Why isn't Reactive Power defined as Q = S - P ?

[vintageplayer]

The instantaneous power in an AC circuit is given by: p = i(t)⋅v(t) = V_rmsI_rmscosφ + V_rmsI_rmscos(2ωt -φ).

The average power P = V_rmsI_rmscosφ is often a useful quantity to know. For example, it can tell me the work being done by a motor.

The apparent power S = V_rmsI_rms is also a useful quantity to know. For example, it can tell me the maximum work the motor could potentially do (when the current and voltage are in phase φ=0).

However... the reactive power Q = V_rmsI_rmssinφ = sqrt(S² - P²) doesn't really tell me anything useful. All I know is that it is some unintuitive measure of the difference between the apparent and average power. Why isn't reactive power simply defined as Q = S - P instead?

 

[Hesch]

Why isn't reactive power simply defined as Q = S - P instead?

Because P and Q have perpendicular directions. Thus Pythagoras must be used.

 

[tech99]

The instantaneous power in an AC circuit is given by: p = i(t)⋅v(t) = V_rmsI_rmscosφ + V_rmsI_rmscos(2ωt -φ).

The average power P = V_rmsI_rmscosφ is often a useful quantity to know. For example, it can tell me the work being done by a motor.

The apparent power S = V_rmsI_rms is also a useful quantity to know. For example, it can tell me the maximum work the motor could potentially do (when the current and voltage are in phase φ=0).

However... the reactive power Q = V_rmsI_rmssinφ = sqrt(S² - P²) doesn't really tell me anything useful. All I know is that it is some unintuitive measure of the difference between the apparent and average power. Why isn't reactive power simply defined as Q = S - P instead?

The definition you give is fine; the reactive power is the energy stored in the system.

 

[vintageplayer]

Because P and Q have perpendicular directions. Thus Pythagoras must be used.
View attachment 94557

This is circular reasoning. We defined it to be that way!

 

[Hesch]

Why isn't reactive power simply defined as Q = S - P instead?

Q = S - P →
S = P + Q , which doesn't make sense.

A phase shifted current, I / φ, can be represented by the complex value , I / φ = I*cos φ + j*I*sin φ.

Thus S = V * I = V * ( I*cos φ + j*I*sin φ ) = P + jQ ,
j = 1 / 90°

So the only "definition" comes from the complex representation of the phase shifted current, I / φ.

 

[jim hardy]

When you study generators
peaks of reactive current and 'real' current occur with rotor directly or perpendicular to stator winding
so it was kinda natural for the old timers' thought processes to go with that complex representaion
figuring it out the first time from physical hands-on is different from trying to back into it from algebra.

 

[vintageplayer]

Q = S - P →
S = P + Q , which doesn't make sense.

S = V_rmsI_rms
P = V_rmsI_rmscosφ
Q = V_rmsI_rms - V_rmsI_rmscosφ

Then S = P + Q would make sense. This definition for reactive power is more intuitive because you can easily tell how much power in Watts is being stored, and how much power you are using for useful work. For example, 100VAr would mean you are missing out on 100W of extra useful power.

Thus S = V * I = V * ( I*cos φ + j*I*sin φ ) = P + jQ ,

You're right that it comes out naturally with: VI* = S = P + jQ as long as you define reactive power to be Q = V_rmsI_rmssinφ. The single complex number contains a lot of information: P = Re{VI*}, Q = Im{VI*}, S = mag{VI*}, φ = arg{VI*} which is probably why Q was defined this way. On the other hand, you're now stuck with an ugly definition for reactive power. For example, 100VAr does not tell you how much power is being stored in Watts.

When you study generators
peaks of reactive current and 'real' current occur with rotor directly or perpendicular to stator winding
so it was kinda natural for the old timers' thought processes to go with that complex representaion
figuring it out the first time from physical hands-on is different from trying to back into it from algebra.

Do you have an example (or link to one) of this?

 

[Hesch]

you're now stuck with an ugly definition for reactive power.

How would you calculate:

100 [W] + 100 [VAr] = 200 [unit ?] ?

 

[jim hardy]

Do you have an example (or link to one) of this?

https://www.physicsforums.com/threads/armature-reaction-drop.826513/#post-5191083

 

[cnh1995]

Cosider a box floating in space in the absence of gravity.

