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]

Davidbeach. Manually calculation up to now didn't get the clear tips, still in doubt, if you cant do go away but dont be negative

 

[afzumannu]

Dear @Argotier and @Anoter can you attach the Excel sheet which you made, now results are coming correctly. thanx

 

[7anoter4]

I'll try

 

[afzumannu]

First of all thanks to all Guys specially @ Anoter, @ beyond @ argotier.
@ beyond calculated the neutral current it was real capacitance parameters i given, after that by mistake i given the capacitance parameters which was typing error and calculated by Mr. anoter and the results is 4.007
actually you guys work hard and only one step remaining to find. based on later parameters which Anoter did the results in Matlab simulink is 4.451 while Mr. Anoter calculated the 4.007
In Matlab simulink there is default source Resistance and source Inductance values. source resistance 0.8929Ω and inductance 16.58mH, But if i put the source inductance value 0 then the Results coming correct 4.007. but the Results are actually matching with Matlab simulink one, we have to use this Source inductance value in Manual calculation then the answer will be fine, waiting from Mr. Anoter Mr. Argotier for reply.

 

[afzumannu]

@ Mr. anoter i check excel file and understand lot of things, but how you define the Equation for IR and IY please clarify, only this i have confusion can you do only these two step by step.

 

[7anoter4]

Sorry ! This is a wrong file [not finished]The new one is more clear, I hope.
If you remark in the right upside corner are the source impedances. If you put 0 instead this you’ll get the
former solution-without source impedance.
The calculation is done using determinants system-one general and one for IR and IY. Finally IR=DIR/DTG
and IY=DIY/DTG.

 

[afzumannu]

@ ANOTER,
IR=[VRY*(ZY+ZB)+VYB*ZY)]/[ZR*(ZY+ZB)+ZY*ZB]
IY=[ZR*VYB-ZB*VRY]/[ZR*(ZY+ZB)+ZY*ZB]
How you make this equation please suggest me step by step.
how this source resistance and source Inductance came ?( 0.8929 & 16.58 mH), From where came please give me Idea, is it default for all the circuits?

 

[7anoter4]

This new file may explain better:

 

[afzumannu]

@ Anoter, now i understand little bit as i am working as a technician it is enough for me, i understand all before but don't have any idea about your equations, now have some idea at least, your typing mistake after Rearrange there should VYB. you did not specify me that source inductance and resistance from where these came? i am still behind this source inductance default value.
@Anoter one circuit with Resistance which you are trying to make in sheet-3 is it possible to calculate it? let me know. thanx

 

[7anoter4]

Thank you afzumannu.
1)The source impedance it is from your post of 27 May 20 14:04
2)You are right.My mistake.It has to be VYB instead of VRB.I got a new laptop with a new office [Office 2019]lately and I am still learning it. It is very different from my old 2010 edition.
3)It was not my intention to attach the sheet no.2 and 3 [Still I used an old file!]
4)However, I will try to calculate the new case. It will take some time, I think.

 

[afzumannu]

@Anoter,
i know that i posted the source impedance, but why it is default and also from where it comes? it was in Matlab default not only in this circuit but both circuit Mr. beyond Put the same default Impedance Value, How i will know that default Value source? anybody let me know.. thanx

 

[7anoter4]

I am not sure what is considered source impedance. In my opinion it is a 25 MVA 10% power transformer [66/33 kV, 440 A].However the resistance is too high that means 5 km OHD Line may be included.

 

[afzumannu]

@ ANOTER,
Actually it is default for all the circuit in Matlab as Mr. Beyond did, our power Transformer is 120 MVA 66/33 KV. it is ( Source Impedance ) required to put in Formula then only the Result Values are closer to real Value.
Are you Calculating circuit with resistance

 

[7anoter4]

The problem is more complicated then the former case since there are 14 elements per one pair of phases instead of 6. So, first I have to transform 4 stars into 4 triangles and then I have to do a lot of other transformations.

 

[7anoter4]

I'll try it again:

 

[7anoter4]

New laptop. Sorry..

 

[afzumannu]

Dear Anoter, Thanx for your very hard work, first 2 file nothing is there, but last file some formulas and diagram i can see, what is result of neutral current, it seems more element(14 no's), you didn't calculated it, is it in progress? sure hard work always fruitful. i am eagerly waiting for your calculation. thanx

 

[7anoter4]

The result-ZSource included-
IR= 449 A ; 56.4 degrees
IY= 450 A ;-63.6 degrees
IB= 450 A ; 177 degrees
Io= 0.635 A; -81 degrees
The result-without ZSource
IR= 405.6 A ; 58 degrees
IY= 406.2 A ;-62 degrees
IB= 406.2 A ; 178 degrees
Io= 0.762 A; -89 degrees

 

[afzumannu]

@ Anoter,
Thank you for your continuous help, the Result is wrong, but I think it is my mistake, today i went to site and checked all the connections and diagram also, I found that it is L= 41.23mH all phases, before i given you 4.062mH is wrong, I am very sorry for my mistake, Please update with new Value.

 

[afzumannu]

@Anoter, rated value is 41.23mH, but measured value is 40.62mH in the site.
The result should be in between 0.20 to 0.25 Amp., if you prepared the excel file, please share it