Question

Let the transmission line of Problem 5.12 be extended to cover a distance of $\mathbf{200}\mathbf{miles}$. Assume conditions at the load to be the same as in Problem 5.12. Determine the (a) sending-end voltage, (b) sending-end current, (c) sending-end real and reactive powers, and (d) percent voltage regulation.

Short answer

(a) The value of sending end voltage $V_S$ is$159.9\angle 29.6°\text{\hspace{0.33em}kV}$.

(b) The value of sending end current $I_S$is$\text{482.93\hspace{0.33em}A}$.

(c) The value of sending-end real power $P_{\text{S},3\varphi }$is $\text{227\hspace{0.33em}MW}$ and reactive power is$Q_{\text{S,3}\varphi }$ is$\text{47\hspace{0.33em}MVAR}$.

(d) The value of percentage voltage regulation of the transmission line is$20.4\%$ .

Step-by-step solution

Write the given data from the question.

The length of given transmission line is$\text{200\hspace{0.33em}miles}$.

The line-to-line voltage of given transmission line is$230-\text{kV}$.

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

The shunt capacitive reactance-to-neutral of$y$ is$185.5\times {10}^3\angle -90°$$185.5\times {10}^3\angle -90°$.

The series impedance transmission line$0.1826+j0.784\Omega /\text{mi}$.

Base MVA of transmission line is $\text{200\hspace{0.33em}MVA}$.

Determine the formula of sending-end voltage, sending-end current, sending-end real and reactive power and percent voltage regulation.

Write the formula ofsending end voltage$V_S$.

${\mathbf{V}}_S={\mathbf{V}}_R\mathbf{+}{\mathbf{I}}_Z{\mathbf{Z}}^{\mathbf{\text{'}}}$…… (1)

Here, ${\mathbf{V}}_R$ is receiving-end voltage, ${\mathbf{I}}_Z$ is impedance current and ${\mathbf{Z}}^{\mathbf{\text{'}}}$ is impedance.

Write the formula ofsending end current$I_S$.

${\mathbf{I}}_S={\mathbf{I}}_Z\mathbf{+}{\mathbf{V}}_S\left(\frac{Y^{\text{'}}}{2}\right)$…… (2)

Here, ${\mathbf{I}}_Z$is impedance current, ${\mathbf{V}}_S$is voltage source and ${\mathbf{Y}}^{\mathbf{\text{'}}}$is admittance.

Write the formula ofsending-end real power and reactive power.

localid="1655894719973" $S_{\mathbf{S}\mathbf{,}\mathbf{3}\varphi }=\mathbf{3}\left({\mathbf{V}}_s\right)\left({\mathbf{I}}_s^{*}\right)$…… (3)

Here,${\mathbf{V}}_S$is voltage source and $I_s^{*}$ is sending-end current.

Write the formula ofof percentage voltage regulation of the transmission line.

localid="1655894805725" $\mathbf{\%}\mathbf{VR}=\left(\frac{{\mathbf{V}}_S\mathbf{-}{\mathbf{V}}_R}{{\mathbf{V}}_R}\right)\left(100\right)$…… (4)

Here, ${\mathbf{V}}_S$ is voltage source and ${\mathbf{V}}_R$is receiving-end voltage.

Step 3:(a) Determine the sending end voltage.

$\begin{array}{c}{F}_2=\frac{0.4038\angle 83.82°}{\left(0.2083\angle 83.44°\right)\left(1.916\angle 0.59°\right)}\\ =\frac{0.4038\angle 83.82°}{0.399\angle 84.04°}\\ =1.012\angle -0.22°\end{array}$

Determine the length of the transmission line in kilo-meters.

localid="1656761120549" $\begin{array}{c}l=\text{200\hspace{0.33em}miles}\\ =\left(1.6093\right)\left(200\right)\text{\hspace{0.33em}km}\\ =321.86\text{\hspace{0.33em}km}\end{array}$

Therefore, the length of the transmission line is greater thanlocalid="1656761124754" $\text{250\hspace{0.33em}km}$, it fails in the category of long transmission line.

Determine the shunt admittance .

localid="1656761129713" $y=\frac{1}{x_c\angle -90°}$

Here, localid="1656761135515" $x_c$is the capacitance reactance-to-neutral.

Substitutelocalid="1656761139677" $185.5\times {10}^3\angle -90°\Omega -\text{mi\hspace{0.17em}per\hspace{0.33em}phase}$ forlocalid="1656761145022" $x_c$
into above equation .

localid="1656761151388" $\begin{array}{c}y=\frac{1}{185.5\times {10}^3\angle -90°}\\ =5.391\times {10}^{-6}\angle 90°\text{\hspace{0.33em}S\hspace{0.33em}per\hspace{0.33em}mile\hspace{0.33em}per\hspace{0.33em}phase}\end{array}$

Determine the total impedance.

localid="1656761156604" $\begin{array}{c}Z=zl\\ =\left(0.1826+j0.784\Omega /\text{mile\hspace{0.33em}per\hspace{0.33em}phase}\right)\left(200\text{\hspace{0.33em}miles}\right)\\ =36.52+j156.8\Omega \text{\hspace{0.33em}per\hspace{0.33em}phase}\\ =160.99\angle 76.89°\Omega \text{\hspace{0.33em}per\hspace{0.33em}phase}\end{array}$

Determine the total shunt admittance.

localid="1656761162448" $\begin{array}{c}Y=yl\\ =\left(5.391\times {10}^{-6}\angle 90°\right)\left(200\right)\\ =1.078\times {10}^{-6}\angle 90°\text{\hspace{0.33em}S}\end{array}$

Consider the circuit diagram of equivalent long transmission line $\pi$ circuit.

