Question

The 500-kV, 60-Hz, three-phase line in Problems 4.20 and 4.41 has a 300-km length. Calculate: (a)$Z_c$, (b)$\left(\gamma l\right)$, and (c) the exact ABCD parameters for this line. Assume a${50}^{\circ }C$ conductor temperature.

Short answer

(a)Therefore, the value of characteristic impedance

$Z_c=264.54\angle -1.{72}^{\circ }\Omega$.

(b)Therefore, the value of $\gamma l$ is$0.0114+j0.381pu$.

(c)Therefore, the value of ABCD parameter are$D=A=0.928\angle 0.{268}^{\circ }pu$,$B=98.4\angle 86.{52}^{\circ }\Omega$and$C=1.406\times {10}^{-3}\angle 89.{96}^{\circ }$.

Step-by-step solution

Given data.

The given supply voltage is 500-kV.

The given frequency is 60-Hz.

The given length of the line I=300km.

The conductor temperature is${50}^{\circ }C$.

Determine the required formula.

Write the formula of characteristic impedance.

$Z_c=\sqrt{\frac{z}{y}}$ …… (1)

Here, $z$ is the impedance and $y$is the admittance of the conductor.

Write the formula of$\gamma l$.

$\gamma l=\left(\sqrt{zy}\right)l$ …… (2)

Here, $\gamma$is the propagation constant, $z$ is impedance, $y$ is admittance and $l$ is the length of line.

Write the formula ofABCDparameter.

$A=D=cosh\left(\gamma l\right)$ …… (3)

Here, $\gamma l$is dimensionless quantity.

$B=Z_{c}sinh\left(\gamma l\right)$ …… (4)

Here,$Z_c$is characteristics impedance and $\gamma l$is dimensionless quantity.

$C=\frac{1}{Z_c}sinh\left(\gamma l\right)$ …… (5)

Here,$Z_c$is characteristics impedance and $\gamma l$is dimensionless quantity.

Determine the characteristic impedance Zc.

(a)

There are three ASCR 1113 kcmil conductors per bundle.

Distance between conductors in the bundle d=0.5m.

Consider the length of the line l=300km.

Consider the length of the line,l=300km, so the line is long transmission line.

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

$D_{eq}=\sqrt[3]{D_{12}{D}_{23}{D}_{31}}$ …… (6)

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 (6).

$$\begin{gathered} D_{eq}=\sqrt[3]{\left(10\right)\left(10\right)\left(20\right)} \\ =\sqrt[3]{2000} \\ =12.6m \end{gathered}$$

Geometric mean radius corresponding to 1113 Kcmil ASCR conductor is$D_s=0.0435ft$.

Calculate the geometric mean radius in meters.

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

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

$D_{SL}=\sqrt[9]{{\left(D_S\times d\times d\right)}^3}$ …… (7)

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 (7).

$$\begin{gathered} D_{SL}=\sqrt[3]{\left(0.0133\right){\left(0.5\right)}^2} \\ =\sqrt[3]{\left(0.0133\right)\left(0.25\right)} \\ =\sqrt[3]{3.315\times {10}^{-3}} \\ =0.149m \end{gathered}$$

Calculate the inductive reactance of the 1113 kcmil ASCR conductor.

$X=j\omega L$ …… (8)

Here, $\omega$is frequency L is length of the line of long transmission line.

Substitute $2\pi \left(60\right)$for $\omega$and $2\times {10}^{-7}In\left(\frac{D_{eq}}{D_{SL}}\right)$into equation (8).

$X=j2\pi \left(60\right)\left(2\times {10}^{-7}In\left(\frac{D_{eq}}{D_{SL}}\right)\right)$

Substitute $D_{eq}$and $D_{SL}$from equation (6) and (7).

$$\begin{gathered} X=j2\mathrm{\pi }\left(60\right)\left(2\times {10}^{-7}\mathrm{In}\left(\frac{12.6}{0.149}\right)\right)\left(\frac{1000\mathrm{m}}{1\mathrm{km}}\right) \\ =\mathrm{j}2\mathrm{\pi }\left(60\right)\left(2\times {10}^{-7}\mathrm{In}\left(84.49\right)\right)\left(\frac{1000\mathrm{m}}{1\mathrm{km}}\right) \\ =\mathrm{j}2\mathrm{\pi }\left(60\right)\left(2\times {10}^{-7}\right)\left(4.436\right)\left(\frac{1000\mathrm{m}}{1\mathrm{km}}\right) \\ =\mathrm{j}0.335\frac{\mathrm{\Omega }}{\mathrm{km}} \end{gathered}$$

Equivalent geometric radius for a capacitor for a bundle of three conductors is

$D_{sc}$.

Determine the value of the capacitance.

$C=\frac{2\mathrm{\pi }\left(8.85\times {10}^{-12}\right)}{In\left({\displaystyle \frac{D_{eq}}{D_{sc}}}\right)}$ …… (9)

Here, $D_{eq}$is equivalent distance and $D_{sc}$is geometric radius for a capacitor.

Substitute 12.6 for $D_{eq}$and 0.16 for $D_{sc}$into equation (9).

$$\begin{gathered} C=\frac{2\mathrm{\pi }\left(8.85\times {10}^{-12}\right)}{In\left({\displaystyle \frac{12.6}{0.16}}\right)} \\ =\frac{2\mathrm{\pi }\left(8.85\times {10}^{-12}\right)}{In\left(78.75\right)} \\ =\frac{2\mathrm{\pi }\left(8.85\times {10}^{-12}\right)}{4.366} \\ =1.274\times {10}^{-11}F/m \end{gathered}$$

Now, determine the admittance of line per km.

$y=j\omega C$ …… (10)

Here, $\omega$is frequency and C is the capacitance.