If you give some velocity 'v' to the block at an angle Φ, it will move horizontally with velocity vcosΦ and vertically with velocity vsinΦ. If you know the vertical velocity vsinΦ and horizontal velocity vcosΦ and you are asked to find the total velocity v, would you say it is vsinΦ+vcosΦ? No, because velocity is a vector, you MUST consider it's direction and therefore, the angle it makes with the reference. Hence, you'll have to use Pythagoras theorem here for vector addition. Similarly, voltage and current are PHASORS. Hence, you must consider phase angles as Hesch has demonstrated in #2 and #5. V_rmsI_rmscosΦ is the power dissipated in the circuit, hence, V_rmsI_rmssinΦ turns out to be the reactive power in the circuit. Apparent power is the phasor addition of the powers and not simple arithmatic addition.
Also, you can verify this by drawing some waveforms. Take a simple RL ac circuit with some convenient component values and plot the waveforms of active power(power dissipated in R) and reactive power(power absorbed by L)separately for a complete cycle and then calculate their corrosponding average values.

 

[Sergey_KGB]

Once, I supervized installation of meteting equipment on Power Substation 220KV in Bashkortostan (Russia). Our device showed that nearby plant consumes only 0.5MW of active power but generates 10MVAR of reactive power (due to heating inductors that were switched off just before the measurement).
As for the question then let's look at the speed in horizontal and vertical directions. No one would mechanically sum them. Formula
V=SQRT(Vh*Vh+Vv*Vv)
is almost obvious.
The situation with active and reactive powers is essentially the same. They are (in specific meaning) orthogonal to each other.
Sorry, haven't read previous answers carefully.

 

[tech99]

But is it not true that power is a scalar quantity and does not therefore have a phase angle, as do voltage and current?

 

[Sergey_KGB]

If {Vn} and {In} are series of samples (voltage and current) then
P - active power is a sclar multiplication P=ΣVn*In
and thus is a scalar value while voltage and current are vectors. Though RMS values of both voltage and current are scalar values (length of respective vectors).

 

[sophiecentaur]

But is it not true that power is a scalar quantity and does not therefore have a phase angle, as do voltage and current?

Yes. Power must be a scalar quantity because Energy is scalar and power is energy per unit time. Imo, "reactive Power" is really an oxymoron which has been introduced into the vernacular for convenience. In a purely reactive component, energy is stored and not dissipated so the 'lost power' must be there because of resistive components in the supply impedance. If people stick with volts and current for their calculations then they can't go wrong and will get the right answers (arithmetic permitting).

 

[tech99]

Agree, so it seems incorrect to show vector diagrams of P and Q.

 

[sophiecentaur]

Agree, so it seems incorrect to show vector diagrams of P and Q.

I'm not a Power Engineer so I would have to be more careful with my 'shorthand', which is what that diagram is. I would do any calculations the 'formal' way and stand a better chance of getting the right answer. A PQ vector diagram has got to be wrong but it 'may' be forgiveable if the people who use it are aware that they are being sloppy.

 

[jim hardy]

Phasors aren't vectors I've been told by folks who are better at maths than i..

I'd have to go back to Steinmetz - i think it was he who came up with idea of complex numbers to represent AC current...

 

[vintageplayer]

Agree, so it seems incorrect to show vector diagrams of P and Q.

Assume a voltage Vcos(wt) across an impedance of R + jX = Z∠φ

The total instantaneous power is:
p(t) = v(t).i(t) = Vcos(wt).Icos(wt - φ) = VI.cos(wt).[cos(wt)cosφ + sin(wt)sinφ] = (VI/2)cosφ[1 + cos(2wt)] + (VI/2).sinφ.sin(2wt)

The expression above is the total power consumed, and it does not tell you the power consumed by the resistor, or the power consumed by the reactance. The power consumed by the resistor is actually NOT (VI/2)cosφ[1 + cos(2wt)], nor is the power consumed by the reactance (VI/2).sinφ.sin(2wt).

To find the power consumed by the individual resistor and reactance, you have to find the voltage across these individually (e.g. by voltage divider).
v_R(t) = (VR/Z).cos(wt - φ) = Vcosφcos(wt - φ)
v_X(t) = (VX/Z).cos(wt + 90 - φ) = Vsinφcos(wt + 90 - φ)

Hence:
p_R = v_R(t).i(t) = Vcosφcos(wt - φ).Icos(wt - φ) = (VI/2).cosφ.[1 + cos(2wt - 2φ)]
p_X = v_X(t).i(t) = Vsinφcos(wt + 90 - φ).Icos(wt - φ) = (VI/2).sinφ.sin(2φ - 2wt)

Note that adding these two terms together gives the total instantaneous power above (after some further trig identities).

The average power consumed is P = V_rmsI_rmscosφ.
The peak power consumed by the reactance is Q = V_rmsI_rmssinφ.

Although both of these quantities are scalars and mean different things (P is the average power, and Q is the peak power consumed by the reactance), you can artificially construct a vector of magnitude V_rmsI_rmsand angle φ. The x-component of this vector will be the average power P, and the y-component will be the reactive power Q.

