HV Breaker Calculation

If you don't know how to do this, hire an electrician who does!

The life you save may be your own.

And you won't be liable for any deaths that do occur due to incompetence.

Thanks Lyn for giving me such pride....it's o.k if you don't want to give answer....thanks again

JustJust want to be more specific

You see, I am not qualified to work with high voltage, and neither are you.

Therefore, if you don't how to cook, stay out of the kitchen.

There are many more attributes needed to be known to properly design a power distribution system than you indicate.

If you are in school trying to find a quick cheat to doing a homework assignment, CR4 is not a cheat site.

If you are trying to review for approval somebody else's design, we cannot teach you everything you need to know in this public forum. Review the aspects of the design (account numbers, quantity ordered, etc.) that you are qualified to review. If the person doing the design has a free moment you might ask them to explain something that you do not understand. Who knows, your inquiry might validly prompt a need for a slight redesign. Every engineer has had moments that they cannot see "the tree in the forest".

If you have been tasked to do this design then immediately quit your job. You are being setup to be fired for incompetence anyway.

The given units do not form a sensible question.

Sorry, I agree with lyn. This is of such an absolute basic nature that if you have to ask, it identifies you as wholly unqualified to be performing this work. Assisting you further is in essence providing aid and support for a murder-suicide pact; if you continue on without getting proper help from a qualified professional experienced in this field, you are likely to get yourself and someone else killed.

First of all I have to agree with the general opinion: you have to hire a specialist if you intend to solve this problem. For your general information, in the open post- except the medium voltage has to be 11 kV [not kVA] most of necessary data are missing: Low voltage for instance. There is not only a "formula" and it is not about calculation only. You have to know your local code [NEC,BS,NF,CEC,DIN/VDE,GOST or else] and experience how to use this. Briefly, it is a profession and it is not a "formula".

As i understood your incoming Voltage is 11k KV not POWER 11KVA

If it is correct than you can calculate the transformer rating for 433 Volt LT by following formula

Vp/Vs=Is/Ip

Vp-Primary Voltage 11000V

Vs-Secondary Voltage 433v

Is- Secondary Current 1000 Amp

Ip -Primary Current to be calculate=39.36Amp

Now transformer Rating In Mva=3.17*Vp*Ip/1000000=0.75 MVA

So you can Select 1 MVA transformer of 11/0.433 Kva and accordingly you can select HT breaker VCB of 630 Amp and LT Breaker ACB of 1200Amp.

Click-click... Boom.

Is this thread something about India's education system, then?

Hey you Crabtree, Yo can't say about INDIA's education in such manner, if want you can say about me then it's acceptable.

Is your education system in your country learnt you to make this kind of loose comments in your country in such open forum.

And about the my question, is is asked without making home work and just for curiosity.

And just for your information I am with Instrumentation background-- not from Electrical.

You can't judge whole country's eduction system by judging only one person, even after you know him well.

Thanks for you compliments...

For unsupervised transformer (Imp up to 6%)

According to NEC 450

Primary volatage >600 V (Fuse = 300%, CB= 600%)

Secondary voltage <=600 V (Fuse / Cb = 125%)

Secondary voltage > 600 V (Fuse = 250%, CB = 300%)

• verCurrent Protection at Primary Side (Primary Voltage >600V):
• Rating of Pri. Fuse at Point A= 300% of Pri. Full Load Current or Next higher Standard size. or
• Rating of Pri. Circuit Breaker at Point A= 600% of Pri. Full Load Current or Next higher Standard size.
• OverCurrent Protection at Secondary Side (Secondary Voltage <=600V):
• Rating of Sec. Fuse / Circuit Breaker at Point B= 125% of Sec. Full Load Current or Next higher Standard size.
• OverCurrent Protection at Secondary Side (Secondary Voltage >600V):
• Rating of Sec. Fuse at Point B= 250% of Sec. Full Load Current or Next higher Standard size. or
• Rating of Sec. Circuit Breaker at Point B= 300% of Sec. Full Load Current.

Example: 750KVA, 11KV/415V 3Phase Transformer having Impedance of Transformer 5%

• Full Load Current At Primary side = 750000/(1.732X11000) = 39A
• Rating of Primary Fuse = 3X39A = 118A, So Standard Size of Fuse = 125A.
• OR Rating of Primary Circuit Breaker = 6X39A = 236A, So standard size of CB = 250A.
• Full Load Current at Secondary side = 750000/(1.732X415) = 1043A.
• Rating of Secondary of Fuse / Circuit Breaker = 1.25X1043A = 1304A, so standard size of Fuse = 1600A

We get a large number of clueless questions from India lately. You are but one example. I suspect that this stems from India being about 17% of the world population and not the education system itself. At the same time any student in or out of a classroom asking a clueless question needs to be corrected. When the topic can endanger anyone the clueless should be dissuaded away from the topic.

POWER IN (-) LOSSES => POWER OUT

P=EI

KVA IN = KVA OUT