Question

Rework Problem 5.2 in per unit using1000 MVA(three-phase) and 230KV(line-to-line) base values. Calculate: (a) the per-unit ABCD parameters, (b) the per-unit sending-end voltage and current, and (c) the percent voltage regulation.

Short answer

$$\begin{gathered} \left(\mathrm{a}\right)\mathrm{The}A,B,C,D\mathrm{parameters}\mathrm{are}0.968\angle 0.{315}^{°}\mathrm{pu},97.32\angle 80.{54}^{°}\mathrm{\Omega }, \\ 6.553\times {10}^{-4}\angle 90.{155}^{°}\mathrm{and}0.968\angle 0.{315}^{°}\mathrm{pu}. \\ \left(\mathrm{b}\right)\mathrm{The}\mathrm{sending}\mathrm{end}\mathrm{voltage}\mathrm{is}269.2\mathrm{kV}\mathrm{and}\mathrm{current}\mathrm{is}0.6353\angle -0.{34}^{°}\mathrm{kA}. \\ \left(\mathrm{c}\right)\mathrm{The}\mathrm{voltage}\mathrm{regulation}\mathrm{is}26.4\%. \end{gathered}$$

Step-by-step solution

Given data.

$$\begin{gathered} \mathrm{The}\mathrm{load}\mathrm{absorbs}P=250MW\mathrm{at}\mathrm{receiving}\mathrm{end}\mathrm{voltage}{V}_R=220kV. \\ \mathrm{The}\mathrm{line}\mathrm{impedance}\mathrm{is}z=0.08+j0.48\frac{\Omega }{km}. \\ \mathrm{The}\mathrm{shunt}\mathrm{admittance}\mathrm{is}y=j3.33\times {10}^{-6}\frac{S}{km}. \\ \mathrm{The}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{is}l=200km. \\ \mathrm{The}\mathrm{power}\mathrm{factor}\mathrm{is}p.f=0.99. \\ \mathrm{The}\mathrm{base}\mathrm{power}\mathrm{is}{S}_{base}=1000\mathrm{MVA}\mathrm{and}\mathrm{the}\mathrm{base}\mathrm{voltage}\mathrm{is}{V}_{base}=230\mathrm{kV}. \end{gathered}$$

Determine the formulas of ABCD parameters, line current, power angle, voltage regulation and sending end voltage.

