Question

Rework Example 14.3 with ${\mathbf{R}}_{\mathbf{L}}\mathbf{=}\mathbf{40}\mathbf{}\raisebox{1ex}{$\mathbf{\Omega }$}\!\left/ \!\raisebox{-1ex}{$\mathbf{phase}$}\right.$, ${\mathbf{X}}_{\mathbf{L}}\mathbf{=}\mathbf{60}\mathbf{}\raisebox{1ex}{$\mathbf{\Omega }$}\!\left/ \!\raisebox{-1ex}{$\mathbf{phase}$}\right.$, and data-custom-editor="chemistry" ${\mathbf{X}}_{\mathbf{C}}\mathbf{=}\mathbf{20}\mathbf{}\raisebox{1ex}{$\mathbf{\Omega }$}\!\left/ \!\raisebox{-1ex}{$\mathbf{phase}$}\right.$.

Short answer

(a) With shunt capacitor bank out of service:

(1) The line current is$0.213\angle -38.{555}^0\raisebox{1ex}{$\mathrm{kA}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{phase}$}\right.$ .

(2) The voltage drop across the line is $1.429\angle 24.{87}^0\mathrm{kV}$.

(3) The line-to-line load voltage is $12.289\angle -4.{86}^0\mathrm{kA}$.

(4) The real power delivered to the load is$3.774\mathrm{MW}$ and reactive power delivered to the load is $2.517\mathrm{Mvar}$.

(5) The power factor is$0.832$ lagging.

(6) The real and reactive power line losses are$0.408\mathrm{MW}$ and$0.817\mathrm{Mvar}$ respectively.

(7) The real, reactive and apparent power are $4.183\mathrm{MW}$,$3.324\mathrm{Mvar}$ and$5.343\mathrm{MVA}$ respectively.

(b) With shunt capacitor bank in service.

(1) The line current is$0.193\angle -7.{94}^0\raisebox{1ex}{$\mathrm{kA}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{phase}$}\right.$ .

(2) The voltage drop across the line is $1.295\angle 55.{492}^{\mathrm{o}}\mathrm{kV}$.

(3) The line-to-line load voltage is $13.347\angle -7.{959}^{\mathrm{o}}\mathrm{kA}$.

(4) The real power delivered to the load is$4.454\mathrm{MW}$ and reactive power delivered to the load is $2.969\mathrm{Mvar}$.

(5) The power factor is$0.832$ lagging.

(6) The real and reactive power line losses are$0.335\mathrm{MW}$ and$0.670\mathrm{Mvar}$ respectively.

(7) The real, reactive and apparent power are $4.789\mathrm{MW}$,$3.639\mathrm{Mvar}$ and$6.015\mathrm{MVA}$ respectively.

The reactive power delivered by the shunt capacitor is $2.969\mathrm{Mvar}$.

(c) The power delivered to the load is increase from$3.775\mathrm{MW}$ to $4.454\mathrm{MW}$.

Step-by-step solution

Write the given data from the question:

The load resistance,${\mathrm{R}}_{\mathrm{LOAD}}=40\raisebox{1ex}{$\mathrm{\Omega }$}\!\left/ \!\raisebox{-1ex}{$\mathrm{phase}$}\right.$

The load inductive reactance,${\mathrm{X}}_{\mathrm{LOAD}}=60\raisebox{1ex}{$\mathrm{\Omega }$}\!\left/ \!\raisebox{-1ex}{$\mathrm{phase}$}\right.$

Load capacitive reactance,${\mathrm{X}}_{\mathrm{C}}=60\raisebox{1ex}{$\mathrm{\Omega }$}\!\left/ \!\raisebox{-1ex}{$\mathrm{phase}$}\right.$

The line resistance,${\mathrm{R}}_{\mathrm{LINE}}=3\mathrm{\Omega }$

The line reactance,${\mathrm{X}}_{\mathrm{LINE}}=\mathrm{j}6\mathrm{\Omega }$

Determine the equation to calculate the line current, voltage drop across the line, load voltage, real and reactive power delivered to the load, load power factor, real and reactive line distribution, substation.

The equation to calculate the total impedance seen from the source side is given as follows.

