Question
A 3-phase, 380 V, 50 Hz , 6 pole, Y-connected induction motor has R1 = 0.1...
A 3-phase, 380 V, 50 Hz , 6 pole, Y-connected induction motor has R1 = 0.1 W , X1 = 0.4 W, R2’ = 0.12 W , X2’ = 0.4 W, Xm = 20 W , Pr = 1 kW. The motor is operated at 970 rpm what is the The ratio of the stator starting current to full-load current (Istart / Ifull-load)?
Answers
[page-1. lalui Given, 3-phase, Y-connected induction niotor: 380V, 50 Hz, 6 pole R, = 0.14 x = 0.42 R2:20.12 , x2 = 0.41 Xm=2o4, Pralko Nm=970rpm Istart RI ix, > Act start Vph 1=1 R² fx2 + you (slip) fig: equivalent circuit for Y connection: Vph =ų _ 380 B 3 - Vph=219.3930 Ist 10.4 0.12 jo.lt I = Iph t 219.3930 3j20 P. Zeq = (j 20) (0.12 +jo.4) zoq = (0.12 tjo.4+j20) Zeq jzox x 0.417(733 20.4/89.66 Zeq = 0.40P [73.64 2 Zeq (0.114 70.391)Ist oil 30-4 > (page 2 1 2q = (0.114 tjo.391) + 219.39 IV KUL) By KULI Ist = 219.393 (0.1tjo.4+0.114 14 tj0:39 Ist = 219.393 219.393 (0.214 tj0.791) 0.819/74-86 Ist = 267.8792-74.86 A Ist 1 = 267.879 A Now, synchronous need (Ms) = 12 of P 120x50 6 Ms=1000rpm Hm=motor speed 9 Torpm slip at full load=&= ls -len 7 st 1000 - 970 f = 0.03 1000 R Ifl R, 20.1 jxFjou - 4 jo.4 219.3934 jxzj 20 ZealZeqi 그 j 20+4+jot > Zeqi [page-37 (jro) (4+jo.4) (20290) (4.019/5.5 (4+j20-4) (20290*) (4.01975-11) (20.788 178.9) = 3.8666 [16.8 is > 24=13.701+j1.117) 2 If 0.1 jo.4 219.393V 16.701+j117) + KUL By KUL: 219.393 Ill 219.393 2 I fl- (0.1 +jo.4 +3.701 4.1.17) 219.393 A 4.09 (21-25 (3.801 +j1517) Ifl = 53.64(-21.75 A 1ff1= 53.644 || F 1 = 53.64A 267.879 53.64 4.994 y 5 - (I start (If I (I start 1 Il -5
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