$$\begin{gathered} \mathrm{Write}\mathrm{the}\mathrm{formulas}\mathrm{of}\mathrm{ABCD}\mathrm{parameters}\mathrm{for}\mathrm{a}\mathrm{medium}\mathrm{line}\mathrm{length}. \\ \mathit{A}\mathbf{=}\mathbf{\left(}\begin{array}{c}\mathbf{1}\mathbf{+}\frac{\mathbf{YZ}}{\mathbf{2}}\end{array}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{\left(}\begin{array}{c}\mathbf{1}\mathbf{+}\frac{\mathbf{\left(}\mathbf{yl}\mathbf{\right)}\mathbf{\left(}\mathbf{zl}\mathbf{\right)}}{\mathbf{2}}\end{array}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{D}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)} \\ \mathit{C}\mathbf{=}\mathit{Y}\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{YZ}}{\mathbf{4}}\mathbf{)}\mathit{S} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{\left(}\mathbf{yl}\mathbf{\right)}\mathbf{\left(}\begin{array}{c}\mathbf{1}\mathbf{+}\frac{\mathbf{\left(}\mathbf{yl}\mathbf{\right)}\mathbf{\left(}\mathbf{zl}\mathbf{\right)}}{\mathbf{4}}\end{array}\mathbf{\right)}\mathit{S}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)} \\ \mathit{B}\mathbf{=}\mathit{z}\mathit{l}\mathit{\Omega } \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{Z}\mathit{\Omega }\mathbf{}\mathbf{}......................\left(3\right) \\ \mathrm{Here},A,B,C,D\mathrm{are}\mathrm{the}\mathrm{transmission}\mathrm{line}\mathrm{parameters},Z\mathrm{is}\mathrm{line}\mathrm{impedance},Y\mathrm{is}\mathrm{line} \\ \mathrm{admittance}\mathrm{and}l\mathrm{is}\mathrm{line}\mathrm{length}. \\ \mathrm{Write}\mathrm{the}\mathrm{formula}\mathrm{of}\mathrm{sending}\mathrm{end}\mathrm{voltage}\mathrm{in}\mathrm{terms}\mathrm{of}\mathrm{transmission}\mathrm{line}\mathrm{parameters}. \\ {\mathbf{V}}_{\mathbf{S}}\mathbf{=}{\mathbf{AV}}_{\mathbf{pu}}\mathbf{+}{\mathbf{Bl}}_{\mathbf{pu}\mathbf{}}.........................\left(4\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Here},V_S\mathrm{is}\mathrm{the}\mathrm{sending}\mathrm{end}\mathrm{voltage},V_R\mathrm{is}\mathrm{the}\mathrm{receiving}\mathrm{end}\mathrm{voltage},\mathrm{and}{l}_R\mathrm{is}\mathrm{the} \\ \mathrm{transmission}\mathrm{line}\mathrm{current}. \\ \mathrm{Write}\mathrm{the}\mathrm{formula}\mathrm{of}\mathrm{line}\mathrm{current}. \\ {\mathbf{l}}_{\mathbf{R}}\mathbf{=}\frac{\mathbf{P}}{\sqrt{\mathbf{3}}\mathbf{\left(}{\mathbf{V}}_{\mathbf{R}}\mathbf{\right)}\mathbf{(}\mathbf{p}\mathbf{.}\mathbf{f}\mathbf{)}}............\left(5\right) \\ \mathrm{Here},S\mathrm{is}\mathrm{apparent}\mathrm{power}\mathrm{consumed}\mathrm{by}\mathrm{load}. \\ \mathrm{Write}\mathrm{the}\mathrm{formula}\mathrm{of}\mathrm{power}\mathrm{angle}. \\ \mathbf{\theta }\mathbf{=}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{-}\mathbf{1}}\mathbf{(}\mathbf{p}\mathbf{.}\mathbf{f}\mathbf{)}..............\left(6\right) \\ \mathrm{Here},\mathrm{p}.\mathrm{f}\mathrm{is}\mathrm{a}\mathrm{power}\mathrm{factor}. \\ \mathrm{Write}\mathrm{the}\mathrm{formula}\mathrm{of}\mathrm{voltage}\mathrm{regulation}. \\ \mathbf{\%}\mathbf{VR}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{RNL}\mathbf{\left(}\mathbf{pu}\mathbf{\right)}}\mathbf{-}{\mathbf{V}}_{\mathbf{RFL}\mathbf{\left(}\mathbf{pu}\mathbf{\right)}}}{{\mathbf{V}}_{\mathbf{RFL}\mathbf{\left(}\mathbf{pu}\mathbf{\right)}}}\mathbf{\times }\mathbf{100} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{{\displaystyle \frac{{\mathbf{V}}_{\mathbf{S}\mathbf{\left(}\mathbf{pu}\mathbf{\right)}}}{\mathbf{A}}\mathbf{-}{\mathbf{V}}_{\mathbf{RFL}\mathbf{\left(}\mathbf{pu}\mathbf{\right)}}}}{{\mathbf{V}}_{\mathbf{RFL}\mathbf{\left(}\mathbf{pu}\mathbf{\right)}}}\mathbf{\times }\mathbf{100}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{7}\mathbf{\right)} \\ \mathrm{Here},V_{RNL}\mathrm{is}\mathrm{the}\mathrm{no}-\mathrm{load}\mathrm{receiving}\mathrm{end}\mathrm{voltage}\mathrm{and}{V}_{FNL}\mathrm{is}\mathrm{the}\mathrm{full}-\mathrm{load}\mathrm{receiving} \\ \mathrm{end}\mathrm{voltage}. \\ \mathrm{Write}\mathrm{the}\mathrm{formula}\mathrm{of}\mathrm{sending}\mathrm{end}\mathrm{current}\mathrm{in}\mathrm{terms}\mathrm{of}\mathrm{transmission}\mathrm{line}\mathrm{parameters}. \end{gathered}$$

