Neutral Current calculation in two star connection in LC Filter circuit 2

[afzumannu]

We have 3 phase LC Filter circuit 21.66mH per Phase also the Capacitance value in all 6 Branches which are Xc1 = 27.2231µF, Xc2 = 27.2499, Xc3 = 26.6247 ( one star group), Xc4 = 26.6499,Xc5 = 26.7749, Xc6 = 26.7747(2nd star group), i want to calculate the Neutral current for unbalance protection. supply voltage is 33 KV. Please suggest me the correct formula for calculation. please see the attached Diagram.

 

[afzumannu]

@ Anoter, below is the real SLD which i taken from Drawing, Unbalance current should be in between 0.20 to 0.25A

 

[7anoter4]

I am sorry afzumannu, but it is hard to do this calculation on excel. If the former task was 6 unknowns per phase on the new one are 14. However there was 22 parameters and now are 100.It is very difficult to find a location of an error. It was very interesting task. As argotier said it was merely "a personal crusade".

 

[afzumannu]

@ Beyond, dear you are true, I can imagine how it is hardwork for someone, I am also thankful for all the guys here are very good especially you.but if you can get any clue or solve it let me know and please do share. Thank you once again.

 

[afzumannu]

@ Beyond till now if you prepared any files regarding this circuit excel file, please do share, let me also study about it. thanx

 

[afzumannu]

Dear
@ Anoter please share the updated file related to Resistance circuit,how much you prepared will be Ok and appreciatable. Thanx

 

[7anoter4]

I'll try.

 

[7anoter4]

Sorry.Due to complexity of the program it is difficult to check the data.I found already that the new excel did not change the sign of the capacitive reactance and so the calculation went wrong from the beginning. You have to change the sign in the "complex(R,-X)".

 

[afzumannu]

Ok sir, please send the updated file..Thanx

 

[7anoter4]

The last..

 

[7anoter4]

If we neglect the resistances, the system is as in the previous case with only 6 parameters on a pair of phases. I calculated the main currents and the neutral current, and the differences are less than 1%.

 

[afzumannu]

@ thanks, please share the file..I want to see that.

 

[7anoter4]

The calculation for ZSource=0

 

[afzumannu]

Dear@Anoter and all the Engineers here, i was bit busy as same Filter Bank trouble in Plant, lot of capacitors blown, this LC Circuit it was unbalance 30% trouble time, but with the help of you ( Anoter) formula we Rearrange the capacitors and bring down to 1%, thanx once again to the entire Eng-tips guys specailly Anoter.
I checked the File for RLC circuit that You mentioned IoA & IoB, but in LC circuit it was only Io also from both side IoA & IoB the calculated value was Equal, But here RLC you are showing different Values i.e.= IoA = 1.2080347 , IoB = 1.2507006, Please clarify?

 

[7anoter4]

In order to simplify the user operation it is attached the program "after last”. Here you have not to do any other operation except change in "Data" sheet [red noted]-if you will need. The result is presented in "Filter R,L,C "sheet.
The new program consists of adding spreadsheets for complex determinants for each phase.

 

[afzumannu]

Dear@ Anoter,
I understand its same file and is for user understanding, but my Question is that RLC circuit why IoA & IoB?
in LC circuit it was Io only,if IoA and IoB also here it should be Equal if i am not wrong same like LC circuit. Finally I am looking for Io ( unbalance current ) in between star points which should be around 0.18 to 0.26 or +-.

 

[7anoter4]

You are right. I did not finish the job. We have to calculate the current passes from point A to point B: IAB=IoA+IoB. Then:
If ZSource=0 then:
If we consider the resistances: IAB=0.1157 A
If we neglect the resistances: IAB=0.03128 A
If ZSource=0.8929+5.208761i [as it was before]
If we consider the resistances: IAB=0.1354 A
If we neglect the resistances: IAB=0.03661 A

 

[7anoter4]

If the total current per phase is 500 A this current represents less than 2-3/10000.
This is the order of the calculation error.
It is not possible to calculate exactly a value that is of the order of magnitude of the calculation error. I am sorry.