Question

Rework Problems 5.9 and 5.16, neglecting the conductor resistance. Compare the results with and without losses.

Short answer

(a) 5.9 as part (A).

Part (A)

(a) For 0.9 lagging power factor the sending-end voltage and percentage voltage regulation are $254.4\angle 14.26°{\mathrm{kv}}_{\mathrm{LL}}\mathrm{and}17.7\%$

(b)For unity power factor the sending-end voltage and percentage voltage regulation are $227.2\angle 17.85°{\mathrm{kV}}_{\mathrm{LL}}\mathrm{and}5.08\%$

(c) For 0.9 leading power factor the sending-end voltage and percentage voltage regulation are $196.2\angle 18.63°{\mathrm{kV}}_{\mathrm{LL}}\mathrm{and}-9.22\%$

(b) 5.16 as part (B).

Part (B)

(a)The value of characteristics impedance is$264\Omega$.

(b)The value of localid="1655468974893" $\gamma I\mathrm{is}j0.38\mathrm{per}\mathrm{unit}$.

(c)The transmission line parameters are localid="1655468979671" $A=0.9285$per unit, localid="1655468985523" $j98.252\mathrm{\Omega }$, localid="1655468990717" $C=j1.408\times {10}^{-3}\mathrm{S}$and localid="1655468996929" $\mathrm{D}=0.9285$per unit.

Step-by-step solution

Given data.

As using reference problem 5.9.

The given load at receiving-end of line is $230-\mathrm{kV}$.

The given load delivers 300 MVA at 218 kV to the receiving-end at full load.

The given frequency of line is $60-\mathrm{Hz}$.

The given length of the line is $100-\mathrm{km}$.

The given conductor temperature is$50°\mathrm{C}$.

Determine the required formula.

Write the formula of sending-end voltage of the line.

${\mathbf{V}}_{\mathbf{s}}\mathbf{=}{\mathbf{AV}}_{\mathbf{R}}\mathbf{+}{\mathbf{BI}}_{\mathbf{R}}$ …… (1)

Here, $\mathbf{A}\mathbf{,}\mathbf{B}$are the parameters, ${\mathbf{V}}_{\mathbf{R}}$receiving-end voltage and${\mathbf{I}}_{\mathbf{R}}$is sending-end voltage.

Write the formula of percentage voltage regulation.

$\mathbf{\%}{\mathbf{V}}_{\mathbf{R}}\mathbf{=}\frac{(V_{\mathrm{RNL}}-V_{\mathrm{RFL}})}{{\mathbf{V}}_{\mathbf{RFL}}}\mathbf{\left(}\mathbf{100}\mathbf{\right)}$ …… (2)

Here,${\mathbf{V}}_{\mathbf{RNL}}$is no-load receiving-end voltage, ${\mathbf{V}}_{\mathbf{RFL}}$is full-load receiving-end voltage.

Write the formula of characteristic impedance.

${\mathbf{Z}}_{\mathbf{c}}\mathbf{=}\sqrt{\frac{\mathbf{z}}{\mathbf{y}}}$ …… (3)

Here, $\mathit{z}$is the impedance and$\mathbf{y}$is the admittance of the conductor.

Write the formula of $\mathit{\gamma }\mathit{I}$.

$\mathbf{\gamma I}\mathbf{=}\mathbf{\left(}\sqrt{\mathbf{zy}}\mathbf{\right)}\mathbf{I}$…… (4)

Here, $\mathit{\gamma }$is the propagation constant, $\mathit{z}$is impedance, $\mathit{y}$is admittance and $\mathit{I}$is the length of line.

Write the formula of ABCDparameter.

$$\begin{gathered} \mathit{A}\mathbf{=}\mathit{D} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathbf{\left(}\mathbf{\gamma I}\mathbf{\right)} \end{gathered}$$ …… (5)

Here,$\mathit{\gamma }\mathit{I}$is dimensionless quantity.

