Question

Let the three-phase lossless transmission line of Problem 5.31 supply a load of $\mathbf{1000}\mathbf{}\mathbf{MVA}$ at $\mathbf{0}\mathbf{.}\mathbf{8}$ power factor lagging and at $\mathbf{500}\mathbf{}\mathbf{kV}$. (a) Determine the capacitance/phase and total three-phase Mvars supplied by a three-phase, $\mathbf{∆}$-connected shunt-capacitor bank at the receiving end to maintain the receiving-end voltage at $\mathbf{500}\mathbf{}\mathbf{kV}$when the sending end of the line is energized at $\mathbf{500}\mathbf{}\mathbf{kV}$. (b) If series capacitive compensation of $\mathbf{40}\mathbf{\%}$ is installed at the midpoint of the line, without the shunt capacitor bank at the receiving end, compute the sending-end voltage and percent voltage regulation.

Short answer

(a) The capacitance per phase is $6.1\mathrm{\mu F}$.

(b) The voltage regulation is $18.06\%$.

Step-by-step solution

Write the given data from the question:

The voltage, $\mathrm{V}=500\mathrm{kV}$

The power, $\mathrm{S}=1000\mathrm{MVA}$

Power factor, $\mathrm{pf}=0.8$

Series capacitive compensation$=40\%$

The length of transmission line, $\mathrm{l}=300\mathrm{km}$

Frequency of the system,$\mathrm{f}=60\mathrm{Hz}$

Series inductance,$0.97\frac{\mathrm{mH}}{\mathrm{km}}$

Shunt capacitance,$\mathrm{C}=0.0115\frac{\mathrm{\mu F}}{\mathrm{km}}$

The result from problem 5.31

The characteristic impedance, ${\mathrm{Z}}_{\mathrm{c}}=290.43\mathrm{\Omega }$

The value of the electrical line, $\mathrm{\beta l}=21.{641}^0$

Determine the formulas to calculate the capacitance per phase and voltage regulation with 40% series capacitive compensation is installed.

The equation to calculate the equivalent line reactance for the losses line is given as follows.

${\mathbf{X}}^{\mathbf{\text{'}}}\mathbf{=}{\mathbf{Z}}_{\mathbf{c}}\mathbf{sinh}\mathbf{\left(}\mathbf{\beta l}\mathbf{\right)}$ …… (1)

The equation to calculate receiving end power is given as follows,

${\mathbf{S}}_{\mathbf{R}}\mathbf{=}\mathbf{S}\mathbf{\angle }{\mathbf{cos}}^{\mathbf{-}\mathbf{1}}\mathbf{pf}$ …… (2)

The equation to calculate the three phase real power is given as follows.

${\mathbf{P}}_{\mathbf{R}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{S}}{\mathbf{V}}_{\mathbf{R}}}{{\mathbf{X}}^{\mathbf{\text{'}}}}\mathbf{sin\delta }$ …… (3)

Here, $\mathbf{\delta }$is the power angle.

The equation to calculate the three-phase receiving end reactive power is given as follows.

${\mathbf{Q}}_{\mathbf{R}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{S}}{\mathbf{V}}_{\mathbf{R}}}{{\mathbf{X}}^{\mathbf{\text{'}}}}\mathbf{cos\delta }\mathbf{-}\frac{{\mathbf{V}}_{\mathbf{R}}^{\mathbf{2}}}{{\mathbf{X}}^{\mathbf{\text{'}}}}\mathbf{cos}\left(\mathrm{\beta I}\right)$ …… (4)

The equation to calculate the required capacitor is given as follows.

${\mathbf{S}}_{\mathbf{C}}\mathbf{=}{\mathbf{Q}}_{\mathbf{R}}\mathbf{-}\mathbf{Q}$ …… (5)

The equation to calculate the capacitive reactance is given as follows.

${\mathbf{X}}_{\mathbf{C}}\mathbf{=}\frac{\mathbf{-}{\mathbf{JV}}_{\mathbf{L}}^{\mathbf{2}}}{{\mathbf{S}}_{\mathbf{C}}}$ …… (6)

The equation to calculate the capacitor per phase is given as follows.

$\mathbf{C}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\tau \tau f}\mathbf{\times }{\mathbf{X}}_{\mathbf{C}}}$ …… (7)

The equation to calculate the series capacitor reactance with compensation is given as follows.

${\mathbf{Z}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{j}\mathbf{}\mathbf{(}{\mathbf{X}}^{\mathbf{\text{'}}}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{4}{\mathbf{X}}^{\mathbf{\text{'}}}\mathbf{)}$ …… (8)

The equation to calculate the equivalent reactance for the nominal circuit is given as follows.

