Protection Riddle No.13 - Setting of earth fault relay
If the circuit has a lot of unbalance load, then we must concern the continuous rating of transformer in addition of its short time withstand currents. The continuous rating power of mentioned grounding transformer is 100 KVA.
The primary current setting of related earth fault relay that installed in neutral point of E.T. is 8 Amp.
This set point current is chosen in base of E.T. continuous rating current and following calculations:

It =100/ (20x1.73) = 2.89 A

In= 3x It = 8.67 A

But the earth fault current calculation and analyses dictate other primary current setting (12.5 A).

Which one current setting is correct? Which one do you choose?

#1
Wed, May 21, 2008 - 07:22
For setting of relay 8.67 Amp. is correct . In fact 12.5 Amp. consists of  two components resistive and capacitive .
12.5 Amp.  is  8.67 *1.44 (root of 2) .
8.67 passes the earthing resistance or earthing transformer and initiates the relay  and this is correct for setting of relay.

#2
Wed, May 21, 2008 - 08:50
I think we must think more and more.

#3
Wed, May 21, 2008 - 10:22
Of course , we think more. In 20 KV system we don't have zero phase  sequence current as any result of unbalace load . Any unbalance load appears as negative phase sequence  , and NPS current doesn't pass  zigzag earthing transformer. This is the question , How 100 KVA or 8.67 Amp.  is achieved ?If neutral point of zigzag earthing transformer is grounded solidly , so the earth fault current won't be 8.67 Amp. It will be much higher. and if it has a secondary wye winding the earth fault depends on probable resistance (if any) and impedance voltage . the earthing scheme  of your idea isn't   clear . It's not clear why the continuous rating is 100 KVA.

#4
Mon, June 23, 2008 - 12:10
Unfortunately, nowadays with our world information transmission system (internet), we usually neglect the basic conceptions and deep thinking about new problems.
With more deep thinking, we can find that the mentioned evident equation (S = 1.73 VI) is not correct for zig-zag earthing transformers.
Indeed, each leg of a grounding transformer carries one-third of the neutral current and has line-to-neutral voltage. So in a grounded wye – delta transformer, the total power rating including all three phases is the neutral current times
The line-to-ground voltage:
S= 1.73 VI
But a zig-zag transformer is more efficient than a grounded wye – delta transformer.
In a zig-zag, each winding has less than the line-to-ground voltage,
by a factor of 1.73, so the bank may be rated lower:
S = 1.73/1.7.3 VI = VI
Therefore, the continuous rating current of E/T is 5 A and rating neutral current is 15A, and we can choose 12.5A as correct earth fault primary current setting .

#5
Mon, June 23, 2008 - 17:08
20000/1.73/1.73 NAMELY 6666.6 V IS THE VOLTAGE OVER HALF WINDNG ' NOT TWO HALVES WIDINGS. IN THIS CASE THE APPARENT POWER OF HALF WINDING IS 16.66 KVA. THE RELATIONSHIP OF S=1.73 VI CAN NOT BE VIOLATED DUE TO ZIGZAG CONFIGURATION . THIS RELATIONSHIP IS STILL VALID . SO THE CURRENT PASSING THE NEUTRAL IS 8.67 Amp. NOT 15 Amp.Sorry I  can't agree to Mr. Hamid's idea.

#6
Sat, June 28, 2008 - 16:54
If we assume , the voltage over winding is half and on the other hand the flux density is half , it means the number of winding is fix and hasn't been reduced.
U=4.44*N*(Φ)*F
In this case we have used 2.3 times of copper in comparison with star winding . On the one hand , we state that the flux density is 1.7 T divided by two , while , this isn't feasible and economical . If we assume the flux density is 1.7 T (as it is usual) , thus the current in one leg can't be 5 AMP. It is 2.88 Amp. And if flux density is 0.85 T , so, it isn't surprising if it can transfer 1.73 S.

#7
Sun, June 29, 2008 - 17:48
Mr. AMIR KHASHAYAR, excuse me I think we must go back to basic conseption of electrical transformer.

#8
Mon, June 30, 2008 - 06:48
Both of us should be referred again not me alone. Flux density is very important in economical design and construction of transformer .Normally this is 1.7 T . when you say the voltage over zigzag is divided by 1.73 with the same number of turns, in fact you are saying that  flux is also divided by 1.73 . in this case your transformer hasn't been designed economically . However it can transfer 1.73 times its rated apparent power . I think both of us must come back university.

#9
Mon, June 30, 2008 - 08:37
Yes, I said "we must go back to basic".