$\mathbf{B}\mathbf{=}{\mathbf{Z}}_{\mathbf{c}}\mathbf{sinh}\mathbf{\left(}\mathbf{gl}\mathbf{\right)}$ …… (6)

Here, ${\mathbf{Z}}_{\mathbf{c}}$is characteristics impedance and$\mathit{\gamma }\mathit{I}$is dimensionless quantity.

$\mathbf{C}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{Z}}_{\mathbf{c}}}\mathbf{sinh}\mathbf{\left(}\mathbf{\gamma I}\mathbf{\right)}$ …… (7)

Here, ${\mathbf{Z}}_{\mathbf{c}}$is characteristics impedance and$\mathit{\gamma }\mathit{I}$is dimensionless quantity.

Determine the rework problem 5.9.

(a)

Rework of problem 5.9.

Calculate the GMR of the conductor from problem 4.18 and appendix A.4.

$$\begin{gathered} {\mathrm{D}}_{\mathrm{s}}=0.0403\mathrm{ft} \\ =\left(0.0435\right)\left(\frac{1\mathrm{m}}{3.28\mathrm{ft}}\right) \\ =0.0123\mathrm{m} \end{gathered}$$

Calculate the cube root of the product of the three-phase spacing (equivalent distance).

$D_{eq}=\sqrt[3]{D_{12}{D}_{23}{D}_{31}}$ …… (8)

Here, $D_{12},D_{23},D_{31}$is the distance between positions, conductor positions are denoted by 1, 2 and 3.

Substitute 8 for $D_{12},D_{23}$and 16 for $D_{31}$into equation (8).

$$\begin{gathered} D_{eq}=\sqrt[3]{\left(8\right)\left(8\right)\left(16\right)} \\ =\sqrt[3]{1024} \\ =10.079\mathrm{m} \end{gathered}$$

Determine the inductance per phase of the conductor.

$L_a=\left(2\times {10}^{-7}\right)\mathrm{In}\left(\frac{{\mathrm{D}}_{\mathrm{eq}}}{{\mathrm{D}}_{\mathrm{S}}}\right)$ …… (9)

Here, ${\mathrm{D}}_{\mathrm{eq}}$is phase spacing between the conductors and ${\mathrm{D}}_{\mathrm{S}}$is GMR of the conductor.

Substitute 10.079m for ${\mathrm{D}}_{\mathrm{eq}}$and 0.0123 m for ${\mathrm{D}}_{\mathrm{S}}$into equation (9).

$$\begin{gathered} L_a=\left(2\times {10}^{-7}\right)\mathrm{In}\left(\frac{10.079}{0.0123}\right)\left(\frac{\mathrm{H}}{\mathrm{m}}\right) \\ =1.342\times {10}^{-6}\mathrm{H}/\mathrm{m} \end{gathered}$$

Determine the inductive reactance per phase at 60 Hz frequency.

$X_a=\left(2\pi f\right)L_a$ …… (10)

Here, $L_a$is inductance per phase of the conductor and $f$is frequency.

Substitute $1.342\times {10}^{-6}\mathrm{for}{L}_a$and 60 Hz for $f$into equation (10).

$$\begin{gathered} X_a=\left(2\mathrm{\pi }\times 60\right)\left(1.342\times {10}^{-6}\right) \\ =\left(5.057\times {10}^{-4}\right)\left(\frac{\mathrm{\Omega }}{\mathrm{m}}\right)\left(\frac{1000\mathrm{m}}{\mathrm{km}}\right) \\ =0.506\mathrm{\Omega }/\mathrm{km} \end{gathered}$$

From reference problem 4.39

Determine the outer diameter of the conductor from appendix A.4

$$\begin{gathered} d=1.196\mathrm{in} \\ r=\left(\frac{1.196}{2}\mathrm{in}\right)\left(\frac{0.0254}{\mathrm{in}}\right) \\ =0.0152\mathrm{m} \end{gathered}$$

Determine the capacitance per phase of a balanced three phase overhead line.

