Question

The $\mathbf{100}\mathbf{}\mathbf{km}\mathbf{,}\mathbf{}\mathbf{230}\mathbf{}\mathbf{kV}\mathbf{,}\mathbf{60}\mathbf{}\mathbf{Hz}\mathbf{,}$three-phase line in Problems 4.18 and 4.39 delivers$\mathbf{300}\mathbf{MVA}$at $\mathbf{218}\mathbf{kV}$to the receiving end at full load. Using the nominal$\mathbf{\pi }$circuit, calculate the ABCDparameters, sending-end voltage, and percent voltage regulation when the receiving-end power factor is (a) $\mathbf{0}\mathbf{.}\mathbf{9}$lagging, (b) unity, and (c)$\mathbf{0}\mathbf{.}\mathbf{9}$leading. Assume a${\mathbf{50}}^{\mathbf{°}}\mathbf{C}$conductor temperature to determine the resistance of this line.

Short answer

Step-by-step solution

Write the given data.

Determine the formulas of ABCD parameters, receiving end current, sending end voltage, and voltage regulation.

Write the formula of ABCD parameters.

$$\begin{gathered} \left(1\right) \\ \mathrm{A}=\mathrm{D} \\ =\left[1+\frac{\mathrm{ZY}}{2}\right] \\ =\left[1+\frac{\left(\mathrm{zl}\right)\left(\mathrm{yl}\right)}{2}\right]..............\left(1\right) \\ \left(2\right) \\ \mathrm{B}=\mathrm{Z} \\ =\left(\mathrm{zl}\right)................\left(2\right) \\ \left(3\right) \\ \mathrm{C}=\mathrm{Y}\left[1+\frac{\mathrm{Zy}}{4}\right] \\ =\left(\mathrm{yl}\right)\left[1+\frac{\left(\mathrm{zl}\right)\left(\mathrm{yl}\right)}{4}\right].................\left(3\right) \end{gathered}$$

Here, zis line impedance per kilometer, y is shunt admittance per kilometer andl is length of line.

Write the formula of receiving end current.

${\mathbf{l}}_{\mathbf{R}}\mathbf{=}\frac{\mathbf{S}}{\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{R}}}\mathbf{}\mathbf{}...........\left(4\right)$

Here, S is load end power and $V_R$is the receiving end voltage.

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

${\mathbf{V}}_{\mathbf{S}}\mathbf{=}{\mathbf{AV}}_{\mathbf{R}}\mathbf{+}{\mathbf{Bl}}_{\mathbf{R}}...............\left(5\right)$

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 voltage regulation.

$$\begin{gathered} \mathbf{\%}\mathbf{VR}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{RNL}}\mathbf{-}{\mathbf{V}}_{\mathbf{RFL}}}{{\mathbf{V}}_{\mathbf{RFL}}}\mathbf{\times }\mathbf{100} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{{\displaystyle \frac{{\mathbf{V}}_{\mathbf{S}}}{\mathbf{A}}}\mathbf{-}{\mathbf{V}}_{\mathbf{RFL}}}{{\mathbf{V}}_{\mathbf{RFL}}}\mathbf{\times }\mathbf{100} \end{gathered}$$ ……. (6)

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

Write the formula of power angle.

$\mathbf{\theta }\mathbf{=}{\mathbf{cos}}^{\mathbf{-}\mathbf{1}}\mathbf{(}\mathbf{p}\mathbf{.}\mathbf{f}\mathbf{)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{7}\mathbf{\right)}$ ……. (7)

Here, p.f is a power factor.