${\mathbf{Z}}_{\mathbf{TOTAL}}\mathbf{=}{\mathbf{R}}_{\mathbf{LINE}}\mathbf{+}{\mathbf{jX}}_{\mathbf{LINE}}\mathbf{+}\frac{\mathbf{1}}{{\displaystyle \frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{LOAD}}}}\mathbf{+}{\displaystyle \frac{\mathbf{1}}{{\mathbf{jX}}_{\mathbf{LOAD}}}}}$ …… (1)

Here, ${\mathbf{R}}_{\mathbf{LINE}}$is the line resistance,${\mathbf{X}}_{\mathbf{LINE}}$ is the line reactance,${\mathbf{R}}_{\mathbf{LOAD}}$ is the load resistance and${\mathbf{X}}_{\mathbf{LOAD}}$ is the load resistance.

The equation to calculate the line current is given as follows.

${\mathbf{I}}_{\mathbf{LINE}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{SLN}}}{{\mathbf{Z}}_{\mathbf{TOTAL}}}$ …… (2)

Here, ${\mathbf{V}}_{\mathbf{SLN}}$ is the sending end line voltage.

The equation to calculate the voltage drop across the line is given as follows.

${\mathbf{V}}_{\mathbf{DROP}}\mathbf{=}\left(R_{\mathrm{LINE}}+{\mathrm{jX}}_{\mathrm{LINE}}\right){\mathbf{I}}_{\mathbf{LINE}}$ …… (3)

The equation to calculate the line-to-line voltage is given as follows.

role="math" localid="1656742738347" ${\mathbf{V}}_{\mathrm{LOAD}}=\sqrt{3}\left({\mathbf{V}}_{\mathrm{SLN}}-{\mathbf{V}}_{\mathrm{DROP}}\right)$ …… (4)

The equation to calculate the real power delivered the load is given as follows.

${\mathbf{P}}_{\mathrm{LOAD}}=\frac{{\mathbf{V}}_{\mathrm{LOAD}}^2}{{\mathbf{R}}_{\mathrm{LOAD}}}$ …… (5)

The equation to calculate the reactive power delivered to the load is given as follows.

${\mathbf{Q}}_{\mathrm{LOAD}}=\frac{{\mathbf{V}}_{\mathrm{LOAD}}^2}{{\mathbf{X}}_{\mathrm{LOAD}}}$ …… (6)

The equation to calculate the load power factor is given as follows.

data-custom-editor="chemistry" $\mathbf{pf}=\mathbf{cos}\left[{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\frac{{\mathbf{Q}}_{\mathrm{LOAD}}}{{\mathbf{P}}_{\mathrm{LOAD}}}\right)\right]$ …… (7)

The equation to calculate the real line losses is given as follows.

${\mathbf{P}}_{\mathrm{LINELOSS}}=\mathbf{3}{\mathbf{I}}_{\mathrm{LINE}}^2{\mathbf{R}}_{\mathrm{LINE}}$ …… (8)

The equation to calculate the reactive line losses is given as follows.

${\mathbf{Q}}_{\mathrm{LINE}}=\mathbf{3}{\mathbf{I}}_{\mathrm{LINE}}^2{\mathbf{X}}_{\mathrm{LINE}}$ …… (9)

The equation to calculate the real power delivered by the distribution substation is given as follows.

${\mathbf{P}}_{\mathrm{SOURCE}}={\mathbf{P}}_{\mathrm{LOAD}}+{\mathbf{P}}_{\mathrm{LINELOSS}}$ …… (10)

The equation to calculate the reactive power delivered by the distribution substation is given as follows.

${\mathbf{Q}}_{\mathrm{SOURCE}}+{\mathbf{Q}}_{\mathrm{LOAD}}+{\mathbf{Q}}_{\mathrm{LINELOSS}}$ …… (11)

The equation to calculate the complex power is given as follows.

${\mathbf{S}}_{\mathrm{SOURCE}}=\sqrt{{\mathbf{P}}_{\mathrm{SOURCE}}^2+{\mathbf{Q}}_{\mathrm{SOURCE}}^2}$ …… (12)

The equation to calculate the total impedance when capacitive reactance is added, is given as follows.

data-custom-editor="chemistry" ${\mathbf{Z}}_{\mathrm{TOTAL}}={\mathbf{R}}_{\mathrm{LINE}}+{\mathbf{jX}}_{\mathrm{LINE}}+\frac{\mathbf{1}}{{\displaystyle \frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{LOAD}}}}\mathbf{+}{\displaystyle \frac{\mathbf{1}}{{\mathbf{jX}}_{\mathbf{LOAD}}}}\mathbf{-}{\displaystyle \frac{\mathbf{1}}{{\mathbf{jX}}_{\mathbf{C}}}}}$ …… (13)

The equation to calculate the reactive power delivered by the shunt capacitor is given as follows.

