Question

For a short transmission line of impedance$\mathbf{R}\mathbf{+}\mathbf{jX}$ ohms per phase, show that the maximum power that can be transmitted over the line is

${\mathbf{P}}_{\mathbf{max}}\mathbf{=}\frac{{{\mathbf{V}}_{\mathbf{R}}}^{\mathbf{2}}}{{\mathbf{Z}}^{\mathbf{2}}}\left(\frac{{\mathrm{ZV}}_s}{V_R}-R\right)$ where $\mathbf{Z}\mathbf{=}\sqrt{{\mathbf{R}}^{\mathbf{2}}\mathbf{+}{\mathbf{X}}^{\mathbf{2}}}$

when the sending-end and receiving-end voltages are fixed, and for the condition

$\mathbf{Q}\mathbf{=}\frac{\mathbf{-}{{\mathbf{V}}_{\mathbf{R}}}^{\mathbf{2}}\mathbf{X}}{{\mathbf{R}}^{\mathbf{2}}\mathbf{+}{\mathbf{X}}^{\mathbf{2}}}$When$\frac{\mathbf{dP}}{\mathbf{dQ}}\mathbf{=}\mathbf{0}$

Short answer

The maximum power that can be transmitted over the line is $\mathrm{P}=\frac{{{\mathrm{V}}_{\mathrm{R}}}^2}{{\mathrm{Z}}^2}\left(\frac{{\mathrm{V}}_{\mathrm{s}}\mathrm{Z}}{{\mathrm{V}}_{\mathrm{R}}}-\mathrm{R}\right)$.

Step-by-step solution

Determine the equation to derive the expression for maximum power.

The equation for the real power is given as follows.

$\mathbf{P}\mathbf{=}{\mathbf{V}}_{\mathbf{R}}{\mathbf{Icos\varphi }}_{\mathbf{R}}$ …… (1)

The equation for the reactive power is given as follows.

$\mathbf{Q}\mathbf{=}{\mathbf{V}}_{\mathbf{R}}{\mathbf{Isin\varphi }}_{\mathbf{R}}$ …… (2)

Derive the expression for maximum power.

Draw the phasor diagram of the short transmission line.

Here, ${\mathrm{V}}_{\mathrm{s}}$is the sending end voltage, ${\mathrm{V}}_{\mathrm{R}}$ is the receiving end voltage, ${\mathrm{I}}_{\mathrm{R}}$is the receiving end current, ${\mathrm{I}}_{\mathrm{S}}$is the sending end current and ${\mathrm{\varphi }}_{\mathrm{R}}$is the receiving end power factor.

The sending end voltage from the phasor diagram,

role="math" localid="1656331938434" $$\begin{gathered} {\left({\mathrm{V}}_{\mathrm{S}}\right)}^2={\left({\mathrm{V}}_{\mathrm{R}}\right)}^2+\left(2{\mathrm{V}}_{\mathrm{R}}\mathrm{I}\right)\left({\mathrm{Rcos\varphi }}_{\mathrm{R}}+{\mathrm{Xsin\varphi }}_{\mathrm{R}}\right)+{\mathrm{I}}^2\left({\mathrm{R}}^2+{\mathrm{X}}^2\right) \\ {\left({\mathrm{V}}_{\mathrm{S}}\right)}^2={\left({\mathrm{V}}_{\mathrm{R}}\right)}^2+2\mathrm{R}\left({\mathrm{V}}_{\mathrm{R}}{\mathrm{Icos\varphi }}_{\mathrm{R}}\right)+2\mathrm{R}\left({\mathrm{V}}_{\mathrm{R}}{\mathrm{Isin\varphi }}_{\mathrm{R}}\right)+{\mathrm{I}}^2\left({\mathrm{R}}^2+{\mathrm{X}}^2\right)...\left(3\right) \end{gathered}$$

Substitute $\mathrm{P}$for ${\mathrm{V}}_{\mathrm{R}}{\mathrm{Icos\varphi }}_{\mathrm{R}}$and $\mathrm{Q}$for ${\mathrm{V}}_{\mathrm{R}}{\mathrm{Isin\varphi }}_{\mathrm{R}}$into above equation.