I could also define another scalar quantity, Q = S - P, and call it the "reactive power", but it would not be the peak power consumed by the reactance. If you want the "reactive power" to be the peak power consumed by the reactance, then it must be equal to V_rmsI_rmssinφ.

 

[sophiecentaur]

(P is the average power, and Q is the peak power consumed by the reactance),

If we are trying to get to the bottom of something as fundamental as this, we could at least get the basics right. How can any power be "consumed" by a reactance? The Current and Volts are in quadrature. The reason that Power Factor is relevant is that the I and V are greater when there are reactive elements in the circuit. The excess current is passing through the Resisitive components of the supply circuit, which 'consume' extra Power. Also, the extra Volts may be relevant.
There is more power consumed when the PF is not unity but there is no way it can be dissipated in reactances. Reactances can only store energy and return it (twice) per cycle. The 'Peak' power involved, (rate that energy is being stored, instantaneously) can be given a value but it is not very meaningful

 

[vintageplayer]

If we are trying to get to the bottom of something as fundamental as this, we could at least get the basics right. How can any power be "consumed" by a reactance?

Power is consumed by a reactance half the time, and released half the time. Just because on the average the power consumed is 0 doesn't mean a reactance doesn't draw power for half the cycle. The peak power it draws is equal to Q.

The 'Peak' power involved, (rate that energy is being stored, instantaneously) can be given a value but it is not very meaningful

That value is the "reactive power" Q agree?

 

[sophiecentaur]

That value is the "reactive power" Q agree?

You can give it a name and you can calculate it but how does that value contribute to understanding the AC losses in power engineering? What counts is the wasted mean power in the resistive parts of the supply circuit. You could just as easily talk about the "peak"power being dissipated in a resistive load but who does? If there were no resistance in the supply chain, your "reactive power" would be irrelevant because there would be no supply losses.

 

[jim hardy]

A picture is generally wort a thousand words.
It was helpful for me to look at sinewaves:
draw two of them one above the other
label one Volts and the other Amps
draw a pair in phase,
a pair 180 degrees out of phase,
a pair 90 degrees out of phase and
a pair 45 degrees out of phase

then consider instantaneous power, product volts X amps

For the in phase sinewaves , product is always positive , P = volts X amps, which represents power factor of 1.0 (cos 0 degrees)
for the pair 180 out, product is always negative, power = volts X amps and one of them is always negative , power factor = -1.0 which is cosine 180 degrees

for the others , power is sometimes positive and sometimes negative , over a complete cycle it will average to VIcosθ,
that's what reactive power does - shuttle back and forth between source and load during subcycle intervals.
At 90 degrees observe positive and negative intervals are equal and their average power is zero, e power factor of zero which is cos 90 degrees. Nothing gets hot or does any work.
At 45 degrees - would someone less awkward copy and paste some sinewaves into Paint and post ? That's been done someplace on PF (i remember doing it) but i was unable to find, and paint frustrates me terribly.

You'll find P = VICosθ and Q = VISinθ , and Pythagoras Rules !
I think that visual exercise is a good learning tool. It'll cement the concept.

old jim

 

[vintageplayer]

You can give it a name and you can calculate it but how does that value contribute to understanding the AC losses in power engineering? What counts is the wasted mean power in the resistive parts of the supply circuit. You could just as easily talk about the "peak"power being dissipated in a resistive load but who does?

You might want to know the reactive power if you ever want to control the voltage on a power grid. Or size your capacitors for power factor correction. Or size your power cables efficiently. Also, peak power is an important design spec for almost all electrical devices...

 

[sophiecentaur]

You might want to know the reactive power if you ever want to control the voltage on a power grid. Or size your capacitors for power factor correction. Or size your power cables efficiently. Also, peak power is an important design spec for almost all electrical devices...

To correct for the power factor of your load you need to know its reactance and then you tune it out as near as you can, using another reactance network. When would you calculate 'Peak Power'? True, the Peak Voltage could be very relevant - but that actually corresponds to the maximum Energy (E = CV²/2) and not to the rate of energy supplied (Power). Sizing your power cables will affect the Resistance and not the Reactance and I've already been into that.
I realize that these terms are all down to usage and it is not likely that they will change. It doesn't make them correct though. "Make it Work' type Engineering is full of slightly dodgy terms and statements - from Water Analogies where they don't apply, to the expression "dB Volts". The field is littered with pitfalls because people got taught these things and pass them on. But Engineers do 'make things work' very successfully and we mostly stuff get away with approximate thinking and that magic stuff called 'experience'.