Substitute$2\mathrm{\pi }\left(60\right)$for $\omega$and $1.274\times {10}^{-11}$for C into equation (10).

$$\begin{gathered} y=j2\mathrm{\pi }\left(60\right)\left(1.274\times {10}^{-11}\right)\left(1000\mathrm{S}/\mathrm{km}\right) \\ =\mathrm{j}4.801\times {10}^{-6}\mathrm{S}/\mathrm{km} \end{gathered}$$

As per given problems 4.20 and 4.41 the resistance associated with 1113 kcmil ASCR conductors per bundle at ${50}^{\circ }C$is$0.0969\Omega /mile$.

Now determine the value of resistance for three bundled conductors in$\Omega /km$

$$\begin{gathered} R=\left(\frac{0.0969}{3}\right)\left(\frac{\Omega }{mile}\right)\left(\frac{1mile}{1.609km}\right) \\ =0.0201\frac{\Omega }{km} \end{gathered}$$

Determine the impedance per km.

z=R+jX …… (11)

Here, R is resistance for conductor and X is inductive reactance.

Substitute 0.0201 for R and 0.335 for X into equation (11).

$$\begin{gathered} z=0.0201+j0.335 \\ =0.336\angle 86.{56}^{\circ }\Omega /km \end{gathered}$$

Now, determine the characteristic impedance.

Substitute $0.336\angle 86.{56}^{\circ }$for z and $4.801\times {10}^{-6}\angle {90}^{\circ }$for y into equation (1).

$$\begin{gathered} Z_c=\sqrt{\frac{0.336\angle 86.{56}^{\circ }}{4.801\times {10}^{-6}\angle {90}^{\circ }}} \\ =\sqrt{69985.42\angle -3.{44}^{\circ }} \\ =264.54\angle -1.{72}^{\circ }\Omega \end{gathered}$$

Determine the value of γl.

(b)

Calculate the value of $\gamma l$.

Substitute $0.336\angle 86.{56}^{\circ }$for z,$4.801\times {10}^{-6}\angle {90}^{\circ }$for y and 300 for l into equation (2).

$$\begin{gathered} \gamma l=\left[\sqrt{\left(0.336\angle 86.{56}^{\circ }\right)\left(4.801\times {10}^{-6}\angle {90}^{\circ }\right)}\right]\left(300\right) \\ =\left[\sqrt{\left(1.613\times {10}^{-6}\angle 176.{28}^{\circ }\right)}\right]\left(300\right) \\ =\left(1.27\times {10}^{-3}\angle 88.{28}^{\circ }\right)\left(300\right) \\ =0.381\angle 88.{28}^{\circ } \end{gathered}$$

As further solve,

$\gamma l=0.0114+j0.381pu$

Determine the value of ABCD parameter.

(c)

Determine the value of $\cos h\left(\gamma l\right)$.

$$\begin{gathered} \cos h\left(\gamma l\right)=\cos h\left(0.0114+j0.381\right) \\ =\frac{e^{0.0114+j0.381}+e^{-0.0114-j0.381}}{2} \\ =\frac{e^{0.0114}{e}^{j0.381}+e^{-0.0114}{e}^{-j0.381}}{2} \\ =\frac{1.0114\angle 0.381+0.988\angle -.381}{2}radians \end{gathered}$$

Further solve as,

$$\begin{gathered} \cos h\left(\gamma l\right)=\frac{1.856\angle 0.00469}{2}radians \\ =0.928\angle 0.{268}^{\circ } \end{gathered}$$

Determine the value of$\sin h\left(\gamma l\right)$.

$$\begin{gathered} \sin h\left(\gamma l\right)=\sin h\left(0.0114+j0.381\right) \\ =\frac{e^{0.0114+j0.381}-e^{-0.0114-j0.381}}{2} \\ =\frac{e^{0.0114}{e}^{j0.381}-e^{-0.0114}{e}^{-j0.381}}{2} \\ =\frac{1.0114\angle 0.381-0.988\angle -0.381}{2}radians \end{gathered}$$

As further solve,

$$\begin{gathered} \sin h\left(\gamma l\right)=\frac{0.744\angle 1.54}{2}radians \\ =0.372\angle 88.{24}^{\circ } \end{gathered}$$

Calculate the value ofAparameter.

$$\begin{gathered} A=\cos h\left(\gamma l\right) \\ =0.928\angle 88.{24}^{\circ }pu \end{gathered}$$

Determine the value ofD.

D=A

$D=0.928\angle 0.{268}^{\circ }pu$

Determine the value ofBparameter.

Substitute $264.54\angle -1.{72}^{\circ }$for $Z_c$and $0.372\angle 88.{24}^{\circ }$for $\sin h\left(\gamma l\right)$into equation (4).

$$\begin{gathered} B=\left(264.54\angle -1.{72}^{\circ }\right)\left(0.372\angle 88.{24}^{\circ }\right) \\ =98.4\angle 86.{52}^{\circ }\Omega \end{gathered}$$

Finally, determine the value of C parameter.

Substitute $0.372\angle 88.{24}^{\circ }$for $\sin h\left(\gamma l\right)$and $264.54\angle -1.{72}^{\circ }$for$Z_c$ into equation (5).

$$\begin{gathered} C=\frac{0.372\angle 88.{24}^{\circ }}{264.54\angle -1.{72}^{\circ }} \\ =1.406\times {10}^{-3}\angle 89.{96}^{\circ }S \end{gathered}$$

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

$$\begin{gathered} A=0.928\angle 0.{268}^{\circ }pu \\ D=0.928\angle 0.{268}^{\circ }pu \\ B=98.4\angle 86.{52}^{\circ }\Omega \\ C=1.406\times {10}^{-3}\angle 86.{96}^{\circ }S \end{gathered}$$