Question

At full load, the line in Problem 5.14 delivers 900 MW at unity power factor and at 475 kV. Calculate: (a) the sending-end voltage, (b) the sending-end current, (c) the sending-end power factor, (d) the full-load line losses, and (e) the percent voltage regulation.

Short answer

(a)The value of sending-end voltage is$300.2\angle 37.64°\mathrm{kV}$.

(b)The value of sending-end current is$1.061\angle 33.18°\mathrm{kA}$.

(c)The value of sending-end power factor is 0.997 lagging.

(d)The value of full-load line losses is 52.74 MW.

(e)The value of percent voltage regulation is 34.32%.

Step-by-step solution

Given data.

At full load, the line delivers 900 MW at unity power factor.

At full load, the line voltage at 475 kV.

Determine the required formula.

Write the formula of sending-end voltage.

$\stackrel{\mathbf{u}\mathbf{u}}{{\mathbf{V}}_{\mathbf{s}}}\mathbf{=}\mathit{A}\stackrel{\mathbf{u}\mathbf{u}}{{\mathbf{V}}_{\mathbf{R}}}\mathbf{+}\mathit{B}\stackrel{\mathbf{u}\mathbf{u}}{{\mathbf{I}}_{\mathbf{R}}}$

Here,$\stackrel{\mathbf{u}\mathbf{u}}{{\mathbf{V}}_{\mathbf{S}}}\mathbf{=}\mathit{A}\stackrel{\mathbf{u}\mathbf{u}}{{\mathbf{V}}_{\mathbf{R}}}\mathbf{+}\mathit{B}\stackrel{\mathbf{u}\mathbf{u}}{{\mathbf{I}}_{\mathbf{S}}}$localid="1655526979816" $\stackrel{\mathbf{uu}}{{\mathbf{V}}_{\mathbf{s}}}$is the sending-end voltage substitution,localid="1655526983867" $\stackrel{uu}{V_R}$is the receiving-end voltage substitution and $\stackrel{\mathbf{uu}}{I_R}$is the receiving-end current substitution.

Write the formula of sending-end current.

localid="1655526987573" $\stackrel{\mathbf{uu}}{{\mathbf{I}}_{\mathbf{s}}}\mathbf{=}\mathit{C}\stackrel{\mathbf{uu}}{{\mathbf{V}}_R}$

Here,$\stackrel{\mathbf{uu}}{I_{\mathbf{s}}}$is the sending-end current substitutionlocalid="1655526992650" $\stackrel{\mathbf{uu}}{V_R}$,is the receiving-end voltage substitution and localid="1655526997519" $\stackrel{\mathbf{uu}}{I_R}$is the receiving-end current substitution.

Write the formula of sending-end power factor.

localid="1655527001120" $\mathit{p}{\mathit{f}}_{\mathbf{s}}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}{\mathbf{\theta }}_{\mathbf{v}\mathbf{s}}\mathbf{-}{\mathbf{\theta }}_{\mathbf{Is}}\mathbf{)}$

