Available Fault Current Calculation Question

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Bugman1400

Senior Member
Location
Charlotte, NC
Phil Corso said:
Bugman, Mivey...

"The Y-D-Y will act as a zero sequence source for whichever wye-side is grounded (or both sides if they both are grounded). The buried delta can circulate zero sequence currents from either side."

"Zero sequence currents will not pass through if either or both sides are ungrounded."

The first is correct, but the second... !

Phil
Click to expand...

I see nothing wrong with the second statement but, as always, you need to make sure your interpretation of the meaning is the same. I interpret the second statement to mean that zero seq currents cannot pass from the primary to the secondary side of a YDY xfmr where either Y side is ungrounded.
 
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Phil Corso

Senior Member
Location
Boca Raton, Fl, USA
Bugman...

You and Mivey are correct! I misinterpreted what was meant! But, I now see Mivey meant that with only one of the Y's grounded, there is no "connection" to the other!

Phil
 
mivey

mivey

Senior Member
Phil Corso said:
Mivey...

So then, the term zero Zero-Sequence Impedance simply means it doesn't exist!

Phil
Click to expand...
No. It means it is zero or close enough to what we are approximating as zero or driving to zero as a limit. For example, if we were to drive one of resistors a-e in the prior illustration to zero (or close enough). On the other hand, Rx from the prior illustration does not exist in the fault equation so its value is irrelevant.

If for some reason you felt compelled to stick a non-existent resistor in the series circuit, then I suppose you would then have to set its value to zero so it would have no impact on the circuit. Can't imagine why you would want to do that.
 
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NewtonLaw

Senior Member
Location
Honesdale, PA USA
Phil Corso said:
Gentlepeople...

In referring to solidly-grounded systems, then I rigorously disagree. Whether or not the magnitude of single-phase to ground (L-G) fault is greater than a 3-phase (L-L-L) is dependent on the magnitude of the Zero-sequence impedance of the network compared to Positive and Negative impedances!

L-G faults are seldom greater if the source is a 2-winding transformer! For alternators it is common to find that the L-G fault magnitude exceeds 3-phase fault magnitude! And, if the network under consideration involves 3-winding transformers, special scrutiny is required because resultant impedance could be less than zero, thus increasing on L-G fault magnitude!

Regards, Phil Corso
Click to expand...

I agree completely with this statement. Calculating single phase line to ground faults at our substations where we supply the 69kV to the Delta wound primary with a 12470 WYE grounded secondary, the single phase fault is always higher than the three phase fault.
 
mbrooke

mbrooke

Batteries Included
Location
United States
Occupation
Technician
GoldDigger said:
Why should it? Any OS usually supports a calculator function, as does Google. Or do you specifically mean a fault current calculator? I could see that being a liability issue. :)

Sent from my XT1585 using Tapatalk
Click to expand...



Not if its worse case... though I do believe that worst case can back fire in arc flash studies because worst case might throw the instantaneous trip portion of an OCPD, but the actual fault current might not- leaving you in the delayed range.

What does the 2017 hold in store?
 
Julius Right

Julius Right

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Occupation
Electrical Engineer Power Station Physical Design Retired
The short-circuit currents are calculated at 480 V side-of course.
In my opinion, the system short-circuit apparent power it is about 200 MVA.
The direct, inverse and zero sequence impedance are equal for transformer and for system too.
No arc resistance is taken in consideration in ground fault case.
 
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Ingenieur

Senior Member
Location
Earth
xfmr data:
1000 KVA
34500-480Y/277
PU %Z 5.75
X1/R1 9.74, X0/R0 9.13
LLL Fault Test Data 19921, LN Fault Test Data 19790

Z base = 480^2 / 1 MVA = 0.2304
Z act = 0.0132 Ohm
Z act 3 ph basis = 0.0247 Ohm

fault MVA = sqrt3 x 480 x 19921 /10^6 = 16.56 MVA
i fault LL based on Z = 480/0.0247 = 19420 (a bit lower than test data)
i fault LN based on Z = 277/0.0132 = 20910 (a bit higher than test data)

note: in this case the pos Z1 and neg Z2 sequence components are usually considered the same since I direction has no real bearing on Zx

for those who care the x/r can be used to derive actual R and X for a detailed model
X1/R1 = 9.74 or X1 = 9.74 x R1

from above Z = 0.0132 = sqrt(R1^2 + X1^2) = sqrt(R1^2 + (9.74 R1)^2) = sqrt(95.87R1^2) = 9.79 R
R1 = 0.0132/9.79 = 0.0013
X1 = 9.74 R1 = 0.0127
Z1 = 0.0013 + 0.0127j = 0.0127 / 84.2 deg Ohms
it can also be done in PU

for reference standard infinite bus method
i fault = 1000000/(480 1.732 0.0575) = 20920
 
