Fault Current

Exactly

If the only life the OP was risking by learning the rules of power distribution, I could understand learning the hard way. Only authorized people should work on power distribution.

this assumes infinite source.

The equation should have been Transformer rated voltage/Rated Impedance = SC

In actual practice we do not have infinite source. There will be a source impedance [or thevenin's impedance ] and source voltage [or thevenin's voltage]

The actual fault current is determined by Thevenin/s voltage divided by thevenin's impedance.

i can see the logic of what you say. However, i have seen people just take the transformer power ÷ pu impedance...for example, 1 MVA tfr,5%impedance, so SC current = 28kA. (415V system)

How wrong is this figure in your opinion/experience ? Is it likely to be 1%, 5%, 10% ... higher than real SC current? Some ballpark figure will be appreciated.

Well I prepared a table of typical values based on my experience with Indian grids.

The first part of the table shows thevenin's impedance on 100 MVA base for typical fault levels in MVA indicated for different kV level.

The second part of the table shows the transformer impedance on 100 MVA base for different MVA rating of the transformer.

The total impedance to be used is the sum of the value of the first part + corresponding value of the second part, depending on actual fault level and the transformer rating.

We will see that when the grid fault level reduces, the thevenin's impedance increases and will have more significant bearing on the accuracy of the calculations.

For example if we consider 11 kV 500 MVA fault level, with Thevenin's impedance of 0.2 per unit on 100 MVA base and 10 MVA transformer with 0.5 per unit impedance on 100 MVA base, we get a total impedance of [0.5+0.2 = 0.7 per unit] on 100 MVA base. Obviously this is significant change in this case , if we ignore thevenin's impedance.

thank a lot Dr. i am erecting a 33 kv / 66 kv substation in my 33 kv feeders how many kamps i have to consider as a fault current for 25 mva transformer . is 33 kv line's lenth , soil's type or resistivity and voltage level are also effecting the fault current ?

redfred - i think it is only by working on anything can one acquire skills required to be authorized . . . i am sure you will ponder a while on this . . .

I've pondered your statement for a reasonable amount of time. I see only two scenarios that explain your statement.

First, you misunderstand the meanings of the phrases, authorized to work on a system and an authority of a system. Authorized to work on a system is synonymous to an individual or group has been granted permission to work on a complicated or dangerous system. Likely this permission will be granted with conditions of safety equipment and/or training to work on the system. This permission will also be granted only when proper coordination with other affected and effected groups get informed. The individual or group that grants permission has the position of authority for that specific system. With this authority comes the responsibility that only individuals with proper training will be permitted to work on the system. Similarly, this authority comes with the responsibility that coordination of all individuals working on this system are aware of their responsibilities and conditions during all phases of operation. Now I do agree that once an individual has been granted authority to work on a dangerous system, they must continue to learn from the experience and be open to others criticism of their work. I also agree that those granted the responsibility of being the system authority should first work somewhere as one seeking authority. But before their hands, tools, design gets installed on a dangerous system they must first get the authority to do this from the proper authority.

My second option that explains your statement should be considered offensive. It is meant to be offensive because lives are at stake. You don't give a damn about the life of anybody foolish enough to ask about a dangerous system that is not asking their questions to the authority in charge of the system. You have no problem giving enough of a plausible answer that they feel empowered enough to circumvent the safety of a well controlled system. If the system is sufficiently robustly maintained that before a fatal incident happens the erroneous intrusion or design gets caught and a fool has been found. What greatly bothers me of this attitude and approach is that the fatality may not be the eager fool but somebody who happens to be at the wrong place at the wrong time.

Before you make an indignant reply, I do not think that you are so callous as my second scenario. I believe that your thoughts were closer to my first scenario. But I hope you appreciate the potential gravity of the second scenario and why I had to bring it up.