Determine the ABCD parameters, sending end voltage and voltage regulation for   lagging power factor.

$$\begin{gathered} \mathrm{Therefore},\mathrm{the}\mathrm{ABCD}\mathrm{parameters}\mathrm{are}0.9918\angle 0.{0999}^{°}\mathrm{pu},51.1\angle 82.{11}^{°}\mathrm{\Omega } \\ 3.216\times {10}^{-4}\angle 90.{033}^{°}\mathrm{S},\mathrm{and}0.9918\angle 0.{0999}^{°}\mathrm{pu}. \\ \mathrm{Determine}\mathrm{the}\mathrm{receiving}\mathrm{end}\mathrm{current}. \\ \mathrm{Substitute}300\mathrm{MVA}\mathrm{for}\mathrm{S}\mathrm{and}\frac{281}{\sqrt{3}}\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{R}}\mathrm{in}\mathrm{equation}\left(4\right) \\ {\mathrm{l}}_{\mathrm{R}}=\frac{300\mathrm{MVA}}{\sqrt{3\times {\displaystyle \frac{281}{\sqrt{3}}}}\mathrm{KV}} \\ =0.7945\mathrm{kA} \end{gathered}$$(a)

Determine ABCD parameters.

$$\begin{gathered} \mathrm{Substitute}0.0701+\mathrm{j}0.506\frac{\mathrm{\Omega }}{\mathrm{km}}\mathrm{for}\mathrm{z},\mathrm{j}3.229\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}\mathrm{for}\mathrm{y},\mathrm{and}100\mathrm{km}\mathrm{for} \\ \mathrm{l}\mathrm{in}\mathrm{equation}\left(1\right) \\ \mathrm{A}=\mathrm{D} \\ =\left\{1+\frac{\left[\left(0.0701+\mathrm{J}0.506\right)\left(100\right)\right]\left[\left(\mathrm{J}3.229\times {10}^{-6}\right)\left(100\right)\right]}{2}\right\} \\ =0.9918\angle 0.{0999}^{°}\mathrm{pu} \end{gathered}$$

Substitute $00701+j0506\frac{\Omega }{\mathrm{km}}$for $Z$and $100\mathrm{km}$for $l$in equation (2).

$$\begin{gathered} B=(0.0701+j0.506)\left(100\right) \\ =51.1\angle 82.{11}^{°}\Omega \end{gathered}$$

$$\begin{gathered} \mathrm{Substitute}0.0701+\mathrm{j}0.506\frac{\mathrm{\Omega }}{\mathrm{km}}\mathrm{for}\mathrm{z},\mathrm{j}3.229\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}\mathrm{for}\mathrm{y},\mathrm{and}100\mathrm{km}\mathrm{for} \\ \mathrm{l}\mathrm{in}\mathrm{equation}\left(1\right) \\ \mathrm{C}=(\mathrm{j}3.229\times {10}^{-6})\left(100\right)\left\{1+\frac{\left[\left(0.0701+\mathrm{j}0.506\right)\left(100\right)\right]\left[\left(\mathrm{j}3.229\times {10}^{-6}\right)\left(100\right)\right]}{4}\right\} \\ =3.216\times {10}^{-4}\angle 90.{033}^{°}\mathrm{S} \end{gathered}$$

localid="1655351749229" $$\begin{gathered} \mathrm{Therefore},\mathrm{the}\mathrm{ABCD}\mathrm{parameters}\mathrm{are}0.9918\angle 0.{0999}^{°}\mathrm{pu},51.1\angle 82.{11}^{°}\mathrm{\Omega }, \\ 3.216\times {10}^{-4}\angle 90.{033}^{°}\mathrm{S},\mathrm{and}0.9918\angle 0.{0999}^{°}\mathrm{pu}. \end{gathered}$$

Determine the receiving end current.

$$\begin{gathered} \mathrm{Substitute}300\mathrm{MVA}\mathrm{for}\mathrm{S}\mathrm{and}\frac{281}{\sqrt{3}}\mathrm{KV}\mathrm{for}{\mathrm{V}}_{\mathrm{R}}\mathrm{in}\mathrm{equation}\left(4\right) \\ {l}_R=\frac{300\mathrm{MVA}}{\sqrt{3\times {\displaystyle \frac{281}{\sqrt{3}}}}\mathrm{KV}} \\ =0.7945\mathrm{kA} \end{gathered}$$

