Fire pump kW to kVA

hhsting

Senior Member
I have fire pump 125HP 208V three phase. Trying to convert 125HP to kVA.

Based on NEC Table 430.250 125HP is 343A at 208V three phase.

343x.208x1.733=123.57kVA.

Now if I use calculator internet to convert assuming 100% efficiency and 0.8 PF then I get 116kVA.

Which is correct or both are incorrect? What should be used to convert when sizing panelboard load calculation?
 

kwired

Electron manager
Location
NE Nebraska
The NEC tables assume about worst case ever made efficiency and power factor to come up with the amps they posted. Most motors you get will have lower amps meaning they are higher efficiency or higher power factor or both.

So your calculation yields what is likely worst case possible for that particular HP rating for general use 1800 and 3600 rpm motors. Lower speed or other specialty motors may not fit within those table values.
 

don_resqcapt19

Moderator
Staff member
Location
Illinois
Occupation
retired electrician
I have fire pump 125HP 208V three phase. Trying to convert 125HP to kVA.

Based on NEC Table 430.250 125HP is 343A at 208V three phase.

343x.208x1.733=123.57kVA.

Now if I use calculator internet to convert assuming 100% efficiency and 0.8 PF then I get 116kVA.

Which is correct or both are incorrect? What should be used to convert when sizing panelboard load calculation?
How can a fire pump be part of a panelboard load calculation? It cannot be fed from a panelboard.
 

David Castor

Member
Location
Washington, USA
Occupation
Electrical Engineer
Assuming 1 kVA per hp is a conservative assumption in most cases. This becomes ever more conservative as the motor size increases. If you need a more precise calculation, just the motor FLA x Nameplate Voltage x 1.732 (if three-phase). Motor is nowhere near 100% efficient so you can't make that assumption.
 

topgone

Senior Member
Assuming 1 kVA per hp is a conservative assumption in most cases. This becomes ever more conservative as the motor size increases. If you need a more precise calculation, just the motor FLA x Nameplate Voltage x 1.732 (if three-phase). Motor is nowhere near 100% efficient so you can't make that assumption.
Agree! The best method is to calculate from the nameplate figures. Get the line amps, the operating voltage multiplied by the indicated efficiency and 1.732.
 

Besoeker3

Senior Member
Location
UK
Occupation
Electrical Engineer
1hp=0.746kW this what I learned about this type of convertion, is that correct?
That is a correct conversion. I used to do that in school around 1960. Now we use SI (or commonly metric). But them I'm a Brit.
 

kwired

Electron manager
Location
NE Nebraska
1hp=0.746kW this what I learned about this type of convertion, is that correct?
Basically that is what the motor is rated to deliver at the output shaft, but inefficiency of the motor (there isn't any that are 100%) means the input will be higher, then you have power factor that makes the input VA even higher.
 

marcosgue

Member
Location
Tampa
Occupation
Electrician
agree, but we can use for general calculations, there're a lot of motors with efficiency close to 100% nowadays
 

kwired

Electron manager
Location
NE Nebraska
agree, but we can use for general calculations, there're a lot of motors with efficiency close to 100% nowadays
in polyphase induction motor category?

I can see maybe closer in some the ECM motors that are usually limited to low HP ratings. All motors are going to have some mechanical losses in efficiency.
 

Besoeker3

Senior Member
Location
UK
Occupation
Electrical Engineer
746W/Hp is a mechanical term, where Hp = Torque x RPM / 5252.

Electrical kW has to include the efficiency of the motor.
kW and HP are just units. That's all. If what you is efficiency you can measure it in kW or HP..
 

gadfly56

Senior Member
Location
New Jersey
Occupation
Professional Engineer, Fire & Life Safety
How can a fire pump be part of a panelboard load calculation? It cannot be fed from a panelboard.
^^^^^This right here. It has to be fed all by its lonesome. You could put a disconnect where it enters the building but it can't be part of the general MDP.
 

texie

Senior Member
Location
Fort Collins, Colorado
Occupation
Electrician, Contractor, Inspector
It seems to me that the OP is trying to get the KVA value for load calculation purposes. He should just use the table value amps in Table 430.250 and convert to KVA and stop complicating it further than that.
 

David Castor

Member
Location
Washington, USA
Occupation
Electrical Engineer
746 watts per hp - correct. BUT - the hp rating of the motor is the mechanical shaft output power. To determine the electrical power (kW) INTO the motor the losses of the motor must be accounted for. To determine the kVA into the motor (which is what the electrical system must provide), the power factor of the motor must also be accounted for. 1 kVA per hp will get you pretty close.
 

Besoeker3

Senior Member
Location
UK
Occupation
Electrical Engineer
746 watts per hp - correct. BUT - the hp rating of the motor is the mechanical shaft output power.
Power IS power. That's it. Whether it is HP or kW. What you do with that power is your choice. For me and the vast majority of the world we use kW.
 
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