Negative Seq. Relay setting for Xfmr 3

[SMB1]

I have 69/13.8KV Xfmr Delta(HV)/Way(LV)
I want to set the –ve seq. relay to coordinate it with OC protection in the LV side

If the L-G fault is 4500 A. What is the value of the –ve sequence in the HV side
I got the result from Aspen Oneliner but I could not get the same result by hand.
So, what is the the formula shall I use?

 

[stevenal]

Okay, I've been patient. I did the calcs, presented the proof and still no star? Either explain it better than me of fork 'em over.

 

[jbartos]

Suggestion: The large fault currents are often suppressed by suitable system grounding. Medium resistance grounding, reactor grounding, etc. might be considered to limit a destruction of vicinity due to large fault current.

 

[jghrist]

jbartos,

By far, the most common grounding system used in US utility MV distribution systems is solid grounding. 9800A would not be considered excessive.

SMB1,
What values do you get by hand calculation and from Aspen? Seems like failure to divide line-to-ground fault current by 3 as noted in stevenal's Sep 2 post is a likely source of error. Also, the relay setting current may not be the negative sequence current. Seems to me I remember SEL using 2 or 3 (I forget which) times the negative sequence current as the setting value.

 

[stevenal]

Jim,

Check SMB1's Sept 3 statement that "(-ve seq. outside the delta = -ve seq. inside the delta)" His error is that he believes negative sequence current exists only in the faulted phase, and doesn't understand that symmetrical components are actually symmetrical and present in all phases. His error is by a factor of sqrt(3) as dpc first indicated.

 

[SMB1]

stevenal (Electrical):

Please refere to my post dated Sep 3, 2003, and let me know where is the error in the prove.

I got the correct furmula from the (Symmetrical Components)book,
But still I believe that during LG fault in the Y side:
(-ve seq. outside the delta = -ve seq. inside the delta)in the HV side.

Thanks and Regards

 

[stevenal]

SMB1

Negative sequence current in each one of the delta connected windings is of equal magnitude and 120 degrees apart. To find the negative sequence current outside the delta, on A phase for example, do node analysis at the A terminal. The negative sequence leaving the A bushing is something like I2A=I2BA-I2AC (check me on this, cause this is the hard way). An easier way is to remember that the line currents in a balanced system have sqrt(3) times the magnitude of the line to line currents and are shifted by 30 degrees. An even easier way is to use the effective turns ratio rather than actual turns ratio as it already includes the sqrt(3) factor, leaving just the phase shift to deal with. Now if you know which way the positive sequence currents shift across the delta wye, just shift the negative sequence currents 30 degrees the other way. Have I earned my star yet? Down below there to the left where it says "Mark this post as a helpful/expert post!"

 

[jghrist]

stevenal,

I'll give you a star. The key thing that SMB1 has to understand is that sequence currents are balanced three phase currents by definition.

 

[stevenal]

Yee Ha, two stars. Thanks.

 

[jghrist]

In his 9/3 post, SMB1 wrote:
any LG fault in Y side will behave as LL fault in the Delta side.
The fault current will flow on one winding (Y side). After applying the (AmpsXTurns) concept the current in the Delta side will flow also on one winding.

So, in the (HV)Delta side during the LG fault in(LV)Y side

The fallacy is in treating sequence currents like fault currents. Even though, during a low side line-to-ground fault, there is fault current in only one low side phase and two high side phases, there is negative sequence current in all three phases of both the low side and the high side. The positive and negative sequence currents are present in all three phases and are equal.

If you had 4500A <0° in secondary Phase A, then on the secondary side:

Voltage ratio=69000/13800=5
Turns ratio=8.66

Ia0=Ia1=Ia2=1500<0°

Ia=4500<0°, Ib=0, Ic=0

on the primary side:

IAC=519.6<0°, IBA=0, ICB=0
IA=IAC-IBA=519.6<0°, IB=IBA-ICB=0, IC=ICB-IAC=519.6<180°
IA0=0, IA1=300<30°, IA2=300<-30°

IA1 and IA2 both flow in Phase B, but from the definition of symmetrical components:

IB=IA0+a²·IA1+a·IA2 where a=1<120°, a²=1<240°
IB=0+300<270°+300<90°
IB=0

 

[busbar]

Negative-sequence overcurrent relaying is applicable on transformer primaries, based on ANSI/IEEE C37.91-1985 Protective Relay Applications to Power Transformers. §5.6.2 “Since these relays do not respond to balanced load or three-phase faults, negative-sequence overcurrent relays may provide the desired overcurrent protection. This is particularly applicable to &#8710;-Y grounded transformers where only 58% of the secondary per-unit phase-to-ground fault current appears in any one primary-phase conductor.”