role="math" localid="1656331985630" $$\begin{gathered} {\left({\mathrm{V}}_{\mathrm{S}}\right)}^2={\left({\mathrm{V}}_{\mathrm{R}}\right)}^2+2\mathrm{RP}+2\mathrm{RQ}+\frac{{\mathrm{P}}^2+{\mathrm{Q}}^2}{{{\mathrm{V}}_{\mathrm{R}}}^2}\left({\mathrm{R}}^2+{\mathrm{X}}^2\right) \\ -{\left({\mathrm{V}}_{\mathrm{S}}\right)}^2+{\left({\mathrm{V}}_{\mathrm{R}}\right)}^2+2\mathrm{RP}+2\mathrm{RQ}+\frac{1}{{{\mathrm{V}}_{\mathrm{R}}}^2}\left({\mathrm{P}}^2+{\mathrm{Q}}^2\right)\left({\mathrm{R}}^2+{\mathrm{X}}^2\right)=0.....\left(4\right) \end{gathered}$$

Differentiate$\mathrm{P}$with respect to $\mathrm{Q}$,

$$\begin{gathered} 2\mathrm{R}\left(\frac{\mathrm{dP}}{\mathrm{dQ}}\right)+2\mathrm{X}+\frac{1}{{{\mathrm{V}}_{\mathrm{R}}}^2}\left({\mathrm{R}}^2+{\mathrm{X}}^2\right)\left(2\mathrm{P}\left(\frac{\mathrm{dP}}{\mathrm{dQ}}\right)+2\mathrm{Q}\right)=0 \\ \frac{\mathrm{dP}}{\mathrm{dQ}}\left[2\mathrm{R}+2\mathrm{P}\left(\frac{1}{{{\mathrm{V}}_{\mathrm{R}}}^2}\right)\left({\mathrm{R}}^2+{\mathrm{X}}^2\right)\right]=-\left[2\mathrm{X}+\frac{{\mathrm{R}}^2+{\mathrm{X}}^2}{{{\mathrm{V}}_{\mathrm{R}}}^2}2\mathrm{Q}\right] \\ \frac{\mathrm{dP}}{\mathrm{dQ}}=\frac{-\left[2\mathrm{X}+{\displaystyle \frac{{\mathrm{R}}^2+{\mathrm{X}}^2}{{{\mathrm{V}}_{\mathrm{R}}}^2}}2\mathrm{Q}\right]}{\left[2\mathrm{R}+2\mathrm{P}\left(\frac{1}{{{\mathrm{V}}_{\mathrm{R}}}^2}\right)\left({\mathrm{R}}^2+{\mathrm{X}}^2\right)\right]} \end{gathered}$$

The condition for the maximum power is $\frac{\mathrm{dP}}{\mathrm{dQ}}=0$,

$$\begin{gathered} 0=\frac{-\left[2\mathrm{X}+{\displaystyle \frac{{\mathrm{R}}^2+{\mathrm{X}}^2}{{{\mathrm{V}}_{\mathrm{R}}}^2}}2\mathrm{Q}\right]}{\left[2\mathrm{R}+2\mathrm{P}\left({\displaystyle \frac{1}{{{\mathrm{V}}_{\mathrm{R}}}^2}}\right)\left({\mathrm{R}}^2+{\mathrm{X}}^2\right)\right]} \\ \left[2\mathrm{X}+\frac{{\mathrm{R}}^2+{\mathrm{X}}^2}{{{\mathrm{V}}_{\mathrm{R}}}^2}2\mathrm{Q}\right]=0 \\ \frac{{\mathrm{R}}^2+{\mathrm{X}}^2}{{{\mathrm{V}}_{\mathrm{R}}}^2}2\mathrm{Q}=-2\mathrm{X} \\ \mathrm{Q}=\frac{-{{\mathrm{V}}_{\mathrm{R}}}^2\mathrm{X}}{{\mathrm{R}}^2+{\mathrm{X}}^2} \end{gathered}$$

Substituterole="math" localid="1656333096907" $\frac{-{{\mathrm{V}}_{\mathrm{R}}}^2\mathrm{X}}{{\mathrm{R}}^2+{\mathrm{X}}^2}$for$\mathrm{Q}$equation (4).

