How to calculate 3 phase apparent power from voltage and current measurements

Thread Starter

wrt54gl

Joined Jul 20, 2015
6
Hello

I will be installing a PLC to monitor a 3 phase delta electrical service. I am installing current transducers to measure Ia, Ib, and Ic, and voltage transducers to measure Vab Vbc and Vca. I am not sure how to calculate the apparent power with these values as the system will not be perfectly balanced.
 

Hymie

Joined Mar 30, 2018
1,081
The mathematics of this are complicated where the loads are not balanced, and if you add phase shifts due to capacitive/inductive loading and variations between the phase voltages, things get extremely complicated.

Through my work I occasionally have need to make 3 phase power measurements using a Fluke power analyser – the instrument cost around £3k. I doubt you could achieve anywhere near reasonable accuracy using a PLC device.

Personally I would make the calculations assuming the loads are balanced and resistive with no phase shift. If you do that, you can consider that the power draw from each phase is the current multiplied by the phase line voltage to neutral.
 

Thread Starter

wrt54gl

Joined Jul 20, 2015
6
I am okay with the inaccuracy and am expecting it. I am only looking for apparent power.
My loads are all connected line to line so in the case a 400V service:
If Ia =100A , Ib = 100 A and Ic = 0A this would calculate to 230x100 + 230x100 + 230 x 0 = 46,000VA but in reality it would be 100A x 400V = 40,000VA.
Is it possible to calculate this with better accuracy than this? I should know, but I've been away from school to long!
 

Hymie

Joined Mar 30, 2018
1,081
In your extreme example of an unbalanced load, you would be reading 15% high using my proposed method.

If you search on-line you will find details of unbalanced 3 phase power calculations, which involves vector diagrams with voltage/current phase angles. I doubt that a PLC based system can make such detailed calculations on the fly, at such a speed to give useful information.

You might consider improving my proposal by measuring the current phase angle (against voltage) for each of supply phases – but it is not a simple mathematical calculation to determine the phase angle when the load (and phase angle) may be varying.
 

profbuxton

Joined Feb 21, 2014
419
In power station use they use a VAR meter to give them the apparent power being shipped out. Not sure how they work. Maybe you could look up some circuits.
 

MrAl

Joined Jun 17, 2014
8,990
Hello,

If you understand complex numbers you can do anything you want with a 3 phase system.
Convert everything to complex quantities, then just use the algebra of complex numbers.
Balanced, unbalanced, doesnt matter anymore.
 

MrAl

Joined Jun 17, 2014
8,990
Hello

I will be installing a PLC to monitor a 3 phase delta electrical service. I am installing current transducers to measure Ia, Ib, and Ic, and voltage transducers to measure Vab Vbc and Vca. I am not sure how to calculate the apparent power with these values as the system will not be perfectly balanced.
Are the three voltages considered constant or is there line impedance?
Also, do you measure any phase anywhere too?
 

Thread Starter

wrt54gl

Joined Jul 20, 2015
6
Are the three voltages considered constant or is there line impedance?
Also, do you measure any phase anywhere too?
I don't measure phase angle anywhere. The loads are all uniform and all connected line to line(Delta). I know the load is very close to unity power factor. The voltages will be more or less constant but be subject to small fluctuations/imbalance just like any electrical service.
The load is greenhouse lighting. All the lights are identical electronic ballasted (unity pf) and connected line to line. So typically we would have 800 lights connected AB, 800 connected BC and 800 connected CA. They are controlled in small stages (9 at a time) so the load fluctuates but is more or less balanced.
 

MrAl

Joined Jun 17, 2014
8,990
With pure resistive load does this make any sense to you:
Vb-Va=((sqrt(3)*%i*A2-A2-2*A1)*R12)/(2*(R12+3*R1))
Vc-Vb=-((sqrt(3)*%i*A3+A3+sqrt(3)*%i*A2-A2)*R12)/(2*(R12+3*R1))
Va-Vc=((sqrt(3)*%i*A3+A3+2*A1)*R12)/(2*(R12+3*R1))

%i is the imaginary operator numerically equal to sqrt(-1) and the percent sign just shows that it is a constant in the software used to derive these equations,
Va-Vc is the delta line to line voltage phase a to c,
Vb-Va is the delta line to line voltage phase b to a,
Vc-Vb is the delta line to line voltage phase c to b,
A1 is the line voltage phase A,
A2 is the line voltage phase B,
A3 is the line voltage phase C,
R12 is the resistance across each phase line to line (for example 14.4 ohms and 120v would give us 1000 watts per phase with line resistance R1=0),
R1 is the series line resistance each phase usually quite low typically 0.1 Ohms on a regular home system but could be a little higher.