Determine the propagation constant.

$\begin{array}{c}\gamma =\sqrt{zy}\\ =\sqrt{\left(0.1826+j0.784\right)\left(j5.391\times {10}^{-6}\right)}\\ =\sqrt{4.34\times {10}^{-6}\angle 166.88°}\\ =2.083\times {10}^{-3}\angle 83.44°\end{array}$

Determine the $\gamma l$.

$\begin{array}{c}\gamma l=\left(2.083\times {10}^{-3}\angle 83.44°\right)\left(200\right)\\ =0.4166\angle 83.44°\\ =0.0476+j0.414\end{array}$

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

$\begin{array}{c}\cosh \left(\gamma l\right)=\frac{e^{\gamma l}+e^{-\gamma l}}{2}\\ =\frac{\left(e^{0.048}\right)\left(e^{j0.414radians}\right)+\left(e^{-0.048}\right)\left(e^{-j0.414radians}\right)}{2}\\ =\frac{\left(e^{0.048}\right)\left[\begin{array}{l}\cos \left(0.414\right)+\\ j\sin \left(0.414\right)\end{array}\right]+\left(e^{-0.048}\right)\left[\begin{array}{l}\cos \left(0.414\right)-\\ j\sin \left(0.414\right)\end{array}\right]}{2}\\ =\frac{\left(1.049\right)\left[0.9155+j0.402\right]+\left(0.953\right)\left(0.9155\_j0.402\right)}{2}\end{array}$

Solve further as

$\begin{array}{c}\cosh \left(\gamma l\right)=\frac{0.96+j0.42+0.872-j0.38}{2}\\ =\frac{1.832+j0.04}{2}\\ =0.916+j0.02\end{array}$

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

$\begin{array}{c}\sinh \left(\gamma l\right)=\frac{e^{\gamma l}-e^{-\gamma l}}{2}\\ =\frac{\left(e^{0.048}\right)\left(e^{j0.414\text{\hspace{0.33em}radians}}\right)-\left(e^{-0.048}\right)\left(e^{-j0.414\text{\hspace{0.33em}radians}}\right)}{2}\\ =\frac{\left(e^{0.048}\right)\left[\begin{array}{l}\cos \left(0.414\right)+\\ j\sin \left(0.414\right)\end{array}\right]-\left(e^{-0.048}\right)\left[\begin{array}{l}\cos \left(0.414\right)-\\ j\sin \left(0.414\right)\end{array}\right]}{2}\\ =\frac{\left(1.049\right)\left[0.9155+j0.402\right]-\left(0.953\right)\left(0.9155-j0.402\right)}{2}\end{array}$

Solve further as

$\begin{array}{c}\sinh \left(\gamma l\right)=\frac{0.96+j0.42-0.872+j0.38}{2}\\ =\frac{0.087+j0.803}{2}\\ =0.0435+j0.4015\end{array}$

Determine the correction factors$F_1$ .

$\begin{array}{c}{F}_1=\frac{\sinh \left(\gamma l\right)}{\left(\gamma l\right)}\\ =\frac{0.0435+j0.4015}{0.0476+j0.414}\\ =0.97+j0.006\\ =0.97\angle 0.37°\end{array}$

Determine the correction factors $F_2$.

$\begin{array}{c}{F}_2=\frac{\tanh \left(\frac{\gamma l}{2}\right)}{\left(\frac{\gamma l}{2}\right)}\\ =\frac{\sinh \left(\gamma l\right)}{\left(\frac{\gamma l}{2}\right)\left(1+\cosh \left(\gamma l\right)\right)}\\ =\frac{0.0435+j0.4015}{\left(\frac{0.4166\angle 83.44°}{2}\right)\left(1+0.916+j0.02\right)}\\ =\frac{0.4038\angle 83.82°}{\left(0.2083\angle 83.44°\right)\left(1.916\angle 0.59°\right)}\end{array}$

Solve further as

$\begin{array}{c}{F}_2=\frac{0.4038\angle 83.82°}{\left(0.2083\angle 83.44°\right)\left(1.916\angle 0.59°\right)}\\ =\frac{0.4038\angle 83.82°}{0.399\angle 84.04°}\\ =1.012\angle -0.22°\end{array}$

Determine the value of$Z^{\text{'}}$.

$\begin{array}{c}{Z}^{\text{'}}=Z{F}_1\\ =\left(160.99\angle 76.89°\right)\left(0.97\angle 0.37°\right)\\ =156.16\angle 77.26°\Omega \end{array}$

Determine the value of $\frac{Y^{\text{'}}}{2}$.