 

[jim hardy]

aha found it

single phase volts, amps, and power
when volts and amps are in phase ie power factor = 1
observe instantaneous power is always positive

single phase volts, amps and instantaneous power
when volts and amps are out of phase by 90 degrees
observe instantaneous power no longer unipolar, in fact averages zero; cos(90degrees) = 0...

now

The peak power it draws is equal to Q.

That value is the "reactive power" Q agree?

no i don't agree.
do we have a definition of terms question here?

Where would you select peak power on that graph? Copy&paste into Paint and add an arrow...
I'd guess Q is average of the orange power wave .

those graphs came from http://www.electrical4u.com/electric-power-single-and-three-phase/11

 

[vintageplayer]

Where would you select peak power on that graph? Copy&paste into Paint and add an arrow...

For a purely reactive component p(t) = (VI/2).sinφ.sin(2φ - 2wt). Q is therefore the peak value of the AC power waveform.

When you have both resistive and reactive components, p(t) = (VI/2)cosφ[1 + cos(2wt)] + (VI/2).sinφ.sin(2wt). Q is the peak of the sine component, but there is no nice way to show it graphically without drawing the sinusoid components separately.

I'd guess Q is average of the orange power wave .

The real power P = (VI/2)cosφ is the average of the power wave, not Q.

 

[Wee-Lamm]

Why isn't reactive power simply defined as Q = S - P instead?

For the math side, let's presume a simple measure where S=33 and P=22.
Q = S - P
Q = 33-22
Q = 11

Whereas,
Q = sqrt(33x33 - 22x22)
Q = sqrt(1089 - 484)
Q = sqrt(605)
Q = 24.596
Certainly a nontrivial difference, imo.

However... the reactive power Q = VrmsIrmssinφ = sqrt(S2 - P2) doesn't really tell me anything useful. All I know is that it is some unintuitive measure of the difference between the apparent and average power.

I think we have to remember that S and P are not in phase with each other nor is either one a constant value, largely due to their interaction with each other.
If we relate this to a musical instrument amplification system, for example, being able to measure reactance and it's impact on the waveform at the output, can give us a better understanding of the "perceived" volume of the output.

 

[sophiecentaur]

aha found it

single phase volts, amps, and power
when volts and amps are in phase ie power factor = 1
observe instantaneous power is always positive

single phase volts, amps and instantaneous power
when volts and amps are out of phase by 90 degrees
observe instantaneous power no longer unipolar, in fact averages zero; cos(90degrees) = 0...

now
no i don't agree.
do we have a definition of terms question here?

Where would you select peak power on that graph? Copy&paste into Paint and add an arrow...
I'd guess Q is average of the orange power wave .

those graphs came from http://www.electrical4u.com/electric-power-single-and-three-phase/11

Good pictures there but your statement here demonstrates the limits of such a graphical approach; you have to infer things that calculus and the complex Maths will automatically prove for you. Furthermore, you would really need to be drawing graphs showing the instantaneous Volts and Current on all R and X elements separately. That would give you some pictorial 'proof' of the situation.
Sorry - I can'tmanage to make the folloing red bits change into black!
It's pretty well accepted that complex notation for Impedance gives the answers for simple circuit questions like this. No one (??) would need to work out the power dissipated in a resistor with an AC applied voltage on a stepwise time basis - at least, not more than once. Likewise, complex arithmetic will give tha answers to all other questions in linear circuits. I would actually disaggree that a picture tells the whole story. Fact is that I am not particularly confident about my Maths; I always go through things several times before believing an an answer. But I basically BELIEVE in it and I accept the results of it. The answer to the main question in this thread is dead straightforward if you draw a realistic equivalent circuit of the system: Voltage Source -> Source R and Source X -> Load R and Load X. You just have to believe that Reactances cannot dissipate Power so wasted power has to go into the source resistance. Instantaneous Power is a totally incidental concept in this question and the calculations do not need to include Phasors, explicitly - why would anyone not believe in such well established procedures?
If the internal resistance and the resistance of supply cables is low enough, compared with the Load Resistance, then the only problem with a non unity PF is the excess volts.

 

[Wee-Lamm]

You just have to believe that Reactances cannot dissipate Power so wasted power has to go into the source resistance.

Wasted power implies that the actual output would be less than the expected average output, whereas an ideal reactance should have 0 resistance and effectively should increase the power output. In terms of sound amplification from a listeners perspective, reactance should make it sound louder.

 

[sophiecentaur]

You just have to believe that Reactances cannot dissipate Power so wasted power has to go into the source resistance.

Wasted power implies that the actual output would be less than the expected average output, whereas an ideal reactance should have 0 resistance and effectively should increase the power output. In terms of sound amplification from a listeners perspective, reactance should make it sound louder.

