3 ray Is = 100x10 Te 100x10} = 144.34A √3x 400 costa 0.9 %) 25.5419 I= 144.34 €25.841 -) = If Lot Te [Rst jxs) E 400 √3 t. 144.34 L28.841..2.051282.87 230.94 +996.041 2108-1741 - 230.94 - 93.248 + j275.46 137.692 692+1 275.44 307.95 (63-441 June line = 533.3694 &= 634410 =) voltage regulation. Eve 나 533.3694-400 Xo 400 33.33%

- phasor diagram Tats ES 8= 63.44° Mats Dash 250ml Is Rs to lo part ② Machine aurs motor with I= 1440 Power factor 1, (aco =) COSA= =)] = 144 Lo mte Rs jus RES Is fromku y = yLi-Is (Rstjes 400-144 Lô 2015 * 282.87" JB 230.94 290.16 L82.87° 230.94 -36-014 - 287.91° 194.899 - 1269.98 f = 347.67 L55.904 855.904 > 4 602.1644 line =

pharos t. Is fa 8=-55.904 Texs Xs a as Ef = Vt LO I E cos$= 0.6 XI = 144A (-36.86 4- 36.86, Lagging 1s (estjxs) 400 144 L-36.86 x 2.01528287° √3 230.94 290.16 L4601 230.94 - 201.525 -3 208.723 29.415 -3 906.723 210.785 L81.97 E 0

f line = 210-785x83 O 365.077 8 = -81.97 phasor Is fa • Lo 36.86 Ist 1 81.97 ç Cosp=0.8 b= 36.86° leading Is=144 L36.86 2) E = 400 - 144 €36.86 x 8.05282.87° 3 230.94 290.16 L 119.73 20.96 930-94-[-143.89 +1251.96) 374.830) 451.642 2-33.9° If line = 451.6424.53 G 728 - 742.2431 6=-339°

phasor 36.86 33.9° L -IPs Tas