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Wound-Rotor Motor. Torque-slip (speed) curves for a wound-rotor motor. Curve 3 – T D,max @ s=1 Curves 4, 5 – T D,max @ s>1 s>1 – “plugging” – reversal of a motor before it comes to rest by interchanging 2 of the 3 line leads going to the stator. n r is negative ! . Example 5.10 page 198.
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Curve 3 – TD,max @ s=1 Curves 4, 5 – TD,max @ s>1 s>1 – “plugging” – reversal of a motor before it comes to rest by interchanging 2 of the 3 line leads going to the stator. nr is negative!
Example 5.10 page 198 • A family of torque-speed curves for a y-connected, 400-hp, 2300-v, 14-pole, 60-Hz, wound-rotor induction motor is shown in Figure 5.10. Curves A and D show the extremes of rheostat adjustment. Determine • a) the range of rotor speeds available by rheostat adjustment, assuming 100% rated torque load on the shaft:
Example 5.10 page 198 (continued) • b) the rheostat resistance required to obtain 260% rated torque when starting. • The ratio of stator turns per phase to rotor turns per phase is 3.8, and the motor parameters, in Ω/phase are R1=0.403, R2=0.317, X1=1.32, X2=1.32, XM=35.46.
Example 5.10 page 198 (continued) • Solution for part (a) • nr = ns(1-s) • Determine ns • ns = 120f/P = (120)(60)/14 = 514.29 r/min • Look at the intersections of the curves with the 100% torque line • for the low-speed curve, D, s = 0.395 • for the high-speed curve, A, s = 0.02
Example 5.10 page 198 (continued) • for the low-speed case, • nr = ns(1-s) = 514.29(1-0.395) = 311 r/min • for the high-speed case, • nr = ns(1-s) = 514.29(1-0.02) = 504 r/min
Example 5.10 page 198 (continued) • for part (b) • Curve B has locked-rotor torque of 260% • TDmax occurs at s = 0.74