Lesson-40 Problems on Alternators A 3phase, 50 Hz, 16 pole generator with star connected winding has 144 slots with conductor per slot is 10. The Flux per pole is 24.8 mWb is sinusoidally distributed. The coils are full pitched. Find the speed, line EMF. [June/July 2015] = P=16, slots=144, conductors/slot= 10,Ф = 24.8mWb, f 50Hz, Kc=1 Slots/phase =144/3 =48 = = 20˚, m = 3(three phase); = = slot angle = Kd = = 0.96 = = = Ns= 375rpm Z=slots× conductors/slot = 144 10=1440 × = =480 Zph= = Tph = 240 Eph= 4.44 Kc Kd f Tph= 4.44 ×1×0.96×24.8×10-3×50×240 Ф =1268.5V EL = = √3 Eph √3×1268.5=2196.2V A 2 pole, 3 phase alternator running at 3000 rpm has 42 armature slots with 2 – – conductors in each slot. Calculate the flux per pole required to generate a line voltage of 2300 V. Distribution factor is 0.952 and pitch factor is 0.956. [June/July 2014] Given P=2, Ns=3000rpm, Eline=2300V, Kd=0.952, Kc=0.956 = Ns= i.e. f= = 50Hz = = Eph= 1327.9056V
Lesson-40 Problems on Alternators A 3phase, 50 Hz, 16 pole generator with star connected winding has 144 slots with conductor per slot is 10. The Flux per pole is 24.8 mWb is sinusoidally distributed. The coils are full pitched. Find the speed, line EMF. [June/July 2015] P=16, slots=144, conductors/slot= 10,Ф = 24.8mWb, f = 50Hz, Kc=1 Slots/phase =144/3 =48 m = 3(three phase); = slot angle = = = = 20˚, Kd = = = 0.96 Ns= = = 375rpm Z=slots× conductors/slot = 144 × 10=1440 Zph= = =480 Tph = = 240 Eph= 4.44 Kc Kd Ф f Tph= 4.44 ×1×0.96×24.8×10-3×50×240 =1268.5V EL = √3 Eph = √3×1268.5=2196.2V A 2 – pole, 3 – phase alternator running at 3000 rpm has 42 armature slots with 2 conductors in each slot. Calculate the flux per pole required to generate a line voltage of 2300 V. Distribution factor is 0.952 and pitch factor is 0.956. [June/July 2014] Given P=2, Ns=3000rpm, Eline=2300V, Kd=0.952, Kc=0.956 Ns= = i.e. f= = 50Hz Eph= = = 1327.9056V