Hi,

We will be installing a de-tuned power factor correction system and I had a few questions regarding the calculation of the line reactor. Our distribution system is 600V, 60Hz and each step for the power factor correction is 60KVAR for a total of 180KVAR on one distribution pannel and 120KVAR on the other one (two 1500KVA TX in parallel, Z=5.83%). Before the reactor is installed, we're able to determine that the value in Farad for the capacitor is : C = KVAR / 2*pi*f* V^2 = 442.1 uF. The supplier we are buying from did the quality analysis over a week period and from the harmonic profile decided to tuned to the 3.78 th harmonic (226.8Hz), which is a value that's pretty standard I guess. After the line reactor is added in the system, the capacitor value is dropped to 411.2 uF and the reactor value is 1.1979mH (7% reactor). The quality factor Q being used in all of this is 40. Now what I am wondering is how come the capacitor value changes after the line reactor is added and how is it being determined? And what values are imposed in all of this? Is the quality factor something determined from the harmonic profile of our load and how is it being applied to our calculations? Any information, formulas and theory on how to make sense out of all of this would be greatly appreciated! Thank you very much for your help!!

# Power factor line Reactor calculation?

Started by Kenny, Oct 22 2005 10:59 PM

3 replies to this topic

### #1

Posted 22 October 2005 - 10:59 PM

### #2

Posted 23 October 2005 - 06:47 AM

Hello Kenny

Welcome top the forum.

Adding detuning reactors inseries with the power factor correction capacitors will reduce the effective capacitance. The reduction in effective capacitance is frequency dependent. At higher frequencies, the reduction is high, in fact the inductive reactance becomes higher than the capacitive and so the inductive reactance reduces the harmonid current flowing into the capacitors. - this is the reason for fitting it!!

Detuning reactors are typpically rated at 5% or 7%. A 5% reactor will reduce the effective capacitive KVAR by 5%, and similarly a 7% reactor will reduce the effecitve capacitive reactance by 7%.

As you reduce the reactance, the reduction in harmonic current through the capacitors is reduced, the reactors are less effective.

When you connect inductance in series with capacitance, you create a tuned circuit. If the Q of the circuit is high, the bandwidth of the tuned circuit is narrow. If the Q is low, the bandwidth of the tuned circuit is wide.

A wide bandwith will allow resonance response to harmonics well off the tuned frequency. This can be undesirable if it is too close to say the third harmonic.

To select a detuning reactor, determine the KVAR of the circuit being detuned. Note, one reactor per unswitched capacitor circuit. If you have an automatic correction unit with six stages and four capacitors per stage, you will need six reactors, one per stage, sized to the total KVAR per stage. You must not alter the capacitive reactance on the output of a detuning reactor.

The greater the harmonics on your supply, the higher the value of inductive reactance you need to reduce the harmonic currents to a suitable level. - High harmonics, use 7% reactor, low harmonics use 5% reactor.

Best regards,

Welcome top the forum.

Adding detuning reactors inseries with the power factor correction capacitors will reduce the effective capacitance. The reduction in effective capacitance is frequency dependent. At higher frequencies, the reduction is high, in fact the inductive reactance becomes higher than the capacitive and so the inductive reactance reduces the harmonid current flowing into the capacitors. - this is the reason for fitting it!!

Detuning reactors are typpically rated at 5% or 7%. A 5% reactor will reduce the effective capacitive KVAR by 5%, and similarly a 7% reactor will reduce the effecitve capacitive reactance by 7%.

As you reduce the reactance, the reduction in harmonic current through the capacitors is reduced, the reactors are less effective.

When you connect inductance in series with capacitance, you create a tuned circuit. If the Q of the circuit is high, the bandwidth of the tuned circuit is narrow. If the Q is low, the bandwidth of the tuned circuit is wide.

A wide bandwith will allow resonance response to harmonics well off the tuned frequency. This can be undesirable if it is too close to say the third harmonic.

To select a detuning reactor, determine the KVAR of the circuit being detuned. Note, one reactor per unswitched capacitor circuit. If you have an automatic correction unit with six stages and four capacitors per stage, you will need six reactors, one per stage, sized to the total KVAR per stage. You must not alter the capacitive reactance on the output of a detuning reactor.

The greater the harmonics on your supply, the higher the value of inductive reactance you need to reduce the harmonic currents to a suitable level. - High harmonics, use 7% reactor, low harmonics use 5% reactor.

Best regards,

Mark Empson | administrator

Skype Contact = markempson | phone +64 274 363 067

LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | LMP Software | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | Soft Starters

Skype Contact = markempson | phone +64 274 363 067

LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | LMP Software | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | Soft Starters

### #3

Posted 23 October 2005 - 10:23 AM

Thank you very much for your help! This is really appreciated!!!

Thanks

Thanks

### #4

Posted 18 September 2010 - 11:51 AM

Hi,

I think the formula for calculating farad will go like this : C=VAR/(V*V*2*pi*f) . Here VAR (KVAR*1000) value will be used.

As per my calculation, C = (120*1000)/(600*600*2*22/7*60) = 0.8838mF for 120kvar bank.

Thanks.

Asif Mansoori

I think the formula for calculating farad will go like this : C=VAR/(V*V*2*pi*f) . Here VAR (KVAR*1000) value will be used.

As per my calculation, C = (120*1000)/(600*600*2*22/7*60) = 0.8838mF for 120kvar bank.

Thanks.

Asif Mansoori

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