# Negative Reactive power

1. Nov 8, 2013

### Physicist3

What would it mean if in a power system with purely inductive and resistive elements (No Capacitance) if you were seeing a negative value of reactive power when there was no load connected? E.g. If the load was disconnected but the transmission line contained both resistance and inductive reactance?

2. Nov 8, 2013

### Averagesupernova

Strange question. Of course a transmission line in the real world always contains resistance. However, transmission lines are funny things. How a transmission line appears from one end depends on how many multiples of 1/4 wave that it is long when it is open at the other end. The definition of a capacitor is 2 conductors that are separated by an insulator. Does a length of transmission line not meet this criteria? So, we cannot have a transmission line that does not contain capacitance by nature. Nor can we have a transmission line that does not contain inductance. By manipulating the length we can make the transmission line look any way we want.

3. Nov 8, 2013

### FOIWATER

Inductive reactances consume negative vars while capacitors consume positive vars.
One way to think about this is that inductors consume reactive power while capacitors provide negative reactive power.
I suppose the negative reactive power could be from the line charging susceptance?
(as an aside, that first sentence is the nature of power factor correction. If inductors consume reactive power, and a capacitor can supply it, the source doesn't have to)

4. Nov 8, 2013

### Baluncore

Strictly speaking real power cannot be negative, but a reactive current can. The unit of reactive power is the VAR (volts * amps_reactive), which is not real power, since the current is in quadrature with the applied voltage.

If a line has an open circuit load, and shows a negative VAR then it is appearing as a capacitor to the generator. An open circuit transmission line will appear to have capacitance if its length is below one quarter of a wavelength. At exactly one eighth of a wavelength it will appear most like a lumped capacitor.

At 60 Hz the wavelength would be 5000 km. One quarter wavelength will be 1250 km. So between 0 and 1250 km an open line will appear to be capacitive, and so have a negative reactive current. A lossless, one eighth wavelength, (625 km @ 60 Hz), open circuit stub, will appear to be a perfect capacitor. The apparent capacitance of a one eighth wavelength open line is a function of the line's characteristic impedance and frequency.

Last edited: Nov 8, 2013