Metering a Single Phase 208 VAC Circuit

[aemla]

Thank you for the detailed explanation. I believe I understand now.
What prompted the question was that I knew that the meter was giving incorrect total apparent power value. So I assumed active power would also be incorrect without understanding how the value is actually obtained.

Per the explanation above, total active power is in fact the correct value.
Thank you again for sticking with me on this one.

 

[aemla]

each phase CT "sees" the same current, but it is "in" one CT and "out" the other. The meter evaluates each, so in one case the real current is shifted 30 Deg Leading (ahead) of the phase voltage and in the other it is shifted 30 deg Lagging(behind) the phase voltage- each is phase shifted from the Phase-Neutral voltage that it is measuring

One additional follow-up: does this mean that with the exact the same setup, removing one of the CTs will result in incorrect total active power measurement?

 

[Averagesupernova]

Think as if you were the watt-hour meter in question. What would you see and based on it what conclusion would you draw?

 

[Windadct]

Again - this question can be worked out with a vector diagram.

Earlier I noted that I was pretty sure your dataset (Post #32) indicated you had a Phase angle of "-105" this is telling me that you have a CT connected backwards. Did you check this ?

Edit - extra credit - which of these would a reversed CT affect and which ones would it not : Apparent, Real and Reactive Power?

 

[aemla]

Hopefully I get this right:
1. Removing one CT.
Second CT is required since addition of the scalar values is needed to get the correct answer.

2. Reversing one CT.
Phase for that CT will be rotated 180 degrees. This will affect active and reactive power (vector quantities) but not apparent power (scalar quantity).

I took another screenshot with corrected wiring:

To calculate total apparent power, Pythagorean Theorem was used with total active and reactive powers. The result would be about 501 VA.

I would think another way to solve would be via Watt's law (P=IV) where I is one of the currents and V is UL1-L2. The results would be about 868 VA.

Don't quite understand why one of these would be incorrect.

 

[Averagesupernova]

The apparent power in the load is quite simply the voltage across it multiplied by the current through it.

Never forget that a voltage across a device multiplied by the current through the device will get you apparent power without fail.

 

[aemla]

How come using Pythagorean Theorem gives a wrong answer?

 

[aemla]

Hmm, seems that previous wiring was incorrect as well.
Changed it and took another set of readings.

Performed calculations for each scenario:

First scenario appears to be the correct one.
Ran a motor with bad bearings with no load, PF seems to be reasonable.

Final phase angle also makes sense since per the image below.

I'm sure some meters with L-L configuration (few previously mentioned) could provide correct results with just one CT. However, I'm glad that correct value can be obtained through 3 phase configuration and 2 CTs.

Thank you again for all your help.