Question

Problem #2: A three-phase synchronous generator is rated 16 kV and 200 MVA, has negligible losses and has reactance of 1.65 pu. The generator is operated on an infinite bus of 15 kV voltage and delivers 100 MVA at 0.8 Power Factor, lagging. a)Determine the internal voltage (E_i), the power angle (5), and the line current of the machine. b)If the field current of the machine is reduced by 10%, while the mechanical power input to the machine is maintained constant, determine the new value of the power angle (0) and reactive power delivered to the system. c)The prime mover is next adjusted without changing the excitation so that the machine delivers zero reactive power to the system. Determine the new power angle (0) and the real power being delivered to the system. d)What is the maximum reactive power that the machine can deliver if the level of excitation is maintained as in part b?

Answer 1

Given Baye voltage = 16kv Base MVA = 200MVA S=3* VI base Bus voltage = 15kV 4 MVA -TOOMVA - Pay Per unit MVH - 100 =0.58.4. actual MVA Pare MVA EMUA 200 V p.u= Vactual - 15-09375 200 MNA= 3 + 16000 & I base bax - To - I base - 4166.666 A Now 100x106_ 3 + 15*10* Iactual lactual 2222.2222 - 20000 Ip.u dactual Ibane 72800 - 0.533P.4 E = + la Z Razo X = 1.650.4 ELS = 0.9375 Zo +0.533 ZOS (0-8) */*65296 = 0.937520't 0.87992 253-13010 EL8 = 1.62581.225.6586° E= 1.6258) S = 25.6586 The Futerual Voltage Power is constant Pa Exsind Pi - Ei sindi E2" sinsz B

E is reduced by 10% E = 0.9E, E fin (26-6586) 0.94 sindr find) = 0.48 111 &2=28.7585 we know that Q=3 EV Q = 3EU COS SE E = 0.9E1 = 0.9* 1.62581 =1.4632 3714632 +0.937500$(28.7585) (6-9378) 1-65 1.65 = 3 + 0.9375 | 1.4632 cos(28.7585) - 0.9375 1.65 Q=0.58844 P.y. The power is zero reactive Q = 3V Floss-u] Écoss av Coss = 0.9375 1.4632 20.640718 8. = 50.15us

a the manimum Power i e reactive lower Q = 34 (Ecoss - V) 10 de o - =3-ESiu8-01-0 sind=0 8=0 substitute & zo in Q = 380-9375 [.463) # los0 - 0-9375] 1665 Q-0.89607 Pu.