Electronic – Are 3 single phases equivalent to a single 3 phase system

measurementthree phase

I am not an electrical engineer (I did work as an automation technologist a long time ago, but please forgive me my electricity is very rusty). I've been given some data that was measured from a 3-phase delta system using current transformers and voltage leads on a pump system.

The person who gave me it claims that the software for the logger was configured incorrectly, and that the power consumed is half of what it's reading. He insists that the correct way to calculate the power in this situation is to use the averages of the 3 phases (\$V_{avg} \times I_{avg} \times pf_{avg} \times \sqrt{3}\$).

The meter however is using the summation of the three individual phase powers (\$V_1 \times I_1 \times pf_1 + … \$).

My question is: should these values not come out to be the same in the end?
Should we not be able to calculate the power using both methods?

ETA: Here is a sample row of the metering data I have been given, to try and make what I'm working with more clear:

Sample Data

Each of the 3 real channels are measuring a phase voltage, current and power factor – so my approach was to just calculate power using each phase measurement and summing them (much like the meter did). This should remain the same regardless of the way the system is wired right? (same calculation for Wye / Delta?)

Best Answer

The average of three line-to-line voltages multiplied by the average of three line currents multiplied by pf and sqrt(3) will give a reasonably accurate results if the phases are reasonably well balanced.

The sum of three wattmeter reading will only be correct if the measurements are based of the line-to-neutral voltages and the line currents or the line-to-line voltages and the delta phase currents.

If two wattmeters measure the power based on any two line currents and the line-to-line voltage between the iines in which the current is measured and the third line, the power is the sum of the two wattmeter readings. The system does not need to be balanced for the two-wattmeter method to be accurate.