Determine the power angle.

$$\begin{gathered} \mathrm{Substitute}0.9\mathrm{for}\mathrm{p}.\mathrm{f}\mathrm{in}\mathrm{equation}\left(7\right) \\ \mathrm{\theta }={\cos }^{-1}\left(0.9\right) \\ =25.{84}^{°} \end{gathered}$$

Determine the sending end voltage for lagging power factor.

$$\begin{gathered} \mathrm{Substitute}0.9918\angle 0.{999}^{°}\mathrm{pu}\mathrm{for}\mathrm{A},\frac{281}{\sqrt{3}}\angle 0^{°}\mathrm{KV}\mathrm{for}{\mathrm{V}}_{\mathrm{R}},51.1\angle 82.{11}^{°}\mathrm{\Omega }\mathrm{for} \\ \mathrm{B}\mathrm{and}0.7945\angle -25.{84}^{°}\mathrm{kA}\mathrm{for}{\mathrm{l}}_{\mathrm{R}}\mathrm{in}\mathrm{equation}\left(5\right). \\ {\mathrm{V}}_{\mathrm{s}}=(0.9918\angle 0.{0999}^{°}\mathrm{pu})\left(\frac{281}{\sqrt{3}}\angle 0^{°}\mathrm{KV}\right)+\left(51.1\angle 82.{11}^{°}\mathrm{\Omega }\right)\left(0.7945\angle -25.{84}^{°}\mathrm{kA}\right) \\ =151.3\angle 12.{98}^{°}\mathrm{kV} \end{gathered}$$

The phase voltage is

$$\begin{gathered} {\mathrm{V}}_{\mathrm{S}}=\sqrt{3}\times 151.3\angle 12.{98}^{°}\mathrm{kV} \\ =262\angle 12.{98}^{°}\mathrm{kV} \end{gathered}$$

Therefore, the sending end voltage is $262\angle 12.{98}^{°}k$.

Determine the voltage regulation.

$$\begin{gathered} \mathrm{Substitute}0.9918\mathrm{for}\mathrm{A},218\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{RNL}}\mathrm{and}262\mathrm{KV}\mathrm{for}{\mathrm{V}}_{\mathrm{S}}\mathrm{in}\mathrm{equation}\left(6\right). \\ \%\mathrm{VR}=\frac{{\displaystyle \frac{262\mathrm{kV}}{0.9918}}-218\mathrm{kV}}{218\mathrm{kV}}\times 100 \\ =21.2\% \\ \mathrm{Therefore},\mathrm{the}\mathrm{voltage}\mathrm{regulation}\mathrm{is}21.2\%. \end{gathered}$$

Determine the sending end voltage and voltage regulation for unity power factor.

(b)

Determine the power angle.

$$\begin{gathered} \mathrm{Substitute}1\mathrm{for}\mathrm{p}.\mathrm{f}\mathrm{in}\mathrm{equation}\left(7\right). \\ \mathrm{\theta }={\cos }^{-1}\left(1\right) \\ =0^{°} \end{gathered}$$

Determine the sending end voltage for unity power factor.

$$\begin{gathered} \mathrm{Substitute}0.9918\angle 0.{999}^{°}\mathrm{pu}\mathrm{for}\mathrm{A},\frac{281}{\sqrt{3}}\angle 0^{°}\mathrm{KV}\mathrm{for}{\mathrm{V}}_{\mathrm{R}},51.1\angle 82.{11}^{°}\mathrm{\Omega }\mathrm{for} \\ \mathrm{B}\mathrm{and}0.7945\angle -25.{84}^{°}\mathrm{kA}\mathrm{for}{\mathrm{l}}_{\mathrm{R}}\mathrm{in}\mathrm{equation}\left(5\right). \\ {\mathrm{V}}_{\mathrm{s}}=(0.9918\angle 0.{0999}^{°}\mathrm{pu})\left(\frac{281}{\sqrt{3}}\angle 0^{°}\mathrm{KV}\right)+\left(51.1\angle 82.{11}^{°}\mathrm{\Omega }\right)\left(0.7945\angle -25.{84}^{°}\mathrm{kA}\right) \\ =136.6\angle 17.2^{°}kV \end{gathered}$$

