SLG Fauly Yg/D transformer 2

[sky2020]

I have a 10MVA 33/4.16kV Yg/D transformer, X=5%. What is the phase-A SLG fault current on secondary delta? I think it should be zero as the transformer zero sequence is open and then there is no current is passing?
If a zigzag transformer is to be installed with x0 = 5%, how to compute the new SLG fault?

 

[waross]

A SLG fault on the primary (wye) side will have several effects.
It will become a line to line fault on the delta side.
The line to line fault on the delta side will set up a circulating current in the delta limited by three times the transformer impedance.
The circulating current will backfeed into the primary fault and add to the available fault current at the fault location.
The backfeed current will be 1/3 of the transformer Available Short Circuit Current.
This may have an impact on the ASCC at switches for 33 kV feeders.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[sky2020]

Thank you Bill.

The fault is in phase A at the 4.16kV - it is outside the delta (i.e. downstream the transformer secondary delta side). I believe there is no SLG fault value as the transformer won't pass the zero sequence. If you think there is a SLG current value, what value it will be?
If a zigzag transformer is to be installed with x0 = 5% downstream of the transformer delta secondary, how to compute the new SLG fault?

 

[waross]

The fault current will be limited by the Neutral Grounding Resistor.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[7anoter4]

Z1=0.05 p.u.; Z2=0.05 p.u.; Zo[of Zztr]=3*X1=0.15 p.u.[usually]
I"k1=sqrt(3)*1.05*4.16/(5*0.05)/Zbase[IEC 60909-0]
Zbase=4.16^2/10=1.73
I"k1=sqrt(3)*1.05*4.16/(5*0.05)/Zbase=17.487 kA
IG=3*I"k1
IG=3*17.487=52.45 kA for 10 sec.
Inzz=3%*IG=1.574 A [IEEE 32/1972]
P=Inzz*VLG/3=1.574*4.16/sqrt(3)/3=1.26 MVA

 

[sky2020]

These calculations are very helpful. Thank you.

The only remaining question shouldn't the SLG fault current at the delta side zero as the transformer does not pass zero sequence current. Why did we install the zigzag, Is that for reducing the voltage LG as it went to sqrt(3)* VLG? or because we wanted to detect the SLG fault? why we need to add this zigzag transformer?

 

[7anoter4]

For delta connection the SLG current depends on the capacitive reactance of the load-including supply cables and overhead line. This current is low and a protection relay is not enough sensible to act.
Usually an ungrounded neutral system is able to function with one phase grounded. The voltage between the ungrounded phases and the ground will rise up to between phases voltage that means 73% more.
The system may function so for short time[seconds] or limited duration [usually up 1 hour] or indefinitely.
That depends on many factors: requirement of unbreakable supply on the consumer part and the system insulation possibility to withstand the overvoltage.
Usually a low resistance grounding will cancel the overvoltage but high resistance grounding may be enough sometime.
In your case –for instance-a 100 A resistor Rn=4160/sqrt(3)/100=24 ohm.
The actual loop impedance includes downstream line impedance up to fault location.
As waross said the transformer and the supply system reactance are negligible in this case.

 

[waross]

An ungrounded system is may have issues with discontinuous ground faults, AKA arcing ground faults.
In the event that a transformer winding fails and starts arcing to ground, it may generate high frequency currents.
The transformer winding may then act as an auto-transformer and impose high frequency, high voltage on the affected phase.
In the days when ungrounded systems were common, more than one industrial plant lost a large number of motors due to an arcing ground fault superimposing several thousand Volts to ground on one phase.
The zig-zag transformer provides a symmetrical ground connection for an otherwise ungrounded system.
You may also provide a symmetrical ground with a small wye/delta transformer bank.
Asymmetrical grounding may be accomplished by corner grounding a delat or by grounding the center tap of a for wire delta bank.
Eg: 120/240 Volt, four wire delta.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[7anoter4]

“The transformer winding may then act as an auto-transformer and impose high frequency, high voltage on the affected phase”.
Thank you waross. I forgot it.

 

[sky2020]

Thank you 7anoter4 and waross. Very good information!

 

[Kiribanda]

Hi Zanoter4,
After seeing your calculation I thought of asking the following question from you.
To size the ZZ transformer donot you need the total zero seq capacitance (XC0) of
the system?

 

[7anoter4]

Sorry for the delay, Kiribanda.You are right. You have to verify if the current you calculated [Inzz] is not less than Ico.