Unbalanced Transformer Fault Currents 2

[cbark]

We had a fault on a feeder off the the secondary of a Delta-Wye transformer that caused the primary overcurrent relays to operate. The relays that operated on the primary are electromechanical relays, so they don't provide any indication as to the fault current values. The secondary relays are SEL and had event captures so we know what the secondary currents were during the fault. I'm trying to determine what the primary currents were based on the secondary fault currents. Below is the system information and fault data.

Transformer Information:
Primary: 138 kV - Delta
Secondary: 13.8 kV - Wye Resistance Grounded (400 A)
Rating: 30 MVA
Z% = 8.2%

Fault Currents on the Secondary:
Ia = 11392 @ 180.8 degrees
Ib = 10411 @ 6.4 degrees
Ic = 1438.6 @ 316.2 degrees

How do I determine what the primary currents are for each phase?

 

[davidbeach]

You have the currents in each secondary winding. The primary winding currents are scaled by the transformer ratio. At each corner of the delta, the primary phase current into the corner plus the two winding currents into the corner vector sum to zero. Hint: one of the winding currents is into the corner and the other is out. Draw it out.

 

[cbark]

That's what I am having problems with. I know that the primary currents should be the secondary currents through the transformer ratio, but I can't figure out the vectors. I want to go through the numbers so that I can also look at other faults and determine what the currents should be.

 

[waross]

Secondary A phase current divided by 17.3 (winding ratio, not transformer ratio) should be the current in the A-B winding of the primary. Did I get that right, David?
Can you take it from there?

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

 

[sibeen]

Knowing whether the transformer is a Dy1 or Dy11 (or other) might also be useful information.

 

[cbark]

Dy1

 

[sibeen]

I get:

IA = 1259A 183.4deg

IB = 552A 13deg

IC = 720A -3.8deg

But I've probably stuffed it somewhere

 

[7anoter4]

It is correct sibeen [neglecting transformer impedance of course].

 

[sibeen]

cbark, I've shown my working in the attached file.

 

[cbark]

Thank you sibeen and 7anoter4. I worked on this over the weekend and I got the same numbers.

Thank you for the help.

 

[davidbeach]

I get numbers very similar to what sibeen got.

 

[ijl]

The currents can be solved without any manual calculations (well, almost), with a suitable power system analysis program. See the attached file.

 

[sibeen]

ijl, I'm sorry but I absolutely hate this sort of thing. It gives the person "working out" the problem no idea on how it is solved. It gives no insight into the problems parameters. It's something that could be performed by anyone with about five minutes training.

If I'm stuck on a site in the middle of the night, with a pen and paper handy, I can work through this quite simple problem. A simple piece of circuit analysis shouldn't require the use of a computer program, unless we're talking the old school meaning of computer and that means me, and the little bit of grey matter I have stuck between my ears.

I'll admit to signs of laziness in that I used a maths program to do the 'complex arithmetic' for me, but to even write the simple equations I had to have a basic understanding at what I was looking at, perchance an understanding of Mr Kirchoff's current law.

Maybe it is my old and curmudgeonly ways but I'm often vexed by dumbing down that can occur in the use of programs in a case such as this. I was a technician well before I became an engineer and this would have been a second year problem in trade school.


*rant off*

 

[davidbeach]

Hear! Hear!