Fault Calculation for DG

Irrespective of the load, the fault current fed by the generator would be same.if the load is connected and if the load is motor load then there will be some contribution of fault current from the motor.

It's a simple I=V/Z relation. If the Z changes, then I will change also.

Fault calculation procedure will be the same but fault current will be different due to load impedence as well as momentary contribution of motor load which acts as generator at the instant of fault.

Thats not true at all , at the point where the fault happens current flows and it flows from the source , since fault doesnt depend on load it depends on source , only if motor is connected load , due to sudden dip in voltage and inertia of the rotor the motor starts acting like a generator for sometime then is such a case fault current will be addtion of fault supplied by the source and this motor acting as a generator.

The fault level in terms of KVA or MVA will be the same and determined the generator capacity and other parameters.However fault current will vary at different locations depending upon the impedance in between the generator and the fault location.

i have 380 KVA DG set make by KIRLOSKAR, i don't Know the impedence of DG but i know the lenth of the cable from DG to LT panel. it is 3.5*400sqmm AL*2R lenth is 50 m each. now tell me what is Fault level of DG breaker in LT panel

plz help me

Generally the % impedancew will be between 10 and 14 %

For 10 % impedance the fault current at the generator terminal would be 4.9 kA and it will be 4.7 kA after 50 met of cable of 2 runs 3.5 x 400 sq.mm Al cable

For 14 % impedance the fault current at the generator terminal would be 3.5 kA and it will be 3.4 kA after 50 met of cable of 2 runs 3.5 x 400 sq.mm Al cable

Can you please tell me the formula used for above calculation?

Thanks

It is important to know the sub transient impedance of D.G. set ie ZSq=Xd'Sq+RaSq.And Isc=Rated voltage/Z.

yes but obvious.............