We know that reactive loads such as inductors and capacitors dissipate zero power, yet the fact that they drop voltage and draw current gives the deceptive impression that they actually do dissipate power. This “phantom power” is called reactive power, and it is measured in a unit called Volt-Amps-Reactive (VAR), rather than watts. The mathematical symbol for reactive power is (unfortunately) the capital letter Q. The actual amount of power being used, or dissipated, in a circuit is called true power, and it is measured in watts (symbolized by the capital letter P, as always). The combination of reactive power and true power is called apparent power, and it is the product of a circuit’s voltage and current, without reference to phase angle. Apparent power is measured in the unit of Volt-Amps (VA) and is symbolized by the capital letter S.

You might like this analogy interesting:

https://www.youtube.com/watch?v=2jM5eTTT8cE 

@Admin Ghaderi, Beer glass analogy is a well utilized analogy to explain active, reactive and apparent power, thank you for the video illustration. But that just gives a visual realization of the ratio of real and reactive power as well as the utilization. What about the concept?  

Any electrical circuit could be presented as combination of, Resistance(R=(resistivity)(length)/(area), Inductance(L=do/di, variation of flux with variation of current),and Capecitence (C=dq/dv, variation of charge with the variation of voltage). In circuit power is consumed by resistance (heat,light ect.), which could not be regained by circuit(like money spended from pocket) and is called active power loss(W). Inductance the current sensitive(series) property of circuit, converts the change of current to flux developed, and would be regain by circuit, similarly the capacitance the voltage sensitive(parallel) property of circuit, converts the change of voltage to charge storage, and would be regained by circuit. Thus L and C would be existed(present), when there is change, L for current and C for voltage, or for A.C, it would not come in picture for D.C (as flux and charge are constant for D.C circuit, as Xl=zero and Xc=infinity for frequency(f) equal to zero,, hence  Z=R), and in A.C is called reactive power(VAR), or a spinning reserve of circuit(like money in pocket).To inductance(L),when voltage applied, voltage appears immediately at output, while current utilized in developing the required field, or current would be lagging the voltage. For capacitance(C) till capacitance is not charged(current flow), but the voltage will not appear at output, or voltage would lag the current. Thus the pure inductor would gave lagging VARs and the capacitance the leading VARs loading to circuit. Hence VA= P+/-jQ., P the active and Q the reactive power. 

reactive power exists when there is a phase difference between voltage and current, so this type of energy can not exist when the current and voltage are in phase.

for a pure resistance, the reactive power is zero for the same sinusoidal regime. In conclusion reactive power is manifested only in the reactive components such as inductors and capacitors.

A good question Chittesh. Many dont dare ask such a question.

An inductor has the property to oppose the change of current through it. It opposes the change by virtue of storing or giving back the energy to the emf source.

Let us consider the positive half cycle. when the current is in the raising period, the inductor tries to oppose the change of current,. It does so by storing the energy in its magnetic field.

During the period when the current falls, (remember we are still in positive half cycle only) the inductor again tries to oppose the change of current, ie., it tries to keep the current constant. it does so by giving the energy that it has stored back to the emf source.

This exchange of energy between the inductor and the source, will not add any thing to the load power (active). But this exchange of energy causes an additional current component to flow in the circuit, which causes losses. It is this exchange of energy, that causes the phase difference between voltage and current. 

Dear all,

Thank you for your generous contribution. @Ingmar Kühl the tutorials you provided are excellent, thank you for that. @Bhupendra Desai, a very good example to prove the headache caused by the reactive power.

Even-after all these explanation my question still stands. Let me explain the question more, we have learned a lot about how to calculate the reactive power, how to compensate for it and we also say that it is required for the system to function. Now how is it required by the system for functioning(other that for voltage control in transmission lines)? What is the effect of this reactive power in the system? Is it causing magnetic field only? or electric field? or some other effects are also there?

If it is required by the motor to run, then what happens in DC motor? 

Finally, if a kid from school asks me "what is reactive power?" how can i explain it to him?

For example we can narrate active power story by using various examples like the heating effect, torque in motor, light from the bulb etc..

The simplest expanation does not need equations.

If you have current and voltage in a circuit and this does not cause power loss (or heat), this is reactive power.