From the three equations far above we can calculate anything like power for a balanced load but where the three voltages could be slightly off. R1 in each phase also allows us to calculate the line current which can get significant if there are serious line voltage differences.

Note that those equations resolve into real numbers in the final calculations.
We can also do unbalanced loads but the equations get bigger. Ultimately we use algebraic software to do all of this anyway because doing it by hand is too hard to accomplish really.

Does any of this make any sense to you? If so you might find this easy to do.

Just to note one line to line voltage with an unbalanced load is:
Va-Vb=((sqrt(3)*%i*A3*R12*R23+A3*R12*R23+2*A1*R12*R23+2*sqrt(3)*
%i*A3*R1*R23+2*A3*R1*R23+sqrt(3)*%i*A2*R1*R23-A2*R1*R23+2*A1*R1*R23+sqrt(3)*%i*
A3*R1*R12+A3*R1*R12-sqrt(3)*%i*A2*R1*R12+A2*R1*R12+4*A1*R1*R12)*R31)/(2*(
R12*R23*R31+2*R1*R23*R31+2*R1*R12*R31+3*R1^2*R31+2*R1*R12*R23+3*R1^2*
R23+3*R1^2*R12))
and there you can see we allow differences for R12, R23, and R31 which are the three line to line load resistances.
 
Last edited:

Hymie

Joined Mar 30, 2018
1,081
One further thought on this matter, to stand any chance of obtaining a reasonably accurate value for the sampled waveforms (voltage and current), you will need to sample at a rate of at least 10x the fundamental frequency (500/sec) – across 6 parameters works out at 3,000/sec (sampling every 0.3ms).
 

Thread Starter

wrt54gl

Joined Jul 20, 2015
6
With pure resistive load does this make any sense to you:
Vb-Va=((sqrt(3)*%i*A2-A2-2*A1)*R12)/(2*(R12+3*R1))
Vc-Vb=-((sqrt(3)*%i*A3+A3+sqrt(3)*%i*A2-A2)*R12)/(2*(R12+3*R1))
Va-Vc=((sqrt(3)*%i*A3+A3+2*A1)*R12)/(2*(R12+3*R1))

%i is the imaginary operator numerically equal to sqrt(-1) and the percent sign just shows that it is a constant in the software used to derive these equations,
Va-Vc is the delta line to line voltage phase a to c,
Vb-Va is the delta line to line voltage phase b to a,
Vc-Vb is the delta line to line voltage phase c to b,
A1 is the line voltage phase A,
A2 is the line voltage phase B,
A3 is the line voltage phase C,
R12 is the resistance across each phase line to line (for example 14.4 ohms and 120v would give us 1000 watts per phase with line resistance R1=0),
R1 is the series line resistance each phase usually quite low typically 0.1 Ohms on a regular home system but could be a little higher.

From the three equations far above we can calculate anything like power for a balanced load but where the three voltages could be slightly off. R1 in each phase also allows us to calculate the line current which can get significant if there are serious line voltage differences.

Note that those equations resolve into real numbers in the final calculations.
We can also do unbalanced loads but the equations get bigger. Ultimately we use algebraic software to do all of this anyway because doing it by hand is too hard to accomplish really.

Does any of this make any sense to you? If so you might find this easy to do.

Just to note one line to line voltage with an unbalanced load is:
Va-Vb=((sqrt(3)*%i*A3*R12*R23+A3*R12*R23+2*A1*R12*R23+2*sqrt(3)*
%i*A3*R1*R23+2*A3*R1*R23+sqrt(3)*%i*A2*R1*R23-A2*R1*R23+2*A1*R1*R23+sqrt(3)*%i*
A3*R1*R12+A3*R1*R12-sqrt(3)*%i*A2*R1*R12+A2*R1*R12+4*A1*R1*R12)*R31)/(2*(
R12*R23*R31+2*R1*R23*R31+2*R1*R12*R31+3*R1^2*R31+2*R1*R12*R23+3*R1^2*
R23+3*R1^2*R12))
and there you can see we allow differences for R12, R23, and R31 which are the three line to line load resistances.
The resistances are not known but the current is. If I knew the resistances I would just use V*V/R for each phase and add the three phases together. In my case I only know the current in each line and the voltage line to line.
 

MrAl

Joined Jun 17, 2014
8,990
The resistances are not known but the current is. If I knew the resistances I would just use V*V/R for each phase and add the three phases together. In my case I only know the current in each line and the voltage line to line.
Well phase angle for a resistance is zero degrees :)
 

DickCappels

Joined Aug 21, 2008
8,710
I am pretty sure he doesn't care about reactance or impedance, phase angle,or distortion if he is measuring RMS voltage and RMS current and calculating apparent power. Somebody correct me if I lost the plot somewhere.
 
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