Question

A 480-V 100-kW 0.85-PF leading 50-Hz six-pole Y-connected synchronous motor has a synchronous reactance of 1.5 and a negligible armature resistance. The rotational losses are also to be ignored. This motor is to be operated over a continuous range of speeds from 300 to 1000 r/min, where the speed changes are to be accomplished by controlling the system frequency with a solid-state drive. (a) Over what range must the input frequency be varied to provide this speed control range? 6-6. (b) How large is E at the motor's rated conditions? (c) What is the maximum power the motor can produce at the rated conditions? (d) What is the largest E could be at 300 r/min? e) Assuming that the applied voltage V, is derated by the same amount as E, what is the maximum power the motor could supply at 300 r/min? How does the power capability of a synchronous motor relate to its speed?

Answer 1

6-6 . 2 of F = PN 120 - 6 x 300 6 x 10oo 120 120 16) 1) For a speed, (300 opm ) Fet a speed , (10007bm) - N= 1207 fzPN = 6x 300 = 6X 1000 +7=1542] → $ = 50 #2/1 The range of frequency must be controlled as 'Is Hz to sotz We know that, E = V - RI -ỷx >0 (Armature IA= P 1 00x103 Силистe v , Cospx4xo. 85 = 141.51A sy=0.85 P = 40590.85 = 31.78 :In = 141.51131.79" AJ [physingen Woha at = 277. 128 V Equation 0 becomes, E - 277.128 Lồ - (o) – I-ra-)(1 44. - ( since RA is negligible = EA = 428.74L-24.88 VV Sresistance) 31.78)

les Marimum power, max = 3 The EA Xo ③ Х - 3x 277.128 x 426.74 - 1.5 237.632 kWh mar Pande = 237.633 kwy (En Det serben 428.74 [edy ) = 128.62 V at 300rpm (e) from equation (2), vq = 1x 277.128 = 83.138 V/ Xs - hex 1:520.452/ o'. Equalcón becomes, 3 X 83.138 x 128.62 man 0.45 14 > Prow = 71.288 kW) From bart (c) Ele) Synchronow motor is varying linearly with the speed of the motor