@Francesc Casanellas, but sir, as we know there is a reactive power current component and wouldn't this current cause heat loss in conductor. Hence can we always say that  reactive power doesn't cause a loss. Yet again if this is true, how does the definition fit as it says it reactive power doesn't provide any work?

You are right, I should have said no power or heat EXCEPT the one caused by the current in resistances (IR2).

You could refer Electrical Installation Guide by Schneider Electric. It has a chapter on Reactive Power that should solve all your doubts. 

http://www2.schneider-electric.com/sites/corporate/en/products-services/product-launch/electrical-installation-guide/electrical-installation-guide.page

Dear Chittesh,

I think you can explain a school kid about reactive power with the attached file. I think the analogy fits.

Sometimes long explanations are better, Bear with me.

These concepts I have explained in my classes for GATE exams based on the basic laws like Conservation, Lenz's, Faraday's law of EMI, Coulombs, etc. and concepts of duality. To be in full alignment with conservation laws, not only must flow of current be opposed ( by R), its rate of change with respect to time must also be limited and the concept of an inductor does this. ( As per duality, a dual can be spoken of regarding change of voltage because of the beautiful concept of power being an amalgamation of both V and I). Any concept in engineering had better have its material equivalent. Hence materials like ferromagnetic materials, even air, in association or linking with a carrier of changing (AC) current respond by aligning the dipoles in a particular direction. Rather they react or a reactive power is required when the dipole alignment has to reverse whenever current direction reverses.  In between a pair of metallic plates, with a potential across it,  the pulling and aligning of charges in a direction then in another, as applicable to AC voltage requires a dual kind reactive power. 

However with DC this requirement of the materials to react is only during the transient condition of switching on or off etc. Because once the alignment is brought into being, there is no need to draw more current or discharge charges. Thus reactive power is more a requirement of the material behaviour surrounding the current carrying conductor or the potential bearing plates.

This is to hope, that such an explanation is helpful?

Reactive power is the measure of (electric or magnetic) energy (power) exchanged between source and load considered in a time interval. Consequently, if there is no both signs of power in an interval there is no exchange and there is no reactive power regardless the instantaneous  power can vary in a very wide band.

Dear sudha gopalan,

The energy that you are talking about, which is used for alignment of dipoles is spent energy. It is called hysteresis loss. This loss will be converted into heat, due to inter molecular forces while alignment. This loss along with i2r loss due to resistance forms the active power in electromagnetic circuit. This loss shouldn't be mistaken to reactive power.

Thank you. To take the discussion forward, if it is not energy that I am talking about, since active power is not implied, but just the motivation, or induction for the dipoles to switch directions, rather the magnetic flux, even in air to change directions, perhaps you'll agree that reactive power adds to the current requirement and drawn by the inductive load but not the active power needs. You are absolutely right that any movement ( actual movement) results in heat and this energy has to be supplied from the source. Looks like I have not worded my response appropriately. 

Dear all, 

once again thank you for providing enlightening concepts of reactive power.

Especially @ Sudha Balagopalan, i feel that your explanation was what i was looking for. It seems to fit perfectly to most of the characteristics that we know about reactive power and with my limited knowledge i would say that your answer is more convincing than that i have come across till now.

First of all, the explanation perfectly fits for AC and DC part, where reactive power is seen only associated with AC unless in DC there is transient.

And the concept of medium is a little different then. That is, its not exactly medium for transmitting  AC power rather it is the effect of transmitting AC power.

Yes this effect will cause heating and i believe that explains the reactive power current component also(which will inturn cause joules heating).

Its required in functioning of motors as they require constant flux change for generating rotating torque.

But i have two question left. First is, under fault condition we know that it demands lot of reactive power, why is that ?  How can we explain that?

Secondly, and most importantly, how to explain the relationship between Voltage and Reactive power with this idea in mind?

If i am wrong in any aspect please feel free to correct me everyone.

@Prakash Kumar the file which you have give is an excellent analogy and is well explained, full marks for that and thank you so much as it is the simplest way to explain reactive power for an school kid. 

Regards

Chittesh VC

Thanks for understanding what I wanted to convey. 

May I answer your first question with confidence- Yes, the fault current is mostly reactive since the fault bypasses the load ( the largest resistive part) and the entire transmission line  is now the load with a high reactance now because of the high frequency ( harmonics especially for unsymmetrical faults) and large transient  current ( higher flux- larger the reactance) flowing. So the fault current is the current drawn by the transmission line which makes this current mostly reactive.  Capacitance effect decreases since the voltage drops.

