# Symmetrical three-phase power flow problem, √3

I have done quite a lot of research regarding the power flow problem for the steady state operation of an electric system. In many papers about the power flow problem the power flow equation are referred to as the network equation:

$$YU=I$$

where $Y$ is the admittance matrix of the network, $U$ the complex vector of bus voltages and $I$ the complex vector of bus currents, and the power equation:

$$S_i=U_i I_i^*$$

where $i$ is one bus of the network and $S_i$ is the complex power in that bus.

Those two equations can be solved e.g. the Newton-Raphson method, when the power equation is substituted into the network equation. However, my question is not about how to solve the power flow problem but about the usage in electric distribution systems with three-phase electric power.

For symmetrical thee-phase power systems, the equation for complex power is:

$$S= \sqrt{3} U I^*$$

where $U$ is the voltage between two conductors. This voltage between conductors is always the value referred to, when talking about the voltage level of an electric distribution network. In theoretical literature about electric distribution networks, this equation with the additional factor $\sqrt{3}$ for three phase current is always used for the power flow problem, whereas in many papers where the power flow problem is solved the equation is used without the factor $\sqrt{3}$. In my opinion, only the second equation is correct for symmetrical three phase systems.

We have developed a program for solving the power flow problem for three-phase electric distribution networks, but we are using the equations above without the factor $\sqrt{3}$. We have compared our result to the commercial software NEPLAN and our results are exactly the same. Also in the documentation of NEPLAN the factor $\sqrt{3}$ is not used in the power flow equation. I am currently writing my master's thesis and I am trying to write about the theoretical background of the power-flow problem for three phase current and I am wondering why the factor $\sqrt{3}$ usually doesn't occur in many papers and why our results are correct, when this factor is not used.

Briefly, my question is, where is the factor $\sqrt{3}$ in the power flow problem for three phase electric distrubtion systems? I it hidden somewhere, maybe in the Admittance matrix?

• I suggest you get familiar with this sites latex formula abilities to make your formulas less ambiguous. Also try to cut off some fluff Oct 27 '17 at 9:43
• If such papers are using per-unit values, that could a reason you don't see the factors. May 16 '20 at 8:24

The $\sqrt(3)$ comes from resolving 3phase power when the voltage is in line voltage NOT phase voltage.

As you wrote:

For symmetrical thee-phase power systems, the equation for complex power is:

$S=\sqrt3UI∗$

where U is the voltage between two conductors. This voltage between conductors is always the value referred to, when talking about the voltage level of an electric distribution network.

The power in a 3phase system is 3*I*V where I is the phase current and V is the phase voltage. Your statement is with regards to line-line voltage. line-line voltage is $\sqrt3$ larger THUS

$P = 3 \frac{U}{\sqrt3}I$ where U is line-line voltage, I is phase voltage

via surds: $\frac{3}{\sqrt3}= \sqrt3$

Thus $P = 3 \frac{U}{\sqrt3}I = \sqrt3UI$

• Thank you for your answer. I am aware of the difference between line-line voltage and phase voltage, but in many power flow studies, there will be neither factor 3 for the phase voltage nor factor $\sqrt{3}$ is used, but only S=UI despite three phase current Oct 27 '17 at 12:22
• the only times I have seen this (and I just cross-referenced with mohan power electronics book) is when referencing per-phase power Oct 27 '17 at 12:46

Basically, the $$\\sqrt{3}\$$ is an simplification of the real calculation:

$$\begin{equation} S_i = [{U_p}_i]^T.[{I_p}_i]^\star \end{equation}$$

where:

• $$\[{U_p}_i]=[U_a\ U_b\ U_c]_i^T\$$ - $$\i-th\$$ node phase voltages;
• $$\[{I_p}_i]=[I_a\ I_b\ I_c]_i^T\$$ - $$\i-th\$$ node phase currents.

Substitutin into the formula yields:

$$S_i = {U_a}_i.{I_a}_i^\star + {U_b}_i.{I_b}_i^\star + {U_c}_i.{I_c}_i^\star$$

In transmission systems, since loads and voltages are ballanced, this procudes:

$$S_i = {U_p}_i.{I_p}_i^\star + {U_p}_i\angle-120^\circ.({I_p}_i\angle-120)^\star + {U_p}_i\angle-240.({I_p}_i\angle-240)^\star$$

Due to the complex conjugate operation, this equation simplifies to:

$$S_i = 3{U_p}_i.{I_p}_i^\star$$

Now, depending on your conection, we have different approaches.

For a $$\\Delta\$$ load, we have that the phase voltage is equal to the line voltage, and the phase current is equal to the line current over $$\\sqrt{3}\$$:

\begin{align} {U_p}_i & = U_i \\ {I_p}_i & = \frac{I_i}{\sqrt{3}} \end{align}

Substituting in our latter equation, we get:

\begin{align} S_i & = 3.U_i.\frac{I_i^\star}{\sqrt{3}} \\ S_i & = \sqrt{3}.U_i.I_i^\star \end{align}

For a $$\Y\$$ connected load, we have the phase voltage being equal to the line voltage over $$\\sqrt{3}\$$ and the phase current is equal to the line current:

\begin{align} {U_p}_i & = \frac{U_i}{\sqrt{3}} \\ {I_p}_i & = I_i \end{align}

Substituting back in our equation also yields the same formula. That is where the $$\\sqrt{3}\$$ comes from in transmission systems.

Since distribution systems have sometimes single phase lines, and since most loads are not balanced, theses simplifications do not apply.As such, the first formula is used.