Electronic – Why does reactive power affect voltage

First of all, your question I think shouldn't just be 'why do reactive loads improve grid voltage stability' - it should be grid stability - not just voltage or current or power. Everything is improved.

Let's go back to the 1970s: the power grid is entirely AC, entirely linear (i.e. AC power is generated at a power plant and multiple linear transformer stages are used to deliver it to the end customer). No DC power lines in between, no inverters, no PFC. The voltage on the power line is fairly precisely timed to allow for the use of timing motors (synchronous motors in e.g. train station clocks), timers and DTMF encoding on power lines, etc.

Most devices that regular households use and that use an appreciable amount of power have a good power factor; they are almost perfect resistive loads. Clothing irons, lightbulbs, ovens. Also, households use a pretty small amount of power (historically between 10 and 15% of electrical power). Now, a large industrial facility turns on its giant motors. Motors are a very inductive machines, i.e. they have a low power factor. One big sewage plant pump system can use as much power as an entire city block, so this has a large effect on the grid.

A perfect grid is very 'rigid': its power lines have no voltage drop, no self-inductance and no propagation delay. In reality of course, power lines do have some elasticity and especially the use of large motors and other devices with a large difference in power factor from the mean can destabilize the local grid. Regard the voltage and current waveforms; in an ideal world these are synchronous sine waves. However, if 50% of the grid is almost perfectly resistive, and the other 50% of the grid has a power factor of say 0.5, the current waveform isn't a sine anymore; current is drawn both on the peak of the voltage waveform and in between the peak and zero crossing. It's more like a block waveform. This increased current 'in between' the peaks combined with the self-inductance of the grid causes voltage spikes.

Not only that; circuit breakers for instance have traditionally always relied on the current zero-crossing for a reasonable amount of time to be able to switch. You cannot switch 100kA off; even if you would literally cut the wire with an axe they would still arc and cause current to keep flowing for way too long to be safe. Thyristors and other solid-state circuit breakers also just keep conducting until the current zero-crosses, even with their gates turned off. The increased edge speeds that distortion on the grid cause may cause big problems with circuit breakers.

So, the old-fashioned way to fix this is to put large capacitor banks as close to the motors as possible. The motors are a reactive 'load', the capacitors are a reactive 'generator' and combined they appear to the grid to be a well-behaved almost-resistive load.

This is very simple and effective, and even though I say it is old fashioned, its low complexity makes it incredibly reliable. It does have disadvantages though; capacitor banks large enough to compensate for a large-ish (100s of kW to MW range) motor are excessively expensive and large. Also, they are only optimal for a specific motor load (this depends on the type of electrical machine). You have to switch capacitors in and out to fine-tune the compensation. Lastly, there is still quite some energy loss in this system.

Note that this is not just done near or on specific machines; sometimes power companies use large capacitor banks on entire branches of their power grid to equalize the effective power factors of different domains.

A more modern approach is to use a frequency controller or inverter to control the motor. Mains power is rectified to a DC voltage, and then chopped (inverted) again to feed into the AC motor. PFC on the rectifier makes sure the drive has a good power factor on the grid. This approach is much more space- and cost-efficient, gives better control over torque and speed and easier on the grid.

The main reason for reactive power compensation is to regulate the voltage magnitude. Note that the compensation might be both positive and negative (reactive power in, or reactive power out). In a transmission system, there is a strong correlation between reactive power and voltage magnitude, whereas the active power is mainly dependent on the voltage angle. Have a look here for a bit more information.

In the transmission system, a branch might have impedance Z = R + jX, where the reactive X is about 10 times the purely resistive R.

I'm assuming you're familiar with the per unit system. Let me know if you're not and I'll explain it closer.