Regarding the 2nd question, are you wondering about the P-f and Q-V decoupling?

Let me join in the wonderful discussion on reactive power. Many of us continue to teach the students with mathematics and abstract concepts. In simple terms, reactive power is energy storage and release. The answer given by Muharem Mehmedovic is the precise answer. Inductance and capacitance are like a mechanical spring which stores energy only to be released later. In AC, this storage and release happen in each of the half cycles whereas in DC, this only happens for Capacitance during the transient period.

Feel free to enhance my knowledge.

@ Sudha Balagopalan, yes i was asking the same. I want to know how reactive power - voltage relationship can be explained by using the dipole alignment theory?

Regards

Chittesh VC

Dear Chittesh, The connection between Q and V appears very obvious, going  by the analogy of a spring or mass equivalent, especially when the reactive power stored over time ( energy) results in tighter compression of the spring or larger pumped storage resulting in a higher potential.

In the alignment interpretation we need to work back from the concept that reactive power is associated mostly with the medium- the permeable one or the dielectric. So more reactive power drawn means more motivation for alignment of dipoles. Then if we look at the linear part of the phi-i curve (between inertia and saturation) we can draw the conclusion that the larger the no. of flux lines set up by unit magnetizing current ( more alignment), larger the voltage induced in any metallic parts, especially the coil, linking with this flux at reversals as per FL of EMI. Here flux in the medium connects between reactive power and voltage in coil. Similarly the electric lines of force (charges aligned) in the dielectric connects between reactive power and voltage. 

Experimentally, if you insert a ferromagnetic material into the coil of an R-L circuit, thereby increasing the impedance of the circuit, for the same current from the source, the voltage drop increases. A look at the impedance triangle and the corresponding power triangle also shows that the flux increase associated with the coil increases the reactance (Nxphi/i) and also the reactive power, without changing the active power.

Hope something clicks into place with your concepts.

Reactive power is simply due to the Faraday Law. If a power system element draws magnetizing current from the source, it causes reactive power comsumption. If you compensate it (magnetizing current) with capacitors, that means you eliminate ractive power and Faraday Law of magnetizing. Therefore, you make power factor correction up to 0.98 not 1 unless the load is pure resistive. 

Great question and this one certainly fits the expression - If you can't explain it to a 6 year old, you don't understand it yourself.  I would argue that none of us could explain it to a six year old  and trust me, I've tried!  When I talk about reactive power or vars, I use many of the same analogies - the beer mug, the boat pulled by a horse and all of the others but they all leave some part that doesn't quite define it precisely. 

I'm presently working on a video to help explain why capacitors don't save significant energy (yes, minor I^2R but not significant) as that is a big issue in the industry (false energy savings claims with capacitors) - in that video, I use actual measurements to display how an unloaded induction motor draws real power to overcome friction but reactive power to create the magnetic field to allow the motor to spin.  Look for that in a few weeks at www.eaton.com/experience.  I think the measurements are the easiest way to show what happens with induction motors that are unloaded (i.e. PF = 0.1) draw low Watts and high vars.  Loading the motor increases Watts but vars increase at a much lower rate until the motor reaches the rated PF of 0.8 or so.

The interesting thing that I have also thought of as I read through everyone's explanations is what about harmonics?  Harmonics are currents drawn by the load but do no useful work (so they are neither kW or kvar - they are sometimes explained as a 3rd dimension on the power triangle).  Similar to kvar, harmonics are difficult to explain but I feel that it is best through measurements of the actual waveforms. 

If you think about it, all electrical concepts are all difficult to "sense" and understand (you can't see electricity, hear it, or smell it unless something goes wrong and you shouldn't taste it - unless you are an old time electrician).  So, thank you to everyone for your interesting insights - I will certainly use some of these thoughts as I present about electrical power systems especially reactive power.

I liked that expression of an explanation to a six year old. I also liked the  fact that you are working against false energy savings claims with capacitor. I have seen some research papers based on such claims with some dimension-ally incompatible equations to support. Regarding harmonics, the higher frequencies that combine with the fundamental frequency results in increase of both eddy current loss and hysterisis loss in the ferromagnetic core ( in a dielectric the ESR is reduced and energy loss increases since Q=Xc/ESR) is again a source of heating ( conceptually) and can be measured. So it goes back to active power loss.

