Question

Name: Student ID: A 208-V, six-pole, Y-connected, 25-hp design class B induction motor is tested in the laboratory, with the following results: No load: 208 V, 24.0 A, 1400 W, 60 Hz Locked rotor: 24.6 V, 64.5 A, 2200 W, 15 Hz dc test: 13.5 V, 64 A a) Find the parameters of the induction motor equivalent circuit. b) Calculate the stator current if the slip is 0.02.

Answer 1

a reactance =52 Xitxu = Vpn = 120.1 Ine from blocked rotor test we can determine stator and color reactance from blocked rotor ZLR= RLRTS XLR = Vph 24.6 0.222 ILR 2200 PLR SLR cous) cos 105 reactance of locked rotor at 60Hz O phase vallare Vph=k 120.14 V = 208 @ from no load test we can determine combined reactance of stator and magnetizing 24 ouis PLR costl VB VIRILR XLR = XLR = 160 (0:132) = 0.5282 frest rotor is class-B type hence the stator and sofor reactance is as function of ter is X=6-4XLR X2-06 XLR =64) (0-528) = (0-6) (6-528) 6:211212 CS$canned with CamScanner Oir= cost of 13 (24-6)(64:5) OLR= 36.87 at 15H3 frequency 2LR = 0.22 [-36-872 RAR -Rere Ri= stator resistance IZRI Coso rotor resistance LR RE = 0.1322 2,+R, -(0-22) 60S (36-87) = 0.1763 X,+ t = ZLR Sinar. (0-22) sin(36.87) = . = 0.31682 14

Xitxm=52 = xm=5-8, -5-0.2112 4.7892 of In DC test Ri 13.5V DC Ri 13.5 Given Ioc = 64 A. 2 R, - Noc Joc 64 =) R, = 0.1054 R+R2 =0:1762 I R₂ = 0.176-0.1054 = 0.07062 Motor equivalent Cawit RI jх, R = 0.0706 f in jo.3168 54789 jym: Istil i m 0:1054 jo.2112 ŚRO Vph 6 Given S=0:02 To find stator lossent. I 0.0706 j0.3168 M 0.1054 jo 2112 3 54-789 120111 Zeq= free) 3:53+00-3168 -3.54(5!28 Zey = 2.208+j1.9607 I = 30-38-j26974 120-1 2-208+j1.9607 Is = 40-632-41.6 1 Is) = 40-63A CSScanned with CamScanner

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