Question

A 3 phase, 440 V, 6 pole, 970 rpm, 50 Hz, Y-connected induction motor has the following parameters referred to the stator Rs = 0.2 Ω, Rr'一0.15 Ω, xs-x-= 0.4 Ω The stator to rotor turns ratio is 3.5 The motor speed is controled by the Static Scherbius Drive. The drive is designed for a speed range of 30% below the synchronous speed. The maximum value of firing angle is 170'. Calculate (i) Turns ratio of the transformer (ii) Torque for a speed of 750 rpm and α = 140 ng angle for half the rated motor torque and a speed of 850 rpm.

Answer 1

"Maximum slip s_m = -a cos d_m for 30% , speed range s_m = 0.3 thus 0.3 = -a cos 170 a = 0.3046 n/m = a 3.5/m = 0.3046 a) for a speed of 750 ref s = 1000 - 750/1000 = 250/1000 = 0.25 v_d1 = 3 root 6/pi sv/n = 3 root 6 /pi times 0.25 times 440root 3/3.5 = 42.4435 v_d2 = 3 root 6/pi v/m cos alpha = 3 root 6/pi times 440/root3/11.4405 times cos 140 = -39.6145 I_d = V_d1 + V_d2/2(sR_3 + R_f) + R_d R^1_s = (0.2) times (0.5)^2 = 0.05 ohm R^1_r = (0.15) times (0.5)^2 = 0.0375 ohm I_d = 42.4435 - 39.6145/2(0.25 times 0.05 + 0.0375) + 0.01 dc link indicator resistance I_d = 25.7182 A T =