THE PHYSICS OF... POWER

[Ingenieur]

reactive power √-1X = not real power, and no cost to poco
but, in real world, dealing with Amps is very real, so thata where jX bites you
but why all the fuss, if the unit has a PF less than 1 just add X_C or X_L as needed and be done with it. build to kW demand and adjust kVAr as needed, install those solenoid cap banks and call it a day, thus the imaginary jX "power" is nothing to worry about.

the use of the term "power" for jX is kinda wrong, but the power folks needed to place a name on it. "imaginary power",

there is cost to poco
not fuel but increased line size (well a bit of fuel due to increased losses)
power factor correction
switching and fault transient mitigation

 

[Ingenieur]

actually no. you have +amps_vector for 50% of the current cycle, and -amps_vector for the other 50% of the current cycle. a pure cap has no real power, we went over this during the last 20 pages or so. the amplitude of the current varies, has same period as voltage, but has angle offset from voltage by 90°

He is correct
work hence real power is expended over a partial cycle
the electrons take force to move over time
power = force x mass / time = work/time
but over a complete cycle the work is returned so the net is zero

 

[Ingenieur]

huh? being a delta t=0 observation does not make it a scalar. at any instant in time the Amps has a amplitude AND direction, Amps is s defintion of charge flow which always has amplitude AND direction, and has none of those when Amps hits zero-crossing

Current has amplitude and SIGN
it is scalar, not a vector
although it's active/reactive components can be expressed as a phasor on the complex plain
BIG difference

are you just talking crazy to elicit a reaction or do you really believe that is how these things work?

 

[FionaZuppa]

but over a complete cycle the work is returned so the net is zero

as you noted, some of the work is returned (real world).

 

[Ingenieur]

i not sure what you mean, where does the current Amps_vector change direction with DC PWM? take a inductor, DC PWM will create a mag field that follows PWM shape/period yet the mag field N/S never swaps to S/N.


still not correct. there is no √-1 Amps. but there is √-1 Power :thumbsup:, its the Amps associated with the √-1 Power that the poco does not like to provide even though their avg of that √-1Power=0 (+ some losses of course).

think of pwm as dc biased ac
it can be constructed as a fourier series of sine waves
and an fft of it will show a fundemental freq plus harmonics
like a boat flowing in a 20 mph river that alternately rows upstream/downstream at 10 mph
it changes direction but has a positive net movement


Absolutely incorrect^10th power

V 480/0 deg
Z j48 or 48/90
I = 480/0 / 48/90 = 10/-90 = -j10
it ALL sqrt-1 as you put it

you do realize that S = P + jQ = VI* ???

 

[Ingenieur]

as you noted, some of the work is returned (real world).

That is exactly what he said and you claimed to be incorrect
glad you see it correctly now

 

[Sahib]

Again, you _must_ distinguish between instantaneous measurements and time averages.

RMS is a type of time average. RMS current doesn't tell you how much current is flowing in a circuit at any given instant; it tells you something about how much current is flowing 'on average'.

All energy is actually energy. But in AC circuits, some energy will flow from the load back to the source for part of the AC cycle.

If you want to describe this using instantaneous measurements, you get graphs such as the one Carultch posted in #243.

If you want to describe this using time average measurements, then you need to use power factor.

-Jon

Again, you have to take into account energy conservation law: How to take into account the 'imaginary' power i.e reactive power along with real power so that the energy conservation law is not violated? ( I know the answer; let me know your point.)

 

[winnie]

Again, you have to take into account energy conservation law: How to take into account the 'imaginary' power i.e reactive power along with real power so that the energy conservation law is not violated? ( I know the answer; let me know your point.)

If you are not going to bother reading what I write, then please don't ask leading questions in an attempt to educate me.

Imaginary power, apparent power, reactive power, power factor, etc. are only needed to describe the fact that if you multiply RMS current * RMS amps the result is not in general the average power delivered to the load.

RMS current is a time average for the current in the circuit
RMS voltage is a time average for the voltage across the circuit

Instantaneous voltage * instantaneous current = instantaneous power
RMS current * RMS voltage != average power

Now, since you know the answer, where in the above do I imply that conservation laws are broken?