$C_{\mathrm{an}}=\frac{2\mathrm{\pi \epsilon }}{\mathrm{In}\left({\displaystyle \frac{{\mathrm{D}}_{\mathrm{eq}}}{r}}\right)}$ …… (11)

Here, ${\mathrm{D}}_{\mathrm{eq}}$is phase spacing between the conductors and $r$is radius of the conductor.

Substitute $8.854\times {10}^{-12}\mathrm{for}\mathrm{\epsilon },10.079\mathrm{for}{\mathrm{D}}_{eq}\mathrm{and}0.0149$for $r$into equation (11).

$$\begin{gathered} C_{an}=\frac{\left(2\mathrm{\pi }\right)\left(8.854\times {10}^{-12}\right)}{\mathrm{In}\left({\displaystyle \frac{10.079}{0.0149}}\right)} \\ =8.532\times {10}^{-12}\mathrm{F}/\mathrm{m} \end{gathered}$$

Determine the admittance-to-neutral of the given arrangement at 60 Hz frequency.

$Y_n=j\omega C_{an}$ …… (12)

Here,$\omega$is frequency and$C_{an}$is capacitance per phase of a balanced three phase overhead line.

Substitute 60 for $\omega$and $8.532\times {10}^{-12}$for $C_{an}$into equation (12).

$$\begin{gathered} Y_n=j\omega C_{an} \\ =j\left(2\mathrm{\pi }\times 60\right)\left(8.532\times {10}^{-12}S/m\right) \\ =\left(j3.214\times {10}^{-9}\frac{\mathrm{S}}{\mathrm{m}}\right)\left(\frac{1000\mathrm{m}}{\mathrm{km}}\right) \\ =j3.214\times {10}^{-6}\mathrm{S}/\mathrm{km} \end{gathered}$$

(a) Determine the sending-end voltage and percentage voltage regulation for 0.9 lagging power factor.

Determine the impedance of the line after neglecting the resistance.

$z=R+j{X}_a$ …… (13)

Here, $R$is resistance for conductor and $X_a$is inductive reactance.

Substitute $0\mathrm{for}R\mathrm{and}0.506\mathrm{\Omega }/\mathrm{km}\mathrm{for}X$for and for into equation (13).

$$\begin{gathered} \mathrm{z}=0+j0.506 \\ =0.506\angle 90°\mathrm{\Omega }/\mathrm{km} \end{gathered}$$

Determine the total series impedance for 100 km line length.

$$\begin{gathered} Z=zI \\ Z=\left(0.506\angle 90°\right)\left(100\right) \\ =50.6\angle 90° \end{gathered}$$

Now, determine the total shunt admittance for 100 km line length.

$Y=yI$ …… (14)

Here, $y$is shunt admittance and $I$is length of line.

Substitute $j3.214\times {10}^{-6}\angle 90°\mathrm{for}y\mathrm{and}100\mathrm{for}I$for and for into equation (14).

$$\begin{gathered} Y=\left(j3.214\times {10}^{-6}\angle 90°\right)\left(100\right) \\ =3.229\times {10}^{-4}\angle 90°\mathrm{S} \end{gathered}$$

Now determine the transmission line parameters.

$A=D=1+\frac{YZ}{2}$ …… (15)

Here, $y$is shunt admittance and $z$is total series impedance.

Substitute $3.299\times {10}^{-4}\angle 90°S\mathrm{for}\mathrm{Y}\mathrm{and}5.06\angle 90°\mathrm{for}\mathrm{Z}$into equation (15).

$$\begin{gathered} A=D+1+\frac{\left[\left(3.229\times {10}^{-4}\angle 90°\right)\left(3.229\times {10}^{-4}\angle 90°\right)\left(50.6\angle 90°\right)\right]}{2} \\ =0.9918\mathrm{per}\mathrm{unit} \end{gathered}$$

$$\begin{gathered} B=Z \\ =50.6\angle 90°\Omega \end{gathered}$$

Solve for parameter C as,

$c=Y\left(1+\frac{YZ}{4}\right)$ …… (16)

Here, $Y$is shunt admittance and $Z$is total series impedance.