Is it temporary storage of electric energy converted to magnetic energy or electrostatic energy in one half cycle which is again released into electric energy in another half cycle ? In DC, can the storage and release of energy occur during a switch on / off cycle ?

Harmonics result in active power loss. It could definitely be treated as active power loss. But as reactive power???? Well ...I have doubts. Main doubt is the frequency. To which frequency can we relate, fundamental or harmonic? But frequency doesn't matter with active power loss. 

Dear Carnovale, I appreciate your idea about video. But the energy stored by capacitor is not i2r energy. 

Energy drawn by capacitor = i2r energy ( lost in the form of heat) + 1/2 ( cv2) energy. Please correct me if I am wrong.

MoreMoreover I have come across one article stating that super capacitors are being used as backup for electric locomotives for a brief period. Sorry to say I don't have reference. Thought it would be helpful. 

What I understood Carnovale was talking about is the loss reduction thro' reduction of reactive power being made much of by sellers of capacitor based circuits, claiming substantial gains which does not happen, as proved by him my measurements since the small gain is only in the loss reduction in the cables leading to the point of consumption. I don't think he talked about energy stored by capacitor. 

With the high capacity of super capacitors and the large torque requirement of traction, the controlled discharge of current from the capacitor may be a very good idea, especially because of the size reduction and low loss inherent in super capacitors and regenerative charging possibilities that can be investigated. 

Correct - my comments were about the I^2R losses in power lines feeding reactive loads and compensated by capacitors.  I didn't mean to steer the conversation so far away from the question but my comment related to the fact that reactive loads can affect real and reactive power.

Ultracaps or super caps are simply more dense capacitors having ratings in the Farad range instead of uF range.  I've used ultra caps to provide "ride-through" for about 2 seconds on a 50 HP VFD - again, we talk about stored energy and real power provided but it is small relativly speaking.

As far as harmonics on PQ meters today - they either put harmonics in the reactive power domain or use a 3D model and use a "Z" axis on the power triangle for harmonics.  Harmonics show up in the difference between TPF (or total PF) and DPF (displacement PF - fundamental PF). 

As explained by Dr. Amlan, the reactive power is the part of total power which is stored inside the reactances in the system. When there is reactances in the system, the current will be out of phase with the voltage. This current have one component in phase with voltage, called active component and other perpendicular to the voltage, called reactive component. Active power is the power produced by the  active component only. This is analogous to the definition of work done in mechanics. "The amount of work done is product of force and displacement in the direction of force". Reactive power is the power produced by the reactive component of current. It is stored in the reactances.

Rower generated from large 3-phase synchronous machines is complete (S), where S = P + jQ. The real power which does mechanical work is P; while the reactive power is largely absorbed by induction motors to set up the magnetic field in such loads. Without the magnetic fields, no work can be done in energy conversion (Faraday's law and flux linkage methods). Cheers.

If a linear load is connected to an ac voltage source, the load draws sinusoidal current, which is expressed by Ohm's Law involving the voltage and the impedance of the load. If the load is purely reactive, ie., purely inductive or purely capacitive, then it does not consume any average power even though there is current flow and there exists instantaneous power (the positive durations cancel with the negative durations). Thus, in such a situation, the reactive load, by virtue of drawing current, is said to be drawing "Reactive Power", although average Joules over a cycle is zero. Resistances, however, draw "Active Power" or "Real Power" since they consume Joules per cycle, converting it to heat energy.

In case the load comprises of a mixture of reactive load and resistive load, then the equivalent reactive component of the load continues to draw "Reactive Power" while the equivalent resistive component of load draws "Active Power" or "Real power". This is reflected into the total input current, which can be split into a component that is "Reactive" ie., that does not contribute to average power drawn, and another component that is "Active", which contributes to average power consumed.

It is the reactive component of load that does not consume average power. However, any current generated by a load (resistive or reactive) flowing through a non-ideal conductor, that is resistive, will cause resistive power loss in that conductor, but not in the load. Hence, it the total system comprising a purely reactive load supplied through non-ideal conductors is viewed (instead of the load alone), it is no more a purely 'reactive' load but a mixed load now. Hence there will be a component of power (or current) that is 'reactive power' for the system (corresponding to load) and a component that is 'active power' corresponding to the conductor resistance. Note that now, although the current through conductor and load is the same, the total system voltage impressed, is different from the voltage across the reactive load, due to drop in the conductor resistance.

