Zero Sequence fault currents 4

[Electoman]

Could anybody please explain the following: I am interested in the current distribution of a single-phase to ground fault on a YY transformer with both neutrals earthed. The PRAG explains that the currents for a A-E fault will be Ia=(2C1 + C0)I0, Ib= Ic= -(C1-C0)I0, where C0 and C1 are zero and positive sequence distribution factors.
The fault is on the secondary side of the TX and I want to know the impact on the primary side.
What I could gather, is that Ia'=2*Io and Ib'= Ic' = -Io.
E.g. if Ia' = 0.5A, Ib'= Ic'= 0.25A.
If this is the right assumption, how is this derived from first principals. The PRAG (Protective Relay Application Guide) is not very clear on this. Any explanation will be much appreciated.

 

[jghrist]

I'm not sure what you are reading in PRAG, but first principles when dealing with a single phase-to-earth fault through a Y-Y transformer should be that the current in the unfaulted phases will be zero.

Ia'=3·Io
Ib'=Ic'=0

 

[ScottyUK]

Are you sure it relates to a YNyn transformer?

It sounds from the little bit you posted that you are reading about a line-ground fault on an earthed star winding transformer, as seen from the delta winding side of a Dyn?? transformer.

You can do a first principles derivation based on ampere-turn balance of each of the individual windings. The unfaulted secondary windings have no current in them, and the primary winding on the same core leg has no current in it either to preserve ampere-turn balance. Using this approach you can determine where currents are flowing and in which direction.




-----------------------------------

Start each new day with a smile.

Get it over with.

 

[stevenal]

These equations are for current distribution in a networked system. For the single branch of a radial system, C1 and C0 should equal 1 giving jghrist's answers above. PRAG jumps right into the middle and treats the subject incompletely and unconventionally. I'd suggest consulting Power System Analysis by Stevenson. A little more money, though.

 

[cuky2000]

From the system point of view, the basic relation for SLG fault are:
1) Va = Vo + V1 + V2 = 0.
2) Ib = Ic =0
3) Ia = 3Io.
4) Zo = R+jXo + 3Zg

For transformer YY solidly grounded in both sides, the equivalent network connection parameters are: R = Zg = 0 => Zo = jXo.

Please notice that the connection zero sequence diagram varies for each configuration (see slide 3 on the enclose site).

http://www.eng.fsu.edu/~tbaldwin/eel4213/public/lecture17.pdf

)

 

[jbartos]

Comment on the previous posting: Please, is the link working?

 

[jbartos]

Comment on jghrist (Electrical) Apr 30, 2004 marked ///\\I'm not sure what you are reading in PRAG, but first principles when dealing with a single phase-to-earth fault through a Y-Y transformer should be that the current in the unfaulted phases will be zero.
///The current in unfaulted phases is zero if the transformer secondary is open. If the transformer is loaded, then there will be currents flowing to the fault from phase b, Ib' and phase c, Ic'. Therefore, the following posted above equations:
Ia'=3·Io
Ib'=Ic'=0
have to account for currents Ib' different from zero and Ic' different from zero. If the load is connected, then 3Iao=Ia'+Ib'+Ic'
See Reference:
William D. Stevenson "Elements of Power System Analysis," 3rd Edition, McGraw-Hill Book Co., 1975.
Chapter 13 Unsymmetrical Faults
Any contribution from the loads having stored energy, e.g. rotating electrical machinery energy stored in dynamics, has to be added to the line to ground short.\\\

 

[Electoman]

Guys, thanks for your replies so far. Let me try to give more information. I am testing a T60 (GE Multilin TX relay). GE provide a TX Diff calculation Excel sheet to help calculate the diff and restraint currents. They advise to unground the star points in the TX to test the pick-ups on a per phase basis. However, when you ground the star points of a star-star (YNyn) TX, and a a-g fault takes place, there are currents in the A,B and C phases as well (primary side). E.g. on 100MVA, 132kV/69kV YNyn, 500/1 CTR and 1000/1 CTR, with Ia = 0.87A (0 deg)[Winding 1] and Ia = 1.57A (180 deg) [Winding 2]. (All current values are secondary values injected into relay). When looking at this spreadsheet, Ia diff = 0.514A (-180 deg), Ib diff = Ic diff = 0.257A (0 deg).