${\mathbf{Y}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{j}\frac{\mathbf{2}}{{\mathbf{Z}}_{\mathbf{C}}}\mathbf{tan}\left(\frac{\mathrm{\beta l}}{2}\right)$ …… (9)

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

$\mathbf{A}\mathbf{=}\mathbf{1}\mathbf{+}\frac{{\mathbf{Y}}^{\mathbf{\text{'}}}{\mathbf{Z}}^{\mathbf{\text{'}}}}{\mathbf{2}}$ …… (10)

The equation to calculate the receiving end current is given as follows.

${\mathbf{l}}_{\mathbf{R}}\mathbf{=}\frac{\mathbf{S}}{\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{R}}}$ …… (11)

The equation to calculate the sending end voltage is given as follows.

${\mathbf{V}}_{\mathbf{S}}\mathbf{=}{\mathbf{AV}}_{\mathbf{R}}\mathbf{+}{\mathbf{BI}}_{\mathbf{R}}$ …… (12)

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

${\mathbf{V}}_{\mathbf{RNL}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{S}}}{\mathbf{A}}$ …… (13)

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

$\mathbf{VR}\mathbf{\%}\mathbf{=}\left(\frac{V_{\mathrm{RNL}}-V_{\mathrm{RFL}}}{V_{\mathrm{RFL}}}\right)\mathbf{\times }\mathbf{100}$ …… (14)

Calculate the capacitance per phase.

(a)

Calculate the equivalent line reactance.

Substitute $290.43\mathrm{\Omega }$for ${\mathrm{Z}}_{\mathrm{C}}$and $21.{641}^0$ for $\mathrm{\beta I}$into equation (1).

$$\begin{gathered} {\mathrm{X}}^{\text{'}}=290.43\sin \left(21.{641}^0\right) \\ {\mathrm{X}}^{\text{'}}=107.11\mathrm{\Omega } \end{gathered}$$

Calculate the receiving end power.

Substitute $1000\mathrm{MVA}$ for S and 0.8 for $\mathrm{pf}$ into equation (2).

$$\begin{gathered} {\mathrm{S}}_{\mathrm{R}}=1000\angle {\cos }^{-1}0.8 \\ {\mathrm{S}}_{\mathrm{R}}=800+\mathrm{j}600\mathrm{MVA} \end{gathered}$$

Calculate the power angle.

Substitute for ${\mathrm{P}}_{\mathrm{R}}$,$500\mathrm{kV}$for ${\mathrm{V}}_{\mathrm{S},}{\mathrm{V}}_{\mathrm{R}}$and $107.11\mathrm{\Omega }$ for ${\mathrm{X}}^{\text{'}}$ into equation (3).

$$\begin{gathered} 800=\frac{500\times 500}{107.11}\mathrm{sin\delta } \\ \mathrm{sin\delta }=\frac{800\times 107.61}{{500}^2} \\ \mathrm{\delta }={\sin }^{-1}(0.342) \\ \mathrm{\delta }=20.{044}^0 \end{gathered}$$

Calculate the receiving end power.

Substitute $500\mathrm{kV}$forrole="math" localid="1655200614600" ${\mathrm{V}}_{{\mathrm{S}}_{,}}{\mathrm{V}}_{\mathrm{R}}$, $107.11\mathrm{\Omega }$for ${\mathrm{X}}^{\text{'}}$and $21.{641}^0$for into equation (4).

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{R}}=\frac{500\times 500}{107.11}\cos \left(20.{044}^0\right)-\frac{{500}^2}{107.11}\cos \left(21.641\right) \\ {\mathrm{Q}}_{\mathrm{R}}=\frac{{500}^2}{107.11}(0.93942-0.92950) \\ {\mathrm{Q}}_{\mathrm{R}}=23.15\mathrm{MVAR} \end{gathered}$$

Calculate the required capacitor .

Substitute$\mathrm{j}600\mathrm{MVAR}$ for Q and $\mathrm{j}23.15\mathrm{MVAR}$ for ${\mathrm{Q}}_{\mathrm{R}}$into equation (5).

$$\begin{gathered} {\mathrm{S}}_{\mathrm{C}}=\mathrm{j}23.15-\mathrm{j}600 \\ {\mathrm{S}}_{\mathrm{C}}=\mathrm{j}576.85\mathrm{MVAR} \end{gathered}$$

Calculate the capacitive reactance.

Substitute $\mathrm{j}576.85\mathrm{MVAR}$ for ${\mathrm{S}}_{\mathrm{C}}$and $500\mathrm{kV}$for ${\mathrm{V}}_{\mathrm{L}}$ into equation (6).

$$\begin{gathered} {\mathrm{X}}_{\mathrm{C}}=\frac{\mathrm{j}{500}^2}{\mathrm{j}576.85} \\ {\mathrm{X}}_{\mathrm{C}}=433.38\mathrm{\Omega } \end{gathered}$$

Calculate the capacitance per phase.

