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Calculating real power and power factor

by pmn
Tags: factor, power, real
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pmn
#1
Jan17-12, 09:08 AM
P: 3
I have been trying to understand how to calculate real power and power factor in an AC circuit when given only circuit voltage and real/true current as measured by a current transformer (CT).

What I (think I) know:

1. Real power P = I^2 * R (but I don't know the circuit resistance)
2. Real power = apparent power * power factor
3. Apparent power S = RMS source voltage * RMS current

I believe I can take (source voltage ) / √2 to get the RMS of the voltage then (measured amperage) / √2 to get the RMS of the amperage and can then calculate the apparent power as the product of those two values but I don't know how to get to real power and power factor from there.

Thanks for any direction you can give me!

Phil
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technician
#2
Jan17-12, 09:41 AM
P: 1,506
Are you familiar with the idea that power is only dissipated in resistance.?
And that no power is dissipated in reactive components (inductance and capacitance)?
If the supply voltage is out of phase with the current in a series circuit then the power dissipated = voltage across R x current
The voltage across R = supply voltage x Cos θ
Cosθ is known as the power factor.
The disadvantage of having the supply voltage out of phase with the current is that a voltage,
V, is being generated but only a fraction (VCosθ) is delivering power
sophiecentaur
#3
Jan17-12, 09:43 AM
Sci Advisor
Thanks
PF Gold
sophiecentaur's Avatar
P: 12,261
'Real Power' is V.I (the dot product of the two phasors) or VI cos (phase).
You can measure it by putting the V and I waveforms into an analogue (four quadrant) multiplier and then an leaky integrator / Low Pass filter meter. This will multiply the instantaneous values of V and I, which is instantaneous power. The value of this will vary and always be greater than zero (and corresponds to I^2R). When you take the mean of this, you will get the average power.
This way, you neither need to know the Resistance nor the Power Factor (you are including it in the cos(phase) term.
I assume that those things the Gas Board supply you with do it this way. There's a current transformer on the consumer meter lead and a wireless link to the unit, which is mains powered and measures the voltage.

Your formula will tell you the power transferred in the resistor (natch) but, as you say, you need to know the value of resistance.

pmn
#4
Jan17-12, 11:03 AM
P: 3
Calculating real power and power factor

Thanks for the replies! I get the concept of reactive power being a function of a shifting of the voltage / current phases. (Thus a power factor of 1 means that the voltage and current rise in unison, correct? Inductance or capacitance in the circuit cause the two waves to become out of phase?) I don't get the statement "'Real Power' is V.I (the dot product of the two phasors) or VI cos (phase)". I do get "which is instantaneous power. The value of this will vary and always be greater than zero (and corresponds to I^2R)".

Since I believe that I can calculate apparent power as Vrms * Irms (where rms is a/√2 and a=peak value), and real power can be derived from an instantaneous measurement of voltage and amperage, then power factor can be calculated as real / apparent?

Ultimately I want to be able to put CTs on incoming mains power and several branch circuits as well as measure the voltage on the circuits and be able to make some statement about how much energy is being consumed in total (mains) and by each branch circuit as well as how close I am to a PF of one.
sophiecentaur
#5
Jan17-12, 01:50 PM
Sci Advisor
Thanks
PF Gold
sophiecentaur's Avatar
P: 12,261
Quote Quote by pmn View Post
Thanks for the replies! I get the concept of reactive power being a function of a shifting of the voltage / current phases. (Thus a power factor of 1 means that the voltage and current rise in unison, correct? Inductance or capacitance in the circuit cause the two waves to become out of phase?) I don't get the statement "'Real Power' is V.I (the dot product of the two phasors) or VI cos (phase)". I do get "which is instantaneous power. The value of this will vary and always be greater than zero (and corresponds to I^2R)".

Since I believe that I can calculate apparent power as Vrms * Irms (where rms is a/√2 and a=peak value), and real power can be derived from an instantaneous measurement of voltage and amperage, then power factor can be calculated as real / apparent?

Ultimately I want to be able to put CTs on incoming mains power and several branch circuits as well as measure the voltage on the circuits and be able to make some statement about how much energy is being consumed in total (mains) and by each branch circuit as well as how close I am to a PF of one.
The dot product is just 'vector speak' and takes you further into the business if you're interested. Have you not heard of using Phasors to describe AC?

This is great stuff as a thought experiment but all this stuff is readily available to buy. Furthermore, it is 'Electrically Safe'.

btw, how were you proposing to find the instantaneous V and I?
pmn
#6
Jan17-12, 03:39 PM
P: 3
> The dot product is just 'vector speak' and takes you further into the business if you're interested. Have you not heard of using Phasors to describe AC?

Short answer is 'no' but.... I understand the vectors in the 'AC triangle' showing the relationships to real, apparent and reactive power. I will dig in a little to understand phasors.

> This is great stuff as a thought experiment but all this stuff is readily available to buy. Furthermore, it is 'Electrically Safe'.

Readily available to buy as in 'purchase a Fluke power quality meter'? :-) I am enjoying the exercise at the moment but at times of weakness I do admit to browsing the Fluke website and lusting over their tools.

> btw, how were you proposing to find the instantaneous V and I?

Using an industrial controller to sample the line voltage and real current via CT. I believe I could probably make a reasonable calculation of the power factor by simply timing the (V)peak to (A)peak. (I have a controller, I don't have a Fluke.)


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