Substitute $3.229\times {10}^{-4}\angle 90°S\mathrm{for}\mathrm{Y}\mathrm{and}50.6\angle 90°\mathrm{for}\mathrm{Z}$into equation (16).

$$\begin{gathered} C=\left(3.229\times {10}^{-4}\angle 90°\right)\times \left[1+\frac{\left[\left(3.229\times {10}^{-4}\angle 90°\right)\left(50.6\angle 90°\right)\right]}{4}\right] \\ =\left(3.229\times {10}^{-4}\angle 90°\right)\times \left(1+0.00409\right) \\ =3.216\times {10}^{-4}\angle 90°S \end{gathered}$$

Therefore, the $A,B,C,D$parameters are,

$$\begin{gathered} \mathrm{A}=\mathrm{D} \\ =0.9918\mathrm{per}\mathrm{unit} \\ \mathrm{B}=50.6\angle 90°\mathrm{\Omega } \\ \mathrm{C}=3.216\times {10}^{-4}\angle 90°\mathrm{S} \end{gathered}$$

Receiving end line-to-line voltage$V_{\mathrm{R}}=218{\mathrm{kV}}_{LL}$

Receiving end power ${\mathrm{P}}_{\mathrm{R}}=300\mathrm{MW}$

Receiving end power factor is$0.9\mathrm{lagging}$

Determine the receiving end current.

$l_R=\frac{{\mathrm{P}}_{\mathrm{R}}\angle {\cos }^{-1}\mathrm{PF}}{\sqrt{3{\mathrm{V}}_{\mathrm{RLL}}}}$ …… (17)

Here, $P_R$is receiving end power, $V_{RLL}$is receiving-end voltage and $PF$is power factor.

Substitute $300\mathrm{for}{\mathrm{P}}_{\mathrm{R}},218\mathrm{for}{\mathrm{V}}_{\mathrm{RLL}}\mathrm{and}0.9\mathrm{for}\mathrm{PF}$into equation (17).

$\begin{array}{l}{\mathrm{l}}_{\mathrm{R}}=\frac{300\angle {\mathrm{coss}}^{-1}(0.9)}{\sqrt{3\left(218\right)}}\\ =\left(\frac{1500\times {10}^6}{831.3\times {10}^3}\right)\angle -25.84°\\ =0.7945\angle -25.84°\mathrm{kA}\end{array}$

Determine the sending-end voltage.

Substitute $0,9918\mathrm{for}{\mathrm{V}}_{\mathrm{R}},50.6\angle 90°\mathrm{for}\mathrm{B}\mathrm{and}0.7945\angle -25.84°\mathrm{for}{\mathrm{l}}_{\mathrm{R}}$into equation (1).

$$\begin{gathered} V_s=\left(0.9918\right)\left(125.9\right)+\left(50.6\angle 90°\right)\left(0.7945\angle -25.84°\right) \\ =146.9\angle 14.26°k{V}_{LN} \end{gathered}$$

Now, determine the line to line sending end voltage.

$$\begin{gathered} V_s=\left(146.9\angle 14.26°k{V}_{LN}\right)\sqrt{3} \\ =654.2kV \end{gathered}$$

Determine the no-load receiving end voltage.

$V_R=\frac{V_s}{A}$ …… (18)

Here, $V_s$is sending-end voltage and$A$is parameter.

Substitute $254.4\mathrm{for}{\mathrm{V}}_{\mathrm{s}}\mathrm{and}0.9918\mathrm{for}\mathrm{A}$into equation (18).

