Neutral Current calculation in two star connection in LC Filter circuit thread238-468572

[afzumannu]

thread238-468572,
Dear Anoter,
due to Affected by Covid-19 last year i stopped posting the reply of the Thread 238-468572,
can we remove the resistance from the attached circuit and Calculate the neutral current same as you did in previous circuit, in place of 12 microfarad values. can we make 6 after removing resistance and make series i.e. R=126.15µf & 44.7343434774307 µf can we calculate in series and get one value, so total 6 value of 3 phase and apply in existing circuit which you solved? waiting for your response.

 

[7anoter4]

The sketch [from your file] is clear [except the R to R connection ].
The phases are connected in star, at first, and then the neutrals are connected together.
Following the new sketch, the result it is:
Star 1 0.25918 A
Star 2 -0.1982 A
The grounding current[ the sum- if any]: 0.06098A
I still don't understand what about 6 in stead of 12µF and resistance removing.
Can you make another complete sketch where the required changes will be included?

 

[afzumannu]

@Anoter sketch is like that, I mentioned the 12 capacitance values Red and Blue colour, Resistance in the circuit, if we calculated in series Star1 R phase R=126.15µf & 44.7343434774307 µf then it will 33.0237242 µf, so for all it will become 6 Values only and we can put in Formula which we solved last year with 6 values as well as we can eliminated The resistance for making the circuit simple, then I0 value we can get? I think this will be ok, waiting for your response.

 

[7anoter4]

We will have to separate the two stars by calculating an equivalent impedance for the input impedance group.
To simplify the calculation we will consider the currents passing through the inductance as equal and then in the calculation of the voltage drop on this inductance we will take the double current and in the formula we will multiply by two the inductive reactance.
For instance ZinR1=R1*j(2*XL-XC1)/[R1+j(2*XL-XC1)]
Now we will complete with the second capacitive reactance on each phase and we will calculate the currents initially considering the sum of all currents =0.
IR1+IY1+IB1=IAo=0
IR2+IY2+IB2=IBo=0
Because the currents are not symmetrical, the neutral has a voltage difference from the neutral of the system VAo for star 1[ A] and VBo for star 2 .
We will have to increase or decrease IAo [keeping IBo=-IAo] until VAo becomes equal to VBo.

 

[7anoter4]

I'll try to upload the excel workbook

 

[afzumannu]

@ Anoter, I prepared excel workbook simply and Eliminate the Resistance, because I am looking Finally for I0 only between 2 star, and as per my excel worksheet you can understand i simplified the circuit and the result also coming nearby correct, last year you also send the file with this result. I think this is the correct calculation, only 6µf parameters used instead of 12. you can see the attached file and comment, I0 is 0.194288283.

 

[7anoter4]

This old excel workbook it is wrong. I don't know why I did so. I think I
did not got the problem then.

 

[afzumannu]

Last year you solved this problem with another circuit and it is correct calculation, it was matching with Matlab result, I am attaching the Last year same file.

 

[7anoter4]

I think in this old excel workbook were other sheets-or data- so it is difficult now to follow the calculations. In my opinion, we have to solve your problem in a simple way.
There are-in my opinion-4 ways :
1) involving all 6 currents in the same time-each one in the respective phase and star.
This way is a sophisticated 6 * 6 matrix requirement in complex numbers and I do not have such an excel template.
2)By treating each star individually using IAo and IBo [as negative sum of star currents] variation until VAo=VBo.
If we use only the absolute value of IAo we get 0.194 A. However, if we use the complex number the value is close to 0.254A.
3) Calculating the total currents per phase and then splitting the currents- according to impedances- for instance:
ZR[equivalent total impedance] and ZRA and ZRB for each star.
I introduced here an input impedance[ZinR1] considering the star A current flowing through the inductance as equal cu corresponding current of star B.
The ratio of these currents is about 0.998 [0.2% difference].
In this case IAo=-IBo=0.263 A.
4) neglecting the resistances and calculating the total current and splitting again according to remaining star impedance
In this case : IAo=-IBo=0.264 A

 

[afzumannu]

Dear Anoter, the Attached file is your last year complete file which you solved with another circuit and parameters and it was correct. same file I am using for this circuit and just eliminated the Resistance circuit and Put other parameters. hopefully it will work fine. please check the last year solved file.

 

[7anoter4]

I found an excel file that seems to be the last calculation in the thread238-468015/17 Apr 20. It is very complicate and the result is the following:
complex number absolute degrees how was calculated
IoA -0.53929605368603-1.13814943411676i 1.259454 -115.353 neglecting resistances
-0.586996721535968-1.04817141256148i 1.201344 -119.25 with resistances
IoB 0.508017835781004+1.13818875981324i 1.246417 65.94694 neglecting resistances
0.513366025973028+1.13739708198629i 1.247885 65.70789 with resistances
The difference from IoA [sum of all three currents at star1[A]-and IoB it is the natural calculation error due to the fact the way was too long.
If we take the old data-the new data are not the same- and put them in the simplified last excel file we get 1.25 A for both stars.
However, introducing the new data we get 0.194 A neglecting the resistances and 0.264A if we include the resistances also-I think you already got the excel workbook.

 

[7anoter4]

In order to be more "simplified" here attached the last one.