Substitute $60\mathrm{Hz}$for $\mathrm{f}$ and $\mathrm{j}433.38\mathrm{\Omega }$ for ${\mathrm{X}}_{\mathrm{C}}$into equation (7).

role="math" localid="1655201939996" $$\begin{gathered} \mathrm{C}=\frac{1}{2\mathrm{\pi }\times 60\times \mathrm{j}433.38} \\ \mathrm{C}=6.1\mathrm{\mu F} \end{gathered}$$

Hence the capacitance per phase is $6.1\mathrm{\mu F}$.

Calculate the percentage voltage regulation.

(b)

Calculate the equivalent impedance for nominal $\mathrm{\pi }$ circuit.

Substitute $107.11\mathrm{\Omega }$ for ${\mathrm{X}}^{\text{'}}$ into equation (8).

$$\begin{gathered} {\mathrm{Z}}^{\text{'}}=\mathrm{j}(107.11-107.11\times 0.4) \\ {\mathrm{Z}}^{\text{'}}=\mathrm{j}64.26\mathrm{\Omega } \end{gathered}$$

Calculate the equivalent admittance for nominal $\mathrm{\pi }$circuit.

Substitute $290.43\mathrm{\Omega }$for ${\mathrm{Z}}_{\mathrm{C}}$ and $21.{641}^0$for $\mathrm{\beta I}$ into equation (9).

$$\begin{gathered} {\mathrm{Y}}^{\text{'}}=\mathrm{j}\frac{2}{290.43}\tan \left(\frac{21.{641}^0}{2}\right) \\ {\mathrm{Y}}^{\text{'}}=\mathrm{j}0.00688\times 0.1911 \\ {\mathrm{Y}}^{\text{'}}=\mathrm{j}0.001316\mathrm{S} \end{gathered}$$

Calculate the parameter A of the line.

Substitute role="math" localid="1655204055113" $\mathrm{j}0.0013116\mathrm{S}$for ${\mathrm{Y}}^{\text{'}}$ and $\mathrm{j}64.26\mathrm{\Omega }$for ${\mathrm{Z}}^{\text{'}}$ into equation (10).

$$\begin{gathered} \mathrm{A}=1+\frac{(\mathrm{j}64.26)(\mathrm{j}0.001316)}{2} \\ \mathrm{A}=1-\frac{0.08456}{2} \\ \mathrm{A}=1-0.0422 \\ \mathrm{A}=0.9577 \end{gathered}$$

Calculate the receiving end current.

Substitute $1000\mathrm{MVA}$ for S ,0.8 for pf and 500 kV for ${\mathrm{V}}_{\mathrm{R}}$ into equation (11).

$$\begin{gathered} {\mathrm{I}}_{\mathrm{R}}=\frac{1000\angle {\cos }^{-1}0.8}{\sqrt{3}\times 500} \\ {\mathrm{I}}_{\mathrm{R}}=1.154\angle 36.{87}^0\mathrm{kA} \end{gathered}$$

Calculate the sending end voltage.

Substitute $0.9577$for A , role="math" localid="1655203045149" $\frac{500}{\sqrt{3}}\mathrm{kV},\mathrm{j}64.26\mathrm{\Omega }$,for B and $1.154\angle 36.{87}^0\mathrm{kA}$for ${\mathrm{I}}_{\mathrm{R}}$ into equation (3).

role="math" localid="1655203342863" $$\begin{gathered} {\mathrm{V}}_{\mathrm{s}}=0.9577\times \frac{500}{\sqrt{3}}+\mathrm{j}64.26\times 1.154\angle 36.{87}^0 \\ {\mathrm{V}}_{\mathrm{s}}=276.464+74.201\angle 53.{13}^0 \\ {\mathrm{V}}_{\mathrm{s}}=326.42\angle 10.{47}^0\mathrm{kA} \end{gathered}$$

Calculate the sending end line to line voltage.

role="math" localid="1655202786072" $$\begin{gathered} {\mathrm{V}}_{\mathrm{s},\mathrm{LL}}=326.42\times \sqrt{3} \\ {\mathrm{V}}_{\mathrm{s},\mathrm{LL}}=565.37\mathrm{kA} \end{gathered}$$

Calculate the no load voltage.

Substitute $565.37\mathrm{kV}$for and $0.9577$for $\mathrm{A}$ into equation (11).

$$\begin{gathered} {\mathrm{V}}_{\mathrm{RNL}}=\frac{565.37}{0.9577} \\ {\mathrm{V}}_{\mathrm{RNL}}=590.34\mathrm{kV} \end{gathered}$$

Calculate the percentage voltage regulation.

Substitute $500\mathrm{kV}$for ${\mathrm{V}}_{\mathrm{RFL}}$and $590.34\mathrm{kV}$for ${\mathrm{V}}_{\mathrm{RNL}}$ into equation (12).

$$\begin{gathered} \mathrm{VR}\%=\left(\frac{590.34-500}{500}\right)\times 100 \\ \mathrm{VR}\%=\left(\frac{90.34}{500}\right)\times 100 \\ \mathrm{VR}\%=18.06\% \end{gathered}$$

Hence the voltage regulation is $18.06\%$.