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Electric Machinery Fundamentals (Solutions Manual) Part 2

Tham khảo tài liệu 'electric machinery fundamentals (solutions manual) part 2', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | a This generator is Y-connected so in this generator is I . At rated conditions the line and phase current L I I TP- N kVA 251 A at an angle of -36.87 V 3 3 2300 V . The internal generated voltage of the machinThe phase voltage of this machine is T V V 3 1328 V is E V I A A S R EA 1328 0 0.15 jX 251 36.87 a j 1.1 251 36.87 a Ea 1537 7.4 V The input power to this generator is equal to the output power plus losses. The rated output power is OUT 1 000 kVA 0 8 800 kW 2 2 CU 3 A P I R 251 A 0.15 28.4 kW W 24 kW cP re 18 kW sP ray assumed 0 IN oup P Cp F pcorP stlap 870.4 kW n _oup 800 kW 100 91.9 100 P 870.4 kW b If the generator is loaded to rated kVA with lagging loads the phase voltage is 1328 0 V and V the internal generated voltage is EA 1537 7.4 V . Therefore the phase voltage at no- load would be V 1537 0 V . The voltage regulation would be 1537 1328 _ . . V ----------- 100 15.7 1328 1328 V and c If the generator is loaded to rated kVA with leading loads the phase voltage is V the internal generated voltage is E V I A A 1328 0 Ea Ea 1217 11. I S R jX 0. V 5 251 36.87 a 1.1 251 36.87 a j o The voltage regulation would be 1217 1328 . . V ------------- 100 8.4 1328 d If the generator is loaded to rated kVA at unity power factor the phase voltage is V 1328 QJ V and the internal generated voltage is E V I A A S R jX 115 1328 0 0.15 1.1 251 0 A Ea Ea 1393 11. 0 oA The voltage regulation would be 1393 1328 . V - ---------- 100 4.9 1328 e For this problem we will assume that conditions and see what happens to the voltage as load increases at 0.8 lagging unity and 0.8 leading power factors. Note that the maximum current will be 251 A in any case. A phasor diagram representing the situation at lagging power factor is shown below the terminal voltage is adjusted to 2300 V at no load E o V By the Pythagorean Theorem 2 cos 2 sin sin s co 2 V R I S S R cos 0 X I D V E A R I 2 in s 0 in c A R I S A phasor diagram representing the situation at leading power factor is shown .