# Three phase rms voltage calculation

I am new in instrumentation and control. I am developing a hardware for a project and the following is my application.

1. I have three phase stepped down voltages and stepped down currents entering my board.
3. I am trying to calculate the three phase rms voltage, three phase frequency using pll, real power, reactive power and the apparent power.

I am able to calculate the real power, reactive power and apparent with ease. I am having some questions regarding the three phase rms voltage calculation. Right now, I am measuring the line to neutral of the phase A and calculating the rms of this measurement. The loads are balanced and the system is also balanced. I would like to know, if this is the right way to calculate the three phase rms (just the rms of one phase) or is there any other way to calculate the three phase rms.

Any help would be appreciated.

• Not sure what is meant here by "three phase rms voltage". Voltages in three phase systems can be measured from line-to-line or line-to-neutral. – user28910 Nov 13 '13 at 17:08
• I apologize for not explaining the need for this "three phase rms voltage". I want to observe the three phases in one measurement to reduce the data being observed. For example if one of the phases goes out, I need to observe this change. It need not have to indicate which phase is out, but this should be reflected in the measurement. I guess I can average the three phase rms voltages(a,b,c rms) to indicate that three is a deviation in one of the phases. – John Nov 14 '13 at 17:23

Any help would be appreciated

You can measure phase-neutral voltage and if the supply and load are balanced then you can infer line voltage by: -

line volts = phase volts$\times\sqrt3$

This should be one of the easier measurements yet you say you can calculate "real power, reactive power and apparent with ease". This does make me think that you have used one of these measurements, and the RMS measurement of current and back-calculated phase/line voltage and that's when you are seeing a discrepency.

If this is so then I suspect your current measurement may be at fault either through incorrect use of a current transformer or some ratio being incorrect. Or, it could be the $\sqrt3$ thing mentioned above.

If you need any further help with this you should consider detailing how you make the other measurements and why you believe voltage to be incorrectly calculated/measured.

I wouldn't use a parameter that averages the RMS values of the three phases in the case of a fault. You should measure them separately. If you want to have an indication of the symmetry try using star point current or voltage (dependent on the grounding). You can calculate it from your data as you have plenty of it. They will give you a good indication on the symmetry and faults.

Look up symetrical components. You could use the direct system voltage as your quasi RMS average.

Edit: Now I have a little time to Elaborate:

Symmetrical components are used for simplification of three phase electrical systems, mainly the power system itself. The advantage of it is that you get three decoupled systems which work in superposition:

1. Direct system - the actual symmetric system, in an electrical machine it creates positive torque
2. Inverse system - usually apears because of inter-phase differences, creates negative torque (practically reducing torque)
3. Zero system - appears most prominently because of earth faults.

To calculate them you will need a transformation matrix: \begin{align} \mathbf A=\begin{bmatrix}1& 1& 1\\ 1 &a^2 &a\\ 1 &a &a^2\end{bmatrix} \end{align} The parameter a is taken as $$a=exp(j120)$$

To get the symmetric components you now only have to do the folowing calculation: \begin{align} \begin{bmatrix}I_0\\ I_d\\ I_i\end{bmatrix}=\mathbf A \begin{bmatrix}I_a\\ I_b\\ I_c\end{bmatrix} \end{align}

This formula works the same way for the voltages, so you can put $$\mathbf V_{0di}=\mathbf A \mathbf V_{abc}$$ To get your values back to the standard form just use the inverse of the $\mathbf A$ matrix. \begin{align} \mathbf A^{-1}=\frac{1}{3}\begin{bmatrix}1& 1& 1\\ 1 &a &a^2\\ 1 &a^2&a\end{bmatrix} \end{align}

If you have a balanced system then the direct voltage will be equal to a the phase voltages, and the other components will be zero.