Solve further as,

$$\begin{gathered} -{\left({\mathrm{V}}_{\mathrm{S}}\right)}^2+{\left({\mathrm{V}}_{\mathrm{R}}\right)}^2+2\mathrm{RP}+2\mathrm{X}\left(\frac{-{{\mathrm{V}}_{\mathrm{R}}}^2\mathrm{X}}{{\mathrm{R}}^2+{\mathrm{X}}^2}\right)+\frac{1}{{{\mathrm{V}}_{\mathrm{R}}}^2}\left({\mathrm{P}}^2+\frac{{{\mathrm{V}}_{\mathrm{R}}}^4{\mathrm{X}}^2}{{\left({\mathrm{R}}^2+{\mathrm{X}}^2\right)}^2}\right)\left({\mathrm{R}}^2+{\mathrm{X}}^2\right)=0 \\ -{\left({\mathrm{V}}_{\mathrm{S}}\right)}^2+{\left({\mathrm{V}}_{\mathrm{R}}\right)}^2+2\mathrm{RP}+2\left(\frac{-{{\mathrm{V}}_{\mathrm{R}}}^2{\mathrm{X}}^2}{{\mathrm{R}}^2+{\mathrm{X}}^2}\right)+\frac{{\mathrm{P}}^2\left({\mathrm{R}}_2^2+{\mathrm{X}}_2^2\right)}{{{\mathrm{V}}_{\mathrm{R}}}^2}+\frac{{\mathrm{V}}_{\mathrm{R}}^2{\mathrm{X}}^2}{\left({\mathrm{R}}^2+{\mathrm{X}}^2\right)}=0 \\ -{\left({\mathrm{V}}_{\mathrm{S}}\right)}^2+2\mathrm{RP}+\frac{{\mathrm{P}}^2\left({\mathrm{R}}_2^2+{\mathrm{X}}_2^2\right)}{{{\mathrm{V}}_{\mathrm{R}}}^2}+{\left({\mathrm{V}}_{\mathrm{R}}\right)}^2\left[1-\frac{2{\mathrm{X}}^2}{{\mathrm{R}}^2+{\mathrm{X}}^2}+\frac{{\mathrm{X}}^2}{{\mathrm{R}}^2+{\mathrm{X}}^2}\right]=0 \end{gathered}$$

Solve further as,

role="math" localid="1656334947046" $$\begin{gathered} -{\left({\mathrm{V}}_{\mathrm{S}}\right)}^2+2\mathrm{RP}+\frac{{\mathrm{P}}^2\left({\mathrm{R}}_2^2+{\mathrm{X}}_2^2\right)}{{\mathrm{V}}_{\mathrm{R}}^2}+{\left({\mathrm{V}}_{\mathrm{R}}\right)}^2\left[\frac{{\mathrm{R}}^2+{\mathrm{X}}^2-2{\mathrm{X}}^2+{\mathrm{X}}^2}{{\mathrm{R}}^2+{\mathrm{X}}^2}\right]=0 \\ -{\left({\mathrm{V}}_{\mathrm{S}}\right)}^2+2\mathrm{RP}+\frac{{\mathrm{P}}^2\left({\mathrm{R}}_2^2+{\mathrm{X}}_2^2\right)}{{\mathrm{V}}_{\mathrm{R}}^2}+\frac{{\mathrm{V}}_{\mathrm{R}}^2{\mathrm{R}}^2}{{\mathrm{R}}^2+{\mathrm{X}}^2}=0 \\ \frac{{\mathrm{P}}^2\left({\mathrm{R}}_2^2+{\mathrm{X}}_2^2\right)}{{\mathrm{V}}_{\mathrm{R}}^2}+2\mathrm{RP}+\frac{-\left({{\mathrm{V}}_{\mathrm{s}}}^2\right)\left({\mathrm{R}}^2+{\mathrm{X}}^2\right)+{\mathrm{V}}_{\mathrm{R}}^2{\mathrm{R}}^2}{{\mathrm{R}}^2+{\mathrm{X}}^2}=0 \end{gathered}$$

Solve the equation for $\mathrm{P}$.

