What size generator do I need for a fire pump?

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blingbling4r

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Location
Seattle, WA
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.
 
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tkb

tkb

Senior Member
Location
MA
You can't just use the FLC x 125% for the motor.
There is a large inrush that you need to account for, up to the LRC

Generators are rated in kW.
KW = E x I x PF x 1.73 \ 1000
Most generators are 0.8pf
 
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H

hillbilly1

Senior Member
Location
North Georgia mountains
Occupation
Owner/electrical contractor
Most genset manufactures will tell you what the maximum HP motor their unit will start, unlike utility power though, you don't have to size it for locked rotor. A way to reduce the size of generator needed is to have a shunt trip breaker drop the non-safety related load when the pump is called for.
 
B

blingbling4r

Member
Location
Seattle, WA
The supply conductors coming from the generator by definition are feeders.
The generator is a standby power source.

The NEC Handbook states that ?alternative source overcurrent protection device(s) for the electric-driven fire pump are not required to be sized for locked rotor current ? the circuit components of the alternative source are permitted to be sized according to Article 430.?

I guess my question is how do you size the generator for fire pump applications?
 
cadpoint

cadpoint

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Location
Durham, NC
How do you install a fire pump on a generator and not consider the aspects of the the first few Article(s) of 700 through 708?
 
iwire

iwire

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Location
Massachusetts
tkb said:
You can't just use the FLC x 125% for the motor.
There is a large inrush that you need to account for, up to the LRC
Click to expand...

TKB brings up a good point.

Your generator has to be able to supply that fire pump with no more than 15% voltage drop at locked rotor current. This is much different than normal motor requirements.
 
erickench

erickench

Senior Member
Location
Brooklyn, NY
That FLC value of 109A is actually the nameplate rating of the motor. It is used to size the overload. According to NEC table 430.250 the actual FLC is 114A. But it's close enough though. The FLC is used to size the Branch Circuit-Short Circuit protection which takes inrush current into account depending on what type of device that you use. It's also used to size the conductors.
 
iwire

iwire

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Location
Massachusetts
erickench said:
That FLC value of 109A is actually the nameplate rating of the motor. It is used to size the overload. According to NEC table 430.250 the actual FLC is 114A. But it's close enough though. The FLC is used to size the Branch Circuit-Short Circuit protection which takes inrush current into account depending on what type of device that you use. It's also used to size the conductors.
Click to expand...

The fact this is a Fire Pump changes the rules greatly

See 695.4 and 695.7 for a start.

If we were to select an overload for a fire pump it would be sized to carry the locked rotor current of the fire pump indefinitely.

Normally a listed fire pump controller is used so selecting an overload deice is a moot point.
 
B

blingbling4r

Member
Location
Seattle, WA
Here's what I got straight from the Cummins Whitepaper on the subject...

Here's what I got straight from the Cummins Whitepaper on the subject...

Sizing the generator set
Background: NEC 695-7 requires that voltage dip no
more than 15% of rated controller voltage at the fire
pump controller line terminals (includes cable drop)
during normal starting of the fire pump motor. This may
translate to oversizing the generator set by a factor of
two or three times to provide required motor starting
kVA compared to when a 30-35% starting voltage dip
is permitted.
Where the fire pump is the only significant load on the
generator set, the starting kVA required will be much
greater than the required running kVA. Since there are
practical limits to the alternator capacity in a generator
set, a larger genset may be required, resulting in a
light load running condition for the engine (less than
the recommended minimum of 30% of rated kW). To
alleviate this, consider adding additional loads with low
starting requirements, such as lighting, or the application
of supplemental load banks, especially during normal
routine system testing.
All fire pump controllers, whether reduced-voltage or
DOL (direct-on-line), full voltage, include an emergency
manual mechanical means to start the fire pump under
full voltage should the starting circuit or contactor coil
malfunction. The exception to NEC 695-7 states that
the 15% voltage dip limit does not apply when using
manual starting emergency means.
Caution: Cummins Power Generation recommends
an analysis of generator set voltage and frequency dip
performance when using the manual DOL starting. This
analysis may indicate a larger generator is required to
achieve desired performance during this condition. This
may be desirable to get assurance that the fire pump
controller does not drop out when automatic reduced
voltage transition from start to run occurs prior to when
the pump achieves near rated speed or when the pump
cannot be accelerated during reduced voltage due to
high operating head pressure.
Cummins GenSizeTM sizing software allows a complete
analysis of fire pump starting requirements. Using a
special fire pump load icon in GenSize for the fire pump
motor, establishes a maximum allowable Peak Voltage
dip of 15% while starting the fire pump load (all fire pump
loads will be included in the peak load calculations) after
all other loads are already running on the generator.
Using GenSize, first size the generator with the starting
means desired (DOL or reduced voltage) using the fire
pump load icon. If the fire pump using DOL starting,
is centrifugal (most are) and is not starting into a
significant head pressure, then check Low Inertia in the
fire pump motor load entry form. This will reduce the
starting kW requirements for the genset. The generator
will be sized to achieve the maximum 15% peak voltage
dip. Then, delete the fire pump load(s) from the project.
You will be asked if you want to reset the maximum
peak voltage dip. Change this to the same value used
for the maximum allowable step voltage dip. Then
replace the fire pump load(s) with a regular motor load,
DOL and check the cyclic box to obtain the peak load
for both motors while allowing the peak voltage dip to
exceed 15% for the emergency operating condition.
Use the largest generator recommendation from these
two calculations.
Note: It is not necessary to size the generator set for
locked rotor current continuously.


