Power factor and VA vs Watts
- Thread starter cyriousn
- Start date
- Status
weressl
Esteemed Member
Oh, c'mon now!!!
Couldn't they afford a real Scotsman?
Nope, even hiring Koreans to act as Scottsman.
weressl
Esteemed Member
Do you have any grey poop on?
LarryFine
Master Electrician Electric Contractor Richmond VA
- Location
- Henrico County, VA
- Occupation
- Electrical Contractor
Why? You missing some?
mivey
Senior Member
But of course.
Hameedulla-Ekhlas
Senior Member
- Location
- AFG
I am not just talking the the concept of watching and obervation. It is the concept of understanding something. According to me, All scientific concepts which is not based and proved by math, are incomplete.
anyway, what is the word of Scott come for this discussion.
Perhaps one actually with the Watts
Don't think that's correct.
For the period where the voltage is positive and the current is negative, the first 30deg for example, the real power is negative. The same applies to those periods where the voltage is negative and the current is positive.
Your chart shows real power (in green) as always positive.
Real power is never negative. Negative power means power is flowing back to the source. Can't be real power if it is returning to the source from the load.
Reactive power on the other hand is both positive and negative, and averages out to zero. Its positve power is stored energy which is then returned to the source in the negative half-cycle of its waveform.
Apparent power (the green line on your graph, magenta on mine) is also positive and negative. Its average is the same as real power. FWIW, instantaneous power represents apparent power.
ggunn
PE (Electrical), NABCEP certified
- Location
- Austin, TX, USA
- Occupation
- Consulting Electrical Engineer - Photovoltaic Systems
Sorry, wrong article quoted.
Last edited:
ggunn
PE (Electrical), NABCEP certified
- Location
- Austin, TX, USA
- Occupation
- Consulting Electrical Engineer - Photovoltaic Systems
"Apparently", apparent power = real power + reactive power.
After scrutinizing my graph more carefully, it was not correct (don't know what I was thinking :roll
This one any better ??? ...
In the instantaneous domain, yes.
But in the Vrms and Irms domain, it is...
(apparent power)? = (real power)? + (reactive power)?
ggunn
PE (Electrical), NABCEP certified
- Location
- Austin, TX, USA
- Occupation
- Consulting Electrical Engineer - Photovoltaic Systems
After scrutinizing my graph more carefully, it was not correct (don't know what I was thinking :roll
This one any better ??? ...
Interesting. So real, apparent, and reactive power are all in phase agreement, right? In your previous graph, real power was shown in phase (though double the frequency) with voltage and reactive power likewise with current.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
100413-1134 EST
Turn the load into an equivalent parallel circuit, resistor plus reactive components. The only power lost is in the resistor. The current thru the resistor is in-phase with the voltage across the resistor. Therefore, the instantaneous power (real power) is of a double frequency in phase with the voltage and is represented by
sin^2 t = 1/2 - (cos 2t)/2.
Logic will tell you that a normal linear resistor can not feed power back to the source at any instant of time. Energy always flows into the resistor and out as heat. Therefore, real instantaneous power with a linear load is always positive.
.
Turn the load into an equivalent parallel circuit, resistor plus reactive components. The only power lost is in the resistor. The current thru the resistor is in-phase with the voltage across the resistor. Therefore, the instantaneous power (real power) is of a double frequency in phase with the voltage and is represented by
sin^2 t = 1/2 - (cos 2t)/2.
Logic will tell you that a normal linear resistor can not feed power back to the source at any instant of time. Energy always flows into the resistor and out as heat. Therefore, real instantaneous power with a linear load is always positive.
.
Um.....
Take a power factor correction capacitor.
It consumes no power (other than losses).
It does have an applied voltage and it does take current.
Their product, times sqrt(3) if it is three phase, gives you VA, or apparent power. It is also entirely reactive - after all that's why you use PFC.
So we have reactive VA and zero W.
If the power can only be positive, as you are suggesting then, it would have to be zero all the time.
But that's not what happens. Over the first quarter of a cycle the capacitor charges to a positive voltage. It takes energy (0.5CV^2 Joules) from the supply over that period. Energy is power times time thus the supply has to provide power. During the next quarter of a cycle the capacitor discharges to zero volts and thus zero energy. Power goes back into the supply. Negative power.
Here it is for one cycle.
One other comment. Real, apparent, and reactive powers are based on RMS values. You cannot display them on a waveform diagram that shows instantaneous values.
I said real power. A capacitor (not accounting for losses?an ideal capacitor) uses no real power. Pavg is zero.
This just proves my point. The [ideal] capacitor uses apparent power, no real power.
Sure I can... and I did. While I agree they are based on RMS values, those would be of voltage and current rms values. What we are after is Pavg.
Yet in the instantaneous realm, a portion of the apparent power is always either real or reactive or a combination thereof... and that is what I show. Note in my graph Pavg of Apparent and Real are identical. Yet a portion of the Apparent Power is negative. This is indicative of reactive power during that half-cycle being returned to the source. What is not so obvious is the positive portion of apparent power that is reactive, but with plotting the real power curve, it can readily be distinguished.
Apparent Power = Volts ? Amperes
Real Power = Apparent Power ? Cosine θ
I agree. The real power is the average value of the instantaneous power. Note that for the pure capacitor, the average, or real, power is zero.
In the instantaneous domain, they have to be.
I was showing the RMS domain on an instantaneous graph (which is what I was thinking, for which I later said I don't know what I was thinking :roll
. This follows with gar's explanation of an equivalent R[C]L circuit.Yes, we associate the power with the value of Pavg... But does it not stand to reason if we use Pavg, apparent, real, or reactive, then there is a waveform associated with it?
- Status