Motor Inrush Current
Motor Inrush Current
(OP)
In order to minimize voltage drop, I want to start various motors in multiple steps.
Step 1: the largest motor will be started under no load (mechanical load will be zero).
Step 2: the second step will start the smaller motors, under full mechanical load.
Step 3: The last step will connect the motor in step one to its mechanical load.
My questions are:
for step 1) What will be the inrush current as a percentage of full load?
for step 3) What will be the inrush current, if any, as a percent of full load? I presume it will be reduced, given the motor is already at rated speed?
Step 1: the largest motor will be started under no load (mechanical load will be zero).
Step 2: the second step will start the smaller motors, under full mechanical load.
Step 3: The last step will connect the motor in step one to its mechanical load.
My questions are:
for step 1) What will be the inrush current as a percentage of full load?
for step 3) What will be the inrush current, if any, as a percent of full load? I presume it will be reduced, given the motor is already at rated speed?





RE: Motor Inrush Current
RE: Motor Inrush Current
RE: Motor Inrush Current
Step 3 is a little different. Motor is at no-load (~sync) speed and you add load, possibly by clutch? In most cases there would be no elevated starting current as long as motor speed remains above full load speed. But then again if load includes a huge inertia, it may result in motor drawing current above FLA, possibly up to starting current for some period of time.
Starting current can be determined from nameplate "starting KVA code class" (NEMA motors), corrected propoportional to voltage. Come to think of it, it's odd that they use the phrase starting KVA since KVA drawn during start is not constant with voltage. Look up starting KVA in the tables, convert it to currrent using motor nameplate voltage, then multiply by ratio actual terminal voltage / nameplate voltage to estimate starting current.
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(2B)+(2B)' ?
RE: Motor Inrush Current
600% to 800% depending on the motor. Some special motors are designed for less. Some draw more, but with no other data on the motor, and on a free internet site, 600% to 800%.
The load has more effect on the duration of the starting current than on the magnitude of the inrush.
Haven't looked at those tables for a few years Pete. Did need them at one time.
Aren't they titled starting KVA per Horse Power?
Bill
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"Why not the best?"
Jimmy Carter
RE: Motor Inrush Current
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(2B)+(2B)' ?