My problem is I want to calculate three phase kilo watt hour from time sampled data of current and voltages.

My doubts

1) How can I calculate the kilowatt hour from time sampled data? Is it any equations is available?

2) Is it need to take the phase shift ? ( How can I calculate the phase shift from sampled data of voltage and current? How I link this to calculating the three phase power.)

3) Please suggest me if some better platform is available for solving my doubt?

4) Is it true when I calculate the kilowatt hour of each phase separately and some up all?

I get the instantaneous sample value (Not continues).(I have some sensors that gives the current and voltage - I convert this to digital for processing). Around 50 samples are got from 1sec.(Is it to be zero when we some up all the power of three phase - due to phase shift of 120) . How can I calculate total three phase kilo watt hour from these sampled value. I am processed my data in arduino.

  • \$\begingroup\$ Instantaneous Power= 3 *Vrms *Irms * cosθ Vrms = phase voltage Irms = phase current θ = the phase different between the voltage & current waveforms You might read this article. tmworld.com/design/characterization/4392053/… Essentially what you are going to do is take many measurements of current and voltage over at least one cycle and integrate them. that will give you average power for the length of time you measured. This can be multiplied to get kw per hour. \$\endgroup\$ – mfarver May 3 '13 at 2:32
  • 3
    \$\begingroup\$ 50 samples per second is not enough. As a minimum for a decent calculation you need 2 x highest harmonic frequency of current. As an example, the 9th harmonic of 50Hz is 450Hz so you'll need at least 1000 samples per second. This assumes the voltage is fairly sinusoidal. \$\endgroup\$ – Andy aka May 3 '13 at 15:18

It depends on your load. If you have a symmetrical load which sinks a sinusoidal current and power changes relatively slow, then what you're doing works. You can calculate the power at each sample by:

$$P = U_1 \cdot I_1 + U_2 \cdot I_2 + U_3 \cdot I_3$$

Sum all samples up and you will get the desired energy \$W\$. For example with 50 samples/s:

$$W = \frac{\Sigma P}{50} ~~[Ws]$$

But if your current is not such a nice sine and your power consumption changes fast (within 1 period) than you need much more samples than 50/s (as Andy aka said).

  • \$\begingroup\$ What about the power factor? If the load were purely inductive there would be voltage and current but the power would be zero. \$\endgroup\$ – tomnexus Feb 14 '17 at 8:00
  • \$\begingroup\$ You don't need the power factor here, as you are sampling current and voltage at the same time. You only need the power factor if you measure the RMS voltage and the RMS current. But when you sample it at the same time you get the right value. For a purely inductive load, the sum is always zero. \$\endgroup\$ – Tim Jan 3 '18 at 11:01

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