I want to make you understand about reactive power by using simple circuit knowledge. Most of the load you see are inductive. But they are not purely inductive, but a mix of resistive and inductive. Similarly a very few load are capacitive, which is also not a pure capacitive load, but a mixing of capacitive and resistive part. We know power taken by a load is called apparent power. apparent power is composed of active and reactive power. The power consumed by the resistive part of a load is known as active power and it is, active power=VIcos(theta) and unit of this power is watt(W). where V is the voltage across the load, I is the current through the load and theta is the angle between current and voltage. The power consumed by the reactive (capacitive or inductive) part of a load is known as reactive power and it is, reactive power=VIsin(theta) and unit of this power is var (volt amp reactive). the difference between both of the equation is only on sine and cosine. So apparent power=sqrt(active power2+reactive power2)=VI and its unit is volt ampere. For your kind convenience I uploaded two photo. reactive power means lost of power. Hope you understand

Hmm.. Good pics and analogy except for the scalar addition, for my ignorance regarding the 'reactive '  aspect in the fizz ( unless the customer reacts at payment time), for the accompanying 90 degrees phase change, for a difference of opinion regarding 'the work done'...

In dc machines magnetising power(reactive power) which is used for energy conversion apply separately through poles and power which is to be converted applied through shaft in generators and through armature in motors. But in induction motors both are from stator only that is only reason to minimise the cost of induction motor,here both working power and magnetising power existing in single winding,further reduction of reactive power by designing machine with air gap as minimum as possible,air gap is required for separation of stator and rotor only.

Hi

http://www.pserc.wisc.edu/documents/publications/special_interest_publications/grid_reliability/Sauer_Reactive_Power_Sep_2003.pdf

http://www.electronics-tutorials.ws/accircuits/reactive-power.html

Power wasted (circulating power)in the inductance or capacitance acting as loads in an AC Circuit is reactive power

n most electrical circuits, Reactive Power comes from the creation of an electro-magnetic field necessary in motors and transformers. This impedes the electrical current, causing Reactive Power to be required.

Reactive power is actually the stored power in the circuit, which is not getting transferred to the load, nor it will be lost as heat. It is due to inductance/capacitance present in the circuit. 

How to explain the reactive power (fundamental, not harmonics) due to a simple dimmer circuit operation?

Hi Vasile,

Actually, I am asking how to explain the angle displacement between the fundamental voltage and the fundamental current without the presence of any energy storage element? If I add a suitable shunt capacitor in the circuit, the power factor becomes unity. So, the fundamental reactive power (and not "distortion power") must be real. How is it possible?

We know that reactive loads such as inductors and capacitors dissipate zero power, yet the fact that they drop voltage and draw current gives the deceptive impression that they actually do dissipate power. This “phantom power” is called reactive power, and it is measured in a unit called Volt-Amps-Reactive (VAR), rather than watts. The mathematical symbol for reactive power is (unfortunately) the capital letter Q. The actual amount of power being used, or dissipated, in a circuit is called true power, and it is measured in watts (symbolized by the capital letter P, as always). The combination of reactive power and true power is called apparent power, and it is the product of a circuit’s voltage and current, without reference to phase angle. Apparent power is measured in the unit of Volt-Amps (VA) and is symbolized by the capital letter S.

Is not the fundamental components of both voltage and current a result of Fourier analysis? If so, then we have no problem extracting the maximum amplitude sinusoid from both, calling them the fundamental components and seeing that there is a displacement angle between the voltage and current sinusoid. And such a situation must have a reactive power component. You must be familiar with the technique used in power electronic circuits where any distorting in the waveform is compensated to improve power factor. Same here. 

Hi Jose,

I do not belong to power electronic group. With whatever the little knowledge I have in power electronics, let me explain. Correct me if I am wrong.

Any thyristor based dimmer varies the out put voltage by changing the firing angle. The shape of the output current waveform is not same for different firing angles. For example, keeping the load same, if you change the firing angle from zero to 180, the shape of the current wave form will not remain same. Right? I am not talking about the amplitude, I am talking about the shape. 