localid="1655527006041" $$\begin{gathered} \mathrm{Here},{\mathbf{pf}}_{\mathbf{S}}\mathbf{}\mathbf{}\mathrm{is}\mathrm{the}\mathrm{sending}-\mathrm{end}\mathrm{power}\mathrm{factor}. \\ \mathrm{Write}\mathrm{the}\mathrm{formula}\mathrm{of}\mathrm{full}\mathrm{load}\mathrm{line}\mathrm{losses}. \\ \mathit{F}\mathit{u}\mathit{l}\mathit{l}\mathit{}\mathit{l}\mathit{i}\mathit{n}\mathit{e}\mathit{}\mathit{l}\mathit{o}\mathit{s}\mathit{s}\mathit{e}\mathit{s}\mathit{=}{\mathit{P}}_{\mathit{S}}\mathit{-}{\mathit{P}}_{\mathit{R}} \\ \mathrm{Here},{\mathbf{P}}_{\mathbf{S}}\mathrm{is}\mathrm{sending}-\mathrm{end}\mathrm{power}\mathrm{and}\mathbf{}{\mathbf{P}}_{\mathbf{R}}\mathrm{is}\mathrm{receiving}-\mathrm{end}\mathrm{power}. \\ \mathrm{Write}\mathrm{the}\mathrm{formula}\mathrm{of}\mathrm{percent}\mathrm{voltage}\mathrm{regulation}. \\ \mathit{V}\mathit{R}\mathbf{\%}\mathbf{=}\mathbf{\left(}\frac{{\mathbf{V}}_{\mathbf{R}\mathbf{N}\mathbf{L}}\mathbf{-}{\mathbf{V}}_{\mathbf{R}\mathbf{F}\mathbf{L}}}{{\mathbf{V}}_{\mathbf{R}\mathbf{F}\mathbf{L}}}\mathbf{\right)}\mathbf{(}\mathbf{100}\mathbf{\%}\mathbf{)} \\ \mathrm{Here},\mathbf{VR}\mathbf{\%}\mathrm{is}\mathrm{the}\mathrm{percent}\mathrm{voltage}\mathrm{regulation} \end{gathered}$$

Determine the sending-end voltage.

(a)

First determine the per phase receiving end voltage.

$$\begin{gathered} \stackrel{ur}{V_{R,L-N}}=\frac{475\times {10}^3}{\sqrt{3}}\angle 0° \\ =274.24\angle 0°\mathrm{kV} \\ \mathrm{Determine}\mathrm{the}\mathrm{line}\mathrm{to}\mathrm{line}\mathrm{receiving}\mathrm{end}\mathrm{current}. \\ \mathrm{Write}\mathrm{the}\mathrm{formula}\mathrm{of}\stackrel{\mathrm{r}}{I_{R\mathit{,}L\mathit{-}L}}. \\ \stackrel{\mathrm{r}}{{\mathrm{I}}_{\mathrm{R}\mathit{,}\mathrm{L}\mathit{-}\mathrm{L}}}=\frac{{\mathrm{P}}_{\mathrm{R}}}{\sqrt{3}\left({\mathrm{V}}_{\mathrm{R},\mathrm{L}-\mathrm{L}}\right)\left(\mathrm{pf}\right)}\angle {\cos }^{-1}\left(\mathrm{pf}\right) \\ \mathrm{Here},p_R\mathrm{is}\mathrm{receiving}-\mathrm{end}\mathrm{power}{\mathrm{V}}_{\mathrm{R},\mathrm{L}-\mathrm{L}}\mathrm{is}\mathrm{line}\mathrm{to}\mathrm{line}\mathrm{receiving}-\mathrm{end}\mathrm{voltage}\mathrm{and}\mathit{}pf\mathit{}\mathrm{is}\mathrm{power}\mathrm{factor}. \\ \mathrm{Substitute}900\times {10}^6\mathrm{for}{\mathrm{P}}_{\mathrm{R}},4.75\times {10}^3\mathrm{for}{\mathrm{V}}_{\mathrm{R},\mathrm{L}-\mathrm{L}},1\mathrm{for}\mathrm{pf}\mathrm{and}0°\mathrm{for}{\cos }^{-1}\left(\mathrm{pf}\right)\mathrm{into}\mathrm{equation}\left(6\right). \\ \stackrel{\mathrm{r}}{{\mathrm{I}}_{\mathrm{R}\mathit{,}\mathrm{L}\mathit{-}\mathrm{L}}}=\frac{900\times {10}^6}{\sqrt{3\left(475\times {10}^3\right)\left(1\right)}}\angle {\cos }^{-1}\left(1\right) \\ =\frac{900\times {10}^6}{822.724\times {10}^3}\angle 0° \\ =1.094\angle 0°\mathrm{kA} \end{gathered}$$

