A $30-\mathrm{km}, 34.5-\mathrm{kV}, 60-\mathrm{Hz}$ three-phase line has a positive-sequence series impedance $z=0.19+j 0.34 \Omega / \mathrm{km}$. The load at the receiving end absorbs 10 MVA at $33 \mathrm{kV}$. Assuming a short line, calculate: (a) the $A B C D$ parameters, (b) the sending-end voltage for a load power factor of 0.9 lagging, (c) the sending-end voltage for a load power factor of 0.9 leading.

Question

Quizplus · Accepted Answer

The Answer is (a)

\bar{A}=\bar{D}=1.0 \angle 0^{\circ} \mathrm{pu} ; \bar{C}=0.0 \mathrm{~S}

\bar{B}=\bar{Ƶ}=(0.19+j 0.34)(30)=11.685 \angle 60.8^{\circ} \Omega

(b)

\bar{V}_{R}=(33 / \sqrt{3}) \angle 0^{\circ}=19.05 \angle 0^{\circ} \mathrm{kV}_{\mathrm{LN}}

\begin{aligned}\bar{I}_{R} & =\frac{S_{R}}{\sqrt{3} V_{R L-L}} \angle-\cos ^{-1}(p f)=\frac{10}{\sqrt{3}(33)} \angle-\cos ^{-1} 0.9 \& =0.1750 \angle-25.84^{\circ} \mathrm{kA} \\bar{V}_{S} & =\bar{A} \bar{V}_{R}+\bar{B} \bar{I}_{R}=1.0(19.05)+\left(11.685 \angle 60.8^{\circ} ight)\left(0.175 \angle-25.84^{\circ} ight) \& =19.05+2.045 \angle 34.96^{\circ}=20.73+j 1.172 \& =20.76 \angle 2.22^{\circ} \mathrm{kV}_{\mathrm{LN}} ; V_{S}=20.76 \sqrt{3}=35.96 \mathrm{kV}_{\mathrm{LL}}\end{aligned}

(c)

\bar{I}_{R}=0.175 \angle 25.84^{\circ} \mathrm{kA}

\begin{aligned}\bar{V}_{S} & =1.0(19.05)+\left(11.685 \angle 60.8^{\circ} ight)\left(0.175 \angle 25.84^{\circ} ight) \& =19.05+2.044 \angle 86.64^{\circ} \& =19.17+j 2.04 \& =19.28 \angle 4.07^{\circ} \mathrm{kV}_{\mathrm{LN}} \V_{S} & =19.28 \sqrt{3}=33.39 \mathrm{kV}_{\mathrm{LL}}\end{aligned}

A 30−km,34.5−kV,60−Hz30-\mathrm{km}, 34.5-\mathrm{kV}, 60-\mathrm{Hz}30−km,34.5−kV,60−Hz Three-Phase Line Has a Positive-Sequence Series Impedance z=0.19+j0.34Ω/kmz=0.19+j 0.34 \Omega / \mathrm{km}z=0.19+j0.34Ω/km

A $30-\mathrm{km}, 34.5-\mathrm{kV}, 60-\mathrm{Hz}$ Three-Phase Line Has a Positive-Sequence Series Impedance $z=0.19+j 0.34 \Omega / \mathrm{km}$