Question

Determine the equivalent circuit for the line in Problem 5.14 and compare it with the nominal circuit.

Short answer

(a) The value of characteristic impedance $Z_c$is$15+j175\Omega$.

(b) The value of $\gamma l$ is$0.0265+j0.6194\text{\hspace{0.33em}pu}$.

(c) The value of $\text{ABCD}$ parameters are $1+\frac{Y{Z}^{\text{'}}}{2}$, $Z^{\text{'}}$, $Y^{\text{'}}\left[1+\frac{Y^{\text{'}}{Z}^{\text{'}}}{4}\right]$ and $1+\frac{Y{Z}^{\text{'}}}{2}$.

Step-by-step solution

Write the given data from the question.

Refer to the problem 5.14.

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 sequence series impedance of given three phase line is$z=0.03+j0.35\Omega /\text{km}$.

The positive sequence shunt impedance of given three phase line is $y=j4.4\times {10}^{-6}\text{\hspace{0.33em}S}/\text{km}$.

Determine the formula of characteristic impedance,   γl and  ABCD parameters.

${\mathbf{Y}}^{\mathbf{\text{'}}}$Write the formula ofvalue of characteristic impedance.

${\mathbf{Z}}_c=\mathbf{zl}$…… (1)

Here, $\mathbf{z}$is impedance and $\mathbf{l}$ is length of line.

Write the formula of$\gamma l$.

$\gamma l=l\sqrt{\mathrm{zy}}$…… (2)

Here, role="math" localid="1656061474081" $\mathbf{z}$ is impedance, $\mathbf{y}$is admittance and $l$is length.

Write the formula of $\mathbf{A}$line parameters of the equivalent $\pi$circuit.

$\mathbf{A}=\mathbf{1}\mathbf{+}\frac{Y^{\text{'}}{Z}^{\text{'}}}{2}$…… (3)

Here, ${\mathbf{Y}}^{\mathbf{\text{'}}}$ is equivalent admittance and ${\mathbf{Z}}^{\mathbf{\text{'}}}$is equivalent impedance.

Write the formula of $\mathbf{B}$line parameters of the equivalent circuit.

$\mathbf{B}={\mathbf{{\rm Z}}}^{\mathbf{\text{'}}}$…… (4)

Here, ${\mathbf{Z}}^{\mathbf{\text{'}}}$ is equivalent impedance.

Write the formula of$\mathbf{C}$line parameters of the equivalent circuit.

$\mathbf{C}={\mathbf{Y}}^{\mathbf{\text{'}}}\left[\mathbf{1}\mathbf{+}\frac{Y^{\text{'}}{Z}^{\text{'}}}{4}\right]$…… (5)

Here, ${\mathbf{Y}}^{\mathbf{\text{'}}}$ is equivalent admittanceand ${\mathbf{Z}}^{\mathbf{\text{'}}}$ is equivalent impedance.

Write the formula of$\mathbf{D}$line parameters of the equivalent circuit.

$\begin{array}{c}\mathbf{D}=\mathbf{A}\\ \mathbf{D}=\mathbf{1}\mathbf{+}\frac{Y^{\text{'}}{Z}^{\text{'}}}{2}\end{array}$…… (6)

Here, ${\mathbf{Y}}^{\mathbf{\text{'}}}$ is equivalent admittance and ${\mathbf{Z}}^{\mathbf{\text{'}}}$ is equivalent impedance.

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

Determine the expression of characteristics impedance$Z_c$.

Substitute $0.03+j0.35$ for $\text{z}$ and $500$ for $l$ into equation (1).

$\begin{array}{c}{Z}_c=\left(0.03+j0.35\right)\left(500\right)\\ =15+j175\Omega \\ =R+jX\end{array}$

Hence, the characteristics impedance is $Z_c$.$15+j175\Omega$

(b) Determine the γl.

Determine the value of $\gamma l$.

Substitute $500$ for $l$, $0.03+j0.35$ for $z$and $j4.4\times {10}^{-6}$for $y$ into equation (2).

$\begin{array}{c}\gamma l=\left(500\right)\left(\sqrt{\left(0.03+j0.35\right)\left(j4.4\times {10}^{-6}\right)}\right)\\ =\left(500\right)\left(\sqrt{\left(0.351\angle 85.1°\right)\left(4.4\times {10}^{-6}\angle 90°\right)}\right)\\ =\left(500\right)\sqrt{1.54\times {10}^{-6}\angle 175.1°}\\ =\left(500\right)\left(1.24\times {10}^{-3}\angle 87.55°\right)\end{array}$

Solve further as

$\begin{array}{c}\gamma l=0.62\angle 87.55°\\ =0.0265+j0.6194\text{\hspace{0.33em}pu}\end{array}$

Hence, the value of $\gamma l$is $0.0265+j0.6194\text{\hspace{0.17em}pu}$.

