Question

Question: Rework Example 13.6 if the overhead line and cable are interchanged. That is,${\mathit{Z}}_{\mathbf{A}}\mathbf{=}\mathbf{100}\mathbf{}\mathit{\Omega }$ ,${\mathit{v}}_{\mathbf{A}}\mathbf{=}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathbf{}\frac{\mathbf{m}}{\mathbf{s}}$ , ${\mathit{l}}_{\mathbf{A}}\mathbf{=}\mathbf{20}\mathbf{}\mathit{k}\mathit{m}$, ${\mathit{Z}}_{\mathbf{B}}\mathbf{=}\mathbf{400}\mathbf{}\mathit{\Omega }$,${\mathit{v}}_{\mathbf{B}}\mathbf{=}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathbf{}\frac{\mathbf{m}}{\mathbf{s}}$ , and ${\mathit{l}}_{\mathbf{B}}\mathbf{=}\mathbf{30}\mathbf{}\mathit{k}\mathit{m}$. The step voltage source ${\mathit{e}}_{\mathbf{G}}\left(t\right)\mathbf{=}\mathit{E}{\mathit{u}}_{\mathbf{-}\mathbf{1}}\left(t\right)$is applied to the sending end of line with ${\mathit{Z}}_{\mathbf{R}}\mathbf{=}\mathbf{2}{\mathit{Z}}_{\mathbf{B}}\mathbf{=}\mathbf{800}\mathbf{}\mathit{\Omega }$, and ${\mathit{Z}}_{\mathbf{G}}\mathbf{=}{\mathit{Z}}_{\mathbf{A}}\mathbf{=}\mathbf{100}\mathbf{}\mathit{\Omega }$, at the receiving end. Draw the lattice diagram for $\mathbf{0}\mathbf{⩽}\mathit{t}\mathbf{⩽}\mathbf{0}\mathbf{.}\mathbf{6}\mathbf{}\mathit{m}\mathit{s}$and plot the junction voltage versus time t.

Short answer

Answer

The Bewley lattice diagram.

The graph between and time .

Step-by-step solution

Write the given data from the question.

The source voltage,$e_G\left(t\right)=E{u}_{-1}\left(t\right)$

The data for single phase losses line A.

The source impedance,$Z_G=100\Omega$

Line impedance,$Z_A=100\Omega$

Velocity of the wave,$v_A=2\times {10}^8\frac{\text{m}}{\text{s}}$

The length of the line,$L_A=20\text{km}$

The data for single phase losses line B.

Receiving end impedance,$Z_R=800\Omega$$Z_R=800\Omega$

Cable impedance,$Z_B=800\Omega$

Velocity of the wave,$v_B=3\times {10}^8\frac{\text{m}}{\text{s}}$

The length of the line,$L_B=30\text{km}$

Determine the equations to draw the Bewley lattice diagram.

The equation to calculate the time taken by the wave to transverse the length of line A is given as follows,

${\mathit{\tau }}_{\mathbf{A}}\mathbf{=}\frac{{\mathbf{L}}_{\mathbf{A}}}{{\mathbf{v}}_{\mathbf{A}}}$ …… (1)

The equation to calculate the time taken by the wave to transverse the length of cable B is given as follows,

${\mathit{\tau }}_{\mathbf{B}}\mathbf{=}\frac{{\mathbf{L}}_{\mathbf{B}}}{{\mathbf{v}}_{\mathbf{B}}}$ …… (2)

The equation to calculate the reflection coefficient for the waves at the sending end is given as follows.

${\mathit{\Gamma }}_{\mathbf{S}}\mathbf{=}\frac{\frac{{\mathbf{Z}}_{\mathbf{G}}}{{\mathbf{Z}}_{\mathbf{A}}}\mathbf{-}\mathbf{1}}{\frac{{\mathbf{Z}}_{\mathbf{G}}}{{\mathbf{Z}}_{\mathbf{A}}}\mathbf{+}\mathbf{1}}$ …… (3)

The equation to calculate the reflection coefficient for the waves at the receiving end is given as follows.

${\mathit{\Gamma }}_{\mathbf{R}}\mathbf{=}\frac{\frac{{\mathbf{Z}}_{\mathbf{R}}}{{\mathbf{Z}}_{\mathbf{B}}}\mathbf{-}\mathbf{1}}{\frac{{\mathbf{Z}}_{\mathbf{B}}}{{\mathbf{Z}}_{\mathbf{B}}}\mathbf{+}\mathbf{1}}$ …… (4)

The equation to calculate the reflection coefficient for the waves arriving at the junction from line A is given as follows.

