Question

A200km , 230 kV , 60 Hzthree-phase line has a positive-sequence series impedance$\mathbf{Z}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{08}\mathbf{+}\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{48}\frac{\mathbf{\Omega }}{\mathbf{km}}$and a positive-sequence shunt admittancelocalid="1655353285431" $\mathbf{y}\mathbf{=}\mathbf{j}\mathbf{3}\mathbf{.}\mathbf{33}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\frac{\mathbf{S}}{\mathbf{km}}$. At full load, the line delivers 250MWat 0.99p.f. lagging and at 220 kV. Using the nominallocalid="1655353289576" $\mathrm{\pi }$circuit, calculate: (a) the ABCDparameters, (b) the sending-end voltage and current, and (c) the percent voltage regulation.

Short answer

$$\begin{gathered} \left(\mathrm{a}\right)\mathrm{The}A,B,C,D\mathrm{parameters}\mathrm{are}0.968\angle 0.{315}^{°}\mathrm{pu},97.32\angle 80.{54}^{°}\mathrm{\Omega }, \\ 6.553\times {10}^{-4}\angle 90.{155}^{°}\mathrm{and}0.968\angle 0.{315}^{°}\mathrm{pu}. \\ \left(\mathrm{b}\right)\mathrm{The}\mathrm{sending}\mathrm{end}\mathrm{voltage}\mathrm{is}269.2\mathrm{kV}\mathrm{and}\mathrm{current}\mathrm{is}0.6353\angle -0.{34}^{°}\mathrm{kA}. \\ \left(\mathrm{c}\right)\mathrm{The}\mathrm{voltage}\mathrm{regulation}\mathrm{is}26.4\%. \end{gathered}$$

Step-by-step solution

Given data.

The load absorbs P = 250 MW at receiving end voltage$V_R=220kV$.

The line impedance is$z=0.08+j0.48\frac{\Omega }{km}.$

The shunt admittance is$\mathrm{Y}=\mathrm{j}3.33\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}$.

The length of the line is l= 200km .

The power factor is p .f = 0.99.

Determine the formulas of ABCD parameters, line current, power angle, voltage regulation and sending end voltage.

Write the formulas of ABCD parameters for a medium line length.

$$\begin{gathered} \mathit{A}\mathbf{=}\left(\begin{array}{c}1+\frac{YZ}{2}\end{array}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\left(\begin{array}{c}1+\frac{\left(yl\right)\left(zl\right)}{2}\end{array}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{D}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)} \\ \mathit{C}\mathbf{=}\mathit{Y}\left(1+\frac{YZ}{4}\right)\mathit{S} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\left(yl\right)\left(\begin{array}{c}1+\frac{\left(yl\right)\left(zl\right)}{4}\end{array}\right)\mathit{S}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)} \\ \mathit{B}\mathbf{=}\mathit{z}\mathit{l}\mathit{\Omega } \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{Z}\mathit{\Omega }\mathbf{}\mathbf{}......................\left(3\right) \end{gathered}$$

Here, A,B,C,Dare the transmission line parameters,Z is line impedance, Y is line admittance andl is line length.

Write the formula of sending end voltage in terms of transmission line parameters.

localid="1655356715369" ${\mathbf{V}}_{\mathbf{S}}\mathbf{=}{\mathbf{AV}}_{\mathbf{R}}\mathbf{+}{\mathbf{Bl}}_{\mathbf{R}}...........\left(4\right)$

Here, $V_S$is the sending end voltage, $V_R$is the receiving end voltage, and $l_R$ is the transmission line current.

Write the formula of line current.

localid="1655356720025" ${\mathbf{l}}_{\mathbf{R}}\mathbf{=}\frac{\mathbf{P}}{\sqrt{\mathbf{3}}\mathbf{\left(}{\mathbf{V}}_{\mathbf{R}}\mathbf{\right)}\mathbf{(}\mathbf{p}\mathbf{.}\mathbf{f}\mathbf{)}}\mathbf{}...............\left(5\right)$

Here, is apparent power consumed by load.

Write the formula of power angle.

$\mathbf{\theta }\mathbf{=}{\mathbf{cos}}^{\mathbf{-}\mathbf{1}}\mathbf{(}\mathbf{p}\mathbf{.}\mathbf{f}\mathbf{)}............\left(6\right)$

Here, p.f is a power factor.

