Starters - Various starting Method of an Induction Motor and the details

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Effect of operating an Induction Motor designed at 415V, 50 Hz in 380V, 60Hz

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Let us consider a situation where a customer is asking your support to deliver a 45KW/4P, 380V, 60Hz motor on immediate basis. The load data is not available. You have one motor readily available which is also 45KW/4P but designed in 415V, 50Hz. Now if we are going to use this motor for 380V, 60Hz operation, let us see what will happen... 1. We know in an Induction motor, the starting torque and the pull out torque is related with voltage as            below                        Tst ( starting torque ) ∝ V^2                             Tpot ( Pull out torque )  ∝ V^2 So if we run the existing 415V motor in 380 V, then the Tst and Tpot at 380V will be calculated as per below Tst ( starting torque ) ∝ (380/415)^2 Tpot ( Pull out torque ) ∝ (380/415)^2 It is very much clear that both the values at 380V wi...
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The concept of Minimum - Base - Maximum speed of a motor in VFD operation

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In general, in an enquiry sheet for motors which will be running with VVVFD, the speed range may read as below 0-500-1200 0-1000-1500 100-750-1000 The meaning of this speed range is that the within the first two speed the motor should provide constant torque, and at the last or the highest speed the motor is expected to provide constant power. For a 50Hz motor, Constant Torque Zone will be limited up to 50Hz. For a self cooled motor it may be from 30Hz to 50Hz and for a forced cooled motor it may be from 10Hz to 50Hz.  Constant power zone is that zone which is beyond motor rated frequency. in this zone Voltage becomes constant and the frequency is increasing. hence V/f ratio will not be maintained and thus the torque will fall. However, upto certain speed the power will be constant. In constant power zone the rated torque of the motor will be  Tnom ∝ 1/ Nop Tnom = nominal or rated torque Nop = operating speed. Hope this is okay to understand. 
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Overload class of Induction Motors and RMS overload calculation

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  This overload cycle is common for mill duty motors. The motor manufactures usually get enquiry from the users with corresponding duty cycle. it is advisable to send such enquiries to the design team for proper selection not only in terms of frame size and KW, but also he temp rise of the motor. for such cases RMS overload needs to be calculate. I am providing example for better understanding of the calculation Condition 1: 160% overload for 60 sec in every 600 sec so in a 300 sec cycle , the motor will run with 100% load for (600-60) = 540 sec So the RMS overload is = √((160^2) x 60 + (100^2) x 540) / 600) ( √ for total calculation to be considered ) = 107.5% Condition 2: 150% overload for 60 sec 175% for 20 sec so 100% loading for a 600 sec cycle will be = (600-60-20)=520 sec So the RMS overload is = √((150^2) x 60 +(175^2) x 20 + (100^2) x 520) / 600) ( √ for total calculation to be considered ) = 109% For any clarification please ask question.
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