Jules: /* Worked Example */

2026-04-23T23:03:25Z

Worked Example

Jules: Created page with "Consider the simple model shown in Figure 1, where a large stiff network supplies a constant power load $\boldsymbol{S} \,$ through an impedance <..."

2020-11-18T14:04:58Z

Created page with "Consider the simple model shown in Figure 1, where a large stiff network supplies a constant power load <math>\boldsymbol{S} \, </math> through an impedance <..."

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Consider the simple model shown in Figure 1, where a [[Infinite Bus|large stiff network]] supplies a constant power load <math>\boldsymbol{S} \, </math> through an impedance <math>\boldsymbol{Z_{s}}</math>:

[[Image:Simple_Power_Flow_Fig1.png|left|frame|250px|Figure 1. Simple power flow model (note that all quantities are in [[Per-unit System|per-unit]])]]
<br clear=all>

Suppose the load power <math>\boldsymbol{S} \, </math> is known and we want to calculate the load bus voltage <math>V_{r} \, </math>. 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 System|per-unit]]. Recall that the load [[Electrical_Power#Complex_Power_from_Phasors|complex power can be calculated from the voltage and current phasors]] as follows:

: <math> \boldsymbol{S} = \boldsymbol{V_{r}} \boldsymbol{I}^{*} \, </math>

:: <math> = V_{r} \angle 0 \left( \frac{1 \angle \delta - V_{r} \angle 0}{\boldsymbol{Z_{s}}} \right)^{*} \, </math>

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

: <math> \boldsymbol{S} = S \angle \phi \, </math>
: <math> \boldsymbol{Z_{s}} = Z \angle \theta \, </math>

Then the power equation can be re-written as:

: <math> S \angle \phi = V_{r} \angle 0 \left( \frac{1 \angle \delta - V_{r} \angle 0}{Z \angle \theta} \right)^{*} \, </math>

Conjugating the terms in brackets and simplifying, we get:

: <math> S \angle \phi \times Z \angle (\theta) = V_{r} \angle (\delta) - V_{r}^{2} \, </math>

: <math> SZ \angle (\phi + \theta) + V_{r}^{2} = V_{r} \angle (\delta) \, </math>

Applying [http://en.wikipedia.org/wiki/Euler%27s_formula Euler's law], we get:

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

Separating the real and imaginary terms of the above equation:

: <math> SZ \cos(\phi + \theta) + V_{r}^{2} = V_{r} \cos(\delta) \, </math>

: <math> SZ \sin(\phi + \theta) = V_{r} \sin(\delta) \, </math>

Squaring both equations and summing them together, we get:

: <math> \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) \, </math>

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

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

This equation can be solved for the load bus voltage <math>V_{r} \, </math> using the [http://en.wikipedia.org/wiki/Quadratic_equation#Quadratic_factorization quadratic formula]:

: <math> V_{r} = \sqrt{\frac{-b \pm \sqrt{b^{2} - 4c}}{2}} \, </math> ... Equ. (1)

Where <math> b = 2SZ \cos(\phi + \theta) - 1 \, </math>

and <math> c = (SZ)^{2} \, </math>

== 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:

: <math>||Z_{s}|| = \frac{V^{2}}{S_{SC}} = \frac{33kV^{2}}{800MVA} = 1.361 \Omega \, </math>

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

: <math>||Z_{s,pu}|| = \frac{||Z_{s}||}{Z_{base}} = \frac{1.361}{4.84} = 0.281 pu \, </math>

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

: <math>\boldsymbol{Z_{s,pu}} = 0.02799 + j0.2799 pu = 0.281 \angle 1.471 \, </math> (angle in radians)

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

: <math>\boldsymbol{S_{pu}} = 0.5 - j0.375 pu = 0.625 \angle -0.6435 \, </math> (angle in radians)

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

: <math>V_r = 0.8480 pu \, </math>

== 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 [[Power Flow|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_Power_Flow|Newton-Raphson algorithm]].

[[Category:Fundamentals]]
[[Category:Modelling / Analysis]]

@@ Line 65: / Line 65: @@
 : <math>||Z_{s}|| = \frac{V^{2}}{S_{SC}} = \frac{33kV^{2}}{800MVA} = 1.361 \Omega \, </math>
-Converted to per-unit values (on a 100MVA base):
+Converted to per-unit values (on a 100 MVA base):
-: <math>||Z_{s,pu}|| = \frac{||Z_{s}||}{Z_{base}} = \frac{1.361}{4.84} =  0.281 pu \, </math>
+: <math>||Z_{s,pu}|| = \frac{||Z_{s}||}{Z_{base}} = \frac{1.361}{10.89} =  0.125 pu \, </math>
 For an X/R ratio of 10, the source impedance is therefore:
@@ Line 77: / Line 77: @@
 : <math>\boldsymbol{S_{pu}} = 0.5 - j0.375 pu = 0.625 \angle -0.6435 \, </math> (angle in radians)
-Plugging the parameters <math>S = 0.625 \, </math>, <math>Z = 0.281 \, </math>, <math>\phi = -0.6435 \, </math> and <math>\theta = 1.471 \, </math> into Equation (1), we get the load bus voltage:
+Plugging the parameters <math>S = 0.625 \, </math>, <math>Z = 0.125\, </math>, <math>\phi = -0.6435 \, </math> and <math>\theta = 1.471 \, </math> into Equation (1), we get the load bus voltage:
-: <math>V_r = 0.8480 pu \, </math>
+: <math>V_r = 0.9433 pu \, </math>
 == Intuition for General Power Flow Solutions ==

Simple Power Flow Example - Revision history

Jules: /* Worked Example */

Jules: Created page with "Consider the simple model shown in Figure 1, where a large stiff network supplies a constant power load \boldsymbol{S} \, through an impedance <..."

Jules: Created page with "Consider the simple model shown in Figure 1, where a large stiff network supplies a constant power load $\boldsymbol{S} \,$ through an impedance <..."