zxing barcode reader java download Therefore, the starting current referred to the primary side of the transformer will be in Software

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Therefore, the starting current referred to the primary side of the transformer will be
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V ,bus Rline + jX line + R1 + jX 1 + RF + jX F 120 0 V I = I A = L 050 + j 035 + 0475 + j 0605 + 0232 + j 0592 I = I A = 612 52 A L I L = I A =
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The voltage at the motor end of the transmission line would be the same as the referred voltage at the terminals of the motor
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V = I A (R1 + jX 1 + RF + jX F )
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V = ( 612 52 A )( 0475 + j 0605 + 0232 + j 0592 )
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V = 850 74 V
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Therefore, the line voltage at the motor end of the transmission line will be 3 ( 85 V ) = 1473 V Note that this voltage sagged by 292% during motor starting, which is less than the 375% sag with case of across-the-line starting Since the sag is still large, it might be possible to use a bigger autotransformer turns ratio on the starter 7-21 In this chapter, we learned that a step-down autotransformer could be used to reduce the starting current drawn by an induction motor While this technique works, an autotransformer is relatively expensive A much less expensive way to reduce the starting current is to use a device called Y- starter If an induction motor is normally -connected, it is possible to reduce its phase voltage V (and hence its starting current) by simply re-connecting the stator windings in Y during starting, and then restoring the connections to when the motor comes up to speed Answer the following questions about this type of starter 142
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(a) How would the phase voltage at starting compare with the phase voltage under normal running conditions (b) How would the starting current of the Y-connected motor compare to the starting current if the motor remained in a -connection during starting SOLUTION (a) The phase voltage at starting would be 1 / conditions
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3 = 577% of the phase voltage under normal running
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(b) Since the phase voltage decreases to 1 / 3 = 577% of the normal voltage, the starting phase current will also decrease to 577% of the normal starting current However, since the line current for the original delta connection was 3 times the phase current, while the line current for the Y starter connection is equal to its phase current, the line current is reduced by a factor of 3 in a Y- starter For the -connection: For the Y-connection:
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I L , = 3 I , I L ,Y = I ,Y
But I , = 3I ,Y , so I L , = 3I L ,Y 7-22 A 460-V 50-hp six-pole -connected 60-Hz three-phase induction motor has a full-load slip of 4 percent, an efficiency of 91 percent, and a power factor of 087 lagging At start-up, the motor develops 175 times the full-load torque but draws 7 times the rated current at the rated voltage This motor is to be started with an autotransformer reduced voltage starter (a) What should the output voltage of the starter circuit be to reduce the starting torque until it equals the rated torque of the motor (b) What will the motor starting current and the current drawn from the supply be at this voltage SOLUTION (a) The starting torque of an induction motor is proportional to the square of VTH ,
start2 VTH2 VT 2 = = start1 VTH1 VT 1 start2 VTH2 VT 2 = = start1 VTH1 VT 1
If a torque of 175 rated is produced by a voltage of 460 V, then a torque of 100 rated would be produced by a voltage of
100 rated VT 2 = 175 rated 460 V VT 2 =
(460 V )2
= 348 V
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