A $150-\mathrm{km}, 230-\mathrm{kV}, 60-\mathrm{Hz}$ three-phase line has a positive-sequence series impedance $z=0.08+j 0.48 \Omega / \mathrm{km}$ and a positive-sequence shunt admittance $y=j 3.33 \times 10^{-6} \mathrm{~S} /$ $\mathrm{km}$. At full load, the line delivers $250 \mathrm{MW}$ at 0.99 p.f. lagging and at $220 \mathrm{kV}$. Using the nominal $\pi$ circuit, calculate: (a) the $A B C D$ parameters, (b) the sending-end voltage and current, and (c) the percent voltage regulation.

Question

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The Answer is

\begin{array}{l} ext { (a) } \begin{array}{l}\bar{A}=\bar{D}=1+\frac{\bar{Y} \bar{Ƶ}}{2}=1+\frac{1}{2}\left(3.33 imes 10^{-6} imes 150 \angle 90^{\circ} ight)(.08+j .48)(150) \\begin{aligned}\bar{A}=\bar{D} & =1+\frac{1}{2}\left(4.995 imes 10^{-4} \angle 90^{\circ} ight)\left(72.99 \angle 80.54^{\circ} ight) \& =1+0.01823 \angle 170.54^{\circ}=0.9820+j 0.002997 \& =0.9820 \angle 0.175^{\circ} ext { per unit }\end{aligned} \\begin{aligned}\bar{B}= & \bar{Ƶ}=72.99 \angle 80.54^{\circ} \Omega \\bar{C}= & \bar{Y}\left(1+\frac{\bar{Y} \bar{Ƶ}}{4} ight)=4.995 imes 10^{-4} \angle 90^{\circ}\left(1+.009115 \angle 170.54^{\circ} ight)\end{aligned} \\begin{aligned}\bar{C} & =\left(4.995 imes 10^{-4} \angle 90^{\circ} ight)(0.991+j 0.00150) \& =\left(4.995 imes 10^{-4} \angle 90^{\circ} ight)\left(0.991 \angle 0.0867^{\circ} ight)=4.950 imes 10^{-4} \angle 90.09^{\circ}\end{aligned}\end{array}\end{array}

(b)

\bar{V}_{R}=\frac{220}{\sqrt{3}} \angle 0^{\circ}=127.02 \angle 0^{\circ} \mathrm{kV} \mathrm{V}_{\mathrm{LN}}

\begin{array}{l}\begin{aligned}\bar{I}_{R}= & \frac{P_{R} \angle-\cos ^{-1}(P . F .)}{\sqrt{3} V_{R L L}(P . F .)}=\frac{250 \angle-\cos ^{-1} .99}{\sqrt{3}(220)(0.99)}=0.6627 \angle-8.11^{\circ} \\bar{V}_{S}= & \bar{A} \bar{V}_{R}+\bar{B} \bar{I}_{R}=\left(0.9820 \angle 0.175^{\circ} ight)\left(127.02 \angle 0^{\circ} ight)+\left(72.99 \angle 80.54^{\circ} ight) \& \quad imes\left(.6627 \angle-8.11^{\circ} ight) \\bar{V}_{S}= & 124.73 \angle 0.175^{\circ}+48.37 \angle 72.43^{\circ} \\bar{V}_{S}= & 139.33+j 46.49=146.9 \angle 18.45^{\circ} \mathrm{kV} \mathrm{LN} \V_{S}= & 146.9 \sqrt{3}= \underline{ 254.4 } \mathrm{kV} \mathrm{LI} \\bar{I}_{S}= & \bar{C} \bar{V}_{R}+\overline{D I}_{R}=\left(4.95 imes 10^{-4} \angle 90.09^{\circ} ight)(127.02)+ \& \quad\left(.9820 \angle 175^{\circ} ight)\left(.6627 \angle-8.6^{\circ} ight) \= & 0.06287 \angle 90.09^{\circ}+0.6508 \angle-7.935^{\circ} \\bar{I}_{S}= & 0.6445-j 0.02697= \underline{\underline{ 0.6750}} \angle-2.396^{\circ} \mathrm{kA}\end{aligned}\end{array}

(c)

\begin{aligned}V_{R V Z} & =\frac{V_{S}}{A}=\frac{254.4}{0.9820}=259.1 \mathrm{kV} \mathrm{~V}_{\mathrm{LL}} \\% V_{r} R_{r} & =\frac{V_{R V Z}-V_{R F L}}{V_{R F L}} imes 100=\frac{259.1-220}{220} imes 100=\underline{\underline{17.8}} \%\end{aligned}

A 150−km,230−kV,60−Hz150-\mathrm{km}, 230-\mathrm{kV}, 60-\mathrm{Hz}150−km,230−kV,60−Hz Three-Phase Line Has a Positive-Sequence Series Impedance z=0.08+j0.48Ω/kmz=0.08+j 0.48 \Omega / \mathrm{km}z=0.08+j0.48Ω/km

A $150-\mathrm{km}, 230-\mathrm{kV}, 60-\mathrm{Hz}$ Three-Phase Line Has a Positive-Sequence Series Impedance $z=0.08+j 0.48 \Omega / \mathrm{km}$