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Detuning Reactors


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#1 submonkey

submonkey

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Posted 09 November 2007 - 12:51 PM

Greetings,

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
reactive power.

3. The apparent kVAr is higher!

Thanks,
Submonkey

#2 marke

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Posted 12 November 2007 - 04:08 AM

Hello submonkey

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,
Best regards,




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