$\begin{array}{c}\frac{Y^{\text{'}}}{2}=\left(\frac{Y}{2}\right)F_2\\ =\left(\frac{1.078\times {10}^{-3}}{2}\angle 90°\right)\left(1.012\angle -0.22°\right)\\ =\left(0.539\times {10}^{-3}\angle 90°\right)\left(1.012\angle -0.22°\right)\\ =0.545\times {10}^{-3}\angle 89.78°\text{\hspace{0.33em}S}\end{array}$

Determine the current through impedance, .$Z^{\text{'}}$

$\begin{array}{c}{I}_{Z^{\text{'}}}=I_{\text{R}}+V_{\text{R}}\left(\frac{Y^{\text{'}}}{2}\right)\\ =\left(\frac{200\times {10}^6}{\sqrt{3}\left(230\times {10}^3\right)}\right)+\left(\frac{230\times {10}^3}{\sqrt{3}}\right)\left(0.545\times {10}^{-3}\angle 89.81°\right)\\ =\left(\frac{200\times {10}^6}{398.37\times {10}^3}\right)+\left(132.79\times {10}^3\right)\left(0.545\times {10}^{-3}\angle 89.81°\right)\\ =502.04+72.37\angle 89.81°\end{array}$

Solve further as

$I_{Z^{\text{'}}}=507.47\angle 8.2°\text{\hspace{0.33em}A}$

Now determine the sending end voltage.

Substitute $132790$for${\text{V}}_{\text{R}}$, $507.47\angle 8.2°$ for $I_{Z^{\text{'}}}$and $156.16\angle 77.26°$for$Z^{\text{'}}$ into equation (1).

$\begin{array}{c}{V}_S=132790+\left(507.47\angle 8.2°\right)\left(156.16\angle 77.26°\right)\\ =132790+79246.92\angle 85.5°\\ =159.9\angle 29.6°\text{\hspace{0.33em}kV}\end{array}$

Therefore,the value of sending end voltage $V_S$ is.$159.9\angle 29.6°\text{\hspace{0.33em}kV}$

Determine the magnitude of the sending end voltage (line to line).

$\begin{array}{c}{V}_{\text{S,L-L}}=\sqrt{3}\left(159.9\times {10}^3\right)\\ =277\text{\hspace{0.17em}kV}\end{array}$

Hence, the magnitude of the sending end line to line voltage $V_{\text{S,L-L}}$ is$\text{277\hspace{0.17em}kV}$.

(b) Determine the sending end current.

Determine the value of sending end current.

Substitute$507.47\angle 8.24°$ for$l_Z$, $159.9\times {10}^3\angle 29.6°$ for $V_S$ and $0.545\times {10}^{-3}\angle 89.78°$ for $\frac{y^{\text{'}}}{2}$ into equation (2).

$\begin{array}{c}{I}_S=507.47\angle 8.24°+\left(159.9\times {10}^3\angle 29.6°\right)\left(0.545\times {10}^{-3}\angle 89.78°\right)\\ =507.47\angle 8.24°+87.145\angle 119.38°\\ =482.93\angle 17.9°\text{\hspace{0.33em}A}\end{array}$

Hence, the magnitude of sending end current, $I_S$ is $\text{482.93\hspace{0.33em}A}$

(c) Determine the value of sending end real and reactive power.

Determine thesending-end real power and reactive power.

Substitute $159.9\times {10}^3\angle 29.6°$ for $V_S$ and $482.93\angle -17.9°$ for $I_S^{*}$ into equation (3).

$\begin{array}{c}{S}_{S,3\varphi }=3\left(159.9\times {10}^3\angle 29.6°\right)\left(482.93\angle -17.9°\right)\\ =231.7\times {10}^6\angle 11.7°\\ =227+j47\text{\hspace{0.17em}MVA}\end{array}$

The total complex power is equal to sum of real and reactive powers.

The value of sending end real power is.$P_{S,3\varphi }=227\text{\hspace{0.33em}MW}$

The value of sending end and reactive power is.$Q_{S,3\varphi }=47\text{\hspace{0.33em}MVAR}$

Hence, the value of sending-end real power $P_{\text{S},3\varphi }$ is $\text{227\hspace{0.33em}MW}$and reactive power $Q_{\text{S,3}\varphi }$is .$\text{47\hspace{0.33em}MVAR}$

(d) Determine the value of percentage voltage regulation of the transmission line.

Determine the line to neutral receiving end voltage.

$\begin{array}{c}{V}_{\text{R,LN}}=\frac{230}{\sqrt{3}}\angle 0°\\ =132.8\angle 0°\text{\hspace{0.33em}kV}\end{array}$

Now, determine the value of percentage voltage regulation of the transmission line.

Substitute $159.9$ for$V_S$ and $132.8\times {10}^3$ for $V_{\text{R}}$ into equation (4).

$\begin{array}{c}\text{\%VR}=\left(\frac{159.9\times {10}^3-132.8\times {10}^3}{132.8\times {10}^3}\right)\left(100\right)\\ =20.4\%\end{array}$

Therefore, the value of percentage voltage regulation of the transmission line is .$20.4\%$