Question

A 480 V three-phase synchronized motor (line to line), 60 Hz, 6 poles, connected in Y....

A 480 V three-phase synchronized motor (line to line), 60 Hz, 6 poles, connected in Y. The equivalent circuit parameters are

R1 = 0.15 W R2 = 0.154 W XM = 20 W

X1 = 0.852 W X2 = 1.066 W

PFW = 400 W PMISC = 150 W PCORE = 400 W

For a slip of 0.02, find

(1) Line current. [3 points]

(2) The power factor of the stator. [1 point]

(3) The power factor of the rotor. [1 point]

(4) The frequency of the voltage induced at the rotor (fr). [1 point]

(5) The copper losses from the stator. [1 point]

(6) The power of the air gap. [1 point]

(7) Power converted from electrical form to mechanical form. [1 point]

(8) The induced torque. [2 points]

(9) The load torque. [1 point]

(10) The overall efficiency of the machine h. [1 point]

(11) Engine speed in revolutions per minute and radians per second.
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Answer #1

Answer Criven V = 480v 60Hz, pole = 6 R = 0.15 h - 0.154 xm = con x - 0.852 r, 72 - 1.0664 Slip こ 0.02 Ns 120x60 120f P 1200ZE 6.932 /27.96 ZF 6. 1227 +7 3.25 + F Rs & TE RE = 6.1227n z R1 + 7 x + ZF 21 11 0.15 + 6.1227 + 7 0.852 + 73.25 2 = 6. 27274) frequency of Rotor voltage st fr 0.02 x 60 fr 1.2 H₂ 5 ) Cu losses in stator 3 £,²R, Pculoss 3x (36.95 x 0,15 Palos 614.388) Induced torque Te Pag 25078.02 وليا 2 K & Ns Te 2507802 x 60 2 K x 1200 Te 200 Nom 9) load torque T output power Pgond Pro1) Efficiency pout Pin Xloo 24.026.56 ny 11 x luo 3XVLL, Costs ns. 290 26.56 x 100 3 x 480x36.95 x 0.8369 ny. - 93.45%. 10) N

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