A$320-\mathrm{km} 500-\mathrm{kV}, 60-\mathrm{Hz}$ Three-Phase Uncompensated Line Has a Positive-Sequence Series Reactance

Question

Examlex · Accepted Answer

The Answer is (a)

\bar{Ƶ}_{C}=\sqrt{\frac{\bar{z}}{\bar{y}}}=\sqrt{\frac{j 0.34}{j 4.5 imes 10^{-6}}}=274.9 \Omega

(b)

\bar{\gamma} l=\sqrt{\bar{z} \bar{y}} l=\sqrt{(j 0.34)\left(j 4.5 imes 10^{-6} ight)}(320)=j 0.3958 \mathrm{pu}

(c)

\bar{\gamma} l=j \beta l ; \quad \beta l=0.3958 \mathrm{pu}

\bar{A}=\bar{D}=\cos \beta l=\cos (0.3958

radians

)=0.9227 \angle 0^{\circ} \mathrm{pu}

\bar{B}=j \bar{Ƶ}_{C} \sin \beta l=j(274.9) \sin (0.3958

radians

)

=j 108.81 \Omega

\bar{C}=j\left(\frac{1}{Ƶ_{C}} ight) \sin \beta l=j \frac{1}{274.9} \sin (0.3958

radians

)

=j 1.44 imes 10^{-3} \mathrm{~S}

(d)

\beta=\beta l / l=0.3958 / 300=1.319 imes 10^{-3}

radians

/ \mathrm{km}

\lambda=2 \pi / \beta=4766 \mathrm{~km}

(e)

\mathrm{SIL}=\frac{V_{ ext {rated } L-L}^{2}}{Ƶ_{C}}=\frac{(500)^{2}}{274.9}=909.4 \mathrm{MW}(3 \phi)

A $320-\mathrm{km} 500-\mathrm{kV}, 60-\mathrm{Hz}$ Three-Phase Uncompensated Line Has a Positive-Sequence Series Reactance