$$\begin{gathered} \mathrm{Determine}\gamma I\mathit{.} \\ \gamma I\mathit{=}\mathit{\left(}\sqrt{\mathit{y}\mathit{z}}\mathit{\right)}I \\ \mathrm{Here},\mathrm{\gamma I}\mathrm{is}\mathrm{dimensionless}\mathrm{quantity}\mathit{,}Z\mathit{}\mathrm{is}\mathrm{series}\mathrm{impedance}\mathit{,}Y\mathrm{is}\mathrm{shunt} \\ \mathrm{admittance}\mathrm{and}I\mathrm{is}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{line}. \\ \mathrm{Substitute}\mathrm{j}4.4\times {10}^{-6}\mathrm{S}/\mathrm{km}\mathrm{for}\mathrm{y},0.03+\mathrm{j}0.35\mathrm{\Omega }/\mathrm{km}\mathrm{for}\mathrm{z}\mathrm{and}500\mathrm{km}\mathrm{for}I \\ \mathrm{into}\mathrm{equation}\left(7\right). \\ \mathrm{\gamma I}=\left(\sqrt{\left(\mathrm{j}4.4\times {10}^{-6}\mathrm{S}/\mathrm{km}\right)\left(0.03+\mathrm{j}0.35\mathrm{\Omega }/\mathrm{km}\right)}\right)\left(500\mathrm{km}\right) \\ =\left(\sqrt{\left(4.4\times {10}^{-6}\angle 90°\mathrm{S}/\mathrm{km}\right)\left(0.35+\angle 85.1°\mathrm{\Omega }/\mathrm{km}\right)}\right)\left(500\mathrm{km}\right) \\ =\sqrt{\left(1.54\times {10}^{-6}\angle 175.1°{\left(1/\mathrm{km}\right)}^2\right)}\left(500\mathrm{km}\right) \\ \mathrm{As}\mathrm{further}\mathrm{solve}, \\ \gamma \mathrm{I}=0.62\angle 87.55° \\ =0.0265+\mathrm{j}0.0619 \\ \mathrm{Now}\mathrm{determine}{\mathrm{e}}^{\mathrm{\gamma }I}. \\ {\mathrm{e}}^{\mathrm{\gamma I}}={\mathrm{e}}^{0.0265+\mathrm{j}0.619} \\ ={\mathrm{e}}^{0.0265}{\mathrm{e}}^{\mathrm{j}0.619} \\ =\left(1.0268\right)(\cos \left(0.619\right)+\mathrm{jsin}\left(0.619\right))\left[\mathrm{angle}\mathrm{in}\mathrm{radius}\right] \\ =\left(1.0268\right)\left(0.8144+\mathrm{j}0.5802\right) \\ \mathrm{As}\mathrm{further}\mathrm{solve}, \\ {\mathrm{e}}^{\mathrm{\gamma I}}=0.836+\mathrm{j}0.596 \end{gathered}$$

$$\begin{gathered} \mathrm{Now}\mathrm{determine}{e}^{\mathit{-}yI} \\ {e}^{\mathit{-}yI}\mathit{=}{e}^{\mathit{-}\mathit{0}\mathit{.}\mathit{0265}\mathit{-}j\mathit{0}\mathit{.}\mathit{619}} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}{e}^{\mathit{-}\mathit{0}\mathit{.}\mathit{0265}}{e}^{\mathit{-}j\mathit{0}\mathit{.}\mathit{619}} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\mathit{(}\mathit{0}\mathit{.}\mathit{9738}\mathit{)}\left[\cos \left(0.619\right)-\mathrm{jsin}\left(0.619\right)\right] \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\mathit{(}\mathit{0}\mathit{.}\mathit{9738}\mathit{)} \end{gathered}$$

