Three phase voltage system of polynomial equations

Question

I'm working with the development of a product in the company where I work. This product measures three phase voltages and currents. I cannot change the circuit because it has been sold for a long time and is well tested.

The circuit gives me only the measure of phase to phase, without neutral. From this point, I have to find each phase to neutral value.

I know how to calculate line-to-line voltages when the line-to-line separation is exactly 120 degrees. The equation is:

$$\left\{\begin{array}{rclr}|V_{ab}| = \sqrt{(|V_{an}| + \frac{|V_{bn}|}{2})^2 + (|V_{bn}|\frac{\sqrt{3}}{2})^2} &\qquad\qquad(1)\\ |V_{bc}| = \sqrt{(|V_{bn}| + \frac{|V_{cn}|}{2})^2 + (|V_{cn}|\frac{\sqrt{3}}{2})^2} &\qquad\qquad(2)\\ |V_{ca}| = \sqrt{(|V_{cn}| + \frac{|V_{an}|}{2})^2 + (|V_{an}|\frac{\sqrt{3}}{2})^2} &\qquad\qquad(3)\\ \end{array}\right.$$

To simplify, I've used $\mathbf x = V_{ab}, y = V_{bc}, z = V_{ca}, a = V_{an}, b = V_{bn}, c = V_{cn}$

So I've tried to resume the equations, as can be seen below:

$$\left\{\begin{array}{rclr}|x|^2 &=& 4|a|^2 + 4|a||b| + 3|b|^2 &(3)\\ |y|^2 &=& 4|b|^2 + 4|b||c| + 3|c|^2 &\qquad\qquad(4)\\ |z|^2 &=& 4|c|^2 + 4|c||a| + 3|a|^2&(5)\\ \end{array}\right.$$

If my thought is correct (has more than 13 years that I've studied three phase circuits and math) I think that solving the system of polynomial equations above to determine a, b and c could also solve my problem.

So, I'd like to know how to solve this system of polynomial equations.

Answer 1

Your system can be written as:

$$ \begin{align} x^2 &= a^2 + |ab| + b^2 \quad (1) \\ y^2 &= b^2 + |bc| + c^2 \quad (2) \\ z^2 &= c^2 + |ca| + a^2 \quad (3) \end{align} $$

It is clear that each equation $i$ specifies some surface $S_i$, using WolframAlpha for a quick impression, the shape seems to look like a diamond with rounded sides:

Graph by WA

The diamond shape rises from absolute value function, it combines my initially forcasted ellipses:

Those general ellipses $E_i^\pm$ are rotated $$ \alpha = \pm 45^\circ $$

The upper right and lower left side of the diamond are taken from $E_i^+$, the other two sides from $E_i^-$.

Which level curve from the contour plot is valid, depends on the value of $|x|$, $|y|$ or $|z|$.

Each of these values selects one of the diamonds. As one variable is free, the diamond is extruded into a tube.

It should be possible to draw the surfaces in $a$-, $b$-, $c$-space, so one can see, if the common intersection

$$ S_1 \cap S_2 \cap S_3 $$

is non-empty. Thus:

This problem is similiar to finding the intersection of three perpendicular diamond-shaped tubes (with possibly different diameters).

Here is a sketch of the easier case with three cylinders of the same radius instead of diamond-shaped tubes:

The situation here is already complicated, see

• Steinmetz Solid from Wikipedia
• Steinmetz Solid from mathworld.wolfram.com:

Some easy solutions:

• $(x,y,z)^t = (0,0,0)^t$ with solution $(a,b,c)^t = (0,0,0)^t$.

Two equations can be fulfilled with non-zero values, and they must have the same radius. The third one enforces its variables to zero.

• $(x,y,z)^t = (\pm R, 0, \pm R)^t$ with $(a,b,c)^t = (\pm R, 0, 0)^t$.

• $(x,y,z)^t = (\pm R, \pm R, 0)^t$ with $(a,b,c)^t = (0, \pm R, 0)^t$.

• $(x,y,z)^t = (0, \pm R, \pm R)^t$ with $(a,b,c)^t = (0, 0, \pm R)^t$.

Another solution:

• $(x, y, z)^t = (\pm R,\pm R,\pm R)^t$ with $(a, b, c)^t = \pm (\frac{R}{\sqrt{3}}, \frac{R}{\sqrt{3}}, \frac{R}{\sqrt{3}})^t$.

Conclusion: First I thought this problem is very hard (I had no real grip yet), then I thought it was boringly easy (because I overestimated my model and made the mistake to think in circles, not in cylinders), now I think it is quite complicated because the intersection looks not that simple to describe. Right now I would stick to a mix of geometrical insight and numerical methods to search for solutions.

Answer 2

According to Kirchhoff's law, the sum of voltages in a closed electrical ciruit path is zero. So: $$V_{ab} + V_{bc} +V_{ca} = 0$$

When drawing these vectors, they form a triangle.
The neutral voltage will a point somewhere within this triangle.
The phase to neutral voltages $V_{an}, V_{bn}, V_{cn}$ you are looking for, are the connection vectors between the neutral point and the triangle corners. Assuming the phase to neutral voltages have a $120$ degrees phase shift (which should be the case in an electrical three phase system), there will be only one solution for the neutral point that fits.
You could draw this easily on a piece of paper, and next do some trigonometry to find the exact mathematical answer.