Now, the fundamental component of the current waveform , in terms of amplitude and its phase angle wrt the fundamental component of voltage, depends on the shape of the waveform. I stress again , the phase angle of the fundamental component of current with the fundamental component of the voltage changes with the firing angle(correct me if I am not correct). 

I think this is the reason why you get a phase difference in a voltage dimmer.

The only "originator" of processes in an electrical circuit is always a source (voltage, current, real).

Part of the energy is converted into heat (directly consumed, resistor) - this corresponds to an active power (supplied by a source).

Part of the energy is stored in magnetic field - inductance - or in an electric field - the capacity. This energy is not lost - it returns "in the game" - this corresponds to a reactive power.

Reactive power is simply due to the Faraday Law. If a power system element draws magnetizing current from the source, it causes reactive power comsumption. If you compensate it (magnetizing current) with capacitors, that means you eliminate ractive power and Faraday Law of magnetizing.

By definition, instantaneous power is p(t) = i(t) x v(t). Assuming a sinusoidal shape signal and that the phase difference between i(t) and v(t) is alpha then the average power, over one period, is P = I x V x cos(alpha) where I and V are the RMS values of i(t) and v(t) respectively. What mathematical operation should I do over p(t) in order to get I x V x sin(alpha) which represents reactive power? I believe that reactive power must come from the same expression of instantaneous power as the active one. Any hint?

You can use "Pythagoras' Theorem", i.e. in a right angled triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In you case it will be: the square of the apparant power S (in VA) is equal to the sum of the squares of the active power P (in Watts) and square of the reactive power Q (in Var): S2=P2+Q2

So, you can made operations as follow:

- calculate instantaneous power p(t) = i(t) x v(t),

- calculate active power P as average value of instantaneous power p(t) over one period,

- calculate RMS values of voltage v(t) and current i(t) as Root Mean Square,

- calculate apparant power as S=VxI

- calculate reactive power Q=(S2-P2)1/2.

The second idea is to make operations as follow:

- becouse of known equotion: sin(alfa)=cos(alfa-90o):

- calculate values q(t)=i(t) x v(t-T/4), where T is period and T/4 is equal to phase shifting of the voltage v(t) by 90 degrees,

- calculate reactive power Q as average value of instantaneous values q(t) over one period.

This second idea is used for measurement reactive power by using classic wattmeter.

Phase shifting circuit called "the Hummel circuit" effects the necessary phase shift 90o. According to this idea a single phase reactive power meters are build.

Both ideas are true only for sinusoidal shapes of voltage an current.

For nonsinusoidal shapes it's a little bit more complicated ...

Typically explanations are based on old concept from AC beginnings. Not very surprised - for pure sin mains frequency, without green energy sources and smart grids they are true. As it was 100 years ago.

But today it's just a particular case - we live in era of electronic converters. Convert from AC to DC or vice versa - no problem. Variable frequency and voltage - no problem. From grid or back to grid - no problem.

The therm "power" - just a parameter how fast the work is done, energy is "the king". And a bit longer explanation:

All electrical devices do conversion as “general task, duty and job”. Electrical motors convert electrical energy to mechanical energy, light bulbs - to visible or invisible light, heaters – to heat and so on.

Let’s take a look on simple AC electrical motor. Rated power X watts, voltage Y volts, current Z amps and cos(60). One can say – very low cosφ and thus huge current in to supply wires. He is right but... cosφ 0,6 mainly depend from design of motor and without that motor simply doesn’t work - or we accept this osφ 0,6 or motor simply stand still. If we accept, then part of energy flowing from motor back to grid must be included in motor efficiency calculation. Here discussion about "active" or "reactive" - why? Some amout of energy in and some amout of that we get on rotor axis.

Motor is just a converter from electrical energy to mechanical energy with some mechanical loses and electrical energy loses. Adding capacitor mean adding extra charging - discharging energy source close to load and current value in to supply wires are reduced, but not in between capacitor and motor.

Negative energy flow is the same "recuperation" in AC circuit if one compare it to bi-directional DC or AC energy flow. Because time scale (20ms or 20 hours) doesn’t matter and such process are never defined under some time frame.