$$\begin{gathered} {\mathbf{l}}_{\mathbf{S}}\mathbf{=}{\mathbf{CV}}_{\mathbf{R}}\mathbf{+}{\mathbf{Dl}}_{\mathbf{R}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{8}\mathbf{\right)} \\ \mathrm{Here},l_S\mathrm{is}\mathrm{the}\mathrm{sending}\mathrm{end}\mathrm{current},V_R\mathrm{is}\mathrm{the}\mathrm{receiving}\mathrm{end}\mathrm{voltage},\mathrm{and}{l}_S\mathrm{is}\mathrm{the}\mathrm{receiving}\mathrm{end}\mathrm{current}. \\ \mathrm{Write}\mathrm{the}\mathrm{formula}\mathrm{of}\mathrm{base}\mathrm{impedance}. \\ {\mathbf{Z}}_{\mathbf{base}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{base}}^{\mathbf{2}}}{{\mathbf{S}}_{\mathbf{base}}}\mathbf{}\dots \dots .\left(9\right) \\ \mathrm{Here},V_{base}\mathrm{is}\mathrm{the}\mathrm{base}\mathrm{voltage},S_{base}\mathrm{is}\mathrm{the}\mathrm{base}\mathrm{power}. \\ \mathrm{Write}\mathrm{the}\mathrm{formula}\mathrm{of}\mathrm{base}\mathrm{admittance}. \\ \mathbf{}{\mathbf{Y}}_{\mathbf{base}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{Z}}_{\mathbf{base}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\dots \dots .\left(10\right) \\ \mathrm{Write}\mathrm{the}\mathrm{formula}\mathrm{of}\mathrm{per}-\mathrm{unit}\mathrm{quantity}. \\ \mathbf{per}\mathbf{-}\mathbf{unit}\mathbf{}\mathbf{value}\mathbf{=}\frac{\mathbf{actual}\mathbf{}\mathbf{value}}{\mathbf{base}\mathbf{}\mathbf{value}}\mathbf{}\dots \dots .\left(11\right) \\ \mathrm{Write}\mathrm{the}\mathrm{formula}\mathrm{of}\mathrm{base}\mathrm{current}. \\ {\mathbf{l}}_{\mathbf{base}}\mathbf{=}\frac{{\mathbf{S}}_{\mathbf{base}}}{\sqrt{\mathbf{3}}\mathbf{\left(}{\mathbf{V}}_{\mathbf{base}}\mathbf{\right)}}\mathbf{}\dots \dots .\left(12\right) \end{gathered}$$

Determine the ABCD parameters.

$$\begin{gathered} \left(a\right) \\ \mathrm{Determine}\mathrm{the}A\mathrm{parameter}. \\ \mathrm{Substitute}0.08+\mathrm{j}0.48\frac{\mathrm{\Omega }}{\mathrm{km}}\mathrm{for}\mathrm{z},\mathrm{j}3.33\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}\mathrm{for}y\mathrm{and}200\mathrm{km}\mathrm{for}l\mathrm{in} \\ \mathrm{equation}\left(1\right) \\ \mathrm{A}=\frac{1+\left(0.08+\mathrm{j}0.48\frac{\mathrm{\Omega }}{\mathrm{km}}\right)\left(200\mathrm{km}\right)\left(\mathrm{j}3.33\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}\right)\left(200\mathrm{km}\right)}{2} \\ =0.968\angle 0.{315}^{°}\mathrm{pu} \\ \mathrm{Determine}\mathrm{the}C\mathrm{parameter}. \\ \mathrm{Substitute}0.08+\mathrm{j}0.48\frac{\mathrm{\Omega }}{\mathrm{km}}\mathrm{for}\mathrm{z},\mathrm{j}3.33\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}\mathrm{for}\mathrm{y}\mathrm{and}200\mathrm{km}\mathrm{for}\mathrm{l}\mathrm{in} \\ \mathrm{equation}\left(2\right) \\ \mathrm{C}=\left(\mathrm{j}3.33\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}\right)\left(200\mathrm{km}\right)\left(1+\frac{\left(0.08+\mathrm{j}0.48\frac{\mathrm{\Omega }}{\mathrm{km}}\right)\left(200\mathrm{km}\right)\left(\mathrm{j}3.33\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}\right)\left(200\mathrm{km}\right)}{4}\right) \\ =6.553\times \angle 90.{155}^{°}\mathrm{S} \end{gathered}$$