$V_{R.NL}=\begin{array}{c}\frac{254.4}{0.9918}\\ =256.5K{V}_{LL}\end{array}$

Determine the voltage regulation when receiving end power factor is 0.9 lagging.

Substitute $256.5\mathrm{for}{\mathrm{V}}_{\mathrm{RNL}}\mathrm{and}218\mathrm{for}{\mathrm{V}}_{\mathrm{RFL}}$into equation (2).

$$\begin{gathered} \%VR=\left(\frac{256.5-218}{218}\right)\left(100\right) \\ =17.7\% \end{gathered}$$

(b) Determine the sending-end voltage and percentage voltage regulation for unity power factor.

Determine the receiving end current.

$l_R=\frac{P_R\angle {\cos }^{-1}PF}{\sqrt{3}{V}_{RLL}}$ …… (19)

Here, $P_R$is receiving end power, $V_{RLL}$is receiving-end voltage and $PF$is power factor.

Substitute $300$for$P_R$,$218$for $V_{RLL}$and unity PF into equation (19).

$$\begin{gathered} l_R=\frac{300}{\sqrt{3}\left(218\right)} \\ =\left(\frac{1500\times {10}^6}{831.3\times {10}^3}\right) \\ =0.7945kA \end{gathered}$$

Determine the sending-end voltage.

Substitute $0.9918\mathrm{for}\mathrm{A},125.9\mathrm{for}{\mathrm{V}}_{\mathrm{R}},50.6\angle 90°\mathrm{for}\mathrm{B}\mathrm{and}0.7945\mathrm{for}{\mathrm{l}}_{\mathrm{R}}$into equation (1).

$$\begin{gathered} V_s=\left(0.9918\right)\left(125.9\right)+\left(50.6\angle 90°\right)\left(0.7945\right) \\ =131.2\angle 17.85°k{V}_{LN} \end{gathered}$$

Now, determine the line to line sending end voltage.

$$\begin{gathered} V_s=\left(131.2\angle 17.85°k{V}_{LN}\right)\sqrt{3} \\ =227.2\angle 17.85°k{V}_{LL} \end{gathered}$$

Determine the no-load receiving end voltage.

$V_{RNL}=\frac{V_s}{A}$ …… (20)

Here, $V_s$is sending-end voltage and $A$is parameter.

Substitute $227.2$for $V_s$and $0.9918$for $A$into equation (20).

$$\begin{gathered} V_{RNL}=\frac{227.2}{0.9918} \\ =229.1k{V}_{LL} \end{gathered}$$

Determine the voltage regulation when receiving end power factor is 0.9 lagging.

Substitute $229.1$for $V_{RNL}$and $218$for $V_{RFL}$into equation (2).

$$\begin{gathered} \%VR=\left(\frac{256.5-218}{218}\right)\left(100\right) \\ =5.08\% \end{gathered}$$

(c) Determine the sending-end voltage and percentage voltage regulation for 0.9 leading power factor.

Determine the receiving end current.

$l_R=\frac{P_R\angle {\cos }^{-1}PF}{\sqrt{3V_{RLL}}}$ …… (21)

Here, $P_R$is receiving end power, $V_{RLL}$is receiving-end voltage and $PF$is power factor.

Substitute $300$forrole="math" localid="1655526372107" $P_R$, $218$for $V_{RLL}$and $0.9$for PF into equation (21).

$$\begin{gathered} l_R=\frac{300\angle {\cos }^{-1}\left(0.9\right)}{\sqrt{3\left(218\right)}} \\ =\left(\frac{1500\times {10}^6}{831.3\times {10}^3}\right)\angle 25.84° \\ =0.7945\angle 25.84°kA \end{gathered}$$

Determine the sending-end voltage.

role="math" localid="1655527110202" $\mathrm{Substitute}0.9918\mathrm{for}\mathrm{A},125.9\mathrm{for}{\mathrm{V}}_{\mathrm{R}},50.6\angle 90°\mathrm{for}\mathrm{B}\mathrm{and}0.7945\angle 25.84°\mathrm{for}{\mathrm{l}}_{\mathrm{R}}$into equation (1).

