Question

A 60-Hz, 230-mile, three-phase overhead transmission line has a series impedance $z=\Omega 0/mi8431\angle 79.{04}^{\circ }$ and a shunt admittance $y=5.105\times {10}^{-6}\angle {90}^{\circ }S/mi$.The load at the receiving end is 125 MW at unity power factor and at 215 kV. Determine the voltage, current, and both real and reactive power at the sending end and the percent voltage regulation of the line. Also find the wavelength and velocity of propagation of the line.

Short answer

The value of sending-end voltage of the line is 238.8kV.

The value of sending-end current of the line is $332.31\angle 26.{33}^{\circ }A$.

The percent voltage regulation of the line is 24.7%.

The wavelength of the line is 3043mi.

The velocity of propagation of line is 182,580mi/s.

Step-by-step solution

Given data.

The given load at receiving-end of line is 125 MW.

The given frequency of line is 60-Hz.

The given length of the line I=230mile.

The given series impedance of line is $z=0.8431\angle 79.{04}^{\circ }\Omega /mi$.

The given shunt admittance of line is $y=5.105\times {10}^{-6}\angle {90}^{\circ }S/mi$.

Determine the required formula.

Write the formula of sending-end voltage of the line.

$YV{I}_S=I{V}_{R}Zcossimhh\left(Y\right)I_{RC}()$ …… (1)

Here, role="math" localid="1654950759076" $V_R$is receiving-end voltage, $I_R$is receiving-end current, $Z_C$is characteristics impedance and role="math" localid="1654951189901" ${\gamma }^I$is dimensionless quantity.

Write the formula of sending-end current of the line.

role="math" localid="1654950990756" $I_S=I_R\cos h\left(gl\right)+\frac{V_R}{Z_C}\sin h\left(gl\right)$ …… (2)

Here, role="math" localid="1654951041743" $V_R$is receiving-end voltage, $I_R$is receiving-end current, $Z_C$is characteristics impedance and role="math" localid="1654951458068" ${\gamma }^I$is dimensionless quantity.

Write the formula of percent voltage regulation of the line.

$VR=\left(\frac{YV{I}_S/-V_{cosh}\left(\right)}{V_R}\right)\left(100\right)$ …… (3)

Here, $V_S$is sending-end voltage, $V_R$is receiving-end voltage and ${\gamma }^I$is dimensionless quantity.

Write the formula of wavelength of line.

$\lambda =\frac{2\pi }{\beta }$ …… (4)

Here, $\beta$is the dimensionless quantity.

Write the formula of velocity of propagation of line.

$Velocityofpropagation=f\lambda$ …… (5)

Here, $f$is frequency and $\lambda$is wavelength.

Determine the sending-end voltage.

Determine the characteristic impedance.

$Z_c=\sqrt{\frac{z}{y}}$ …… (6)

Here, z is the impedance and y is the admittance of the conductor.

Substitute $0.8431\angle 79.{04}^{\circ }$for z and $5.105\times {10}^{-6}\angle {90}^{\circ }$for y into equation (6).

$$\begin{gathered} Z_c=\sqrt{\frac{0.{8431}^{\circ }}{5.105\times {10}^{-6}}}\angle \left(79.{04}^{\circ }-{90}^{\circ }\right)/2 \\ =406.4\angle -5.{48}^{\circ }\Omega \end{gathered}$$

Here, $Z_c$is the characteristic impedance.

Determine the value of ${\gamma }^I$.

${\gamma }^I=\left(\sqrt{zy}\right)I$ …… (7)

Here, $\gamma$is the propagation constant, z is impedance, y is admittance and l is the length of line.

Substitute $0.8431\angle 79.{04}^{\circ }$for z,$5.105\times {10}^{-8}\angle {90}^{\circ }$for y and 230 for l into equation (7).

$$\begin{gathered} {\gamma }^I=230\left(\sqrt{0.843\times 5.105\times {10}^{-6}}\right)\angle \left(79.{04}^{\circ }+{90}^{\circ }\right)/2 \\ =0.4772\angle 84.{52}^{\circ } \\ =0.0456+j0.475 \end{gathered}$$

Determine the line to neutral receiving end voltage.

$$\begin{gathered} V_R=\frac{215}{\sqrt{3}} \\ =124.13\angle 0^{\circ }kV/ph \end{gathered}$$

Determine the receiving end current.