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Julius Right

Julius Right

Senior Member
Occupation
Electrical Engineer Power Station Physical Design Retired
I agree with you Ingenieur, however it could be simpler as following:
Take Sgrid=200 MVA[on 34.5 kV].
Take Xgrid/Rgrid=10
will get Z1grid=1.146E-4+i11.46E-4 ohm
Transformer Z1xfm=13.04E-4+i127E-4
Transformer Zoxfm=13.9E-4+i127E-4
If you'll take Z1grid=Z2grid=Zogrid and Z2xfm=Z1xfm you'll get:
I"k3=0.48/sqrt(3)/sqrt((13.04+1.146)^2+(11.07+127)^2)*10^4=19.97 kA [0.3% more]
I"K1=sqrt(3)*0.48/sqrt((3*1.14+2*13.04+13.9)^2+(3*11.46+2*127+127)^2)*10^4=19.90 [0.6% more]
Usually, a correction of transformer reactance is required[IEC 60909-0 for instance]:
KT=0.95*1.1/(1+0.6*xT%)=0.95*1.1/(1+0.6*5.75/100)=1.01
However, it was neglected here, I think.
 
steve66

steve66

Senior Member
Location
Illinois
Occupation
Engineer
Ingenieur said:
xfmr data:

i fault LL based on Z = 480/0.0247 = 19420 (a bit lower than test data)
i fault LN based on Z = 277/0.0132 = 20910 (a bit higher than test data)

for reference standard infinite bus method
i fault = 1000000/(480 1.732 0.0575) = 20920
Click to expand...

Which takes us back to the fact that the L-L and L-N faults are about equal, And again, the L-N fault current will fall off quicker than the L-L current, so by the time we get to the main SWBD or OCP, its probably less. So for a simple typical service, there is probably no point in trying to calculate the single phase fault current in addition to the 3 phase fault current.

That's pretty much what I said in post #8. :slaphead:
 
mbrooke

mbrooke

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Location
United States
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Technician
Julius Right said:
This paper is very interesting indeed. However in our case it is not much useful.
Do you mean to verify “our” calculation against this, do you?
The Appendix Part IV.Analitical Techniques it is close to our ways, I think.
Click to expand...



Why would it not be of use though? Or I guess I am asking: how do you compute fault current to meet the 2017 NEC requirements?
 
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rian0201

Senior Member
Location
N/A
Julius Right said:
This paper is very interesting indeed. However in our case it is not much useful.
Do you mean to verify “our” calculation against this, do you?
The Appendix Part IV.Analitical Techniques it is close to our ways, I think.
Click to expand...

Why not useful? This is how fault calculations should be done.. But you have update the standards used in that paper.




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Ingenieur

Senior Member
Location
Earth
summarizing what has been said:

L-G
I0 = I1 = I2 = Vf / (Z0 + Z1 + Z2 + 3Zf)
Ia = 3 I0

L-L
I0 = 0 (obviously)
I1 = -I2 = Vf / (Z1 + Z2 + Zf)
Ia = -Ib = -j sqrt3 I1

double L-G
I1 = Vf / (Z1 + ((Z2 (Z1 + 3Zf)) / (Z0 + Z2 + 3Zf))
I2 = -I1 ((Z0 + 3Zf) / (Z0 + 3Zf + Z2))
I0 = -I1 ((Z2) / (Z0 + 3Zf + Z2))

[I ph abc] = [A] [Iseq] (matrices)
A = a Identity matrix
basically:
Ia = I0 + I1 + I2
Ib = I0 + a^2 I1 + a I2
Ic = I0 + a I1 + a^2 I2
where a = 1/120 deg

with the utility supplied information all 3 can be calculated
this is why they give you the info they do
Zf can usually assumed to be 0 for the worse case

Where:
I0, I1, I2 = 0, Pos, Neg, seq domain currents
Z0,, Z1, Z2 = 0, P, N seq impedances (the P and N are the same for a balanced network, ie, Z1 = Z2)
Zf = fault impendence
Vf = voltage across fault
Ia = phase domain current

for reference:
wye
Z0 = Zy + 3 Zn (or Zg)
Z1 = Zy
Z2 = Z1 = Zy

delta
Z0 = infinity or open
Z1 = Zd/3
Z2 = Z1 = Zd/3
 
I

Ingenieur

Senior Member
Location
Earth
Just for the heck of it
all pu
Vf = 1
Z0 = Z1 = Z2 = 0.1
Zf = 0

L-G
I0 = 1/(0.3) = 3.33
Ia = 10

L-L
I1 = -I2 = 1/(0.2) = 5
Ia = 8.66 j

2xL-G
I1 = 6.67
I2 = -3.33
I0 = -3.33

Ia = 0
Ib = 10/-120
Ic = 10/120

check my math, lol, hand calcs
 
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