Register now or log in to join your professional community.
<p>A500 kVA transformer full load current is721.7 Amps. If all18 no of induction motor starts simultaneously current becomes441.7 Amps. For DoL starting starting current of4416.7 Amps at once is needed. With a4% of short circuit impedance, short circuit current of transformer is18042 Amps. Clealy, the transformer will be overloaded by more than6 times, but it will not reach s.c. current value. What to do?</p>
Best thing is relay, and if that much starting current is needed and your transformer can get trip, then it is better to delay rather than trying to start motors at once. We may take small break between starting manually or use programmed relay that we can see in our normal small stabilizers and transformers of fridge and other accessories. It is not a wise decision to start all the motors at once with low capacity of transformer.
if I am right , the total load of motors is441.7 and starting current is4416.7 , there is no problem , so far the adjustment of short circuit tripping time exceeds the starting time of motors , you can refer to the vendors of motors to know the starting time and adjust the s.c time setting of the protection relay to be more than starting time of motors , don't worry the transformer can withstand this starting current , if you like you can contact the transformer manufacturer to assure this .
Thanks
