ISOLATOR RATING (according to NFPA 79)

Hi, I know this might be not quite the place to ask a question like this but I need to confirm how I should rate an isolator for machinery being shipped to the U.S. You guys in the U.S may be able to shed some light on this.



According to NFPA 79 5.3.3.1(6)(b) "where rated in horsepower the horsepower rating shall be at least equal to that which is defined by table 430.251(b) of NFPA 70 NEC for a locked rotor current equal to the largest sum resulting from the locked rotor currents of any combination of motors that can be started simultaneously and the full load current of the remaining motor and non motor loads that can be operated at that time."



So does that mean that my 40HP (45.5A @ 460V) motor (the only motor/load) according to table 430.251(B) having a locked rotor current of 290A @ 460V will require the main isolator to be rated at 290A @ 460V?

As long as it runs when it get here that will be fine ..... just kidding

I have no idea, but you may want to ask here http://forums.mikeholt.com/ if you don't get an answer

My first thought is what are you terming as an isolater?

Do you mean a disconnect switch? Is it located at the machine and is it unfused ie just a local disconnect?
Is it between teh controller and the motor? If this is the case I believe it has to only be able to interrupt running current

Will there be another disconnecting means at the switchboard with overcurrent protection? This one would have to be sized for full current ie starting locked rotor and all that.

I THINK - would have to look it up.

Dan Bentler

Dan Bentler

I think you're on the right track, Dan. I'm not familiar (yet) with the NFPA 79 document. But I also think 69Firebird's isolator is what we call in the US the main disconnect. I also think what the NFPA79 article is looking for us an 'Amps Interrupting' rating. With the numbers quoted, the isolator needs to be able to interrupt (stop) the possible 290 amps of current, as determined by an accepted testing laboratory, i.e. UL - again here in the US.

Does this sound right 69 Firebird?

Hi Guys, I have had a look and posted article on Mike Halts forum as well. First response is that the isolator ("local disconnect" mounted in the control panel for the entire machine) needs to have a "breaking capacity" / "amps interrupting" rating to cover isolation during a locked rotor period and that by design the HP rating will take this into account as it will be designed to cover a disconnect at locked rotor currents. so i think I'm onto it. If I rate the isolator as per the regs with respect to HP's of the motors then the "breaking capacity" / "amps interrupting" should be covered for their locked rotor currents by design.

Sound right?

Yes, sounds right to me. The amps interrupting capacity should be printed on the disconnect/isolator body somewhere, or in the papers that come with the device, to verify.

So does that mean that my 40HP (45.5A @ 460V) motor (the only motor/load) according to table 430.251(B) having a locked rotor current of 290A @ 460V will require the main isolator to be rated at 290A @ 460V?

Click to expand...

No, you may be confusing the ampere interrupting capacity (AIC) with the isolator ampere current-carrying rating.

The ampere rating and interrupting capacity of the main isolator (we call it the motor main disconnect switch) is covered by the the US National Electrical Code Article 430.110.

That article says that "the disconnecting means for motor circuits shall have a ampere rating [current-carrying rating] of at least 115 percent of the full-load current rating for the motor."

Your 40 horsepower motor will have a Full-Load Amp (FLA) of about 52 amps. 52 X 1.15 = 59.8 Amperes The next higher standard size switch or breaker here would be a 60-Amp switch.

However if your isolator switch is a circuit breaker or has fuses, then it will need to be larger to handle the motor starting current. NEC Article 430.52 allows the motor branch circuit and short-circuit protection device to be rated 250 percent of motor FLA for a circuit breaker, or 175 percent for a dual-element fuse. You might want to use an 80 Amp (or larger) circuit breaker for the 40 HP motor, depending on the type of load.

Now, that same isolator switch must have a current interrupting rating that is higher than the highest motor locked-rotor current. If the "isolator" is rated for use with a 40 horsepower or larger motor, then you are good to go. Otherwise, you must check the Ampere Interrupting Capacity (AIC) rating of the switch or breaker to see that it is greater than 290 Amperes. In other words, the switch must be able to break the circuit under locked-rotor load conditions without failing or burning up. If the switch has a rating of "40 Horsepower" then it already has been tested to be able to interrupt the 290 amps. It will still trip at about 80 amps of current, but will not fail if the current is as high as 290 amperes.

Thanks Lancie1. The isolator is purely an isolator no fuses etc. OK so I size for the motor full load current x 115% and pick the isolator with that current rating (or next up size) now if the isolator has a HP rating that covers the HP of the motor then inherently by design it will have the ability to interrupt locked rotor currents.

One last question what about 2 motors on a control panel as opposed to 1? I assume as according to NFPA 79 5.3.3.1(6)(b) that it will be the sum of FLI x 115% and then sized accordingly.

Thanks

Firebird,

Yes, the isolator must handle 1.15 X Sum of Motor FLA.

Also, for 2 or more motors on same isolator, the isolator switch must have enough interrupting capacity to handle the "largest sum of the Locked-Rotor Currents of all motors that can be started simultaneously AND the Full Load Currents of any other concurrent loads."