MCB

Many people have seen, hopefully you will get a useful answer soon. Meanwhile.....

1. "MCB" at 1434 A? You must be meaning "MCCB". Better you use "ACB" if you are using it on the outgoing of the trafo.

2. Thermal protection should be set at the rated current.... in your case, 1434A if you can set it so accurately. Otherwise, the nearest lower value you can set.

3. Transformers have a high, but short-time, inrush current, maybe 30X of rated current. You may need to consult the trafo supplier for their recommendation. Diffucult to say what to set, generally ok to set 10X.

The release/relay of the circuit breaker is meant to trip and protect the transformer.Better have a nuisance trip than no protection. Better be safe than sorry.

thank you , but still your answer missing the Magnetic current , but any how ,let us sumarize ,

Transformer Load = 994 KVA

assume short circuit is infinity= 5.75 % , input voltage =415 V

assume short circuit is infinity= 5.75 % , V=415

MVAs/c= 994kva/5.75/100= 17.3 MVA

therefore short circuit current on the secondary side is :

Is/c= 17.3MVA/(1.73*400=25KA

thermal Current

I = 1434 A, Ir = I approx. = 1500 A in order to set it accurately

up to here every thing is clear to me but the problem is how to get

Isd (current of the magnetic protection ))????

If the rated current is 1434 A, my recommendation would be that you set slightly lower, say 1400A if possible. Otherwise, the transformer will be somewhat underprotected in the zone 1434 - 1500 A. Nothing catastrophic will happen, but the insulation will degrade faster than if the transformer load was within its rating.

i mentioned that the transformer will have a magnetising inrush current which will be for a short duration. You need to set the magnetic trip threshold high enough not to nuisance-trip during switching on, but not so high that it will not trip during short-circuit.

However, i expect the 400V is the output side. So maybe no worry about the switching transient. If this is the case, you can set it as low as the CB permits, you need to decide what constitutes a safe-but-no-nuisance-trip level. i would set at 5 X

The 1500 A assumed only for Ir, but as In=2500 A , and you put 5x

Isd= 5*1500= 7500 A

but still can not understand how did you get or set 5x ?

i am confused a little here...do you mean you are using a 2500A CB? If so, good. But remember to set the thermal trip based on 1400A.

The 5x was an off-the-cuff casual answer. Not the best way to do proper coordination between the CB and downstream devices. i request you to download and study this paper : schneider-electric/ect167.pdf

It will tell you how to set the trips based on a proper discrimination system. You will, of course, not be able to decide your own system just by reading this paper. You may need to interact with the technical experts of the company whose CB you are using.

i suppose you do have a full distribution system downstream of this CB?

actually I did read it already before , but again I could not understand how to calculate it

You need to decide what system you want to use : Time-based, Current-based, energy-based etc. After this, the company will tell you how to select and set the CBs for proper coordination. You can google for "Coordination Tables" of the CB manufacturer.

It seems you are not contented with the fine answers of KVS. Let me try to put it in "your language". The magnetic trip of an ACB Protection Release must be set as minimum as possible, but at the same time, set so as to avoid nuisance tripping due to transient load switching conditions like motor starting; and it must not be set beyond the fault level at the location of the breaker.

I assume from your earlier comments that the Magnetic Trip is an instantaneous one. If it has a time delay setting, then the time setting of the magnetic trip has to be set as per co-ordination requirements with the downstream SCPDs (Short Circuit Protecticve Devices), if any.