North America Split phase formula

cjcar · 2021-01-25T23:59:36.824Z

I tried really really hard to find the answer to this (on this forum and elsewhere) but could not.

Looking at Learn | OpenEnergyMonitor it seems that the usual approach is to attach a current transformer (CT) to each leg of the incoming service ensuring that the two CTs are ‘facing’ in the opposite direction. That page then goes on to say that CT1 will see I1 + I3 and CT2 will see I2 + I3. Shouldn’t that be I2 - I3? This seems to be the only way to make the final formula
Total Power = ½ × V × ICT1 + ½ × V × ICT2
work.
I think I can test this if I can shut down all the other circuits in the house and turn on our electric stove which is a (except for the electronic controls) pure 240V appliance. Then I should see the raw input as I3 on (in my case) emontx:power1 and -I3 on emontx:power2.

Can someone please confirm/comment/correct as appropriate?
…Chris

Robert.Wall · 2021-01-26T00:17:08.492Z

cjcar:

Shouldn’t that be I2 - I3?

If you’re referring to Fig.2, no. Look at the directions of the currents in relation to the phase dots.

(Hint: What’s the current and the voltage for the 240 V load in relation to the portion of its current and voltage that you measure? And if you still aren’t convinced, substitute the currents into the expression for total power and arrange it in terms of each load.)

cjcar · 2021-01-26T01:12:48.637Z

Thanks so much for your quick response. I continue to be very impressed by the support offered by you and others in this community.

I guess I’m still not expressing myself very well. Looking at Fig.3a and the formulas thereafter…
If
ICT1 = I1 + I3
and
ICT2 = I2 + I3
how can
Total Power = ½ × V × ICT1 + ½ × V × ICT2
?
Please connect the dots for me rather than leaving it as ‘an exercise for the reader’.
…Chris

Robert.Wall · 2021-01-26T11:16:56.912Z

Look where the voltage measurement is in Fig 3a. Each 120 V load gets ½V, the 240 V load gets all of V.
P = V × I.

The power in the ‘top’ 120 V load is ½V × I₁
The power in the ‘bottom’ 120 V load is ½V × I₂
The power in the 240 V load is V × I₃

Is that clear so far?

Add them together:

Total Power = ½V × I₁ + ½V × I₂ + V × I₃

Now here’s the really tricky part – you need to recognise that you can write:

V × I₃ = ½V × I₃ + ½V × I₃

So substitute that into the total power:

Total Power = ½V × I₁ + ½V × I₂ + ½V × I₃ + ½V × I₃

Now collect up the parts with ‘I_x’ in that make up I_CT1 and I_CT2

Total Power = (½V × I₁ + ½V × I₃) + (½V × I₂ + ½V × I₃)

and put the common terms outside the brackets:

Total Power = ½V × (I₁ + I₃) + ½V × (I₂ + I₃)

and those terms inside the brackets are I_CT1 and I_CT2, are they not?

That’s all very, very basic algebra - the stuff I learned about 60 years ago in school. If that doesn’t help, I’m at a loss to know what your problem is.

cjcar · 2021-01-26T17:34:21.515Z

Thanks so much for helping me with this. I had to stare at that page for quite a while before I finally realised that ‘VT’ is a voltage transformer and that in Fig. 2 it is measuring (nominally) 120V just as it is when using the standard setup supplied by the Open Energy Monitor store. So it all makes sense to me now. And most importantly I understand that in the setup as depicted in Fig.2 the total power is in fact the simple sum (for me) of node emontx1/power1 and emontx1/power2.

Robert.Wall · 2021-01-26T19:05:14.924Z

cjcar:

I finally realised that ‘VT’ is a voltage transformer

The symbol is actually a voltmeter – which is exactly the function that it is performing.

cjcar:

the setup as depicted in Fig.2 the total power is in fact the simple sum (for me) of node emontx1/power1 and emontx1/power2.

You never mentioned that you were using an emonTx, so as far as I knew, it was a purely theoretical question.