$I_R=\frac{S}{3V_R}\angle {\cos }^{-1}PF$ …… (8)

Here, S is complex power, $V_R$is receiving-end voltage and PF is power factor.

Substitute $125\times {10}^3$for,$124.13\angle 0^{\circ }$for $V_R$and 0 for PF into equation (8).

$$\begin{gathered} I_R=\frac{125\times {10}^3}{3\times 124.13\angle 0^{\circ }}\angle 0^{\circ } \\ =335.7\angle 0^{\circ }A \end{gathered}$$

Determine the value of ABCD parameters.

Determine the value of $\cos h\left({\gamma }^I\right)$.

$$\begin{gathered} \cos h\left({\gamma }^I\right)=\cos h\left(0.0456+j0.475\right) \\ =\frac{e^{0.0456}\angle 27.{22}^{\circ }+e^{-0.0456}\angle -27.{22}^{\circ }}{2} \\ =0.8904\angle 1.{34}^{\circ }A \end{gathered}$$

Determine the value of.$\sin h\left({\gamma }^I\right)$

$$\begin{gathered} \sin h\left(\gamma I\right)=\sin h\left(0.0456+j0.475\right) \\ =\frac{e^{0.0456}\angle 27.{22}^{\circ }-e^{-0.0456}\angle -27.{22}^{\circ }}{2} \\ =0.4597\angle 84.{93}^{\circ } \end{gathered}$$

Now, determine the sending-end voltage of line.

Substitute 124.13 for$V_R$, $0.8904\angle 1.{34}^{\circ }$for $\cos h\left(\gamma I\right)$, 0.3357 for$I_R$, $406.4\angle -5.{48}^{\circ }$for $Z_C$and $0.4597\angle 84.{93}^{\circ }$for $\sin h\left(\gamma I\right)$into equation (1).

$$\begin{gathered} V_S=\left(124.13\times 0.8904\angle 1.{34}^{\circ }\right)+\left(0.3357\times 406.4\angle -5.{48}^{\circ }\angle 84.{93}^{\circ }\right) \\ =137.86\angle 27.{77}^{\circ }kV \end{gathered}$$

Now, determine the line to line sending end voltage.

$$\begin{gathered} V_S=\left(137.86\right)\sqrt{3} \\ =238.8kV \end{gathered}$$

Determine the sending-end current.

Determine the sending-end current of line.

Substitute 124.13 for$V_R$, $0.8904\angle 1.{34}^{\circ }$for$\cos h\left(\gamma I\right)$, 0.3357 for$I_R$, $406.4\angle -5.{48}^{\circ }$for $Z_C$and $0.4597\angle 84.{93}^{\circ }$for $\sin h\left(\gamma I\right)$into equation (1).

$$\begin{gathered} I_S=\left(335.7\times 0.8904\angle 1.{34}^{\circ }\right)+\frac{124.130}{406.4\angle -5.{48}^{\circ }}\times 0.4597\angle 84.{93}^{\circ } \\ =332.31\angle 26.{33}^{\circ }A \end{gathered}$$

Determine the percent voltage regulation of the line.

Substitute 238.8 for$V_S$,$0.8904\angle 1.{34}^{\circ }$for $\cos h\left(\gamma I\right)$and 124.13 for $V_R$into equation (3)

$$\begin{gathered} VR=\left(\frac{\left(238.8/0.8904\angle 1.{34}^{\circ }\right)-124.13}{124.13}\right)\left(100\right) \\ =24.7\% \end{gathered}$$

Determine the wavelength of the line.

Since, $\beta =0.475/230$the value in radian per min is,

$\beta =0.002065\frac{rad}{mi}$

Substitute 0.002065 for $\beta$into equation (4).

$$\begin{gathered} \lambda =\frac{2\pi }{0.002065} \\ =3043mi \end{gathered}$$

Determine the velocity of propagation of line.

Substitute 60 for f and 3043 for $\lambda$into equation (5).

$$\begin{gathered} Velocityofpropagation=60\times 3043 \\ =182,580\frac{mi}{s} \end{gathered}$$