Rework Test Bank Problem 5.2 in per-unit using 100-MVA (three-phase) and 230-kV(line-to-line) base values. Calculate (a) the per-unitABCDparameters, (b) the per-unitsending-end voltage and current, and (c) the percent voltage regulation.

Question

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The Answer is

\begin{array}{l}Z_{ ext {base }}=\frac{V_{ ext {base }}^{2}}{S_{ ext {base }}}=\frac{(230)^{2}}{100}=529 \Omega \Y_{ ext {base }}=1 / Ƶ_{ ext {base }}=1.890 imes 10^{-3} \end{array}

(a)

\begin{array}{l}\bar{A}_{p u}=\bar{D}_{p u}=0.9820 \angle 0.175^{\circ} ext { per unit } \\bar{B}_{p u}=\frac{\bar{B}}{Z_{ ext {base }}}=\frac{72.99}{529} \angle 80.54^{\circ}=0.1380 \angle 80.54^{\circ} ext { per unit } \\bar{C}_{p u}=\frac{\bar{C}}{Y_{ ext {base }}}=\frac{4.950 imes 10^{-4}}{1.890 imes 10^{-3}} \angle 90.09^{\circ}=0.2619 \angle 90.09^{\circ} ext { per unit }\end{array}

$\begin{array}{l} Z_{ ext {base }}=\frac{V_{ ext {base }}^{2}}{S_{ ext {base }}}=\frac{(230)^{2}}{100}=529 \Omega \ Y_{ ext {base }}=1 / Ƶ_{ ext {base }}=1.890 imes 10^{-3} \end{array} (a) \begin{array}{l} \bar{A}_{p u}=\bar{D}_{p u}=0.9820 \angle 0.175^{\circ} ext { per unit } \ \bar{B}_{p u}=\frac{\bar{B}}{Z_{ ext {base }}}=\frac{72.99}{529} \angle 80.54^{\circ}=0.1380 \angle 80.54^{\circ} ext { per unit } \ \bar{C}_{p u}=\frac{\bar{C}}{Y_{ ext {base }}}=\frac{4.950 imes 10^{-4}}{1.890 imes 10^{-3}} \angle 90.09^{\circ}=0.2619 \angle 90.09^{\circ} ext { per unit } \end{array} \begin{array}{l} \bar{V}_{S P U}= \bar{A}_{P U} \bar{V}_{R P U}+\bar{B}_{P U} \bar{I}_{R P U}=\left(0.982 \angle 0.175^{\circ} ight)\left(0.9565 \angle 0^{\circ} ight)+ \ \quad\left(0.1380 \angle 80.54^{\circ} ight)\left(2.64^{\circ} \angle-8.11^{\circ} ight) \ = 0.9393 \angle 0.175^{\circ}+0.3643 \angle 72.43^{\circ}=1.049+j 0.3501 \ \bar{V}_{S P U}= \underline{\overline{1.106} } \angle 18.45^{\circ} ext { per unit }\ \bar{I}_{S P U}=\bar{C}_{P U} \bar{V}_{R P U}+\bar{D}_{P U} \bar{I}_{R P U}=\left(0.2619 \angle 90.09^{\circ} ight)(0.9565)+ \ \quad\left(0.982 \angle 0.175^{\circ} ight)\left(2.64 \angle-8.11^{\circ} ight) \ \bar{I}_{S P U}= \underline{\underline{2.569}} \angle-2.398^{\circ} ext { per unit } \end{array} (c) \begin{array}{l} V_{R N L_{p u}}=V_{S p u} / A_{p u}=1.106 / 0.982=1.1263 ext { per unit } \ \% V . R .=\frac{V_{R N L_{p u}}-V_{R F L p u}}{V_{R F L p u}} imes 100=\frac{1.126-.9565}{.9565} imes 100=\underline{\underline{17.8}} \% \end{array}$

\begin{array}{l} \bar{V}_{S P U}= \bar{A}_{P U} \bar{V}_{R P U}+\bar{B}_{P U} \bar{I}_{R P U}=\left(0.982 \angle 0.175^{\circ} ight)\left(0.9565 \angle 0^{\circ} ight)+ \\quad\left(0.1380 \angle 80.54^{\circ} ight)\left(2.64^{\circ} \angle-8.11^{\circ} ight) \= 0.9393 \angle 0.175^{\circ}+0.3643 \angle 72.43^{\circ}=1.049+j 0.3501 \\bar{V}_{S P U}= \underline{\overline{1.106} } \angle 18.45^{\circ} ext { per unit }\\bar{I}_{S P U}=\bar{C}_{P U} \bar{V}_{R P U}+\bar{D}_{P U} \bar{I}_{R P U}=\left(0.2619 \angle 90.09^{\circ} ight)(0.9565)+ \\quad\left(0.982 \angle 0.175^{\circ} ight)\left(2.64 \angle-8.11^{\circ} ight) \\bar{I}_{S P U}= \underline{\underline{2.569}} \angle-2.398^{\circ} ext { per unit }\end{array}

(c)

\begin{array}{l}V_{R N L_{p u}}=V_{S p u} / A_{p u}=1.106 / 0.982=1.1263 ext { per unit } \\% V . R .=\frac{V_{R N L_{p u}}-V_{R F L p u}}{V_{R F L p u}} imes 100=\frac{1.126-.9565}{.9565} imes 100=\underline{\underline{17.8}} \%\end{array}