7
\$\begingroup\$

In my power engineering course, we were introduced to reactive power. I believe I understand what reactive power is in essence: there is energy transfer between the source and inductor/capacitor. This makes sense to me, and I can do many calculations just fine. However, I am failing to understand what the value of reactive power represents. From my notes:

$$\begin{align*} P=VI\cos\psi \\ Q=VI\sin\psi \end{align*}$$ where P and Q are active and reactive power respectively. I understand the formula for P, and its derivation is presented in my notes. Physically, P represents the amount of energy dissipated through the resistor. What does Q represent?

Considering the graphs of power vs time in capacitor/inductor only circuits, the average power is 0. Shouldn't this mean that on average reactive power is just 0?

Apologies if the question is a bit unclear, I am having a difficult time just explaining my confusion.

\$\endgroup\$
  • \$\begingroup\$ So Q is a quantification of the power transferred between source and load. Positive, power is absorbed by device, negative, returned due to collapse of magnetic or electric field, \$\endgroup\$ – StainlessSteelRat Apr 18 '17 at 16:15
  • 1
    \$\begingroup\$ You can also look at this from an energy view point. Reactive power is the work that is done to establish and collapse magnetic and electric fields in components (specifically inductors and capacitors). This power is "not real" in the sense it does no useful work, but is required to be put into the system to make it operational to begin with. \$\endgroup\$ – Envidia Apr 18 '17 at 16:50
  • \$\begingroup\$ Just to add another analogy. Think of walking across a long trampoline. If you go by leaps and bounds, you will be transferring energy between sprung weight on the trampoline (call that electrostatic potential stored in capacitance) and potential energy in the air when out of contact with the surface (call that magnetic potential stored in inductance.) Either way, the energy needed to get to the end is the same, whether you walk or bounce your way. But the headroom you need is different, depending. \$\endgroup\$ – jonk Apr 18 '17 at 16:57
  • 3
    \$\begingroup\$ Reactive power makes for a great beer analogy \$\endgroup\$ – Mast Apr 18 '17 at 22:12
  • 1
    \$\begingroup\$ 0 on average is very different from just 0. AC voltage and current are also 0 on average. \$\endgroup\$ – Dmitry Grigoryev Apr 19 '17 at 8:44
15
\$\begingroup\$

Let's ignore the power aspect for a second, and think about what reactance really is.

You know a math and theory, you can talk about things in abstract theoretical constructs using complex numbers, phasors, and all that. But abstract models are exactly that: abstract models. The math can model a thing, but it won't really help you understand the physical system it is modeling except in other abstract ways.

So let's ignore the math for a second, and talk about what reactance really is.

Actually, let's talk about resistance. The real component to impedance. Ultimately, resistance represents energy loss. Resistance consumes some of the kinetic energy of electrons moving through the circuit, and that manifests as the familiar ohmic voltage drop we see across any resistive load. Electrons smack into stuff, set them vibrating, and the resistive load heats up as joules are lost by the electrons and transferred into the load. The faster the rate of energy that moves through this load, the faster the rate of power lost, and the harder you have to push to make that happen.

But that's just one side of the coin. Besides simply dissipating energy to the environment, there is another option that can happen: energy can be stored. Capacitance and inductance are often talked about as being 'duals' of each other because they're both measures of energy storage. Capacitance is a measure of energy stored in an electric field, while inductance is a measure of energy stored in a magnetic field.

Energy being stored looks just like energy being dissipated, at least at first. In both cases, energy that was in the circuit is no longer present. The only difference between resistance and reactance here is that with resistance, that energy is gone for good, but reactance will eventually return that energy back to the circuit at a later time. Well, and of course as a measure of storage, they eventually reach a maximum amount of storage given a static circuit. A capacitor will need a higher voltage to store more energy, an inductor likewise will need higher current to store more energy. This is the 'reactance' aspect. As energy is stored, less power is seemingly dissipated by this reactance until it vanishes entirely. If the power starts going down, the stored energy is released back into the circuit.

So what is apparent power? It's simply the rate at which a circuit or part of a circuit (depending on what you're calculating for/looking at) is storing energy, or, if the magnitude is opposite, the rate at which it is releasing energy. That's all. It's not weird, or strange, and it is a real, physical, quantifiable thing. If you charge a massive capacitor bank, from a battery, it will consume joules from that battery, and it will do so at a certain rate, one that is highest at first, but will eventually fall to zero. This is, technically, reactive power. But it's still measured in watts, and watts are, well, always watts. You just are measuring the rate at which something stored joules, rather than the rate at which it simply dissipated them.

