# Are 3 single phases equivalent to a single 3 phase system?

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 x I_avg x PF_avg x sqrt(3)). The meter however is using the summation of the three individual phase powers (V1 x I1 x PF1 + ... ).

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?

Any help would be seriously appreciated!!!

EDIT:

WOW Thank you everyone for your quick response! I figured I would attach a sample row of the metering data I have been given, to try and make what I'm working with more clear:

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?) Again, THANK YOU ALL SO MUCH!

• Some other sources of error are lack of RMS current vs peak or average current and phase. Realtime VIcos(theta) or 3phi watt meter is best. Others measure avg I over x cycles and avg V which may be ok or not if high pulse unbalanced currents. – Tony Stewart Sunnyskyguy EE75 May 26 '17 at 4:13
• Yes. I assume Power 5 is your line numbers. – StainlessSteelRat May 26 '17 at 13:12

1, 2 and 3 are delta phase voltages. $cos\theta$ is fancy electrotech speak for power factor.

$$P_T = P_1 + P_2 + P_3$$ $$P_T = V_1 I_1 cos {\theta}_1 + V_2 I_2 cos {\theta}_2 + V_3 I_3 cos {\theta}_3$$

Three-phase balanced delta, so phase voltage, currents and pf are the same. $$\color {red} {P_T = 3 V_{Phase} I_{Phase} cos {\theta}_{Phase}}$$

For a delta, $V_{Line} = V_{Phase}$ and $I_{Line} = \sqrt {3}I_{Phase}$, so using line values.

$$P_T = 3 V_{Line} \frac {I_{Line}} {\sqrt {3}} cos {\theta}_{Line}$$ $$\color {red} {P_T = \sqrt {3} V_{Line} I_{Line} cos {\theta}_{Line}}$$

So you are correct. Whether you use line quantities or phase quantities, the answers will be the same. This works for the load being connected up in wye or delta.

You can use averages for your line quantities. Not sure if you have access to power factor.

Now, you have to reconcile the math and measurements with your co-worker.

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.