# What is reactive energy?

We know the typical equation for instantaneous power (which can be proven): $$\p(t)=v(t)i(t)\$$. In sinusoidal steady state (let's ignore harmonics for simplicity), $$\v(t)=\sqrt{2} V_{\text{rms}} \cos{(\omega t + \phi_v)}\$$ and $$\i(t)=\sqrt{2} I_{\text{rms}} \cos{(\omega t + \phi_i)}\$$, where $$\\omega\$$ and $$\T = 2 \pi / \omega \$$ are respectively the angular frequency and period of $$\v\$$ and $$\i\$$. From this, letting $$\ \theta = \phi_v - \phi_i\$$, it can be shown that

\ \begin{align} p(t) &= \underbrace{\left[ V_{\text{rms}} I_{\text{rms}} \cos{(\phi_v - \phi_i)} \right]}_{\text{DC component}} + \underbrace{\left[ V_{\text{rms}} I_{\text{rms}} \right] \cos{(2 \omega t + \phi_v + \phi_i)}}_{\text{AC component}} \tag*{} \\ &= \underbrace{\left[ V_{\text{rms}} I_{\text{rms}} \cos{\theta} \right]}_{\text{unidirectional}} + \underbrace{\left[ V_{\text{rms}} I_{\text{rms}} \cos{(\phi_v + \phi_i)} \right] \cos{2 \omega t} - \left[ V_{\text{rms}} I_{\text{rms}} \sin{(\phi_v + \phi_i)} \right] \sin{2 \omega t}}_{\text{bidirectional}} \end{align} \

where $$\2 \omega\$$ and $$\T' = 2 \pi / 2 \omega = T/2\$$ are respectively the angular frequency and period of $$\p\$$. Now, since $$\p(t) \overset{\text{def}}{=} dw(t)/dt\$$, then the energy transferred in an integer multiple $$\n\$$ of the period of $$\p\$$ is

\ \begin{align} W &= \displaystyle\int_0^{nT'} p(t) \, dt \tag*{} \\ &= \underbrace{P \displaystyle\int_0^{nT'} \, dt}_{\text{net energy}} + \underbrace{V_{\text{rms}} I_{\text{rms}} \cos{(\phi_v + \phi_i)} \displaystyle\int_0^{nT'} \cos{2 \omega t} \, dt - V_{\text{rms}} I_{\text{rms}} \sin{(\phi_v + \phi_i)} \displaystyle\int_0^{nT'} \sin{2 \omega t} \, dt}_{\text{no net energy}} \\ &= P \cdot n \cdot T' + 0 \end{align} \

It's clear that there's no net energy transferred due to the reactive component of a load impedance. However, I've seen three documents (this webpage in section 1.7, this PDF in page 2, this PDF,) talk about reactive energy and their corresponding meters. Since, as shown above, there's no net energy transferred due to the reactance of loads, what energy do those meters actually read? If it's "reactive energy", what do they mean by that? Two documents I found define reactive energy mathematically as

$$\ \dfrac{1}{T} \displaystyle\int_0^{T} v(t) i \left( t+\dfrac{T}{4} \right) \, dt = Q \tag*{} \$$

which is non-sense; they're defining reactive energy as reactive power, which is a different quantity; energy has units of joules, yet the right-hand side of the previous relation has units of joules per second (or watts or VArs; they're dimensionally the same). Why do they define it as such? I read these three questions (1, 2, 3), but they don't really address the ones I've asked.

EDIT: My questions are not directly about reactive power, but about reactive energy.

• There's actually a pretty funny picture that give a good analogy of what the different types of power are... That is, if you drink beer (or probably even soda), you'd enjoy this. – user103380 Jun 13 '20 at 5:16
• Reactive power has no physical meaning in terms of a direct measurement, it's just one side of the power triangle of which the hypotenuse is apparent power ($V_{RMS}\cdot I_{RMS}$) and the other side is real power. – Andy aka Jun 13 '20 at 8:19
• Hi everyone. Just to be clear, the question is about reactive energy, not reactive power. – Alejandro Nava Jun 13 '20 at 14:36
• meruspower.fi/benefits-improving-power-factor-part-1 – sarthak Jun 13 '20 at 21:46
• @sarthak The questions is not directly about power factor. – Alejandro Nava Jun 13 '20 at 22:35

In the AC power industry, reactive energy is electrical energy that is stored rather than converted to some other form of energy and thus "used" or "consumed." Reactive power is the rate of transfer of reactive energy from one storage component to another.

