Jump to content



Recommended Posts



I am having problems understanding the principals of operation of a de-tuned filter for power factor correction in an industrial network polluted with harmonics? The fifth and the seventh are pretty high on the network I've checked out. Now a reactor of 7% was used and tuned to the 3.78 (228Hz) harmonic, below the fifth to reduce the effect of resonnance. Global compensation will be 300KVAR with 60KVAR steps. Now what I have done is found the equivalent impedance for one phase of the network in parallel with a 60KVAR bank and plotted it. Impedance will attain a maximum value when capacitive and inductive reactance are equal. Now when that happens, I can solve for frequency and find out that the resonnant frequency (when capacitive reactance from power factor unit is equal to inductive reactance of the load) is 440Hz, which matches the peak value of the curve plotted for impedance versus frequency. This is for one 60KVAR step. This is fine since the de-tuned frequency is 228Hz. When all 300KVAR are are on the load has changed and so has the impedance. Plotting the same data will result in a peak impedance occuring at roughly 200Hz? How can I make sense out of all this? When the load changes, so will the resonnant frequency of the network. A de-tuned reactor is suppose to be set at a frequency that is safely below the expected resonnance points so that it is still safe even with large load changes. Maybe my calculations are wrong? I don't know...


Thanks for your help!

Link to comment
Share on other sites

Hello Kenny


If you consider the case where no detuning reactors are fitted, then the resonant frequency will be dependant on the load and supply reactance, and the power factor correction capacitance. - note you need to consider the supply reactance as well as the load reactance as a shunt circuit. Also consider the resistive component of the supply that is in shunt with the reactance and dropping the Q to a very low figure.


When you connect the detuning reactors in series with the capacitors, the Q of the circuit comprising the cpacitor and the detuning reactor will be higher than the Q without the detuning reactor. The supply impedance will have some affect on the resonant frequency, but it will be small relative to the effect of the detuning reactor.


Best regards,

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
  • Create New...