Primary Side Fuse Blown Transformer

[MeinKaun11]

Hi,

I have exhausted all resources on the internet and books that could help me understand the phenomena in the subject above. I owe this to may be my poor understanding of phasors specifically related to primary and secondary winding of transformers.
And I need your help.

I wanted to program a digital relay to detect a High side primary fuse detection in a Delta Start medium voltage transformer rated at 13.8 kv/4.16 KV, 7.5 MVA. SEL has application guides which suggest to use 27 function and I get it.

But here lies the question, I have tried my best to transfer voltages primary to secondary that and how does the mathematics works in vein. I have attached a picture and please explain how simply this can be explained.

On Primary side, if B phase is lost: Vca will be 1 p.u. while Vab and Vbc will be .5 P.U. and It is also mentioned that Vbc and Vab will be opposite of each other. This is what bugs me to satisfy kirchoofd voltage law Vca= Vab + Vbc, so how Van and Vbc can be opposite to each other.

 

[waross]

I'll give you a hint on how to visualize this.
Vac remains at the same voltage.
Vabc is equivalent to Vac, and Vab + Vbc = Vac.
Vab rotates 60 degrees clockwise.
Vbc rotates 60 degrees counterclockwise.
Now look at the secondary wye diagram and rotate the vectors that correspond to Vab primary and Vbc primary in the same direction and amount.

The secondary will be a straight line Vcn[ab]
Vcn = 1pu
Vna in parallel with Vnb = 0.5 pu.

It is also mentioned that Vbc and Vab will be opposite of each other

Vab and Vbc implies primary. Secondary will be Van and Vbn
If Vab and Vbc were opposite the resultant would be zero.
The voltage from the healthy phase, Vac is applied across Vabc.
They had better be in the same direction.
However the secondarys, Van and Vbn are opposite to Vcn as you will see when you rotate the secondary vectors. (Anchored by the "n" connection.)

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

 

[MeinKaun11]

Hi Waross,
I just want to get some of my facts straight about 3 phase transformers so that I am not wasting my time going overboard in trying to get something in to head for which I am not ready.
3 Phase. 10 MVA 27.6 kV/4.16 kV transformer
Primary A phase voltages are
VA= 16< 0 degrees
VB= 16<-120 degrees
Vc=16<+120 degrees
Now, if I have to transfer these voltages directly to secondary, how would I do it:
Va= 2.4<0 degrees
Vb=2.4<-120 degree
Vc=2.4<+120 degrees
Vab= Va-Vb= Vab<30 degrees Now what does this mean in reference to VAB(primary)
Here is the question, when we talk about angular displacement about Delta to Star conversion or Star to Delta conversion, what does that actually means: Here is what I know:
VAB(primary-Delta) leads Vab (Secondary-Star) by 30 degress- Did I get this right or it could either or- Leading or Lagging and has to be specified based on the design.
I would appreciate your input here then I would trouble you with explaining further on my post on High side fuse blown application.

 

[MeinKaun11]

Hi Waross
I have done the calculation, please let me know, if I got it right.

 

[arrehman1]

@MeinKaun11

Take a look at this video tutorial:

http://gpac.link/1zOr4XD

 

[stevenal]

I think 0.5 pu works okay if the remaining load is perfectly balanced. When the B fuse opens the B corner of the delta will move to a point on the line between A and C, but will float along that line (like an open neutral on split phase single phase) based on the balance of the remaining load on the two associated wye phases. Login required

 

[VTer]

Attached is what I came up with, assuming no load on transformer.

"Throughout space there is energy. Is this energy static or kinetic! If static our hopes are in vain; if kinetic ù and this we know it is, for certain ù then it is a mere question of time when men will succeed in attaching their machinery to the very wheelwork of nature". û Nikola Tesla

 

[MeinKaun11]

Hi Vter
thanks for putting the effort but based on the literature two line to line voltages are 1.5 pu that is 87 % of 1.73 pu.

This is what I Am trying to figure out how ?

 

[stevenal]

Start with 1 pu line to line. With that as your base, your normal l-n is 1 pu/sqrt(3). Now cut one of your normal l-n phasors in half and rotate it 60 degrees so it opposes the healthy phase. Your new line to line voltage is 1 pu/sqrt(3) +0.5 pu/sqrt(3) = 0.87 pu. The other l-n phasor is also cut in half and rotated the other direction so that the voltage between the two rotated phasors is 0 pu.

If you look at it using l-n as your base, the healthy phase stays at 1 pu while the affected phases go to 0.5 pu.

I see too much mixing of line to line and line to neutral voltages above. Pick one base and stick with it.

 

[VTer]

What literature are you referring and what two line to line voltages?

"Throughout space there is energy. Is this energy static or kinetic! If static our hopes are in vain; if kinetic ù and this we know it is, for certain ù then it is a mere question of time when men will succeed in attaching their machinery to the very wheelwork of nature". û Nikola Tesla

 

[MeinKaun11]

Thanks gentleman I got it.
besides mixing line to line and line to neutral I was mixing phasor as well In my head.

I will create a calc sheet and post it here.

This forum is the best.

 

[MeinKaun11]

Just also curious to know how would secondary behave when only one phase is applied to primary of transformer ?

Will any voltage be reflected ?

 

[VTer]

Attached is the revised sketch. Secondary voltages are on the line-to-neutral V base.

If two phases are open and only one phase is available on the primary of the delta, you would have 0 volts on secondary - there is no complete circuit on primary.

"Throughout space there is energy. Is this energy static or kinetic! If static our hopes are in vain; if kinetic ù and this we know it is, for certain ù then it is a mere question of time when men will succeed in attaching their machinery to the very wheelwork of nature". û Nikola Tesla