ohm, Rr'=0.258

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Reduced parameters of 480V, 60 Hz, 6 pole, 3 phase, delta connected induction motor to stator are Rs= 0.461 ohm, Rr'=0.258 ohm, Xs=0.507 ohm, Xr'= 0.309 ohm, Rc=100 ohm and Xm=30.74 ohm. given. The motor rotates at 1170 rpm. According to the given values; a) Find the input power (Pin). b) Find the output power (Pout).
21:53
ull ?
S=0.015
S=2.5%
Step2
b)
© Find the Statos phase Cureut
So.
Is R= 0 4bln Xe=jo-5072
R= 0-L58 - 10-322
eelle
Vphase 4P0v
Re= 10on
.742)
Pazanel lomrehon
z!_ loox (j30-74)
= (8. 634+j28.1)n
100+j30.74
Is (o-461+jo-soa)s
ナ1
480(v
pazallel connet'02
z"- (8-61u4]28:1) ( 10 32+jo:30)
8.634tjeg17 10-32fjo-s09
C30.42tj292.66)
18-954+jes.vo0)
Step3
c)
Transcribed Image Text:21:53 ull ? S=0.015 S=2.5% Step2 b) © Find the Statos phase Cureut So. Is R= 0 4bln Xe=jo-5072 R= 0-L58 - 10-322 eelle Vphase 4P0v Re= 10on .742) Pazanel lomrehon z!_ loox (j30-74) = (8. 634+j28.1)n 100+j30.74 Is (o-461+jo-soa)s ナ1 480(v pazallel connet'02 z"- (8-61u4]28:1) ( 10 32+jo:30) 8.634tjeg17 10-32fjo-s09 C30.42tj292.66) 18-954+jes.vo0) Step3 c)
21:53
ull ?
(80.42+j292.66)
tjeavor)
/ 8-954+)
Step3
c)
-"- &.4354fj2:793)
2"- 8.887L18-345
480V
480 Lo°
Is =
coiyertjosos) + (8.vs5utj2294)
480/0°
Is=
3.49 (20.370
^y80
o°-20-37
9.49
Is= 50.58/-20.37°
statos phate Cursent
Transcribed Image Text:21:53 ull ? (80.42+j292.66) tjeavor) / 8-954+) Step3 c) -"- &.4354fj2:793) 2"- 8.887L18-345 480V 480 Lo° Is = coiyertjosos) + (8.vs5utj2294) 480/0° Is= 3.49 (20.370 ^y80 o°-20-37 9.49 Is= 50.58/-20.37° statos phate Cursent
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