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

[Wee-Lamm]

If "inside the device" was all that counted, the PF of any appliance would have no effect on the system. But the effect of PF of multiple appliances on the system is additive.
You seem to be rather preoccupied with finding a good model for 'what's really going on. I am being more pragmatic and looking at the overall effect (cost) of PF. It seems to me that PF is only of concern to the supplier, as long as out Energy Meters just measure Energy and charge us for Watts only. The actual effect of a particular PF and a particular Load will vary from place to place because the supply equipment is very much a part of the (£$) equation. If the load demand on an isolated power station were to be reduced (say all the factories in a town closed down) then the PF of the remaining houses and equipment would be pretty well irellevant.

Yes, we are not both speaking from the same domain or at least, we haven't yet agreed on what a system is. I tend to not see everything as simply 1 system but rather, as a huge collection of systems, where each system can contain several devices or sub-systems. Can we view the power grid as a system of systems, to allow for a common reference point?

I still stand on my opinion that; my device does not care if the power it draws from the supply system, contains reactive or only active power. My device cannot tell the difference unless I design it to do so.

Having read a little more as encouraged by your comments, I agree that if my device includes components that create Induction or capacitance, which will put the E&I out of phase within my device, I should include properly valued Capacitors or Inductors to put them back in phase, at the exit point of my device. While this will avoid having the "power, returned off cycle" from being viewed as consumed power, it will also avoid the off-cycle overloading of the neutral side of the supply system. I could mention switching capacitors and such, but I suspect you are already ahead of me in that respect.

Being a good neighbor, by having a PF as close to 1 as possible, is definitely beneficial to anyone connected to the same grid and especially to the stability and safety of the grid itself, just as I would benefit if everyone else is a good neighbor and keeps everything at the interface, in the same cycle-phase relationship. Clearly if nobody else is a good neighbor, I will potentially suffer from periodic under-supply until the neutral line catches up, but I still cannot tell if the power I am receiving is made from burning coal, an induction network, or generated otherwise. As long as E & I are in phase, my local device will only see it as power.

 

[vintageplayer]

Reactive power is defined as (VI/2).sinφ. This is the amplitude of the second component in the equation:
p(t) = v(t).i(t) = (VI/2)cosφ[1 + cos(2wt)] + (VI/2).sinφ.sin(2wt)

But the above equation can also be written as:
(VI/2)cosφ + (VI/2).cos(2wt -φ)Why not define reactive power as the amplitude of the second component in the above equation instead?

Having thought about this some more, defining reactive power in the first way is best because:

1. Both Q and P are conservative at every node, whereas VI/2 isn't conservative. This allows a conservation of power approach to circuit analysis (much like a conservation of energy approach in physics).
   
2. It ensures reactive power is not consumed by resistive loads. Reactive power is only consumed by reactive components. Even though you have sloshing of power across a purely resistive load, you might not necessarily want to call this a "reactive power".
3. When φ is negative the reactive power is negative. This allows you to think of capacitors as supplying reactive power, and inductors as consuming reactive power.
4. Q can be expressed as the imaginary component of VI*.

 

[Wee-Lamm]

Having thought about this some more, defining reactive power in the first way is best because:

1. Both Q and P are conservative at every node, whereas VI/2 isn't conservative. This allows a conservation of power approach to circuit analysis (much like a conservation of energy approach in physics).
   
2. It ensures reactive power is not consumed by resistive loads. Reactive power is only consumed by reactive components. Even though you have sloshing of power across a purely resistive load, you might not necessarily want to call this a "reactive power".
3. When φ is negative the reactive power is negative. This allows you to think of capacitors as supplying reactive power, and inductors as consuming reactive power.
4. Q can be expressed as the imaginary component of VI*.

Q = VI Sinφ //Note: I > I/2
or
VAR =√ (VA2 – P2) //Where P = VI Cosθ and Q = VAR

The image of the Power Triangle should show that V & I are 90 degrees out of phase, in Reactive power.
http://www.electricaltechnology.org/2013/07/active-reactive-apparent-and-complex.html

1. I'm not sure if this applies to Reactive power, it's not so much conserved but time shifted.
2. I think Reactive power is still consumed by resistive loads, but the measurement must account for the phase differential.
3. A Capacitor creates an Electric field which causes E to lag behind I, whereas an Inductor creates a Magnetic field which causes I to lag behind E. You could say that a Capacitor "absorbs" voltage and that an Inductor "absorbs" current, but each one releases it again on the down-sweep of it's respective wave, but neither one consumes energy or power. (Except perhaps minimal heat loss of the materials the inductor or capacitor are made from).

I don't think φ can be negative. It indicates which angle is being measured in the power triangle.

4. Q is not imaginary, it simply has the V & I existing at different time points on the curve. φ is needed so we can measure the angle of shift, to determine how many degrees it has been shifted. Sin and Cos determine which angle we should measure to determine if it was E or I which has been held back in the cycle.

 

[jim hardy]

Q is not imaginary,

good for you

we call it imaginary because it's current that is shifted 90 degrees wrt voltage

Operator j shifts a sinewave 90 degrees and is called i for imaginary because that's equivalent to √-1, and multiplying a sine by j twice makes it negative implying j is √-1 .
But j must be imaginary because everybody knows negative numbers don't have square roots !
So we call the in phase and out of phase components real and imaginary ... and name the axes on our phasor diagram real and imaginary.
There had to be abundant lab humor back in late 1800's when this stuff was being first worked out from lab experiments. . Here's how Sylvanus P Thompson began his calculus book

Sylvanus P. Thompson, Calculus Made Easy
Considering how many fools can calculate, it is surprising that other fools think it is difficult...

Maybe some wag in late 1800's figured out that those out-of-phase volt-amps make no heat so he called them "imaginary watts" ?

 

[Wee-Lamm]

good for you

we call it imaginary because it's current that is shifted 90 degrees wrt voltage

Operator j shifts a sinewave 90 degrees and is called i for imaginary because that's equivalent to √-1, and multiplying a sine by j twice makes it negative implying j is √-1 .
But j must be imaginary because everybody knows negative numbers don't have square roots !
So we call the in phase and out of phase components real and imaginary ... and name the axes on our phasor diagram real and imaginary.

Thank you for the primer, it lead to further educational reading. :-)

There had to be abundant lab humor back in late 1800's when this stuff was being first worked out from lab experiments. . Here's how Sylvanus P Thompson began his calculus book

Maybe some wag in late 1800's figured out that those out-of-phase volt-amps make no heat so he called them "imaginary watts" ?

Or a Whig. :D

 

[sophiecentaur]

Or a Whig. :D

We don't get many political jokes (or historical ones) on PF. A double whammy.

 

[anorlunda]

If I can toot my own horn, I can cast light on the origin of the term "imaginary power". Just today my new Insights article The Case for Learning Complex Math appeared. It discusses that topic.

The short answer is, blame Gottfried Wilhelm (von) Leibniz (1646-1716)