Question

Rated line voltage is applied to the sending end of the line in Problem 5.26. Calculate the receiving-end voltage when the receiving end is terminated by (a) an open circuit, (b) the surge impedance of the line, and (c) one-half of the surge impedance. (d) Also calculate the theoretical maximum real power that the line can deliver when rated voltage is applied to both ends of the line.

Short answer

(a)The value of receiving-end voltage when the receiving end is terminated byan open circuit is$536.5\mathrm{kV}$.

(b)The value of receiving-end voltage when the receiving end is terminated bythe surge impedance of the line is$500\mathrm{kV}$.

(c)The value of receiving-end voltage one-half of the surge impedance of the line$423.4\mathrm{kV}$.

(d)The value of maximum real power of the line when rated voltage is applied to both ends of the line is$2508\mathrm{MW}$.

Step-by-step solution

Given data.

As using reference problem 5.26, we know that

A given length of the line is$I=350-\mathrm{km}$.

A given rated line voltage is$V_s=500\mathrm{kV}$.

A given frequency of line is 60-Hz.

A positive-sequence series reactance of three-phase uncompensated line is$X=0.34\Omega /\mathrm{km}$.

A positive-sequence shunt admittance of three-phase uncompensated line is$y=j4.5\times {10}^{-6}\mathrm{S}/\mathrm{km}$.

Determine the required formula.

Write the formulaof receiving-end voltage when the receiving end is terminated byan open circuit.

${\mathbf{V}}_{\mathbf{s}}\mathbf{=}{\mathbf{V}}_{\mathbf{s}}\mathbf{/}\mathbf{A}$

Here, ${\mathbf{V}}_{\mathbf{s}}$is sending-end voltage and $\mathbf{A}$is the parameter.

Write the formula ofreceiving-end voltage when the receiving end is terminated bythe surge impedance of the line.

${\mathbf{V}}_{\mathbf{R}\mathbf{}}\mathbf{=}{\mathbf{V}}_{\mathbf{S}}\mathbf{/}\mathbf{A}$

Here, role="math" localid="1655417334353" ${\mathbf{V}}_{\mathbf{R}}$is the sending-end voltage and $\mathbf{A}$is the parameter.

Write the formula of receiving-end voltage ofsurge impedance of the line.

${\mathbf{V}}_{\mathbf{3}}\mathbf{=}\mathbf{cos}\left(\mathrm{\beta I}\right)\mathbf{.}{\mathbf{V}}_{\mathbf{R}}\mathbf{+}\left({\mathrm{jZ}}_C\mathrm{sin\beta I}\right)\left(\frac{V_R}{{\displaystyle \frac{1}{1}{Z}_C}}\right)$

Here, ${\mathbf{V}}_{\mathbf{S}}$is the sending-end voltage,${\mathbf{V}}_{\mathbf{R}}$is the receiving-end voltage and is the characteristics impedance

Write the formula ofreal power of the line.

${\mathbf{P}}_{\mathbf{max}\mathbf{3}\mathbf{f}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{S}}{\mathbf{V}}_{\mathbf{R}}}{{\mathbf{X}}^{\mathbf{\text{'}}}}$

Here, ${\mathbf{V}}_{\mathbf{S}}$is the sending-end voltage, ${\mathbf{V}}_{\mathbf{R}}$is the receiving-end voltage and$\mathbf{X}\mathbf{\text{'}}$ is the reactance of the line.

Determine the receiving-end voltage.

(a)

Determine the characteristic impedance.

$Z_c=\sqrt{\frac{z}{y}}$

Here,$Z$is series impedance and$y$is shunt admittance.

Substitute $\mathrm{j}0.34\mathrm{for}\mathrm{z}\mathrm{and}\mathrm{j}4.5\times {10}^{-6}\mathrm{for}\mathrm{y}$into equation (5).

$$\begin{gathered} Z_c=\sqrt{\frac{j0.34}{j4.5\times {10}^{-6}}} \\ =274.9\angle 0°\Omega \end{gathered}$$

Determine the value of $y/$

$y/=\left(\sqrt{zy}\right)/$

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

Substitute$j0.34\mathrm{for}\mathrm{z}\mathrm{and}j4.5\times {10}^{-6}\mathrm{for}\mathrm{y}\mathrm{and}300\mathrm{for}/$into equation.

$$\begin{gathered} y=\sqrt{\left(j0.34\right)\left(j4.5\times {10}^{-6}\right)}\left(300\right) \\ =j0.3711\mathrm{pu} \end{gathered}$$

Initially determine the value of $\beta I$

Calculate the value of $yI$.

