Calculating real power and power factor

In summary, the conversation discusses the concept of reactive power and its relationship to voltage and current phases in an AC circuit. The formula for real power is also mentioned, along with the idea that power can only be dissipated in resistance and not in reactive components. The use of phasors and dot products in describing AC is briefly mentioned, and the conversation ends with a mention of using an industrial controller to measure instantaneous voltage and current in order to calculate real power and power factor.
  • #1
pmn
3
0
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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  • #2
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
 
  • #3
'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.
 
  • #4
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.
 
  • #5
pmn said:
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?
 
  • #6
> 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.)
 

1. What is real power and how is it calculated?

Real power is the actual power consumed by an electrical device or system. It is calculated by multiplying the voltage and current of the circuit, and then multiplying that value by the power factor.

2. What is power factor and why is it important?

Power factor is the ratio of real power to apparent power. It represents the efficiency of an electrical system. A higher power factor means that the system is using power more efficiently, while a lower power factor indicates inefficiency and wasted energy.

3. How do you measure power factor?

Power factor can be measured using a power factor meter or by using a multimeter to measure the voltage and current of the circuit. The power factor can then be calculated by dividing the real power by the apparent power.

4. What is a good power factor value?

A power factor of 1.0 is considered ideal, as it means that all of the electrical energy is being used efficiently. A power factor of 0.9 or above is typically considered acceptable for most electrical systems.

5. How can power factor be improved?

Power factor can be improved by using power factor correction techniques, such as adding capacitors to the electrical system. These devices help to balance the reactive power and increase the power factor, resulting in more efficient energy usage.

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