Question

The$\mathbf{500}\mathbf{}\mathbf{kV}\mathbf{}\mathbf{,}\mathbf{60}\mathbf{}\mathbf{Hz}$three-phase line in Problems 4.20 and 4.41 has a$\mathbf{180}\mathbf{}\mathbf{km}$length and delivers$\mathbf{1600}\mathbf{}\mathbf{MW}$at$\mathbf{475}\mathbf{}\mathbf{kV}$and at$\mathbf{0}\mathbf{.}\mathbf{95}$power factor leading to the receiving end at full load. Using the nominal$\mathit{\tau }\mathit{\tau }$circuit, calculate the (a) ABCDparameters, (b) sending-end voltage and current, (c) sending-end power and power factor, (d) full-load line losses and efficiency, and (e) percent voltage regulation. Assume a$\mathbf{50}\mathbf{°}\mathit{C}$conductor temperature to determine the resistance of this line.

Short answer

(a) Therefore, the ABCD parameters are$0.9739\angle 0.0912°\mathrm{pu},60.48\angle 86.6°\mathrm{\Omega }8.54\times {10}^{-4}\angle 90.05°\mathrm{S},\mathrm{and}0.9739\angle 0.0912°\mathrm{pu}$.

(b) Therefore, the sending end voltage is$457.9\angle 27.02°\mathrm{kV}$and sending end current is$2.079\angle 24.42°\mathrm{kA}$.

(c) Therefore, the three phase power is$1646\mathrm{MW}$and the source power factor is$0.5766$.

(d) Therefore, the line losses are $46\mathrm{MW}$and the efficiency is$97.2\%$.

(e) Therefore, the voltage regulation is$-1\%$.

Step-by-step solution

Write the given data.

From problem 4.20 and 4.41, the line impedance is $z=0.0201+j0.335\frac{\Omega }{\mathrm{km}}$and shunt admittance is$y=j4.807\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}$.

The line-to-line receiving end voltage is $475\mathrm{kV}$and load absorbs$P=1600\mathrm{MW}$

The line length is$I=180\mathrm{km}$.

The power factor is.

Determine the formulas of ABCD parameters, receiving end current, sending end voltage, sending end current, source power, source power factor and voltage regulation.

Write the formula of ABCD parameters.

(i)

$$\begin{gathered} \mathbf{A}\mathbf{=}\mathbf{D} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\left(1+\frac{\mathrm{ZY}}{2}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\left(1+\frac{\left(\mathrm{zI}\right)\left(\mathrm{yI}\right)}{2}\right) \end{gathered}$$ ……. (1)

(ii)

$$\begin{gathered} \mathbf{B}\mathbf{=}\mathbf{Z} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\left(\mathrm{zI}\right) \end{gathered}$$ ……. (2)

(iii)

$$\begin{gathered} \mathbf{C}\mathbf{=}\mathbf{Y}\left(1+\frac{\mathrm{ZY}}{4}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\left(\mathrm{yI}\right)\left(1+\frac{\left(\mathrm{zI}\right)\left(\mathrm{yI}\right)}{4}\right) \end{gathered}$$ ……. (3)

Here,$Z$is line impedance per kilometer, $y$is shunt admittance per kilometer and $I$is length of line.

Write the formula of receiving end current.

${\mathbf{I}}_{\mathbf{R}}\mathbf{=}\frac{\mathbf{P}}{\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{R}}\mathbf{.}\mathbf{}\mathbf{p}\mathbf{.}\mathbf{f}}$ ……. (4)

Here,$S$is load end power and$V_R$is the receiving end voltage.

Write the formula of sending end voltage in terms of transmission line parameters.

${\mathbf{V}}_{\mathbf{S}}\mathbf{=}{\mathbf{AV}}_{\mathbf{R}}\mathbf{+}{\mathbf{BI}}_{\mathbf{R}}$ ……. (5)

Here, $V_S$is the sending end voltage, $V_R$is the receiving end voltage, and$I_R$is the transmission line current.

Write the formula of sending end current.

${\mathbf{I}}_{\mathbf{s}}\mathbf{=}{\mathbf{CV}}_{\mathbf{R}}\mathbf{+}{\mathbf{DI}}_{\mathbf{R}}$ ……. (6)

Write the formula of voltage regulation.

