Question

Given the uncompensated line of Problem 5.18, let a three-phase shunt reactor (inductor) that compensates for $\mathbf{70}\mathbf{\%}$ of the total shunt admittance of the line be connected at the receiving end of the line during no-load conditions. Determine the effect of voltage regulation with the reactor connected at no load. Assume that the reactor is removed under full-load conditions.

Short answer

The voltage regulation is $14.84\%$and the use of shunt reactor improve the voltage regulation from $24.74\%$to$14.84\%$.

Step-by-step solution

Write the given data from the question.

The length of the transmission line,$\mathrm{I}=230\mathrm{mile}$

The voltage $215\mathrm{kV}$

The series impedance,$\mathrm{z}=0.8432\angle 79.{04}^0\frac{\mathrm{\Omega }}{\mathrm{mi}}$

The shunt admittance,$\mathrm{y}=5.105\times {10}^{-6}\angle {90}^0\frac{\mathrm{S}}{\mathrm{mi}}$

Load at receiving end $\mathrm{P}=125\mathrm{MW}$

Shunt reactor compensate for $70\%$of total shunt admittance.

Data from problem 5.18,

The voltage regulation from is $\mathrm{VR}\%=24.74\%$

Series impedance,${\mathrm{Z}}^{\text{'}}=186.68\angle 79.{42}^0\mathrm{\Omega }$

The sending end voltage,${\mathrm{V}}_{\mathrm{S}}=238.8\mathrm{kV}$

Determine the formula to calculate the effect on the voltage regulation.

The equation to calculate the shunt admittance is given as follows.

$\mathbf{Y}\mathbf{=}\mathbf{yl}$ …… (1)

The equation to calculate the correction factor${\mathrm{F}}_2$ to convert Y for nominal p-circuit to ${\mathrm{Y}}^{\text{'}}$for equivalent$\mathrm{\pi }$ circuit is given as follows.

${\mathbf{F}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{cosh}\left(\mathrm{\gamma l}\right)\mathbf{-}\mathbf{1}}{\left({\displaystyle \frac{\mathrm{\gamma l}}{1}}\right)\mathbf{sinh}\left(\mathrm{\gamma l}\right)}$ …… (2)

The equation to calculate the admittance for equivalent p-circuit.

${\mathbf{Y}}^{\mathbf{\text{'}}}\mathbf{=}{\mathbf{YF}}_{\mathbf{2}}$ …… (3)

The equation to equivalent series impedance is given as follows.

${\mathbf{Z}}_{\mathbf{eq}}\mathbf{=}{\mathbf{Z}}^{\mathbf{\text{'}}}$ …… (4)

The equation to calculate the parameter A is given as follows.

$\mathbf{A}\mathbf{=}\mathbf{1}\mathbf{+}\frac{{\mathbf{Y}}_{\mathbf{eq}}{\mathbf{Z}}_{\mathbf{eq}}}{\mathbf{2}}$ …… (5)

The equation to calculate the no load voltage at receiving end is given as follows.

${\mathbf{V}}_{\mathbf{RNL}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{s}}}{\mathbf{A}}$ …… (6)

The equation to calculate the voltage regulation is given as follows.

$\mathbf{VR}\mathbf{\%}\mathbf{=}\left(\frac{V_{\mathrm{RNL}}-V_{\mathrm{RFL}}}{V_{\mathrm{RFL}}}\right)\mathbf{\times }\mathbf{100}$ …… (7)

Calculate the effect on the voltage regulation.

Calculate the shunt admittance.

Substitute $5.105\times {10}^{-6}\angle {90}^0\frac{\mathrm{S}}{\mathrm{mi}}$for y and 230 mile for l into equation (1).

$$\begin{gathered} \mathrm{Y}=5.105\times {10}^{-6}\angle {90}^0\times 230 \\ \mathrm{Y}=1.174\times {10}^{-3}\angle {90}^0\mathrm{S} \end{gathered}$$

Calculate the correction factor ${\mathrm{F}}_2$.

