Question

A $\mathbf{350}-\mathbf{km}$ , $\mathbf{500}-\mathbf{kV}$, $\mathbf{60}-\mathbf{Hz}$,three-phase uncompensated line has a positive sequence series reactance$\mathbf{x}=\mathbf{0}\mathbf{.}\mathbf{34}\mathbf{\Omega }/\mathbf{km}$ and a positive-sequence shunt admittance $\mathbf{y}=\mathbf{j}\mathbf{4}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{-\mathbf{6}}\mathbf{S}/\mathbf{km}$. Neglecting losses, calculate: (a) ${\mathbf{Z}}_c$, (b)$\gamma l$, (c) the$\mathbf{ABCD}$ parameters, (d) the wavelength $\lambda$of the line in kilometres, and (e) the surge impedance loading in MW

Short answer

(a) The value of characteristic impedance $Z_c$ is$274.9\Omega$.

(b) The value of given $\gamma l$ is$j0.3711\text{\hspace{0.33em}pu}$.

(c) The value of given $\text{ABCD}$ line parameters are$\text{0.9319\hspace{0.33em}pu}$, $j99.68\Omega$, $j1.32\times {10}^{-3}\text{\hspace{0.33em}S}$ and $\text{0.9319\hspace{0.33em}pu}$.

(d) The value of wavelength $\lambda$ of the line in kilometres $\lambda$ is$\text{5079\hspace{0.33em}km}$.

(e) The value of surge impedance loading in MW $\text{SIL}$is$\text{909.4\hspace{0.33em}MW}$ .

Step-by-step solution

Write the given data from the question.

The length of given three phase line is$350-\text{km}$.

The voltage of given three phase line is$500-\text{kV}$.

The frequency of given three phase line is$60-\text{Hz}$.

The positive series reactance of given three phase line is$\text{x}=\text{0.34\hspace{0.33em}Ω}/\text{km}$.

Determine the formula of symmetrical components of the line voltages.

Write the formula ofvalue of characteristic impedance.

${\mathbf{Z}}_c=\sqrt{\frac{z}{y}}$…… (1)

Here, localid="1655965483230" $\mathbf{z}$is impedance and $\mathbf{y}$is admittance.

Write the formula of$\gamma l$.

$\gamma l=l\sqrt{\mathrm{zy}}$…… (2)

Here, $\mathbf{z}$is impedance, $\mathbf{y}$ is admittance and $l$ is length.

Write the formula of$\text{ABCD}$line parameters.

$\mathbf{A}=\mathbf{cosh}\left(\gamma l\right)$ …… (3)

Here, $\gamma l$is a dimensionless quantity.

$\mathbf{B}={\mathbf{jZ}}_c\mathbf{sin}\left(\mathbf{\beta }l\right)$ …… (4)

Here, ${\mathbf{Z}}_c$is characteristic impedance and $\beta l$is lossless line.

$\mathbf{C}=\mathbf{j}\left(\frac{1}{{\mathbf{Z}}_c}\right)\mathbf{sin}\left(\mathbf{\beta }l\right)$ …… (5)

Here, ${\mathbf{Z}}_c$is characteristic impedance and $\beta l$ is lossless line.

localid="1655965910181" $\begin{array}{c}\mathbf{D}=\mathbf{A}\\ \mathbf{D}=\mathbf{cosh}\left(\gamma l\right)\end{array}$ …… (6)

Here, $\gamma l$is a dimensionless quantity.

Write the formula of wavelength $\lambda$of the line in kilometres.

localid="1655965995694" $\lambda =\frac{\mathbf{2}\pi }{\beta }$ …… (7)

Here,$\beta$ is complex quantity.

Write the formulaof surge impedance loading.

$\mathbf{SIL}=\frac{{\mathbf{V}}_{\mathrm{rated}}^2}{{\mathbf{Z}}_c}$ …… (8)

Here, ${\mathbf{V}}_{\mathrm{rated}}^2$ are rated voltage of three phase line and $\mathbf{Z}$ is impedance.

Step 3:(a) Determine the characteristic impedance  Zc of three phase line.

Determine the expression of characteristics impedance${\mathbf{Z}}_c$.

Substitute $j0.34\Omega /\text{km}$ for $\text{z}$and $j4.5\times {10}^{-6}\text{S}/\text{km}$ for $y$ into equation (1).

$\begin{array}{c}{Z}_c=\sqrt{\frac{j0.34}{j4.5\times {10}^{-6}}}\\ =\sqrt{75.55\times {10}^3}\\ =274.9\Omega \end{array}$

Therefore, the characteristics impedance ${\mathbf{Z}}_c$is $274.9\Omega$ .

(b) Determine the γl.

Determine the value of$\gamma l$.

Substitute$300$ for$l$, $j0.34$ for $z$ and $j4.5\times {10}^{-6}$ for $y$ into equation (2).