${\mathit{\Gamma }}_{\mathbf{A}\mathbf{A}}\mathbf{=}\frac{\frac{{\mathbf{Z}}_{\mathbf{B}}}{{\mathbf{Z}}_{\mathbf{A}}}\mathbf{-}\mathbf{1}}{\frac{{\mathbf{Z}}_{\mathbf{B}}}{{\mathbf{Z}}_{\mathbf{A}}}\mathbf{+}\mathbf{1}}$ …… (5)

The equation to calculate the refraction coefficient for the waves arriving at the junction from line A is given as follows.

${\mathit{\Gamma }}_{\mathbf{B}\mathbf{A}}\mathbf{=}\frac{\mathbf{2}\frac{{\mathbf{Z}}_{\mathbf{B}}}{{\mathbf{Z}}_{\mathbf{A}}}}{\frac{{\mathbf{Z}}_{\mathbf{B}}}{{\mathbf{Z}}_{\mathbf{A}}}\mathbf{+}\mathbf{1}}$ …… (6)

The equation to calculate the reflection coefficient for the waves returning at the junction from cable B is given as follows.

${\mathit{\Gamma }}_{\mathbf{B}\mathbf{B}}\mathbf{=}\frac{\frac{{\mathbf{Z}}_{\mathbf{A}}}{{\mathbf{Z}}_{\mathbf{B}}}\mathbf{-}\mathbf{1}}{\frac{{\mathbf{Z}}_{\mathbf{A}}}{{\mathbf{Z}}_{\mathbf{B}}}\mathbf{+}\mathbf{1}}$ …… (7)

The equation to calculate the reflection coefficient for the waves returning at the junction from cable B is given as follows.

${\mathit{\Gamma }}_{\mathbf{A}\mathbf{B}}\mathbf{=}\frac{\mathbf{2}\frac{{\mathbf{Z}}_{\mathbf{A}}}{{\mathbf{Z}}_{\mathbf{B}}}}{\frac{{\mathbf{Z}}_{\mathbf{A}}}{{\mathbf{Z}}_{\mathbf{B}}}\mathbf{+}\mathbf{1}}$ …… (8)

Th equation to calculate the voltage wave reaching the junction from A is given as follows.

${\mathit{V}}_{\mathbf{s}\mathbf{s}}\left(s\right)\mathbf{=}{\mathit{E}}_{\mathbf{G}}\left(s\right)\left(\frac{Z_A}{Z_A+Z_G}\right)$ …… (9)

Draw the Bewley lattice diagram.

Calculate the time taken by the wave to transverse the length of line A.

${\text{Substitute20kmforL}}_{A}and2\times {10}^8\frac{m}{s}for{V}_{A}intoequation\left(1\right).$

$$\begin{gathered} {\tau }_A=\frac{20\times {10}^3}{2\times {10}^8} \\ {\tau }_A=0.1\times {10}^{-3}\text{s} \end{gathered}$$

Calculate the time taken by the wave to transverse the length of line B.

${\text{Substitute30kmforL}}_{A}and3\times {10}^8\frac{m}{s}for{V}_{A}intoequation\left(2\right).$

$$\begin{gathered} {\tau }_B=\frac{30\times {10}^3}{3\times {10}^8} \\ {\tau }_B=0.1\times {10}^{-3}\text{s} \end{gathered}$$

Calculate the reflection coefficient for the waves at the sending end

).${\text{Substitute100ΩforZ}}_{G}and{Z}_{A}intoequation\left(3\right).$

localid="1656308253451" $$\begin{gathered} {\Gamma }_S=\frac{\frac{100}{100}-1}{\frac{100}{100}+1} \\ {\Gamma }_S=\frac{1-1}{1+1} \\ {\Gamma }_S=0 \end{gathered}$$$$\begin{gathered} {\Gamma }_S=\frac{\frac{100}{100}-1}{\frac{100}{100}+1} \\ {\Gamma }_S=\frac{1-1}{1+1} \\ {\Gamma }_S=0 \end{gathered}$$

Calculate the reflection coefficient for the waves at the receiving end.

localid="1656308259242" ${\text{Substitute800ΩforZ}}_{R}and400\Omega for{Z}_{AB}intoequation\left(4\right).$

localid="1656308264093" $$\begin{gathered} {\Gamma }_S=\frac{\frac{100}{100}-1}{\frac{100}{100}+1} \\ {\Gamma }_S=\frac{1-1}{1+1} \\ {\Gamma }_S=0 \end{gathered}$$

Calculate the refraction coefficient for the waves arriving at the junction from line A.

localid="1656308271534" ${\text{Substitute100ΩforZ}}_{A}and400\Omega for{Z}_{B}intoequation\left(5\right).$

localid="1656308288194" $$\begin{gathered} {\Gamma }_{AA}=\frac{\frac{400}{100}-1}{\frac{400}{100}+1} \\ {\Gamma }_{AA}=\frac{4-1}{4+1} \\ {\Gamma }_{AA}=\frac{3}{5} \\ {\Gamma }_{AA}=0.6 \end{gathered}$$