The way I understand this is: When both of the star windings are grounded, there is a path for zero sequence current on the grounded side. So in a case of a a-g fault on the grounded star side (69kV), Ia = Io + I1 + I2 = 3Io. Through the electromagnetic linkage, the 3Io appear on the neutral connection on the primary side of the TX. Each phase will now have Io flowwing back to the source. Therefore, Ia' = 2*Io, Ib' = Ic' = Ic' = -Io. This explains the difference in diff currents on the Excel spreadsheet.

Question: Is this explanation correct? If so, where can I find more information to prove that the Io will split in this relation of 2:1:1 for a YNyn TX?

 

[jghrist]

Electoman,

There is something basically wrong with your explanation. If an a-g fault occurs, there is no current in b or c. That's why it is called a single phase-to-ground fault. If the transformer is YNyn, then the current in the primary side of phase b is the current in the secondary side times the turns ratio. Same with phase c. So, there is no current in the primary phase b or c either.

Full load amps on the 132 kV side would be 437A, or 0.875A on a wye connected 500:1 CT secondary. On the 69 kV side, this would be 837A, or 0.837A on the secondary of a wye connected 1000:1 CT secondary. I'm not sure where your 1.57A comes from. In any case, 1.57 - 0.87 does not equal 0.514, so I can't figure where your Ia diff comes from.

 

[stevenal]

The spreadsheet in question can be found at

http://www.geindustrial.com/products/software/t60/t60diffsimulation.xls

I find the results puzzling with Iop appearing in the unfaulted phases. But the spreadsheet is meant to simulate the algorithm of a microprocessor relay. Maybe they use an unconventional approach. Note that the highest Iop is in the faulted phase and relay correctly decided to trip. Do your test results validate the spreadsheet Iop values? Otherwise, the spreadsheet may be flawed.

Your scenario was a little bit strange in that winding 1 remained fully loaded while winding 2 faulted.
Try injecting quantities that are the same multiple of Inom Prim and Inom Sec. With IAW2 at -180 degrees with respect to IAW1, Iop should be zero with a no trip condition (through fault). Then zero out either the W1 or W2 value to simulate a more realistic single source internal fault. For a double source fault, increase both currents, but alter one phase angle by 180.

 

[stevenal]

Okay, I'm a little slow this morning and so's Jghrist. In the pre-microprocessor relay days, you would connect your CTs in delta for this transformer. This has the effect of filtering out the zero sequence quantities, sending only positive and negative sequence currents to the three relays. For a single phase fault, one relay sees twice the current that the others do. Although you are actually connecting your CTs in wye, the relay and spreadsheet are simulating this delta connection to get the Iop and Ires quantities.

 

[stevenal]

Correction: A SLG fault with delta connected CTs would provide operate and restraint quantities to only two of the three relays. Still a mystery.

 

[stevenal]

Sorry to hog the forum, but no one else is picking this up. I was sort of on the right track before, but relay and spreadsheet do not simulate delta connected CTs. They just filter out the zero sequence components, and calculate Idiff and Ires per phase from the filtered wye connected CTs. See row 1, column 2 of Table 5-4, page 5-51 of the instruction manual. The superscript p values are positive and negative sequence values only. 2/3rds the total ground fault current on the faulted phase, 1/3 on each of the unfaulted phases.

 

[jghrist]

Stevenal,

That makes sense. So the answer to Electoman's query is that the values he is using are not the CT secondary current, but the filtered relay current (with zero-seq taken out).

I_a = I_a1 + I_a2 = 2·I₀

I_b = a²·I_a1 + a·I_a2 = -I₀

I_c = a·I_a1 + a²·I_a2 = -I₀

The sum of the three filtered phase currents is the zero-sequence current and equals zero (since it has been filtered out).

 

[stevenal]

Jghrist,

Yeah, those are the restraint quantities. Sum the corresponding restraint quantities from each winding to obtain the differential quantities. SEL uses a similar calculation if winding compensation number 12 is used (360 degree phase shift.) If a zero degree shift is selected, no filtering is done, similar to GE's ungrounded case. New stuff for me, all ours are delta wye.

 

[rsherry]

Sequence network analysis is often (as in the lecture notes quoted) looked at only for the no load condition. The Ib=Ic=0 statement for a fault in phase a is only true for the phase conductors if there is no load on the system, as jbartos pointed out earlier. For an on load system only the fault current itself has Ib=Ic=0 and not the currents in the phase conductors (as measured by relays etc).
An on load system fault is usually calculated using the Helmholtz (Thevenin) method of superposition of sequence currents from two calculations - one calculation at the point of fault for the faulted phase (where Ib=Ic=0 for the fault) and one 'calculation' for the pre-fault load condition. The sequence currents are then added before back-calculating the actual phase currents seen on the system during the fault. Ib and Ic in the unfaulted phases will not be zero (due to the loads) and may rise due to the added component of positive sequence current due to the fault.
I think the GRAP is applying a general rule in using zero and positive sequence distribution factors to arrive at the on load currents without the detailed calculation.

