What size generator do I need for a fire pump?

[ohmhead]

Well i agree with Bob the pump is going to burn on the pad before it stops !

Heres a fire pump controller we had on one job http://i611.photobucket.com/albums/tt195/stringking/428.jpg

The feed from the generator was 500 mcm one run.

The feed from the poco transformer was 500 mcm .

The fire pump was a 200 hp 3 phase 480 v .

The 6 wire to the motor was 3/0 .

The brk in the genset was a 400 amp .

And no over loads used .

This was a ul listed fire pump controller and transfer switch in the fire pump room most power companys now ask to install a disconnect at transformer but we never do it its just a direct connection to the transfer switch .

But keep this in mind if you running this in conduit its underground or outside of the building in rigid pipe its a un protected feeder from that transformer .

 

[erickench]

The article that you gentlemen referred me to says that the conductors are sized 125% of full load amps and not the locked rotor current. How could an overload device of six times the FLA be used to protect conductors of this smaller ampacity? The locked rotor current has values similar to what you would calculate the short-circuit protection using NEC Table 430.52. These values range anywhere from 150% to 1100%. In fact the mean value would be 625%. That's pretty close to 600%. Besides the definition for overcurrent protection includes both short circuit and over load. This article of the NEC state's that power conductors shall not have overload protection but only short circuit protection.

 

[sceepe]

You need to use manufacturer's sizing software. It will anaylze load and determine the % voltage drop. If you have loads other than Fire pump (maybe emerg lights) this becomes even more important. Also, sizing software will select generator componnents such as PMG that could be critical to successful starting of generator. Be prepared %voltage drop limits result in oversized alternators. Also, if the CAT software says you need a 75kw, that does not garantee a 75Kw Onan will work.

 

[iwire]

The article that you gentlemen referred me to says that the conductors are sized 125% of full load amps and not the locked rotor current.

Apparently we are talking about different code sections.

How could an overload device of six times the FLA be used to protect conductors of this smaller ampacity?

They can't protect the conductors, they do not have to, it's a fire pump.

We seem to be having a communication problem.:smile:

The rules for fire pumps are unusual.

 

[blingbling4r]

You need to use manufacturer's sizing software. It will anaylze load and determine the % voltage drop. If you have loads other than Fire pump (maybe emerg lights) this becomes even more important. Also, sizing software will select generator componnents such as PMG that could be critical to successful starting of generator. Be prepared %voltage drop limits result in oversized alternators. Also, if the CAT software says you need a 75kw, that does not garantee a 75Kw Onan will work.

Forgive my newb-ness what is a PMG? Also what other components are you aware of that would affect the voltage drop? Is my assumption correct that every genset will have different factors (slip inertia, torque, etc) that will affect every situation differently?

 

[weressl]

I was wondering how to size a generator for a fire pump. The code on the required power supply for a fire pump is a little vague as far as the KW rating of the generator except for that it must be "adequate" for supplying the load (fire pump and additional loads). The generator is for backup power as the main source of power for the fire pump will be tapped off the new service that will be feeding the building.

Here are the specs on the 208VAC 40 HP fire pump

FLC is rated at 109A
LRC is 587A
Standby current is 40A

I figured a 600 A for the OCPD is adequate based off of Art 695. I figured the minimum capacity for the generator would be calculated as follows:

[(Fire pump FLC) x 125%] + [(Additional loads) x 100%] = Generator Current

[(208 VAC) x (Generator Current) x (1.73)]/100 = Minimum KVA required to supply the fire pump system

I used the tables from Art 430 to give me a higher FLC based off the HP to give me a little more fudge on the size.

FLC for 40 HP 208V Fire Pump Motor = 114
FLC for 1 HP 208 V Jockey Pump Motor = 4.6


The jockey pump may not even be on the generator system but I added it again for fudge factor.

So to use the formula above I got

114A x 1.25 = 142.5A
4.6A x 1.25 = 5.75A (I factored at 1.25 for safety)

Total Amperage = 148.25 (1/0 THHN wire would be used)

(208 VAC x 148.25 Total Amps x 1.73)/1000 = 53.34 KVA

So I contacted a 3rd party Cummins dealer and got a quote for a 56 KVA generator.

I also contacted the Cummins Direct dealer here in Seattle and when they put the fire pump specs (no jockey pump) through their system they recommended a 112 KVA generator to supply the fire pump. This is a significant difference.

So basically I calculated what it would be for a transformer from Art 695 to figure the minimum supply to the fire pump load. Is this the correct way to do it or is the Cummins rep correct when he "plugged in the numbers" into his application. If so what am I missing here and any code references would be greatly appreciated.

It may be more cost effective to procure a diesel PUMP than a generator.