Or what to say in case of smart AC grids utilising electric car batteries? Or, form old days, squirrel cage motor as generator? What must be defined as "reactive" energy and what - as "recuperated" energy? In one case "reactive", in other -"regenerative", but energy flow direction is the same. New mismatch or new Imperial and Metric? But screw and nut do the same regardless Imperial or Metric...:)

Some mismatch of terminology and understanding today.

Very serious mismatch, I would like to say.

@ Eligiusz Pawlowski

Maybe I was not clear in my question. If I assume the existence of reactive power and its relation with active power then the math you present is obvious. What I'm saying is how the reactive power unveils mathematically from the application of the power definition to an sinusoidal based voltage and current. The active power comes, in a straightforward manner, from after applying the concept of average power over a period. My question is, if I was unaware of the existence of reactive power, I was expecting that the average power over one period was of the form P + j Q. And, by observing this latter expression, I could postulate the existence of two different kind of power over a load...

Your question was:

"What mathematical operation should I do over p(t) in order

to get I x V x sin(alpha) which represents reactive power?"

Maybe I didn't understand you properly.

So, what is your question exactly?

Are you looking for "any hint" about what?

p(t) = i(t) x v(t) -- this is the definition of instantaneous power

if

i(t) = sqrt(2) I cos(wt + alpha)

and

v(t) = sqrt(2) V cos(wt)

then

(1/T) integral_(from 0 to T) of p(t) = I V cos(alpha)

which is known as active power

my question is that the reactive component should also emerge from the same definition of power since there are not multiple definitions for electric power. I should expect that the average power along a period would lead to

IV(cos(alpha) + j sin(alpha)) where the first part is the active power and the second the reactive power. The modulus of this quantity would lead to the apparent power.

The pq theory [AKAGI 2007] proposes a precise mathematical modeling for the definition of the powers in GENERIC three-phase electrical systems (that may contain harmonics and unbalance between phases). This theory allows a differentiated mathematical treatment for such generic systems in which the classical theory based on phasors [CLOSE 1966] can not explain with accuracy, causing errors of interpretation on the real behavior of the flows and exchanges of energy in three-phase systems containing linear and non-linear elements, as the modern systems containing FACTS equipment and HVDC links.

In the system shown in the Figure, the three-phase load R1 ≠ R2 ≠ R3 causes unbalance between the system phases and the SVC causes network harmonics. In this figure the real and imaginary powers are indicated, where "~" refer to oscillating power and "-" refer to the constant power (average value). The imaginary powers (in red) correspond to the energies exchanged between the 3 phases of the system and they do not contribute in any moment to the energy transfer between the source and the load [AKAGI 2007].

Precisely for this reason, in systems containing HVDC links, it is incorrect the frequent affirmation of engineers that the network "supplies" reactive power to the AC/DC converter, or that the reactive power is "consumed" by the converter. The real three-phase power (in blue) supplied to the resistive load p3φ is equal to the sum of the real power p flowing through the three phases with the zero sequence power p0 flowing through the neutral conductor. It is also observed that the imaginary power q1 which circulates in delta connected SVC capacitors is different from the imaginary power q2 which circulates in the TCR terminals.

References: 

[AKAGI 2007] Akagi, H., Watanabe, E. H., Aredes, M., "Instantaneous Power Theory and Applications to Power Conditioning", New Jersey: IEEE Press / Wiley-Interscience, 2007.

[CLOSE 1966] Close, C.M., "Analysis of Linear Circuits", Harcourt College Pub, 1966.

It is the energy that is useless and lost. It is the power of electrical machines, such as transformers and motors, that work with coil and magnetic field. This power that we have lost actually has a duty as important as his work. The charges with the coil and the capacitor are generated by changing the magnetic field required to exchange power from the network and this is called reactive energy. Due to this magnetic field, there is a certain amount of unnecessary load on the transmission lines and no work is done with this load.

Some loads require to store energy to do actual work. If that energy is not dissipated and is sent back to the source, it will take up the transmission line capacity and increase the demand on the source (generator) without resulting in any work in the classical physics sense. As an example, inductive motors have to store energy in their magnetic fields to be able to do their job of converting the electric energy to torque (primarily). That energy is send back and forth between the source and the load (motor) at every AC cycle. This back-and-forth flow of energy strains the capacity of the source as well as the transmission line. A capacitor bank can store and provide this energy to the motor, at least partially.