$\mathbf{\%}\mathbf{VR}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{RNL}}\mathbf{-}{\mathbf{V}}_{\mathbf{RFL}}}{{\mathbf{V}}_{\mathbf{RFL}}}\mathbf{\times }\mathbf{100}$

localid="1655900674649" $\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{A}}{{\mathbf{V}}_{\mathbf{R}\mathbf{F}\mathbf{L}}}\mathbf{\times }\mathbf{100}$ ……. (7)

Here,$V_{RNL}$is the no-load receiving end voltage and$V_{FNL}$is the full-load receiving end voltage.

Write the formula of power angle.

$\mathit{\theta }\mathbf{=}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{-}\mathbf{1}}\left(p.f\right)$ ……. (8)

Here, $p.f$is a power factor.

Write the formula of sending end power.

${\mathit{P}}_{\mathbf{S}}\mathbf{=}\mathbf{3}{\mathbf{I}}_{\mathbf{S}}{\mathbf{V}}_{\mathbf{S}}\mathit{p}\mathbf{.}\mathit{f}$ ……. (9)

Write the formula of sending end power factor.

$\mathbf{p}\mathbf{.}\mathbf{f}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{R}}{\mathbf{cos\varphi }}_{\mathbf{R}}\mathbf{+}{\mathbf{I}}_{\mathbf{R}}\mathbf{R}}{{\mathbf{V}}_{\mathbf{S}}}$ ……. (10)

Determine the ABCD parameters.

(a)

Determine ABCD parameters.

Substitute$0.0201+j0.335\frac{\Omega }{\mathrm{km}}\mathrm{for}\mathrm{z},\mathrm{j}4.807\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}\mathrm{for}\mathrm{y},\mathrm{and}180\mathrm{km}\mathrm{for}\mathrm{I}$in equation (1).

$$\begin{gathered} A=D \\ =\left\{1+\frac{\left[\left(0.0201+J0.335\right)\left(180\right)\right]\left[J4.807\times {10}^{-6}\right]\left(180\right)}{2}\right\} \\ =0.9739\angle 0.0912°\mathrm{pu} \end{gathered}$$

Substitute$0.0201+j0.335\frac{\Omega }{\mathrm{km}}\mathrm{for}\mathrm{z}\mathrm{and}180\mathrm{km}\mathrm{fo}/$in equation (2).

$$\begin{gathered} B=\left(0.0201+j0.335\right)\left(180\right) \\ =60.48\angle 86.6°\Omega \end{gathered}$$

Substitute$0.0201+j0.335\frac{\Omega }{\mathrm{km}}\mathrm{for}\mathrm{z},j4.807\times {10}^{-6}\frac{\mathrm{S}}{\mathrm{km}}\mathrm{for}\mathrm{y},\mathrm{and}180\mathrm{km}\mathrm{for}/$in equation (3).

$$\begin{gathered} C=\left(j4.807\times {10}^{-6}\right)\left(100\right)\left\{1+\frac{\left[\left(0.0201+j0.335\right)\left(100\right)\right]\left[\left(j4.807\times {10}^{-6}\right)\left(100\right)\right]}{4}\right\} \\ =8.54\times {10}^{-4}\angle 90.05°\mathrm{S} \end{gathered}$$

Therefore, the ABCD parameters are$0.9739\angle 0.0912°\mathrm{pu},60.48\angle 86.6°\mathrm{\Omega }8.54\times {10}^{-4}\angle 90.05°\mathrm{S},\mathrm{and}0.9739\angle 0.0912°\mathrm{pu}$.

Determine the sending end voltage and current.

(b)

Determine the receiving end current.

Substitute$1600\mathrm{MW}\mathrm{for}P,0.95\mathrm{for}p.f\mathrm{and}\frac{475}{\sqrt{3}}\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{R}}$in equation (4).

$$\begin{gathered} I_R=\frac{1600\mathrm{MW}}{\sqrt{3}\times {\displaystyle \frac{475}{\sqrt{3}}}\mathrm{kV}\times 0.65} \\ =2.047\mathrm{kA} \end{gathered}$$

Determine the power angle.

Substitute$0.95\mathrm{for}p.f$in equation (8).

$$\begin{gathered} \theta ={\cos }^{-1}\left(0.95\right) \\ =18.9° \end{gathered}$$

Determine the sending end voltage for leading power factor.