 

[Electoman]

Thanks guys,
You helped me a lot.
Much appreciated.

 

[jbartos]

Suggestion to jghrist (Electrical) May 3, 2004 marked ///\\ If an a-g fault occurs, there is no current in b or c. That's why it is called a single phase-to-ground fault.
///Please, would you provide a reference for this statement or your justification in view of the fact that there are single phase-to-ground fault on unloaded system and many many more on loaded system?\\\

 

[cuky2000]

[COLOR=red yellow]
FROM THE ORIGINAL QUESTION: [/color red yellow]
Could anybody please explain the following: ……for a A-E (SLG) fault will be :
Ia’=(2C1 + C0)I0,
Ib’= Ic’= -(C1-C0)I0.[/color blue]

Here is a possible explanation that may require validation
======================================================
Sequence branch circuit parameters

Branch Phase Currents: Ia’ , Ib’ and Ic’ are phase currents in an arbitrary branch due to a normalized fault at point “x” (an arbitrary point in the network).

Branch Impedance: Consider a sequence network reduced as follow:

- Branch “pq” sequence Impedances: Zpq)0 , Zpq)1 & Zpq)2
- Equivalent parallel sequence impedances of all branches except “pq”: Zeq)0, Zeq)1 & Zeq)2.

a) Sequence curent and voltages for SLG fault:
a.1 )Sequence current Relationship:
Ipq)o =Ipq)1 =Ipq)2 =Io Eq (A)
a.2 ) Phase voltages as a function of the sequence voltages:

Ea= E1+ E2+ E0
Eb= a^2 E1+ aE2+ E0
Ec= a E1+ a^2E2+ E0 Eq (B)

b) Transforming Eq(B) as a function of the sequence current and impedances.

Zpq.Ia’ = Zpq)1.Ipq)1+Zpq)2.Ipq)2 +Zpq)0.Ipq)0

Zpq.Ib’ = a^2.Zpq)1.Ipq)1+a.Zpq)2.Ipq)2 +Zpq)0.Ipq)0

Zpq.Ic’ = a.Zpq)1.Ipq)1+a^2.Zpq)2.Ipq)2 +Zpq)0.Ipq)0 Eq (C)


c) Sequence Distribution Factors
C1= Zpq)1/Zpq; C2= Zpq)2/Zpq and Co= Zpq)o/Zpq Eq (D)
_{NOTE: For typical transmission system, the positive and negative sequence impedances are very close. Therefore, C1 =C2}

d) Branch current of sequence parameters I'=f(C_i,I_i)

Dividing Eq(C ) by Zpq, replacing sequence currents by eq (A) and substituting by eq (D)the branch currents are:

Ia’ = (C1+C2+C0)Io = (2C1+Co).Io
Ib’= (a^2.C1+aC2+Co)Io =[(a^2+a)C1+Co]Io=-(C1-C0)Io
Ic’= (a.C1+a^2.C2+Co)Io =[(a^2+a)C1+Co]Io=-(C1-C0)Io

 

[jbartos]

Suggestion to Electoman (Electrical) Apr 30, 2004 marked ///\\\The PRAG explains that the currents for a A-E fault will be Ia=(2C1 + C0)I0, Ib= Ic= -(C1-C0)I0, where C0 and C1 are zero and positive sequence distribution factors.
///See above Cuky2000 posting for derivations.\\The fault is on the secondary side of the TX and I want to know the impact on the primary side.
What I could gather, is that Ia'=2*Io and Ib'= Ic' = -Io.
E.g. if Ia' = 0.5A, Ib'= Ic'= 0.25A.
///If you consider C1=1 in p.u. and C0=0 in p.u., then it is correct. However, the Y-Y solidly grounded has some final C0 different from 0. It is probably very small and it can be approximated by C0=0 in p.u.\\\
If this is the right assumption, how is this derived from first principals.
///It is reasonably correct assumption. See Cuky2000 derivations for the rest.\\ The PRAG (Protective Relay Application Guide) is not very clear on this.