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primary resistance reduced voltage motor starting question 1

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apprentice345

Electrical
Feb 24, 2014
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Having a little trouble finding a solution to this question.

The DOL starting current of a motor is 90A at 400V. If a primary resistance starter is used, what must the voltage across the motor fall to if the starting current is to be reduced to 30A?

Any help would be much appreciated.
 
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Let's say Ist=VL_L/sqrt(3)/sqrt[(Rst+R'rot/slip)^2+(Xsst+X'srot)^2] .At start slip=1 .
Where VL-L it is the line voltage Rst is the stator resistance R’rot it is the rotor resistance referred to stator, Xsst is the stator leakage magnetic flux reactance and X’srot it is the rotor leakage reactance referred to stator.
If the frequency is constant then the denominator is constant and the current is proportional with the supply voltage.
In order to produce 30 A then the supply voltage has to be 400/3=167 V.
But the motor produced torque will drop with the square of voltage. If at rated voltage the start toque is 1.5 times the rated torque now will be 1.5/9=0.167 times the rated. I wonder if the motor could start now.
Only if you would reduce the frequency in the same time you could maintain the torque as was before.
 
We'll, any time you have Reduced Voltage Starting, you have reduced torque. That's a given. Considering RV starting when you need full torque is a non-starter in all cases. The concept of using RV starting is for when you do NOT need Locked Rotor Torque to accelerate the load.

"Will work for (the memory of) salami"
 
At locked rotor, the motor looks pretty much like an inductance. To go from 90 A to 30 A, I think you'd need to have about 33% of the nameplate voltage. That would give you about 11% of normal starting torque.
 
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