${\mathbf{Q}}_{\mathrm{C}}=\frac{{\mathbf{V}}_{\mathrm{LOAD}}^2}{{\mathbf{X}}_{\mathrm{C}}}$ …… (14)

Calculate the line current, voltage drop across the line, load voltage, real and reactive power delivered to the load, load power factor, real and reactive line distribution, substation when shunt capacitor banks out of service.

Calculate the total resistance.

Substitute $3\mathrm{\Omega }$for ${\mathrm{R}}_{\mathrm{LINE}}$, $40\mathrm{\Omega }$for ${\mathrm{R}}_{\mathrm{LOAD}}$, $\mathrm{j}6\mathrm{\Omega }$for ${\mathrm{X}}_{\mathrm{LINE}}$and $\mathrm{j}60\mathrm{\Omega }$for ${\mathrm{X}}_{\mathrm{LOAD}}$into equation (1).

localid="1656745017926" $$\begin{gathered} {\mathrm{Z}}_{\mathrm{TOTAL}}=3+\mathrm{j}6+\frac{1}{{\displaystyle \frac{1}{40}+\frac{1}{\mathrm{j}60}}} \\ {\mathrm{Z}}_{\mathrm{TOTAL}}=3+\mathrm{j}6+\frac{\mathrm{j}2400}{{\displaystyle 40+\mathrm{j}60}} \\ {\mathrm{Z}}_{\mathrm{TOTAL}}=3+\mathrm{j}6+27.692+\mathrm{j}18.462 \\ {\mathrm{Z}}_{\mathrm{TOTAL}}=30.692+\mathrm{j}24.462\raisebox{1ex}{$\mathrm{\Omega }$}\!\left/ \!\raisebox{-1ex}{$\mathrm{phase}$}\right. \end{gathered}$$

Convert the total impedance from cartesian from to polar form.

${\mathrm{Z}}_{\mathrm{TOTAL}}=39.248\angle 38.{55}^{\mathrm{o}}\raisebox{1ex}{$\mathrm{\Omega }$}\!\left/ \!\raisebox{-1ex}{$\mathrm{phase}$}\right.$

(1)

Calculate the sending end voltage with above the rated voltage.

$$\begin{gathered} {\mathrm{V}}_{\mathrm{SLN}}=\left(100+5\right)\%\mathrm{of}\frac{13.8}{\sqrt{3}} \\ {\mathrm{V}}_{\mathrm{SLN}}=\frac{105}{100}\times \frac{13.8}{\sqrt{3}} \\ {\mathrm{V}}_{\mathrm{SLN}}=8.365\mathrm{kV} \end{gathered}$$

Calculate the line current.

Substitute$8.365\mathrm{kV}$ for${\mathrm{V}}_{\mathrm{SLN}}$ and$39.248\angle 38.{55}^{\mathrm{o}}\raisebox{1ex}{$\mathrm{\Omega }$}\!\left/ \!\raisebox{-1ex}{$\mathrm{phase}$}\right.$ for${\mathrm{Z}}_{\mathrm{TOTAL}}$ into equation (2).

$$\begin{gathered} {\mathrm{I}}_{\mathrm{LINE}}=\frac{8.365\times {10}^3}{39.248\angle 38.{55}^{\mathrm{o}}} \\ {\mathrm{I}}_{\mathrm{LINE}}=0.213\angle -38.{55}^{\mathrm{o}}\mathrm{kA}/\mathrm{phase} \end{gathered}$$

Hence the line current is$0.213\angle -38.{555}^{\mathrm{o}}\raisebox{1ex}{$\mathrm{kA}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{phase}$}\right.$ .

(2)

Calculate the voltage drop across the line.

