Ground faults on the secondary of transformer as seen by the primary 3

[gdeep]

Hi,

I have a 12.47kV/4.16kV 5MVA delta-wye(10A,10s resistive grounded system) feeding a 5kV switchgear. The NGR is rated for 10A, 10s and I am using SEL-751A protection relays.

1) A ground fault on the secondary of the transformer would be seen by the primary as L-L fault?
2) How would this 10A ground fault current be reflected on the primary? Would be there be a considerable increase in currents on two phases of the XFMR primary and would I be able to see negative sequence currents( I am using 800:5 CT)
3) Are there any good references I could use? I have tried simulating a model in PSCAD and I could see currents go up in two phases of the primary but I cannot verify the magnitude of the currents just the relative increase to the pre fault currents.

Any inputs would really be appreciated
Thanks

 

[waross]

A ground fault will be at 2.4 kV.
Your ratio will be 12.47kV/2.4kV = 5.2
Your increase in primary current will be 10A/5.2 or 1.9A
That's 00.2375% of the CT rating. I am surprised you see anything at all. A 10A ground fault on this transformer in the field would probably not be detectable on the primary switchboard instruments.
{Rant-on}Why do we get the feeling that more and more, schools are focusing on teaching the software and spending little time on the basic principles behind the software. This is not any criticism of you, qdeep. It is more a comment on the direction that training seems to be going.{Rant-off}

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

 

[gdeep]

Thanks for the info.
But would the current seen on the primary depend on the turns ratio of the XFMR ( or would it be very close to 5.2)
Also wouldn't the primary single phase see only 58% of the secondary L-G fault currents ( IEEE Std C37.91)

 

[waross]

The transformer winding voltages are 12.47kV to 2.4kV. The 2.4kV windings are star connected to develop 4.16kV. A ground fault will be from a phase to neutral at 2.4kV The winding ratio is 12.47kV/2.4kV, even though the apparent transformer ratio is 12.47kV/4.16kV
The difference between the transformer ratio and the winding ratio accounts for the 58% factor.

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

 

[gdeep]

Thank you very much

 

[dpc]

Bill,

Great explanation - thanks.

dpc

 

[Mila15]

I agree...this is very good explained! I have a question though. Since we're using winding ratios (not XFMR ratios), should we convert the 1.9 A to line current?

 

[waross]

It is line current. Unless I have made a mistake, a 10 Amp line to ground/neutral fault at 2.4 kV will increase the line current on two primary phases by 1.9 Amps.
10 Amp x 2.4 kV / 12.47 kV = 1.925 Amps.

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

 

[Mila15]

I might be missing something, but here is how I understand it...

(Winding Voltage)p = 12.47KV
(Winding Voltage)s = 2.4KV

(Winding Current)p = (Winding Current)s x (Winding Voltage)s / (Winding Voltage)p
= 10A x 2.4/12.47 = 1.92A
Then, (Line Current)p would be 1.92 x 1.732 = 3.33A

 

[LiteYear]

Funny how a seemingly simple question can reveal hidden complexities. I like waross' derivation, but there's one two many conversions from winding current to phase current for me to be confident in the result. To the simulation I say!

As you can see in the attachment I simulated a 12.47kV/4.16kV delta-wye transformer with about 4MVA of load and some nominal line impedances. I simulated long enough (20s) for transients to settle down and then introduced a 10A ground fault on phase C. The primary currents in amps RMS were:

Before fault/During fault
A: 182.36 / 184.29
B: 182.42 / 182.41
C: 182.40 / 183.40

So two primary phases increased by 1.95A and 1.00A. The numbers we're interested in are so close to the simulation error noise that I wouldn't like to draw a conclusion only based on this. Instead, I bumped the ground fault current up to 100A and ran it again. Results this time:

Before fault/During fault
A: 182.36 / 201.60
B: 182.40 / 182.42
C: 182.40 / 192.99

Negative sequence current rose from close enough to zero to about 12A while zero sequence stayed at approximately zero.

So this time the two primary currents increased by 19.24A and 10.59A. By waross' formula we'd expect a primary current increase of 100 Amp x 2.4 kV / 12.47 kV = 19.25 Amps. Why the other phase only increases by about half that I'm not sure.

