Question

Consider the transmission line shown in Figure 10.48 with series impedance${\mathit{Z}}_{\mathbf{L}}$, negligible shunt admittance, and load impedance ${\mathit{Z}}_{\mathbf{R}}$at the receiving end. (a) Determine ${\mathit{Z}}_{\mathbf{R}}$for the given conditions of ${\mathit{V}}_{\mathbf{R}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathit{p}\mathit{e}\mathit{r}\mathbf{}\mathit{u}\mathit{n}\mathit{i}\mathit{t}$and$\mathbf{}{\mathit{S}}_{\mathbf{R}}\mathbf{=}\mathbf{2}\mathbf{+}\mathit{j}\mathbf{0}\mathbf{.}\mathbf{8}\mathbf{}\mathit{p}\mathit{e}\mathit{r}\mathbf{}\mathit{u}\mathit{n}\mathit{i}\mathit{t}$. (b) Construct the impedance diagram in the R-X plane for ${\mathit{Z}}_{\mathbf{L}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{+}\mathit{j}\mathbf{0}\mathbf{.}\mathbf{3}$. (c) Find$\mathbf{}{\mathit{Z}}_{\mathbf{S}}$for this condition and the angle $\mathit{\delta }$between ${\mathit{Z}}_{\mathbf{S}}$and${\mathit{Z}}_{\mathbf{R}}$.

Figure 10.48

Short answer

Answer

(a) The value of load impedance $Z_R$for the given receiving-end voltage$V_R$is$Z_R=\left(0.431+j0.1724\right)p.u$.

(b) Construct the impedance diagram in the R-X plane for$Z_L=0.1+j0.3$.

(c) The value of total impedance $Z_3$is $0.71\angle 41.66°$and the angle between $Z_L$and $S_R$is$\delta =19.66°$.

Step-by-step solution

Write the given data from the question.

Consider the given following series impedance is $Z_L$of transmission line.

Consider the given following load impedance is $Z_R$of transmission line.

Consider receiving-end voltage of transmission line $V_R=1.0perunit$.

Consider power of transmission line $S_R=2+j0.8perunit$.

Determine the formula of given expression to find  and of long transmission line.

Write the formula of load impedance$Z_R$.

${\mathit{Z}}_{\mathbf{R}}\mathbf{=}{\mathit{R}}_{\mathbf{r}}\mathbf{+}{\mathit{X}}_{\mathbf{r}}$ …… (1)

Here, ${\mathit{R}}_{\mathbf{r}}$is resistive component and ${\mathit{X}}_{\mathbf{r}}$is the reactive component.

Write the formula of total impedance$Z_S$.

${\mathit{Z}}_{\mathbf{S}}\mathbf{=}{\mathit{Z}}_{\mathbf{r}}\mathbf{+}{\mathit{Z}}_{\mathbf{L}}$ …… (2)

Here, ${\mathit{Z}}_{\mathbf{r}}$is load impedance and ${\mathit{Z}}_{\mathbf{L}}$is series impedance.

Write the formula of angle $\delta$ between $Z_L$and$Z_R$.

$\mathit{\delta }\mathbf{=}{\mathit{\theta }}_{\mathbf{s}}\mathbf{-}{\mathit{\theta }}_{\mathbf{R}}$ …… (3)

Here, ${\mathit{\theta }}_{\mathbf{s}}$is the phase angle of impedance ${\mathit{Z}}_{\mathbf{S}}$and ${\mathit{\theta }}_{\mathbf{R}}$is phase angle of impedance${\mathit{Z}}_{\mathbf{r}}$

(a) Determine the load impedance ZR for the given receiving-end voltageVR.

Draw a circuit diagram of following transmission line.

Figure 1

Determine the resistive component of the impedance.

$R_r=\frac{V_r\times R_R}{\left|S\right|}$

Here, $V_R$is receiving-end voltage, $P_R$is receiving power and$\left|S\right|$is mode of receiving-end power.

Substitute 1 for$V_R$, 2 for $P_R$and (2+j0.8) for$\left|S\right|$into above equation.

$$\begin{gathered} R_r=\frac{{\left(1\right)}^2\times 2}{\left(2^2+0.8^2\right)} \\ =0.434p.u. \end{gathered}$$

Determine the reactive component of the impedance.

$X_r=\frac{V_r\times Q_R}{\left|S\right|}$

Substitute 1 for $V_r$, 0.8 for$Q_R$and $2+j0.8$for $\left|S\right|$into above equation.

$$\begin{gathered} X_r=\frac{{\left(1\right)}^2\times 0.8}{\left(2^2+0.8^2\right)} \\ =0.1724p.u. \end{gathered}$$

Now determine the load impedance$Z_R$.

Substitute 0.431 for $R_r$and $X_r$for $X_r$into above equation (1).

$Z_R=\left(0.431+j0.1724\right)p.u$

Therefore, the value of load impedance$Z_R$for the given receiving-end voltage$V_R$is$Z_R=\left(0.431+j0.1724\right)p.u$.

(b) Construct the impedance diagram in the R-X plane.

Draw a circuit diagram of series impedance as shown below.

Figure 2

(c) Determine the total impedance ZS and the angle δ between ZL and ZR.

Determine the value total impedance$Z_S$.

Substitute 0.431 for $Z_r$and j0.1724 for $Z_L$into equation (2).

$$\begin{gathered} Z_S=\left(0.431+j0.1724\right)p.u+\left(0.1+j0.3\right)p.u \\ =\left(0.531+j0.4724\right)p.u \\ =0.71\angle 41.66° \end{gathered}$$

Determine the angle $\delta$between $Z_L$and$Z_R$.

Substitute $41.66°$for ${\theta }_S$and role="math" localid="1656334530300" ${\tan }^{-1}\left(\frac{0.1724}{0.421}\right)$for ${\theta }_R$into equation (3).

$$\begin{gathered} \delta =41.66°-{\tan }^{-1}\left(\frac{0.1724}{0.421}\right) \\ =19.66° \end{gathered}$$

Therefore, the value of total impedance $Z_s$is $0.71\angle 41.66°$and the angle between $Z_L$and $Z_R$is$\delta =19.66°$.