Substitute $0.0456+\mathrm{j}0.475$for $\mathbb{y}$into equation (2).

Calculate the equivalent admittance localid="1655210943367" ${\mathrm{Y}}^{\text{'}}$.

Substitute localid="1655210951479" $1.174\times {10}^{-3}\angle {90}^0\mathrm{S}$for Y and localid="1655210957809" $1.0218\angle -{18}^0\mathrm{pu}$for localid="1655210966303" ${\mathrm{F}}_2$into equation (3).

localid="1655210973008" $$\begin{gathered} {\mathrm{Y}}^{\text{'}}=1.174\times {10}^{-3}\angle {90}^0\times 1.0218\angle -0.8^0\mathrm{pu} \\ {\mathrm{Y}}^{\text{'}}=3.7687\times {10}^{-6}+\mathrm{j}1.1996\times {10}^{-3}\mathrm{S} \end{gathered}$$

Calculate the equivalent shunt admittance with localid="1655210984146" $70\%$
compensation as,

localid="1655210992521" $$\begin{gathered} {\mathrm{Y}}_{\mathrm{eq}}=3.7687\times {10}^{-6}+\mathrm{j}1.1996\times {10}^{-3}\left(1-\frac{70}{100}\right) \\ {\mathrm{Y}}_{\mathrm{eq}}=1.1306+\mathrm{j}3.5988\times {10}^{-4} \\ {\mathrm{Y}}_{\mathrm{eq}}=3.5988\times {10}^{-6}\angle 89.{82}^0\mathrm{S} \end{gathered}$$

Calculate the equivalent series impedance.

Substitute localid="1655211000863" $186.68\angle 79.{42}^0\mathrm{\Omega }$for localid="1655211013121" ${\mathrm{Z}}^{\text{'}}$into equation (4).

localid="1655211020209" $Z_{eq}=186.68\angle 79.{42}^0\mathrm{\Omega }$

Calculate the parameter A.

Substitute localid="1655211026808" $3.5988\times {10}^{-6}\angle 89.{82}^0\mathrm{S}$ for localid="1655211039257" ${\mathrm{Y}}_{\mathrm{eq}}$
and localid="1655211046881" $186.68\angle 79.{42}^0\mathrm{\Omega }$ for localid="1655211087748" ${\mathrm{Z}}_{\mathrm{eq}}$into equation (5).

localid="1655211095724" $$\begin{gathered} {\mathrm{A}}_{\mathrm{eq}}=1+\frac{(3.5988\times {10}^{-6}\angle 89.82\mathrm{S})(186.68\angle 79.{42}^0)}{2} \\ {\mathrm{A}}_{\mathrm{eq}}=1+\frac{0.0671\angle 169.{24}^0}{2} \\ {\mathrm{A}}_{\mathrm{eq}}=1+0.0355\angle 169.{24}^0 \\ {\mathrm{A}}_{\mathrm{eq}}=0.9671\angle 0.37\mathrm{pu} \end{gathered}$$

Calculate the no load receiving end voltage.

localid="1655211110391" $$\begin{gathered} {\mathrm{V}}_{\mathrm{RNL}}=\frac{238.8}{0.9671} \\ {\mathrm{V}}_{\mathrm{RNL}}=246.92\mathrm{kV} \end{gathered}$$

Calculate the percentage voltage regulation.

Substitute localid="1655211121404" $246.96\mathrm{kV}$forlocalid="1655211130091" ${\mathrm{V}}_{\mathrm{RNL}}$and 215 kV into equation (7).

localid="1655211140565" $$\begin{gathered} \mathrm{VR}\%=\left(\frac{246.92-215}{215}\right)\times 100 \\ \mathrm{VR}\%=\frac{31.92}{215}\times 100 \\ \mathrm{VR}\%=14.84\% \end{gathered}$$

Hence the voltage regulation is localid="1655211153976" $14.84\%$and the use of shunt reactor improve the voltage regulation from localid="1655211161849" $24.74\%$to localid="1655211170333" $14.84\%$.