So I repeat my earlier comment. Only authorized people should work on power distribution.

of course i understand and appreciate where you are coming from . . . maybe my statement came from a third premise and was intended to elicit responses quite like yours . . . after all there can be 11 types of people in the world . . . thanks for writing in to expand our knowledge and understanding of serious and important concerns . . .

http://cr4.globalspec.com/thread/62152

http://cr4.globalspec.com/thread/56152/fault calculation

Formula in post # 3 (KVS post) can be utilized (together with 'parallel' combination of source MVA) to arrive at the fault current with 'finite source' as well. As per KVS post # 3/6, for 1.0 MVA, Z= 5% trafo, S.C. for a through fault is 28 kA @ 415V sec. (or 20 MVA). This is assuming infinite source at the trafo prim.

Let us now consider 'finite source' with below assumed S.C. MVA values:

1. 50 MVA

2. 25 MVA

3. Infinite MVA source (also included for comparison purpose)

For a fault on sec. side of trafo, S.C. MVA will be:

For 1: = (50x20)/70 = 14.28 MVA (down to about 71.4%)

For 2: = (25x20/)/45 = 11.11 MVA (down to about 55.5%)

For 3: = 20 MVA (100%)

Thus it can be seen that there is a significant change in fault current based on "strength/weakness of source" as rightly mentioned by previous post # 5/8 of Raghunath.

My thinking is

....is it not safer to use the 'infinite source' value since any 'finite source' can only deliver a lower fault? If one does an actual case study, one may be surprised that while the arguments are interesting in the abstract, reality is quite different, and that when you want a 36 kA CB, you will actually find a 50kA CB is not more expensive. So, rather than have an uneasy feeling that the CB may be somewhat underrated if the Thevenin calculations are awry, better to be safer with a higher capacity CB.....

when you use a 50kA CB for a system with a fault current of only 35kA - instead of using a 36kA CB - you have to be cautious while setting the same . . . else you will compromise protection . . . this in the field is often overlooked even by experienced engineers . . .

Higher kA rating is safer but expensive.

Normally fault level in the grid continuously increases with addition of new transmission lines, and power generators as in India [India is supposed to have robust growth plans over next decade] and the higher rating chosen now may become insufficient in about 10 to 20 years time.

I have come across situations, where we had to put reactors in series to limit the calculated fault levels.

Higher kA rating of the breakers will not affect the protective relay settings. Relay settings are primarily dependent on [for overcurrent relays]

a. normal operational currents for which they should not trip

b. Fault currents [above normal operational currents] for which they should trip. This is independent of kA rating of the breaker and depends on system impedance. Also thermal withstand limits before which they should trip.

Typical industrial systems will have few levels of transformers and voltage levels and fault levels at lowest voltage level usually remain constant. The concern is only at the voltage level that is directly connected to the grid, where variation of fault level may influence the choice of ratings.

I agree.

As far as "equipment selection/sizing" is concerned, it may be fair to a certain extent, to use conservative values. However, for "relay setting", as reqd. by OP, the co-ordination should satisfy for both 'Max'. as well as 'Min.' fault conditions.

Referring to LV CBs only, 36 or 50 kA doesn't make any difference ... please step back and think .... this is CURRENT-LIMITING zone. A LVCB, seeing a 36kA(~70kA peak) or 50kA(105kA peak) fault, will start opening very, very rapidly.....long before any peak is reached.

Again going back to #9, if the O/C relay is set to operate at 20 MVA (28 kA) 'conservative' fault current, it may not 'coordinate' properly for the 'actual' fault current of say 11.11 MVA (15 kA).

This is why, for relay setting, it is reqd. to do the setting at the actual fault current.

We are not talking about the same thing obviously, so here is my final post on the subject of LV current-limiting circuit breakers, and their coordination....

Above 2400A in this case, the CB goes into "Reflex trip" mode, and operates irrespective of any setting on the relay.

Granted that i am only talking about one small part of the overall distribution system. Perhaps the OP is asking about 400kV systems, in which case, all my posts are not applicable.