$\begin{array}{c}\gamma l=\left(300\right)\left(\sqrt{\left(j0.34\right)\left(j4.5\times {10}^{-6}\right)}\right)\\ =\left(300\right)\left(j1.237\times {10}^{-3}\right)\\ =j0.3711\text{\hspace{0.33em}pu}\end{array}$

Therefore, the value of $\gamma l$ is$j0.3711\text{\hspace{0.17em}pu}$ .

(c) Determine the value of given  ABCD line parameters.

Determine thevalue of given$\text{ABCD}$ line parameters.

Write the expression for$\gamma l$.

$\begin{array}{c}\gamma l=\alpha l+j\beta l\\ =j0.3711\end{array}$

Therefore,

$\beta l=0.3711\text{\hspace{0.33em}pu}$

Determine the expression of$\cosh \left(\gamma l\right)$.

$\begin{array}{c}\cosh \left(\gamma l\right)=\frac{e^{\gamma l}+e^{-\gamma l}}{2}\\ =\frac{e^{j\beta l}+e^{-j\beta l}}{2}\end{array}$

Since,$\gamma l=j\beta l$

$\begin{array}{c}\cosh \left(\gamma l\right)=\frac{\cosh \left(\beta l\right)+j\sinh \left(\beta l\right)+\cosh \left(\beta l\right)-j\sin \left(\beta l\right)}{2}\\ =\cos \left(\beta l\right)\end{array}$

Similarly,

Determine the expression of$\sinh \left(\gamma l\right)$.

$\begin{array}{c}\sinh \left(\gamma l\right)=\frac{e^{\gamma l}-e^{-\gamma l}}{2}\\ =\frac{e^{\beta l}-e^{-\beta l}}{2}\\ =\frac{\cos \left(\beta l\right)+j\sin \left(\beta l\right)-\left[\cos \left(\beta l\right)-j\sin \left(\beta l\right)\right]}{2}\\ =j\sin \left(\gamma l\right)\end{array}$

Determine the values of $A$ parameters.

Substitute $\cos \left(\beta l\right)$ for $\cosh \left(\gamma l\right)$ into equation (3).

$\begin{array}{c}A=\cos \left(\beta l\right)\\ =\cos \left(0.3711\right)\text{\hspace{0.33em}radians}\\ =0.9319\text{\hspace{0.17em}pu}\end{array}$

Determine the values of $B$ parameters.

Substitute $274.9$ for$Z_c$ and $\text{0.3711\hspace{0.33em}radians}$ for$\beta l$ into equation (4).

$\begin{array}{c}B=j\left(274.9\right)\sin \left(0.3711\text{\hspace{0.33em}radians}\right)\\ =j\left(274.9\right)\left(0.3626\right)\\ =j99.68\Omega \end{array}$

Determine the value of $C$ parameters.

Substitute $274.9$ for $Z_c$ and $\text{0.3711\hspace{0.17em}radians}$ for $\beta l$into equation (5)

$\begin{array}{c}C=j\left(\frac{1}{274.9}\right)\sin \left(0.3711\text{\hspace{0.17em}radians}\right)\\ =j\left(3.64\times {10}^{-3}\right)\left(0.3626\right)\\ =j1.32\times {10}^{-3}\text{\hspace{0.33em}S}\end{array}$

Determine the values of $D$ parameters.

Both $A$and $D$parameters are same for transmission line therefore,

Substitute $\text{0.9319\hspace{0.17em}pu}$ for $A$ into equation (6).

$D=0.9319\text{\hspace{0.33em}pu}$

Therefore, the values of given $\text{ABCD}$ line parameters are $\text{0.9319\hspace{0.33em}pu}$, $j99.68\Omega$, $j1.32\times {10}^{-3}\text{\hspace{0.33em}S}$ and$\text{0.9319\hspace{0.33em}pu}$ .

(d) Determine the value of wavelength λ  of the line in kilometres.

Determine the value of$\beta$.

$\begin{array}{c}\beta =\frac{0.3711}{300}\\ =1.237\times {10}^{-3}\text{\hspace{0.33em}radians}/\text{km}\end{array}$

Determine thevalue of wavelength $\lambda$ of the line.

Substitute $1.237\times {10}^{-3}$ for $\beta$ into equation (7)

$\begin{array}{c}\lambda =\frac{2\pi }{1.237\times {10}^{-3}}\text{\hspace{0.33em}km}\\ =\text{5079\hspace{0.33em}km}\end{array}$

Therefore, the value of wavelength $\lambda$ of the line in kilometres $\lambda$is$\text{5079\hspace{0.33em}km}$ .

(e) Determine the value of surge impedance loading in MW.

Determine the surge impedance loading in MW.

Substitute${\left(500\times {10}^3\right)}^2$ for $V_{rated}^2$ and $274.9$ for $Z_c$ into equation (8).

$\begin{array}{c}\text{SIL}=\frac{{\left(500\times {10}^3\right)}^2}{274.9}\\ =\frac{250\times {10}^9}{274.9}\\ =909.4\text{\hspace{0.33em}MW}\end{array}$

Therefore, the value of surge impedance loading in MW $\text{SIL}$ is$\text{909.4\hspace{0.33em}MW}$ .