$yI=j\left(\beta I\right)$

Here, $yI$is dimensionless quantity and$\beta I$is also dimensionless quantity.

Substitute the value of$yI$in equation.

$$\begin{gathered} j\left(\beta I\right)=j0.3711\mathrm{pu} \\ \left(\mathrm{\beta I}\right)=\mathrm{j}0.3711\mathrm{pu} \end{gathered}$$

Now, determine the $ABCD$parameters.

Write the formula of $\mathrm{A}$parameter.

$A=D=\cos \left(\beta I\right)$

Substitute the value of$\beta I$into equation.

$$\begin{gathered} A=D=\cos \left(0.3711\mathrm{radians}\right) \\ =0.9319\angle 0°\mathrm{pu} \end{gathered}$$

Write the formula of $B$parameter.

$B=j{Z}_C\sin \left(\beta I\right)$

Here, $Z_C$is characteristics impedance and $\beta I$is also dimensionless quantity.

Substitute the value of $Z_C$and role="math" localid="1655419149241" $\beta I$into equation.

$$\begin{gathered} B=j\left(274.9\right)\sin \left(0.3711\mathrm{radians}\right) \\ =j99.68\Omega \end{gathered}$$

Write the formula of $C$parameter.

$C=j\left(\frac{1}{Z_c}\right)\sin \left(\beta I\right)$

Here, $Z_C$is characteristics impedance and $\beta I$is also dimensionless quantity.

Substitute the value of$Z_C$and$\beta I$into equation.

$$\begin{gathered} C=j\left(\frac{1}{274.9}\right)\sin \left(0.3711\mathrm{radians}\right) \\ =j1.319\times {10}^{-3}S \end{gathered}$$

Now, determine thevalue of receiving-end voltage when the receiving end is terminated byan open circuit.

$V_R=V_S/A$

Here, $V_S$is sending-end voltage and $A$is the parameter

Substitute the value of$V_S$and$A$into equation.

$$\begin{gathered} V_R=500/0.9319 \\ =536.5\mathrm{kV} \end{gathered}$$

Determine the receiving-end voltage of the surge impedance line.

(b)

Calculate the value of receiving-end voltage when the receiving end is terminated bythe surge impedance of the line then sending end voltage is equal to receiving-end voltage.

Write the formula of receiving-end voltage.

$V_R=V_S$

Here, $V_S$is sending-end voltage.

Substitute $500$for $V_S$into equation (12).

$V_R=500\mathrm{kV}$

Determine the receiving-end voltage one-half of the surge impedance.

(c)

Write the formula of receiving-end voltage.

$$\begin{gathered} V_S=\cos \left(\beta I\right)V_R+\left(j{Z}_C\sin \beta I\right)\left(\frac{V_R}{{\displaystyle \frac{1}{1}}{V}_C}\right) \\ {V}_S=V_R\cos \left(\beta I\right)+jZ\sin \beta I \\ {V}_R=\frac{V_S}{\cos \left(\beta I\right)+jZ\sin \beta I} \end{gathered}$$

Here,$V_S$is sending-end voltage, $Z$is impedance and $\beta I$is also dimensionless quantity.

Substitute $500\mathrm{for}{\mathrm{V}}_{\mathrm{s}},2\mathrm{for}\mathrm{Z}\mathrm{and}0.3711\mathrm{radians}\mathrm{fo}\beta i$into equation.

$$\begin{gathered} V_R=\frac{500}{\left|\cos 0.3711\mathrm{radians}+j2\sin 0.3711\mathrm{radians}\right|} \\ =\frac{500}{1.18} \\ =423.4\mathrm{kV} \end{gathered}$$

Determine the value of maximum real power of the line.

(d)

Write the formula of real power.

$P_{max3\varphi }=\frac{V_S{V}_R}{X^{\text{'}}}$ …… (14)

Here,$V_S$is the sending-end voltage,$V_R$is the receiving-end voltage and $X^{\text{'}}$is the reactance of the line.

Substitute $500\mathrm{for}{\mathrm{V}}_{\mathrm{s}},500\mathrm{for}{\mathrm{V}}_{\mathrm{R}}\mathrm{and}99.68\mathrm{for}{\mathrm{X}}^{\text{'}}$into equation (14).

$$\begin{gathered} P_{max}=\frac{\left(500\right)\left(500\right)}{99.68} \\ =2508\mathrm{MW} \end{gathered}$$