Fire Pump Locked Rotor Protection

[Schwatzernov]

Hi,

I have read NFPA20 which requires that the thermal overload is not required to protect the motor but the locked rotor protection is required.

I have question about the electric fire pump locked rotor protection. Is it necessary that the protective relay curve needs to lie below motor safe stall time(hot and cold)in order to protect the motor from locked rotor protection ?

Thank you,

 

[waross]

Consider adding a zero speed switch to prove rotation in addition to the protective relay. This will give you added protection at high slip and locked rotor conditions by limiting the maximum time that the rotor may be energized without spinning up.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[Schwatzernov]

Hi Waross,

I commented to fire pump vendor that their setting protection curve is above hot stall time and as such the motor is not really protected from locked rotor condition which will cause damage to motor.

However they said that the fire pump set is approved as UL listed and as such they cannot change the setting of curve even though the protection curve is above locked rotor safe stall time.

I wonder that whether this setting violate NFPA20 or not ?

If this setting condition complies with NFPA20,this means that there is no meaning to protect the motor against locked rotor condition but why the NFPA20 still require and has clause regarding to locked rotor protection ?

 

[jraef]

The only purpose I know of for providing locked rotor protection for the motor on a fire pump is to prevent another fire starting in the pump room. Pump protection is anethema, a fire pump is EXPECTED to run to destruction and anything that might interfere with it running is defeating the stated purpose. Generally the locked rotor protection is done with the magnetic trips of a circuit breaker or seriously over sized fuses.

But to be blunt, if you have to ask this question and you don't know where to get the answer or how to interpret what is being said, you probably have no business taking on this task. Please consider that this is not about general electrical engineering, this is about life and safety. If you were in a fire and the sprinklers failed to come on because the fire pump motor tripped off line because of a protective relay setting being too sensitive, you or your survivors are not going to be happy.

But their lawyers will be.

"Dear future generations: Please accept our apologies. We were rolling drunk on petroleum."
— Kilgore Trout (via Kurt Vonnegut)
For the best use of Eng-Tips, please click here -> faq731-376

 

[davidbeach]

What jraef said.

 

[electricpete]

Generally the locked rotor protection is done with the magnetic trips of a circuit breaker or seriously over sized fuses.

I didn’t quite understand this part. I interpret magnetic to mean instantaneous… which is set above locked rotor current and therefore cannot provide locked rotor protection (in the general case where we want to protect for locked rotor).

I agree it doesn’t sound like standard protection… sounds like a code question. Maybe power engineering or code forum have insights.


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(2B)+(2B)' ?

 

[davidbeach]

Locked rotor shouldn't trip anything. Fault currents above locked rotor are allowed to trip the motor.

 

[dpc]

I'd refer to Article 695 in the NEC for specific fire pump motor requirements.

The overcurrent protection on the fire pump feeder must sized to carry locked rotor current indefinitely (to motor destruction) without tripping. There is only short circuit protection.

And not only is thermal overload protection not required, it is not ALLOWED.

Fire pump requirements trump motor protection and pretty much everything else.

 

[LionelHutz]

I agree with the others who have already made some very good points. A fire pump motor is expected to operate to destruction and only trip the breaker when the windings have become a phase to phase or phase to ground short.

Besides, you don't set the instantaneous trip of the breaker below the locked-rotor current or expected inrush current for any motor. Doing so would cause the breaker to trip when you started the motor.