Substitute$$\begin{gathered} 0.9739\angle 0.0912°\mathrm{pu}\mathrm{for}\mathrm{A},\frac{475}{\sqrt{3}}\angle 0°\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{R}},60.48\angle 86.6°\mathrm{\Omega }\mathrm{for}Band \\ 2.047\angle 18.9°\mathrm{kA}\mathrm{for}{I}_R \end{gathered}$$in equation (5).

$$\begin{gathered} V_S=\left(0.9739\angle 0.0912°\mathrm{pu}\right)\left(\frac{475}{\sqrt{3}}\angle 0°\mathrm{kV}\right)+\left(60.48\angle 86.6°\Omega \right)\left(2.047\angle 18.9°\mathrm{kA}\right) \\ =264.4\angle 27.02°\mathrm{kV} \end{gathered}$$

The phase voltage is,

$$\begin{gathered} V_{\mathrm{S}}=\sqrt{3}\times 264.4\angle 27.02°\mathrm{kV} \\ =457.9\angle 27.02°\mathrm{kV} \end{gathered}$$

Determine the sending end current.

Substitute $$\begin{gathered} 8.54\times {10}^{-4}\angle 90.05°S\mathrm{for}\mathrm{C},\frac{475}{\sqrt{3}}\angle 0°\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{R}}, \\ 0.9739\angle 0.0912°\mathrm{pu}\mathrm{for}D\mathrm{and}2.047\angle 18.9°\mathrm{kA}\mathrm{for}{I}_R \end{gathered}$$in equation (6).

$$\begin{gathered} I_S=\left(8.54\times {10}^{-4}\angle 90.05°\right)\left(\frac{475}{\sqrt{3}}\angle 0°\mathrm{kV}\right)+\left(0.9739\angle 0.0912°\right)\left(2.047\angle 18.9°\mathrm{kA}\right) \\ =2.079\angle 24.42°\mathrm{kA} \end{gathered}$$

Therefore, the sending end voltage is$457.9\angle 27.02°\mathrm{kV}$and sending end current is$2.079\angle 24.42°\mathrm{kA}$.

Determine the sending end power and power factor.

(c)

Substitute $475\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{R}}\left(90°-18.9°\right)\mathrm{for}{\mathrm{\varphi }}_{\mathrm{R}},2.079\mathrm{kA}\mathrm{for}{I}_R,\frac{457.9\mathrm{kV}}{\sqrt{3}}\mathrm{for}{\mathrm{V}}_{\mathrm{s}}\mathrm{and}0.0201\mathrm{\Omega }\mathrm{for}R$in equation (9).

$$\begin{gathered} p.f=\frac{\left(475\mathrm{kV}\right)\cos \left(90°-18.9°\right)+\left(2.079\mathrm{kA}\right)\left(0.0201\Omega \right)}{{\displaystyle \frac{457.9\mathrm{kV}}{\sqrt{3}}}} \\ =0.5766 \end{gathered}$$

Substitute$0.5766\mathrm{for}p.f,2.079\mathrm{kA}\mathrm{for}{I}_R,\mathrm{and}457.9\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{s}}$in equation (10).

$$\begin{gathered} {\mathrm{P}}_{\mathrm{S}}=3\left(457.9\mathrm{kV}\right)\left(2.079\right)\left(0.5766\right) \\ =1646\mathrm{MW} \end{gathered}$$

Therefore, the three phase power is$1646\mathrm{MW}$and the source power factor is$0.5766$.

Determine the full-load line losses and efficiency.

(d)

Determine the losses.

$$\begin{gathered} P_L=P_S-P_R \\ =1646\mathrm{MW}-1600\mathrm{MW} \\ =46\mathrm{MW} \end{gathered}$$

Determine the efficiency.

$$\begin{gathered} \eta =\frac{{\mathrm{P}}_{\mathrm{R}}}{{\mathrm{P}}_{\mathrm{S}}}\times 100 \\ =\frac{1600\mathrm{MW}}{1646\mathrm{MW}}\times 100 \\ =97.2\% \end{gathered}$$

Therefore, the line losses are$46\mathrm{MW}$and the efficiency is$97.2\%$.

Determine the voltage regulation.

(e)

Determine the voltage regulation.

Substitute$0.9739\mathrm{for}\mathrm{A},475\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{RNL}}\mathrm{and}457.9\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{s}}$in equation (7).

$$\begin{gathered} \%VR=\frac{{\displaystyle \frac{457.9\mathrm{kV}-}{0.9739}}-475\mathrm{kV}}{475\mathrm{kV}} \\ =-1\% \end{gathered}$$

Therefore, the voltage regulation is$-1\%$.