The phase voltage is,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{S}}=\sqrt{3}\times 151.3\angle 12.{98}^{°}\mathrm{kV} \\ =236.6\angle 17.2^{°}\mathrm{kV} \end{gathered}$$

Therefore, the sending end voltage is$236.6\angle 17.2^{°}kV.$

Determine the voltage regulation.

$$\begin{gathered} \mathrm{Substitute}0.9918\mathrm{for}\mathrm{A},218\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{RNL}}\mathrm{and}236.6\mathrm{KV}\mathrm{for}{\mathrm{V}}_{\mathrm{S}}\mathrm{in}\mathrm{equation}\left(6\right). \\ \%\mathrm{VR}=\frac{{\displaystyle \frac{236\mathrm{kV}}{0.9918}-218\mathrm{kV}}}{218\mathrm{kV}}\times 100 \\ =9.43\% \\ \mathrm{Therefore},\mathrm{the}\mathrm{voltage}\mathrm{regulation}\mathrm{is}9.43\%. \end{gathered}$$

Determine the sending end voltage and voltage regulation for   leading power factor.

(c)

Determine the power angle.

Substitute $0.9$for $p.f$in equation (7).

$$\begin{gathered} \theta ={\cos }^{-1}\left(0.9\right) \\ =25.{84}^{°} \end{gathered}$$

Determine the sending end voltage for unity power factor.

$$\begin{gathered} \mathrm{Substitute}0.9918\angle 0.{999}^{°}\mathrm{pu}\mathrm{for}\mathrm{A},\frac{281}{\sqrt{3}}\angle 0^{°}\mathrm{KV}\mathrm{for}{\mathrm{V}}_{\mathrm{R}},51.1\angle 82.{11}^{°}\mathrm{\Omega }\mathrm{for} \\ \mathrm{B}\mathrm{and}0.7945\angle -25.{84}^{°}\mathrm{kA}\mathrm{for}{\mathrm{l}}_{\mathrm{R}}\mathrm{in}\mathrm{equation}\left(5\right). \\ {\mathrm{V}}_{\mathrm{s}}=(0.9918\angle 0.{0999}^{°}\mathrm{pu})\left(\frac{281}{\sqrt{3}}\angle 0^{°}\mathrm{KV}\right)+\left(51.1\angle 82.{11}^{°}\mathrm{\Omega }\right)\left(0.7945\angle -25.{84}^{°}\mathrm{kA}\right) \\ =118.1\angle 19.1^{°}kV \end{gathered}$$

The phase voltage is,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{S}}=\sqrt{3}\times 118.1\angle 19.1^{°}\mathrm{kV} \\ =205.9\angle 19.1^{°}\mathrm{kV} \end{gathered}$$

Therefore, the sending end voltage is$=205.9\angle 19.1^{°}\mathrm{kV}$.

Determine the voltage regulation.

$$\begin{gathered} \mathrm{Substitute}0.9918\mathrm{for}\mathrm{A},218\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{RNL}}\mathrm{and}205.9\mathrm{KV}\mathrm{for}{\mathrm{V}}_{\mathrm{S}}\mathrm{in}\mathrm{equation}\left(6\right). \\ \%\mathrm{VR}=\frac{{\displaystyle \frac{205.9\mathrm{kV}}{0.9918}-218\mathrm{kV}}}{218\mathrm{kV}}\times 100 \\ =-5.6\% \\ \mathrm{Therefore},\mathrm{the}\mathrm{voltage}\mathrm{regulation}\mathrm{is}-5.6\%. \end{gathered}$$