Your confusion I think is that you've actually sort of arrived at the answer already without realizing it. If you have a circuit with only capacitors and inductors, then there is no 'P' as there is no energy being dissipated at some number of joules per second. There is only energy being stored, and it will eventually get released, and so yes, it averages to 0. Reactive power always does. It's ultimately just storage, not consumption, so yes, the it will always always always average to 0. Those joules got loaned out, but inductors and capacitors have terrific credit ratings and always pay you back eventually, so you didn't actually lose any money/joules in the long run.

So, you don't need to talk about this in terms of math at all. In fact, if you understand that reactive power is simply the rate energy is being stored and then released, measured in joules per second or watts like anything else involving power, then the behavior and math should all just make logical sense, because that's ultimately what you are modeling with said math.

Now, one might wonder why reactive power even matters if it averages to zero.

Let's talk power factor real quick. Power factor is, of course, the ratio of real to apparent power. This might seem like a rather strange or pointless thing to have a ratio of. I mean, who cares? The apparent power isn't actually being lost, why even measure it?

The issue is that this energy storage is never (excepting in the case of perhaps superconductors) totally efficient. Electrons have to move onto the negative plate of a capacitor, while an equal number of electrons are pushed off the positive plate. Moving charge is current. Conductors (again, except in the case of superconduction) always have some resistance, so you have losses. In the context of alternating current, where storing energy will have a profound effect in this regard, you wind up having electrons flowing in and out over and over, uselessly storing energy for no reason. So even though the energy is getting returned to the circuit, you're still suffering losses in the form of current flowing but without it doing any work. Really, the idea of current and voltage phase is just a way of looking at how reactance is effectively dropping voltage but because it is storing energy, or maintaining voltage (or increasing it to maintain current instead), by releasing energy.

Never forget that important concept, that all of this is ultimately just different sort of abstract ways of looking at or modeling one true, physical process going on, which is actually very simple at its core. The storage of energy, and that there are two different fields with which it can be stored. From that concept, everything else can be derived.

\$\endgroup\$
  • \$\begingroup\$ So does adding a thermo-electric generator next to a resister turn it into a reactor? \$\endgroup\$ – JDługosz Apr 19 '17 at 6:00
  • 3
    \$\begingroup\$ This talk of energy storage, while correct, can be a bit misleading in this context. You really should point out that the energy is being stored and unstored twice per cycle. This is not what is meant by storage on the power grid. It's not like pumping water uphill during low demand times, then using it to run generators during peak times, for example. \$\endgroup\$ – Olin Lathrop Apr 19 '17 at 10:38
8
\$\begingroup\$

When you have a complex load, its associated power can be modeled with a complex number:

$$S = P+jQ$$

Where S is the complex power, usually measured in VA, P is the real power, measured in W, and Q is the reactive power, measured in VAR.

As op noted, the only "useful" power is the real power P, since it is the only power dissipated in the load that can do useful work.

But what about delivering the power to the load?

Suppose you have a purely capacitive load, and a sinusoidal power source, and you need to size the conductors from the source to the capacitor. You calculate the real power, and it is zero! You use very thin wires to connec the cap and... BANG they vaporize. What happened?

The reactive power Q, even if it is not dissipated, must be carried to and from the load and must be taken in account when sizing all the components making a power transmission line. Since usually \$P\neq0\$, to size a power line power engineers use apparent power, i.e. the magnitude of S, or \$\sqrt{P^2+Q^2}\$.

To answer your question, Q physically represents the power that is continuously "bounced" between the source and the load, and it is fundamental to size conductors, transformers, switches and everything that constitutes a power line.

\$\endgroup\$
6
\$\begingroup\$

"Reactive power" is one of several possible ways of reconciling the fact that in AC systems, the voltage times the current isn't the average power.

The reason rms(V) x rms(I) is not ave(V x I) is phase shift. When V and I have the same phase, then rms(V) x rms(I) is ave(V x I), which is the power delivered. As the phase between V and I increases, the power delivered goes down while rms(V) and rms(I) stay the same.

This can be accounted for by tracking the phase angle, looking at V or I as a phaser, or treating the 0° and 90° parts separately.