So again it begs the question...

How do you know what that 15% voltage drop is going to be without actually testing the generator/fire pump combination? It looks like I need to get my hands on the Cummins PowerSuite software :)
 
roger

roger

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Fl
Occupation
Retired Electrician
Just curious, why would you want to bear the liability of sizing this generator?

In any case, there are more variables than just electrical numbers to consider, a couple of more would be fuel type and altitude.

Even if you had a program to work with this would be better left to a PE or a generator manufacturer.

Roger
 
B

blingbling4r

Member
Location
Seattle, WA
Generator sizing

Generator sizing

I'm just curious as to how the process goes. I needed to get a size for the generator for a bid that I needed to get done.


The Cummins dealer is swamped with larger orders then mine so I got swept to the back burner.


I tried to come close to get a rough budget number for the generator and the labor involved.

Once we get the job we were going to send it out for engineering and try to adjust up or down whatever the price is. I bidded it to half the size of what the Cummins rep told me over the phone after bid time so I had to sweet talk the G.C. to let me float my number a little bit to compensate for this and concede any extra money if the generator is smaller.

Basically I'm just a fiend for knowledge and want to be better at my job for estimating and as a wireman.
 
erickench

erickench

Senior Member
Location
Brooklyn, NY
iwire said:
If we were to select an overload for a fire pump it would be sized to carry the locked rotor current of the fire pump indefinitely.
Click to expand...

iwire, I just looked at NEC 695.6(D). It state's:

Power circuits shall not have automatic protection against overloads. Branch-circuit and feeder conductors shall be protected against short circuit only.

The locked rotor current for this case would be 641A for letter B,C, & D motors.
I guess this value would be used for branch circuit-short circuit protection as indicated in NEC 695.4(B)(1).
 
erickench

erickench

Senior Member
Location
Brooklyn, NY
Yes but NEC 695.4(B)(1) in this case mean's overcurrent as short-circuit protection. Look how high the locked rotor current is compared to the name plate rating. Almost six times as much. You would'nt have to apply any of the percentages in NEC Table 430.52 to size the short circuit protection. Overload protection could'nt be this high.
 
iwire

iwire

Moderator
Staff member
Location
Massachusetts
erickench said:
Yes but NEC 695.4(B)(1) in this case mean's over current as short-circuit protection.
Click to expand...

IMO If it did mean that then they would call it short circuit and ground fault protection. Just like Part IV of 430 does.


Look how high the locked rotor current is compared to the name plate rating. Almost six times as much.
Click to expand...

Yes it is.

You wouldn't have to apply any of the percentages in NEC Table 430.52 to size the short circuit protection.
Click to expand...

430 has very little if anything to do with fire pumps.


Overload protection couldn't be this high.
Click to expand...

Sure it can be. As I understand it the idea is they want a fire pump to run, even if it runs to self destruction.
 
bob

bob

Senior Member
Location
Alabama
erickench said:
Yes but NEC 695.4(B)(1) in this case mean's overcurrent as short-circuit protection. Look how high the locked rotor current is compared to the name plate rating. Almost six times as much. You would'nt have to apply any of the percentages in NEC Table 430.52 to size the short circuit protection. Overload protection could'nt be this high.
Click to expand...

Iwire is correct. The logic is that you don't want the fire pump to shut off while the building is still burning. No need to save the fire pump at the expense of the building. The pump is to run until it is destroyed.
 
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