Question

A ,$\mathbf{230}\mathbf{kV}\mathbf{,}\mathbf{}\mathbf{100}\mathbf{km}\mathbf{,}\mathbf{}\mathbf{60}\mathbf{Hz}\mathbf{,}$three-phase overhead transmission line with a rated current of $\mathbf{900}\mathbf{A}$per phase has a series impedance $\mathbf{z}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{088}\mathbf{+}\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{465}\frac{\mathbf{\Omega }}{\mathbf{km}}$and a shunt admittance $\mathbf{y}\mathbf{=}\mathbf{j}\mathbf{3}\mathbf{.}\mathbf{524}\frac{\mathbf{\mu S}}{\mathbf{km}}$(a) Obtain the nominal $\mathbf{\pi }$equivalent circuit in normal units and in per unit on a base of 100MVA (three phase) and 230kV (line-to-line). (b) Determine the three-phase rated MVA of the line. (c) Compute the ABCD parameters. (d) Calculate the SIL.

Short answer

(a) The nominal$\pi$network of the transmission is shown below.

(b) The rated three phase MVA of the line is$358.5\mathrm{MVA}$.

(c) The values of parameter of transmission line are

$$\begin{gathered} \mathrm{A}=0.9918\angle 0.1° \\ \mathrm{B}=47.32\angle 73.93°\mathrm{\Omega } \\ \mathrm{C}=0.35\angle 90.04°\mathrm{mS} \\ \mathrm{D}=0.9918\angle 0.1 \end{gathered}$$

(d) The surge impedance loading is $144.3\mathrm{MVA}.$

Step-by-step solution

Write given data from the question:

The voltage of transmission line, ${\mathrm{V}}_{\mathrm{L},\mathrm{Base}}=230\mathrm{kV}$

The Frequency of the system,$\mathrm{f}=60\mathrm{Hz}$

The Current per phase in transmission line, ${\mathrm{l}}_{\mathrm{L},\mathrm{Base}}=900\mathrm{A}$

The Impedance of the transmission line,$\mathrm{z}=0.088+\mathrm{j}0.465\frac{\mathrm{\Omega }}{\mathrm{km}}|$

The admittance of transmission line,$\mathrm{y}=\mathrm{j}3.524\frac{\mathrm{\mu S}}{\mathrm{km}}$

Length of the transmission line,$\mathrm{l}=100\mathrm{km}$

Base MVA,$\mathrm{S}=100\mathrm{MVA}$

Determine the formulas to calculate the nominal equivalent circuit, three phase base rated MVA, ABCD parameter and SIL.

The equation to calculate the total impedance of the line is given as follows.

$\mathbf{Z}\mathbf{=}\mathbf{zl}$ …… (1)

The equation to calculate the total shunt admittance of transmission line is given as follows.

$\frac{\mathrm{Y}}{2}=\frac{\mathrm{yl}}{2}$ …… (2)

The equation to calculate the base impedance of the system is given as follows.

${\mathrm{Z}}_{\mathrm{base}}=\frac{{\mathrm{V}}_{\mathrm{L},\mathrm{Base}}^2}{{\mathrm{S}}_{3\mathrm{\varphi Base}}}$ …… (3)

Here, ${\mathrm{V}}_{\mathrm{L},\mathrm{Base}}$is the line-to-line base voltage and${\mathrm{S}}_{3\mathrm{\varphi Base}}$ is the base MVA of the system.

The equation to calculate the per unit value is given as follows.

${\mathrm{Z}}_{\mathrm{p}.\mathrm{u}}=\frac{\mathrm{Z}}{{\mathrm{Z}}_{\mathrm{base}}}$ …… (4)

The equation to calculate the per unit admittance is given as follows.

${\mathrm{Y}}_{\mathrm{p}.\mathrm{u}}=\frac{{\displaystyle \frac{{\displaystyle \frac{\mathrm{Y}}{2}}}{1}}}{{\mathrm{Z}}_{\mathrm{p}.\mathrm{u}}}$ …… (5)

The equation to calculate the base MVA of the system is given as follows.

${\mathrm{S}}_{3\mathrm{f},\mathrm{rated}}=\sqrt{3}{\mathrm{V}}_{\mathrm{L},\mathrm{Base}}|\mathrm{L},\mathrm{base}$ …… (6)

The equations to calculate the parameter of the transmission line are given as follows.

localid="1655457770270" $$\begin{gathered} \mathrm{A}=\mathrm{D} \\ \mathrm{A}=1+\left(\frac{\mathrm{ZY}}{2}\right)\dots \dots \left(7a\right) \\ \mathrm{B}=\mathrm{Z}\dots \dots \left(7\mathrm{b}\right) \\ \mathrm{C}=\mathrm{Y}\left(1+\frac{\mathrm{YZ}}{4}\right)\dots \dots \left(7\mathrm{c}\right) \end{gathered}$$

The equation to calculate the characteristics impedance is given as follows.