You have to be careful to specify the exact situation you are describing. You have to be making the right comparison between systems. There will always be wasted power in supply cables etc. That can reduced by reducing the quadrature current flowing through them (that what PF correction is for). . The presence of extra reactances will also alter the power delivered to the load but the consumer pays for what's used - so it's 'fair'.
Your example of audio amplification is the same. An amplifier with zero output impedance (ideal but not far from what you can achieve) will deliver more power to a speaker if there is correct tuning out of reactive elements in the speaker (same for a transmitting antenna).

 

[jim hardy]

I'd guess Q is average of the orange power wave .

Well ! That was sure a false cue !

big ten-oops , what was i thinking

still I'm confused by your use of term PEAK. I'm accustomed to it meaning the top of a wave as opposed to average or RMS.

 

[Wee-Lamm]

There will always be wasted power in supply cables etc

I was ignoring line loss as it would be present whether there was an Inductance or Capacitance in the circuit or not, and would not change the formula, but only the values used.

We also know there are 2 types of Reactance that can be made to occur in an AC circuit, Inductance and Capacitance.

Inductance is aimed at maintaining a more constant current flow over a period of time, in that it creates a magnetic field in which it stores more energy as input voltage increase and releases more energy as input voltage decreases. This process of storing/releasing energy causes the current and voltage cycles to be 90 degrees out of phase and ultimately results in a current flow that rises and falls less dramatically than it would without inductor. Ignoring the heat loss of the materials used in the coil and core, the inductor itself does not result in a decrease in power but rather, it results in a higher overage output of power over time and there is less loss resulting from the cycling nature of ac current.

Capacitance is aimed at maintaining a more constant voltage output over a period of time, in that it creates a static electrical field in which it stores energy more rapidly as current increases, and releases energy as input current decreases. Capacitors are more prone to failure and leakage, and do tend to generate more loss through heat by virtue of the materials they are made of, but the act of storing the energy itself does not create a lost of power, it merely stores it to be released as voltage during a period of reduced current input. Capacitors do have the added benefit of blocking DC current while allowing AC to pass through, but that is trivial to this discussion.

In this regard, and ignoring losses that relate only to the materials used to make Inductors and Capacitors, the storage and release of energy itself 'should not' create a loss of power. I suggest that, when measured over a period of time, they each increase the amount of power, in presuming that 'Power' is the measure of how much work can be done by the energy being exchanged.

 

[jim hardy]

Good pictures there but your statement here demonstrates the limits of such a graphical approach; you have to infer things that calculus and the complex Maths will automatically prove for you.

i like to work the math until i get same answer 3 times in a row
and i can see it agree with a picture

but i didn't do my due diligence for that post...

 

[sophiecentaur]

still I'm confused by your use of term PEAK.

I'm not really in favour of 'Peak Power" as applied in this thread because it's been used in strange ways. The Energy going into and out of a reactance will vary during the cycle and there was the implication that its maximum value is somehow relevant. The "peak Power" into a Capacitor is not at the voltage peak, because that's when the energy is not increasing any more (P=dW/dt) and it's not at a zero crossing, where the current is at its max but the volts are zero. That'w why I have been objecting to the relevance of Power in a Reactance. What counts, surely, is Mean Power which is what gets things hot and costs the money. Power is something associated with Resistance (or the work done by an electric motor). The extra currents and volts, due to reactive load (and transmission) components will cost you because that current flows through actual resistances and can stress components beyond what they're designed for.
In any case, I think this thread has shown that the shorthand formula it started with, has no meaning and is not valid. Treating scalars as if they are vectors really can't ever make sense.

 

[sophiecentaur]

in presuming that 'Power' is the measure of how much work can be done by the energy being exchanged.

Power is defined as rate of energy transfer per unit time. If it's transferred in the load, it will be useful. If it's transferrend in the supply chain, it's loss and will cost everyone, unnecessasarily. A Power source with no resistance (impossible) will not dissipate any power. All generators have finite resistance; they are made with as low resistance as possible (at a cost). Likewise the cables.
I think you are getting too preoccupied with the mechanisms of Reactance for the purpose of this thread. The mathematical description of their behaviour with an AC signal is all that's needed here. The theory is quite good enough for this discussion.

 

[lychette]

To correct for the power factor of your load you need to know its reactance and then you tune it out as near as you can, using another reactance network. When would you calculate 'Peak Power'? True, the Peak Voltage could be very relevant - but that actually corresponds to the maximum Energy (E = CV²/2) and not to the rate of energy supplied (Power). Sizing your power cables will affect the Resistance and not the Reactance and I've already been into that.
I realize that these terms are all down to usage and it is not likely that they will change. It doesn't make them correct though. "Make it Work' type Engineering is full of slightly dodgy terms and statements - from Water Analogies where they don't apply, to the expression "dB Volts". The field is littered with pitfalls because people got taught these things and pass them on. But Engineers do 'make things work' very successfully and we mostly stuff get away with approximate thinking and that magic stuff called 'experience'.

mmmm...can you give some examples of water analogies that are useful...I use them all the time...How can we stop people who use "dB Volts" from using and communicating in this way.

