Simple Power Flow Example

Consider the simple model shown in Figure 1, where a large stiff network supplies a constant power load ${\boldsymbol {S}}\,$ through an impedance ${\boldsymbol {Z_{s}}}$:

Figure 1. Simple power flow model (note that all quantities are in per-unit)

Suppose the load power ${\boldsymbol {S}}\,$ is known and we want to calculate the load bus voltage $V_{r}\,$. Unfortunately, this cannot be computed in a straightforward manner because the load is of constant power and thus the load current and impedance are voltage dependent, i.e. the load draws more current as voltage decreases. As a result, the load bus voltage is non-linearly related to the load itself.

Derivation of the Load Bus Voltage

Note that in this derivation, all quantities are in per-unit. Recall that the load complex power can be calculated from the voltage and current phasors as follows:

${\boldsymbol {S}}={\boldsymbol {V_{r}}}{\boldsymbol {I}}^{*}\,$

$=V_{r}\angle 0\left({\frac {1\angle \delta -V_{r}\angle 0}{\boldsymbol {Z_{s}}}}\right)^{*}\,$

Suppose that we represent both the load power ${\boldsymbol {S}}\,$ and impedance ${\boldsymbol {Z_{s}}}$ in polar coordinates, i.e.

${\boldsymbol {S}}=S\angle \phi \,$

${\boldsymbol {Z_{s}}}=Z\angle \theta \,$

Then the power equation can be re-written as:

$S\angle \phi =V_{r}\angle 0\left({\frac {1\angle \delta -V_{r}\angle 0}{Z\angle \theta }}\right)^{*}\,$

Conjugating the terms in brackets and simplifying, we get:

$S\angle \phi \times Z\angle (\theta )=V_{r}\angle (\delta )-V_{r}^{2}\,$

$SZ\angle (\phi +\theta )+V_{r}^{2}=V_{r}\angle (\delta )\,$

Applying Euler's law, we get:

$SZ\left[\cos(\phi +\theta )+j\sin(\phi -\theta )\right]+V_{r}^{2}=V_{r}\left[\cos(\delta )+j\sin(\delta )\right]\,$

Separating the real and imaginary terms of the above equation:

$SZ\cos(\phi +\theta )+V_{r}^{2}=V_{r}\cos(\delta )\,$

$SZ\sin(\phi +\theta )=V_{r}\sin(\delta )\,$

Squaring both equations and summing them together, we get:

$\left[SZ\cos(\phi +\theta )+V_{r}^{2}\right]^{2}+(SZ)^{2}\sin ^{2}(\phi +\theta )=V_{r}^{2}\cos ^{2}(\delta )+V_{r}^{2}\sin ^{2}(\delta )\,$

Simplifying and re-arranging the equation above, we can get the following homogenous equation:

$V_{r}^{4}+\left[2SZ\cos(\phi +\theta )-1\right]V_{r}^{2}+(SZ)^{2}=0\,$

This equation can be solved for the load bus voltage $V_{r}\,$ using the quadratic formula:

$V_{r}={\sqrt {\frac {-b\pm {\sqrt {b^{2}-4c}}}{2}}}\,$ ... Equ. (1)

Where $b=2SZ\cos(\phi +\theta )-1\,$

and $c=(SZ)^{2}\,$

Worked Example

Suppose that the source bus has a nominal voltage of 33kV and a short circuit level of 800MVA at X/R ratio of 10. What is the voltage at the load bus if it is supplying a constant 50MW load at 0.8pf (lagging)?

The source impedance is:

$||Z_{s}||={\frac {V^{2}}{S_{SC}}}={\frac {33kV^{2}}{800MVA}}=1.361\Omega \,$

Converted to per-unit values (on a 100 MVA base):

$||Z_{s,pu}||={\frac {||Z_{s}||}{Z_{base}}}={\frac {1.361}{10.89}}=0.125pu\,$

For an X/R ratio of 10, the source impedance is therefore:

${\boldsymbol {Z_{s,pu}}}=0.02799+j0.2799pu=0.281\angle 1.471\,$ (angle in radians)

The 50MW load converted to per unit is (by convention, a load with lagging power factor has a negative reactive power):

${\boldsymbol {S_{pu}}}=0.5-j0.375pu=0.625\angle -0.6435\,$ (angle in radians)

Plugging the parameters $S=0.625\,$, $Z=0.125\,$, $\phi =-0.6435\,$ and $\theta =1.471\,$ into Equation (1), we get the load bus voltage:

$V_{r}=0.9433pu\,$

Intuition for General Power Flow Solutions

We saw in this simple example that the power flow problem for constant power loads is non-linear (i.e. quadratic) and cannot be solved with linear techniques. This intuition can be extended to more general power flow problems. While a closed form solution was found for this simple case, the general power flow problem is typically solved using iterative numerical methods, for example a Newton-Raphson algorithm.