mivey
Senior Member
Probably not a bad one in most cases.
Generator short circuit.
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mivey
Senior Member
The pdf in the same post? Must be some missing pages.
Or just use the actual conductor sizes rather than complicate the issue?
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mivey
Senior Member
The best I can figure is that you have confused yourself or do not understand what you are doing and just copy/pasting some stuff you read somewhere.
your post:
Ingenieur said:
is wrong. The pdf linked does not state that and I would be suprised if it did. The correct formulas are:
isc" = V/(sqrt3 X"d)
isc' = V/(sqrt3 X'd)
or in p.u. reactance:
isc" = I/X"d
isc' = I/X'd
I have the nec which is based on awg
did not feel like looking up metric sizing
but I note you did niether, contributed nothing but criticism
only kibitz
I can only surmise that the subject matter has confused you
are you actually a licensed graduate engineer?
no need to be defensive or to feel inferior and lash out
the link does state that
you are wrong now and will be again
it's ok
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except in this case
yes, 2 is conservative but not real
peaking factor
n = R/X0 = resist/surge imped
surge imped is sqrt(L/C) and since L > C in most t&d lines n ~1.5
the extra 0.5 may add 33% to the aic for no good reason
if you have ckt parameters n can be looked up on graphs for various systems
a fault is no more than a switching operation
close switch on sc
open switch, ie, cb trips
the dc decay is of the form ~ e^-(t/TC) ignoring damping
so yes, it will be less at 3 TC vs 1
TC = L/R or RC depending on what you are looking at
TC ....%
0.1 90%
0.5 60%
1 37%
3 5%
mivey
Senior Member
How about you just show where your link stated what you posted. Perhaps I keep reading over it. Page and paragraph would help me find it.
mivey
Senior Member
Maybe. Only the OP can say. When someone looks up the generator X'd it is almost always in the same place as the X''d. Very few exceptions.
mivey
Senior Member
Let me save you the trouble of looking because you are not going to find it.
The known electrical engineering world uses I=E/X (for negligent values of R). Not sure where you get your info but you need to refresh your recollection.
I had to wait till I got in because I could not remember the general formula format for sudden faults for a generator but it is:
I = E/Xd + (E/X'd - E/Xd)e^(-t/T'd) + (E/X''d - E/Xd)e^(-t/T''d)
Note the similarity for the transient and subtransient portions.
A /sqrt(3) term would be a voltage coversion from L-L to L-G voltage and would apply to both transient and subtransient equations.
As I stated before, your prior formulas are incorrect. Look it up.
take it up with the author in the link
the 3 term equation is in a link posted in the other fault thread
mivey
Senior Member
The "AIC Calculations & It's Significance" thread? One of your posts?
the first link ref is in this thread
the second in a post I made in the thread you ref, it's title 'fault current calculations'
mivey
Senior Member
Well, not there either that I could find. At any rate, there are three possibilities:
1) You are referencing some other link I haven't found where the author has a typo.
2) You are referencing some other link I haven't found where you did not follow the author.
3) You are referencing a link I found where you did not follow the author.
The fact remains (see most any EE text) that the equation is I=E/X for transient and subtransient. That fact means there is nothing to "take up with the author". Until you prove that wrong you will stay wrong and any author who says the same will also be wrong.
Any EE worth their salt will recognize text errata or will look into it if there is a question. Just part of engineering nature to solve mysteries.
mivey
Senior Member
The paper by Dr Jaffari? Nice paper.
However he does not say
as you seem to have concluded.
He in fact states:
which is correct.
From your data, you don't have to worry about the short circuit capacities of your cables. 90mm2 cables have a short-circuit capacity of 13.6kA for 1 second. Since you are using 8 cables per phase (a total capacity of 8 x 298 = 2,384A), you are assured your cables can withstand any short-circuit at your MSB.
mivey
Senior Member
Ingenieur,
Are you trying to say something about the DC offset giving a reason to use X'd instead of X''d to find the maximum fault?
You would have to have X'd < X''d which still makes no sense. Consider this:
At the maximum DC offset of sqrt(3) for the subtransient, you would have
I = I'' * (1+sqrt(3)) = 2.732 * I'' = 2.732 * V/X''d
compared to the decayed DC result for the transient of
I = I' * (1+D*sqrt(3)) = (1+D*1.732) * I' = (1+D*1.732) * V/X'd, where D=decay factor (a value between 0 and 100%)
At zero DC offset, you would have
I = I'' * (1+0) = I'' = V/X''d
compared to transient of
I = I' * (1+0) = I' = V/X'd
Which means X'd could at best be equal to X''d for no DC offset or otherwise will have to be smaller than X''d (which makes no sense).
Add: I'm too tired to continue trying to figure you out tonight so I'm calling it quits.
Are you trying to say something about the DC offset giving a reason to use X'd instead of X''d to find the maximum fault?
You would have to have X'd < X''d which still makes no sense. Consider this:
At the maximum DC offset of sqrt(3) for the subtransient, you would have
I = I'' * (1+sqrt(3)) = 2.732 * I'' = 2.732 * V/X''d
compared to the decayed DC result for the transient of
I = I' * (1+D*sqrt(3)) = (1+D*1.732) * I' = (1+D*1.732) * V/X'd, where D=decay factor (a value between 0 and 100%)
At zero DC offset, you would have
I = I'' * (1+0) = I'' = V/X''d
compared to transient of
I = I' * (1+0) = I' = V/X'd
Which means X'd could at best be equal to X''d for no DC offset or otherwise will have to be smaller than X''d (which makes no sense).
Add: I'm too tired to continue trying to figure you out tonight so I'm calling it quits.
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Julius Right
Senior Member
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I agree with you, mivey. However, if we take formula 4-58 from IEEE 399 :
I = E/Xd + (E/X'd - E/Xd)e^(-t/T'd) + (E/X''d - E/X'd)e^(-t/T''d)
In this case if t=0 I=E/X"d [VLL/SQRT(3)/X"d] as maximum rms short-circuit current.:weeping:
I = E/Xd + (E/X'd - E/Xd)e^(-t/T'd) + (E/X''d - E/X'd)e^(-t/T''d)
In this case if t=0 I=E/X"d [VLL/SQRT(3)/X"d] as maximum rms short-circuit current.:weeping:
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