Substitute$3\mathrm{\Omega }$ for ${\mathrm{R}}_{\mathrm{LINE}}$,$\mathrm{j}6\mathrm{\Omega }$ for${\mathrm{X}}_{\mathrm{LINE}}$ and$0.213\angle -38.{555}^{\mathrm{o}}\raisebox{1ex}{$\mathrm{kA}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{phase}$}\right.$ for${\mathrm{I}}_{\mathrm{LINE}}$ into equation (3).

role="math" localid="1656746756778" $$\begin{gathered} {\mathrm{V}}_{\mathrm{DROP}}=\left(3+\mathrm{j}6\right)\left(0.213\angle -38.{555}^{\mathrm{o}}\right) \\ {\mathrm{V}}_{\mathrm{DROP}}=\left(6.708\angle 63.{435}^{\mathrm{o}}\right)\left(0.213\angle -38.{555}^{\mathrm{o}}\right) \\ {\mathrm{V}}_{\mathrm{DROP}}=1.429\angle 24.{87}^{\mathrm{o}}\mathrm{kV} \end{gathered}$$

Hence the voltage drop across the line is$1.429\angle 24.{87}^{\mathrm{o}}\mathrm{kV}$ .

(3)

Calculate the line-to-line load voltage.

Substitute the$8.365\mathrm{kV}$ for${\mathrm{V}}_{\mathrm{SLN}}$ and$1.429\angle 24.{87}^{\mathrm{o}}\mathrm{kV}$ for${\mathrm{V}}_{\mathrm{DROP}}$ into equation (4).

$$\begin{gathered} {\mathrm{V}}_{\mathrm{LOAD}}=\sqrt{3}\left(8.365-1.429\angle 24.{875}^{\mathrm{o}}\right) \\ {\mathrm{V}}_{\mathrm{LOAD}}=\sqrt{3}\left(7.093\angle -4.{86}^{\mathrm{o}}\right) \\ {\mathrm{V}}_{\mathrm{LOAD}}=12.29\angle -4.{86}^{\mathrm{o}}\mathrm{kA} \end{gathered}$$

Hence the line-to-line load voltage is $12.289\angle -4.{86}^{\mathrm{o}}\mathrm{kA}$.

(4)

Calculate the real power delivered to the load.

Substitute$12.289\mathrm{kV}$ for${\mathrm{V}}_{\mathrm{LOAD}}$ and$60\mathrm{\Omega }$ for${\mathrm{X}}_{\mathrm{LOAD}}$ into equation (5).

$$\begin{gathered} {\mathrm{P}}_{\mathrm{LOAD}}=\frac{12.{289}^2}{40} \\ {\mathrm{P}}_{\mathrm{LOAD}}=3.775\mathrm{MW} \end{gathered}$$

Calculate the reactive power delivered to the load.

Substitute$12.289\mathrm{kV}$ for${\mathrm{V}}_{\mathrm{LOAD}}$ and$60\mathrm{\Omega }$ for${\mathrm{X}}_{\mathrm{LOAD}}$ into equation (6).

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{LOAD}}=\frac{12.{289}^2}{60} \\ {\mathrm{Q}}_{\mathrm{LOAD}}=2.517\mathrm{Mvar} \end{gathered}$$

Hence the real power delivered to the load is $3.774\mathrm{MW}$and reactive power delivered to the load is $2.517\mathrm{Mvar}$

(5)

Calculate the power factor.

Substitute$3.774\mathrm{MW}$ for${\mathrm{P}}_{\mathrm{LOAD}}$ and$2.517\mathrm{Mvar}$ for${\mathrm{Q}}_{\mathrm{LOAD}}$ into equation (7).

$$\begin{gathered} \mathrm{pf}=\cos \left[{\tan }^{-1}\left(\frac{2.517}{3.775}\right)\right] \\ \mathrm{pf}=\cos \left[33.{394}^{\mathrm{o}}\right] \\ \mathrm{pf}=0.832 \end{gathered}$$

Hence the power factor is 0.832 lagging.

(6)

Calculate the real power line losses.

Substitute $0.213\raisebox{1ex}{$\mathrm{kA}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{phase}$}\right.$for ${\mathrm{I}}_{\mathrm{LINE}}$and $3\mathrm{\Omega }$for ${\mathrm{R}}_{\mathrm{LINE}}$into equation (8)

$$\begin{gathered} {\mathrm{P}}_{\mathrm{LINELOSS}}=3{\left(0.213\right)}^2\times 3 \\ {\mathrm{P}}_{\mathrm{LINELOSS}}=0.408\mathrm{MW} \end{gathered}$$

Calculate the reactive line losses.