-Jon

 

[Sahib]

If you are not going to bother reading what I write, then please don't ask leading questions in an attempt to educate me.

I read and understood your posts and I attempt to improve on them.

In that vein.......

Imaginary power, apparent power, reactive power, power factor, etc. are only needed to describe the fact that if you multiply RMS current * RMS amps the result is not in general the average power delivered to the load.

Imaginary power, apparent power, reactive power etc. are physical reality and not mathematical constructs as you claim.

Instantaneous voltage * instantaneous current = instantaneous power........(1)
RMS current * RMS voltage != average power

Insert Imaginary power or reactive power suitably in the above equation (1) for further discussion, if you are pleased.

 

[Besoeker]

Imaginary power,

Isn't power.

 

[Ingenieur]

Isn't power.

sure it is
that is why it's called reactive POWER

 

[Besoeker]

sure it is
that is why it's called reactive POWER

It isn't power. Power is expressed in Watts.
It isn't Watts.
I don't which part of that you don't get.

 

[Ingenieur]

It isn't power. Power is expressed in Watts.
It isn't Watts.
I don't which part of that you don't get.

I get the part that you are wrong
watt = joule/sec = v x a = joule/coulomb x coulomb/sec = joule/sec = watt = va
var is the reactive component

 

[Besoeker]

I get the part that you are wrong
watt = joule/sec = v x a = joule/coulomb x coulomb/sec = joule/sec = watt = va
var is the reactive component

VA is not a measure of power.
Nor is VAr.
And watt = va is just nonsense.
And I don't soak bees.

 

[Ingenieur]

VA is not a measure of power.
Nor is VAr.
And watt = va is just nonsense.
And I don't soak bees.

sure they are
power transformer, capacity 1 MVA, not 1 MW
S = P + jQ, P and Q, active and reactive power being components of the total power
EE 101
active power
reactive power

watt = joule/sec = v x a = joule/coulomb x coulomb/sec = joule/sec = watt = va
basic physics

 

[Besoeker]

sure they are
power transformer, capacity 1 MVA, not 1 MW

You are obviously confused by capacity and power.

watt = joule/sec = v x a = joule/coulomb x coulomb/sec = joule/sec = watt = va
basic physics

It's nonsense. I thought you might have known better.

 

[Ingenieur]

You are obviously confused by capacity and power.


It's nonsense. I thought you might have known better.

I am not the one 'confused'
it is the capacity rating of the power it can deliver, be it active or reactive, or the combination
or total POWER S

you SHOULD know better
this may help you, a good primer
http://www.aptsources.com/resources/pdf/True vs. Apparent Power.pdf

in phasor notation
S = VI* = P + jQ
P = VI* cos ph ang
Q = VI* sin ph ang
note that cos and sin are unitless making S, P and Q all the same units
W, va and var are used as convention for differentiation of different forms of electrical power

 

[Besoeker]

W, va and var are used as convention for differentiation of different forms of electrical power

Power is power. VA and VAr are not.
I don't know why I bother......

 

[Ingenieur]

Power is power. VA and VAr are not.
I don't know why I bother......

then every text written on POWER engineering needs changed lol

please reread slowly, again

in phasor notation
S = VI* = P + jQ
P = VI* cos ph ang
Q = VI* sin ph ang
note that cos and sin are unitless making S, P and Q all the same units
W, va and var are used as convention for differentiation of different forms of electrical power

watt = joule/sec = v x a = joule/coulomb x coulomb/sec = joule/sec = watt = va

no matter how many times you repeat 'power is power' (weth that means) you are still lacking a true understanding of the subject matter

 

[K8MHZ]

Power is power. VA and VAr are not.
I don't know why I bother......

Wiki says:

Engineers use the following terms to describe energy flow in a system (and assign each of them a different unit to differentiate between them):

• Active power,^[2] P, or real power:^[3] watt (W)
• Reactive power, Q: volt-ampere reactive (var)
• Complex power, S: volt-ampere (VA)
• Apparent power, |S|: the magnitude of complex power S: volt-ampere (VA)

The above is from a pretty good Wiki article:

https://en.wikipedia.org/wiki/AC_power#Apparent_power