Pump Starting Current

What does the pump supplier recommend?

i dont know what the rule is there but here in india we should have 3 pumps as we have -

1. Jokey Pump (10HP) which is just maintain pressure when all the hydrant post is closed.

2. Main Pump (60HP) which will maintain pressure when will you open the any hydrant post

(both the above pumps take power from government supply which we generally cut in case of fire)

3. Same rating main pump (60HP) but its directly coupled by engine only (no need of alternator)

(this pump will use in case of fire when we cut the main raw power (government supply))

What I understand from this is , first 2 pumps (10hp & 60hp) are working on electric power. Third is a stand by , which will work from engine (not from electric motor) when there is no power available. Is it correct ?

yes exactly

because generally we have to cut the electricity from the main L.T. or H.T. Panel in case of heavy fire.

So, is no-one considering direct diesel-driven fire pumps? - the ones that run either to fuel-out, to water-out or to destruction without assistance from the local electricity supply?

Mr. ateeff has a different case . His pump is electric pump -150hp

Please check the name plate for full load ampere. OR (150x746)/(1.73x380x0.85)=200A

Consider the starting current for selection of generator.

For me 250kVA generator is ok for this.

So the effect of a generator supply fuse or a residual current device or some other electrical safety feature operating while the generator is running and supplying power to the pump is to be ignored, then?

OK.

We normally use MCCB in generator panel. The starting current is not causing MCCB to trip, since it is for very short duration. MCCB is of thermal trip type.

- OR MCCB I USE MAY BE NOT WORKING -

what,s abut NEMA standard code G and what,s your opinion for the following calculations

1. motor's HP - 150HP, kVA/HP - code G (5.96), motor's PF - 0.84
Starting kW : 150HP x 5.96 = 894kVA x 0.84 = 750kW

2. By starting star, starting kW is reduced by one third.
750kW x 1/3 = 250kW

3. When load over 75% is instantaneously loaded to either engine or alternator,
by character of engine, rpm drops and higher voltage drops.
accordingly, in order to start 150HP motor, engine power must have 33% more than 250kW.

thus, in this case, gen. set power of 332kW is needed.
otherwise, motor's starting kVA/HP have to be considered to code E (4.7).
we suggest that motor's starting method should be with inverter starting.

It is better check with local statutory conditions.

I suggested 250kVA , because

1) Its a fire pump. Its not going to be in use normally, only when emergency/fire condition it will be on operation.

2) If slight drop of voltage , once only when fire happens is allowed by the system

3) I have used 62.5kVA generator to run 40hp motor - it was starting on no load, then loading to full load.

when we operated this pump with new generator 32o KVA the pump not operated and the starting current was 430 A

150 HP = 112kW at full speed (delta),

at star, as you write very realistically, there should be left 1/3-d of the full power, or 37kW. Why 37kW have produced so much current - 430 A ? Is the motor/its starting installation all right?

can any body describe here what is in the NEMA STANDARD CODE G?

Now study on the NEMA

Mr. Ateeff your multiply factor may be wrong (5.96) its is 6.2 go through the below link -

http://www.joliettech.com/nema_codes-ac_motors.htm

http://wiki.answers.com/Q/What_is_locked_rotor_current

is it a back up generator and engine driven?If so by the time it picks up and generator can take the load fire would have caused enough damage.A diesel engine driven back up pump is essential as mentioned .I do not understand why a back up gen is needed as if mains fail then diesel driven pump will come on.can u explain the system?

Locked rotor current!