Reactive power is basically a bookkeeping method that accounts for the 0° and 90° components of the current separately. Since the phase shift of voltage and current is only relative to each other, you can pick one as the reference and consider the other phase shifted relative to that reference. In the reactive power scheme, the voltage is the reference and the current is shifted in phase relative to the voltage.

Note that the phase shift of the current can be expressed as a angle, or equivalently, as the 0° and 90° components. Real power is the voltage times the 0° current component, and reactive power the voltage times the 90° current component.

You are right in that reactive power doesn't actually deliver any power. The confusion is that the word "power" is applied. Calling it the "reactive VA" might be more accurate, but right or wrong, the industry has converged on the term "reactive power".

This bookkeeping scheme has some useful properties. By thinking of and accounting for the real and reactive power separately, they can more easily be dealt with separately. For example, capacitors are reactive power generators. This may sound odd at first glance, but once you buy into the definition of reactive power, that's what capacitors are. Note that inductors are therefore reactive power consumers. The electric grid usually looks a little inductive, so "reactive power" has been defined to be positive to feed this inductance.

Capacitors really do offset the phase shift caused by inductors regardless of what bookkeeping convention is used. When dealing with the electric grid, the "reactive power" bookkeeping method has proven to be useful in quickly evaluating what is going on and what is required to offset reactive loads.

For example, you could describe a situation as 101.5 MVA with 10° lagging phase, or you could describe it as 100 MW real power and 17.4 MW reactive power. The two are equal mathematically. However, the latter is usually a more immediately useful expression when dealing with this situation in the power utility world.

\$\endgroup\$
  • \$\begingroup\$ Thank you very much for the response. The mathematical equivalence you mentioned at the end makes a lot of sense now. One thing I don't see is is why the reactive power can be positive or negative, since it refers to a zero-sum transfer of energy between the load and the source? If it is going back and forth in equal amounts, what makes an inductive load 'positive' and a capacitive load 'negative'? I will try to understand this, thanks very much! \$\endgroup\$ – masiewpao Apr 18 '17 at 17:05
  • \$\begingroup\$ @masi: Again, it's not really power, more like VA. Inductors and capacitors have 90 deg phase shift in current with respect to voltage, but each with opposite sign. Since reactive power is only concerned with the current component along the 90 deg axis, capacitors and inductors will "produce" opposite polarities of reactive power. By convention, capacitors "generate" positive reactive power and inductors negative. That's another way of saying inductors are reactive power loads. This polarity convention is used because that's the most common situation on real power lines. \$\endgroup\$ – Olin Lathrop Apr 18 '17 at 17:11
  • 4
    \$\begingroup\$ I strongly disagree with this answer. Reactive power is a measure of a real, physical process and is not at all a 'book keeping method' and does not exist to 'reconcile' anything. Reactive power is a measure of the rate of energy storage (or energy release when it is negative). It takes time to store energy, and we measure that in watts, and they are watts like any other watts. Unless you're saying no energy is stored in electric fields or magnetic fields, then you cannot say that apparent power is just some sort of theoretical convenience. \$\endgroup\$ – metacollin Apr 18 '17 at 17:49
  • \$\begingroup\$ @meta: But this energy storage you talk about gets filled and unfilled twice per power line cycle. The storage is so short term, it's really more of a sloshing of energy back and forth. I didn't say reactive power isn't real, but it is one of several possible ways to describing the same thing. Presenting it as a bookkeeping scheme I think is a useful way to help people understand it. Your bookkeeping scheme could be based on phase angles and be just as valid, but in this case, less immediately useful to the utility power industry. \$\endgroup\$ – Olin Lathrop Apr 18 '17 at 18:19
  • \$\begingroup\$ @Olin: Thank you, I think I understand what you mean. Essentially we have chosen a convenient sign convention. \$\endgroup\$ – masiewpao Apr 18 '17 at 18:50
4
\$\begingroup\$

Reactive power \$Q\$ is a measure of the energy \$E\$ that is oscillating between the generator and the load without loss. At any instant energy might be moving from load to generator or the other way, and the energy stored in the load will vary over time. But the sum of the energy in both is constant, since by definition there's no loss.

Specifically:

$$ E = {Q \over 2 \pi f} $$

or

$$ Q = 2\pi fE $$

where \$f\$ is the frequency.

If you follow the convention of using a negative reactance for capacitors, you might end up with a "negative power". But this is just a mathematical convention to distinguish between a 90° or a -90° phase shift; either way the total energy oscillating between the load and generator is the same.


Derivation:

Let's say the load is just an ideal inductor.