 

[LiteYear]

And the attachment...

 

[dpc]

EasyPower agrees with Bill: For a 10 A grounding resistor on the secondary side, 1.9 A in two phases of the primary, 0 A in the other phase.

 

[gdeep]

Hi,

In Waross explanation, I believe we can assume the transformer impedance to have minimal effect on the fault currents ( 240 ohm NGR is the most dominant impedance). As we decrease the NGR resistance ( 240 ohm to 24 ohms), the SLG fault currents increases to 100A and so do the losses due to transformer impedance. The losses will not be the same across all phases. If you increase the SLG fault current to 1000A, the difference between the two phase currents on the primary will be more. You can also test this by varying the impedance of the transformer. In model I simulated, I used a 7.5% impedance transformer. When I increased the transformer impedance, the difference between the phase currents was more.

This is my reasoning. Please correct me if I am wrong.

 

[dpc]

For a line-to-ground fault on the wye side, there will always be equal resulting current in two terminals on the delta primary - in one and out the other. Bear in mind that the windings on each side of transformer are directly related to each other, regardless of the external connection. The transformer impedance does begin to have a greater effect as the grounding impedance is decreased. This is straightforward - two impedances in series - one much larger than the other. I'm not clear on the difference in phase current you are referring to.

 

[gdeep]

Thanks dpc, my mistake, I was referring to the "line currents" in two terminals on the transformer delta side.

 

[LiteYear]

After a few paper calculations I agree - the unusual sharing of the phase current in my simulation is probably more of a distraction than a help. There are large variables associated with the primary reactive component and the transformer impedance that are too heavy an influence to draw out the principles.

Certainly if, as dpc suggests, you just consider the current in the windings, a secondary ground fault causes additional current in one secondary winding. Since the effect on primary and secondary voltages is negligible, there must be a corresponding increase in current in the primary winding (divided by the turns ratio). If everything is balanced, this increase must be provided for by a change in the two primary phases that connect directly to that winding. Since the phases of the primary currents are 30 degrees to the currents in the windings, balancing the vector equation shows the two phases increase by root 3 times the increase in the winding - recall that line current in a delta configuration is root 3 times the winding current.

So in an ideal world, my money is actually with MICHAELUSA:

line current = 10A x 2.4/12.47 x 1.732 = 3.33A

What happens in the real world, I really don't know.

dpc, what's the winding ratio in your simulation?

 

[sibeen]

My two cents worth.

I get 1.926 amps.

 

[sibeen]

Oops, I put a root 3 in where I shouldn't.

Now get 3.33.

 

[LiteYear]

sibeen, your transformer ratio, Tr, is wrong - should be 12.47/2.4 = 5.2. See Bill's 24 Jan 13 17:10 post for explanation.

But you appear to have undone the error in your I1, I2 and I3 formulas by using Tr*root(3) as the ratio.

The three IA:=I1-I2 equations don't appear to match the sign conventions used in your diagram. They should be:

IA:=I2-I1 IB:=I3-I2 IC:=I1-I3

But I think your reasoning is spot on! I was misled trying to add the ground fault current in the same phase as the primary line current. That need not happen at all, as your derivation shows. The additional primary line current due to the ground fault will actually be in phase and anti-phase with the fault current.

I think the attached vector diagram makes that clear. I've adopted the sign conventions from your diagram, so everything works out according to the three equations above. The solid lines are the phase (bold) currents and winding (not bold) currents due to the load. The small dashed lines are due to the ground fault. The ground fault causes a little more current to be added to the winding current I1, in ratio with the winding ratio, but at a phase determined by the nature of the fault (drawn here as in phase with the load). The IA and IC line currents compensate - one simply adds the extra winding current and one subtracts.

Of course, the magnitude change in primary line currents could be anything depending on the relative phase of the load current to the fault current. This diagram shows the result for a purely resistive load and fault. But the relative magnitude change is indeed 10A x 2.4/12.47 = 1.92A. Sorry for doubting those that were right all along!

 

[sibeen]

*face palm*

Yeah, LiteYear, I had it right the first time.

As to the sign convention *cough* I used an old diagram I had lying around on turbocad. Used that, wrote the equations, noticed I had the currents the wrong way around, and thought, bugger...no-one will notice that