Calculate the reflection coefficient for the waves returning at the junction from cable B.

localid="1656308330702" ${\text{Substitute100ΩforZ}}_{A}and400\Omega for{Z}_{B}intoequation\left(6\right).$

localid="1656308337208" $$\begin{gathered} {\Gamma }_{BA}=\frac{2\left(\frac{400}{100}\right)}{\frac{400}{100}+1} \\ {\Gamma }_{BA}=\frac{8}{4+1} \\ {\Gamma }_{BA}=\frac{8}{5} \\ {\Gamma }_{BA}=1.6 \end{gathered}$$

Calculate the reflection coefficient for the waves returning at the junction from cable B.

localid="1656308321789" ${\text{Substitute100ΩforZ}}_{A}and400\Omega for{Z}_{B}intoequation\left(7\right).$

localid="1656308359277" $$\begin{gathered} {\Gamma }_{BB}=\frac{\frac{100}{400}-1}{\frac{100}{400}+1} \\ {\Gamma }_{BB}=\frac{\frac{1}{4}-1}{\frac{1}{4}+1} \\ {\Gamma }_{BB}=-\frac{3}{5} \\ {\Gamma }_{BB}=-0.6 \end{gathered}$$

Calculate the voltage wave reaching the junction from A.

localid="1656308346654" ${\text{Substitute100ΩforZ}}_{A}and400\Omega for{Z}_{B}intoequation\left(8\right).$

localid="1656308372667" $$\begin{gathered} {\Gamma }_{BA}=\frac{2\left(\frac{400}{100}\right)}{\frac{400}{100}+1} \\ {\Gamma }_{BA}=\frac{8}{4+1} \\ {\Gamma }_{BA}=\frac{8}{5} \\ {\Gamma }_{BA}=1.6 \end{gathered}$$

Substitute for , for and into equation (9).localid="1656308352983" $Substitute\frac{E}{S}for{E}_G,100for{Z}_{A}and{Z}_{B}intoequation\left(9\right).$

localid="1656308365409" $$\begin{gathered} V_1\left(s\right)=\frac{E}{s}\left(\frac{100}{100+100}\right) \\ {V}_1\left(s\right)=\frac{E}{s}\left(\frac{100}{200}\right) \\ {V}_1\left(s\right)=\frac{E}{2s} \end{gathered}$$

Calculate the reflected wave at junction A.

localid="1656308380144" $V_2\left(s\right)={\Gamma }_{AA}{V}_1\left(s\right)$

Substitute for and for into above equation.

localid="1656308387365" $$\begin{gathered} V_2\left(s\right)=\left(0.6\right)\left(\frac{E}{2s}\right) \\ {V}_2\left(s\right)=\frac{3E}{10s} \end{gathered}$$

Calculate the refracted wave at junction A.

localid="1656308394994" $V_3\left(s\right)={\Gamma }_{BA}{V}_1\left(s\right)$

Substitute for and for into above equation.

localid="1656307380097">V3s=1.6E2sV3s=4E5s

Calculate the reflected wave at receiving end.

$V_4\left(s\right)={\Gamma }_R{V}_3\left(s\right)$

Substitute for and for into above equation.

$$\begin{gathered} V_4\left(s\right)=\left(0.333\right)\left(\frac{4E}{5s}\right) \\ {V}_4\left(s\right)=\frac{4E}{15s} \end{gathered}$$

Calculate the refracted wave at coming from receiving end.

$V_5\left(s\right)={\Gamma }_{AB}{V}_4\left(s\right)$

Substitute for and for into above equation.

$$\begin{gathered} V_5\left(s\right)=-0.6\left(\frac{4E}{15s}\right) \\ {V}_5\left(s\right)=-\frac{4E}{25s} \end{gathered}$$

Draw the Bewley lattice diagram.\

Calculate the steady state value of the voltage .

$V_{ss}\left(s\right)=E_G\left(\frac{Z_R}{Z_R+Z_G}\right)$

$Substitute\frac{E}{S}for{E}_G,800\Omega for{Z}_{R}and100\Omega for{Z}_{G}intoequation\left(9\right).$

$$\begin{gathered} V_{ss}\left(s\right)=\frac{E}{s}\left(\frac{800}{800+100}\right) \\ {V}_{ss}\left(s\right)=\frac{E}{s}\left(\frac{800}{900}\right) \\ {v}_{ss}\left(s\right)=\frac{8E}{9s} \end{gathered}$$

The final value reaches to zero.

Plot the graph between and time .