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Anonymous Poster #1

Fault Level of Substation

04/18/2020 3:25 PM

We have a substation which has 4 transformer. 1×33/11kv, 3×33/6kv. All are 5mva each. Supply coming from a government bigger substation located 5 km away. What is the fault level of this substation?

I do not know how to calculate. Searched internet but not found something. Please help.

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#1

Re: Fault level of substation

04/18/2020 3:46 PM

There is a theory that says it takes a chain of only six telephone calls to reach anyone on the planet. So, start making calls until the team running the <...government bigger substation...> can be asked the question and give a satisfactory answer.

Simples ξ.

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#2

Re: Fault level of substation

04/19/2020 12:57 AM

I agree with PWSlack if you know these 6 numbers and could trust them...However, in this Covid-19 era..I agree with you.

When I was 22 and MS in Electrical Machines, my first job was -in a utility distribution unit- to calculate a voltage drop. After days of deep study I did it!

That remember me the Baudelaire "l'Albatros": His big wings bother him to walk.

A vague appreciation-let's say "proportions"-you can get from IEC 60076-5 Power Transformer Ability to withstand short circuit-for instance: for 33kV system 1000 MVA short-circuit should be the limit.

Zsys=33^2/1000=1.089 Ω

From ABB Switchgear Manual vol.3 Fig. 3-14 an overhead line of 36 kV 70 mm^2 ACSR presents Z1=0.38 Ω/km that means total 5*0.38= 1.9 Ω.

If we take 33 kV as base voltage the three phase short-circuit apparent power at transformer high voltage terminals will be: Ssh=33^2/(1.089+1.9)=364.3 MVA Isc=364.3/sqrt(3)/33=6.37 kA

Transformer uk%=7% [minimum according to IEC 60076-5]

At secondary transformer terminals [keeping the same reference voltage 33 kV] Ztrf=33^2/5*7%=15.25 Ω

Ztot33kV=1.089+1.9+15.246=18.235 Ω

Sshtrf=33^2/18.235=59.7 MVA

So for 6 kV Isc will be 59.7/sqrt(3)/6=5.74 kA and 3.13 kA for 11 kV.

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#3

Re: Fault Level of Substation

04/20/2020 12:02 AM

The good information you received is half of the information you need to calculate the available fault energy. The other half is the amount of time for any upstream protective device (probably a fuse) to clear the fault. If you can't find the clearing time then you can use the anticipated amount of time for a person to escape the area after a fault (crawl, walk, be blown through the air, or whatever)--this usually is 2-10 seconds. The available fault energy is a function of I2t. Since all of this is presently guesswork you need to follow the advice also given--to ask the supervisors or engineers at the substation.

--JMM

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