$$\begin{gathered} V_s=\left(0.9918\right)\left(125.9\right)+\left(50.6\angle 90°\right)\left(0.7945\angle 25.84°\right) \\ =113.3\angle 18.63°k{V}_{LN} \end{gathered}$$

Now, determine the line to line sending end voltage.

$$\begin{gathered} V_s=\left(113.3\angle 18.63°k{V}_{LN}\right)\sqrt{3} \\ =196.2\angle 18.63°k{V}_{LL} \end{gathered}$$

Determine the no-load receiving end voltage.

$V_{RNL}=\frac{V_s}{A}$ …… (22)

Here, $V_s$is sending-end voltage and $A$is parameter.

Substitute $196.2$for $V_s$and $0.9918$for $A$into equation (22).

$$\begin{gathered} V_{RNL}=\frac{196.2}{0.9918} \\ =197.9k{V}_{LL} \end{gathered}$$

Determine the voltage regulation when receiving end power factor is 0.9 lagging.

Substitute $197.9$for role="math" localid="1655527773426" $V_{RNL}$and $218$for $V_{RFL}$into equation (2).

role="math" localid="1655527869980" $$\begin{gathered} \%VR=\left(\frac{197.9-218}{218}\right)\left(100\right) \\ =-9.22\% \end{gathered}$$

Determine the rework problem 5.16.

Rework of problem 5.16.

Calculate the GMR of the conductor from problem 5.16 and appendix A.4.

$$\begin{gathered} D_{\mathrm{s}}=0.0435\mathrm{ft} \\ =\left(0.0435\right)\left(\frac{1\mathrm{m}}{3.28\mathrm{ft}}\right) \\ =0.0133\mathrm{m} \end{gathered}$$

Calculate the cube root of the product of the three-phase spacing (equivalent distance).

$D_{eq}=\sqrt[3]{D_{12},D_{23}{D}_{31}}$ …… (23)

Here, $D_{12},D_{23},D_{31}$is the distance between positions, conductor positions are denoted by 1, 2 and 3.

Substitute $10$for localid="1655528438357" $D_{12},D_{23}$and $20$for $D_{31}$into equation (23).

$\begin{array}{l}{D}_{eq}=\sqrt[3]{\left\{10\right)\left(10\right)\left(20\right)}\\ =\sqrt[3]{2000}\\ =12.599\mathrm{m}\end{array}$

Determine the geometric mean radius (GMR) for inductor for a three conductor bundle.

$D_{SL}=\sqrt[9]{{\left(D_s\times d\times d\right)}^3}$ …… (24)

Here, $D_s$is geometric mean radius and $d$is distance between conductors.

Substitute $0.0133$for $D_s$and $0.5$for $d$into equation (24).

$$\begin{gathered} D_{SL}=\sqrt[3]{\left(0.0133\right){\left(0.5\right)}^2} \\ =\sqrt[3]{\left(0.0133\right)\left(0.5\right)} \\ =\sqrt[3]{3.315\times {10}^{-3}} \\ =0.1493m \end{gathered}$$

Determine the inductance per phase of the conductor.

$L_a=\left(2\times {10}^{-7}\right)In\left(\frac{D_{eq}}{D_{SL}}\right)$ …… (25)

Here, $D_{eq}$is phase spacing between the conductors and $D_{SL}$is GMR of the conductor.

Substitute $10.079\mathrm{m}$for $D_{eq}$and $0.0123\mathrm{m}$for $D_s$into equation (25).

$L_a=\left(2\times {10}^{-7}\right)\mathrm{In}\left(\frac{12.599}{0.1493}\right)\left(\frac{\mathrm{H}}{\mathrm{m}}\right)\left(\frac{1000}{1\mathrm{km}}\right)$

Determine the inductive reactance per phase at $60\mathrm{Hz}$frequency.

