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Reactive Power Absorption by Series Reactors

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Tinh123

Electrical
Nov 16, 2006
27
Is there an equation which would explain the reactive power absorption of a series reactor with impedance X. Assume that the series reactor between point 1 and point 2. Thanks for any help.
 
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Use Ohm's law to determine the current. (Voltage divided by impedance. E/Z)
Use Ohm's law to determine the reactive component of the current. (Current times reactance. IX rather than the more familiar IR)
respectfully
 
Actually, the voltage used for that Ohms Law solution would have to be the voltage drop across the series reactor. If you could determine that, then you would have to multiply it by the reactive current, not the total current(x sqrt3 for 3-phase), to get the reactive power.

You should be able to determine the current flow through the reactor based on the downstream load (or the FLA of the reactor), and then use I^2(R + jX)=(P + Q), or just I^2X=Q, Q being Reactive Power.
 
I have heard quite a bit about series reactors on high voltage line were intentionally use to mitigate potential high fault current, but inadvertently, block (absorbs) excessive VARs in the system. How does this actually work? and by what magnitude (proportionally) to the impedance value of the Series reactor? Is it true that the voltage drop corresponding to the series reactor the same per unit factor as the impedance of the series reactor (i.e. a SR with .02 per unit will suggest a voltage drop of .02 p.u as the effect of this series reactor.) Thanks for the help.
 
It's just an inductance. Put it in the circuit and do the calcs. There is nothing magic about it. Just another circuit element.
 
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