The diagram below shows the typical transfer of power from the electrical grid to a point of use. The source voltage is supplied to the user and is assumed to be an ideal single-phase AC voltage source. The load can be represented as a resistor in parallel with an inductor. The source voltage is the voltage across both components of the load.

The resistor current is in phase with the source voltage. The instantaneous resistor power waveform is the product of the resistor current multiplied by the source voltage. That minimum points on that curve lie on the X axis. The power is positive at all times, indicating that all of the power is transferred from the source to the resistor. The area under the curve represents the energy received by the resistor and dissipated as heat.

The inductor current lags the source voltage by 90 degrees. The product of the source voltage and the inductor current is a sine wave that has positive and negative values that average zero. Since that does not represent real power, is called "volt-amperes, reactive" or "VARs." There are equal areas above and below the curve indicating energy received from the source and returned to the source. That is the reactive energy.

Shown as an ideal circuit, the average and the net reactive energy transfer is zero. However there is real energy that is constantly going back and forth. In an ideal system, the reactive energy is generated when the load is connected, transfers back and forth as long as the load is connected and is given back to the source when the load is turned off. In actuality, something like 7% of the energy is lost in every transfer between the load and the generator. The utility will put some capacitor storage in local substations or even on transmission poles. Using their rate structure, the utilities encourage large users to furnish their own capacitors.

The total volt-amperes (VA) is the sum of the power (watts) and the reactive volt-amperes (VARs). That is shown as a sine wave that dips below the zero axis. Circuit Data For Above

Supply voltage: 240 Vrms, 339.4 Vpeak

Resistor current: 200 Arms (282.8 pk)

Inductor current: 150 Arms (212.1 pk)

Supply current: 250 Arms (353.8 pk)

Power: 48 kW (96 pk-pk)

Reactive power: 36 kVAR (72 pk-pk)

Apparent power: 60 kVA (120 pk-pk)

there's no net energy transferred due to the reactive component of a load impedance.

There is no net energy, but that is only because energy is transferred in both directions.

... reactive energy and their corresponding meters ... what energy do those meters actually read?

They read the rate of energy transfer back and forth.

If it's "reactive energy", what do they mean by that?

See above.

... Why do they define it as such?

VARs are called VARs to distinguish energy that is transferred back and forth from energy that is "consumed." Energy that is "consumed" has a much higher cost than energy that is just transferred back and forth, but VARs still have a cost.

Utility Metering

The unit of measurement that us used for utility energy billing is the kilowatt-hour. That is the area under the power curve integrated over the billing cycle. For fossil fuel generation, energy measured by a kilowatt-hour meter is equal to the energy content of the input fuel plus the losses incurred in generating, transmitting and distributing the energy. Most of those losses are directly proportional to the energy generated.

Utilities may also measure the kilovar-hours. That is the area under the var curve integrated over the billing cycle without regard to the direction of energy flow. Although the net reactive energy transferred is zero, the losses incurred in transmitting and distribution vars is directly proportional to the total vars transferred. There is also an associated capital cost of generation, transmission, and distribution equipment that is proportional to the the total vars transferred.

Billing formulae and metered quantities used are determined by individual utility companies. The basics are generally similar, but a variety of specific methods are used.

References

The basic information presented is presented similarly in text books that cover AC circuits. Here are some references that are specific to the electric power industry:

Edison Electric Institute,Handbook For Electricity Metering

Michael Bearden, Understanding Power Flow and Naming Conventions In Bi-directional Metering Applications

• Hi Charles. Based on the first paragraph before the first period, and the fourth paragraph after the second last period, did you mean that reactive energy is the energy stored in inductors and capacitors, i.e. $w(t) = 0.5 L i^2(t) + 0.5 C v^2(t)$? – Alejandro Nava Jun 14 '20 at 0:18
• While my questions aren't directly about power, a few things don't seem correct to me: 1) the "Inductor VAR" curve, which is the $p(t)$ of the inductor, should have a different name, because the $Q$ of the inductor is independent of time; 2) the "Total VA" curve, which is the sum of the $p(t)$ of the resistor and the inductor, should have a different name, because apparent power isn't $P+Q$ (nor $p_R(t) + p_L(t)$), it's $\sqrt{P^2+Q^2}$.) – Alejandro Nava Jun 14 '20 at 0:18
• 1st comment: Yes, I believe that is what I mean. 2nd comment: 1) I don't think so. 2) I intended it to be the VA or apparent power as a function of time, but I did it wrong. I will try to fix it tomorrow. I will also try to look at everything again tomorrow. – Charles Cowie Jun 14 '20 at 3:18
• 2) I double checked that and found it to be correct. Instantaneous VA as a function of phase angle is equal to W as a function of phase angle plus VA reactive as a function of phase angle. I revised my answer to show the data fro the diagram. – Charles Cowie Jun 14 '20 at 21:10
• "They read the rate of energy transfer back and forth" but that's (reactive) power, not energy. Then they're confusing one term with the other. – Alejandro Nava Jun 14 '20 at 23:07

