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current in 3 phase cap's

current in 3 phase cap's

current in 3 phase cap's

(OP)
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

RE: current in 3 phase cap's

The MVAR rating for the cap bank is based on operation at the nameplate voltage and frequency.  So rated current would be the kVAR rating divided by the kV rating (line-to-line voltage) divided by the square root of 3 for a three-phase bank.

But your current will be slightly different due to the series reactors.  If you need a precise value, you will need to compute the capacitive reactance (from voltage and current above) and then create an equivalent circuit and solve that.

RE: current in 3 phase cap's

And if you are actually sizing the feeder for the cap bank, there will be code requirements that require the feeder to be rated for more than the rated current of the cap bank.  Check the NEC or your local code.  

RE: current in 3 phase cap's

(OP)
thank you thank you thank you
very much

RE: current in 3 phase cap's

current inside capacitor bank doesnt depend on reactor size, and equals 384,000kVAR / 1.73 / 230 = 960A. The reactor size affects total current fed to the capacitor|reactor assembly.

RE: current in 3 phase cap's

mykh,

Your calculation of current is correct if the voltage at the capacitor terminals is exactly equal to the nameplate voltage rating of the capacitors.  The cap bank is a constant impedance load.  As the voltage changes, so does the current (and the kVARs vary as the square of the voltage.)

If rated voltage is applied to the input **series** reactor, there will be voltage across the reactor that has to be accounted for if an exact solution is desired.  

 

RE: current in 3 phase cap's

I concur with dpc's last post and add that the voltages across the capacitors and the reactor will be 180 deg. out of phase. The voltage across the capacitors can be higher than line voltage, depending on the values of the components. With the high ratio between the KVAR capacitive and KVAR inductive there will probably not be a significant difference but best to wait for another post or two.
Be patient, someone will probably do the calculations for you. I haven't done these calculations for so many years that I'm afraid I may make some silly mistake.

If this is an assembly, it's worth a call to the manufacturer to ask for their recomendations.
yours

RE: current in 3 phase cap's

I would be careful if there is any harmonic distortion of the voltage.  If my calculations are correct, there is a resonance point at about the 11th harmonic.  Any 11th harmonic distortion could cause a high 11th harmonic current through the series combination and a high voltage across the capacitors.  It is essentially a tuned harmonic filter.  Filter capacitors have to be designed especially for this service.

RE: current in 3 phase cap's

I agree with Jghirst. The reactor shall dampen any inrush frequency currents and also sink 11th Harm.

RE: current in 3 phase cap's

current with reactor 972 amps
current without reactor 964 amps
In comparison to other effects such as voltage change, harmonics and capacitor tolerance, the reactor does not affect current much.

However, do be aware to include the reactor for calculations on a voltage differential scheme.  We have a 115 kV bank that can be either 45, 67 or 112 MVAR with a 5 mH reactor.  When it is switched from one size to another the drop in current flow through the reactor will cause a lower the voltage measured across the midpoint of the capacitor bank.  Going from memory, it seems like the voltage change is about 1/4th that of a blown fuse.  Not enough to cause a misoperation by itself, but it does eat into the safety factor.

RE: current in 3 phase cap's

(OP)
thank you all so much for all the information.
it has help a bunch.

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?

RE: current in 3 phase cap's

Well bacon4life,

I think u meant without reactor 972 Amps and 964 otherwise.

I think dropped voltage across reactor lowers the Mvar capacity of the bank.

THX

Patrick7: In absence of any other information i would think that could be a Surge Arrester. Please check the manual, that should surely have respective data

RE: current in 3 phase cap's

Bacon4life has got it right;

973A with reactor
964A without reactor.

It looks like the capacitor current and voltage both rise just under 1% due to the reactor. And, the others are right in that the reactor does create a trap at about the 10.2nd harmonic.

RE: current in 3 phase cap's

I dont seem to understand this voltage rise due to reactor in series. Can u please explain elaborately. I thought that the Capacitor current was completely voltage dependent.

Please explain.

Thanks

RE: current in 3 phase cap's

Draw the equivalent circuit.  There is an inductive reactance in series with a capacitive reactance along with the unavoidable series resistance of the conductors.  Each circuit element produces its own voltage drop.  

The voltage seen by the capacitor is the phasor sum of the source voltage and the voltage drops across the circuit resistance and the circuit inductance.  Change the inductance and the voltage across the capacitor will change.  

RE: current in 3 phase cap's

To find the current through the series combination, you add the impedances of the two elements.  The inductive reactance is positive (2·pi·f·L)j and the capacitive reactance is negative -1j/(2·pi·f·L).  The series combination (neglecting resistance) will be zero ohms if f=1/[2·pi·sqrt(L·C)] and the current will be infinite.  This is the resonant point.  Since the voltage across either element is I·X, the voltage is also infinite (neglecting resistance which will keep the current finite).

RE: current in 3 phase cap's

(OP)
could you please describe the current through the reactor calk?
thank you

RE: current in 3 phase cap's

I concur with jqhrist up to the last sentence.
The current is not infinite, but is limited by the resisance of the inductor and the impedance of the source.
As the reactor resistance is very low, the current will be very high to the point that it is limited by the impedance of the source. In many instances, you may be able to assume the reactor impedance is zero with no serious errors in calculating the current limiting by the source impedance.
A small inductor in an electronic circuit with a fairly high resistance will have a very finite current and the resistance must be included in the calculations.
yours

RE: current in 3 phase cap's

Look at jghrist's post.

The capacitive reactance is a negative number, call it -jX.

The inductive reactance of a tuning or detuning reactor is a positive number, call it jY, with an impedance (absolute value) lower than -jX.

The capacitor and reactor are in series so they add together, -jX + jY = -jZ. This new -jZ is a smaller impedance than the capacitor impendance alone so the series L-C circuit will draw more current than the C circuit.

More current through the same -jX capacitor impedance means there must be more voltage across the capacitor.

I hope this makes sense.

RE: current in 3 phase cap's

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

Thanks

RE: current in 3 phase cap's

(OP)
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

RE: current in 3 phase cap's

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


RE: current in 3 phase cap's

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

XC=-j137.8·(60/f)
XL=j1.32·(f/60)

At f/60=10.1951
XC=-j13.46
XL=j13.46
series combination is 0 ohms

RE: current in 3 phase cap's

(OP)
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

RE: current in 3 phase cap's

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.

RE: current in 3 phase cap's

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.

RE: current in 3 phase cap's

(OP)
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.
 

RE: current in 3 phase cap's

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

RE: current in 3 phase cap's

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.

RE: current in 3 phase cap's

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.

RE: current in 3 phase cap's

(OP)
Does anyone have those formulas handy?

RE: current in 3 phase cap's

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?

RE: current in 3 phase cap's

(OP)
thank you for all  you most valuable information and help

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