$$\begin{gathered} \mathrm{Determine}\mathrm{the}B\mathrm{parameter}. \\ \mathrm{Substitute}0.08+\mathrm{j}0.48\frac{\mathrm{\Omega }}{\mathrm{km}}\mathrm{for}\mathrm{z},\mathrm{and}200\mathrm{km}\mathrm{for}\mathrm{l}\mathrm{in}\mathrm{equation}\left(3\right) \\ \mathrm{B}=\left(0.08+\mathrm{j}0.48\frac{\mathrm{\Omega }}{\mathrm{km}}\right)\left(200\mathrm{km}\right) \\ =\left(16+\mathrm{j}96\right)\mathrm{\Omega } \\ =97.32\angle 80.{54}^{°}\mathrm{\Omega } \\ \mathrm{Determine}\mathrm{the}\mathrm{base}\mathrm{impedance}. \\ \mathrm{Substitute}1000\mathrm{MVA}\mathrm{for}{S}_{Base},\mathrm{and}230\mathrm{kV}\mathrm{for}{V}_{base}\mathrm{in}\mathrm{equation}\left(9\right). \\ {\mathrm{Z}}_{\mathrm{base}}=\frac{{\left(230\mathrm{kV}\right)}^2}{1000\mathrm{MVA}} \\ =52.9\mathrm{\Omega } \\ \mathrm{Determine}\mathrm{the}\mathrm{base}\mathrm{admittance}. \\ \mathrm{Substitute}52.9\mathrm{\Omega }\mathrm{for}{\mathrm{Z}}_{\mathrm{base}}\mathrm{in}\mathrm{equation}\left(10\right). \\ {Y}_{base}=\frac{1}{52.9\Omega } \\ =1.89\times {10}^{-3}\mathrm{S} \end{gathered}$$

$$\begin{gathered} \mathrm{Determine}\mathrm{the}\mathrm{per}-\mathrm{unit}\mathrm{value}\mathrm{o}fB\mathrm{parameter}\mathrm{using}\mathrm{equation}\left(11\right). \\ \mathrm{B}=\frac{97.32\angle 80.54°\mathrm{\Omega }}{52.9\mathrm{\Omega }} \\ =0.184\angle 80.54°\mathrm{pu} \\ \mathrm{Determine}\mathrm{the}\mathrm{per}-\mathrm{unit}\mathrm{value}\mathrm{of}\mathrm{C}\mathrm{parameter}\mathrm{using}\mathrm{equation}\left(11\right). \\ \mathrm{C}=\frac{6.553\times {10}^{-4}\angle 90.155°\mathrm{S}}{1.89\times {10}^{-3}\mathrm{S}} \\ \mathrm{Therefore}\mathrm{the}\mathrm{A},\mathrm{B},\mathrm{C},\mathrm{D}\mathrm{parameters}\mathrm{are}0.968\angle 0.315°\mathrm{pu},0.184\angle 80.54°\mathrm{pu}, \\ 0.3467\angle 90.155°\mathrm{pu}\mathrm{and}0.968\angle 0.315°\mathrm{pu}. \end{gathered}$$

Determine the sending end voltage for lagging power factor and the value of the current.

