Once we have solved the economic dispatch problem to find g1* and g2*, we can use our three-node power flow model from Lesson 2 to calculate the flows on the transmission lines that arise from the economic dispatch.

#### Example:

We'll use the economic dispatch from our first example, g1* = 30 MWh and g2* = 20 MWh, and we'll use the three node network shown in the figure below, in which the resistances on all branches are identical.

We start by calculating the distribution factors for each line due to the power injected by the generator at node 1:

$$\text{D}\left(\text{1,3;G1}\right)\text{=}\frac{{\text{dF}}_{\text{13}}}{\text{dG1}}\text{=}\left(\frac{1/\text{R}}{1/\text{R}\text{+}1/\text{2R}}\right)\text{=+}2/3$$

$$\text{D}\left(\text{1,2;G1}\right)\text{=}\frac{{\text{dF}}_{\text{12}}}{\text{dG1}}\text{=}\left(\frac{1/\text{2R}}{1/\text{R}\text{+}1/\text{2R}}\right)\text{=+}1/3$$

$$\text{D}\left(\text{2,3;G1}\right)\text{=}\frac{{\text{dF}}_{23}}{\text{dG1}}\text{=}\left(\frac{1/\text{2R}}{1/\text{R}\text{+}1/\text{2R}}\right)\text{=+}1/3$$

The distribution factors due to the power injected by the generator at node 2 are:

$$\text{D}\left(\text{1,3;G2}\right)\text{=}\frac{{\text{dF}}_{13}}{\text{dG2}}\text{=}\left(\frac{1/\text{2R}}{1/\text{R}\text{+}1/\text{2R}}\right)\text{=+}1/3$$

$$\text{D}\left(\text{1,2;G2}\right)\text{=\u2013}\frac{{\text{dF}}_{12}}{\text{dG2}}\text{=\u2013}\left(\frac{1/\text{2R}}{1/\text{R+}1/\text{2R}}\right)\text{=\u2013}1/3$$

$$\text{D}\left(\text{2,3;G2}\right)\text{=}\frac{{\text{dF}}_{23}}{\text{dG2}}\text{=}\left(\frac{1/\text{R}}{1/\text{R}\text{+}1/\text{2R}}\right)\text{=+}2/3$$

The resulting power flows are thus:

$${\text{F}}_{\text{13}}\text{=30}\times \text{D}\left(\text{1,3;G1}\right)\text{+20}\times \text{D}\left(\text{1,3;G2}\right)\text{=30}\times \text{}\frac{2}{3}\text{+20}\times \text{}\frac{1}{3}\text{=26}\frac{2}{3}\text{MWh}$$

$${\text{F}}_{12}\text{=30}\times \text{D}\left(\text{1,2;G1}\right)\text{+20}\times \text{D}\left(\text{1,2;G2}\right)\text{=30}\times \text{}\frac{1}{3}\text{+20}\times \text{\u2013}\frac{1}{3}\text{=3}\frac{1}{3}\text{MWh}$$

$${\text{F}}_{\text{23}}\text{=30}\times \text{D}\left(\text{2,3;G1}\right)\text{+20}\times \text{D}\left(\text{2,3;G2}\right)\text{=30}\times \text{}\frac{1}{3}\text{+20}\times \text{}\frac{2}{3}\text{=23}\frac{1}{3}\text{MWh}$$

#### Here is one for you to try on your own

Using the three-node network. Assume that C_{1}(g1) = 15 + 5g1+ 3g1^{2} and C_{2}(g2) = 5 + 45g2 + 2g2^{2}. Total demand for electricity is 100 MWh and the resistances on all branches of the three-node network are identical. Show the following:

- g1* = 44 MWh, g2* = 56 MWh
- λ
_{sys}= $269/MWh - Total System Cost = $14,840
- Flows are: F
_{13}= 48 MW; F_{23}= 52 MW; F_{12}= -4 MW