$$\begin{gathered} \mathrm{Further}\mathrm{solve}\mathrm{as}, \\ {\mathrm{e}}^{-\mathrm{\gamma }I}=0.793-\mathrm{j}0.565 \\ \mathrm{Now},\mathrm{finally}\mathrm{determine}\mathrm{the}\mathrm{characteristic}\mathrm{impedance}\mathit{}{Z}_c\mathit{.} \\ \mathit{}{Z}_c\mathit{=}\sqrt{\frac{z}{y}} \\ \mathrm{Here},\mathrm{z}\mathrm{is}\mathrm{series}\mathrm{impedance}\mathrm{and}\mathrm{y}\mathrm{is}\mathrm{shunt}\mathrm{admittance}. \\ \mathrm{Substitute}0.03+\mathrm{j}0.35\mathrm{for}\mathrm{z}\mathrm{and}\mathit{}j\times 4.4\times {10}^{-6}\mathrm{for}y\mathit{}\mathrm{into}\mathrm{equation}. \\ {\mathrm{Z}}_{\mathrm{c}}=\sqrt{\frac{0.03+\mathrm{j}0.35}{\mathrm{j}4.4\times {10}^{-6}}} \\ =\sqrt{\frac{0.03+\mathrm{j}0.35}{\mathrm{j}4.4\times {10}^{-6}}} \\ =\sqrt{\frac{0.351\angle 85.1°}{4.4\times {10}^{-6}\angle 90°}} \\ =\sqrt{79.84\times {10}^3\angle -4.9°} \end{gathered}$$

Further solve as,

$Z_C=282.6\angle -2.45°\Omega$

Define the expressions for the ABCD parameters.

$$\begin{gathered} A=D \\ A=\cos h\left(yI\right)\mathrm{pu} \\ B=Z_C\sin \mathrm{h}\left(yI\right)\Omega \\ C=\frac{1}{Z_C}\sin \mathrm{h}\left(yI\right)\mathrm{S} \end{gathered}$$

Calculate the parameters Aand D.

$\stackrel{u}{A=}\stackrel{u}{D}$

Rewrite as,

$\stackrel{\mathrm{u}}{\mathrm{A}=}\cosh \left(yI\right)$

Here, $yI$is dimensionless quantity.

$\frac{e^{yI}+e^{yI}}{2}=\frac{0.836+j0.596+0.793-j0.565}{2}$

$$\begin{gathered} =\frac{1.629+j0.031}{2} \\ =0.8145+j0.0155 \end{gathered}$$

As further solve,

$\stackrel{u}{A=}0.815\angle 1.09°$

Determine the value of$\sin \mathrm{h}\left(yI\right)$

$$\begin{gathered} \sin \mathrm{h}\left(yI\right)=\frac{{\mathrm{e}}^{\mathrm{yI}}-{\mathrm{e}}^{-\mathrm{yI}}}{2} \\ =\frac{0.836+j0.596-\left(0.793-j0.565\right)}{2} \\ =\frac{0.043+j1.161}{2} \\ =0.0215+j0.58 \end{gathered}$$

As further solve,

$\sin \mathrm{h}\left(yI\right)=0.581\angle 87.9°$

Determine the parameter B.

$B=Z_C\sin \mathrm{h}\left(yI\right)$

Here,$Z_C$is characteristics impedance and$yI$is dimensionless quantity.

$$\begin{gathered} B=\left(282.6\angle 85.45°\right)\left(0.581\angle 87.9°\right) \\ =164.2\angle 85.45°\Omega \end{gathered}$$

Determine the parameter C.

$C=\frac{1}{Z_c}\sin \mathrm{h}\left(yI\right)$

Here,$Z_C$is characteristics impedance and $yI$is dimensionless quantity.

Substitute $282.6\angle -2.45°\mathrm{for}{\mathrm{Z}}_{\mathrm{c}}\mathrm{and}0.581\angle 87.9°\mathrm{for}yI$into equation.

$$\begin{gathered} c=\left(\frac{1}{282.6\angle -2.45°}\right)\left(0.581\angle 87.9°\right) \\ =\left(3.54\times {10}^{-3}\angle 2.45°\right)\left(0.581\angle 87.9°\right) \\ =2.056\times {10}^{-3\angle 90.35°}S \end{gathered}$$

Now, finally determine the sending-end voltage by substituting the values of parameters.