Reactive power is calculated or indirectly measured by applying an operation on the waveform before measuring. Reactive power is not directly measurable:

• Calculated: for instance, we can measure the active power P, the current I and the voltage V. From V and I we get the apparent Power Pa=V.I and hence the reactive power is calculated as: • Indirectly measured: for instance, we shift one of the waveforms (voltage or current) by 90° and then we multiply the two waveforms (or measure the resulting active power). We shift by 90° because cos x = sin (PI/2 + x). And 90° represents a delay of T/4 (quarter period). That is why we get: The Work is exclusively produced by the active power. For instance, a diesel engine produces only Active power (its power is only expressed in kW or hp never in kVA, you cannot think of voltage and current when dealing with mechanical energy). This power can drive an alternator which generates an electrical power. Now, depending on the load profile we can have a phase shift between voltage and current so, for different loads, we can have the same Active Power with different currents and hence different reactive powers.

Reactive energy is additionally billed in some cases. Why do we have to pay unused energy? It is not a matter of inventing new payment. In fact, consider an energy provider generating electrical power and consuming fuel depending on the mechanical power (active power).Although he is paying the active power (fuel consumption needed for the mechanical power) but he must distribute this power through transformers and transmission lines (we are not interested now in the conversion efficiency and losses). To simplify let us only consider transmission lines. His active power is the sum of active power consumed by clients and the active power that heats the transmission lines depending on the current. So As I mentioned, since for the same active power we can have different currents, the energy provider consumption increases with current even if the real work of his clients is still the same. The reactive energy is then a justified billing argument. It is usually imposed to incite the customer to use a power factor corrector and hence reducing the reactive energy.

• Hi Paul. The question is about reactive energy, not reactive power. In regards to the secons bullet, I posted that equation for reactive energy, but my questions are what does that mean, and why do they do it. – Alejandro Nava Jun 13 '20 at 14:39
• The second equation is non-sense because of the following. The left-hand side says "reactive energy", and so it should have units of joules or watt-hours. But if you compute the integral of the right-hand side, you get reactive power, $Q = V_{\text{rms}} I_{\text{rms}} \sin{(\phi_v -\phi_i)}$, which has units of VArs or watts or joules per second. The units of each side of the equation are not the same, so it's invalid. – Alejandro Nava Jun 14 '20 at 23:28
• @AlejandroNava thank you for this interesting question and interesting feedback. Reactive energy is confused to reactive power in many documents. it is in fact the reactive energy per unit time. I updated my answer and added additional information. I hope it will be helpful. – Paul Ghobril Jun 15 '20 at 6:12
• Great. I was realizing the same (reactive energy being confused with reactive power) after searching a bit more. – Alejandro Nava Jun 15 '20 at 16:23

I guess many of us would be capable enough to invent new payments. Some of the introduced payments can be created to control the behaviour of people.It's surely useful for power companies to persuade people to have power factors as near 1 as possible.

Reactive energy got by integrating the absolute value of the reactive power can be a good behaviour controlling money collecting criteria which can be justified. It's a single number that has a price and people can make their number smaller by using properly designed devices. It can be presented as a hard fact when people have only one electricity supplying network available. It can also be explained as a compensation of losses - that much energy your actions have forced us to transfer to your devices and back through the wires and every transferred pulse has dissipated a part. Pay it!

• LOL. If they're going to create some new quantity for payment purposes, they should instead charge by the power factor, which has a physical meaning, not the reactive energy meaningless (to me so far) thing. – Alejandro Nava Jun 13 '20 at 15:16
• That is in use. I have seen "maximum momentary reactive power (or actually the maximum of about 1 second average) registering devices" The price of used energy was multiplied by a number which increased as the max peak reactive power during the billing period increased. People saw it as injustice. – user287001 Jun 13 '20 at 16:43