 

[vintageplayer]

If there were no resistance in the supply chain, your "reactive power" would be irrelevant because there would be no supply losses.

Not everything is about calculating supply losses. The reactive power Q is related to the rate of change of energy stored in your capacitors (1/2⋅CV^2) or inductors (1/2⋅LI^2). One utility of Q is that it allows you to calculate the maximum energy stored in your reactive elements during the cycle. In some contexts (e.g. reactive power control) it is also convenient to talk in terms of reactive power for controlling a voltage level.

I realize that these terms are all down to usage and it is not likely that they will change. It doesn't make them correct though. "Make it Work' type Engineering is full of slightly dodgy terms and statements

There is nothing incorrect about a reactive power. It is the maximum rate of change of energy across a capacitor or inductor. It is a valid measure of power not a "dodgy approximate".

 

[sophiecentaur]

mmmm...can you give some examples of water analogies that are useful...I use them all the time...How can we stop people who use "dB Volts" from using and communicating in this way.

Your analogies may well be valid. The problem is that 'other people' tend to be too literal in their interpretation and their memory of what you may have told them. My experience is that, here on PF (even) there are statements which are intended to be helpful, made by people who do not cross their fingers or give caveats when 'explaining' things using such analogies. The poor, unsuspecting recipient grabs it with both hands and takes the statement and treats it as gospel and passes it on - and so it goes on and on.
As for the dreaded 'dB Volts', it's probably a lost cause but people actually believe that a transformer is an amplifier, because of it.

 

[sophiecentaur]

Not everything is about calculating supply losses.

I agree that component stress is also relevant but normal circuit analysis will tell you the peak volts and current in a reactive load. I do not have experience in Power Engineering but have not come across the specification for a component in terms other than maximum working voltage and 'ripple current' etc.. How does one measure this Reactive Power quantity? If you were to do the sums and arrive at a value for the Reactive Power, I wonder how it could be applied to a circuit with more than one reactive element and aportion the component stress, without going back to square one and doing a formal analysis of the circuit.
If reactive power is a well defined quantity and has been explained in a reference then perhaps you would give a link and I can then read someone (authoritative) else's take on it. I am just quoting stuff that you can find in any circuit analysis reference that you may have handy.

 

[Wee-Lamm]

Treating scalars as if they are vectors really can't ever make sense.

Agreed. :-)

 

[Wee-Lamm]

Not everything is about calculating supply losses. The reactive power Q is related to the rate of change of energy stored in your capacitors (1/2⋅CV^2) or inductors (1/2⋅LI^2). One utility of Q is that it allows you to calculate the maximum energy stored in your reactive elements during the cycle. In some contexts (e.g. reactive power control) it is also convenient to talk in terms of reactive power for controlling a voltage level.
There is nothing incorrect about a reactive power. It is the maximum rate of change of energy across a capacitor or inductor. It is a valid measure of power not a "dodgy approximate".

Reactive power is a "limit" to the maximum rate of change as a function of design, but it is more akin to a power conditioner in that it extracts and stores power from the circuit during periods of peak supply, then releases the stored energy during periods of lowering supply. Reactive power should not cause any power loss to the measure of average or real power, but it's nature of being out of phase with reference to real power, should reduce the valleys of low power, throughout each cycle.

For reference;
An inductor drops current and passes voltage through, where a capacitor drops voltage and passes current through.

I also feel that there is a difference between a 'functional perspective that considers the math' and a 'math perspective that considers function'. With this in mind, I agree with your comment that it isn't always about measuring supply losses. Supply loss is indeed a nontrivial matter but it should not change the amount of impact a reactive load has on a circuit.

 

[vintageplayer]

I agree that component stress is also relevant but normal circuit analysis will tell you the peak volts and current in a reactive load. I do not have experience in Power Engineering but have not come across the specification for a component in terms other than maximum working voltage and 'ripple current' etc.. How does one measure this Reactive Power quantity? If you were to do the sums and arrive at a value for the Reactive Power, I wonder how it could be applied to a circuit with more than one reactive element and aportion the component stress, without going back to square one and doing a formal analysis of the circuit

Imagine you have an electrical device that is largely inductive that you need to connect to the grid. Let's say the average real power it consumes is 1MW. Guess what is going to happen if the power station only burns 1MW worth of coal? Your device is not going to run because it's not getting enough power supplied to it. The power station still needs to supply extra power to your load for the inductance, even though it gets this energy back a second later. Power stations need to know how much reactive power to supply as well as average real power. Reactive power is much nicer to talk about in this context than the voltage and current waveforms.