$X_a=\left(2\pi f\right)L_a$ …… (26)

Here, $L_a$is inductance per phase of the conductor and $f$is frequency.

Substitute $8.871\times {10}^{-4}$for $L_a$and $60\mathrm{Hz}$for $f$into equation (26).

$$\begin{gathered} X_a=\left(2\pi \times 60\right)\left(8.871\times {10}^{-4}\right) \\ =0.335\mathrm{\Omega }/\mathrm{km} \end{gathered}$$

From reference problem 4.41

Determine the outer diameter of the conductor from appendix A.4

$d=1.293\mathrm{in}$

Determine the radius.

$$\begin{gathered} r=\left(\frac{1.293}{2}\mathrm{in}\right)\left(\frac{0.3\mathrm{m}}{\mathrm{in}}\right) \\ =0.0161\mathrm{m} \end{gathered}$$

The bundle spacing$d=0.5\mathrm{m}$.

Determine the radius of each conductor.

${\mathrm{D}}_{SC}=\sqrt[3]{r{d}^2}$ …… (27)

Here, $r$is radius of conductor and $d$is diameter of each conductor.

Substitute $0.0161$for $r$and $0.5$for $d$into equation (27).

$$\begin{gathered} D_{SC}=\sqrt[3]{\left(0.0161\right){\left(0.5\right)}^2} \\ =\sqrt[3]{0.004025} \\ =0.1591\mathrm{m} \end{gathered}$$

Calculate the cube root of the product of the three-phase spacing (equivalent distance).

$Deq=\sqrt[3]{D_{12},D_{23}{D}_{31}}$ …… (28)

Here, $D_{12},D_{23}{D}_{31}$is the distance between positions, conductor positions are denoted by 1, 2 and 3.

Substitute $10$for $D_{12},D_{23}$and $20$for $D_{31}$into equation (28).

$\begin{array}{l}{D}_{eq}=\sqrt[3]{\left\{10\right)\left(10\right)\left(20\right)}\\ =\sqrt[3]{2000}\\ =12.599\mathrm{m}\end{array}$

Determine the capacitance per phase to neutral of a balanced three phase overhead line.

$C_{an}=\frac{2\pi }{\mathrm{In}\left({\displaystyle \frac{D_{eq}}{D_{sc}}}\right)}$ …… (29)

Here, $D_{eq}$is phase spacing between the conductors and $r$is radius of the conductor.

Substitute $8.854\times {10}^{-12}\mathrm{for}\mathrm{\epsilon },12.599\mathrm{for}{\mathrm{D}}_{\mathrm{eq}}\mathrm{and}0.1591\mathrm{for}{\mathrm{D}}_{\mathrm{SC}}$into equation (29).

$$\begin{gathered} C_{an}=\frac{\left(2\mathrm{\pi }\right)\left(8.854\times {10}^{-12}\right)}{\mathrm{In}\left({\displaystyle \frac{12.599}{0.1591}}\right)} \\ =12.718\times {10}^{-12}\mathrm{F}/\mathrm{m} \end{gathered}$$

Determine the admittance-to-neutral of the given arrangement at $60\mathrm{Hz}$frequency.

$Y_n=j\omega C_{an}$ …… (30)

Here, $\omega$is frequency and $C_{an}$is capacitance per phase of a balanced three phase overhead line.

Substitute $60\mathrm{for}\mathrm{\omega },12.718\times {10}^{-12}\mathrm{for}{\mathrm{C}}_{\mathrm{an}}$into equation (30).

$$\begin{gathered} Y_n={\mathrm{j\omega C}}_{\mathrm{an}} \\ =\mathrm{j}\left(2\mathrm{\pi }\times 60\right)\left(12.718\times {10}^{-12}\mathrm{S}/\mathrm{km}\right) \\ =\left(\mathrm{j}4.807\times {10}^{-9}\frac{\mathrm{S}}{\mathrm{m}}\right)\left(\frac{1000\mathrm{m}}{\mathrm{km}}\right) \\ =\mathrm{j}4.807\times {10}^{-6}\mathrm{S}/\mathrm{km} \end{gathered}$$

(a) Determine the characteristic impedance.

