# Voltage regulation with reactive power

Supposing to have a generator located in the electric grid providing a voltage regulation controlling its reactive power output. If the voltage on the grid increases near the generator, should the generator inject reactive power or absorb it in order to reduce the voltage?

I am doing a thesis on this and I am studying how distributed generators can help the grid controlling their reactive power. My professor told me that when the voltage increases the generator should absorb reactive power, but I think this is not true because we already have reactive power in the grid and we should decrease it injecting reactive power and so reducing the current circulating and so also the voltage. But I don't know if I an right or not.

Start with the two basic approaches to raising/lowering bus voltage. The application of capacitor banks (raise) and reactor banks (lower). It all hinges on the system being predominately inductive.

In the figure below we have a bus for which we want to control the voltage magnitude. The system is represented as an infinite source and the Thévenin equivalent impedance. Since it is predominately inductive we will assume it is all $$\j\$$.

Using voltage division we can see that when the capacitor bank is closed in the bus voltage, $$\V_{bus}\$$, will be,

$$V_{bus}=V_{sys}\frac{X_C}{X_C+X_L}$$

Assuming $$\X_L=j0.05 \$$ pu and a target $$\V_{bus}\$$ of 1.05 pu we can solve for the required $$\X_C\$$ as,

$$1.05 = 1.0\frac{X_C}{X_C+j0.05}$$

then

$$X_C=\frac{-j0.053}{0.05}=-j1.05\text{ pu}$$

The current flowing into the capacitor bank from the bus will be,

$$I_C =\frac{1.05}{-j1.05}= j1.0$$

So, the reactive power flowing into the capacitor bank is $$\-1.05\$$ pu ($$\S=V_{bus}I_C^*=-j1.05\$$).

Now, if we want to replace our capacitor bank with a generator to do the same thing then our circuit would look like the following where $$\jX_d\$$ is the direct axis synchronous reactance of the machine and $$\V_G\$$ is the voltage behind that reactance. We can control this voltage by raising/lowering field current. We can control the angle on this voltage (as compared to the system) by increasing/decreasing power (torque) applied to the shaft.

Now, lets replace the capacitor bank with a generator. To calculate the required generator $$\V_G\$$ to result in that same capacitive current flowing into the generator ($$\j1.0\$$) we can use our $$\Q\$$ flow equation and assume the angle $$\\delta\$$ is zero.

So, the reactive flow equation is,

$$Q = \frac{E^2-EVcos\delta}{X}$$

where $$\\delta\$$ is the angle by which $$\E\$$ leads V. So,

$$-1.05 = \frac{1.05^2-1.05V_G}{1.0}$$

$$V_G = 2.05\text{ pu}$$

If you want to lower the bus voltage you will change the conditions so the current coming into the generator has a negative angle (e.g. $$\1\angle-90°\$$) making the machine look like a reactor to the system.

The basic principle of this is simple - to raise the bus voltage you drag a capacitive current across the system equivalent impedance. The resulting voltage drop will add to $$\V_{sys}\$$ and result in lowering $$\V_{bus}\$$. The drop is "in-phase" with $$\V_{sys}\$$ if you plot on phasor diagram.

To lower the bus voltage you drag an inductive current across the system equivalent impedance. The resulting voltage drop will subtract from the $$\V_{sys}\$$ and result in lowering $$\V_{bus}\$$. The drop is "out-of-phase" with $$\V_{sys}\$$ if you plot on phasor diagram.

If your machine is also putting out real power then you will need to account for both the voltage $$\V_G\$$ and the angle $$\\delta\$$. In that case you can still work the problem but requires more work (or use a load flow to solve).

• should I consider V constant to understand if E increases or not changing Q? Because also V=f(Q), therefore changing Q both E and V change Feb 11 at 7:51
• thank you very much! Now it is all clear! Thank you! Feb 12 at 8:03

If the grid voltage increases the alternator has to reduce its own excitation to reduce its terminal voltage and at the same time reduce the bus voltage. If the grid voltage decreases the alternator terminal voltage has to increase to offset the reduced bus voltage. It is said that in the first case the alternator absorbs reactive power, while in the second case it injects reactive power to the grid.