# 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 100MVA base):

$||Z_{s,pu}|| = \frac{||Z_{s}||}{Z_{base}} = \frac{1.361}{4.84} = 0.281 pu \,$

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

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

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

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

Plugging the parameters $S = 0.625 \,$, $Z = 0.281 \,$, $\phi = -0.6435 \,$ and $\theta = 1.471 \,$ into Equation (1), we get the load bus voltage:

$V_r = 0.8480 pu \,$

## 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.