${\mathrm{Z}}_{\mathrm{c}}=\sqrt{\frac{\mathrm{Z}}{\mathrm{y}}}$ …… (8)

The equation to calculate the surge impedance loading is given as follows.

$\mathrm{SIL}=\frac{{\left({\mathrm{V}}_{\mathrm{L},\mathrm{Base}}\right)}^2}{{\mathrm{Z}}_{\mathrm{c}}}$ …… (9)

Obtain the nominal equivalent circuit in normal units and in per unit on a base of (three phase) and (line-to-line).

(a)

Calculate the total impedance of transmission line.

$$\begin{gathered} \mathrm{Substitute}100\mathrm{km}\mathrm{for}\mathrm{l}\mathrm{and}0.088+\mathrm{j}0.465\frac{\mathrm{\Omega }}{\mathrm{km}}\mathrm{for}\mathrm{z}\mathrm{into}\mathrm{equation}\left(1\right). \\ \mathrm{Z}=(0.088+\mathrm{j}0.465)\times 100 \\ \mathrm{Z}=8.8+\mathrm{j}46.5\mathrm{\Omega } \\ \mathrm{Calculate}\mathrm{the}\mathrm{total}\mathrm{admittance}\mathrm{of}\mathrm{transmission}\mathrm{line}. \\ \mathrm{Substitute}\mathrm{j}3.524\frac{\mathrm{\mu S}}{\mathrm{km}}\mathrm{for}\mathrm{y}\mathrm{and}100\mathrm{km}\mathrm{for}\mathrm{l}\mathrm{into}\mathrm{equation}\left(2\right). \\ \frac{\mathrm{Y}}{2}=\frac{\mathrm{j}3.524\times {10}^{-6}\times 100\times {10}^3}{2} \\ \frac{\mathrm{Y}}{2}=\mathrm{j}0.1762\mathrm{mS} \\ \mathrm{Calculate}\mathrm{the}\mathrm{base}\mathrm{impedance}. \\ \mathrm{Substitute}230\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{L}.\mathrm{Base}}\mathrm{and}100\mathrm{MVA}\mathrm{for}{\mathrm{S}}_{3\mathrm{\varphi },\mathrm{Base}}\mathrm{into}\mathrm{equation}\left(3\right). \\ {\mathrm{Z}}_{\mathrm{base}}=\frac{{230}^2}{100} \\ {\mathrm{Z}}_{\mathrm{base}}=529\mathrm{\Omega } \\ \mathrm{Calculate}\mathrm{per}\mathrm{unit}\mathrm{impedance}. \\ \mathrm{Substitute}8.8+\mathrm{j}46.5\mathrm{\Omega }\mathrm{for}\mathrm{Z}\mathrm{and}529\mathrm{\Omega }\mathrm{for}{\mathrm{Z}}_{\mathrm{base}}\mathrm{into}\mathrm{equation}\left(4\right). \\ {\mathrm{Z}}_{\mathrm{p}.\mathrm{u}}=\frac{8.8+46.5}{529} \\ {\mathrm{Z}}_{\mathrm{p}.\mathrm{u}}=0.0166+\mathrm{j}0.088\mathrm{pu} \\ \mathrm{Calculate}\mathrm{the}\mathrm{per}\mathrm{unit}\mathrm{admittance}. \\ \mathrm{Substitute}\mathrm{j}0.1762\mathrm{S}\mathrm{for}\frac{\mathrm{Y}}{2}\mathrm{and}0.0166+\mathrm{j}0.088\mathrm{pu}\mathrm{for}{\mathrm{Z}}_{\mathrm{pu}}\mathrm{into}\mathrm{equation}\left(5\right). \\ {\left(\frac{\mathrm{Y}}{2}\right)}_{\mathrm{pu}}=\frac{{\displaystyle \frac{\mathrm{j}0.1762}{1}}}{0.529} \\ {\left(\frac{\mathrm{Y}}{2}\right)}_{\mathrm{pu}}=\mathrm{j}0.09321\mathrm{pu} \\ \mathrm{The}\mathrm{nominal}\mathrm{\pi }\mathrm{network}\mathrm{of}\mathrm{the}\mathrm{transmission}\mathrm{is}\mathrm{shown}\mathrm{below}. \end{gathered}$$

Calculate the three-phase rated MVA of the line.