The reactance \$X\$ will tell us the ratio of RMS voltage \$V\$ to current \$I\$:

$$ X = {V/I} \tag 1 $$

And reactive power \$Q\$ is just the product of voltage and current since our load is purely reactive:

$$ Q = V I \tag 2 $$

Reactance is related to frequency \$f\$ and inductance \$L\$:

$$ L = {X \over 2 \pi f} \tag 3 $$

Combine (1) and (2) to get:

$$ I = \sqrt{Q/X} \tag 4 $$

Energy \$E\$ stored in an inductor is a function of the inductance and current \$I\$:

$$ E = 1/2\: LI^2 \tag 5 $$

When current is at a peak, so too is the stored energy in the load inductor. We can convert RMS current to a peak instantaneous current like so:

$$ I_\text{peak} = \sqrt{2} \cdot I_\text{RMS} \tag 6 $$

Let's combine these equations, starting with (5):

$$ E = {1\over 2}\: LI_\text{peak}^2 $$

Sub in equations 3 and 6:

$$ E = {1\over 2}\: {X \over 2 \pi f}\: (\sqrt{2}\: I_\text{RMS})^2 $$

Simplify:

$$ E = {1\over 2}\: {X \over 2 \pi f}\: 2I_\text{RMS}^2 $$

Ohm's law or (4):

$$ E = {1\over 2}\: {X \over 2 \pi f}\: 2(\sqrt{Q/X})^2 $$

Simplify:

$$ \require{cancel} E = {1\over 2}\: {X \over 2 \pi f}\: {2Q\over X } \ = {Q \over 2\pi f} $$

Again, that's the peak energy stored in our load inductor, a purely reactive load.

By laws of energy conservation it should be apparent that when the energy stored in the inductor is at a maximum, energy stored in the generator is zero, and energy can't go anywhere else, so this is also at any time the total energy in the system.

\$\endgroup\$
  • 1
    \$\begingroup\$ I bet there's a relation between reactive power and the rate of change of stored energy in the inductor as well, though I don't have time to derive that. \$\endgroup\$ – Phil Frost Apr 18 '17 at 18:49
3
\$\begingroup\$

The simplest way to think about reactive power is to realize that some kinds of a load will spend a portion of each cycle taking energy from the line in excess of what they'll use, and then spend another portion of each cycle giving the excess energy back. If a load consists of a passive network of resistors, inductors, and capacitors, then any time the reactive component of instantaneous power is positive, the device will be absorbing energy that it is ultimately going to return, and whenever the reactive component is negative it will be returning power that it had previously stored up.

\$\endgroup\$
  • \$\begingroup\$ That makes sense! In the RLC circuit, if the source has constant frequency, then won't the phase angle be constant? Then from equation Q=VI*(psi), Q will be a constant value, either positive or negative. I'm getting confused because the energy is meant to be returned (or absorbed), and this is associated with the sign of Q. But if Q is constant and not changing, how does it account for the energy 'sloshing' between the source and load! \$\endgroup\$ – masiewpao Apr 18 '17 at 18:56
  • 1
    \$\begingroup\$ @masiewpao: Both P and Q describe long-term averages or, if preferred, an average over any integer number of cycles, or the energy used or transferred in any one of many equal cycles. \$\endgroup\$ – supercat Apr 18 '17 at 19:11
  • \$\begingroup\$ I thought I understood this, but I've looked at it again and I don''t think I do. The long term average of P is just the integral of vi over 2pi. But in terms of Q, because the power curve is a sine wave, any average over its cycle comes out to zero. I tried to derive equation for Q but integrating from 0 to pi/2 instead (I thought this may be valid since Q represents the rate of energy absorption/release, so integrating over half a cycle made sense to me). The result however is still a cos term, and not a sin term like it should be. Any idea why? \$\endgroup\$ – masiewpao Apr 19 '17 at 12:38
  • 1
    \$\begingroup\$ @masiewpao: I should have clarified; since P is in-phase voltage times in-phase current, the instantaneous value is always positive, and the long-term average will be half the peak value. For Q, what's of interest is the average absolute value, since as noted the positive portion and negative portion will precisely cancel. \$\endgroup\$ – supercat Apr 19 '17 at 14:08
2
\$\begingroup\$

Physically represent while still taking about electricity is a tough one. If you can accept an analogy, this is my favorite explination of reactive power it you look at it from the utility companies perspective:

Beer

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.