Question

2) A 480 V, 60 Hz, 400 hp 0.8 PF-lagging eight-pole A-connected synchronous motor has a synchronous reactance of 0.6 and negligible armature resistance. Ignore the mechanical and core losses. The magnitude of the internal machine voltage is 480 V when the field current is 4 A. Assume magnetic linearity. Initially, the motor operates at the rated conditions. a) Find the speed of this motor. b) Find the armature and field currents. c) What is the developed torque? What is the torque angle? How near is this value to the maximum possible induced torque of the motor for this field current setting? d) If the field current is increased by 30%, what is the new armature current? What is the motor's new power factor? e) With the field current set as in part (d), the load is removed from the motor, find the armature current. 3) Fig. 1(a) shows a small industrial plant supplied by an external 480 V three-phase power supply. The plant includes three main loads, as shown in the figure. The synchronous motor is rated at 100 hp, 460 V, and 0.8 PF-leading. The motor's synchronous reactance is 2.51 12, and the armature resistance is 0.023 2. The OCC for this motor is shown in Fig. 1(b). a) If the switch on the synchronous motor is open, how much real, reactive, and apparent power is being supplied to the plant? What is the current in the transmission line? The switch is now closed, and the synchronous motor is supplying rated power at rated power factor. ECE332 Assignment 7 b) What is the field current in the motor? c) What is the torque angle of the motor? d) What is the power factor of the motor? e) How much real, reactive, and apparent power is being supplied to the plant now? What is the current in the transmission line? Now suppose that the field current is increased to 3.0 A. f) What is the real and reactive power supplied to the motor? g) What is the torque angle of the motor? h) What is the power factor of the motor?

Answer 1

Soln: 15x60 = goo x porn and 120 x 60 N = 1204 -GQHz, → NE Ispeed k V = 4801__60Hz, _pe 400mpos lag BC Pole) = 8 Xo = obeh = 298. 28kW E = 480 @ L=4A 2 po 53 V P = 3 v → Ion For A V in = Uphase IL LOS Q" Iph Lose 298,28 x 103 3x 480 x 0.8 = 258.924. = 258.92 L-vos (0.8) A. bis 258.92 | -36.87° A = 480 7 258.92 (-36.87 x 4; 0.6) 406.2- 406.2 TIREA 480 es Scanned with Camscanner

298.28x103 = 298.2€ x 10° 2 We 2 x 1201 3.164.85N 0800 10.00 Torque angle & = 125 17.810 _Tomax 13 y = 3619.83 Nm! 3 we (d) I' = 41.3 If → Ephe 1.3 Eph __ = 528.02 13.00 → Generator (Overexcited I 14.00 SEP sin 6 = = se sin"| 298.28x 103 1 15.00 3 x 528. O2 X 480 0.6 Xs NOOT 17.00 = 13.613° = Feb= 528./13.6130 Iph - Eph = {214.3 1-14.90 A jXg -pt = cos (14.9) = 10. 966 oad removed so Eph = 528 v In = 528-480 = 5; 80 A * Scanned with CamScanner 18.00 L = 528-48 A jo.6 NOV