current in 3 phase cap's 1

[patrick7]

i need some help calculating the current in a 3 phase cap bank
the circuit is 230 kv with a three phase reactor in series with a 3 phase cap. the reactor is on the source side of the cap
reactor size is 2700 kva 1500 amp 3.2 mh
cap size ia 384 mvar
thank you

 

[amps21]

I meticulously agree, i was mislead somehow, but when put on paper, it proved.

Thanks

 

[patrick7]

I understand the calculation of the cap alone. and I come up with the same value of current. but for some reason I am at a loss for the reactor/cap calk. I really would like to run through the calculation with the figures to come up with the current. I follow better with numbers in the theory. Could someone run through the figures with me? Thank you

 

[LionelHutz]

The capacitor impendance is -j*V^2/kVAR = -j137.8 ohms

The reactor impedance is j*2*pi*60*3.2mH = j1.32 ohms

The series combination is -j138 + j1.32 = -j136.5 ohms

The new current is then 230kV/sqrt(3)/-j136.5 = 972 amps

 

[jghrist]

And since capacitor impedance is inversely proportional to frequency and reactor impedance is directly proportional to frequency:

X[sub]C[/sub]=-j137.8·(60/f)
X[sub]L[/sub]=j1.32·(f/60)

At f/60=10.1951
X[sub]C[/sub]=-j13.46
X[sub]L[/sub]=j13.46
series combination is 0 ohms

 

[patrick7]

thank you very much.
the numbers i am getting are as follows.

cap current w/o reactor = 384,000,000/(230,000*sqrt3)
which is 963.92amps

then the cap imp is
cap imp. = -j(230,000)(230,000)/384,000,000
which is -j137.76 ohms

then the reactor imp is
reac. imp = j 2*3.14159*60*3.2e-3 = 1.206 ohms

then the current through the series combination is
cap imp + reac. imp = -j137.76+j1.206
which is -j136.55 ohms

so the current would be
230,000/(1.73)(-j136.55) = 972.4 amps

then setting the cap and reactor imped equal to each other and solving for the freq. yields a resonant freq. of 641.26
or the 10.687 harmonic.

correct?

thank you all sooooo much

 

[LionelHutz]

Yes, that looks correct. I messed up a little and used 3.5mH in the calculation even though I wrote 3.2mH in the formula above.

 

[Jensgisla]

Patrick7:"...there is another question that i have.
we have a 230 kv abb pmi typer circuit breaker.
the drawings show what appears to be an additional
bushing bolted alongside of the breaker bushing.
any information on what it really is?"

This is a capacitor. It helps the arc suppressing by by-passing the high frequency. If you have these connected on both breaker poles (assuming a Y-pole breaker), you will have some voltage flowing over open CB. Remember to open disconnectors before grounding.

 

[patrick7]

Well here we go! I hope that this discussion will spark some interest and benefitial information exchange. I am in the process of writing a new standard for our company. The new standard deals with dampening cable inside horizontal Al. bus in electric substations.

First off here is a list of the current sizes that we have as min. diameter.
bus size min size of acsr
2 266.8
2.5 266.8
3 266.8
3.5 397.5
4 795
5 1431
6 1590


There are several questions that come up. Your professional opinions and career experiences will be of great help.

First, Do we have to use ACSR? Or will AAC work?
What about the newer conductors available ACCC?

Second, Are the above min. sizes ok? and what is the max.
size?

Third, Is the dampening characteristic of the conductor
within the bus a function of the size ie the cross
sectional area, or weight of the conductor or both?

Fourth, How long is the max bus length that does not
require dampening?

Thank you for all your input and help.

 

[patrick7]

Another question that i have come accross is the inrush current. How would you calculate the inrush current for this combinations.
thank you

 

[amps21]

I think although u can calculate the inrush by hand using formulaes given in IEEE 39.99-.... the most realistic switching transients can only be deciphered from EMTP modelling.

 

[JL123]

It is interesting to know how to calculate the LC filter.

Any kowledge of how to calculate resistors applied in the LCR filter circuit on a damping network and what purpose it would serve.

 

[patrick7]

Does anyone have those formulas handy?

 

[amps21]

try

http://gilbertelectrical.com/library/document_library.htm

 

[JL123]

Thank you Amps21 for the valuable informaion. However, it does not show the application of the resistors in the circuit. My understanding is that when a LC filter is tuned to a harmonic frequency, in my case 3rd(150Hz), there will be a flow of current during other harmonic frequencies. To limit this resistors are applied in parallel to the filter. How do you determine the value of the resister?

 

[patrick7]

thank you for all you most valuable information and help