$$\begin{gathered} \mathrm{Determine}\mathrm{the}\mathrm{per}-\mathrm{unit}\mathrm{current}\mathrm{using}\mathrm{equation}\left(11\right). \\ {I}_{pu}\mathit{=}\frac{0.6627\mathrm{kA}}{2.51\mathrm{kA}} \\ \mathit{}\mathit{}\mathit{}\mathit{}=0.264\mathrm{pu} \\ \mathrm{Determine}\mathrm{the}\mathrm{power}\mathrm{angle}. \\ \mathrm{Substitute}0.99\mathrm{for}\mathrm{p}.\mathrm{f}\mathrm{in}\mathrm{equation}\left(6\right). \\ \mathrm{\theta }={\cos }^{-1}\left(0.99\right) \\ =8.11° \\ \mathrm{Determine}\mathrm{the}\mathrm{per}-\mathrm{unit}\mathrm{voltage}\mathrm{at}\mathrm{receiving}\mathrm{end}\mathrm{using}\mathrm{equation}\left(11\right). \\ {\mathrm{V}}_{\mathrm{pu}}=\frac{220\mathrm{kV}}{230\mathrm{kV}} \\ =0.9565\mathrm{pu} \\ \mathrm{Determine}\mathrm{the}\mathrm{sending}\mathrm{end}\mathrm{voltage}\mathrm{for}\mathrm{lagging}\mathrm{power}\mathrm{factor}. \\ \mathrm{Substitute}0.968\angle 0.315°\mathrm{pu}\mathrm{for}\mathrm{A},0.9565\mathrm{pu}\mathrm{for}{\mathrm{V}}_{\mathrm{PU}},0.184\angle 80.54°\mathrm{pu}\mathrm{for}\mathrm{B} \\ \mathrm{and}0.264\angle -8.11°\mathrm{pu}\mathrm{for}{I}_{pu\mathit{}}\mathrm{in}\mathrm{equation}\left(4\right). \\ {\mathrm{V}}_{\mathrm{s}\left(\mathrm{pu}\right)}=\left(0.968\angle 0.315°\mathrm{pu}\right)\left(0.9565\mathrm{pu}\right)+\left(0.184\angle 80.54°\mathrm{pu}\right)\left(0.264\angle -8.11°\mathrm{pu}\right) \\ =1.17\angle 23.58°\mathrm{pu} \end{gathered}$$(b)

Determine the line current.

Substitute 250MV for P,0.09 for $p.f$and 220 kV for $V_R$in equation (5).

Determine the sending end current for lagging power factor.

localid="1655372069855" $$\begin{gathered} \mathrm{Substitute}0.361\angle 90.155°\mathrm{pu}\mathrm{for}C,0.9565\mathrm{pu}\mathrm{for}{\mathrm{V}}_{\mathrm{PU}},0.968\angle 0.315°\mathrm{pu}\mathrm{for}D \\ \mathrm{and}0.264\angle -8.11°\mathrm{pu}\mathrm{for}{I}_{pu\mathit{}}\mathrm{in}\mathrm{equation}\left(8\right). \\ {\mathrm{I}}_{\mathrm{s}\left(\mathrm{pu}\right)}=\left(0.3467\angle 90.155°\mathrm{pu}\right)\left(0.9565\mathrm{pu}\right)+\left(0.968\angle 0.315°\mathrm{pu}\right)\left(0.264\angle -8.11°\mathrm{pu}\right) \\ =2.531\angle -0.34°\mathrm{pu} \\ \mathrm{Therefore},\mathrm{the}\mathrm{sending}\mathrm{end}\mathrm{voltage}\mathrm{is}1.17\angle 23.58°\mathrm{pu}\mathrm{and}\mathrm{current}\mathrm{is}2.531\angle -0.34°\mathrm{pu}. \end{gathered}$$

Determine the percent voltage regulation.

(c)

Determine the voltage regulation.

Substitute 0.968 for A , 0.9565pu for $V_{RNL}$and 1.17pu for$V_{S\left(pu\right)}$in equation

(7).

$$\begin{gathered} \%VR=\frac{{\displaystyle \frac{1.17pu}{0.968}}-0.9565\mathrm{pu}}{0.9565\mathrm{pu}}\times 100 \\ =26.4\% \end{gathered}$$

Therefore, the voltage regulation is 26.4% .