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Tool Talk Discussion Forum

Current to start motor

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DaveInMI

10-09-2003 04:37:07




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What percentage of ID plate amperage does it take to start an electric motor under no load. I just want to know if the motor will run. It is 3-phase and my rotary converter is only about half the HP rating as the motor I want to start.




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Kevin (OH)

10-10-2003 14:57:08




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 Re: Current to start motor in reply to DaveInMI, 10-09-2003 04:37:07  
Here is an accurate and technical response to your psot.

If your 3-phase motor is of recent vintage and is of a NEMA design, the nameplate will give you this information..... .indirectly.

There should be a block that lists the motor "CODE" on the nameplate. This will be a letter, something from A to V.

This letter corresponds to a ratio. This ratio is the Locked Rotor Code, per KVA horsepower. LR/KVAHP is figured as follows for 3 phase equipment:
VOLTS*LOCKED ROTOR AMPS*1.732/(1000*HP)

If you know the Letter rating, the operating volts and the rated hosepower, and have a copy of NEMA MG1-10.36 which is the table that lists the ratios for the code letters, you can use some simple algebra to calculate the starting amps.

Give us the name plate data for the operating volts (230, 460, 575, etc, the rated horsepower and the Motor code, and I will help you calculate the starting currect for your particular motor.

Rarely is 150% of nameplate (or Service Factor) Amps enough to properly start a 3 phase motor. My experience is that MOST NEMA frame size 56 to 184 motors require from 5-7 times the rated running amps to start. Although The KVA/LRHP codes do range from 3.15 to more than 22 times.

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DaveInMI

10-10-2003 02:46:22




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 Re: Current to start motor in reply to DaveInMI, 10-09-2003 04:37:07  
Thanks. I'll experiment a little.



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Vern-MI

10-09-2003 10:04:39




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 Re: Current to start motor in reply to DaveInMI, 10-09-2003 04:37:07  
Take a look at the additional wattage required to start motor driven devices on the Honda generator wattage calculator. Wattage is calculated by multiplying motor voltage times the current. i.e, 115 volts times 6 amps = 690 watts.



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F/F

10-09-2003 15:40:09




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 Re: Re: Current to start motor in reply to Vern-MI, 10-09-2003 10:04:39  
Vern, you are on the right track but the question involved a three phase motor. So the formula would be voltage times current times 1.73. Or as in your example 115 x 6 x 1.73 = 1193.7 watts. Correct me if I am wrong.



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Vern-MI

10-10-2003 04:25:20




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 Re: Re: Re: Current to start motor in reply to F/F, 10-09-2003 15:40:09  
You are pretty close F/F. Just multiply that times the power factor of the motor (about .8 to .9) and it will be right on. I.E., 115 x 6 x 1.73 x .8 = 954.96 watts.

I don't know where to find the starting or inrush current for a motor without a load however.



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buickanddeere

10-10-2003 07:08:50




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 Re: Re: Re: Re: Current to start motor in reply to Vern-MI, 10-10-2003 04:25:20  
Inrush current is the same loaded or unloaded. Just the duration of the inrush varies.



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F/F

10-10-2003 05:33:24




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 Re: Re: Re: Re: Current to start motor in reply to Vern-MI, 10-10-2003 04:25:20  
Vern, it has been a while since I had this theory stuff but if I remember right the power factor would have no bearing on the current draw of the motor only on the work performed by the motor. The power co. would charge you for the full amount (E x I x 1.73) The inefficacy of the motor is due to heat loss, friction, etc. And all this happens at the motor after all the current has gone thru the starter and the overloads so the full draw would be E x I x 1.73. Like I said it has been a while and that is what my memory tells me.

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buickanddeere

10-10-2003 12:37:06




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 Re: Re: Re: Re: Re: Current to start motor in reply to F/F, 10-10-2003 05:33:24  
On a running motor, the closer the PF is to unity/1.0 the lower the line amperage will be. Reactive current is just current that rushes into the motor to build a magnetic field and is pushed out again when the magnetic field collaspes while cutting through the windings to induce voltage. It doesn't do work, it just circlates requiring larger wire,swtich,fuse/breaker,transformer and power generator. Also wastes power as heat with I2R losses. Hence the desire by utilites for equipment with a high power factor. Put a 3 phase motor on the bench without a load and measure running current. It will draw about 1/4 nameplace current without doing any work. A single phase without capacitors is even worse with nearly 1/2. That current is circulating but doing no work. A reason why installing a motor larger than required is a waste. A motor during start-up will have more starting torque with a high resistance rotor as the rotor's power factor will also be nearer 1.0, keeping voltage and current closer in phase. When running there will be more rotor heating and I2R losses. Initial inrush current is pretty much voltage divided by resistance for the 1st cycle. There is very little inductive resistance to limit current the 1st cycle. More inductive resistance once the stator has a magnetic field but until the rotor achieves near syncronous rpm , back emf to limit stator current will be reduced.

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VaTom

10-09-2003 09:15:51




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 Re: Current to start motor in reply to DaveInMI, 10-09-2003 04:37:07  
Dave, my experience leads me to believe you can probably start your motor with your phase converter. But if you only want to test the motor, all you need to do is spin it and then switch on your 240v to 2 of the legs, without dealing with your phase converter. If the motor is good, it'll sit there and spin, just like my "phase converter" does (I use a chain drive to kick-start mine). Then if you wire another 3 phase motor in parallel, you'll have a larger phase converter if that's what you wanted.

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Robert

10-09-2003 06:12:37




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 Re: Current to start motor in reply to DaveInMI, 10-09-2003 04:37:07  
"Inrush" or starting current for 3 phase motors can be as high as 150% of data plate "running" amperage, depending on motor efficiency, condition, construction, etc...
I don't believe your converter will be up to the task.....



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buickanddeere

10-09-2003 07:49:29




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 Re: Re: Current to start motor in reply to Robert, 10-09-2003 06:12:37  
Starting inrush varies from 4 to 15 times nameplate amperage. Varies with stability of the volatge supply/, rotor design/resistance, stator construction/winding and the amount/quality of the magnetic flux conducting iron in the stator. An unloaded motor should start in your application but the rotary covertor will slow and the pahse angle of the applie dvolatge/current will be distorted even worse than usual. A three phase motor on a rotary covertor can be made to run but never up to full rated power or efficiency.

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Robert

10-09-2003 11:01:14




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 Re: Re: Re: Current to start motor in reply to buickanddeere , 10-09-2003 07:49:29  
Very Interesting----
I've been using 150% for calculations for the last 30 years....I've been wrong for a long time it appears.....



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Jon H

10-09-2003 17:12:40




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 Re: Re: Re: Re: Current to start motor in reply to Robert, 10-09-2003 11:01:14  
A few years ago I found a big (75 HP) rotaty phase converter in a scrap yard. There was no control box with it,but it was very cheap. It looked and smelled good,but I wanted to run it to check it out. I have only single ph power on the farm so I cobbled up a 3 hp 3ph motor for a phase converter to power the big units spool up motor.
I used a starting capacitor to start the little 3 hp and connected it to the spool up motor which after an min or so got the big converter up to speed where its cooling fan screamed like an idling jet engine. I connected 2 legs of the converter windings to 240v sp where it ran nicley drawing about 40 amps no load. I set it away in the storage shed where it has been for 20+ years. Sure wish I had a use for the thing.:)

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