KVA to Amps

<...725 KVa DG set,how much load...>

No more than 725kVA

<...the KVA factor conversion in to amps...>

Divide by the voltage.

Assuming a normal motor power factor of say 0.80 (the load power factor you should know approximately) - the load will be 580 KW that is if you mean this to be the load.

The other part is already as in #1

KVA = 1000 x V x I

It is three phase load.factor 1.732 is missing

You are right

Todays DG sets- Sold in India can safely handle upto 0.8 times the rated load, meaning 0.8*725= 580KVA.

DG Sets can also perform at full rated power for upto 60 minutes without any issue, but you would rather be safe and run the same UPTO 580KVA.

Am not an Elec Engr, but do run more DGs than most people have- and this thumb rule works for me.

KD

For 725KVA Diesel Generator, the total output current that is available is approximately 1000Amp base on 415V and assuming that the efficiency of the generator is 100% efficiency. The above calculation is in ideal stage.

The formula is as follows;

KVA=Voltage x Current (I) x 1.732 x Eff %

In actual, a new diesel generator can generated about 96% of the total KVA due to the losses in the generator such as electrical losses and mechanical losses. Therefore, you will get about 94% to 95% of the efficiency of the generator output. As the generator get older the efficiency will also decrease and the output current will also become less.

I agree with this & would also like to add that the type of load is important. For example, a genset that feeds a UPS (a non-sinusoidal type of UPS) should be derated according to the level of harmonics generated by the UPS. As a general rule, a derating factor of 1.5 is safe for all situations (this is a very broad estimation). Eg. Generator is calculated to be able to supply (at 100% duty cycle) 1MVA. However, it could only safely supply about 500KVA to a harmonic load.

Remember the type of load you are supplying will also have an influence on how much load you will supply with the unit.

If you are supplying motors only you need to consider start up currents. Depending on what percentage load you are already applying to the GD, you may run into a situation when attempting to start a motor the motor may not reach full speed.

During the start up of this motor (7 X Full load current) you may overload the GD, and the output voltage of the gen set will drop resulting in a torque loss in the motor ( (Vn/Vr)³, see previous threads on this). Should the inertia of the load on the motor be high, the motor will not reach full speed with the reduced torque available. The motor should eventually trip by the protection relay. This scenario may also happen should you attempt to start a large(400kW) motor, without any other load applied to the GD.

If you have a combination of motor and resistive loads (lighting) you want to do motor starting first and then switch on the resistive loads (no starting currents) to avoid the above scenario.