Now, determine the characteristic impedance.

Substitute $j0.335\Omega /\mathrm{km}\mathrm{for}\mathrm{z}\mathrm{and}j4.807\times {10}^{-6}\mathrm{S}/\mathrm{km}\mathrm{for}\mathrm{y}$into equation (3).

$$\begin{gathered} Z_c=\sqrt{\frac{j0.335}{j4.807\times {10}^{-6}}} \\ =264\Omega \end{gathered}$$

(b) Determine the value ofyl.

Determine the quantity $yl$for a line length of 300 km.

Substitute $j0.335\mathrm{for}\mathrm{z},j4.807\times {10}^{-6}\mathrm{for}\mathrm{y}\mathrm{and}300\mathrm{for}\mathrm{I}$into equation (4).

$$\begin{gathered} yl=\left(\sqrt{\left(j0.335\right)\left(j4.4807\times {10}^{-6}\right)}\right)\left(300\right) \\ =\left(j1.27\times {10}^{-3}\right)\left(300\right) \\ =j0.38\mathrm{per}\mathrm{unit} \end{gathered}$$

(c) Determine the transmission line parameters.

Determine the value of hyperbolic function of$\cosh \left(yl\right)$.

$$\begin{gathered} \cosh \left(\mathrm{yl}\right)=\cosh \left(\mathrm{j}0.38\right) \\ =\frac{{\mathrm{e}}^{\mathrm{j}0.381}+{\mathrm{e}}^{-\mathrm{j}0.381}}{2} \\ =\frac{\left(0.9284+\mathrm{j}0.3716\right)+\left(0.9284-\mathrm{j}0.3716\right)}{2} \\ =0.9284 \end{gathered}$$

Determine the value of$\left(yl\right)$.

$$\begin{gathered} \sinh \left(yl\right)=\sinh \left(j0.381\right) \\ =\frac{e^{jo.381}-e^{-jo.381}}{2} \\ =\frac{e^{jo.381}-e^{-jo.381}}{2} \\ =\frac{\left(0.9284-j0.3716\right)-\left(0.9284-j0.3716\right)}{2} \end{gathered}$$

As further solve,

$si\mathrm{nh}\left(yl\right)=j0.3716$

Calculate the value of A parameter.

$$\begin{gathered} A=\cosh \left(yl\right) \\ =0.9285\mathrm{per}\mathrm{unit} \end{gathered}$$

Determine the parameter D.

$D=A$

$D=0.9285\mathrm{per}\mathrm{unit}$

Determine the value of B parameter.

Substitute $264\mathrm{for}{\mathrm{Z}}_{\mathrm{c}}\mathrm{and}j0.3716\mathrm{for}\sinh \left(\mathrm{yl}\right)$into equation (6).

$$\begin{gathered} B=264x\left(j0.3716\right) \\ =j98.252\Omega \end{gathered}$$

Finally, determine the value of C parameter.

Substitute $j0.3716\mathrm{for}\sinh \left(\mathrm{yl}\right)\mathrm{and}264\mathrm{for}{\mathrm{Z}}_{\mathrm{c}}$into equation (7).

$$\begin{gathered} C=\frac{j0.716}{264} \\ =j1.408\times {10}^{-3}S \end{gathered}$$

Therefore, the ABCD parameters of the long transmission line are,

$$\begin{gathered} A=0.9285\mathrm{per}\mathrm{unit} \\ B=\mathrm{j}98.252\mathrm{\Omega } \\ C=\mathrm{j}1.408\times {10}^{-3}\mathrm{S} \\ D=0.9285\mathrm{per}\mathrm{unit} \end{gathered}$$