(b)

Calculate rated three phase MVA of the line.

$$\begin{gathered} \mathrm{Substitute}230\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{L},\mathrm{Base}}\mathrm{and}900\mathrm{A}\mathrm{for}{\mathrm{l}}_{\mathrm{A},\mathrm{Base}}\mathrm{into}\mathrm{equation}\left(6\right). \\ {\mathrm{S}}_{3\mathrm{\varphi },\mathrm{Base}}=\sqrt{3}\times 230\times {10}^3\times 900 \\ {\mathrm{S}}_{3\mathrm{\varphi },\mathrm{Base}}=358.5\mathrm{MVA} \\ \mathrm{Hence},\mathrm{the}\mathrm{rated}\mathrm{three}\mathrm{phase}\mathrm{MVA}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{is}358.5\mathrm{MVA}. \end{gathered}$$

calculate the ABCD parameter of the transmission line.

(c)

Calculate the A and D parameter.

$$\begin{gathered} \mathrm{Substitute}8.8+\mathrm{j}46.5\mathrm{\Omega }\mathrm{for}\mathrm{Z}\mathrm{and}\mathrm{j}0.1762\mathrm{mS}\mathrm{for}\frac{\mathrm{Y}}{2}\mathrm{into}\mathrm{equation}\left(7\mathrm{a}\right). \\ \mathrm{A}=\mathrm{D} \\ \mathrm{A}=1+\left(8.8+\mathrm{J}46.5\right)\left(\mathrm{j}0.1762\times {10}^{-3}\right) \\ \mathrm{A}=0.9918\angle 0.1° \\ \mathrm{Calculate}\mathrm{B}\mathrm{parameter}\mathrm{of}\mathrm{transmission}\mathrm{line}. \\ \mathrm{Substitute}8.8+\mathrm{j}46.5\mathrm{\Omega }\mathrm{for}\mathrm{Z}\mathrm{into}\mathrm{equation}\left(7\mathrm{b}\right). \\ \mathrm{B}=8.8+\mathrm{j}46.5 \\ \mathrm{B}=47.32\angle 73.93°\mathrm{\Omega } \\ \mathrm{Calculate}\mathrm{the}\mathrm{parameter}\mathrm{C}, \\ \mathrm{Substitute}8.8+\mathrm{j}46.5\mathrm{\Omega }\mathrm{for}\mathrm{Z}\mathrm{and}\mathrm{j}0.1762\mathrm{mS}\mathrm{for}\frac{\mathrm{Y}}{2}\mathrm{into}\mathrm{equation}\left(7\mathrm{c}\right). \\ \mathrm{C}=2\times \mathrm{j}0.1762\left(1+\frac{\left(8.8+\mathrm{j}46.5\right)\left(\mathrm{j}0.1762\times {10}^{-3}\right)}{2}\right) \\ \mathrm{Hence},\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{parameter}\mathrm{of}\mathrm{transmission}\mathrm{line}\mathrm{are} \\ \mathrm{A}=0.9918\angle 0.1° \\ \mathrm{B}=47.32\angle 73.93°\mathrm{\Omega } \\ \mathrm{C}=0.35\angle 90.04°\mathrm{mS} \\ \mathrm{D}=0.9918\angle 0.1 \end{gathered}$$

Calculate the surge impedance loading.

(d)

Calculate characteristics impedance.

$$\begin{gathered} \mathrm{Substitute}0.088+\mathrm{j}0.465\frac{\mathrm{\Omega }}{\mathrm{km}}\mathrm{for}\mathrm{z}\mathrm{and}\mathrm{j}3.524\frac{\mathrm{\mu S}}{\mathrm{km}}\mathrm{for}\mathrm{y}\mathrm{into}\mathrm{equation}\left(8\right) \\ {\mathrm{z}}_{\mathrm{c}}=\sqrt{\frac{0.088+\mathrm{j}0.465}{{\mathrm{j}}_3.524}} \\ {\mathrm{z}}_{\mathrm{c}}=366.4\times {10}^3\mathrm{\Omega } \\ \mathrm{Calculate}\mathrm{Surge}\mathrm{impedance}\mathrm{loading}. \\ \mathrm{Substitute}366.4\times {10}^3\mathrm{\Omega }\mathrm{for}{\mathrm{Z}}_{\mathrm{c}}\mathrm{and}230\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{L},\mathrm{Base}}\mathrm{into}\mathrm{equation}\left(9\right). \\ \mathrm{SIL}=\frac{{\left(230\times {10}^3\right)}^2}{366.4\times {10}^3\mathrm{\Omega }} \\ \mathrm{SIL}=144.3\mathrm{MVA} \end{gathered}$$