Substitute$0.213\raisebox{1ex}{$\mathrm{kA}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{phase}$}\right.$ for${\mathrm{I}}_{\mathrm{LINE}}$ and$6\mathrm{\Omega }$ for${\mathrm{X}}_{\mathrm{LINE}}$ into equation (9)

$$\begin{gathered} {\mathrm{X}}_{\mathrm{LINELOSS}}=3{\left(0.213\right)}^2\times 6 \\ {\mathrm{X}}_{\mathrm{LINELOSS}}=0.817\mathrm{Mvar} \end{gathered}$$

Hence the real and reactive power line losses are 0.408 MW and 0.817 Mvar respectively.

(7)

Calculate the real power delivered to distribution substation.

Substitute 3.775MW for P_LOAD and 0.408 MW for P_LINELOSS into equation (10).

$$\begin{gathered} {\mathrm{P}}_{\mathrm{SOURCE}}=3.775+0.408 \\ {\mathrm{P}}_{\mathrm{SOURCE}}=4.183\mathrm{MW} \end{gathered}$$

Calculate the reactive power delivered to distribution substation.

Substitute 2.517 Mvar for Q_LOAD and 0.817 Mvar for Q_LINELOSS into equation (11).

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{SOURCE}}=2.517+0.817 \\ {\mathrm{Q}}_{\mathrm{SOURCE}}=3.324\mathrm{Mvar} \end{gathered}$$

Calculate the complex power.

Substitute 4.183MW for P_SOURCE and 3.324 Mvar for Q_SOURCE into equation (12).

$$\begin{gathered} {\mathrm{S}}_{\mathrm{SOURCE}}=\sqrt{4.{183}^2+3.{324}^2} \\ {\mathrm{S}}_{\mathrm{SOURCE}}=\sqrt{25.486} \\ {\mathrm{S}}_{\mathrm{SOURCE}}=5.343\mathrm{MVA} \end{gathered}$$

Hence the real, reactive and apparent power are 4.183 MW , 3.324 Mvar and 5.343 MVA respectively.

Calculate the line current, voltage drop across the line, load voltage, real and reactive power delivered to the load, load power factor, real and reactive line distribution, substation when shunt capacitor banks in service.

Calculate the total resistance.

Substitute $3\mathrm{\Omega }$for R , for , for , for and for into equation (1).

$$\begin{gathered} {\mathrm{Z}}_{\mathrm{TOTAL}}=3+\mathrm{j}6+\frac{1}{{\displaystyle \frac{1}{40}}+{\displaystyle \frac{1}{\mathrm{j}60}}-{\displaystyle \frac{1}{\mathrm{j}60}}} \\ {\mathrm{Z}}_{\mathrm{TOTAL}}=3+\mathrm{j}6+40 \\ {\mathrm{Z}}_{\mathrm{TOTAL}}=43+\mathrm{j}6 \\ {\mathrm{Z}}_{\mathrm{TOTAL}}=43.42\angle 7.{94}^{\mathrm{o}}\raisebox{1ex}{$\mathrm{\Omega }$}\!\left/ \!\raisebox{-1ex}{$\mathrm{phase}$}\right. \end{gathered}$$

(1)

Calculate the sending end voltage with above the rated voltage.

$$\begin{gathered} {\mathrm{V}}_{\mathrm{SLN}}=\left(100+5\right)\%\mathrm{of}\frac{13.8}{\sqrt{3}} \\ {\mathrm{V}}_{\mathrm{SLN}}=\frac{105}{100}\times \frac{13.8}{\sqrt{3}} \\ {\mathrm{V}}_{\mathrm{SLN}}=8.365\mathrm{kV} \end{gathered}$$

Calculate the line current.

Substitute for and for into equation (2).

$$\begin{gathered} {\mathrm{I}}_{\mathrm{LINE}}=\frac{8.365\times {10}^3}{43.43\angle 7.{94}^{\mathrm{o}}} \\ {\mathrm{I}}_{\mathrm{LINE}}=0.193\angle -7.{94}^{\mathrm{o}}\raisebox{1ex}{$\mathrm{kA}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{phase}$}\right. \end{gathered}$$

Hence the line current is .

