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

[sophiecentaur]

reactive "power" isn't power at all

There we have it: from the horse's mouth.
This has been a great, if rather inefficient way for me to get into this stuff.

 

[vintageplayer]

reactive "power" isn't power at all it's just Volt-Amps-Reactive and is wattless.

What SI fundamental units does reactive power have? What SI fundamental units does average power have?

 

[vintageplayer]

I still must say, it strikes me as really strange that a diagram is drawn which involves a Mean Power and a Maximum instantaneous VI value

Reactive Power gives a general idea about the various stresses that poor PF subjects the supply equipment to. I find it interesting that such a simple formula gives such an aparently useful parameter for Engineers to work with.

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(2ωt -φ)

Why not define reactive power as the amplitude of the second component in the above equation instead? Q = (VI/2). Wouldn't this make more sense because it would tell you how much the power is fluctuating around the actual supplied real power?

 

[jim hardy]

Power is a flow of energy from a source to a load.
Power is measured by work over time.

The standard unit for power is the watt (abbreviated W) which is a joule per second.

"Reactive power" does no work because it only shuttles energy back and forth between load and source.
So the quotient work/time for reactive power averages to zero
The work done by "Reactive power" is zero. So it's not power at all. It's wattless. It requires no torque from a prime mover.

This calculation is only for cases where the force is in the direction of the velocity, and there are many cases where that is not so. Then for instantaneous power, you just multiply the product of force and velocity by the cosine of the angle between them to get the power. In the more general cases where everything varies, one often calculates the work first and then divides by the time to get the average power.

ibid

Why not define reactive power as the amplitude of the second component in the above equation instead? Q = (VI/2). Wouldn't this make more sense because it would tell you how much the power is fluctuating around the actual supplied real power?

Because VIsinθ works just fine
and we use RMS not peak or instantaneous values because that's what our meters indicate
and VARS do tell you exactly how much reactive power is many reactive volt-amps are circulating,
Our VAR meter is prominently located on the panel near the voltage regulator knobs that control VARS.

Why not define reactive power as the amplitude of the second component in the above equation instead? Q = (VI/2).

"Q = (VI/2)" ?
That's "S" per your first post. We'd use RMS and call it KVA, of course multiplied by √3 for three phase.

 

[Wee-Lamm]

No. Not me. Reactive power does represent a 'demand' on the system - otherwise no one would be bothered by it.

Now that's a strange comment. You yourself have made the point that Reactive Power doesn't come from the supply. The light bulb, itself is merely supplied with a Voltage and it 'chooses' how much current to take. If the tubne itself is purely resistive then the power that it consumes is V.I and that's what the power station needs to provide it with. The 'demand' that results from reactive elements in the load is not an Energy Demand; it just represents an overhead that's involved in extra current or volts associated with the generator (+ all the rest of the supplystuff).
I still must say, it strikes me as really strange that a diagram is drawn which involves a Mean Power and a Maximum instantaneous VI value. The coal that's shovelled into the boiler is somehow treated as the same as the extra spec needed for the components. Little wonder that 'they' tell you just to use the formula and 'get over it', without encouraging too much through about what it all actually represents.

If it is a Power Meter then it will only measure the in phase components of V and I. If your appliances happen to be very reactive, it will not be aware but still tell you the Energy dissipated per second.

It sounds like we are 'almost' saying the same thing much of the time but, I'll pause briefly to be sure I understand if and how PF affects my position.

I think though that we are not both thinking from the same reference point where my reference is inside my device whereas you are speaking from within the distribution network. If I am right on that point, then everything within the distribution/transmission realm would be part of that device.

 

[vintageplayer]

Power is a flow of energy from a source to a load.
Power is measured by work over time.

"Reactive power" does no work because it only shuttles energy back and forth between load and source.
So the quotient work/time for reactive power averages to zero
The work done by "Reactive power" is zero. So it's not power at all. It's wattless. It requires no torque from a prime mover.
ibid

OK I understand what you're saying here. I wasn't aware that the term "Watts" was reserved only for time rate change of work and not energy.

and VARS do tell you exactly how much reactive power is many reactive volt-amps are circulating,

Isn't this reasoning circular? Reactive volt-amps are by definition VARS?

"Q = (VI/2)" ?
That's "S" per your first post.

I know it's S. If S is capable of quantifying the circulating volt-amps on top of your average power, what is the utility of defining an additional quantity, Q and calling it the reactive power?

Because VIsinθ works just fine

Sure it works fine, but couldn't P and S do everything we need? What is the real reason for working with Q?

 

[sophiecentaur]

What SI fundamental units does reactive power have? What SI fundamental units does average power have?

I'm afraid that's not a valid argument. I could ask you what are the units for Work and Torque and a dimensional argument would tell us they are the same. They are not - and the difference is very much along the lines of the difference between W and 'Reactive Power'.

 

[sophiecentaur]

my reference is inside my device

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.

 

[sophiecentaur]

and we use RMS not peak or instantaneous values because that's what our meters indicate

That could be real reason.

 

[jim hardy]

I know it's S. If S is capable of quantifying the circulating volt-amps on top of your average power, what is the utility of defining an additional quantity, Q and calling it the reactive power?

No, S is NOT capable of that
and maybe there's the miscommunication
what do you mean by "circulating" ? To us power guys "circulating amps" are reactive because they circulate between machines.
If you've never had control of two machines in parallel you cannot imagine the sense of wonder when tweaking one voltage regulator makes amps go up or down on both machines. The load doesn't need to accept those amps, they just circulate between the generators.S doesn't separate out of phase volt-amps from in phase volt amps
S is the hypotenuse as pointed out by Hesch in post #2

in my power plant,
P i control by how much steam i admit to Mr Turbine and my wattmeter shows me P
Q i control by how much field current i apply to Mr Generator and my varmeter shows me Q
I don't have a S meter but i do have ammeters in series with the armature windings, which, at constant voltage are an indication of S ,
and i have voltmeters between phases ,
and for a sanity check on my instruments
at constant load (constant megawatts)
...i can adjust excitation to give minimum armature amps which means there's no reactive current
... then i verify varmeter reports zero and that's my zero check on Q meter (varmeter)
...and i verify my wattmeter reports VI√3, that's my check on S .
but of course i'd have to co-ordinate that testing with system dispatcher and nearby power plants so as to not upset the grid.

That's my utility guy perspective. The lingo is well established and is the dialect that's understood in my circles.
Reactive demand on the system is set by what is connected to the system
generators supplying that reactive demand share the reactive current amongst themselves , each supplying its share,
by means of adjusting their field currents individually
raising one machines' field current will affect armature amps on all nearby machines

Changing the language will not be well received. Utilities are just that way.

 

[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)