Start-up amps

[rick carpenter]

I'm hoping someone can enlighten me on start-up amps. I have a Husky 12.5A compressor with Chinese motor that a user here said pulls 60A at start. For each question, is there a standard formula or does it depend on the motor?

1. How long is the motor in an elevated amperage start-up phase and how do the wiring/breakers/receptacles/plugs withstand this? My garage service is 20A.
2. Would 60A be correct? I'd never have guessed it was this high.

 

[Evan(CA)]

The start up load is going to vary motor to motor. The amps will spike and immediately start to fall as the motor begins to turn. The circuit breakers are designed to allow for this but there are "soft start" kits available when the start up load is tripping the breaker.

 

[wyliesdiesels]

Its called in-rush current which usually lasts any where from 50ms-2-300ms...

In rush currents can be 4-6x FLA or more...

 

[MTW]

For general use, the code rule for motor circuits, to allow for starting and some overloading (to allow enough time for the thermal overload to act) is to take the code listed FLA value for the motor and add 25% minimum and up to 175% maximum for the circuit overcurrent protection device rating.

For your motor in question which is probablay a special as most small compressors are is an in between size from the code tables about 3.8HP.

12.5A x 230V = 2875W, 2875W / 745W/HP = 3.85 HP

12.5A x 1.25min = 15.6 Min circuit size
So your 20A circuit should be sufficient provided there's not too much voltage drop from a long run to the outlet.

As to the inrush current, rule of thumb is 6x the FLA of the motor rating for less than a second when it is starting properly.

12.5 FLA x 6 = 75A inrush current.

Every motor and connected load is different on the inrush currents and durations for getting up to speed. Therefore if you follow the minimum code value of 25% added to the motor full load current for the overcurrent device that should cover the inrush currents for most situations and loads, there are exceptions, but this will cover most applications.

MTW Ω

 

[Jlarson]

Inrush current is in the neighborhood of 1000-1800% (can be higher in some cases) of the FLA. This happens for about 2 cycles right when the motor is energized, at this time the windings are essentially a short until magnetized. The impedance of the supply limits this current.

Start-up current follows at about 500-700% (high efficiency designs can be much higher as well) of FLA until the motor reaches 90% of full speed. This time is going to be dependent on load characteristics.

 

[rick carpenter]

Thanks all, I'm not an electrician and I didn't know the formulas. The comp's specs are 12.5A, 120V, 1.5hp, with a stock 14/3 power cord & 5-15 plug engineered for a typical homeowner's garage. My garage has 20A and had 5-15 receptacles. (I changed the receptacles to 20A gfci and changed the comp's power cord to 12/3 but retained a 5-15 plug. So no matter the 20A available and the 20A capacity of the 12/3, it's still effectively a rated 15A max cord/plug like when stock.)

The math...
12.5A x 120V = ~2hp & 12.5A x 110V = ~1.8hp (hmmm, remember the 1.5hp rating)
12.5 FLA x 1.25min = 15.6A
12.5 FLA x 1.75max = 21.9A
12.5A x 6 = 75A

So finally, it looks like I can assume that since in-rush currents are 4-6x FLA and this comp was engineered for 120V with a 5-15 plug, that 5-15 plugs & receptacles and 14/3 wire appear to be able to withstand higher-than-rated amps (this example gives up to 15.6A-21.9A overcurrent and up to 75A in-rush for very very short durations).

 

[DenisG]

There was a similar question posed here and discussion about the "fault curves" of breakers.

 

[rick carpenter]

Thanks again to all! I read the thread DenisG referenced which seemed to wrap up what y'all have said. Like I said I'm not an electrician and I don't like to assume much about electricity.