Let's just review a few basic relationships first:

\begin{align*} S = V \cdot I^*\\ => I = (S/ V)^*\\ \Delta V = I^2\cdot Z\\ Z = (R + jX) \end{align*}

Let's assume we have a very simple power system that looks like this:

G ---|------------------|------------------|----->
     3   Z = R + jX     2   Z = R + jX     1   Load               

• G is the generator
• The verticals lines are buses, labeled 1 - 3
• The load is at the end of the radial.
• The voltage at bus 1 is assumed to be 1pu with angle 0 degrees.
• The load is (1 + j0.2) pu. (If S_base = 100MVA, this would be equal to 100MW + 20MVAr)
• Z = 0.01 + j0.1

The current necessary to supply the load is given by:

\begin{align*} I = (S/ V)^* =((1 + j0.2) / 1)^* = 1 - j0.2\\ \end{align*}

No compesation:

The voltage at bus 2 is give by the voltage at bus 1 plus the voltage rise over the cable (seen from 1 to 2):

\begin{align*} V_2 = V_1 + I^2 \cdot Z = (1-j0.2)\cdot(0.01 + j0.1) = 1.054\angle 5.01 ^{\circ}\;\text{pu} \end{align*}

This means the power injection into the cable between 1 and 2 is:

\begin{align*} S_2 = V_2 \cdot I^* = (1.031 + j0.302) \;\text{pu} \end{align*}

The voltage at V3 is:

\begin{align*} V_3 = V_2 + I^2 \cdot Z = 1.11\angle 9.50^{\circ}\;\text{pu} \end{align*}

Now we can find the power output from the generator by using the first equation:

\begin{align*} S_{Gen} = V_3 \cdot I^* = (1.062 + j0.404) \;\text{pu} \end{align*}

With compensation:

Let us add a capacitor that injects 0.3pu reactive power at bus 2.

The voltage at bus 2 is still given by the voltage at bus 1 and the voltage rise over the cable, so it's still \$\underline{1.054 \angle5.01^{\circ}\;\text{pu}}\$.

Now, the reactive power injection of 0.3pu will give a current injection of:

\begin{align*} I_{inj} = (Q / V_2)^* = 0.285 \angle{-85.0}^{\circ}\;\text{pu} \end{align*}

The current through cable 1-2 is equal to the current through cable 2-3 plus the current injection, so:

\begin{align*} I_3 = I_2 - I_{inj} = 0.979\angle 4.90^{\circ}\;\text{pu} \end{align*}

You see that the current magnitude is lower than it was without compensation. So, let's have a look at the voltage at bus 3:

\begin{align*} V_3 = V_2 + {I_3}^2 \cdot Z = 1.06\angle10.22^{\circ}\;\text{pu} \end{align*}

Now we can find the power output from the generator by using the first equation:

\begin{align*} S_{Gen} = V_3 \cdot I_3^* = (1.037 + j0.096)\;\text{pu} \end{align*}

So, to summarize:

       W/O comp:   W comp:
|V1|   1.000       1.000
|V2|   1.054       1.054
|V3|   1.115       1.060

       W/O comp:         W comp:
Gen    1.062 + j0.404    1.033 + j0.096

As you can see from the above results, the voltage is much more stable with compensation. The current gets lower through the cable resulting in lower active losses.

The reason why the reactive power is needed in the first place is because it accounts for the magnetization of the equipment. If there's no reactive power, transformers, generator rotors/stators, machines etc. have no magnetic field. With no magnetic field, there is no torque, no magnetic coupling in the transformer etc. So, a lot of equipment have to consume reactive power in order to work. If there's too little reactive power available the equipment will try to draw more current to compensate. This will lead to higher voltage drops, which in the end might cause voltage collapse.

As Andy points out, it can also be used as power factor correction for large industrial loads. However when we're talking about reactive power compensation it's most often because of what I've described above.

In a meshed grid it can also be used to control power flow. This works because the active power flow through a cable is mainly given by the voltage angle difference over it. If you inject reactive power, the voltage and currents angles will change, thus it will affect the power flow. If you inject the right amount at the right place you can redistribute the power flow the way you want (but only to a small extent).

Hope this answers your question!