Shunt Reactors for Voltage Control

[jegues]

I was reading through an article on, "Shunt Reactors for Voltage Control" and I am slightly confused. (See link below for article)

http://www.onegrid.com.au/wp-content/uploads/2012/03/BR-EN-TH16-02_2006-Shunt_reactors_for_voltage_control.pdf

The article reads, "When the network load and particularly the high voltage network drops, the voltages on all the busbars increase due to the capacitive current in the transmission line."

It then goes on to say, "The voltage sensitivity to changes at one busbar due to small changes in real and reactive power at another busbar can be determined from the stability analysis expression."

dV_{i} = \frac{\partial V_{ij}}{\partial P} \dot dP_{j} + \frac{\partial V_{ij}}{\partial Q} \dot dQ_{j}

"A simplified expression that is quite accurate in most cases is"

\Delta V_{i} = \frac{\Delta Q_{i}}{S_{SC}}

What do they mean by capacitive currents, and how does that increase the voltage on the busbar?

How does adjusting the reactive power change the voltage levels?

 

[Baluncore]

AC power distribution systems have a very low output impedance so it is convenient to think of the AC voltage on the line as a reference.

A resistive load will cause a line current to flow that is in phase with the voltage. That is real energy flow. It is measured in watts, usually kW or MW.

A capacitive load will cause a current to flow that leads the voltage by 90°.
An inductive load will cause a current to flow that lags the voltage by 90°.

Those reactive currents do not transfer energy because the voltage times current is zero due to the phase shift. That is a circulating energy known as “Volts Amps Reactive” (VAR). Usually specified as kVAR or MVAR.

The circulating reactive current is the difference between the capacitive and the inductive currents. Ideally that will be close to zero so there will be no unnecessary circulating current.

The real current flowing on the line is the vector sum of the real and the reactive currents. It should have a phase angle close to the voltage reference.

There is a voltage drop due to the line's series resistance multiplied by the VAR, so it is taking a real amount of power to circulate the reactive current. The circulating current therefore influences the voltage magnitude since the line drop is in series with the supply.