If reactive power is a well defined quantity and has been explained in a reference then perhaps you would give a link and I can then read someone (authoritative) else's take on it. I am just quoting stuff that you can find in any circuit analysis reference that you may have handy.

Pretty much any electrical textbook defines reactive power pretty precisely... Q = Im{VI*}

 

[sophiecentaur]

Pretty much any electrical textbook defines reactive power pretty precisely... Q = Im{VI*}

Not having a textbook available, I looked at Wiki and other sources and noticed that they all seem to rely on a diagram that implies the energy is a vector quantity. That is pretty nonensical as the basis for strict treatment of the topic. Reactive Power may be a useful number to use when describing 'good' or 'bad' loads but it seems to be based on a very dodgy first step in its derivation. Reactive components do not dissipate power so they cannot, in themselves, consume any power from the supply.
I found a very revealing paragraph http://www.allaboutcircuits.com/textbook/alternating-current/chpt-11/true-reactive-and-apparent-power/:
"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".
He actually comes clean about it. What's being described is not Power - it's all in a name that's been (I would say mis-)applied to make the practicalities a bit more approachable. It's an analogy, in fact.

When you bring the idea of supplying coal to a power station you are begging a lot of questions and assuming that the term is fully justified. If you take a simple voltage source / load model and change the impedance of the load from resistive to partly reactive then the power dissipated by the load will be affected and the power supplied will be affected by the same amount. No Loss would be involved; the power delivered by the Voltage source will change by the same amount as the load power changes. So what's the difference between an ideal voltage source and a power station? An alternator winding has resistance and so will the supply cable. If a load is partly reactive then, in order to get the same Power transferred to the load, it resistance would have to be reduced (or the alternator voltage increased). That would increase the peak current values in the supply, which would increase the dissipation in the Resistive Parts of the supply circuit. The economics of supplying Electricity are crucial and supply circuit resistance is highly relevant. If the peak current increases by 5%, the lost power will be 10% more. That's not a 'Reactive Loss'; it's a Resistive Loss, brought about by the presence of reactive elements.
The increased stress on components, when PF is not unity can also be relevant but the "phantom power" is not relevant - it's the increased Voltage or Current.

 

[vintageplayer]

Not having a textbook available, I looked at Wiki and other sources and noticed that they all seem to rely on a diagram that implies the energy is a vector quantity. That is pretty nonensical as the basis for strict treatment of the topic. Reactive Power may be a useful number to use when describing 'good' or 'bad' loads but it seems to be based on a very dodgy first step in its derivation. Reactive components do not dissipate power so they cannot, in themselves, consume any power from the supply.
I found a very revealing paragraph http://www.allaboutcircuits.com/textbook/alternating-current/chpt-11/true-reactive-and-apparent-power/:
"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".
He actually comes clean about it. What's being described is not Power - it's all in a name that's been (I would say mis-)applied to make the practicalities a bit more approachable. It's an analogy, in fact.

Read a reliable source?

Reactive components do not dissipate power so they cannot, in themselves, consume any power from the supply

They draw instantaneous power from the supply given by:

p(t) = v(t).i(t) = Vsinφcos(wt + 90 - φ).Icos(wt - φ) = (VI/2).sinφ.sin(2φ - 2wt)

Reactive power is defined as the maximum of this value Q = (VI/2).sinφ.

If you worked in a power station, your plant would need to be constantly running to supply and absorb this power. This means your generator will be constantly spinning, and producing a maximum of Q Watts. You could perpetually recoup this energy back from the grid (the power you send out is always returned) but you will regardless need to be outputting Watts. This is one example why reactive power is important, because it equates to extra Watts needed to be produced by a power station.

 

[Wee-Lamm]

Imagine you have an electrical device that is largely inductive that you need to connect to the grid. Let's say the average real power it consumes is 1MW. Guess what is going to happen if the power station only burns 1MW worth of coal? Your device is not going to run because it's not getting enough power supplied to it. The power station still needs to supply extra power to your load for the inductance, even though it gets this energy back a second later. Power stations need to know how much reactive power to supply as well as average real power. Reactive power is much nicer to talk about in this context than the voltage and current waveforms.

Reactive power is not part of the supply equation, but is a part of the load equation. More specifically, reactive power is a design characteristic of the endpoint device and not specifically provided by the supply. Supply is typically provided in a predetermined format, that is the voltage and current have maximum and minimum quantity thresholds, which are a nature of the design characteristics of the supply medium.

