1HP = 746watts
400hp=298400watts
=298kw
KW + root3 x V x I x cos phi x n
I = 298Kw divided by root3x400x.85x.85
= approx 597 amps full load current per phase
( approx figures for cosphi and n used )
starting I between 1200 and 3500 A depending on starting method and dwell starting.
if you choose these direct starting methods you will have a problem with electrical supply company as they will have a limit on the starting surge allowed for high hp motor starting .
you will probably have to use some sort of soft starting method.
not sure what you mean about 1.5 service factor
i can't believe that this is a 1.5 overload condition. or is it a 75% power utilization factor
DOL starting you will have a fuse rating of 800 amps
star delta starting you will have a fuse rating of 630 amps
for VDE0636 fuses ( gL )
If a US-design, the data is available on the nameplate, in the former of a multiplying factor. The factor has a code letter, i.e., ... D, F, G, ... etc, which is listed in the NEC. It wll reveal the inrush kVA per Hp.
If for general information, then the rms starting current may be 5 to 6 times rated (operating) current.
Starting current, starting torque, and starting power-factor, are all independent of load!
Although you didn't ask, runup time is dependent on the available accelerating torque (that produced by the motor minus that required by the load) and the moment of inertia of both the motor and load.
....
> Starting current, starting torque, and starting power-factor, are all independent of load!
....
Starting current is not independent of Load. You never start Blowers driven by AC motor on full load it will take more starting current. YOu always trottle the inlet valve and start the system once it stablizes the you increase the valve output.
The starting current, as well as breakaway torque, are functions of only the motor's design parameters. However, the available acceleration
torque determines the time to accelerate the driven machine from standstill to full speed. Acceleration torque is a function of the torque developed by the motor minus that required by the driven machine.
Closing a fan's inlet or outlet louvres, or closing a pump's inlet or discharge valves, or the unloading of a compressor, simply reduces the
torque required by the driven machine. These operations performed on the driven machine serve to increase net torque, not change the motor's
electrical starting current, nor its starting (breakaway) torque, nor its runup torque. Increasing available acceleration torque, then,
decreases starting time.
BTW, the starting torque or (current) is no-more nor no-less than the locked-rotor torque, with or without a load coupled to the motor. And,
neither can be altered using DOL starting. If you would like to discuss pertinent equations, let me know.
