# What is three phase power

## How do you calculate a three-phase kilowatt-hour from time-sampled data of the voltage and current of each phase?

My problem is that I want to calculate a three phase kilowatt hour from time sampled data of current and voltage.

My doubts

1) How can I calculate the kilowatt hour from time-controlled data? Are there any equations?

2) Does the phase shift have to be carried out? (How can I calculate the phase shift from the sampled data of voltage and current? How do I relate this to the calculation of three phase power.)

3) Please suggest to me if there is a better platform available to resolve my doubts.

4) Is it true if I calculate the kilowatt hour of each phase separately and some in total?

I get the instantaneous sample (does not continue) (I have some sensors that show the current and voltage - I convert this to digital for processing). Approximately 50 samples are obtained from 1 second. (Should it be zero when we increase the total three-phase power - due to a phase shift of 120). How can I calculate the total three-phase kilowatt-hour from this sample? I process my data in Arduino.

### mfarver

Instantaneous power = 3 * Veff * Irms * cosθ Vrms = phase voltage Irms = phase current θ = the phase that differs between the voltage and current waveforms You can read this article. tmworld.com/design/characterization/4392053/… Essentially, you will be taking and integrating many measurements of current and voltage over at least one cycle. This will give you an average performance for the time you have measured. This can be multiplied to get kW per hour.

### Andy aka

50 samples per second is not enough. For a decent calculation, you need at least 2 times the highest harmonic current frequency. For example, the 9th harmonic of 50 Hz is 450 Hz, so you need at least 1000 samples per second. This assumes that the voltage is fairly sinusoidal.

### Tim

It depends on your load. If you have a symmetrical load that is sinking a sinusoidal current and the power changes relatively slowly, what you're doing works. You can calculate the power for each sample by:

P. = U.1⋅ich1 + U.2⋅ich2 + U.3⋅ich3

Combine all the samples and you will get the desired energy W.. For example with 50 samples / s:

W. = ΣP.50 [W.s]

But if your power isn't that nice and your power usage changes quickly (within a period of time) you need a lot more samples than 50 / s (as Andy aka said).

### Tomnexus

What about the power factor? If the load were purely inductive, there would be voltage and current, but the power would be zero.

### Tim

You do not need a power factor here, as you are sampling current and voltage at the same time. You only need the power factor if you are measuring the rms voltage and the rms current. But if you try it at the same time you will get the right value. With a purely inductive load, the sum is always zero.