$\stackrel{uu}{V_S}=A\stackrel{uu}{V_R}+B\stackrel{uu}{I_R}$

Here,$\stackrel{uu}{V_R}$is parameter, is receiving-end voltage,$B$is parameter and$\stackrel{uu}{I_R}$is receiving-end current.

Substitute $0.815\angle 1.09°\mathrm{for}\mathrm{A},274.24\times {10}^3\angle 0°\mathrm{for}\stackrel{\mathrm{uu}}{{\mathrm{V}}_{\mathrm{R}}},164.2\angle 85.45°\mathrm{for}\mathrm{B}\mathrm{and}1.094\times {10}^3\angle 0°\mathrm{for}\stackrel{uu}{I_r}$into equation.

As further solve,

${\stackrel{uu}{V}}_S=300.2\angle 37.64°kV$

Calculate the magnitude of the line to line sending end voltage.

$V_{S,L-L}=\sqrt{3}{V}_{S,L-N}$

Here, $V_{S,L-N}$is line to neutral sending-end voltage.

$$\begin{gathered} \mathrm{Substitute}300.2\times {10}^3\mathrm{for}{\mathrm{V}}_{\mathrm{S},\mathrm{L}-\mathrm{N}}\mathrm{into}\mathrm{equation}. \\ {\mathrm{V}}_{\mathrm{S},\mathrm{L}-\mathrm{N}}=\sqrt{3}\left(300.2\times {10}^3\right) \\ =\sqrt{3}\left(300.2\times {10}^3\right) \\ =520\mathrm{KV} \end{gathered}$$

Determine the sending-end current.

(b)

Write the formula of sending-end current $\stackrel{uu}{I_S}$.

$$\begin{gathered} \stackrel{uu}{I_S}=C\stackrel{uu}{V_R}+D\stackrel{uu}{I_R} \\ \mathrm{Here},\stackrel{\mathrm{uu}}{{\mathrm{V}}_{\mathrm{R}}}\mathrm{is}\mathrm{the}\mathrm{receiving}-\mathrm{end}\mathrm{voltage}\mathrm{substitution}\mathrm{and}\stackrel{\mathrm{uu}}{{\mathrm{I}}_{\mathrm{R}}}\mathrm{is}\mathrm{the}\mathrm{receiving}-\mathrm{end}\mathrm{current}\mathrm{substitution}. \\ \mathrm{Substitute}2.056\times {10}^{-3}\angle 90.35°\mathrm{for}\mathrm{C},274.2\times {10}^3\angle 0°\mathrm{for}\stackrel{\mathrm{uu}}{{\mathrm{V}}_{\mathrm{R}}},0.815\angle 1.09°\mathrm{for} \\ \mathrm{D}\mathrm{and}1.094\times {10}^3\angle 0°\mathrm{for}\stackrel{\mathrm{uu}}{{\mathrm{I}}_{\mathrm{R}}}\mathrm{into}\mathrm{equation}. \\ \stackrel{\mathrm{uu}}{{\mathrm{I}}_{\mathrm{S}}}=\left(2.056\times {10}^{-3}\angle 90.35°\right)\left(274.2\times {10}^3\angle 0°\right)+\left(0.815\angle 1.09°\right)\left(1.094\times {10}^3\angle 0°\right) \\ =563.75\angle 90.3°+891.61\angle 1.09° \\ =-3.44+\mathrm{j}563.74+891.44+\mathrm{j}16.96 \\ =888.+\mathrm{j}580.7 \\ \mathrm{Further}\mathrm{solve}\mathrm{as}, \\ \stackrel{\mathrm{uu}}{{\mathrm{I}}_{\mathrm{S}}}=1.061\angle 33.18°\mathrm{kA} \end{gathered}$$

Determine the sending-end power factor.

(c)

Write the formula of sending-end power factor ${\mathrm{pf}}_{\mathrm{S}}$.