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

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

$\begin{array}{c}\sinh \left(\gamma l\right)=\sinh \left(0.0265+j0.6194\right)\\ =\frac{e^{0.0265+j0.6194}-e^{-0.0265-j0.6194}}{2}\\ =\frac{e^{0.0265}{e}^{j0.6194}-e^{-0.0265}{e}^{-j0.6194}}{2}\\ =\frac{1.0268\left(0.814+j0.581\right)-0.9738\left(0.814-j0.581\right)}{2}\end{array}$

Solve further as

$\sinh \left(\gamma l\right)=0.582\angle 87.7°$

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

$\begin{array}{c}\cosh \left(\gamma l\right)=\cosh \left(0.0265+j0.6194\right)\\ =\frac{e^{0.0265+j0.6194}+e^{-0.0265-j0.6194}}{2}\\ =\frac{e^{0.0265}{e}^{j0.6194}+e^{-0.0265}{e}^{-j0.6194}}{2}\\ =\frac{1.0268\left(0.814+j0.581\right)+0.9738\left(0.814-j0.581\right)}{2}\end{array}$

Solve further as

$\cosh \left(\gamma l\right)=1.629\angle 1.08$

Determine the value of equivalent impedance

$Z^{\text{'}}=\frac{Z_c\sinh \left(\gamma l\right)}{\left(\gamma l\right)}$

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

Substitute $15+j175\Omega$ for $Z_c$and $0.0265+j0.6194\text{\hspace{0.33em}pu}$ for $\gamma l$into above equation$Z^{\text{'}}$.

$\begin{array}{c}{Z}^{\text{'}}=\frac{\left(15+j175\right)\left(0.582\angle 87.7°\right)}{0.62\angle 87.55°}\\ =13.65+j164.31\Omega \\ =164.88\angle 85.2°\end{array}$

Determine the value of characteristics admittance.

$Y_c=yl$

Here, $y$ is admittance and $l$ is length of line.

Substitute $j4.4\times {10}^{-6}$for$y$ and $500$ for $l$into above equation$Y_c$.

$\begin{array}{c}{Y}_c=\left(j4.4\times {10}^{-6}\right)\left(500\right)\\ =j0.0022\text{\hspace{0.33em}S}\\ =\text{j2.2\hspace{0.33em}mS}\end{array}$

Determine the value of equivalent admittance

$\frac{Y^{\text{'}}}{2}=\frac{G^{\text{'}}+j{B}^{\text{'}}}{2}=\frac{Y_c\tanh \left(\gamma l/2\right)}{2\left(\gamma l/2\right)}$

Here, $Y_c$ is characteristics admittance and $\gamma l$is dimensionless quantity.

Substitute $\text{j2.2\hspace{0.33em}mS}$for $Y_c$ and $0.62\angle 87.55°$for $\gamma l$ into above equation $Y^{\text{'}}$.

$\begin{array}{c}\frac{Y^{\text{'}}}{2}=\left(\frac{Y}{\gamma l}\right)\left(\frac{\cosh \left(\gamma l\right)-1}{\sinh \left(\gamma l\right)}\right)\\ =\left(\frac{j0.0022}{0.62\angle 87.55°}\right)\left(\frac{1.629\angle 1.08°-1}{0.582\angle 87.7°}\right)\\ =5.04\times {10}^{-4}-j3.804\times {10}^{-3}\text{\hspace{0.33em}S}\end{array}$

Draw the circuit diagram of nominal $\pi$circuit.

Draw a circuit diagram of equivalent $\pi$ circuit.

Differentiate the nominal $\pi$circuit and equivalent $\pi$circuit parameters.

The value of resistance $R$ is $15\Omega$ and value of equivalent resistance $R^{\text{'}}$is $13.65\Omega$.

Hence, the value of resistance $R^{\text{'}}$is $9\%$smaller than the nominal resistance $R$1.

The value of reactance $X$is $j175\Omega$ and value of equivalent reactance $X^{\text{'}}$is$j164.31\Omega$

Hence, the value of equivalent reactance $X^{\text{'}}$is $5.94\%$smaller than the reactance$X$.

The value of Susceptance $\frac{B^{\text{'}}}{2}$ is $3.804\times {10}^{-3}\text{\hspace{0.33em}S}$andvalue of admittance $\frac{y}{2}$is $1.1\times {10}^{-3}$

Hence, the value of Susceptance $\frac{B^{\text{'}}}{2}$is $2.46\%$larger than value of admittance $\frac{y}{2}$.

The value of Shunt conductance $\frac{G^{\text{'}}}{2}$is$5.04\times {10}^{-4}\text{\hspace{0.33em}S}$.