It is the device which must conform to the format of the power it can receive from the supply, then it may/will manipulate that energy to a format that suit it's own internal needs. In relating this to your scenario of a 1MW device receiving a 1MW supply, we must first acknowledge that the supply rates do fluctuate within predetermined tolerance levels. Voltage and current from the supply can fluctuate by even 10% +/- of the average rate and not have a tremendous impact on most devices, as it is expected that over time there will emerge an average supply to which the device should have been designed.

Devices that require more consistent supply characteristics, such as non-fluctuating voltage and current, can implement Reactive Power Networks which present themselves as the Power Supply 'interface' to the device. Simply, it makes more sense to have the unique endpoint device conform to the average supply presented, than to expect the supply to conform to many millions of possible unique endpoint requirements.

Edit: To adjust quote tag so my comment did not appear as part of the quote.

 

[sophiecentaur]

Read a reliable source?

Did you read the whole of that link I referenced?

They draw instantaneous power from the supply given by:

p(t) = v(t).i(t) = Vsinφcos(wt + 90 - φ).Icos(wt - φ) = (VI/2).sinφ.sin(2φ - 2wt)

And that expression is positive for half a cycle and negative for the other half - integrating to zero over a cycle. So it doesn't contribute to wasting energy. That is what I cannot get my head round.

I must try to find some source that doesn't just jump in, feet first with the 'power triangle'. There must be a lead in that justifies it better than just making assertions. Perhaps I shall have to reach for my pencil and paper.

 

[Wee-Lamm]

Not having a textbook available, I looked at Wiki and other sources and noticed that they all seem to rely on a diagram that implies the energy is a vector quantity. That is pretty nonensical as the basis for strict treatment of the topic.

Let's define a vector as "a quantity having direction as well as magnitude, especially as determining the position of one point in space relative to another. I think then; that Reactive power, requires us to view them as a vector quantity in order to understand how and where, or rather when, they fit within the device; with regard to it's power needs.

Capacitance and Induction are ways of reshaping the waveform presented at the supply, in ways that cannot be accomplished with purely resistive loads, for a simple example, converting a square wave into a sine wave. The nature of Reactive power implies that that portion of the power is somewhat time portable in that we can manipulate it to occur in a time reference from the one they were extracted from. This in turn requires us to know which point along the wave we are measuring as it relates to the mean power, what direction it is moving, and how it's qualities change in reference to it's other qualities as well as those of the original wave.

 

[vintageplayer]

And that expression is positive for half a cycle and negative for the other half - integrating to zero over a cycle. So it doesn't contribute to wasting energy. That is what I cannot get my head round.

Over the cycle it doesn't waste energy. The power station (or voltage source) always gets back what it puts out. The problem is that with a power station, this still amounts to the generator running even if you are capable of recouping all the energy back. This wears your plant and costs money because you have to pay staff. You also have maximum outputs on your generators so the more reactive power you put out, the less real power you can put out (and real power is what the power stations get paid for).

 

[jim hardy]

Reactive power is defined as the maximum of this value Q = (VI/2).sinφ.

(VrmsIrms)sinθ is way more familiar to me. I do concede that's same as (VpeakIpeak/2)sinθ

If you worked in a power station, your plant would need to be constantly running to supply and absorb this power.

true enough,, if the generator is not turning it makes no volts.

This means your generator will be constantly spinning, and producing a maximum of Q Watts.

We in the plant would say Q Volt-Amps

You could perpetually recoup this energy back from the grid (the power you send out is always returned) but you will regardless need to be outputting Watts.

We DO recoup the energy with every sub-revolution of the shaft
and we don't call them Watts because over any time interval perceptible to a human they average zero

This is one example why reactive power is important, because it equates to extra Watts needed to be produced by a power station.

That is a very strange dialect to impose on the jargon for a discussion with old power plant guys. Plants deal in watt-hours not watt-milliseconds or watt-degrees. Watts are Watts and Volt-Amps Reactive are Vars. Energy is Watt-Hours or Megawatt-Hours.Even though Q includes some Joules that do shuttle back and forth between generator and loads on a subcycle basis,
we consider reactive current to be Wattless since it won't turn the disc on a watt-hour meter.

If you want to be understood in practical circles , use terms in the way to which practical people are accustomed.

Trivia question
does reactive current cause torque pulsations in a 3 phase machine?
If not, then the prime move is oblivious to them...

 

[jim hardy]

You also have maximum outputs on your generators so the more reactive power you put out, the less real power you can put out (

Only when you run up against the generator capability curve. Stator amps or field amps...
http://www.eecs.ucf.edu/~tomwu/course/eel4205/notes/17%20Capability%20Curve.pdf