Posted 09 November 2007 - 12:51 PM
The the webpage http://www.lmphotonics.com/pwrfact.htm states:
(under "Supply Harmonics")
> Adding the inductance in series with the cpacitors will reduce their
> effective capacitance at the supply frequency. Reducing the resonant
> or tuned frequency will reduce the the effective capacitance further.
(under "Detuning Reactors")
> Using detuning reactors results in a lower KVAR, so the
> capacitance will need to be increased for the same level of correction.
I'm not convinced! Please consider the reasoning below. I have used
'w' to represent system frequency in radians per second.
The detuned reactor is connected in series with the
capacitors. The impedance of the reactor is:
Zr = jwL
The impedance of the capacitor is:
Zc = 1/jwC = -j(1/wC)
The total impedance of the reactor and capacitor is:
Zt = jwL - j(1/wC)
Note that the effect of adding the detuned reactor is to
reduce the combined impedance, which in turn results
a higher current. The current still "looks" purely
capacitive, since (1/wC) is larger than wL.
I conclude that:
1. Adding a detuned reactor makes the capacitor bank
draw more capacitive current.
2. A higher capacitive current means a greater supply of
3. The apparent kVAr is higher!
Posted 12 November 2007 - 04:08 AM
Welcome to the forum.
Yes you are correct, the impedance is reduced and therefore the current is increased. This will of course increase the KVAR of the circuit. - I will correct the comment on the web page. Thank you for picking that up.
No one else has seen this!! It is always a concern that no matter how often you read something, you can miss the obvious.
Have a great day,
Mark Empson | administrator
Skype Contact = markempson | phone +64 274 363 067
LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | AuCom | Soft Starters
0 user(s) are reading this topic
0 members, 0 guests, 0 anonymous users