${\mathrm{pf}}_{\mathrm{S}}=\cos \left({\theta }_{v_s}-{\theta }_{I_s}\right)$

Here, ${\theta }_{v_s}$is phase angle sending-end voltage and ${\theta }_{I_s}$is phase angle sending-end current.

substitute$$\begin{gathered} 37.64°for{\theta }_{V_s}\mathrm{and}33.18°\mathrm{for}{\theta }_{I_s}\mathrm{into}\mathrm{equation}\left(15\right). \\ p{f}_s=\cos \left(37.64°-33.18°\right) \\ =\cos \left(4.46°\right) \\ =0.997\mathrm{lagging} \end{gathered}$$

Determine the full-load line losses

(d)

Calculate the sending-end power.

$$\begin{gathered} p_s=\sqrt{3}{V}_S{I}_S\left(pf\right) \\ \mathrm{Here},{\mathrm{V}}_{\mathrm{S}}\mathrm{is}\mathrm{sending}-\mathrm{end}\mathrm{voltage},I_{\mathrm{S}}\mathrm{is}\mathrm{sending}-\mathrm{end}\mathrm{current}\mathrm{and}\mathrm{pf}\mathrm{is}\mathrm{a}\mathrm{power} \\ \mathrm{factor}. \\ \mathrm{Substitute}520\times {10}^3\mathrm{for}{\mathrm{V}}_{\mathrm{s}},1.061\times {10}^3\mathrm{for}{\mathrm{I}}_{\mathrm{s}}\mathrm{and}0.997\mathrm{for}\mathrm{pf}\mathrm{into}\mathrm{equation}. \\ {\mathrm{p}}_{\mathrm{s}}=\sqrt{3}\left(520\times {10}^3\right)\left(1.061\times {10}^3\right)\left(0.997\right) \\ =952.74\mathrm{MW} \end{gathered}$$

Determine the percentage voltage regulation.

(e)

Calculate the no load receiving end voltage.

$$\begin{gathered} V_{R,NL}=\frac{V_S}{A} \\ \mathrm{Here},{\mathrm{V}}_{\mathrm{S}}\mathrm{is}\mathrm{sending}-\mathrm{end}\mathrm{voltage}\mathrm{and}\mathrm{A}\mathrm{is}\mathrm{parameter}. \\ \mathrm{Substitute}520\times {10}^3\mathrm{for}{\mathrm{V}}_{\mathrm{S}}\mathrm{and}0.815\mathrm{for}\mathrm{A}\mathrm{into}\mathrm{equation}. \\ {\mathrm{V}}_{\mathrm{R},\mathrm{NL}}=\frac{520\times {10}^3}{0.815} \\ =638\mathrm{kV} \\ \mathrm{Now},\mathrm{determine}\mathrm{the}\mathrm{percentage}\mathrm{voltage}\mathrm{regulation}. \\ \mathrm{VR}\%=\left(\frac{{\mathrm{V}}_{\mathrm{R},\mathrm{NL}}-{\mathrm{V}}_{\mathrm{R},\mathrm{FL}}}{{\mathrm{V}}_{\mathrm{R},\mathrm{FL}}}\right)\left(100\%\right) \\ \mathrm{Here},{\mathrm{V}}_{\mathrm{R},\mathrm{NL}}\mathrm{is}\mathrm{no}-\mathrm{load}\mathrm{receiving}-\mathrm{end}\mathrm{voltage}\mathrm{and}{\mathrm{V}}_{\mathrm{R},\mathrm{FL}}\mathrm{is}\mathrm{full}\mathrm{load}\mathrm{receiving}-\mathrm{end}\mathrm{voltage}. \\ \mathrm{Substitute}638\times {10}^3\mathrm{for}{\mathrm{V}}_{\mathrm{R},\mathrm{NL}}\mathrm{and}475\times {10}^3\mathrm{for}{\mathrm{V}}_{\mathrm{R},\mathrm{FL}}\mathrm{into}\mathrm{equation}. \\ \mathrm{VR}\%=\left(\frac{638\times {10}^3-475\times {10}^3}{475\times {10}^3}\right)\left(100\%\right) \\ =\left(0.3432\right)\left(100\%\right) \\ =34.32\% \end{gathered}$$