(2)

Calculate the voltage drop across the line.

Substitute for , for and for into equation (3).

Hence the voltage drop across the line is .

(3)

Calculate the line-to-line load voltage.

Substitute the for and for into equation (4).

$$\begin{gathered} {\mathrm{V}}_{\mathrm{LOAD}}=\sqrt{3}\left(8.365-1.295\angle 55.{492}^{\mathrm{o}}\right) \\ {\mathrm{V}}_{\mathrm{LOAD}}=\sqrt{3}\left(7.705\angle -7.{959}^{\mathrm{o}}\right) \\ {\mathrm{V}}_{\mathrm{LOAD}}=13.347\angle -7.{959}^{\mathrm{o}}\mathrm{kA} \end{gathered}$$

Hence the line-to-line load voltage is.

(4)

Calculate the real power delivered to the load.

Substitute for and for into equation (5).

$$\begin{gathered} {\mathrm{P}}_{\mathrm{LOAD}}=\frac{13.{347}^2}{40} \\ {\mathrm{P}}_{\mathrm{LOAD}}=4.454\mathrm{MW} \end{gathered}$$

Calculate the reactive power delivered to the load.

Substitute for and for into equation (6).

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{LOAD}}=\frac{13.{347}^2}{60} \\ {\mathrm{Q}}_{\mathrm{LOAD}}=2.969\mathrm{Mvar} \end{gathered}$$

Hence the real power delivered to the load is and reactive power delivered to the load is

(5)

Calculate the power factor.

Substitute for and for into equation (7).

$$\begin{gathered} \mathrm{pf}=\cos \left[{\tan }^{-1}\left(\frac{2.969}{4.454}\right)\right] \\ \mathrm{pf}=\cos \left[33.{394}^{\mathrm{o}}\right] \\ \mathrm{pf}=0.832 \end{gathered}$$

Hence the power factor is lagging.

(6)

Calculate the real power line losses.

Substitute for and for into equation (8)

$$\begin{gathered} {\mathrm{P}}_{\mathrm{LINELOSS}}=3{\left(0.193\right)}^2\times 3 \\ {\mathrm{P}}_{\mathrm{LINELOSS}}=0.335\mathrm{MW} \end{gathered}$$

Calculate the reactive line losses.

Substitute for and for into equation (9)

$$\begin{gathered} {\mathrm{X}}_{\mathrm{LINELOSS}}=3{\left(0.193\right)}^2\times 6 \\ {\mathrm{X}}_{\mathrm{LINELOSS}}=0.670\mathrm{Mvar} \end{gathered}$$

Hence the real and reactive power line losses are and respectively.

(7)

Calculate the real power delivered to distribution substation.

Substitute for and for into equation (10).

$$\begin{gathered} {\mathrm{P}}_{\mathrm{SOURCE}}=4.454+0.335 \\ {\mathrm{P}}_{\mathrm{SOURCE}}=4.789\mathrm{MW} \end{gathered}$$

Calculate the reactive power delivered to distribution substation.

Substitute for and for into equation (11).

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{SOURCE}}=2.969+0.670 \\ {\mathrm{Q}}_{\mathrm{SOURCE}}=3.639\mathrm{Mvar} \end{gathered}$$

Calculate the complex power.

Substitute for and for into equation (12).

$$\begin{gathered} {\mathrm{S}}_{\mathrm{SOURCE}}=\sqrt{4.{789}^2+3.{639}^2} \\ {\mathrm{S}}_{\mathrm{SOURCE}}=\sqrt{36.176} \\ {\mathrm{S}}_{\mathrm{SOURCE}}=6.015\mathrm{MVA} \end{gathered}$$

Hence the real, reactive and apparent power are , and respectively.

Calculate the reactive power delivered by the shunt capacitor.

Substitute for and for into equation (14).

Hence the reactive power delivered by the shunt capacitor is .

Comparison between the result of the part (a) and (b).

From the results of the part (a) and (b), it can be concluded that the power delivered to the load is increase from to .

The following are the benefits of having the shunt capacitor.

(i) The line current decreases

(ii) The real and reactive line losses decrease

(iii) The voltage drops across the line decreases

(iv) The reactive power delivered by the source decreases

(v) The load voltage increases