Question

6. The speed of a 150 HP, 650 V, 1750 rpm, separately excited DC motor is controlled by a 3- phase full converter. The converter is operating from 3-phase, 460 V, 50 Hz supply. The rated armature current of the motor is 170 A. The motor parameters are Ra = 0.099 2, La = 0.73 mH, and K.D=0.33 V/rpm. Neglect the losses in converter system. Determine i. No-load speed at firing angles a = 0º and a = 30°. Assume that at no-load, the armature current is 10% of the rated current and is continuous. ii. the firing angle to obtain rated speed of 1750 rpm at rated motor current. Also, compute the supply power factor. iii. The speed regulation for the firing angle obtained in (ii).

Answer 1

Geven, Ratings of motor (separately P = 150 Hp = 111.855 kW exčited) V = 650V N=1750 rpm 1 used: 3-0 full Converter → AC Supply VS - 4 G0V, f=50 Hz → Rated armature Current, Ia=170 A Ra=0.0992, La=0.73mHd Kam • 0-33v/RPM - (A) N No-load at d=0, and d = 30 Ta' = 10 Ia 100 EIX170 =17A 10 For Full Converter, vo 33 Um Cost 0 ㅠ ove uph 12 VL V3 7 v 460 13 = 375.59 v ata=0 from @ Vo 33 X375.59x1 IT = GRI22V we know that Vo = Eb + Tára Kaon't Idra

621.22 - 033 X N' + 17% 0-099 N'E = 1877. 38 rpm For x = 30 Vo kaon't Tara Vo = 0.33 XN'+ 1780-099 Vo 33 X 375.59 X 605 36 TT 537.97 V Vo = 0.33 N' +17 %0.099 53797 = 0.33N' + 17X0.099 - 1625.13 rpm ® A at N = 1750 rpm, eo Ia = 170 A Vo KaØN+ IaRa 333X 375.59X005 = 0.33x1750 +170 X0-099 IT → 621-220SQ = 594.33 ; Cose = 0.956 7 ..d = 16.92 N no load ©) speed regulation NO full load full load N'N No-load Bat & = 16.92 fromparte (6) VO kaøn't falra at No load Ta'= 10 Ia =17A → 313,375.598 C0916.92 = 0.33xN'+ 17X0.099 N = 1795.87 YPM TT

Speed regulation - N- Nrated Nrated 1795.87-1750 こ 1750 - 0.0262 Speed regulation (lo) - 2.62 0/0

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