Write the formula of voltage regulation.

localid="1655354135906" $$\begin{gathered} \mathbf{\%}\mathbf{VR}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{RNL}}\mathbf{-}{\mathbf{V}}_{\mathbf{RFL}}}{{\mathbf{V}}_{\mathbf{RFL}}}\mathbf{\times }\mathbf{100} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{{\displaystyle \frac{{\mathbf{V}}_{\mathbf{S}}}{\mathbf{A}}}\mathbf{-}{\mathbf{V}}_{\mathbf{RFL}}}{{\mathbf{V}}_{\mathbf{RFL}}}\mathbf{\times }\mathbf{100}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{7}\mathbf{\right)} \end{gathered}$$

Here, $V_{RNL}$is the no-load receiving end voltage and $V_{FNL}$is the full-load receiving end voltage.

Write the formula of sending end current in terms of transmission line parameters.

${\mathbf{l}}_{\mathbf{S}}\mathbf{=}{\mathbf{CV}}_{\mathbf{R}}\mathbf{+}{\mathbf{Dl}}_{\mathbf{R}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{8}\mathbf{\right)}$

Here, $l_S$is the sending end current, $V_R$is the receiving end voltage, and $l_R$is the receiving end current.

Determine the ABCD parameters.

(a)

Determine theAparameter.

$$\begin{gathered} \mathrm{Substitute}0.08+\mathrm{j}0.48\frac{\mathrm{\Omega }}{\mathrm{km}}\mathrm{for}\mathrm{z},\mathrm{j}3.33\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}\mathrm{for}y\mathrm{and}200\mathrm{km}\mathrm{for}\mathrm{l}\mathrm{in} \\ \mathrm{equation}\left(1\right) \\ \mathrm{A}=\frac{1+\left(0.08+\mathrm{j}0.48\frac{\mathrm{\Omega }}{\mathrm{km}}\right)\left(200\mathrm{km}\right)\left(\mathrm{j}3.33\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}\right)\left(200\mathrm{km}\right)}{2} \\ =0.968\angle 0.{315}^{°}\mathrm{pu} \end{gathered}$$

Determine the C parameter.

$$\begin{gathered} \mathrm{Substitute}0.08+\mathrm{j}0.48\frac{\mathrm{\Omega }}{\mathrm{km}}\mathrm{for}\mathrm{z},\mathrm{j}3.33\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}\mathrm{for}\mathrm{y}\mathrm{and}200\mathrm{km}\mathrm{for}\mathrm{l}\mathrm{in} \\ \mathrm{equation}\left(2\right) \\ \mathrm{C}=\left(\mathrm{j}3.33\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}\right)\left(200\mathrm{km}\right)\left(1+\frac{\left(0.08+\mathrm{j}0.48\frac{\mathrm{\Omega }}{\mathrm{km}}\right)\left(200\mathrm{km}\right)\left(\mathrm{j}3.33\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}\right)\left(200\mathrm{km}\right)}{4}\right) \\ =6.553\times \angle 90.{155}^{°}\mathrm{S} \end{gathered}$$

Determine the B parameter.

$$\begin{gathered} \mathrm{Substitute}0.08+\mathrm{j}0.48\frac{\mathrm{\Omega }}{\mathrm{km}}\mathrm{for}\mathrm{z},\mathrm{and}200\mathrm{km}\mathrm{for}\mathrm{l}\mathrm{in}\mathrm{equation}\left(3\right) \\ \mathrm{B}=\left(0.08+\mathrm{j}0.48\frac{\mathrm{\Omega }}{\mathrm{km}}\right)\left(200\mathrm{km}\right) \\ =\left(16+\mathrm{j}96\right)\mathrm{\Omega } \\ =97.32\angle 80.{54}^{°}\mathrm{\Omega } \\ \mathrm{Therefore},\mathrm{the}A,B,C,D\mathrm{parameters}\mathrm{are}0.968\angle 0.{315}^{°}\mathrm{pu},97.32\angle 80.{54}^{°}\mathrm{\Omega } \\ 6.553\times {10}^{-4\mathrm{pu}}\angle 90.{155}^{°}\mathrm{S}\mathrm{and}0.968\angle 0.{315}^{°} \end{gathered}$$

Determine the sending end voltage for lagging power factor and the current.

