Question

Modeling the transmission lines as inductors, With${\mathbf{S}}_{\mathbf{ij}}\mathbf{=}{\mathbf{S}}_{\mathbf{ji}}^{\mathbf{*}}\mathbf{,}$Compute${\mathbf{S}}_{\mathbf{13}}\mathbf{,}{\mathbf{S}}_{\mathbf{31}}\mathbf{,}{\mathbf{S}}_{\mathbf{23}}\mathbf{,}{\mathbf{S}}_{\mathbf{32}}$and${\mathbf{S}}_{\mathbf{G}\mathbf{3}}$in Figure 2.25. (Hint: complex power Balance holds good at each bus, satisfying KCL).

Short answer

Therefore, the complex power $S_{13},S_{31},S_{23},S_{32}$and $S_{G3}$are$(-0.4+j1.8)$,$(-0.4+j1.8)$,$(0.1-j0.7)$,$(-0.1-j0.7)$and$(-0.5-j1.1)$respectively.

Step-by-step solution

Calculate the complex powerS13&S31.

Consider the transmission line diagram.

Apply Kirchhoff’s current law at the bus bar 1.

$$\begin{gathered} S_{13}=S_{G1}-S_{D1}-(0.4+j0.2) \\ {S}_{13}=(1+j1)-1-j1)-(0.4+0.2j) \\ {S}_{13}=-0.4+j1.8 \end{gathered}$$

Since$S_{31}$is the complex conjugate of$S_{13}$. So, $S_{31}$can be calculated as

localid="1655181600671" $$\begin{gathered} S_{31}=S_{13}^{*} \\ {S}_{13}={(-0.4+j1.8)}^{*} \\ {S}_{13}=-0.4-j1.8 \end{gathered}$$

Hence the complex power of $S_{13}$is $(-0.4+j1.8)$and $S_{31}$is $(-0.4-j1.8)$

Calculate the complex power S23&S32.

Apply Kirchhoff’s current law at the bus bar 2.

$$\begin{gathered} S_{23}=S_{G2}-S_{D2}-(0.4+j0.2) \\ {S}_{23}=(0.5+j0.5)-(1+j1)-(0.4+j0.2) \\ {S}_{23}=0.1-j0.7 \end{gathered}$$

Since $S_{32}$is the complex of $S_{23}$. So, $S_{32}$ can be calculated as,

$$\begin{gathered} S_{23}=S_{32}^{*} \\ {S}_{23}={(-0.1-j0.7)}^{*}{S}_{23}=-0.1+j0.7 \end{gathered}$$

Hence the complex power of $S_{23}$is $(0.1-j0.7)$and $S_{32}$is $(-0.1+j0.7)$.

Calculate the complex power SG3. 

Apply Kirchhoff’s current law at the bus bar 3.

The $S_{G3}$is the sum of the $S_{31}\&S_{32}$.

role="math" localid="1655182526090" $S_{G3}=S_{31}+S_{32}$

Substitute the values and solve as

$$\begin{gathered} S_{G3}=(-0.4-j1.8)+(-0.1+j0.7) \\ {S}_{G3}=-0.5-j1.1 \end{gathered}$$

Hence the complex power of $S_{G3}$is $(-0.5-j1.1)$.