$\begin{array}{rcl}\mathrm{P}& =& \frac{-2\mathrm{R}\pm \sqrt{4{\mathrm{R}}^2-4\left({\displaystyle \frac{{\mathrm{R}}^2+{\mathrm{X}}^2}{{\mathrm{V}}_{\mathrm{R}}^2}}\right)\left({\displaystyle \frac{-\left({{\mathrm{V}}_{\mathrm{s}}}^2\right)\left({\mathrm{R}}^2+{\mathrm{X}}^2\right)+{{\mathrm{V}}_{\mathrm{R}}}^2{\mathrm{R}}^2}{{\mathrm{R}}^2+{\mathrm{X}}^2}}\right)}}{2\left({\displaystyle \frac{{\mathrm{R}}^2+{\mathrm{X}}^2}{{{\mathrm{V}}_{\mathrm{R}}}^2}}\right)}\\ & =& \frac{-2\mathrm{R}\pm \sqrt{{\displaystyle \frac{4{\mathrm{R}}^2{\mathrm{V}}_{\mathrm{R}}^2\left({\mathrm{R}}^2+{\mathrm{X}}^2\right)+4{\mathrm{V}}_{\mathrm{S}}^2\left({\mathrm{R}}^2+{\mathrm{X}}^2\right)-4{\mathrm{R}}^2{\mathrm{V}}_{\mathrm{R}}^2\left({\mathrm{R}}^2+{\mathrm{X}}^2\right)}{{\mathrm{V}}_{\mathrm{R}}^2\left({\mathrm{R}}^2+{\mathrm{X}}^2\right)}}}}{2\left({\displaystyle \frac{{\mathrm{R}}^2+{\mathrm{X}}^2}{{{\mathrm{V}}_{\mathrm{R}}}^2}}\right)}\\ & =& \frac{-2\mathrm{R}\pm \sqrt{{\displaystyle \frac{4{\mathrm{V}}_{\mathrm{S}}^2\left({\mathrm{R}}^2+{\mathrm{X}}^2\right)}{{\mathrm{V}}_{\mathrm{R}}^2\left({\mathrm{R}}^2+{\mathrm{X}}^2\right)}}}}{2\left({\displaystyle \frac{{\mathrm{R}}^2+{\mathrm{X}}^2}{{{\mathrm{V}}_{\mathrm{R}}}^2}}\right)}\\ & =& \frac{-2{\mathrm{RV}}_{\mathrm{R}}^2\pm {\mathrm{V}}_{\mathrm{R}}\sqrt{4{\mathrm{V}}_{\mathrm{S}}^2\left({\mathrm{R}}^2+{\mathrm{X}}^2\right)}}{2\left({\mathrm{R}}^2+{\mathrm{X}}^2\right)}\end{array}$

Solve further as,

$\begin{array}{rcl}\mathrm{P}& =& \frac{-2{\mathrm{RV}}_{\mathrm{R}}^2\pm {\mathrm{V}}_{\mathrm{R}}\left(2{\mathrm{V}}_{\mathrm{S}}\mathrm{Z}\right)}{2{\mathrm{Z}}^2}\\ & =& \frac{-{\mathrm{RV}}_{\mathrm{R}}^2\pm {\mathrm{V}}_{\mathrm{R}}{\mathrm{V}}_{\mathrm{S}}\mathrm{Z}}{{\mathrm{Z}}^2}\end{array}$

Take positive sign for the maximum power,

role="math" localid="1656336405520" $\begin{array}{rcl}\mathrm{P}& =& \frac{{\mathrm{V}}_{\mathrm{R}}^2}{{\mathrm{Z}}^2}\left(-\mathrm{R}+\frac{{\mathrm{V}}_{\mathrm{S}}\mathrm{Z}}{{\mathrm{V}}_{\mathrm{R}}}\right)\\ \mathrm{P}& =& \frac{{\mathrm{V}}_{\mathrm{R}}^2}{{\mathrm{Z}}^2}\left(\frac{{\mathrm{V}}_{\mathrm{S}}\mathrm{Z}}{{\mathrm{V}}_{\mathrm{R}}}-\mathrm{R}\right)\end{array}$

Hence, the maximum power that can be transmitted over the line is data-custom-editor="chemistry" $\mathrm{P}=\frac{{\mathrm{V}}_{\mathrm{R}}^2}{{\mathrm{Z}}^2}\left(\frac{{\mathrm{V}}_{\mathrm{S}}\mathrm{Z}}{{\mathrm{V}}_{\mathrm{R}}}-\mathrm{R}\right)$.