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QA.3 A 380 V, 50 Hz, 6-pole, 960 rpm, star connected induction motor has the following parameters: stator and rotor resistances are 1.25 2 each, and stator and rotor leakage reactance are 1.5 2 each (magnetising reactance can be neglected). The motor is controlled by simultaneous stator voltage and frequency variation. Voltage boost is not used and the motor is controlled using V/f = constant law. The motor is supplied from a three-phase voltage source inverter.

QA.3 A 380 V, 50 Hz, 6-pole, 960 rpm, star connected induction motor has the following parameters: stator and rotor resistances are 1.25 2 each, and stator and rotor leakage reactance are 1.5 2 each (magnetising reactance can be neglected). The motor is controlled by simultaneous stator voltage and frequency variation. Voltage boost is not used and the motor is controlled using V/f = constant law. The motor is supplied from a three-phase voltage source inverter.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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A 380 V, 50 Hz, 6-pole, 960 rpm, star connected induction motor has the following parameters: stator and rotor resistances are 1.25 each, and stator and rotor leakage reactance are 1.5 each (magnetising reactance can be neglected). The motor is controlled by simultaneous stator voltage and frequency variation. Voltage boost is not used and the motor is controlled using V/f constant law. The motor is supplied from a three-phase voltage source inverter. The motor operates at 30 Hz. Calculate the ratio of the maximum torque at this frequency with respect to the one at rated frequency, for both motoring and generation. b) For the same 30 Hz operating frequency, calculate ratio of the starting current at this frequency with respec to the starting current at rated frequency. c) Assume that the voltage boost of 20 V (phase voltage) is used. Calculate new value of the phase voltage that would be applied in this case for the operating frequency of 30 Hz. Repeat part b) with this new voltage applied. d) A variable-speed electric drive can operate in two distinct speed control regions. Name these regions, explain the principal difference between them and specify which quantity is kept constant in each of the two. Induction motor torque dependence on the slip can be given for rated frequency, rated voltage operation as: Rr S Te(s) = 3P V₂² 2π fn R. (R₁ + +(Xls + Xır)² Maximum torque under the same conditions can be expressed as: 1 Tem = Te(s = ±Sm) = 3P 4πfn • V₂2². Rs+√√R² + X² QA.3 2
Transcribed Image Text:A 380 V, 50 Hz, 6-pole, 960 rpm, star connected induction motor has the following parameters: stator and rotor resistances are 1.25 each, and stator and rotor leakage reactance are 1.5 each (magnetising reactance can be neglected). The motor is controlled by simultaneous stator voltage and frequency variation. Voltage boost is not used and the motor is controlled using V/f constant law. The motor is supplied from a three-phase voltage source inverter. The motor operates at 30 Hz. Calculate the ratio of the maximum torque at this frequency with respect to the one at rated frequency, for both motoring and generation. b) For the same 30 Hz operating frequency, calculate ratio of the starting current at this frequency with respec to the starting current at rated frequency. c) Assume that the voltage boost of 20 V (phase voltage) is used. Calculate new value of the phase voltage that would be applied in this case for the operating frequency of 30 Hz. Repeat part b) with this new voltage applied. d) A variable-speed electric drive can operate in two distinct speed control regions. Name these regions, explain the principal difference between them and specify which quantity is kept constant in each of the two. Induction motor torque dependence on the slip can be given for rated frequency, rated voltage operation as: Rr S Te(s) = 3P V₂² 2π fn R. (R₁ + +(Xls + Xır)² Maximum torque under the same conditions can be expressed as: 1 Tem = Te(s = ±Sm) = 3P 4πfn • V₂2². Rs+√√R² + X² QA.3 2
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