Question

Expand the summations in (6.12.14) for$\mathit{N}\mathbf{=}\mathbf{2}$, and verify the formula for $\frac{\mathbf{\partial }{\mathbf{P}}_{\mathbf{L}}}{\mathbf{\partial }{\mathbf{P}}_{\mathbf{i}}}$given by (6.12.15). Assume${\mathbf{B}}_{\mathbf{i}\mathbf{J}}\mathbf{=}{\mathbf{B}}_{\mathbf{J}\mathbf{i}}\mathbf{.}$

Short answer

The value of$\frac{\partial {\mathrm{P}}_{\mathrm{L}}}{\partial {\mathrm{P}}_{\mathrm{i}}}\mathrm{is}2\sum _{\mathrm{j}-1}^2{\mathrm{P}}_{\mathrm{j}}{\mathrm{B}}_{\mathrm{iJ}}.$

Step-by-step solution

Write the given data from the question.

The value of $N=2$for given summations in reference (6.12.14).

Assume ${\mathrm{B}}_{ij}={\mathrm{B}}_{ji}$

Determine the formula for transmission line losses and derivatives.

Write the formula of total transmission losses.

${\mathbf{P}}_{\mathbf{L}}\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{-}\mathbf{1}}^{\mathbf{N}}\mathbf{\sum }_{\mathbf{J}\mathbf{=}\mathbf{1}}^{\mathbf{N}}{\mathbf{P}}_{\mathbf{j}}{\mathbf{B}}_{\mathbf{ij}}{\mathbf{P}}_{\mathbf{j}}$ …… (1)

Here,${\mathit{P}}_{\mathbf{i}}$is real power output,${\mathit{B}}_{\mathbf{i}\mathbf{j}}$is a loss coefficient and${\mathit{p}}_{\mathbf{j}}$is imaginary power output.

Write the formula of total derivative with respect to${\mathit{P}}_{\mathbf{1}}$.

$\frac{\mathbf{\partial }{\mathbf{P}}_{\mathbf{L}}}{\mathbf{\partial }{\mathbf{P}}_{\mathbf{1}}}\mathbf{=}\mathbf{2}{\mathbf{P}}_{\mathbf{1}}{\mathbf{B}}_{\mathbf{11}}\mathbf{+}\mathbf{2}{\mathbf{P}}_{\mathbf{2}}{\mathbf{B}}_{\mathbf{12}}$ …… (2)

Here,${\mathit{P}}_{\mathbf{1}}$is real power output,${\mathit{B}}_{\mathbf{11}}$and ${\mathit{B}}_{\mathbf{12}}$is a loss coefficient and${\mathit{P}}_2$is imaginary power output.

Determine the partial derivative ∂PL∂Pi.

The total transmission losses are expressed as a quadratic function of unit output power is,

Substitute N=2 into equation (1).

$$\begin{gathered} P_L=P_1{B}_{11}{P}_1+P_1{B}_{12}{P}_2+P_2{B}_{21}{P}_1+P_2{B}_{22}{P}_2 \\ =P_1^{2+}{B}_{11}+P_1{P}_2{B}_{12}+P_2{P}_1{B}_{21}+P_2^{2+}{B}_{22} \\ \mathrm{Consider}\mathrm{the}\mathrm{expression}. \\ {B}_{ij}\mathit{=}{B}_{ji} \\ \mathrm{Therefore}, \\ {\mathrm{B}}_{12}\mathit{=}{\mathrm{B}}_{21} \\ \mathrm{Then}, \\ {P}_{\mathit{1}}^{\mathit{2}\mathit{+}}{B}_{\mathit{11}}\mathit{+}\mathit{2}{P}_{\mathit{1}}{P}_{\mathit{2}}{B}_{\mathit{12}}\mathit{+}{P}_{\mathit{2}}^{\mathit{2}\mathit{+}}{B}_{\mathit{22}}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}......\left(3\right) \\ \mathrm{Here},\mathrm{Here},P_{\mathit{1}}\mathit{}\mathrm{is}\mathrm{real}\mathrm{power}\mathrm{output},B_{\mathit{11}}\mathrm{and}{B}_{\mathit{12}}\mathrm{is}\mathrm{a}\mathrm{loss}\mathrm{coefficient}\mathrm{and}{P}_{\mathit{2}}\mathrm{is} \\ \mathrm{imaginary}\mathrm{power}\mathrm{output}. \\ \mathrm{Taking}\mathrm{partial}\mathrm{derivative}\mathrm{of}\mathrm{equation}\left(3\right)\mathrm{with}\mathrm{respect}\mathrm{to}{P}_{\mathit{1}}. \\ \frac{\mathit{\partial }{\mathit{P}}_{\mathit{L}}}{\mathit{\partial }{\mathit{P}}_{\mathit{1}}}\mathit{=}\mathit{2}{P}_{\mathit{1}}{B}_{\mathit{11}}\mathit{+}\mathit{2}{P}_{\mathit{2}}{B}_{\mathit{12}} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}=2({\mathrm{P}}_1{\mathrm{B}}_{11}+2{\mathrm{P}}_2{\mathrm{B}}_{12}) \\ =2\underset{\mathrm{j}-1}{\overset{2}{\sum {\mathrm{P}}_{\mathrm{j}}{\mathrm{B}}_{\mathrm{ij}}}} \\ \mathrm{Hence},\mathrm{it}\mathrm{is}\mathrm{verified}. \end{gathered}$$