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current in 3 phase cap's 1
- Thread starter patrick7
- Start date
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.
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.
- Thread starter
- #4
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1
- #5
This thread could be useful:
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.
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.
- Moderator
- #8
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
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
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.
bacon4life
Electrical
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.
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.
- Thread starter
- #12
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?
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?
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
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
LionelHutz
Electrical
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.
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.
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.
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.
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).
- Thread starter
- #18
- Moderator
- #19
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
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
LionelHutz
Electrical
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.
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.
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