Question

A 3-phase, 100 MVA, 11 kV, 60 Hz, 200 rpm synchronous generator has synchronous reactance of 12 2 and negligible stator resistance. a. The generator is connected to an infinite bus and delivers 80 MW at 0.8 leading power factor. Determine the power angle and excitation voltage EA. Draw the phasor diagram. b. The field current of the generator is now slowly reduced until the generator reaches its static stability limit (5=0). For this condition, determine the power angle, excitation voltage, the machine current and power factor. Draw the phasor diagram.

Answer 1

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801 VT (terminal voltage) 30, 100MVA, 11 KV GOH cost=0.8 (leading) Synchronous reactance 0 = 36.86° Xs = 12 r. Stator Resistance, Ra=on Power output of generator per phase, log = 3 EAVT (05(02-8) - Viral Zg2 LO 0 Zs = Ratixs 2 = 0 + 712 Zs = 124900 | .. O2 = tant x = tan' 12 = tan co Ra Oz=90 (a) from eqo, Pog = 80mw (given) for 34 80x10=> y en 190-1) =o) (va replace - you ) 80x10 = 3 XEAX 1X10² 13 cos/90-6) marr - 12 80x10° x1 2 = 3 X EA x Il X10% sin s, - ©

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I where Ia = a 5 - 3 x V (line) 100X108 BX11000 5248.63 A or Fa, - 5 248-63 1836.86 A (leading given) or Tag = 5.248KA (36.860

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: EA, =) ( VF cos & + Tapa} + (We sin p-Taxs) (for leading case, EN = 1000 x0*8+ Ta xo) + ( you x 066 –5208643 X12) 2 TEA, 59.39 KV Using ea, 80x100x 12 = √3x EAX 11x163xsins, putting value of EA, . sins, - 80x100x 12 3x11x 103 x 59.39x10 3 S- 58.031 = 0.8484 Phasor Viagram EA, - 59.39KV/58.03. = 5.248 KA /+3686 30+36-86 36.86 V7 = 11000 - 6351 10° = 6.35kV Li ce Henc, excitation voltage, EA, - 59.39 KV (perphase) & power angle, S = 58.03.

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b) : P EAVI sing (for Ra=0) xs when : de EAVT COSE de when de-o EAVT Cos8=0 XS COSS=0 18-8, = 90° - Sratic stability limit. Power angle , 8 = 960 ]

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= constant ao P=EAVE sing EAsinde constant El2 sind. Efisins, Effax sin go = 59.39 x sin (58. 03) JEA2 = 50.38 k Vlth This is the new excitation voltage.

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