(b)

Determine the line current.

$$\begin{gathered} \mathrm{Substitute}250\mathrm{MW}\mathrm{for}\mathrm{P},0.99\mathrm{for}p.f\mathrm{and}220\mathrm{k}\mathrm{for}{\mathrm{V}}_{\mathrm{R}}\mathrm{in}\mathrm{equation}\left(5\right). \\ {\mathrm{l}}_{\mathrm{R}}=\frac{250\mathrm{W}}{\sqrt{3}\left(220\mathrm{kV}\right)\left(0.99\right)} \\ =0.6627\mathrm{kA} \\ \mathrm{Determine}\mathrm{the}\mathrm{power}\mathrm{angle}. \\ \mathrm{Substitute}0.99\mathrm{for}p.f\mathrm{in}\mathrm{equation}\left(6\right). \\ \mathrm{\theta }={\cos }^{-1}\left(0.99\right) \\ =8.{11}^{°} \\ \mathrm{Determine}\mathrm{the}\mathrm{sending}\mathrm{end}\mathrm{voltage}\mathrm{for}\mathrm{lagging}\mathrm{power}\mathrm{factor}. \\ \mathrm{Substitute}0.968\angle 0.{315}^{°}\mathrm{pu}\mathrm{for}\mathrm{A},\frac{220}{\sqrt{3}}\angle 0^{°}\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{R}},97.32\angle 80.{54}^{°}\mathrm{for} \\ B\mathrm{and}0.6627\angle -8.{11}^{°}\mathrm{kA}\mathrm{for}{l}_R\mathrm{in}\mathrm{equation}\left(4\right). \\ {V}_S=\left(0.968\angle 0.{315}^{°}\mathrm{pu}\right)\left(\begin{array}{c}\frac{220}{\sqrt{3}}\angle 0^{°}\mathrm{kV}\end{array}\right)+\left(97.32\angle 80.{54}^{°}\mathrm{\Omega }\right)\left(0.6627\angle -8.{11}^{°}\mathrm{kA}\right) \\ =155.4\angle 23.{58}^{°}\mathrm{kV} \end{gathered}$$

$$\begin{gathered} \mathrm{The}\mathrm{sending}\mathrm{end}\mathrm{voltage}\mathrm{is}, \\ {V}_S=\sqrt{3}\times 155.4\angle 23.{58}^{°}\mathrm{kV} \\ =269.2\angle 23.{58}^{°}\mathrm{kV} \\ \mathrm{Determine}\mathrm{the}\mathrm{sending}\mathrm{end}\mathrm{current}\mathrm{for}\mathrm{lagging}\mathrm{power}\mathrm{factor}. \\ \mathrm{Substitute}6.553\times {10}^{-4}\angle 90.{155}^{°}\mathrm{S}\mathrm{for}C,\frac{220}{\sqrt{3}}\angle 0^{°}\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{R}}, \\ 0.968\angle 0.{315}^{°}\mathrm{pu}\mathrm{for}D\mathrm{and}0.6627\angle -8.{11}^{°}\mathrm{kA}\mathrm{for}{l}_R\mathrm{in}\mathrm{equation}\left(8\right). \\ {l}_s=(6.553\times {10}^{-4}\angle 90.{155}^{°}\mathrm{S})\left(\frac{220}{\sqrt{3}}\angle 0^{°}\mathrm{kV}\right)+\left(0.968\angle 0.{315}^{°}\mathrm{pu}\right)(0.6627\angle -8.{11}^{°}\mathrm{kA}) \\ \mathrm{Therefore},\mathrm{the}\mathrm{sending}\mathrm{end}\mathrm{voltage}\mathrm{is}269.2\mathrm{kV}\mathrm{and}\mathrm{current}\mathrm{is}0.6353\angle -0.{34}^{°}\mathrm{kA}. \end{gathered}$$

Determine the percent voltage regulation.

$$\begin{gathered} \left(\mathrm{c}\right) \\ \mathrm{Determine}\mathrm{the}\mathrm{voltage}\mathrm{regulation}. \\ \mathrm{Substitute}0.968\mathrm{for}A,220\mathrm{Kv}\mathrm{for}{V}_{RNL}\mathrm{and}269.2\mathrm{kV}\mathrm{for}{V}_S\mathrm{in}\mathrm{equation}\left(7\right) \\ \%VR=\frac{{\displaystyle \frac{269.2\mathrm{KV}}{0.968}}-220\mathrm{kV}}{220\mathrm{kV}}\times 100 \\ =26.4\% \\ \mathrm{Therefore},\mathrm{the}\mathrm{voltage}\mathrm{regulation}\mathrm{is}26.4\%. \end{gathered}$$