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).